Project Management. Project Mangement. ( Notes ) For Private Circulation Only. Prof. : A.A. Attarwala.

Project Mangement ( Notes ) For Private Circulation Only. Prof. : A.A. Attarwala. Page 1 of 380 26/4/2008

Syllabus 1. Total Project Management Concept, relationship with other function and other organizations, organizing for project management matrix organization the project manager as an entrepreneur. 2. Project identification Scounting for project ideas and promoters identification of investment opportunities, basis of governmental regulatory framework, various acts and laws affecting project identification. 3. Location Decisions Objectives, factors affecting location, concept of industrially backward areas, incentives available for appropriate location. 4. Project Appraisal Market Appraisal, Demand estimation and forecasting, Technical Appraisal Raw materials Technology Project Mix plant capacity distribution channels. 5. Project financing Basic concepts of cost of project, profitability analysis, Means of financing, raising capital, asserting tax burdens and using financial projections, Appraisal criteria used by lending institutions. 6. Risk analysis of project measures of risk, use of subjective probabilities, mathematical analysis, sensitivity analysis, simulation analysis, decision tree analysis. 7. Project Planning, monitoring and control network techniques, GATT charts, network cost system, resource allocation and scheduling, progress reports, updating, management information system for project. 8. Use of computer in network analysis project management packages choosing and using them. 9. Case Study University Examination : 100 Marks Paper Time : 4 hrs Professor : Prof. A A Attarwala Page 2 of 380 26/4/2008

Reference Books : Book 1 Project Management - Prasanna Chandra Material Required Scientific Calculator Pencil & Sharpnor Page 3 of 380 26/4/2008

4 th February, 2008. Lecture I. Define Project Management By Project Management Institute ( PMI ) A Project is an entity which is : a) A Temporary effort ( Unique ) b) Constrains by Time & Resources c) Has definite Start & End. Examples : Flyovers, Launching a new product ( marketing ), recruitment, Installing a plant. Anything which satisfy the above a,b,c is project. Project is a logical activity in order, each activity has the duration. How the project is presented In form of Model Network diagram Techniques for Project Presentation. CPM PERT - Critical Path Method. Program Evaluation & Review Technique. In CPM/PERT method a project is exhibited as a NETWORK. In CPM/PERT we define a project as a group of activities which are of Interdependent. Arrow Diagram PERT Network Network What is Activity? An activity is a physical task which consumes Time, Money, Resources, Equipment, Manpower in any combination. Page 4 of 380 26/4/2008

Activity=Task=Job. How activity is Represented? Hunt For Chief Guest Start Activity Activity Label Event Label 1 Fixed up the Chief Guest ( A ) 5 Days 2 Event Label Tail / Start Event Head / End Event 1. First Method A 1 2 5 2. Second Method A 5 1 2 3. Third Method A 5 1 2 Page 5 of 380 26/4/2008

In-Correct ways to Represent the Network diagram. 1. A 1 1 5 Both Start & End Labels are Same. 2. A 1 2 5 Cannot go from High Label to Low Label 3. A 5 No label names are given 4. 1 A 5 5. 6. No tail is defined. No Head is Defined. A A 5 2 5 1 2 7. Extra Event is Defined. A 1 2 5 Arrow is shown in the Centre. Page 6 of 380 26/4/2008

Tips for Drawing the Network Diagram. 1. Suppose activity A & B starts together. A 2 1 B 3 Or B 2 1 A 3 1 A 2 1 B 3 This is Wrong, Because No event can be Twice 2. Common Start. A 2 1 B C 3 4 Page 7 of 380 26/4/2008

3. Interdependent Suppose A & B Starts together, C starts after A is over, D & E starts after B is over. This can be mathematically represented as below A B C D E Seceding - - A B B Presiding Points to be noted here - means does not have any preceding activity A is the preceding event C is the succeeding event A 2 C 4 1 B D 5 3 E 6 Very Important : 1 6 Must be from higher to lower Must be from left to right And upper to lower. Question : What is the difference between arrow and node diagram? Page 8 of 380 26/4/2008

4. R starts after P & Q are over. 4 Q P 6 R 7 5 Start of R depends upon when BOTH P & Q are over. 5. A & B starts together, C starts after A is over D & E starts after B is over F starts after C & E are over A B C D E F - - A B B E 1 A B 2 C E C 4 F 6 3 D 5 Page 9 of 380 26/4/2008

Problem 1. Consider the Activities required to complete the processing of Consumer s order : Activities Preceding Activities (s) A. Receipt of order, checking credit rating, etc - B. Preparation of Material Specification, availability of Materials etc. A C. Inspection Packing etc. B D. Arrangement of transports facilities etc. A E. Delivery C,D Draw the arrow Diagram Solution : First put the Mathematical equation. A B C D E - A B A C,D Check what is combining two. E C, D There can be 3 ways by this can be drawn. First way - D 3 C E 5 A 1 2 B 4 Page 10 of 380 26/4/2008

Second Way 1 A 2 B 3 C 4 E 5 D Third Way B 3 C A 1 2 D 4 E 5 All three ways are correct. Page 11 of 380 26/4/2008

Problem 2. The owner of a chain of fast-food restaurants is considering a new computer system for accounting and inventory control. A computer company send the following information about the system installation. Activity Activity Description Immediate Identification Predecessor A Select the Computer Model - B Design Input / Output system A C Design Monitoring System A D Assemble Computer System B E Develop the main Programs B F Develop Input / output routines C G Create Data base E H Install the System D,F I Test and Implement G,H Construct the PERT network for this Problem. Solution : A B C D E F G H I - A A B B C E D,F G,H E 5 1 A 2 B C 3 4 F D 6 H G I 7 8 Page 12 of 380 26/4/2008

Problem 3. Problem 3.1. Activity P Q,R S T U V Preceding activity - P Q R S,T U Solution : Q 3 S 1 P 2 R 5 U 6 V 7 4 T Problem 3.2. Activity A,B,C D E F G H I K Preceding activity - A B C D E F G,H,I Solution : A 2 D 5 G 1 B C 3 4 E F 6 7 H I 8 K 9 Page 13 of 380 26/4/2008

Problem 3.3. Activity A B C D E,F G H I J K L Preceding Activity - A A A B E F G,H C D I,J,K Solution : F 6 H 3 E G 8 B 7 I 1 A C 2 4 D J K 9 L 10 5 Problem 3.4. Solution : Activity A B C D E F G Linking Events 1-2 1-3 2-4 3-4 3-5 4-5 5-6 A 2 C 1 B D 4 3 F E 5 G 6 Page 14 of 380 26/4/2008

Problem 3.5. Activity A B C D E F G H I J K L Linking 1-2 1-3 2-4 3-4 3-5 4-9 5-6 5-7 6-8 7-8 8-10 9-10 Events Solution : A 2 C 1 B D 4 F 9 L 3 10 E G 6 I K 5 H J 8 7 Page 15 of 380 26/4/2008

Dummy Activity It is an activity which neither consumes any time nor resources. It is denoted by Dotted Line. It is used for logical purpose. e.g Dummy 1 2 0 Note : If dotted line is not drawn then it will be wrong. Two places where dummy is used. FIRST. Suppose A & B start together C starts after both A & B are over. i.e. A B C - - A,B Observe that A & B have Same Tail Events & Head Events. Shown below Three representations are not allowed. A 1 B 2 C 3 A 1 2 B C 3 A 1 2 B C 3 Page 16 of 380 26/4/2008

Correct Representations A 2 1 B D 1 3 C 4 A 3 1 B D 2 2 C 4 SECOND. Suppose A is a starting activity A B C D E F - - A B C,D D Observe that Head / End Events of D is in E as well as F C E D F This is wrong because F can be only after D is over. Page 17 of 380 26/4/2008

Correct is A 2 C E 5 6 1 B D 1 3 D F 5 6 Page 18 of 380 26/4/2008

Problem 3.6. Job A B C D E F G H I J K Immediate - A B C D E D,F E H G,I J Predecessors Solution : 8 H I A B C D E F G 1 2 3 4 5 6 7 9 J D 1 10 K 11 OR 7 H I A B C D 1 2 3 4 5 E 6 9 J 10 F G K 8 11 Page 19 of 380 26/4/2008

Problem 3.7. Job A B C D E F G H J K L M N O Preceding Activity - - A B C C D D E E,F G,H H J,K L,M,N Solution : A 2 C 4 E F D1 6 7 K J 10 N 1 9 L 11 O 12 B 3 D 5 G D2 M H 8 Page 20 of 380 26/4/2008

5 th February, 2008. Lecture II. Problem 3.8. Task A B C D E F G Immediate Activity - - B B B E A,D,C Solution : Note that, Two activity starts together & ends together C,D. Hence this requires a dummy. 1 A 4 B C D1 G 2 D 3 6 E F 5 Very Important Note : There are two ends of the project ( DANGLER ) It is not allowed by default, F & G should meet at one place as they are the final activities Page 21 of 380 26/4/2008

Problem 3.9. A and B are start jobs A controls C,D and E B controls F and K G depends on C H depends on D E and F control J and M L depends on K M is also controlled by L G,H,J and M are last jobs. Solution : Job A B C D E F K G H J M L Preceding Activity - - A A A B B C D E,F E,F,L K Observer that, G,H,J,M are the Final activities E,F C 4 G A 2 D 5 H 9 E J 1 B 3 F 6 D1 M K L 8 7 Page 22 of 380 26/4/2008

Problem 3.10. (a) (b) (c) (d) (e) (f) (g) A,C,D Can start immediately E > B,C F,G > D H,I > E,F J > I,G K > H B > A For Home work Page 23 of 380 26/4/2008

Problem 3.11. (a) (b) (c) (d) (e) (f) (g) (h) A and B start Immediately C,D > A E > B,C F,H > E G > D,F J > G I,K > H L > J,I Solution : Job A B C D E F H G I J K L Preceding Activity - - A A B,C E E D,F G H H J,I No Dummy required from both the rules A 2 D 5 G 6 J 1 B C 3 E F 4 H 7 I 8 K L 9 Page 24 of 380 26/4/2008

Problem 3.12. Activity A B C D E F G H Predecessor - - A A B B C,E D,F Solution : Crossing is allowed. C 4 G A 2 H 6 1 B D F 5 3 Page 25 of 380 26/4/2008

Critical Path / Critical Activity / Project Duration. Consider a given below project. 3 B E 7 11 A D G 2 1 2 5 6 6 C F 5 4 1 4 Now to find out the Critical Path, first define the various Paths & Durations Paths Durations Total Duration 1-2, 2-3, 3-4, 4-5, 5-6 6 + 7 + 2 + 1 + 5 = 21 ➊ Or 1-2, 2-3, 3-5, 5-6 6 + 7 + 11 + 5 = 29 ➋ Or 1-2, 2-4, 4-5, 5-6 6 + 4 + 1 + 5 = 16 ➌ Now out of the three available paths Path II is most critical as it has longest duration, that is 29. Critical path should be shown as below. 3 B E 7 11 A D G 2 1 2 5 6 6 C F 5 4 1 4 Remember Duration of Critical path is the duration of project, therefore above project will be completed in 29 days. Page 26 of 380 26/4/2008

Problem 4. The following Tables lists the activities of a maintenance project. Activity 1-2 1-3 1-4 2-5 3-6 3-7 4-7 5-8 6-8 7-9 8-9 Duration (in Months) 2 2 1 4 5 8 3 1 4 4 3 (i) Draw the project network. (j) Find the Critical path and duration of the project. ( 15 Months ) Solution : 4 2 5 1 2 4 8 3 2 5 1 3 6 9 1 8 5 4 3 7 Paths Durations Total Duration 1-2, 2-5, 5-8, 8-9 2 + 4 + 1 + 3 = 10 ➊ 1-3, 3-6, 6-8, 8-9 2 + 5 + 4 + 3 = 14 ➋ 1-3, 3-7, 7-9 2 + 8 + 4 = 15 ➌ 1-4, 4-7, 7-9 1 + 3 + 5 = 9 ➍ The critical path is 1-3, 3-7, 7-9 (➌) as it has 15 months duration, which is highest. Page 27 of 380 26/4/2008

Problem 5. Draw the network for the following dependencies and identify critical path. Also find the project duration. Activity 1-2 1-3 1-4 2-3 2-6 3-5 3-6 4-5 5-6 5-7 6-7 Duration (in Months) 8 7 3 6 8 6 4 12 0 6 8 Solution : ( 28 Months ) 2 8 8 6 4 6 8 1 7 3 6 0 6 7 3 12 5 4 Tip : Start with Longest route first. Paths Durations Total Duration 1-2, 2-3, 3-5, 5-6, 6-7 8 + 6 + 6 + 0 + 8 = 28 1-2, 2-3, 3-5, 5-7 8 + 6 + 8 + 6 = 28 1-2, 2-3, 3-6, 6-7 8 + 6 + 4 + 8 = 26 1-2, 2-6, 6-7 1-3, 3-5, 5-6, 6-7 1-3, 3-5, 5-7 1-3, 3-6, 6-7 1-4, 4-5, 5-6, 6-7 1-4, 4-5, 5-7 There are more than ONE Critical Paths. Page 28 of 380 26/4/2008

Problem 6. (i) Draw the Network for the following data and determine the critical path and project duration. Activity Preceding Activity Duration ( Days ) A - 4 B - 5 C A 2 D A 3 E B. C 3 F B, C 4 G D, E 5 H F 2 (Ii) If activity F has to precede activity G, will the critical path Changes? If it does change then draw your new network and calculate the length of new critical path. This problem is for Home work. Page 29 of 380 26/4/2008

Time Estimates Time estimates of an Activity Associated with each activity, there are 4 time estimates, They are : i) Earliest Start Time ( EST ) Represented as ( ) EST is also knows as Forward Pass ii) Latest Start Time ( LST ) iii) Earliest Finish Time ( EFT ) iv) Latest Finish Time ( LFT ) Represented as ( ) LFT is also knows as Backward Pass Out of the above Four EST & LFT are shown in the Network Diagram. Page 30 of 380 26/4/2008

Question : Explain forward pass ( EST ) and Backward pass ( LFT ). This was asked as a Theory question. Answer : Theory Plus explanation. Suppose process G starts after E & G are completed, E gets over on 13th, F gets over on 15th. What is the EST of G. 5 6 E 13 F 15 G 7 8 G will start at 15 i.e. after 15 days. Since G starts from Event 7 Therefore EST of G is written near its tail event in a box Like Explain : Forward pass don t explain theory, Explain with the help of an example as follows: Compute EST 10 7 15 3 E 0 6 B 21 25 11 A 4 D 2 G 1 2 5 6 6 4 C F 3 5 4 12 Page 31 of 380 26/4/2008

Suppose we start project at time t=0, [ t = 0 means end of 0 th day, beginning of first day ] Then all activities from event 1 will start at time t = 0 Therefore activity A will start at t = 0 i.e. EST of activity A is Zero. This we write near the tail event of activity A in a Box Like 1 0 Activities B and C start after activity A is completed. There for EST of B and C is : 0 + 6 = 6 Means EST of A + Duration of A. This we write near the tail event of B & C in a Box like Similarly, EST of D & E is 2 6 6 + 3 = 9 EST of F As F starts after C & D are completed, therefore via C EST of F will be 6 + 3 = 9 Via D EST of F will be 10 + 2 = 12 Which will be 12, This we write near the tail event 4 of D Like 12 Now suppose there is another project which starts immediately after G is completed, Then what is EST of that Project? 21 Its EST is + 4 = 25 To compute EST, we move from the first event to the last event. This is called Forward Pass. Page 32 of 380 26/4/2008

Question 1. 4 C 9 3 E A 1 3 2 B 4 3 D 2 G 6 5 F 4 6 1 H 7 I 8 7 Compute the EST. Solution : 16 4 7 C 9 3 E 19 0 3 1 A 3 2 B 4 3 D 2 G 6 9 5 F 4 6 1 H 20 27 7 I 8 7 First Calculate the EST. Then draw the critical path. Total project duration is 27. Page 33 of 380 26/4/2008

Question 2. 3 4 6 1 2 2 5 3 5 4 7 4 7 8 9 3 6 6 Find out the EST. Solution : 7 3 0 3 4 6 13 1 2 2 5 3 5 4 7 20 29 4 7 8 9 9 3 6 6 12 There are more than two Critical Paths exists. Page 34 of 380 26/4/2008

LFT ( Latest Finish Time ) or Backward Pass. i) LFT Requires EST ii) iii) LFT is written at Head Event of each activity in a to Compute LFT we move from the Last Event to the First Event, therefore it is called the Backward Pass. How to find out LFT? 3 B E 7 11 A D G 2 1 2 5 6 6 C F 5 4 1 4 First Find out the EST 10 3 E 0 6 B 21 25 11 A 4 D 2 G 1 2 5 6 6 4 C F 3 5 4 12 EST = 25. Suppose we want to complete the project by 25 th Day then, all activities which get over at the last event must get over latest by 25 th Day. Therefore G must get over latest by 25 th day. Therefore LFT of G is 25. Page 35 of 380 26/4/2008

This we write in 25 triangle near Head Event of G. At 5 E & F get over when G starts, therefore LFT of both E & F is 25-4 = 21 LFT of G Duration of G 21 5 This we write in like near the head event of E & F At 3 10 D & F start after B is over. Therefore LFT of B is 16 LFT of D - 2 = 14 16 Via E - 11 = 10 So that 10. 3 10 This we write near the head event of B in a like Now the Network Diagram can be drawn like given below 10 10 0 6 3 21 25 E 0 6 B 21 25 4 11 A D 2 G 1 2 5 6 6 4 C F 3 5 4 12 16 Page 36 of 380 26/4/2008

Question 3. 4 C 9 3 E A 1 3 2 B 4 3 D 2 G 6 5 F 4 6 1 H 7 I 8 7 Compute EST & LFT. Solution : First find the EST of all the activities. ( Refer above answer of question No 1 ). 16 4 7 C 9 3 E 19 0 3 1 A 3 2 B 4 3 D 2 G 6 9 5 F 4 6 1 H 20 27 7 I 8 7 So EST is 27. LFT of at begin & end will be same i.e. 0 27 & Is 5 9 15 Page 37 of 380 26/4/2008

15 LFT of D - 2 = 13 20 LFT of G - 6 = 14 19 LFT of E - 3 = 16 There fore For a Critical Path Activity LFT = EST EST of G 3 + 6 = 9 EST of D 7 + 2 = 9 EST of F + 4 = 13 3 Therefore for a Critical Activity EFT = LST Now the Network diagram can be drawn as shown below. 16 16 4 7 19 7 C E 3 9 15 19 3 6 0 3 D 9 20 27 F H 0 3 B 2 5 4 1 20 27 4 A G 1 2 I 7 8 3 6 7 Page 38 of 380 26/4/2008

Problem 7. Consider the following project. Activity A B C D E F G H I J K L Duration 3 2 4 6 3 2 6 4 4 2 2 4 Immediate Predecessor None A A A B E C D G D D F,H,I Compute EST/LFT of each activity of the Project. Solution : This problem is for Homework. Page 39 of 380 26/4/2008

9 th February, 2008. Lecture III. 5 D 11 5 F A 1 6 2 B 7 C 3 2 E 6 G 3 I 7 4 3 H J 4 8 9 9 6 Find & This can be done with help of table shown below. Start Finish Activity Duration Earliest Latest Earliest Latest ( EST ) ( LST ) ( EFT ) (LFT) A 1-2 6 0 6 6 6 B 2-3 7 6 7 13 13 C 2-4 3 6 7 9 13 D 3-5 11 13 11 24 24 E 3-6 2 13 2 15 15 F 5-7 5 24 5 29 29 G 6-7 3 15 14 18 29 H 4-8 9 9 24 18 33 I 7-8 4 29 4 33 33 J 8-9 6 33 6 39 39 Page 40 of 380 26/4/2008

Steps followed for doing the above, First Find out the EST & LFT, draw them on the Path as shown below and then put the values in the table. Draw the Critical Path. 24 24 0 6 0 6 1 A 6 2 B 3 7 C 5 13 29 13 D F 5 11 26 29 3 7 E 15 G 2 6 3 I 4 24 33 39 9 33 39 4 H 8 J 9 6 9 Second Now Calculate the EFT & LST. EFT = EST + Duration. LST = LFT Duration. Important Points : Difference between two start times is equal to two finish times, that is LST EST = LFT EFT Page 41 of 380 26/4/2008

Another Concept. Slack of an Event. For e.g. Slack of event 4 4 24 9 - = 15 Slack of event 6 6 26 15 - = 11 Slack of event 2 2 6-6 = 0 Slack of 6 9 6 24 C 2 4 3 So Slack of Tail event of C C = Slack of event 6 = - = 0 2 6 So Slack of Head event of C = 24 - = 15 4 9 Page 42 of 380 26/4/2008

Problem No 8 For the following project of a repair work, find : (a) (b) (c) (d) (e) (f) (g) (h) (i) Earliest Start Time. Latest Start Time. Earliest Finish Time. Latest Finish Time. Total Float. Free Float. Independent Float. Interfering Float. Project Duration. Task 1-2 2-3 2-4 3-5 4-5 4-6 4-7 5-7 6-7 7-8 Immediate Activity 1 5 3 4 2 5 9 4 2 2 Solution : 6 6 0 1 3 10 0 1 4 10 5 1 2 5 5 14 1 4 4 14 3 2 4 7 9 2 16 16 8 5 6 2 12 9 Steps followed. 1. Draw the Network Diagram. 2. First Find the EST & LFT, plot them on the Network diagram 3. Find EFT & LST, Next Slack of Tail & Head Event, then find others as per the chart. Page 43 of 380 26/4/2008

Activity Duration Start Earliest ( EST ) Latest ( LST ) Earliest ( EFT ) Finish Latest (LFT) Slack of Tail Event Slack of Head Event Total Float Free Float Independ ent Float Interfeari ng Float ( 1 ) ( 2 ) ( 3 ) ( 4 ) ( 5 ) ( 6 ) ( 7 ) ( 8 ) ( 9 ) ( 10 ) (11) (12) 1-2 1 0 0 1 1 0 0 0 0 0 0 2-3 5 1 1 6 6 0 0 0 0 0 0 2-4 3 1 2 4 5 0 1 1 0 0 1 3-5 4 6 6 10 10 0 0 0 0 0 0 4-5 2 4 8 6 10 1 0 4 4 3 0 4-6 5 4 7 9 12 1 3 3 0 1 ~ 0 3 4-7 9 4 5 13 14 1 0 1 1 0 0 5-7 4 10 10 14 14 0 0 0 0 0 0 6-7 2 9 12 11 14 3 0 3 3 0 0 7-8 2 14 14 16 16 0 0 0 0 0 0 Earliest Finish Time ( EFT ) = EST + Duration = (3) + (2) Latest Start Time ( LST ) = LFT Duration = (6) (2) Total Float = LFT EST or LST EST = (4) (3) or (6) (5) Free Float = Total Float Slack of Head Event = (9) (8) Independent Float = Free Float Slack of Tail Event = (10) (7) Interfering Float = Total Float Free Float ( 9 ) (10 ) Page 44 of 380 26/4/2008

Problem No 9 Listed below is a set of activities, sequences of activities and activity duration times. Construct a network and determine the minimum project time. Also computer EST, EFT, LST, LFT for each of the activities and calculate the various floats values of the activities. Tabulate your results. Activity A B C D E F G H I J Immediate - - B A,C B D A.C D E,H F,G,I Predecessor Duration ( Days ) 15 15 3 8 5 14 12 1 3 14 Solution : This problem is for Home work. Page 45 of 380 26/4/2008

Problem No 10 For the following network, find the earliest Start Time, Latest Start Time, Earliest Finish Time. Latest Finish Time, Total Float, Independent Float, Interfering Float Critical Path & Project duration. Activity A B C F D G K E H J I Linking Event 1-2 1-3 1-4 2-6 3-4 3-5 3-6 4-6 5-6 5-7 6-7 Duration 7 8 6 5 0 4 6 3 5 16 10 Solution : 13 7 2 0 A 7 15 5 F 18 28 0 8 17 28 B E I 1 4 6 7 6 3 10 8 C D 0 K 6 H 5 J 16 8 8 3 G 4 5 12 12 Page 46 of 380 26/4/2008

Activity Duration Start Earliest ( EST ) Latest ( LST ) Earliest ( EFT ) Finish Latest (LFT) Slack of Tail Event Slack of Head Event Total Float Free Float Independ ent Float Interfeari ng Float ( 1 ) ( 2 ) ( 3 ) ( 4 ) ( 5 ) ( 6 ) ( 7 ) ( 8 ) ( 9 ) ( 10 ) (11) (12) A 1-2 7 0 6 7 13 0 6 6 0 0 6 C 1-3 8 0 0 8 8 0 0 0 0 0 0 B 1-4 6 0 9 6 15 0 7 9 2 2 7 F 2-6 5 7 13 12 18 6 1 6 5-1 ~ 0 1 D 3-4 0 8 15 8 15 0 7 7 0 0 7 G 3-5 4 8 8 12 12 0 0 0 0 0 0 K 3-6 6 8 12 14 18 0 1 4 3 3 1 E 4-6 3 8 15 11 18 7 1 7 6-1 ~ 0 1 H 5-6 5 12 13 17 18 0 1 1 0 0 1 J 5-7 16 12 12 28 28 0 0 0 0 0 0 I 6-7 10 17 18 27 28 1 0 1 1 0 0 Earliest Finish Time ( EFT ) = EST + Duration = (3) + (2) Latest Start Time ( LST ) = LFT Duration = (6) (2) Total Float = LFT EST or LST EST = (4) (3) or (6) (5) Free Float = Total Float Slack of Head Event = (9) (8) Independent Float = Free Float Slack of Tail Event = (10) (7) Interfering Float = Total Float Free Float ( 9 ) (10 ) Page 47 of 380 26/4/2008

Interpretation of Floats. Question : What are the Floats available in Numerical Analogy & what are their uses? Basically there no difference between slack and float, slack refer to an event. Float refers to an activity. A critical activity will have Zero Slack and Zero Float. A non Critical activity will have at least one float positive. All slack may be zero. Float is measured in units of Time. e.g. 4 Days Float, 2 Weeks Float, 1 Month Float etc. Float is a spare time available in a non critical activity which can be utilized either by delaying the activity or by extending its duration. Float is utilized for smoothening of resources. The resources are equipments and manpower. Total Float. Consider an activity F 12 18 here now : 7 17 F 2 6 5 its LFT = 18 its EFT = 7 So maximum time available is - = 11 And its actual duration is = 5 18 7 Page 48 of 380 26/4/2008

There fore Spare time = 11 5 = 6 days. This spare time of 6 days is F, if absorbed, then neither it changes the critical path nor the project duration. This is known a total float of F. Thus Total Float ( TF ) of an activity is TF = LFT EST Duration. Note : Critical activity has no Float. Remarks : (i) (ii) (iii) If TF = 0 then rest of the floats are zero Independent Float Free Float Total Float An activity is critical if and only if its total float is zero. Activity Critical < = > TF = 0. Free Float : Consider activities F and I. F is preceding activity and I is a succeeding activity. 13 18 28 7 17 F 2 6 5 I 10 28 7 Total Float of F = 6 Total Float of I = 1 17 7 Free float of F = EST of I - EST of F 5 ( Duration of F ) = 17 7 5 = 5. Free Float is the spare time available in a preceding activity, if absorbed then it will not delay the start of the succeeding activity. FF EST of the succeeding activity EST of activity ( whose FF we are finding ) Duration of the activity. Page 49 of 380 26/4/2008

Interfering Float : Free Float + Interfering Float = Total Float or Interfering Float = Total Float Free Float e.g. TF of F = 6, FF of F = 5. therefore out of 6 days of F, 5 days is allowed to be absorbed. 1 day is disallowed to be absorbed. This disallowed float of 1 day of F is known as interfering float of F. Independent Float : Consider activity K in the network shown below 0 18 28 0 17 28 1 6 F 5 7 5 F F 5 3 8 8 It has a succeeding activity I, it has a preceding activity C, if K has not to disturb I, then K should allow I to start at the EST 17. If K has not to disturb its preceding activity C then K should all C to get over at its LFT - - 6 = 3. 17 8 8. If K does so, then its independent float is Thus independent float of an activity = EST of the succeeding activity LFT of the preceding activity its duration. Page 50 of 380 26/4/2008

16 th February, 2008. Lecture IV. Crashing. Crashing is based on only one Principle. Maximum Reduction in Duration & Minimum Increase in Cost. To Crash means to reduce the duration of the project. Consider an activity A A 1 2 7 Its normal duration is 7 days and its normal cost is Rs 500/-, its crash duration is 4 days and crash cost is Rs. 620/- Crash Cost - Normal Cost Cost Slope = ---------------------------------------------- Normal Duration Crash Duration 620-500 = --------------------- = 40 7 4 i.e. to reduce the duration of activity A by 1 day, we have to increase the cost by Rs. 40. Y - axis 620 ( 4, 620 ) Cost 500 ( 7, 500 ) 4 7 X - axis Duration Page 51 of 380 26/4/2008

From the Following find the Cost slope. Activity Duration Activity Cost Normal Crash Normal Crash 1-2 5 2 600 900 2-4 6 3 700 1000 1-3 4 2 100 200 3-4 7 4 400 800 4-7 9 5 600 920 3-5 12 3 1600 1960 4-6 10 6 1500 1800 6-7 7 4 400 490 7-9 6 4 300 420 5-9 12 7 400 850 Solution : Crash Cost - Normal Cost Cost Slope = ---------------------------------------------- Normal Duration Crash Duration 900-600 = --------------------- = 100 5 2 Activity Duration Activity Cost Cost Normal Crash Normal Crash Slope 1-2 5 2 600 900 100 2-4 6 3 700 1000 100 1-3 4 2 100 200 50 3-4 7 4 400 800 133.333 4-7 9 5 600 920 80 3-5 12 3 1600 1960 40 4-6 10 6 1500 1800 75 6-7 7 4 400 490 30 7-9 6 4 300 420 60 5-9 12 7 400 850 90 Page 52 of 380 26/4/2008

Tips for Crashing. Consider the following Network. 3 E B 11 A 9 G 1 2 D 3 5 6 7 5 C F 6 5 4 Path Duration Rank i) 1 2 3 4 5 6 25 ➋ ii) 1 2 3 5 6 32 ➊ iii) 1 2 4 5 6 22 ➌ Critical Path 1 2 3 5 6 Normal Duration 32 Days 1) Usually we Crash activities ( CRITICAL ) in order of their cost Slope Suppose activity C ( 2 4 ) has the lease cost Slope Further it is told that it can be crashed by 3 days, if we do so then, i) Duration of ➊ : 32 ii) Duration of ➋ : 28 iii) Duration of ➌ : 19 Therefore New project duration is 32 days, i.e. there is no change in the project duration. This is because activity C is Non Critical. Therefore in order to reduce the project duration, ALWAYS CRASH A CRITICAL ACTIVITY. Page 53 of 380 26/4/2008

2) Suppose critical activity E has the least cost slope. Further it is told that it can be crashed by 6 days. If we do so then. i) Duration of ➊ : 26 ii) Duration of ➋ : 28 iii) Duration of ➌ : 22 therefore new project duration is 28 days, As the original project duration was 32 days, the new project duration after crashing a critical activity by 6 days is 28 days. Even though we crash a critical activity by 6 days, the overall project duration reduces by only 4 days. Therefore there is no point of crashing activity E by 6 days. Crash it by only 4 days. TIP : Crash a critical activity in such a way that the original path remains critical. In the process, if some another path becomes critical, then it OK. 3) Crash Critical Activity E only by 4 days. Now path ➋ also becomes critical. Therefore whenever there are two critical paths, then either crash A Common Activity OR Two Un-common activities by the same duration. Page 54 of 380 26/4/2008

Problem 1 The following table gives normal & crash times as well as normal and crash costs for the activities of a project. Draw the network diagram and find the critical path. In case the project duration is required to be crashed by 2 days, which activity will get the priority in crashing? Activity Duration Activity Cost Normal Crash Normal Crash 1-2 5 2 600 900 2-4 6 3 700 1000 1-3 4 2 100 200 3-4 7 4 400 800 4-7 9 5 600 920 3-5 12 3 1600 1960 4-6 10 6 1500 1800 6-7 7 4 400 490 7-9 6 4 300 420 5-9 12 7 400 850 Solution : First draw the Network diagram. 6 2 6 4 10 7 1 5 7 9 7 4 3 12 5 12 6 9 The Total Paths available are : Paths Duration 1 2 4 6 7 9 = 39 ➊ 1 2 4 7 9 = 26 ➍ 1 3 4 6 7 9 = 34 ➋ 1 3 4 7 9 = 26 ➍ 1 3 5 9 = 28 ➌ Page 55 of 380 26/4/2008

There are two Critical paths viz. 1 2 4 6 7 9 1 3 4 6 7 9 On both the critical paths there are 3 activities which are common activities namely, Cost Slope 4 6 = 75 6 7 = 30 7 9 = 60 Among the three common activities, activity 6 7 has a least cost slope with just 3 days. Therefore if you want to reduce the project duration by 2 days, then we crash activity 6 7 by 2 days. Therefore activity 6 7 will get priority. Page 56 of 380 26/4/2008

Problem - 2 Following table gives the data on normal time and cost and crash time and cost for a project. Activity Duration ( Weeks ) Total Cost ( Rs ) Normal Crash Normal Crash 1-2 6 5 3000 4500 2-3 6 6 750 750 2-4 10 8 2000 3000 2-5 7 7 1200 1200 3-4 7 4 1000 1900 4-6 6 5 900 1300 5-6 6 4 600 1100 (i) (ii) Draw the network and find out the critical path and the normal project duration Find out the optimum duration by crashing and the corresponding project cost. Solution : First find out the Cost Slope Activity Maximum Crashing Possible Cost Slope 1-2 6-5 = 1 (4500-3000 ) / ( 6-5 ) = 1500 2-3 6-6 = 0 0 2-4 10-8 = 2 500 2-5 - 0 3-4 7-4 = 3 300 4-6 6-5 = 1 400 5-6 6-4 = 2 250 Now draw the Network diagram Page 57 of 380 26/4/2008

3 6 7 1 6 2 10 4 6 6 7 6 5 The Total Paths available are : Paths Duration 1 2 3 4 6 = 25 ➊ 1 2 4 6 = 22 ➋ 1 2 5 6 = 14 ➌ ➊ 1 2 3 4-6 25 / 22 / 21 / 20 Critical Activity Max Crashing Cost Rank Remarks Available slope 1 2 1 1500 ❸ ( iii ) 2 3 - - 3 4 3 300 ❶ ( i ) 4 6 1 400 ❷ ( ii ) ➋ 1 2 4 6 22 / 22 / 21 / 20 1 2 1 1500 ❸ ( iii ) 2 4 2 500 ❷ 4 6 1 400 ❶ ( ii ) ➌ 1 2 5 6 19 / 19 / 19 / 18 1 2 1 1500 ( iii ) 2 5 - - 5 6 2 250 Page 58 of 380 26/4/2008

Step i Step ii Step iii Usually we crash critical activity in order of their Cost Slope. On CP (➊) Activity 3 4 has the least slope & it is available for crashing for 3 days. So if we crash it by 3 days then duration of ➊ = 22 ➋ = 22 ➌ = 19 Therefore New project duration will be 22 days And in cost 3 x 300 = 900. Now path ➋ also becomes critical. On both the Critical paths crash the common activity 4-6 by one day. ➊ = 21 ➋ = 21 ➌ = 19 Therefore New project duration 21 days. And in cost 1 x 400 = 400. No other Path becomes critical. Now on both the critical path crash activity 1 2 by 1 day. ➊ = 20 ➋ = 20 ➌ = 18 New Project duration 20 days. And in cost 1 x 1500 = 1500/-. Page 59 of 380 26/4/2008

Now since all activities on Critical Path (➊) are already crashed therefore crashing is over. Conclusion : Normal Path Duration = 25 Days Normal Cost = Rs. 9450/- Crash Path Duration = 20 Days Crash Cost = 9,450 /- + 900 /- + 400 /- + 1,500 /- ------------ 12,250/- Page 60 of 380 26/4/2008

Problem - 3 The following table gives time ( in days ) and cost of activities ( in Rs ) of a project. Activity 1-2 1-3 1-4 2-4 3-4 4-5 Normal Crash Time ( Days ) Cost ( Rs ) Time ( Days ) Cost ( Rs ) 12 1,250 9 1,490 11 1,125 8 1,335 14 1,200 10 1,500 7 1,050 5 1,170 12 800 8 920 3 1,100 1 1,200 ( i ) Draw the network for the project ( ii ) Find the critical path and the normal project length. Determine the minimum cost and the corresponding duration in days. Solution : Activity Maximum Crashing Possible Cost Slope 1-2 12-9 = 3 (1490-1250 ) / ( 12-9 ) = 80 1-3 11-8 = 3 70 1-4 14-10 = 4 75 2-4 7-5 = 2 60 3-4 12-8 = 4 30 4-5 3-1 = 2 50 3 11 12 1 14 4 3 5 12 7 2 Page 61 of 380 26/4/2008

The Total Paths available are : Paths Duration 1 3 4 5 = 26 ➊ 1 4 5 = 17 ➌ 1 2 4 5 = 22 ➋ ➊ 1 3 4-5 26 / 22 / 20 / 17 Critical Activity Max Crashing Cost Rank Remarks Available slope 1 3 3 70 ❸ ( iii ) 3 4 4 30 ❶ ( i ) 4 5 2 50 ❷ ( ii ) ➋ 1 2 4 5 22 / 22 / 20 / 17 1 4 3 80 ❸ 2 4 2 60 ❷ ( iii ) 4 5 2 50 ❶ ( ii ) ➌ 1 4 5 17 / 17 / 17 / 17 1 4 4 75 ( iii ) 4 5 2 50 Step i On CP (➊) Activity 3 4 has the least slope & it is available for crashing for 4 days. So if we crash it by 3 days then duration of ➊ = 22 ➋ = 22 ➌ = 17 Page 62 of 380 26/4/2008

Step ii Step iii Therefore New project duration will be 22 days And in cost 4 x 30 = 120/-. Now path ➋ also becomes critical. On CP (➊) & (➋) activity 4 5 is common activity. Hence crash 4 5 by one day. ➊ = 20 ➋ = 20 ➌ = 17 Therefore New project duration 20 days. And in cost 2 x 50 = 100/-. No other Path becomes critical. On CP (➊) Crash Activity 1 3 by One Day On CP (➋) Crash Activity 2 4 by Two Days On CP (➌) Crash Activity 1 2 by One Day ➊ = 17 ➋ = 17 ➌ = 17 New Project duration 17 days. And in cost 3 x 70 = 210/- + 2 x 60 = 120/- + 1 x 80 = 80/- = 310/- Now since all activities on Critical Path (➊) are already crashed crashing is over. Page 63 of 380 26/4/2008

Conclusion : Normal Path Duration = 26 Days Normal Cost = Rs. 6525/- Crash Path Duration = 17 Days Crash Cost = 6,525 /- + 120 /- + 100 /- + 310 /- ------------ 6,935/- Page 64 of 380 26/4/2008

23 th February, 2008. Lecture V. Problem - 4 The following table gives the activities in a construction project and other relevant information. Activity 1-2 1-3 1-4 2-4 3-4 4-5 Preceding Activity Normal Time Crash Time Normal Crash ( Days ) ( Days ) Cost ( Rs ) Cost ( Rs ) - 20 17 600 720-25 25 200 200 1-2 10 8 300 440 1-2 12 6 400 700 1-3 2-3 5 2 300 420 2-4 3-4 10 5 300 600 ( a ) Draw the Activity Network for the project ( b ) Find the critical path ( c ) Using the above information crash or shorten the activity step by step until the shortest duration is reached. Solution : 2 20 12 1 10 4 10 5 25 5 3 Page 65 of 380 26/4/2008

The Total Paths available are : Paths Duration 1 2 3 4-5 = 45 ➊ 1 2 4-5 = 42 ➋ 1 3 4 5 = 40 ➌ Activity Maximum Crashing Possible Cost Slope 1-2 20-17 = 3 (720-600 ) / ( 20-17 ) = 40 1-3 25-25 = Nil 0 2-3 10-8 = 2 70 2-4 12-6 = 6 50 3-4 5-2 = 3 40 4-5 10-5 = 5 60 ➊ 1 2 3 4-5 45 / 42 / 39 / 34 / 32 Critical Activity Max Crashing Cost Rank Remarks Available slope 1 2 3 40 ❶ ( i ) 2 3 2 70 ❸ ( iv ) 3 4 3 40 ❷ ( ii ) 4 5 5 60 ❹ ( iii ) ➋ 1 2 4 5 42 / 39 / 39 / 34 / 32 1 2 3 40 ( i ) 2 4 6 2 50 ( iv ) 4 5 5 60 ( iii ) ➌ 1 3-4 5 40 / 40 / 37 / 32 / 32 1 3 -- 0 3 4 3 40 ( ii ) 4 5 5 60 ( iii ) Page 66 of 380 26/4/2008

Step i Step ii Step iii On CP (➊) & CP (➋) Crash Activity 1 2 by 3 days. ➊ = 42 ➋ = 39 ➌ = 40 New project duration will be 42 days, And in cost 3 x 40 = 120/-. No other path becomes critical On CP (➊) & CP (➌) crash activity 3 4 by 3 days. ➊ = 39 ➋ = 39 ➌ = 37 New project duration will be 39 days, And in cost 3 x 40 = 120/-. Now both CP (➊) & CP (➋) are critical. On CP (➊) Crash Activity 4 5 by 5 days. ➊ = 34 ➋ = 34 ➌ = 32 New Project duration 34 days. And in cost 5 x 60 = 300/- No other path becomes critical Page 67 of 380 26/4/2008

Step iv On CP (➊) Crash Activity 2 3 by 2 days & CP (➋) Crash Activity 2 4 by 2 days ➊ = 32 ➋ = 32 ➌ = 32 New Project duration 32 days. And in cost 2 x 70 = 140/- + 2 x 50 = 100. i.e. 240/- Now since all activities on CP (➊) are crashed, Crashing is over. Conclusion : Normal Path Duration = 45 Days Normal Cost = Rs. 2100/- Crash Path Duration = 32 Days Crash Cost = 2,100 /- + 120 /- + 120 /- + 300 /- + 240 /- ------------ 2,880/- Page 68 of 380 26/4/2008

Problem - 5 The table below provides cost and time estimates of seven activities of a project. Activity 1-2 1-3 2-4 3-4 3-5 4-6 5-6 Time Estimates ( Weeks ) Direct cost Estimates ( Rs in Thousands ) Normal Crash Nomal Crash 2 1 10 15 8 5 15 21 4 3 20 24 1 1 5 7 2 1 8 15 5 3 1 16 6 2 1 36 ( i ) Draw the project network corresponding to the normal time. ( ii ) Determine the critical path and the normal duration and normal cost of the project. ( iii ) Crash the activities so that the project completion time reduces to 9 weeks with minimum additional cost. Solution : 2 1 8 2 3 4 1 4 5 6 2 5 6 The Total Paths available are : Paths Duration 1 2 4-6 = 11 ➌ 1 3 4-6 = 14 ➋ 1 3 5 6 = 16 ➊ Page 69 of 380 26/4/2008

Activity Maximum Crashing Possible Cost Slope 1-2 2-1 = 1 (15-10 ) / ( 2-1 ) = 5 1-3 8-5 = 3 2 2-4 4-3 = 1 4 3-4 1-1 = 0 0 3-5 2-1 = 1 7 4-6 5-3 = 2 3 5-6 6-2 = 4 6 ➊ 1 3 5-6 16 / 13 / 11 / 9 Critical Activity Max Crashing Cost Rank Remarks Available slope 1 3 3 2000 ❶ ( i ) 3 5 1 7000 ❸ ( iii ) 5 6 4 / 2 6000 ❷ ( ii ) ➋ 1 3 4 6 14 / 13 / 11 / 9 1 3 3 2000 ( i ) 3 4 0 0 ( iii ) 4 6 2 3000 ( ii ) ➌ 1 2-4 6 11 / 11 / 11 / 9 Step i 1 2 1 5000 2 4 1 4000 4 6 2 3000 ( iii ) On CP (➊) Crash Activity 1 3 by 3 weeks. ➊ = 13 ➋ = 11 ➌ = 11 New project duration will be 13 weeks, And in cost 3 x 2000 = 6000/-. No other path becomes critical. Page 70 of 380 26/4/2008

Step ii Step iii On CP (➊) crash activity 5 6 by 2 weeks. ➊ = 11 ➋ = 11 ➌ = 11 New project duration will be 11 days, And in cost 2 x 6000 = 12,000/-. Now all paths are critical. On CP (➊) Crash Activity 5 6 by 2 weeks. CP (➋) Crash Activity 4 6 by 2 weeks CP (➌) Crash Activity 4 6 by 2 weeks ➊ = 9 ➋ = 9 ➌ = 9 New Project duration 11 days. And in cost 2 x 6000 + 2 x 3000 + 2 x 3000 = 18,000/- Now since all activities on CP (➊) are crashed, Crashing is over. Conclusion : Normal Path Duration = 16 weeks Normal Cost = Rs. 82,000/- Crash Path Duration = 11 weeks Crash Cost = 82,000 /- + 6,000 /- + 12,000 /- + 18,000 /- -------------- 1,18,000/- Page 71 of 380 26/4/2008

Problem - 6 The Basic Time Data for the Jobs in a project are as follows : Activity A B C D E F G H Normal Crash Time ( Days ) Cost ( Rs ) Time ( Days ) Cost ( Rs ) 3 140 2 210 6 215 5 275 2 160 1 240 4 130 3 180 2 170 1 250 7 165 4 285 4 210 3 290 3 110 2 160 The activity (Job) dependencies are as follows : (i) (ii) (iii) (iv) (v) A,B,C are starting activities Activities D,F,E can start when A is completed. Activity G can start after B and D are completed. Activity H can start after C and E are completed. Activity F,G and H are the final activities. (a) Draw the Network and indicate critical path ( A D G ) (b) What is the total time required to complete the project (c) If the project is to be completed in 9 days, what is the minimum cost to be incurred? (d) What is the least cost schedule? Solution : A B C D E F G H - - - A A A B,D C,E 3 B 6 D 4 4 G 1 A 3 2 F 7 5 2 C E 2 H 6 4 Page 72 of 380 26/4/2008

Paths Duration 1 3 5 = 10 ➌ 1 2 5 = 10 ➋ 1 4 5 = 5 ➎ 1 2 4 5 = 8 ➍ 1 2 3 5 = 11 ➊ Activity Maximum Crashing Possible Cost Slope 1-2 3-2 = 1 70 1-3 6-5 = 1 60 1-4 2-1 = 1 80 2-3 4-3 = 1 50 2-4 2-1 = 1 80 2-5 7-4 = 3 120 3-5 4-3 = 1 80 4-5 3-2 = 1 50 Critical Activity Max Crashing Cost Rank Remarks Available slope ➊ 1 2 3-5 11 / 10 / 9 / 8 1 2 1 70 ❷ ( iii ) 2 3 1 50 ❶ ( i ) 3 5 1 80 ❸ ( ii ) ➋ 1 2 5 10 / 10 / 9 / 8 1 2 1 70 ( iii ) 2 5 3 2 120 ( ii ) ➌ 1 3-5 10 / 10 / 9 / 8 1 3 1 60 ( iii ) 3 5 1 80 ( ii ) ➍ 1 2 4-5 8 / 8 / 8 / 8 1 2 1 70 2 4 1 80 4 5 1 50 / Page 73 of 380 26/4/2008

➎ 1 4-5 5 / 5 / 5 / 5 Step i 1 4 1 80 4 5 1 50 On CP (➊) Crash Activity 2 3 by 1 day. ➊ = 10 ➋ = 10 ➌ = 10 New project duration will be 10 days, And in cost 1 x 50 = 50/-. No other paths ➋ & ➌ becomes critical. Now there are 2 critical activities which are common on 2 paths, that is : Activity 1 2 is common on CP (➊) & CP (➋) 3 5 is common on CP (➊) & CP (➌) So, Option 1. Option 2. If we crash on CP (➊) & CP (➋), 1 2 by 1 day, in cost Rs. 70/- 1 3 by 1 day, in cost Rs. 60/- --------- 130/- If we crash on CP (➊) & CP (➌), 3 5 by 1 day, in cost Rs. 80/- 2 5 by 1 day, in cost Rs. 40/- --------- 120/- As the cost is less on option 2, we prefer option 2. Page 74 of 380 26/4/2008

Step ii On CP (➊) & CP (➌) 3 5 by 1 day, CP (➋) 2 5 by 1 day New project duration 9 days. in cost Rs. 80/- + 40/- = 120/- Note : Double crashing is not permitted and once a path is made critical it should remain critical throughout. No other path becomes critical. Step iii On CP (➊) & CP (➋) Crash Activity 1 2 by 1 day and CP (➌) Crash Activity 1 3 by 1 day New Project duration 8 days. And in cost 1 x 90 + 1 x 60 = 130/- Now since all activities on CP (➊) are crashed, therefore Crashing is over. Conclusion : Normal Path Duration = 11 Days Normal Cost = Rs. 1,300/- Crash Path Duration = 8 Days Crash Cost = 1,300 /- + 50 /- + 120 /- + 130 /- -------------- 1,600 /- Page 75 of 380 26/4/2008

Problem - 7 Normal Crash time and Cost are given below for a plant expansion project. Activity A B C D E F G H Preceding Activity Normal Time Crash Time Normal Cost Crash Cost ( Months ) ( Months ) ( Rs 000 ) ( Rs 000 ) - 3 2 40 50 A 6 4 200 300-3 2 20 35 C 2 1 20 32 A 1 1 20 20 D 5 3 150 190 D,E 7 6 120 150 B,F,G 4 3 160 195 If the company has Rs. 7,76,000/- available for this project, how should the funds be allocated to minimize overall completion time, to the nearest 0.1 month? What is the minimum completion time? Solution : A B C D E F G H - A - C A D D,E B,F,G 2 A 3 1 E 6 B 1 5 G 7 H 6 4 7 3 C D 0 0 F 5 3 D 2 4 Page 76 of 380 26/4/2008

Total available paths are Paths Duration 1 2 5 6-7 = 15 ➋ 1 2 6-7 = 13 ➍ 1 3 4-5 6-7 = 16 ➊ Critical Path 1 3 4 6 7 = 14 ➌ Critical Activity Max Crashing Cost Rank Remarks Available slope ➊ 1 3 4 5 6-7 16 / 15 / 14 / 13.7 1 3 1 15 ❷ ( ii ) 3 4 1 12 ❶ ( i ) 4 5 - - 5 6 1 / 0.7 30 ❸ ( iii ) 6 7 1 35 ❹ ➋ 1 2 5 6-7 15 / 15 / 14 / 13.7 1 2 1 10 ( ii ) 2 5 - - 5 6 1 / 0.7 30 ( iii ) 6 7 1 35 ➌ 1 3-4 6-7 14 / 14 / 12 / 12 1 3 1 60 ( ii ) 3 4 1 80 4 6 2 60 6 7 1 60 ➍ 1 2 6-7 13 / 13 / 13 / 13 1 2 1 10 2 6 2 50 6 7 1 35 Page 77 of 380 26/4/2008

Step i On CP (➊) Crash Activity 3 4 by 1 month. New project duration will be 15 months, And in cost 1 x 12,000 = 12,000/-. Now there are two options. Option 1. Option 2. Leftover is 46,000 12,000 = 34,000 Now path ➋ also becomes critical. If we crash Un common activities that is, Crash on CP (➊) activity C ( 1 3 ) by 1 month and on CP (➋) activity A ( 1-2 ) by 1 month in cost will be 1 x 15,000 + 1 x 10,000 = 25,000/- If we crash common activities that is, So option 1 is preferred. Step ii Crash on CP (➊) & CP (➋) activity C ( 6 7 ) by 1 month. in cost will be 1 x 35,000 = 35,000/- On CP (➊) activity C ( 1 3 ) by 1 month and CP (➋) activity A ( 1-2 ) by 1 month New project duration 14 Months in cost will be 1 x 15,000 + 1 x 10,000 = 25,000/- No other path becomes critical. Now leftover fund will be 34,000 25,000 = 9,000/- Page 78 of 380 26/4/2008

Step iii Now activity G ( 5 6 ) has the least cost slope and which is common to both critical paths. But for crashing 5 6 by 1 month in cost will be Rs 30,000/- however leftover fund is Rs. 9,000/-. 9,000 So 1 month x ---------- = 0.3 months 30,000 So crash 5 6 by 0.3 months New Project duration 13.7 months. And in cost 9,000/- Now since No fund is available for crashing, no further crashing is possible. Conclusion : Normal Path Duration = 16 months Normal Cost = Rs. /- Crash Path Duration = 13.7 months Leftover fund is nil Page 79 of 380 26/4/2008

Problem 8 The time-cost estimates for the various activities of a project are given below : Activity A B C D E F G Preceding Activity Time ( In weeks ) Cost ( In Rs ) Normal Crash Normal Crash - 8 6 8,000 10,000-7 5 6,000 8,400 A 5 4 7,000 8,500 B 4 3 3,000 3,800 A 3 2 2,000 2,600 D.E 5 3 5,000 6,600 C 4 3 6,000 7,000 The project manger wishes to complete the project in the minimum possible time. However he is not authorised to spend more than Rs. 5,000/- on crashing. Suggest the least Cost schedule for achieving the objective of the project manager. Assume that there is no indirect or utility cost. Solution : A B C D E F G - - A B A D,E C C 5 4 G 4 6 1 A 8 B 7 2 3 E 3 D 4 5 F 5 Page 80 of 380 26/4/2008

Total available paths are Paths Duration 1 2 4 6 = 17 ➊ Critical Path 1 2 5-6 = 16 ➋ 1 3 5-6 = 16 ➌ Critical Activity Max Crashing Cost Rank Remarks Available slope ➊ 1 2 4 6 17 / 16 / 15 / 14 A 1 2 2 1 1,000 ❶ ( ii ) ( iii ) C 2 4 1 1,500 ❸ G 4 6 1 1,000 ❷ ( i ) ➋ 1 2 5 6 16 / 16 / 15 / 14 A 1 2 2 1 1,000 ❸ ( ii ) ( iii ) E 2 5 1 600 ❶ F 5 6 1 300 ❷ ➌ 1 3-5 6 16 / 16 / 15 / 14 / / B 1 3 1 1,200 ❸ ( iii ) D 3 5 1 800 ❶ ( ii ) F 5 6 2 800 ❷ We can crash either A or C as cost slope are disadvantage of A. Only part crashing can crash only by 1 day. ➊ & ➋ become critical advantage of G all 3 becomes critical and more combination allowed full crashing possible. Step i On CP (➊) Crash Activity G ( 4 6 ) by 1 week. New project duration will be 16 weeks, And in cost 1 x 1,000 = 1,000/-. Now path ➋ & ➌ also becomes critical. Leftover fund will be Rs, 4000/- Page 81 of 380 26/4/2008

Now there are two options. Option - 1. On CP (➊) crash 1 2 by 1 week Rs 1,000/- CP (➌) crash 3 5 by 1 week Rs 800/- ----------- 1,800/- Option - 2. On CP (➋) crash 5 6 by 1 week Rs 800/- CP (➌) crash 2 4 by 1 week Rs 1,500/- ----------- 2,300/- So choose option 1. Step ii On CP (➊) activity C ( 1 2 ) by 1 week and CP (➋) activity A ( 3-5 ) by 1 week New project duration 15 weeks in cost will be Rs. 1800 /- Now leftover fund will be Rs 4,000 Rs 1,800 = Rs 2,200/- Now again we have two options. Option - 1. On CP (➊) crash 1 2 by 1 week Rs 1,000/- CP (➌) crash 1 3 by 1 week Rs 1,200/- ----------- 2,200/- Option - 2. On CP (➋) crash 5 6 by 1 week Rs 800/- CP (➌) crash 2 4 by 1 week Rs 1,500/- ----------- 2,300/- So option 1 is suitable. Page 82 of 380 26/4/2008

Step iii CP (➊) crash 1 2 CP (➌) crash 1 3 by 1 week by 1 week New Project duration 14 weeks. And in cost 2,200/- Now since No fund is available for crashing, no further crashing is possible. Conclusion : Page 83 of 380 26/4/2008

1 st March, 2008. Lecture VI. Problem - 9 The table below gives durations, costs of activities of a project. Activity 1-2 1-3 2-4 3-4 3-5 4-6 5-6 Time in days Cost in Rs. Normal Crash Normal Crash 2 1 100 150 8 5 150 210 4 3 200 280 1 1 70 70 2 1 80 150 5 3 100 190 6 2 120 360 (a) Draw a network for the project and determine the critical path (b) What are the normal project duration and the associated cost? (c) Crash the activities of the project to find the minimum cost and the corresponding duration given that indirect cost per day is Rs. 40/-. Solution : 2 1 4 2 3 4 1 4 5 6 2 5 6 Total available paths are Paths Duration 1 2 4 6 = 11 ➌ 1 3 4-6 = 14 ➋ 1 3 5-6 = 16 ➊ Critical Path Page 84 of 380 26/4/2008

Critical Activity Max Crashing Cost Rank Remarks Available slope ➊ 1 3 5 6 16 / 13 / 15 / 14 1 3 3 20 ❶ ( i ) 3 5 1 70 ❷ 5 6 4 60 ❸ ➋ 1 3 4 6 14 / 13 / 15 / 14 1 3 3 20 ❷ ( ii ) ( iii ) 3 4 - - 4 6 2 25 ❶ ➌ 1 2-4 6 11 / 1 2 2 4 4 6 ( iii ) ( ii ) Duration Days Direct Cost Indirect Cost Total Cost 16 820 16 x 40 = 640 1460 15 820 + 20 ( 1 3) = 840 15 x 40 = 600 1440 14 840 + 20 ( 1 3) = 860 14 x 40 = 560 1420 13 860 + 20 ( 1 3) = 880 13 x 40 = 520 1400 12 In complete problem. Page 85 of 380 26/4/2008

Problem - 10 The detailed activities in building project are given below : Activity A B C D E F G H Preceding Normal Crash Activity Time ( In Days) Cost ( In Rs ) Time ( In Days) Cost ( In Rs ) - 9 12,000 6 18,000 A 14 14,000 4 24,000 A 4 2,000 3 2,400 C 6 44,000 4 56,000-14 1,600 13 1,800 E 6 4,000 6 4,000 B,C 5 4,000 3 4,800 F,G 2 12,000 1 14,000 (i) Obtain critical Path (ii) Indirect Cost Per day Rs. 5,000/- (iii) Find the Minimum time schedule and corresponding cost. Page 86 of 380 26/4/2008

Problem - 11 A Small marketing project consists of the jobs in the table given below. With each job is listed its normal time and minimum or crash time (in days). The cost (in Rupees per day) of crashing each job is also given. Job 1-2 1-3 1-4 2-4 3-4 4-5 Normal Duration ( Days ) Minimum ( Crash ) Duration ( Days ) Cost of Crashing ( Rs. Per Day ) 9 6 20 8 5 25 15 10 30 5 3 10 10 6 15 2 1 40 (a) What is the normal project length and the minimum project length? (b) Overhead costs total Rs. 35/- per day, what is the optimal length schedule? Solution : 2 9 5 1 15 4 2 5 8 10 3 Total available paths are Paths Duration 1 2 4 5 = 16 ➌ 1 4-5 = 17 ➋ 1 3 4-5 = 20 ➊ Critical Path Page 87 of 380 26/4/2008

Critical Activity Max Crashing Cost Rank Remarks Available slope ➊ 1 3 4 5 20 / 17 / 16 / 15 / 12 1 3 3 25 ❷ ( iv ) 3 4 4 / 1 15 ❶ ( i ) ( iii ) 4 5 1 40 ❸ ( ii ) ➋ 1 4 5 17 / 17 / 16 / 15 / 12 1 4 5 4 1 30 ❶ ( iii ) ( iv ) 4 5 1 40 ❷ ( ii ) ➌ 1 2-4 5 16 / 16 / 15 / 15 / 12 / / / 1 2 3 2 20 ❷ ( iv ) 2 4 2 10 ❶ ( iv ) 4 5 1 40 ❸ ( ii ) Step i On CP (➊) Crash Activity 3 4 by 3 days. New project duration will be 17 days, And in cost 3 x 15 = Rs 45/-. Now path ➋ also becomes critical. Step ii On Both Critical Paths crash activity 4 5 by 1 day. New project duration 16 days. in cost will be Rs. 1 x 40 = Rs 40/- No other Path becomes critical. Page 88 of 380 26/4/2008

Step iii CP (➊) crash 3 4 CP (➋) crash 1 4 by 1 day by 1 day Step iv New Project duration 15 weeks. And in cost will be 1 x 15 + 1 x 30 = Rs 45/- Now path ➌ also becomes critical. On CP (➊) Crash Activity 1 3 by 3 days & CP (➋) Crash Activity 1 4 by 2 days & CP (➌) Crash Activity 2 4 by 2 days & CP (➌) Crash Activity 1 2 by 1 day New Project duration 12 days. And in cost 3 x 25 + 3 x 30 + 2 x 10 + 1 x 20 = Rs 205/-. Now since all activities on CP (➊) are crashed, Crashing is over. Conclusion : Normal Path Duration = 20 days Normal Cost = Rs. 5,000/- ( Around ) Crash Path Duration = 12 days. Crash Path Cost = 5,000 /- + 45 /- + 40 /- + 45 /- + 205 /- -------------- 5,335 /- In step 4 we did 3 days crashing from 15 to 12 days. Page 89 of 380 26/4/2008

15 to 14 Days On CP (➊) 1 3 1 25/- CP (➋) 1 4 1 30/- CP (➌) 2 4 1 10/- ------- 65/- 14 to 13 Days On CP (➊) 1 3 1 25/- CP (➋) 1 4 1 30/- CP (➌) 2 4 1 10/- ------- 65/- 13 to 12 Days On CP (➊) 1 3 1 25/- CP (➋) 1 4 1 30/- CP (➌) 1 2 1 20/- ------- 75/- Steps i ii iii iv Duration Days Direct Cost Indirect Cost Total Cost 20 5000 ( around ) 35 x 20 = 700 5700 19 5000 + 15 ( 3 4 ) = 5015 35 x 19 = 665 5680 18 5015 + 15 ( 3 4 ) = 5030 35 x 18 = 630 5660 17 5030 + 40 ( 4 5 ) = 5045 35 x 17 = 595 5640 ➈ 16 5000 + 45 ( 4 5 ) = 5085 35 x 16 = 560 5645 15 5085 + 45 ( 3 4 ) 15 ( 1 4 ) 30 = 5030 14 5130 + 65 ( 1 3 ) 25 ( 1 4 ) 30 ( 2 4 ) 10 = 5195 13 5195 + 65 ( 1 3 ) 25 ( 1 4 ) 30 ( 2 4 ) 10 = 5260 12 5260 + 75 ( 1 3 ) 25 ( 1 4 ) 30 ( 1 2 ) 20 = 5335 35 x 15 = 525 5655 35 x 14 = 490 5685 35 x 13 = 455 5715 35 x 12 = 420 5755 Page 90 of 380 26/4/2008

Optimum cost is 5,640/- Optimum duration = 17 days. Page 91 of 380 26/4/2008

Problem - 12 List of activities for constructing canteen in a factory is given below with other relevant details. Job A must precede all other jobs while Job E must follow all other jobs apart from the jobs can run concurrently. Code A B C D E Job Description Duration ( Days ) Normal Cost ( Rs. ) Crash Duration ( Days ) Cost ( Rs. ) Lay foundation & build Walls 5 3000 4 4000 Tile roofing 6 1200 2 2000 Install Electricity 4 1000 3 1800 Installl Plumbing 5 1200 3 2000 Connect sevices to Finish 3 1000 3 1600 i) Draw the Network & identify critical Path ii) iii) Crash the Network fully to find out minimum duration If indirect costs are Rs 300/- per day. Determine time Cost trade off for the project. Solution : A B C D E - - A A B,C,D 3 B 6 0 D 1 1 A 5 2 C 4 5 E 3 6 D 5 D 2 0 4 Page 92 of 380 26/4/2008

Paths Duration 1 2 5 6 = 12 ➌ 1 2-3 5 6 = 14 ➊ Critical Path 1 2-4 5 6 = 13 ➋ Critical Activity Max Crashing Cost Rank Remarks Available slope ➊ 1 2-3 5 6 14 / 13 / 12 / 11 / 10 1 2 1 1000 ❷ ( iii ) 2 3 4 / 3 / 2 / 200 ❶ ( i ) ( ii ) ( iv ) 3 5 - - ( ii ) 5 6 - - ➋ 1 2-4 5 6 13 / 13 / 12 / 11 / 10 1 2 1 1000 ❷ ( iii ) 2 4 2 / 1 / 400 ❶ ( ii ) ( iv ) 4 5 - - 5 6 - - ➌ 1 2-5 6 12 / 12 / 12 / 11 / 10 Step i 1 2 1 1000 ❷ ( iii ) 2 5 1 800 ❶ ( iv ) 5 6 - - On CP (➊) Crash Activity 2 3 by 1 day. New project duration will be 13 days, And in cost 1 x 200 = Rs 200/-. Now path ➋ also becomes critical. Step ii On CP (➊) Crash Activity 2 3 by 1 day. CP (➋) Crash Activity 2 4 by 1 day. Page 93 of 380 26/4/2008

Step iii Step iv New project duration 12 days. in cost will be Rs. 1 x 200 + 1 x 400 = Rs 600/- Now all 3 paths are critical. Crash on all paths activity 1 2 by 1 day New Project duration 11 days. And in cost will be 1 x 1000 = Rs 1,000/- On CP (➊) Crash Activity 2 3 by 1 day & CP (➋) Crash Activity 2 4 by 1 day & CP (➌) Crash Activity 3 5 by 1 day New Project duration 10 days. And in cost 1 x 200 + 1 x 400 + 1 x 800 = Rs 1,400/-. Now since all activities on CP (➌) are crashed, Crashing is over. Conclusion : Normal Path Duration = 14 days Normal Cost = Rs. 8,000/- Crash Path Duration = 10 days Crash Cost = Rs. 11,200/- Duration Direct Cost Indirect Cost Total Cost Days 14 8000 14 x 300 = 4,200/- 12,200 13 12,200 + 13 x 300 = 3,900/- ( 2 3 ) 200 = 14,400/- 12 11 10 Page 94 of 380 26/4/2008

Problem - 13 The following tables gives the activities in a construction project and other relevant information. Activity A B C D E F G Immediate Predecessors Time ( Days ) Direct Cost ( Rs ) Normal Crash Normal Crash - 4 3 60 90-6 4 150 250-2 1 30 60 A 5 3 150 250 C 2 2 100 100 A 7 5 115 175 D,B,E 4 2 100 240 Indirect Costs vary as follows : Days 15 14 13 12 11 10 9 8 7 6 Cost ( Rs. ) 600 500 400 250 175 100 75 50 35 25 (a) (b) Draw an arrow diagram for the project. Determine the project duration which will result in minimum total project cost. Solution : A B C D E F G - - - A C A D,B,E 1 2 A C 4 2 B 6 D 5 E 2 F 7 4 G 4 5 3 Page 95 of 380 26/4/2008

Paths Duration 1 2 4 5 = 13 ➊ Critical Path 1 2-5 = 11 ➋ 1 3-4 5 = 8 ➍ 1-4 5 = 10 ➌ Critical Activity Max Crashing Cost Rank Remarks Available slope ➊ 1 2-4 5 13 / 12 / 10 / 10 1 2 1 30 ❶ ( i ) 2 4 2 50 ❷ ( ii ) 4 5 2 70 ❸ ( iii ) ➋ 1 2 5 11 / 10 / 10 / 10 1 2 1 30 ❶ ( i ) 2 5 2 30 ❷ ( iii ) ➌ 1 4 5 10 / 10 / 10 / 10 Step i 1 4 2 50 ❶ 4 5 2 70 ❷ ( iii ) On CP (➊) Crash Activity 1 2 by 1 day & CP (➋) Crash Activity 1 2 by 1 day New project duration will be 12 days, And in cost 1 x 30 + 1 x 30 = Rs 60/-. No other path becomes critical. Step ii On CP (➊) Crash Activity 2 4 by 2 days. New project duration 10 days. in cost will be Rs. 2 x 50 = Rs 100/- Now all 3 paths are critical. Page 96 of 380 26/4/2008

Step iii Conclusion : On CP (➊) Crash Activity 4 5 by 2 days & CP (➋) Crash Activity 2 5 by 2 days New Project duration 8 days. And in cost will be 2 x 70 + 2 x 30 = Rs 200/- Normal Path Duration = 13 days Normal Cost = Rs. 705/- Crash Path Duration = 8 days. Crash Path Cost = Rs 1035/- Duration Days Direct Cost Indirect Cost Total Cost 13 705 400 1105 12 705 + ( 1 2 ) 30 = 735 250 985 11 735 + Page 97 of 380 26/4/2008

7 th March, 2008. Lecture VII. Problem - 14 Consider the following activities, associated normal time and cost together with extra cost of saving a day on selected activities. Activity Normal Cost of Activities Normal Time in Days Extra Cost of Reducing Normal Time by one Day ( Rs ) A to B A to C A to D B to C B to E B to F C to D C to F D to F E to F 10000 4 3000 12000 6 2000 5000 2 800 6000 5 600 9000 2 2000 5000 7 900 4000 4 700 3000 3 200 5000 7 2200 6000 12 500 No single activity can be reduced by more than one day. In addition site costs will be incurred at the rate of Rs. 3,500 for every day that the work is in progress on the site irrespective of how many activities are in progress. Calculate and state : (a) (b) (c) The Critical path using normal times, the time taken and the total cost of completing the project. The shortest time in which the project can be completed the associated cost and the critical path. The lowest cost for which the project can be completed and associated time. Page 98 of 380 26/4/2008

Solution : B 8 E 5 10 6 12 A 7 C 3 F 9 4 7 D Paths Duration A B C D F = 26 ➊ A B C F = 18 ➌ Critical Path A B - E F = 25 ➋ A B F = 11 A C - D F = 18 A C F = 10 A D F = Critical Activity Max Crashing Cost Rank Remarks Available slope ➊ A B - C D F 26 / 25 / 24 / 23 / 22 A B 1 3000 ❹ ( iii ) B C 1 600 ❶ ( i ) C D 1 700 ❷ ( ii ) D F 1 2200 ❸ ( iv ) ➋ A B E F 25 / 25 / 24 / 23 / 22 A B 1 3000 ❸ ( iii ) B E 1 2000 ❷ ( iv ) E F 1 500 ❶ ( ii ) Page 99 of 380 26/4/2008

Step i On CP (➊) Crash Activity B C by 1 day & New project duration will be 25 days, And in cost 1 x 600 = Rs 600/-. Now both path will be critical. Step ii On CP (➊) Crash Activity C D by 1 day & CP (➋) Crash Activity E F by 1 day New project duration will be 24 days, And in cost 1 x 700 + 1 x 500 = Rs 1,200/-. No other path becomes critical. Step iii Step iv On CP (➊) & (➋) Crash common Activity A B by 1 day New Project duration 23 days. And in cost will be 1 x 3000 = Rs 3,000/- No other path becomes critical. On CP (➊) Crash Activity D F by 1 day & CP (➋) Crash Activity B E by 1 day New Project duration 22 days. And in cost will be 1 x 2,200 + 1 x 2,000 = Rs 4,200/- Now since all activities on CP (➊) are crashed crashing is over. Conclusion : Normal Path Duration = 26 days Normal Cost = Rs. 65,000/- Crash Path Duration = 22 days. Crash Path Cost = Rs 74,000/- Page 100 of 380 26/4/2008

Duration Days Direct Cost Indirect Cost Total Cost 26 65,000/- 26 x 3,500/- = 91,000/- 1,56,000/- 25 65000 + ( B C ) 600 = 65,600/- 25 x 3,500/- = 87,500/- 1,53,100/- 24 65600 + ( C D ) 700 + ( E F ) 500 = 66,800/- 24 x 3,500/- = 84,500/- 1,50,800/- 23 22 Optimum Duration is 23 Days. Optimum Cost Rs 1,50,300/- Page 101 of 380 26/4/2008

Problem - 15 The time cost estimates of different activities of a project and their precedence relationship are given below. Activity A B C D E F Preceding Activity Time ( In weeks ) Cost ( Rs. ) Normal Crash Normal Crash - 6 4 10000 14000-2 1 5000 8000 A 4 3 4000 5000 B 8 3 1000 6000 B 14 6 9000 13000 C,D 6 2 7000 8000 Overhead costs amount to Rs. 1,000 per week. It is stipulated that the contractor will have to pay a penalty of Rs 2,000/- per week for completing the project beyond 12 weeks. (i) Draw the Network (ii) What is the duration of the critical path? (iii) Crash the project so that the project including penalty is minimum. Solution : 2 A 6 4 C 1 2 B 4 F D 6 8 3 E 5 14 Page 102 of 380 26/4/2008

Paths Duration 1 2 4 5 = 16 ➌ 1 3 4 5 = 16 ➊ 1 3 5 = 16 ➋ All 3 are the Critical Paths Critical Activity Max Crashing Cost Rank Remarks Available slope ➊ 1 2-4 5 16 / 12 / 11 / 9 1 2 2 2000 ❸ ( iii ) 2 4 1 1000 ❷ ( ii ) 4 5 4 250 ❶ ( i ) ➋ 1 3 4 5 16 / 12 / 11 / 9 1 3 1 3000 ❸ 3 4 5 / 4 / 2 1000 ❷ ( ii ) ( iii ) 4 5 4 250 ❶ ( i ) ➌ 1 3 5 16 / 12 / 11 / 9 Step i 1 3 1 3000 ❷ 3 5 8 4 3 1 500 ❶ ( i ) ( ii ) ( iii ) / / / On CP (➊) & CP (➋) Crash Activity 4 5 by 4 weeks & CP (➌) Crash Activity 3 5 by 4 weeks New project duration will be 12 weeks, And in cost 4 x 250 + 4 x 500 = Rs 3,000/-. Page 103 of 380 26/4/2008

Step ii On CP (➊) Crash Activity 2 4 by 1 week & CP (➋) Crash Activity 3 4 by 1 week & CP (➌) Crash Activity 3 5 by 1 week New project duration will be 11 weeks, And in cost 1 x 1000 + 1 x 1000 + 1 x 500 = Rs 2,500/-. Step iii On CP (➊) Crash Activity 1 2 by 2 weeks & CP (➋) Crash Activity 3 4 by 2 weeks & CP (➌) Crash Activity 3 5 by 2 weeks New Project duration 9 weeks. And in cost will be 2 x 2000 + 2 x 1000 + 2 x 500 x = Rs 7,000/- Now since all the Activities on CP (➊) are crashed crashing is over. Conclusion : Normal Path Duration = 16 weeks Normal Cost = Rs. 36,000/- Crash Path Duration = 9 weeks. Crash Path Cost = Rs 36,000/- + 3,000/- + 2,500/- + 7,000/- --------------- Rs 47,500/- Page 104 of 380 26/4/2008

Duration weeks 16 36,000/- Direct Cost 15 36,000 + ( 4 5 ) 250 + ( 3 5 ) 500 = 36,750/- 14 36,750 + ( 4 5 ) 250 + ( 3 5 ) 500 = 37,500/- 13 37,500 + ( 4 5 ) 250 + ( 3 5 ) 500 = 38,250/- 12 38,250 + ( 4 5 ) 250 + ( 3 5 ) 500 = 39,000/- 11 39,000 + ( 2 4 ) 1000 + ( 3 4 ) 1000 + ( 3 5 ) 500 = 41,500/- 10 41,500 + ( 1 2 ) 2000 + ( 3 4 ) 1000 + ( 3 5 ) 500 = 45,000/- 9 45,000 + ( 1 2 ) 2000 + ( 3 4 ) 1000 + ( 3 5 ) 500 = 48,500/- Indirect Cost 16 x 1,000/- = 16,000/- 15 x 1,000/- = 15,000/- 14 x 1,000/- = 14,000/- 13 x 1,000/- = 13,000/- 12 x 1,000/- = 12,000/- 11 x 1,000/- = 11,000/- 10 x 1,000/- = 10,000/- 9 x 1,000/- = 9,000/- Penalty @ Rs 2,000/- per week after delay of 12 Weeks ( 16-12 ) = 4 weeks i.e. Rs 8,000/- ( 15-12 ) = 3 weeks i.e. Rs 6,000/- ( 14-12 ) = 2 weeks i.e. Rs 4,000/- ( 13-12 ) = 1 weeks i.e. Rs 2,000/- Total Cost Rs. 60,000/- Rs. 57,750/- Rs. 55,500/- Rs. 53,250/- Rs. 51,000/- Rs. 52,500/- Rs. 55,000/- Rs. 57,500/- Optimum Duration is 12 weeks Optimum cost is Rs. 51,000/- Page 105 of 380 26/4/2008

Network A Network is a graphic representation of all the tasks that must be completed to achieve the end object of a project, showing their interdependence and interrelation. The components, which make up a network are events and activities. PERT In 1958, the United States Navy needed a way to monitor and control the Prolaris Missile Programme. It especially needed a method for minimizing the conflicts, delays and interruptions that so frequently plague accomplish that, the Navy developed the Program Evaluation and Review Technique (PERT). ACTIVITY An Activity is performance of work between any two events. It is a time consuming and resource consuming element in a Network Plan. An Activity is represented by means of an arrow. DUMMY ACTIVITY There is another component of Network, which is called ZERO time activity. It is an activity which consumes neither time nor resources. This is known as Dummy Activity. It is represented by means of broken arrow in a Network Diagram. CPM E I DU PONT DE NUMERIOUR & CO a huge chemical company, sought to devise a way to determine accurate time and cost estimates for the construction of several chemical plants. The firm needed some way to ascertain whether work was falling behind schedule at any given point in time and whether to action to bring the work back on schedule. Between 1956 and 1958, it developed the famous Critical Path Method (CPM). EVENT An event is specific definable accomplish in a project, recognizable at a particular instant of time. An event does not consume either time or response. It is a point in time and not a passage of time. Examples of events on a Project Network are either a commencement point or a completion point of a task to be performed. Page 106 of 380 26/4/2008

RULES FOR DEVELOPING A NETWORK 1. An event can not occur until all the activities leading to it are completed. 2. An activity can not start unless and until its preceding event has occurred. 3. An event once having occurred can not occur again. 4. Time flows from left to right. 5. Arrow flows from left to right. 6. Every activity will have to be completed to reach end event of the Network. 7. Length of arrows has no significance and they do not represent the duration. CPM i) CPM is activity oriented ii) The duration of each activity is assumed to be constant. iii) A Single time estimates is established for each activity. iv) CPM places dual emphasis on time and cost and evaluates the trade-off between project cost and project time. v) CPM is widely used in construction projects. PERT i) PERT is event oriented ii) PERT allows uncertainly in the duration. iii) In PERT each activity is assigned three time estimates i.e. optimistic, most likely and pessimistic. iv) PERT is primarily concerned with time. It describes basic network technique Which includes, Planning, Monitoring, and control of Projects. v) PERT is widely used in R & D Projects Page 107 of 380 26/4/2008

PERT (1) Optimistic Time ( a or t o ) : This is the shortest time an activity can take to complete. For this estimate, no provision for daily or setbacks are made. It represents an ideal estimates. (2) Most Likely Time ( m or t m ) : This refers to the time that would be expected to occur most often if the activity where frequently repeated under exactly the same conditions. It assumes that things go in a normal way with few setbacks. It is the modal time. (3) Pessimistic Time ( b or t p ) : This is the longest time the activity could take to finish. If everything went wrong and abnormal situations prevailed, this would be the time estimate. Using the values of a, b and m the expected time of various activities and their standard deviations are calculated as follows. The times estimates are reduced into a single expected time ( t e ) with the weighted average formula. te = a + 4m + b 6 The standard deviation σ i of the completion time of an activity is calculated as follows. σ i b a = 6 From this, the variance = σ = b 6 2 a i 2 is calculated. Once the expected time of the activities are obtained, the critical path of the project network is determined using these time estimates. Having found the critical path, the PERT methodology assumes that the sum of the mean times and the summation of the variance of critical jobs would yield the expected project duration and its variance. Page 108 of 380 26/4/2008

To solve the problem on PERT following steps are important. Step - ➊ Arrange the activities in order of their events. e.g. So write them as 1 2 1 2 2 4 2 3 5 6 2 4 2 3.. 5 6 Or B 2 1 A C 3 4 Step - ➋ So write them as, B 1 2 A 1 3 C 1 4 Arrange 3 time estimates in order of a / m / b Step - ➌ Compute te a + 4m + b = and 6 σ = b 6 2 a i 2 Step - ➍ for each activity ( prepare table ) Draw Network considering the t e value of each activity find ONLY critical path. Compute expected project completion time ( ) & Standard deviation ( ) Page 109 of 380 26/4/2008

Problem 1 The following table gives the relevant data of the activities in a PERT project. (i) Draw an arrow diagram of the project. (ii) Calculate the Expected Duration and Variance of the critical path. (iii) Assess the probability that the project will take more than 41 days. (iv) What is the Probability that project will be completed in 31 days or less time? Activity 1-2 7-8 2-3 4-5 3-5 5-8 1-6 2-4 6-7 Duration ( Days ) Optimistic Most Likely Pressimistic 3 6 15 4 19 28 6 12 30 3 6 15 5 11 17 1 4 7 2 5 14 2 5 8 3 9 27 Solution : Activity 1-2 1-6 2-3 2-4 3-5 4-5 5-8 6-7 7-8 2 a + 4m + b b a Duration ( Days ) = = te σi 6 6 a m b 3 6 15 7 4 2 5 14 6 4 6 12 30 14 16 2 5 8 5 1 5 11 17 11 4 3 6 15 7 4 1 4 7 4 1 3 9 27 11 16 4 19 28 8 16 2 Page 110 of 380 26/4/2008

3 14 11 2 5 1 7 5 4 7 4 6 6 7 8 11 18 Paths Duration 1 2 3 5 8 = 36 Critical Path 1 2 4 5-8 = 23 1 6 7 8 = 35 Critical Activity t e σ 2 i 1 2 7 4 2 3 14 16 3 5 11 4 5 8 4 1 ---------- ------------ = 36 σ 2 i = 25 Expected project completion time = = 36 Variance of the Critical path = σ 2 i = 25 Standard deviation of Critical path = = 25 = 5 Days Page 111 of 380 26/4/2008

= 36 = 5 Now let x be the project duration. P (x > 41 ) Z = x µ = σ x 36 5 So when x = 41 z = 41 36 5 = 1 P ( x > 41 ) = P ( z > 1 ) Now 1 will lye on the right side of the bell curve. P1 P2 Now P1 + P2 = 0.5 P2 = 0.5 0.3413 = 0.1587 P ( x > 41 ) = 0.1587 = 15.87 % 0 There is 16 % chance that project will be extended. Page 112 of 380 26/4/2008

Same way what is the probability that the project will be completed by 31 days. P ( x 31 ) Z = x µ = σ x 36 5 So when x = 31 z = 31 36 5 = -1 P ( x 41 ) = P ( z -1 ) P1 P2 0 Now P1 + P2 = 0.5 P2 = 0.5 0.3413 = 0.1587 P ( x > 41 ) = 0.1587 = 15.87 % Project may be competed in 31 days has16 % chance. Page 113 of 380 26/4/2008

12 th March, 2008. Lecture VIII. Problem - 2 The data for PERT network is displayed in the table. Determine the critical path and the expected duration of completion of the entire project. Give Answers to the Followings : (i) What is the probability that the project will exceed 60 days? (ii) What is the chance of completing the project between 45 days and 54 days? (iii) If it becomes known later that the duration of the three time estimates for activity 4 6 has to be revised to 14-20-32. What impact does this have on project completion time? What is the probability that the project can now be completed on or before 46 days? Activity 1-2 1-3 1-4 2-3 2-5 3-4 3-6 3-6 5-6 Time Duration ( Days ) a b c 2 4 6 6 6 6 6 12 24 2 4 8 11 14 23 15 24 45 3 6 9 9 15 27 4 10 16 ( a = optimistic, b = most likely, c = pessimistic ) Page 114 of 380 26/4/2008

Solution : Activity 1-2 1-3 1-4 2-3 2-5 3-4 3-6 4-6 5-6 Duration ( Days ) a m b te 2 a + 4m + b b a = = σi 6 6 2 4 6 4 4/9 6 6 6 6 0 6 12 24 17 9 2 5 8 5 1 11 14 23 15 4 15 24 45 26 25 3 6 9 6 1 9 15 27 16 9 4 10 16 10 4 2 2 5 15 4 5 10 1 3 6 6 6 26 13 16 4 Paths Duration 1 2 3 4 6 = 51 Critical Path 1 2 3 6 = 15 1 2 5 6 = 29 1 3 4 6 = 48 1 3 6 = 12 1 4 6 = 29 Page 115 of 380 26/4/2008

Critical Activity t e σ 2 i 1 2 4 4/9 2 3 5 1 3 4 26 25 4 6 16 9 ---------- ------------ = 51 σ 2 i = 35.44 = 51 = 35, 44 = 5.95 So let the = 51 & = 5.95 Expected project duration = 51 days. Standard diviation of the Critical Path is 5.95 days (i) Find probability that Project Duration exceeds 60 days. Now let x be the project duration. P (x > 60 ) Z = x µ = σ x 51 5.95 So when x = 60 z = 60 51 5.95 = 1.51 P ( x > 60 ) = P ( z > 1.51 ) P1 = 0.4345 P2 = 1.51 0 Page 116 of 380 26/4/2008

As 1.51 will left side ( Table value 0.4345 ) Now we have P1 + P2 = 0.5 So probability P ( x > 60 ) = 0.0655 = 6.55 % P2 = 0.5 0.4345 = 0.0655 (ii) Find probability that Project gets completed between 45 and 54 Days. Now let x be the project duration. Let P ( 45 x 54 ) Z = x µ = σ x 51 5.95 So when x = 45 z = 45 51 5.95 = -1.01 So when x = 54 z = 54 51 5.95 = 0.50 P ( 45 x 54 ) = p ( -1.01 z 0.50 ) P2 = 0.3438 P1 = 0.1915 Z = -1.01 0 Z = 0.5 Page 117 of 380 26/4/2008

The required probability = 0.3438 + 0.1915 = 0.5353 = 53.53 % (iii) Sensitivity Analysis. Please note that the activity 4 6 is on the Critical path. Activity 4-6 Duration ( Days ) a m b 2 a + 4m + b b a = t e = σi 6 6 9 15 27 16 9 old 14 20 32 21 9 New 2 Now Critical Activity t e σ 2 i 1 2 4 4/9 2 3 5 1 3 4 26 25 4 6 21 9 ---------- ------------ = 56 = 35. 44 = 5.95 = 56 σ 2 i = 35 + 4/9 Now what is the probability that project can now be completed on or before 46 days? Let P ( x 46 ) Z = x µ = σ x 56 5.95 Page 118 of 380 26/4/2008

Z = 46 56 5.95 = - 1.68 So p ( x 46 ) = p ( x -1.68 ) P1 P2 Z = -1.68 0 P2 = 0.5 0.4535 = 0.465 p ( x 46 ) = 0.465 = 4.65 % Page 119 of 380 26/4/2008

Problem - 3 A project is characterized by the following activity time : Activity Optimistic Time to Days Pessimistic Time to Days Most Likely Time to Days 1-2 1-3 2-4 3-4 3-5 4-5 4-6 5-6 1 5 3 3 7 5 4 8 10 9 11 3 1 5 12 10 20 5 5 13 6 5 9 5.5 a) Find the critical path and the project completion time (34 days) b) Find the standard deviation of the expected project length (1.94) c) What is the probability that the project will be completed in 35 days. (69.85 %) d) What is the expected project completion time if you are allowed to qualify the same with a chance of 95%. (37.18 Days) Solution : Activity 1-2 1-3 2-4 3-4 3-5 4-5 4-6 5-6 Duration ( Days ) a m b t e 2 a + 4m + b b a = = σi 6 6 1 3 5 3 0.44 3 5 7 5 0.44 4 6 8 6 0.44 9 10 11 10 0.11 1 3 5 3 0.44 10 12 20 13 2.78 5 6 13 7 1.78 5 5.5 9 6 0.44 2 Page 120 of 380 26/4/2008

3 2 6 1 10 4 7 5 3 13 6 3 5 6 Paths Duration 1 2 4 6 = 16 1 2 4 5 6 = 28 1 3 4 6 = 22 1 3 4 5 6 = 34 Critical Path 1 3 5 6 = 14 Critical Activity t e σ 2 i 1 3 5 0.44 3 4 10 0.11 4 5 13 2.78 5 6 6 0.44 ---------- ------------ = 34 σ 2 i = 1.94 days (i) Find probability that Project will be completed in 35 days. Now let x be the project duration. P (x 35 ) Z = x µ = σ x 34 1.96 So when x = 35 z = 35 34 1.96 = 0.52 Page 121 of 380 26/4/2008

P ( x 35 ) = P ( z 0.52 ) P2 = 0.5 P1 P1 = 0.1985 0 Z = 0.52 Required probability = 0.5 + 0.1985 = 0.6985 = 69.85 % Note : If the problem says that what is the probability that project gets completed on 35th Day then the answer is Zero. (ii) What is the expected project completion time if you are allowed to qualify the same with a chance of 95% Now let x be the project duration. P (x K ) = 0.95 0.95-0.50 Standard Deduction ---------- 0.45 Z = 1.64 Z = x 34 1.94 Page 122 of 380 26/4/2008

1.64 = x 34 1.94 x 34 = 1.94 x 1.64 x = 37.18 So chance that 95 % the project will be completed in 37.18 days. Page 123 of 380 26/4/2008

Problem - 4 A small project is composed of eight activities whose time estimate are given below : Activity 1-2 1-3 2-4 2-5 3-5 4-5 4-8 5-6 Duration ( Days ) Pressimistic Most Likely Optimistic 21 7.5 3 27 8 3 8 8 8 3.5 2 0.5 10 10 10 1.7 1 0.3 9 7.5 3 5 3 1 i) Draw the project network and identify all paths through it. ii) What is the expected project completion time? (24 weeks) iii) What is the expected project completion time with a confidence of 90 %. Solution : Activity 1-2 1-3 2-4 2-5 3-5 4-5 4-6 5-6 Duration ( Days ) a m b te 2 a + 4m + b b a = = σi 6 6 3 7.5 21 9 9 3 8 27 10 16 8 8 8 8 0 0.5 2 3.5 2 0.25 10 10 10 10 0 0.3 1 1.7 1 0.05 3 7.5 9 7 1 1 3 5 3 0.44 2 Page 124 of 380 26/4/2008

4 7 8 1 6 2 1 9 2 5 3 10 3 10 Paths Duration 1 2 4 6 = 24 Critical Path 1 2 4 5 6 = 21 1 2 5 6 = 15 1 3 5 6 = 23 Critical Activity t e σ 2 i 1 2 9 9 2 4 8 0 4 6 7 1 ---------- ------------ = 24 σ 2 i = 10 days Standard deviation of Critical path = = 10 = 3.16 Days (ii) What is the expected project completion time with a confidence of 90 %. Now let x be the project duration. P (x K ) = 0.90 0.90-0.50 Standard Deduction ---------- 0.40 Page 125 of 380 26/4/2008

Z = 1.28 Z = x 24 3.16 1.28 = x 24 3.16 x 24 = 3.16 x 1.28 x = 28.04 So chance that 90 % the project will be completed in 28.04 days. Page 126 of 380 26/4/2008

Problem 13 ( University Paper ) A project consists of eight activities with following relevant information. Activity Immediate Estimated Duration ( Days ) Predecessor Optimistic Pessimistic Most likely A - 2 4 3 B - 8 8 8 C - 7 11 9 D A 6 6 6 E B 10 18 14 F C 4 6 5 G D,E 1 1 1 H F,G 3 5 4 i) Draw the PERT network and find expected project completion time. ii) What duration will have 97.7% chance of completion? Solution : ( Area from Z = 0 to Z = 2 is 0.477 ) Activity A B C D E F G H Duration ( Days ) a m b t e 2 a + 4m + b b a = = σi 6 6 2 3 4 3 0 8 8 8 8 0 7 9 11 9 0 6 6 6 6 0.00 10 14 18 14 2 4 5 6 5 0.11 1 1 1 1 0 3 4 5 4 0.11 2 A B C D E F G H - - - A B C D,E F,G Page 127 of 380 26/4/2008

3 B 8 14 E 1 A 3 2 D 6 G 5 1 6 H 4 7 9 C F 5 4 Paths Duration 1 3 5 6 7 = 27 Critical Path 1 2 5-6 7 = 14 1 4 6 7 = 18 Critical Activity t e σ 2 i 1 3 8 0 3 5 14 2 5 6 1 0 6 7 4 0.11 ---------- ------------ = 27 σ 2 i = 2.11 days Standard deviation of Critical path = = 2. 11 = 1.45 Days Now let x be the project duration. P (x K ) = 0.9770 0.9770-0.5000 Standard Deduction ------------- 0.4770 Page 128 of 380 26/4/2008

Z = 2 Z = x 27 1.45 2 = x 27 1.45 x 27 = 2 x 1.45 x = 29.9 So chance that 97.7 % the project will be completed in 29.9 days. Page 129 of 380 26/4/2008

Problem 14 ( University Paper ) A project has the following activities and other characteristics. Activity Immediate Estimated Duration ( Days ) Predecessor Optimistic Pessimistic Most likely A - 2 14 2 B - 2 14 3 C - 4 16 4 D A 2 2 2 E B 4 28 10 F C 4 16 10 G D,E 6 30 12 H F,G 2 26 8 i) Draw the PERT network and identify the critical path. ii) Find the expected duration and variance of the project. iii) Find the probability that the project duration will exceed 36 days iv) What is the expected project completion time with a chance of 90%?. Solution : Activity A B C D E F G H 2 a + 4m + b b a = Duration ( Days ) = te σi 6 6 a m b 2 2 14 4 4 2 8 14 8 4 4 4 16 6 4 2 2 2 2 0 4 10 28 12 16 4 10 16 10 4 6 12 30 14 16 2 8 26 10 16 2 A B C D E F G H - - - A B C D,E F,G Page 130 of 380 26/4/2008

3 B 8 12 E 1 A 4 2 D 2 G 5 14 6 H 10 7 6 C F 10 4 Paths Duration 1 3 5 6 7 = 44 Critical Path 1 2 5-6 7 = 30 1 4 6 7 = 26 Critical Activity t e σ 2 i 1 3 8 4 3 5 12 16 5 6 14 16 6 7 10 16 ---------- ------------ = 44 σ 2 i = 52 days Standard deviation of Critical path = = 52 = 7.21 Days Now let x be the project duration. P (x > 36 ) Z = x µ = σ x 44 7.21 So when x = 60 z = 36 44 7.21 = -1.11 P1 = 0.4345 Page 131 of 380 26/4/2008

P1 = 0.3685 P2 = 0.5 Z = -1.11 0 Required probability Now we have P1 & P2 P2 = 0.5 + 0.3685 = 0.8665 So probability P ( x > 36 ) = 0.8665 = 86.65 % (iv) P (x K ) = 0.90 0.90-0.50 Standard Deduction ------------- 0.40 Z = 1.28 Z = x 44 7.21 1.28 = x 44 7.21 x 44 = 1.28 x 7.21 x = 53.23 So chance that 90 % the project will be completed in 53.23 days. Page 132 of 380 26/4/2008

Problem 5 A project is composed of eight activities whose time estimates are given below : Immediate Estimated Duration ( Days ) Activity Predecessor Activity Optimistic Most likely Pessimistic A - 1 3 5 B - 2 4 6 C A 3 6 15 D A 5 6 7 E C 5 7 9 F D 6 8 10 G B 7 9 11 H E,F,G 2 3 4 i) Draw the project and identify all the paths through it. ii) What is expected project completion time? ( 20 days ) iii) Find the probability of completing the project no more than 4 weeks later than expected. ( There are two critical path : A-C-E-H and A-D-F-H with respective variance 5 and 10 / 9 so we choose 5. The required probability =.9633 ) Solution : Activity A B C D E F G H Duration ( Days ) a m b te 2 a + 4m + b b a = = σi 6 6 1 3 5 3 4/9 2 4 6 4 4/9 3 6 15 7 4 5 6 7 6 1/9 5 7 9 7 4/9 6 8 10 8 4/9 7 9 11 9 4/9 2 3 4 3 1/9 2 A B C D E F G H - - A A C D B E,F,G Page 133 of 380 26/4/2008

4 1 A 3 B 4 2 3 C 6 7 D G 9 E 7 5 F H 6 8 3 7 Paths Duration 1 2 4-6 7 = 20 A-G-E-H Critical Path 1 2 5 6 7 = 20 A-D-F-H 1 3 6 7 = 16 Critical Activity t e σ 2 i CP ( 1 2 4 6 7 ) A-G-E-H 1 2 3 4/9 2 4 7 4 4 6 7 4/9 6 7 2 4/9 ---------- ------------ = 20 σ 2 i = 5 days Critical Activity t e σ 2 i CP ( 1 2 5 6 7 ) 1 2 3 4/9 2 5 6 1/9 5 6 8 4/9 6 7 3 1/9 ---------- ------------ = 20 σ 2 i = 10/9 days A-D-F-H Now to select the CP which has larget variance with is 5 days. Page 134 of 380 26/4/2008

Now = 20 & Standard deviation of Critical path = = 5 = 2.24 Days Now let x be the project duration. Find probability that Project will be completed in 4 weeks. As the problem has mentioned 4 weeks means 24 days. Now let x be the project duration. P (x 24 ) Z = x µ = σ x 20 2.24 So when z = 24 24 = x 20 2.24 x = 1.79 P ( x 24 ) = P ( z 1.79 ) P2 = 0.5 P1 P1 = 0.4633 0 Z = 1.79 Required probability = 0.5 + 0.4633 = 0.9623 = 96.23 % Page 135 of 380 26/4/2008

15 th March, 2008. Lecture IX. Problem - 6 (i) (ii) (iii) (iv) (v) A PERT activity has optimistic, most likely and pessimistic duration of 6 weeks, 15 weeks and 30 weeks respectively. Find the expected duration of this activity and its variance. One of the activities in a PERT project has an estimated average duration of 16 weeks, with a standard deviation of 4 weeks. The most likely time estimates of this activity is 15 weeks. What will be the optimistic & pessimistic time estimates of this activity. One of the activities in a PERT project has an expected duration of 12 weeks with a standard deviation of 2 weeks. The most likely time estimates of this activity is 12 weeks calculate the optimistic and pessimistic time estimates for this activity. If in a PERT network the expected duration of the critical path is 36 months and sum of the variances of the activities on the critical path is 25 ( months ) 2 than what is the probability that the project will be completed not earlier than 30 months and not later than 42 months. PERT calculation yields the critical path length of a project to be 24 months with a variance of 9. What is the probability of its completion of 20 months? with in how many months would you expect the project to be completed with probability of 0.99? Solution : (i) a = 6 m = 15 b = 30 t e =? & σ 2 i =? te a + 4m + b = = 6 6 + 4(15) + 30 6 = 16 Page 136 of 380 26/4/2008

b 6 2 a σ = = i 2 2 30 6 6 = 16 (ii) Homework (iii) t e = 0 σ = 2 i m = 12 a =? b =? te = 12 = a + 4m + b 6 a + 4(12) + b 6 12 x 6 = a + b + 48 a + b = 24 ( ➊ ) σ i b a = 6 b a 2 = 6 2 x 6 = b - a b a = 12 ( ➋ ) Page 137 of 380 26/4/2008

Now, a + b = 12 b a = 12 ---------------------------- 2 b = 36 b = 18 Now replacing with Value of b to find out the a. a + b = 24 a + 18 = 24 a = 6 So, a = 6 b = 18 (iv) = 36 2 = 25 = 5 P ( 30 x 42 ) Z = x µ = σ x 36 5 So when x = 30 Z = 30 36 5 So when x = 42 = -1.12 Z = 42 36 5 = 1.2 Page 138 of 380 26/4/2008

P ( 30 x 42 ) = P ( -1.2 z 1.2 ) P2 = 0.3849 P1 P1 = 0.3849 Z = -1.12 0 Z = 1.2 P ( 30 x 42 ) = 0.3849 + 0.3849 = 0.7698 = 76.98 % (v) = 24 2 = 9 = 3 P ( x 20 ) Z = x µ = σ x 24 3 So when x = 20 20 24 Z = = -1.33 3 P ( x 20 ) = P ( z -1.33 ) P1 = 0.4082 P2 Z = -1.33 0 Page 139 of 380 26/4/2008

Now P1 + P2 = 0.5 P2 = 0.5 0.4082 = 0.0918 P ( x 20 ) = 0.0918 = 9.81 % Now with in how many months would you expect the project to be completed with probability of 0.99? To find K such that, P ( x k ) = 0.99 Z = x µ = σ x 24 3 ( ➊ ) 0.99-0.50 ------ 0.49 P1 = 0.49 0 z =? z = 2.33 ( ➋ ) x 24 3 = 2.33 x - 24 = 2.33 x 3 x = 30.99 so with probability of 0.99 the project would get completed on or before 31 months. Page 140 of 380 26/4/2008

Problem 7. The Arcot Machinery Company has been offered a contract to build and deliver nine extruding process to the Home Bottling Company. The Contract price negotiated is contingent upon meeting a specified delivery time with a bonus offered for early delivery. The marketing department has established the following cost and time information. Normal Crash Crash Normal Time ( Weeks ) Activity Cost Time Cost ( a ) ( b ) ( c ) Rs ( Weeks ) Rs. 1 2 1 5 3 5,000 1 9,000 2 3 1 7 4 8,000 3 14,000 2 4 1 5 3 4,000 2 6,000 2 5 5 11 8 5,000 7 6,000 3 6 2 6 4 3,000 2 5,000 4 6 5 7 6 2,000 4 3,000 5 7 4 6 5 10,000 4 14,000 6 7 1 5 5 7,000 1 10,000 Normal delivery time is 16 weeks for a contact price of Rs. 62,000/-. On the basis of the calculated profitability for each delivery time specified in the following table, what delivery schedule do you recommend that the company implement? Contact Delivery Time Contract Amount ( Weeks ) ( Rs ) 15 62,500 14 65,000 13 70,000 12 72,500 ( Here a = Optimistic Time, b = pessimistic time, c = most Likely time ) Solution : This problem is for Homework. Page 141 of 380 26/4/2008

Problems on CPM / PERT ( University Questions ) For Homework. Problem 1. ( University Paper ) Draw the network for the following dependencies. Activity A B C D E F G H Preceding Activity - A A,B - A,D B E,F D Problem 2. ( University Paper ) The Project information for the customer order project of the Air Control Compnay is presented here. Draw a project network for the Project. ID Activity Predecessor Time A Order Review None 2 B Order Standard Parts A 15 C Produce Standard Parts A 10 D Design Custom Parts A 13 E Software Development A 18 F Manufacture Custom Hardware C,D 15 G Assemble B,F 10 H Test E.G 5 Draw the network and find the Critical path and project duration, also compute various floats of the activities. Tabulate your results. Problem 3. ( University Paper ) Activities of a project are described in the table below. Calculate the following. a) Total time for completing the project. b) Critical path c) Early start, Early Finish, Late Start & Late Finish time for all activities d) Total Float and Free Float available for Non-Critical activities. Page 142 of 380 26/4/2008

Activity Estimated Duration Immediate Predecessors A 3 - B 1 A C 2 B D 7 - E 8 D F 3 B G 1 E,F H 2 D Problem 4. ( University Paper ) Referring to the list of activities below, draw a network in Arrow Diagram (AOA) convention. Find out the Total Time for project completion. Critical Path and Total Float available on non critical path. Activity Duration ( Weeks ) Immediate Predecessors A 3 - B 5 A C 7 A D 4 B E 6 C F 4 C G 5 D,E H 8 G,F K 2 G L 4 H,K Page 143 of 380 26/4/2008

Problem 5. ( University Paper ) The following tables gives normal & crash times as well as Normal & Crash Costs for the activities of the Project. Draw the Network Diagram and find the critical path. In case, the project duration is required to be crashed by 2 days, which activities will get the priority in crashing? Activity Estimated Activity Cost Normal Crash Normal Crash 1 2 5 2 600 900 2 4 6 3 700 1000 1 3 4 2 100 200 3 4 7 4 400 800 4 7 9 5 600 920 3 5 12 3 1600 1960 4 6 10 6 1500 1800 6 7 7 4 400 490 7 9 6 4 300 420 5 9 12 7 400 850 Problem 6. ( University Paper ) A marketing project consists of 7 activities A to G with the following sequences, normal duration, normal cost, crash time and crash cost. Activity Preceding Normal Time Crash Time Normal Crash Cost Activity ( Days ) ( Days ) Cost ( Rs. ) ( Rs ) A None 4 3 60 90 B None 6 4 150 250 C None 2 1 30 60 D A 5 3 150 250 E C 2 2 100 100 F A 5 3 115 175 G B,D,E 4 1 100 310 Draw the network diagram for the project and find the normal duration and normal cost of the project. Also find the critical path. Find the corresponding cost of the project on crashing the project by 3 days. Page 144 of 380 26/4/2008

Problem 7. ( University Paper ) The details of project activities are given in the table below : Activity Predecessor Duration ( Days ) Crash Cost Normal Crash ( Per day ) A - 9 6 Rs. 20,000/- B - 8 5 Rs. 25,000/ C - 15 10 Rs. 30,000/ D A 5 3 Rs. 10,000/ E B 10 6 Rs. 15,000/ F C,D,E 2 1 Rs. 40,000/ Draw the project network using Activity on Arrow convention and find out. i) Normal, Crash and Optimum duration of the project. ii) Amount of maximum savings in project cost at optimum duration. Assume indirect cost = Rs. 35/- per day. Problem 8. ( University Paper ) The table below defines the activities for a small project. Activity Immediate Duration ( Days ) Cost ( Rs. 000 ) Predecessors Normal Crash Normal Crash A - 6 4 24 34 B - 4 3 12 22 C A 5 3 20 28 D A 7 4 29 47 E B 6 5 26 34 F B 8 5 34 52 G C,E 10 6 27 47 H D,F 9 7 34 48 The project attracts a penalty of Rs. 10,000/- per day for project completion beyond 18 days. Draw the project network by precedence diagramming method ( i.e Activity Node Method ) and find out the number of days delay intended beyond the schedule date to minimize total project cost. Page 145 of 380 26/4/2008

Problem 9. ( University Paper ) The table below gives activities, their normal estimated duration and their relationships for a small project. It also indicates the minimum possible durations if crashed and extra costs incurred for such crashing. The management desires the project to be completed in a shorter duration than the normally expected. But the extra funds available for this purpose are limited to Rs. 5,00.000/-. Only what would be shortest of the project after the crashing under the budget limitations? Activity Immediate Duration ( Days ) Cost ( Rs. 000 ) Predecessors Normal Crash Normal Crash A - 5 3 400 600 B - 5 1 300 500 C A 10 5 400 700 D B 7 2 400 600 E A 6 2 300 500 F C,D 11 5 600 930 G C,D 6 4 300 600 H E,F 5 1 200 400 I G 4 1 200 500 Problem 10. ( University Paper ) The following table provides normal and crash times as well as normal and crash costs for the activities of a project. Activity Normal Time Normal Cost Crash Time Crash Cost ( weeks ) ( Rs. Lakhs ) ( Weeks ) ( Rs. Lakhs ) 1-2 3 5 1 9 2-3 4 8 3 14 2 4 3 4 2 6 2 5 8 5 7 6 3 6 4 3 2 5 4 6 6 2 4 3 5 7 5 10 4 14 6 7 3 7 1 10 i) Draw the Network diagram and Find critical Path ii) Using the above information, crash or shorten the activities step by step until the shortest duration is reached. Page 146 of 380 26/4/2008

Problem 11. ( University Paper ) The following table gives data on Normal time, Crash Time, Normal Cost and Crash Cost for a project. Activity Normal Time Crash Time Normal Cost Crash Cost ( Days ) ( Days ) ( Rs. ) ( Rs. ) A 1 2 3 2 8,000 10,000 B 1 3 3 2 4,000 7,000 C 2 5 1 1 4,000 4,000 D 2 6 6 4 40,000 60,000 E 3 4 2 1 4,000 6,400 Dummy 4 5 - - - - F 4 6 5 3 30,000 38,000 G 5 6 7 6 24,000 30,000 H 6 7 4 3 32,000 39,000 i) Draw the network and find out the critical path and the Normal project duration. ii) What is the minimum length of the project and the corresponding cost when all the critical activities are crashed to the maximum extent? iii) If the indirect costs are Rs. 6,000/- per day then what is the project Cost Trade Off Point of the Project? Page 147 of 380 26/4/2008

Problem 12. ( University Paper ) Following table gives the list of project activities, estimated duration of individual activities and relationships by defining their successor activities. Activity Estimated Duration ( weeks ) Successor Activities A 5 D,E B 4 F C 6 G D 4 I E 4 J F 6 J G 5 H,G H 2 J I 5 L J 11 N K 4 M L 6 N M 9 N N 5 - i) Identify the critical paths and the project completion time. ii) Find the project completion time in following cases : a) If activity I is delayed by Three weeks. b) If activity K is delayed by Two weeks. c) If activity M is crashed by One Week. d) If activity G is crashed by Three weeks. Page 148 of 380 26/4/2008

PERT Related Problems. Problem 13 & 14. ( University Paper ) Already solved above. Problem 15. ( University Paper ) Following table lists various details of the project activities. Activity Predecessors Duration in Weeks Optimistic = t o Most Likely = t m Pessimistic = t p A - 1 1 7 B - 1 4 7 C - 2 2 8 D A 1 1 1 E B 2 5 14 F C 2 5 8 G D,E 3 6 15 i) Draw the project network using Activity convention and Identity and Identify all paths though it. ii) Find expected ( average ) duration and variance for each activity. iii) What is the expected duration of project with 50 % chance of completion? iv) What should be your level of confidence to accept the project deadline of 20 weeks? ( Note : A normal distribution curve covers 50.0%, 84.1% and 97.7% area to its left for 0 + 1 and +2 standard deviations from the mean respectively ) Page 149 of 380 26/4/2008

Problem 16. ( University Paper ) Referring to the activity table given below. Find the expected time for completion of the project with 84.1% probability. Activity Estimation Activity Duration t o t m t p 1 2 3 6 15 1 6 2 5 14 2 3 6 12 30 2 4 2 5 8 3 5 5 11 17 4 5 3 6 15 6 7 3 9 27 5 8 1 9 7 7 8 4 19 28 Note : A standard deviation of +1.0 to mean value under Normal Distribution Curve Covers 84.1% of area to its left. Page 150 of 380 26/4/2008

Problem 17. ( University Paper ) A new pharmaceutical product is to be released on a deadline for which 44 days are left. Activities involved in Product launch with their interdependence and probabilistic time for completion are given in the table. Draw a project network and find the probability of completing the project in time. Activity Predecessors Time ( In Days ) Activities Optimistic Most Likely Pessimistic A - 6 10 14 B A 1 2 3 C A 16 20 30 D B 3 5 7 E D 2 3 4 F C 7 10 13 G D 1 2 3 H G 1 3 5 I C,G 2 2 2 J I 2 3 4 K H 1 1 1 L J,K 1 2 3 Page 151 of 380 26/4/2008

Resource Allocation. Introduction To be able to schedule the resource requirements for a project the following details are required. (a) (b) (c) (d) The customary activity times, descriptions and sequences as previously described. The resource requirements for each activity showing the classification of the resource and the quantity required. The resources in each classification that are available to the project. If variations in availability are likely during the project life, these must also be specified Any management restrictions that need to be considered e.g. which activities may or may not be split or any limitations on labour mobility. Resource Allocation Resource Smoothing Time is the constraint but resources are not the constraints. Resource Leveling Resource are the constraints while the time is available in plenty Resource Smoothing : It is a network technique used for smoothing peak resource requirements during different periods of the project network. Under this technique the total project duration is maintained at the minimum level. For example if the duration of a project is 15 days, then the project duration is maintained, but the resources required for completing different activities of a project are smoothened by utilizing floats available on non critical activities. These non-critical activities having floats are rescheduled or shifted so that a uniform demand on resources is achieved. in other words, the constraint in the case of resource smoothing operation would be on the project duration time Resource smoothing is a useful technique for business managers to estimate the total resource requirements for various project activities. Page 152 of 380 26/4/2008

In resource smoothing, the time scale diagram of various activities and their floats (if any), along with resource requirements are used. The periods of maximum demand for resources are identified and non-critical activities during these periods are staggered by rescheduling them according to their floats for balancing the resource requirements. Resource Leveling : It is also a network technique which is used for reducing the requirement of a particular resource due to its paucity. The process of resource leveling utilizes the large floats available on non-critical activities of the project and thus cuts down the demand on the resource. In resource leveling the maximum demand of a resource should not exceed the available limit at any point of time. In order to achieve this, non critical activities are re-scheduled by utilizing their floats. Sometimes, the use of resource leveling may lead to enlarging the completion time of the project. (Delay) Rules for Scheduling : (a) (b) (c) (d) Assign the resource to the activity that has the least float If activities have equal float, give the resources to the longest activity. Where there is conflict between activities, give the resources to the activity that uses the largest amount of the resource. Sequence code. Updating the network The progress of various activities in a project network is measured periodically. Normally, either most of the activities are ahead or behind the schedule. It is therefore, necessary to update or redraw the network periodically to know the exact position of completion of each activity of the project. The task of updating the network may be carried out once in a month. Sometimes the updating of the network may provide useful information to such an extent that it may demand the revision of even those very activities which have not started. Even the logic may also change i.e. some of the existing activities may have to be dropped and new activities may be added up. In brief, the network should be amended accordingly in the light of new developments. Updating is done by assigning zero values to the duration of the completed activities. Partially completed activities are assigned times equivalent to their unfinished portions. Changes in the arrow diagram such as addition or deletion of any future activities must also be made. By repeating the usual computations on the arrow diagram with its new time elements, we can determine the new time schedule and possible changes in the duration of the project. Such information is used until it is again necessary to update the time schedule. In real situations many revisions of the time schedule are usually required at the early stages of the execution phase. A stable period follows in which little revision of the current schedule may be required. Page 153 of 380 26/4/2008

3 E B 11 A 7 G 1 2 D 5 5 6 6 5 C F 2 3 4 Find EST, LST & CP 13 13 0 6 3 24 29 E 0 6 B 24 29 7 11 A D 5 G 1 2 5 6 6 5 C F 2 3 4 18 21 Page 154 of 380 26/4/2008

Question 1. 6 E 5 11 G A 1 6 2 B 7 C 3 3 6 4 F 7 H 5 I 7 8 4 K 9 L 10 8 2 D J 2 5 18 18 0 6 0 6 1 A 6 2 B 7 13 13 3 C 3 D 2 E 5 F 6 4 9 6 26 19 7 26 11 H 5 G J 29 29 8 I 7 2 4 K 33 41 33 41 9 L 10 8 5 21 18 Page 155 of 380 26/4/2008

Total Float = LFT EST Duration At C total Float = - 6-3 = 17 26 TC of D = 31 - - 2 = 23 6 TC of F = - 13-6 = 5 24 Note : All Critical activity will have ZERO total Float. Problem 1. (a) Draw network of the following activities. (i) (ii) (iii) (iv) (v) Activities A and B start simultaneously. C follows A D and E can only commence when B and C are completed. D gives rise to F and E gives rise to G and is the final activity. H is dependent on F and is the final activity (b) Find also the critical path if the duration of activities is as follows : Activity A B C D E F G H Duration ( Weeks ) 6 8 3 12 9 7 15 8 (C) At the end of 12 th week, project progress and review have taken place. Following are the observations. (i) (ii) (iii) (iv) Activities 1-2, 1-3, 2-3 are completed. Activity 3-5 has been progressing for 3 weeks and needs 8 more weeks for completion. Activity 3-4 has been in progress for 3 weeks and since a new machine has been commissioned, the present estimate is that it can be completed in 6 more weeks. Re-assessment of activity 6-7 has revealed that it can be completed in 7 weeks. On the basis of these information, update the network & indicate change in critical path if there by any. Page 156 of 380 26/4/2008

Solution : 6 6 2 0 0 1 A 6 B 8 // // 12 9 C // 3 3 // 3 3 // 9 D 8 E 12 6 21 21 4 21 18 5 28 18 28 25 F 6 7 20 G 15 8 H 7 36 7 0 35 How it is Solved? 1. Draw the network diagram. 2. Workout LST EFT. 3. Find the CP. 4. Mentioned in BLUE is the Work Remaining & GREEN is work Completed. 5. Work out new EFT 6. Now the Non critical path will become Critical. Hence New Project duration will be 35 Days. New Critical path will be 1-3 5 7. Page 157 of 380 26/4/2008

Another Method to Solve this Problem. 6 2 A 0 1 6 B 8 C 3 D 9 18 25 4 F 6 7 7 H 35 3 7 9 11 E 20 5 G 15 How it is Solved? 1. Draw the network diagram. 2. Write the Duration considering the New duration ( Mentioned in BLUE ) Page 158 of 380 26/4/2008

Problem 2. The following table lists the activities of a maintenance project : Activity 1-2 1-3 1-4 2-5 3-6 3-7 4-7 5-8 6-8 7-9 8-9 Duration ( Month ) 2 2 1 4 5 8 3 1 4 5 3 (i) (ii) (iii) Draw the project network. Find the critical path and duration of the project. Suppose we require to employ, a special piece of equipment on activities 1-3, 3-6, 2-5, 5-8 and 8-9 one at a time. Will it affect duration of the project? Explain. Solution : 7 11 2 7 5 6 11 2 4 5 1 0 5 2 8 1 12 0 2 2 1 7 1 11 12 1 2 1 3 7 1 8 5 6 4 8 10 10 3 1 15 15 4 7 3 5 9 Now as given in the problem, special piece of equipment on activities 1-3, 3-6, 2-5, 5-8 and 8-9 one at a time. Hence, Page 159 of 380 26/4/2008

At Time t = 0, we start activity 1 3, with special equipment & complete it at t = 2, ( because duration of 1 3 is 2 ). At t = 2 there is a tie, both the activities 2 5 and 3 6 requires the special equipment & they have same EST i.e. 2 5 1 Now total Float of 2 5 is and that of 3 6 is therefore we delay 2 5 by 5 days & start 3 6 at t = 2 with special equipment and compute it at t = 7. At t = 7 start 2 5 with special equipment and complete it at t = 11. At t = 11 we start 5 8 with special equipment and complete it at t = 12. Finally we start 8 9 at t = 12 with special equipment and complete it at t = 15. The other activities will start at their EST and finish at there EFT. There by utilizing total float at 2 5 we can manage with the special equipment. There is NO DELAY in Project. Activities EST EFT Remarks 1 3 0 2 With Special Equipment 3 6 2 7 With Special Equipment, delay 2 5 as it has Largest TF of 5 day. 2 5 7 11 With Special Equipment 5 8 11 12 With Special Equipment 8 9 12 15 With Special Equipment Page 160 of 380 26/4/2008

Problem 3. A protect has the following time schedule. Activity Time (in months) Activity Time (in months) 1-2 2 3-7 5 1-3 2 4-6 3 1-4 1 5-8 1 2-5 4 6-9 5 3-6 8 7-8 4 8-9 3 Construct PERT Network and compute : (i) Total float for each activity (ii) Critical path and its duration. Also find the minimum number of cranes the project must have for its activities 2-5, 3-7 and 8-9 without delaying the project. Does any change require in PERT network? If so, indicate the name. Solution : 7 11 2 5 6 0 5 2 2 4 5 8 1 5 12 0 2 2 1 7 1 11 1 2 6 1 3 7 1 4 7 8 5 4 10 8 10 6 6 3 5 3 1 15 9 15 Page 161 of 380 26/4/2008

Activities EST EFT Remarks 3 7 2 7 Start at t = 2, 3 7 With Crane Machine & Delay 2 5 as it has larger TF of 5 days. 2 5 7 11 With Crane Machine. 5 8 11 12 With Crane Machine. 8 9 12 15 With Crane Machine. Activities will be as per the schedule. Page 162 of 380 26/4/2008

18 th March, 2008. Lecture X. Problem 4. For a project consisting of several activities, the duration and required resources to carry out each of the activities and their availabilities are given below :- (a) (b) Draw the network, identify the critical path and compute the total float for each of the activities. Draw the time Scale Diagram: Activity Duration in Days Resources Require ( Crew Size ) 1-2 6 30 1-6 7 20 2-3 12 30 2-4 5 20 3-5 4 30 4-5 6 20 5-7 3 30 6-7 9 20 7-8 5 20 Resource Availability : - No. of Crew member = 50 Solution : 18 18 6 3 22 6 12 4 22 0 0 1 6 2 5 5 16 25 30 11 3 25 30 6 4 7 8 5 16 7 7 9 6 Page 163 of 380 26/4/2008

Total Nos of Paths : 1 2 3 5 7 8 = 30 Critical path 1 2 4 5 7 8 = 25 1 6 7 8 = 21 Activities EST EFT - - - Duration 1 2 0 6 0 1 6 0 7 16 0 7 = 9 2 3 6 18 0 2 4 6 11 16 6 5 = 5 3 5 18 22 0 4 5 11 17 22 11 6 = 5 5 7 22 25 0 6 7 7 16 25 7 9 = 9 7 8 25 30 0 Note : For Critical Path Float will be Zero. Page 164 of 380 26/4/2008

Time Scale Diagram 20 20 4 5 6 30 30 30 30 20 1 2 3 5 6 8 6 12 4 3 5 20 20 6 7 9 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 50 50 50 50 50 50 70 70 70 70 70 70 70 70 70 70 50 30 30 30 30 30 30 30 30 20 20 20 20 20 Class Intervals Nos of Workers 0 6 50 6 17 70 16 17 50 17 25 30 25 30 20 Page 165 of 380 26/4/2008

Gantt Chart 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 1-2 6 1-6 2-3 2-4 3-5 4-5 5-7 6-7 7-8 7 6 18 6 11 18 22 11 17 22 25 8 16 25 30 50 50 50 50 50 50 70 70 70 70 70 70 70 70 70 70 50 30 30 30 30 30 30 30 30 20 20 20 20 20 Page 166 of 380 26/4/2008

Histrogram 75 70 65 60 55 50 45 40 35 30 25 20 15 10 5 0 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32 34 Page 167 of 380 26/4/2008

Resource Graph. 75 70 65 60 55 50 Maximaum Available 50 45 40 35 30 25 20 15 10 5 0 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32 34 Page 168 of 380 26/4/2008

Problem 5. The following table gives for each activity of a project. its duration and corresponding resources requirement as well as total availability of each resources : Activity Duration Resources Required Machines Men 1-2 7 2 20 1-3 7 2 20 2-3 8 3 30 2-4 6 4 30 3-6 9 2 20 4-5 3 2 20 5-6 5 2 20 Maximum Resource Available 4 40 (a) Draw the network, compute earliest start time, latest finish time for each activity, total float for each activity and identify the critical path assuming that there are no resource constraints. (b) Draw the Time Scale Diagram along with resource accumulation table. Comment on demand for the machines and men for the entire project duration. Solution : 7 16 7 3 13 0 0 1 7 7 // 2 // 8 15 15 6 4 3 3 19 16 5 5 3 24 24 3 // 9 6 Page 169 of 380 26/4/2008

Activities EST EFT - - - Duration 1 2 0 7 0 1 3 0 7 15 0 7 = 8 2 3 7 15 0 2 4 7 13 16 7 6 = 3 3 6 15 24 0 4 5 13 16 19 13 3 = 3 5 6 16 21 24 16 5 = 3 Page 170 of 380 26/4/2008

Time Scale Diagram 4-30 2-20 4 5 2-20 3-30 1 2 3 6 7 2-20 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 Machines 4 4 4 4 4 4 4 7 7 7 7 7 7 5 5 4 4 4 4 4 4 2 2 2 Men 40 40 40 40 40 40 40 60 60 60 60 60 60 50 50 40 40 40 40 40 40 20 20 20 Demand Exceeds Supply Page 171 of 380 26/4/2008

Gantt Chart 1-2 1-3 2-3 2-4 3-6 4-5 5-6 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 2-20 7 2-20 7 3-30 7 15 4-30 7 13 2-20 15 22 2-20 13 16 2-20 16 21 Machines 4 4 4 4 4 4 4 7 7 7 7 7 7 5 5 4 4 4 4 4 4 2 2 2 Men 40 40 40 40 40 40 40 60 60 60 60 60 60 50 50 40 40 40 40 40 40 20 20 20 Page 172 of 380 26/4/2008

Problem 6. A small project consists of seven activities for which the relevant data are given below : Activity Preceding Activity Activity Duration Days A - 4 B - 7 C - 7 D A,B 5 E A,B 7 F C,D,E 6 G C,D,E 5 (i) Draw the network and find the project completion time (ii) Calculate total float for each of the activities. (iii) Draw Time Scaled Diagram Solution : 7 14 20 4 12 20 2 4 6 0 0 1 A B 7 4 D1 0 3 C 7 7 7 D 5 E 7 D2 0 14 14 5 F 6 G 5 D3 0 7 20 20 Page 173 of 380 26/4/2008

Activities EST EFT Total Float 1 2 0 4 3 1 3 0 7 0 1 5 0 7 7 2 3 4 4-3 4 7 12 2 3 5 7 14 0 4 5 12 12-5 6 14 20 0 5 7 14 19 1 6 7 20 20 - Critical Path : 1 3 5 6 7 that is B E F D 3 Time Scale Diagram A 2 D 4 4 5 D1 D2 B E F 1 3 5 6 7 7 6 D3 C 7 5 G 7 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 Page 174 of 380 26/4/2008

Gantt Chart 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 A 4 B C D E F G 7 7 7 12 7 14 15 20 16 19 Page 175 of 380 26/4/2008

Problem 7. Draw a 'Gantt Chad' and 'Resource Graph' for the project schedule based on the table below and find out the following : (i) Total time for completion. (ii) Days of over allocation of resources if maximum workers available are 20. ACTIVITY IMMEDIATE PREDECESSORS DURATION ALLOCATED WORKERS A - 8 4 B - 7 8 C A 6 5 D B 8 4 E B 4 8 F B 8 6 G C,D 5 5 H E 6 4 I F 6 5 J G,H,I 10 6 Solution : 10 8 0 0 A 8 2 6 C 16 15 4 1 7 B 7 7 D 8 16 11 5 G 21 21 31 31 3 E 4 5 H 5 7 J 8 10 8 F 15 15 I 6 6 Page 176 of 380 26/4/2008

Activities EST EFT Total Float A 1 2 0 8 2 B 1 3 0 7 0 C 2 4 8 14 2 D 3 4 7 15 1 E 3 5 7 11 4 F 3 6 7 15 0 G 4 7 15 20 1 H 5 7 11 17 4 I 6 7 15 21 0 J 7 8 21 31 0 Critical Path : 1 3 6 7 8 that is B F I J ( Rest for Home work. ) Page 177 of 380 26/4/2008

Problem 8. The following table gives the activity schedule and resource requirement for a project. Draw a Gantt Chart representing the schedule and resource loading chart. What is the peak requirement of the resource and on which day it occurs? ACTIVITY DURATION START FINISH RESOURCE A 3 1 3 3 B 2 2 3 2 C 4 4 7 4 D 5 4 8 5 E 3 7 9 3 F 2 8 9 2 G 2 10 11 6 H 3 10 12 3 I 2 12 13 2 J 3 12 14 5 K 1 15 15 3 Note : Assume Start at the beginning and Finish at the end of the day numbers mentioned in the table. Solution : Since NO LOGICAL connections are given there is no need of Network diagram hence start directly with activity. ( Resource is for per day. ) Activities EST EFT Total Float A 0 3 3 B 1 3 2 C 3 7 4 D 3 8 4 E 6 9 3 F 7 9 2 G 9 11 2 H 9 12 3 I 11 13 2 J 11 14 3 K 11 15 1 ( Now Draw the Gantt Chart ) Home work. Page 178 of 380 26/4/2008

Problem 9. The table below gives the project time table that is to commence in May 2001. The budgeted costs for various activities as indicated are to be spent uniformly over their duration. ACTIVITY DURATION MONTH START 1 FINISH MONTH MONTH ACTIVITY COST A 4 May 01 Aug. 01 Rs.80,000 B 2 Sep. 01 Oct. 01 Rs.60,000 C 6 Sep. 01 Feb. 02 Rs.90,000 D 9 Mar. 02 Nov.02 Rs.45,000 E 2 May 01 June 01 Rs.50,000 F 5 Jul. 01 Nov. 01 Rs.1,00,000 G 3 Nov. 01 Jan. 02 Rs.75,000 H 7 Jul. 01 Jan. 02 Rs.70,000 I 10 Feb.02 Nov. 02 Rs.1,50,000 J 1 Dec. 02 Dec. 02 Rs.40,000 Note : Assume the activity starting at the beginning and finishing at the end of the month. (a) Draw a "Gantt Chart' for project schedule and work out the monthly aid cumulative cash flows for the project. (b) Draw the 'S' curve for cash flows. What does this curve indicate? Solution : Activity Cost is for full activity, hence first workout Page 179 of 380 26/4/2008

Gantt Chart 05-01 06-01 07-01 08-01 09-01 10-01 11-01 12-01 01-02 02-02 03-02 04-02 05-02 06-02 07-02 08-02 09-02 10-02 11-02 12-02 A 20 B 30 C 15 D 5 E 25 F 20 G 25 H 10 I 15 J 40 Page 180 of 380 26/4/2008

Problem 10. The direct Cost estimates for various activities of a project network are as given below : Activity Duration in Direct Cost ( Rs ) Months Per Month A 1-2 4 80,000 B 1-3 7 40,000 C 2-6 3 40,000 D 2-4 3 80,000 E 3-4 2 120,000 F 3-5 2 60,000 G 4-5 2 60,000 H 5-6 3 20,000 (i) (ii) (iii) Draw the network, find critical path and project duration. Compile monthly and cumulative monthly cash flow requirements for the early start schedule. Determine a near uniform cash flow schedule. Solution : 11 14 4 14 C 2 3 6 0 0 1 A 4 3 D 9 9 4 H 3 7 B 7 7 E 2 2 G 11 3 F 5 11 2 Page 181 of 380 26/4/2008

Total Nos of Paths : 1 2 4 5 6 = 12 1 2 6 = 7 1 3 4 5-6 = 14 Critical path 1 3 5 6 = 12 Activities EST EFT Total Float A 0 4 2 B 0 7 0 C 4 7 2 D 4 7 7 E 7 9 0 F 7 9 2 G 9 11 0 H 11 14 0 Page 182 of 380 26/4/2008

Gantt Chart 1 2 3 4 5 6 7 8 9 10 11 12 13 14 A 80 4 B 40 7 C D E F G H 40 4 7 80 4 7 120 7 9 60 7 9 60 9 11 80 11 14 Monthly Cash flow 120 120 120 120 160 160 160 180 180 60 60 80 80 80 Cummulative Cash Flow 120 240 360 480 600 720 840 960 1080 1200 1320 1440 1560 1680 Page 183 of 380 26/4/2008

S Curve Cummulative Cash Flow 170 160 150 140 130 120 110 100 90 80 70 60 50 40 30 20 10 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 ( Duration ) Page 184 of 380 26/4/2008

Problem 11. The direct estimates for various activities of a pipeline project are as given below: Activity ( I J ) Duration ( Months ) Direct Cost ( Rs ) 1 2 13 26,00,000 1 3 12 60,00,000 2 4 2 20,00,000 3 4 8 20,00,000 2 5 15 15,00,000 4 5 2 15,00,000 Project overheads are Rs.3.00,000/- per month (i) Draw network. find critical path, project duration and total float of each activity. (ii) On graph paper compile monthly, cumulative monthly, cash flow requirement for early start and late start schedule. (iii) Draw cumulative cash flow requirements on graph for early and late start schedule. (iv) The government will release the amount in the following manner. 1st Year Rs.69,00,000 for Direct + Rs.36,00,000 Overhead. 2nd Year Rs.68,00,000 far Direct + Rs.36,00,000 Overhead. 3rd Year RS.19,00,000 for Direct + Rs.12,00,000 Overhead Unspent amount if any will lapse and hence cannot be carried forward Schedule the activities on graph to match release of funds It is possible to schedule the project without extending project duration. (Assume funds for an activity are required uniformly throughout its time duration) Page 185 of 380 26/4/2008

Solution : 13 28 13 28 2 15 5 0 0 13 2 26 20 2 1 4 12 18 12 8 3 Total Nos of Paths : 1 2 4 5 = 17 1 2 5 = 28 Critical path 1 3 5-6 = 22 Activities EST EFT Total Float 1 2 0 13 0 1 3 0 12 6 2 4 13 15 11 3 4 12 20 6 2 5 13 28 0 4 5 20 22 6 Page 186 of 380 26/4/2008

Time Scale Diagram 2 L 1 L 1 2 5 13 15 5 L 2.5 L 7.5 L 3 4 12 8 8 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 7 7 7 7 7 7 7 7 7 7 7 7 4.5 13.5 13.5 3.5 3.5 3.5 3.5 3.5 8.5 8.5 1 1 1 1 1 1 7 14 21 28 35 42 49 56 63 70 77 84 88.5 102 115.5 119 122.5 126 129.5 133 141.5 150 151 152 153 154 155 156 Add Rs 3 L Every Month Add Rs 3 L Every Month Page 187 of 380 26/4/2008

Gantt Chart 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 1-2 2 L 13 1-3 2-4 2-5 3-4 4-5 5 L 12 10 L 13 15 1 L 13 28 2.5 L 12 20 7.5 L 20 22 Monthly Cash flow Cumm. Cash Flow Add Rs 3 L Add Rs 3 L Page 188 of 380 26/4/2008

Problem 12. The following table gives the activity schedule and manpower requirement for a project. Draw the Gnat Chart representing the schedule and resource loading chart. What is the peak requirement of the resource and on which day (s) it occurs? Activity Preceding Activity Duration Total Man Power Days Required A - 3 60 B - 4 84 C A 2 22 D A 3 36 E B 6 72 F D,E 3 18 G C 5 35 H F 6 42 I G,H 7 56 Solution : For Homework. Page 189 of 380 26/4/2008

29 th March, 2008. Lecture XI. Project Control Variance Analysis Approach The traditional approach to project control involves a comparison of the actual cost with the budgeted cost to determine the variance. An example of variance analysis is as follows : Activity A Activity B Budgeted Cost in the period 50,000 30,000 Cumulative budget to date 200,000 75,000 Actual Cost in the Period 55,000 28,000 Cumulative actual Cost to Date 240,000 80,000 Variance for the period (5,000) 2,000 Cumulative variance to date (40,000) (5,000) The variance analysis approach is inadequate for project control for the following reasons : 1. It is backward looking rather than forward looking. It tells only what happened in the past but does not answer the following questions: What will happen in future? Is the rate of work accelerating or decelerating? 2. It does not use the data effectively to provide integrated control. The traditional variance analysis shows whether in the time period under analysis more or less resources were expended than budgeted. However, it does not indicate the value of work done. This information is vital for purposes of control. Page 190 of 380 26/4/2008

Performance Analysis: Modern Approach to Control Effective control over a project requires systematic 'performance analysis' This calls for answering the following questions : Is the project as a whole (and its individual parts) on schedule, ahead of schedule, or behind schedule? If there is' a variation, where did it occur, why did it occur, who is responsible for it, and what would be its implications? Has the cost of the project as a whole (and its individual parts) been as per budget estimates, less than the budget estimates, or more than the budget estimates? If there is a variation, where did it occur, why did it occur, who is responsible for it, and what would be its implications? What is the trend of performance? What would be the likely final cost and completion date for the project and its individual parts? For small and simple projects, the project managers would do performance analysis for the project as a whole, or for its major components. As the project becomes larger and more complex, performance analysis needs to be done for individual segments of the projects which are referred to as 'cost accounts'. Method of Analysis For analyzing the performance at cost account and higher levels of the work breakdown structure, we employ a method of analysis which takes into account the value of work that has been done. In the traditional method of analysis, the project manager measured the actual progress against the predetermined schedule and the actual cost against the budget estimate. This did not enable him to know systematically whether the expenditure incurred was commensurate with progress. He perhaps relied on subjective estimates. Earned Value Analysis / Earned Value Management A method for measuring project performance. It compares the amount of work that was planned with what was actually accomplished to determine if cost and schedule performance is as planned. It helps us to determine how much volume and how much value of work has been done on a particular day of an activity / project. Page 191 of 380 26/4/2008

It is a cost performance measure which helps the project manager to understand the total cost performance. It is achieved by integrating the scope, cost and the schedule measures of work. In EVA, we calculate : (i) Budget Cost (ii) Actual Cost (iii) Budgeted cost estimate (iv) Cost Variance. Earned value is the percentage of the total cost budgeted against the total cost incurred. The Budgeted Cost of each activity is considered as its value for the project. hence completion of an activity is considered to contribute equivalent to its value to the total value of the project. In other words, an activity earns value, equivalent to its budgeted cost on its completion. An activity that is partially completed on a particular date is also considered to have earned value equivalent to its percentage completion on the given date multiplied by its total budgeted cost. In case Actual Expenditure incurred for completion of an activity exceeds its budgeted cost, the project does 'not earn any additional benefit since activity parameters are well specified. The variance therefore indicates cost over-run. The total value earned for the project on any given date is the total of earned values of all the completed activities till date as well as the total of earned values to date for partially completed activities. Performance Analysis seeks to remove this subjectivity by employing an analytical frame work based on the following terms : (1) SCWS (Budgeted Cost of Work Schedule) [ Schedule Expenditure] (Planned Value) ( PV) : It is the budgeted cost of work to be completed till given date if the project runs on schedule. (2) SCWP (Budgeted Cost of Work Performed) [Earned Value] (EV) : It is the budgeted cost for the completed work. When any of the activity is completed, it is considered to have "earned value" equivalent to its budgeted cost. Therefore the total of budgeted cost for the work performed to date is the total "earned value" of the project. Page 192 of 380 26/4/2008

(3) ACWP (Actual Cost of Work Performed) [ Project Expenditure] (Actual Cost) (AC) : It is the actual cost incurred in completing the work. It covers total cost of work done, goods received and services used, whether: or not these have been paid for. (4) BCTW (Budgeted Cost for Total Work) This is simply the total budgeted cost for the entire project work. (5) ACC (Additional Cost for completion) This represents the estimate for the additional cost required for completing the project. Given the above terms, the project may be monitored along the following lines : Cost Variance : BCWP - ACWP ( if ACWP > BCWP, it indicates cost over run) OR Cost variance = Difference between the estimated cost of the activity and the actual cost of the activity. Schedule Variance (in cost terms) : BCWP BCWS Any difference between the schedule completion of an activity and the actual completion of that activity. If BCWP < BCWS, it indicates schedule delay. BCWP Cost Performance index (CPI) = --------- ACWP BCWP Schedule Performance Index (SPI) = --------- BCWS Page 193 of 380 26/4/2008

Theory Experienced project managers utilize a technique called earned value management (EVM) to assess a project's progress over time and allow project teams to understand the health and performance 01 their projects EVM is also a good metric to share with management. According to the Project Management Body of Knowledge: Earned Value Management (EVM) is a method for integrating scope, schedule, and resources for measuring project performance. It compares the amount of work or effort that was planned with what was actually earned and spent to determine if cost and schedule performance are as planned. By comparing planned value (the ideal progress of the project) to the earned value (the value of the project to date based on work or effort expanded), a project manager can detect early if the project is going awry. If the schedule performance index is less than 1.0, then the project is in danger of going over schedule. If the cost performance index is less than 1.0, then the project is in danger of going over budget. By monitoring and reviewing these metrics, a project manager can report these statistics to management so that they can determine whether to continue with the project: The process may be revised for similar projects by learning from the statistics and modifying expectations. Page 194 of 380 26/4/2008

Problem 1. A company has obtained a fixed cost contract for the supply, installation, testing and commissioning of 200 computers of the same specification at a cost of RS.600 Lakhs. The company had estimated that it could supply, install, test and commission 10 computers per day so that the entire work can be completed in 20 days. The project status was reviewed after the completion of 16 days. It was noted at the time of review that 120 computers have been installed and the cost incurred was RS.380 Lakhs. It was estimated at the time of the review that a sum of RS.260 Lakhs would be required for completion of the pending work i.e. installation of the remaining 80 computers. Find :-. Activity A Activity B Budgeted cost in the period 50,000 30,000 Cumulative budget to date 200,000 75,000 Actual cost in the period 55,000 1,28,000 Cumulative actual cost to date 240,000 80,000 Variance for the period (5,000) 2,000 Cumulative variance to date (40,000) (5,000) (i) (ii) (iii) (iv) (v) (vi) (vii) Budgeted Cost of Work Scheduled (BCWS) Budgeted Cost of Work Performed (BCWP) Actual Cost of Work Performed (ACWP) Cost Variance Schedule Variance Cost Performance Index (CPI) Schedule Performance Index (SPI) Solution : 1. Project : To install 200 computers in 20 days @ 10 computers per day. Project Duration 20 days. Rs, 600 Lakhs 2. Cost per computer = -------------------- = Rs. 3 Lakhs 200 Page 195 of 380 26/4/2008

3. Review Period : After 16 days 4. Budgeted Cost of Work Schedule (BCWS) (Schedule Expenditure). [Planned Value (PV)] = 16 Days x 10 Computers per day x RS.3 Lakhs per computer = Rs.480 Lakhs. = It is the budgeted cost of work to be completed till given date if the project runs on schedule. 5. Budgeted Cost of Work Performed (BCWP) [ Earned Value (EV)] = 120 Computers x RS.3 Lakhs / Computer = RS.360 Lakhs It is the budgeted cost for the completed work. When any of the activity is completed it is. considered to have "earned value" equivalent to its budgeted cost. Therefore the total of budgeted cost of the work performed to date is the total "earned value" of the project. 6. Actual Cost of Work Performed [ACWP] [Project Expenditure] [ Actual Cost] = Rs.380 Lakhs It is the actual cost in completing the work. It covers total cost of work done, goods received and serviced used, whether or not these have been paid for. 7. Budgeted Cost for Total Work (BCTW) This is simply the total budgeted cost for the entire project work: RS.600 Lakhs. 8. Additional Cost for Completion (ACC) (200-120) x RS.3 Lakhs per computer Computers yet to be installed = RS.240 Lakhs This represents the estimate for the additional cost required for completing the project. In the problem, it is given to be Rs.260 Lakhs. Page 196 of 380 26/4/2008

9. Cost Variance (CV) : = BCWP - ACWP = Rs.360 Lakhs - Rs.380 Lakhs = (- Rs.29 ) Lakhs if CV is negative, it indicates cost over run i.e. over the budget. If CV is positive then it means' it is under the budget 10. Schedule Variance (SV) (in cost terms) = BCWP - BCWS = Rs. 360 Lakhs - Rs. 480 Lakhs = ( - RS.120 Lakhs ) i.e. if SV is negative then it indicates that the project is behind the schedule, if it is positive then the project is ahead of the schedule. 11. Cost Performance Index (CPI) BCWP 360 = --------- = ------ = 0.9474 = 0.95 ACWP 380 For everyone rupee spent, we have done 95 paise worth of the work done or achieved or for every one rupee spent, we received Re.0.95 worth of cost performance. A value > 1 indicates that the work is being completed better than planned, whereas value < 1 indicates means that work is costing more than and / ore proceeding more slowly than planned. Page 197 of 380 26/4/2008

12. Schedule Performance Index (SPI) BCWP 360 = --------- = ------ = 0.75 BCWS 480 It is a measure of schedule efficiency. [ Project is progressing at 75% of the rate originally planned ]. Or For every rupee spent, we are only receiving Re.0.75 of schedule performance. 13. Estimated Cost at Completion Budgeted Cost for total work (BCTW) = -------------------------------------------------- CPI 600 600 = ------ = ------------ = Rs. 633.33 Lakhs 0.95 360 / 380 Cost over run = RS.633.33 - RS.600 = RS.33.33 Lakhs. Or ACC = RS.33.33 Lakhs 14. Estimated Time at Completion Project Duration 20 Days = ---------------------- = ----------- = 26.67 Days SPI 0.75 Time over run = 26.67-20 = 6.67 Days. Page 198 of 380 26/4/2008

Problem 2. A project was begun on 1 st January 2005 and was expected to be completed by 30 th September 2005. The project is being reviewed on 30th June 2005 when the following information has been developed. ( Rs ) Budgeted cost for work scheduled (BCWS) 15,00,000 Budgeted cost for work performed (BCWP) 14,00,000 Actual cost of work performed (ACWP) 16,00,000 Budgeted cost for total work (BCTW) 25,00,000 Additional cost for completion (ACC) 12,00,000 Solution : Cost variance : = BCWP - ACWP = (1,400,000-1,600,000) = - 200,000 Schedule variance in cost terms : = BCWP - BCWS = (1,400,000-1,500,000) = - 100,000 Cost performance index (CPI) : = BCWP / ACWP = 1,400,000 / 1,600,000 = 08075 Schedule performance index (SPI) : = BCWP / BCWS = 1,400,000 / 1,500,000 = 0.933 Page 199 of 380 26/4/2008

Problem 3. A project has begun on 1 st April 2006 and was expected to be completed by 31 st December 2006. The project is being reviewed on 30th September 2006 when the following information has been developed : Budgeted cost for work scheduled (BCWS) : Rs.6,000,000 Budgeted cost for work performed (BCWP) : Rs.5,500,000 Actual cost of work performed (ACWP) : Rs.5,800,000 Budgeted cost for total work (BCTW) : Rs.10,000,000 Additional cost for completio'n (ACC) : Rs.5,00,000 Determine the following : i) Cost variance ii) Schedule variance in cost term iii) Cost performance index iv) Schedule performance index Solution : This Problem is for Homework. Page 200 of 380 26/4/2008

Problem 4. A project with a total budget cost of Rs. 250 crores is scheduled to be completed in 80 weeks. A periodic review taken at the end of 50 weeks after commencement indicate the following : Earned value BCWP = Rs. 170 Crores Actual Expenditure ACWP = Rs. 180 Crores Scheduled earned value BCWS = Rs. 187 Crores What do you interpret about the project progress till date with regard to time and cost on the basis of following calculations. (A) (B) (C) (D) Cost Performance Index (CPI) Schedule Performance Index (SPI) Estimated Cost to complete Estimated time to complete Solution : This Problem is for Homework. Page 201 of 380 26/4/2008

Problem 5. A project comprising of 10 activities with total expected duration of 35 weeks was reviewed on completion of 20 weeks after start The physical progress of various activities as assessed and actual costs incurred till date are compared with expected progress and budgeted figures as under : Activity % Completion Cost (in Rs. Lakhs) Schedule Actual Budgeted Total Actual Till date 1 100 100 10.0 12.0 2 100 100 12.0 12.5 3 70 60 18.0 12.0 4 55 50 25.0 13.0 5 30 25 20.0 6.0 6 10 0 15.0 0.0 7 0 0 10.0 0.0 8 0 0 8.5 0.0 9 0 0 6.5 0.0 10 0 0 5.0 0.0 TOTAL 0 0 130.00 55.5 If the cost are incurred linearly in proportion to activity completion, find out the : i) Budgeted Cost of Work Scheduled (BCWS) ii) Budgeted Cost of Work Performed (BCWP) iii) Cost Variance iv) Cost Performance Index (CPI) v) Schedule Performance Index (SPI) Page 202 of 380 26/4/2008

Solution : Activity Schedule Completion % Actual Completion % Budget Cost Actual Cost Till date BCWS Budget Schedule Cost * Schedule Competion % BCWP Budget Schedule Cost * Actual Competion % 1 100 100 10 12 10 x 100 % = 10 10 x 100 % = 10 2 100 100 12 12.5 12 x 100 % = 12 12 x 100 % = 12 3 70 60 18 12 18 x 70 % = 12.6 18 x 60 % = 10.8 4 55 50 25 13 25 x 55 % = 13.75 25 x 50 % = 12.5 5 30 25 20 6 20 x 30 % = 6 20 x 25 % = 5 6 10 0 15 0 15 x 10 % = 1.5 0 7 0 0 10 0 0 0 8 0 0 8.5 0 0 0 9 0 0 6.5 0 0 0 10 0 0 5.0 0 0 0 55.85 ( BCWS ) 50.3 ( BCWP ) Project Duration ( PD ) = 35 Weeks BCTW = 130 BCWS = 55.85 BCWP = 50.3 Review period = 20 Weeks Page 203 of 380 26/4/2008

Cost Variance Cost Performance Index Estimated Cost At Competion Additional Cost at Completion Remarks = BCWP - ACWP BCWP CPI = ---------- ACWP BCTW = ---------- CPI = 50.3-55.5 50.3 = ------- 55.5 130 = ---------- 0.91 = 142.86-130 There is a Cost Overrun = ( - 5. 2 ) = 0.91 = 142.86 = 12.86 Schedule Variance Schedule Performance Index Estimated Time At Competion Additional Duration Remarks = BCWP - BCWS BCWP SPI = ---------- BCWS P D = ---------- SPI = 50.3-55.85 50.3 = ---------- 55.85 35 = ---------- 0.9 = 38.89-35 There is a DELAY in the Project = ( - 5. 55 ) = 0.9 = 38.89 = 3.89 Page 204 of 380 26/4/2008

Problem 6. The progress observed at the end of the Seventeenth day from the beginning of a 12 day duration project is as given in the following table. The actual cost incurred till date is reported to be Rs.3,100. Draw a Gantt chart for the project and find the project performance on the basis of Cost and Schedule performance Indices. ( Assume the activity costs are incurred uniformly over its duration ). Activity Solution : Immediate Predecessors Estimated Duration in Days Budgeted Cost of Activity Actual % completion at the end of day 7 A - 3 500 100 % B - 1 200 100 % C A 4 800 75 % D B 4 700 100 % E B 5 500 95 % F D 2 200 80 % G E 3 500 50 % H C 4 400 0 % I F 2 600 0 % J G 3 300 0 % A B C D E F G H I J - - A B B D E C F G 0 0 A 3 2 3 4 C 4 8 5 4 7 8 10 7 H 4 12 12 1 1 B 1 1 D 4 5 F 2 7 I 2 9 J 3 E 5 6 6 6 9 3 9 G 8 3 Page 205 of 380 26/4/2008

Activity Schedule Completion % Actual Completion % Budget Cost Actual Cost Till date BCWS Budget Schedule Cost * Schedule Competion % BCWP Budget Schedule Cost * Actual Competion % A 100 100 600 600 x 100 % = 600 600 x 100 % = 600 B 100 100 200 200 x 100 % = 200 200 x 100 % = 200 C 100 75 800 800 x 100 % = 800 800 x 75 % = 600 D 100 100 700 700 x 100 % = 700 700 x 100 % = 700 E 100 95 500 500 x 100 % = 500 500 x 95 % = 475 F 100 80 200 200 x 100 % = 200 200 x 80 % = 160 G 33.33 50 500 500 x 33.33 % = 166.65 500 x 50 % = 250 H 0 0 400 0 0 I 0 0 600 0 0 J 0 0 300 0 0 3166.65 ( BCWS ) 2985 ( BCWP ) Project Duration ( PD ) = 12 Days BCTW = 4,700 BCWS = 3,166.65 BCWP = 2,985 Review period = 7 Days Page 206 of 380 26/4/2008

Cost Variance Cost Performance Index Estimated Cost At Competion Additional Cost at Completion Remarks = BCWP - ACWP BCWP CPI = ---------- ACWP BCTW = ---------- CPI = 2985-3100 2985 = ------- 3100 4700 = ---------- 0.96 = 4895-4700 There is a Cost Overrun = ( - 115 ) = 0.96 = 4895.83 = 195 Schedule Variance Schedule Performance Index Estimated Time At Competion Additional Duration Remarks = BCWP - BCWS BCWP SPI = ---------- BCWS P D = ---------- SPI = 2985-3166.85 2985 = ---------- 3166.85 12 = ---------- 0.94 = 12.76-12 There is a DELAY in the Project = ( - 181. 85 ) = 0.94 = 12.76 0.76 Page 207 of 380 26/4/2008

Problem 7. The following information is available at the end of 40 th Day of a New Plant eraction project. Determine, if the project is under control base on the named value evaluation system. If not what is the likely extent of cost and time over runs at completion. Activity Predecess Duration Total Budget for Activity ( Rs 000 ) Actual Cost Till Date Actual % completion till Date A - 10 330 250 100 B A 8 400 450 100 C A 2 350 380 100 D C 0 0 0 0 E B.D 18 405 400 70 F E 16 460-0 1905 1480 Solution : This is for Home work Page 208 of 380 26/4/2008

Problem 8. A project has a budget of Rs 5,00,000 and is scheduled to be completed in one year. The following table shows cumulative values of p!anned costs. Earned value and actual costs at the end of each of the first four months. Month No Planned cost Earned Value Actua! Cost (Rs.'000) (Rs.'000) (Rs.'000) 1 20 24 23.5 2 60 58 62 3 110 95 105 4 120 190 205 Calculate following values : i) Cost performance index for each of the Four Months ii) Schedule performance index for each of the four months iii) Estimated Cost at complete based on the performance at the end of the fourth month. iv) Estimated time to complete based on the performance at the end of the fourth month. Solution : This is for Home work Page 209 of 380 26/4/2008

Problem 9. A project with a total budget cost of Rs.25 crores is scheduled to be completed in 80 weeks. A periodic review taken at the end of 50 weeks after commencement indicate the following : Earned Value BCWP = Rs.17 Crores Actual Expenditure ACWP = Rs.18 Crores Schedule Earned Value BCWS = Rs.19 Crores Compute and interpret the following : (i) (ii) (iii) (iv) Cost Performance Index (CPI) Schedule Performance Index (SPI) Estimated Cost to Complete Estimated time to-complete. Solution : This is for Home work Page 210 of 380 26/4/2008

Problem 10. A project consisting of eight activities was revised on completion of 12 days after its start. Find the project performance on the basis of cost and schedule performance indices. What is the estimated duration of the project? Activity Duration ( In days ) Budgeted Cost of Activity ( Rs. 000 ) Actual Cost of Activity ( till date ) Actual % completion at the end of day 12 A 1 2 5 60 62 100 % B 2 3 7 70 70 100 % C 2 4 5 73 73 100 % D 2 5 7 82 70 90 % E 3 6 6 69 0 0 % F 4 6 8 54 10 20 % G 5 7 6 50 0 0 % H 6 7 5 40 0 0 % Solution : This is for Home work Page 211 of 380 26/4/2008

PROJECT APPRAISAL (Theory Notes) Project appraisal is an appraisal of project objective, project design, project finance, Economic, Technical and Commercial feasibility and project s cost, benefit and profit. It is a complete and systematic review of all aspects of a project. It is a second look at the project feasibility report by a person other than one who is associated with its concepts and preparations. Market Appraisal Market and commercial appraisal focuses on the following issues : (1) Scope of Market Is the demand - supply gap substantial in growing market? (2) Completion Will there be new entrants, expansion of existing ones, which will enlarge the competition? (3) Prices Have the prices trends been favourable? Going up? (4) Exports Does the product have prospects of entering in to export market? (5) Customers List of principal customers, their requirements. nature and selling arrangement made with them. (6) Selling Distribution Arrangement The arrangement made / proposal to be made by the company is to be reviewed for its adequacy and efficiency. (7) Government's Role Is there any government control on pllce / distribution / allocation / imports or exports? Page 212 of 380 26/4/2008

Technical Appraisal Technical evaluation has direct bearing on the financial viability of the project: Field / Areas covered by technical appraisal are : (1) Location : - Investigation with respect to suitability of location. - Proximity to sources of raw material and markets, availability of power, fuel, water and transport facilities, manpower availability and adequacy of arrangement of disposal of wastes etc. are some of the critical elements to be evaluated. (2) Appropriate Technology : - Is technology chosen appropriate and compatible with eauipments and raw material to be used? - Is technology suitable from the point of view of size of plant and economic operation of the plant? - Is technical - know how tie up adequately done and is providing necessary guarantees? - In case of new process, has the project sponsor arranged the participation of process supplier as a party in the project. - Design and layout aspects of the projects are also examined. Will it suit to product for maximum changes? - Will it be modifiable without much changes for accommodating new product introduction. Can it be a Multi product design? - The design and layout is examined to see if minimum time is taken in handling equipment, raw material, consumable and stores? - Technical appraisal also focuses on suitability and adequacy of the plant and machinery, basis of selection, reasonabliness of cost, reputation and ability of supplier, performance guarantee and operational efficiency of the equipments. Page 213 of 380 26/4/2008

Finance Appraisal Financial appraisal consists of issues like : (1) Whether means of financing are adequately tied up and assured. (2) Whether the structuring of the capital of the project is done in such a way that acceptable debt: equity norm is used. The project should have capacity to generate revenues and not cash flow such that debt is serviced. (3) Whether cash flows are estimated with reasonable underlying a assumptions. (4) Applying the break-even point and studying margin of safety. (5) Employing discounting criteria for acceptability of the project. (a) (b) (c) NPV criterion Benefit cost ratio Internal rate of return. (6) Use of non - discounting criteria : (a) (b) (c) Payback period Accounting period Urgency Page 214 of 380 26/4/2008

APPRAISAL TECHNIQUES OTHER THAN FINANCIAL Management and Organizational Appraisal Not quantifiable, but relies on judgment. Evaluation of technical and managerial competence, knowledge of the project, integrity and past records. Do the promoters have adequate financial resources for the project? Do the promoters / company envisage and plan an organizational structure suitable for the proper management of the project during operational phase? Information required for managerial appraisal will be through analysis of financial position of existing business company / associate / sister companies / Bankers' report etc. (from credit's angle) Performance and growth records of associated companies with which promoters are connected. It shows entrepreneurial and managerial capability of promoters. Page 215 of 380 26/4/2008

Economic Appraisal It involves the following questions : Would the project contributed any sector's development significantly? Would the project contribute to the industrial development of the concerned region? \/Vhether project outlay will improve the socio - economic status of people, by providing jobs. Does project have a favourable impact on export promotion, import substitution and BOP status? Some of the measures used to evaluate economic viability are as follows : (1) Economic Rate of Return (ERR) (2) Domestic Resource Cost (DRC) (3) Effective Rate of Production (ERP) (4) Investment of Employment Ratio (IER) Page 216 of 380 26/4/2008

Environmental Analysis The growing concern of environment, resource depletion and pollution have forced the planners, policy makers and prc~ect implements to take care of impacts of the project on environment : The appraisal therefore evaluates project impact on : (1) Air (2) Water (3) Monumental resources (4) Land (5) Sound (6) Human inhabitation nearby (7) Animals and Birds The appraisal often relies upon environmental impact assessment (EIA) conducted by independent body Such studies are conducted to reveal whether there is any impact of project in a long run on environment If it is revealed that there will be no harmful change in various social economic and physical attributes of environment of the project then the project will be considered favourably. Page 217 of 380 26/4/2008

Forecasting Techniques A time series is the data on any variable recorded over a constant time interval. The time-frame of data recording may be an hour, a day, a week, a month, or a year, depending on the type of event the data refer to. Hourly temperature in a particular city, weekly earnings of workers in an industrial town, monthly prices of wheat, and annual production of rice may be cited as examples. All time series are subject to variations whose pattern and amplitude differ from one period to another. Given this, the object of dealing with a time series is to study and anlayse these variations with a view to knowing its past behaviour. A knowledge about the past behaviour is useful for drawing inferences about its future course. In an ever changing business and economic environment, it is necessary to have an idea about what the future is going to bring forth. The anlaysis of time series is a means to do so, especially as it acts as the basis to make a forecast. Economic and business forecasts are made for being used by the policy makers to decide the future course of action to survive and grow in a competitive environment. The analysis of time series is useful not only for individual business concerns, it is equally important for the Central and State Government for devising appropriate future growth strategies. A close understanding of time series data on major national aggregates such as population, national income, capital formation, etc. not only provides crucial information about the strength or weakness of strategies adopted in the past, but also help in setting future growth targets. Time series is made up of four components Trend (T) represents the long term behavior of a time series. This would tell whether the time series data reveal a steady upward or downward movement. Seasonal Variation (S) represents variation caused by season. Typically, this shows variation in demand during peak and lean season. For example, demand for snow tires will be at its peak during winter in USA. Cyclical Variation (C) represents the typical business cycles that occur periodically in several years. For example, in stock market, you will witness cycle of boom and cycle of recession that occur once in a while between many years. Random Variation (R) represents irregular variations that occur by chance having no assignable cause. Random variation cannot be predicted. Page 218 of 380 26/4/2008

METHODS OF FORECASTING Method - I ( FORECASTING BASED ON MOVING AVERAGES ) The moving averages method uses the average of the most recent n data values in the time series as the forecast for the next period. Mathematically. ( Most resent n data Values ) Moving average = ------------------------------------------. ( 1 ) N The term moving indicates that, as a new observation becomes available for the time series, it replaces the oldest observation in equation (1), and a new average is computed. As a result, the average will change, or move, as new observations become available. To illustrate the moving averages method, consider the 12 weeks of data presented in Table-1 These data show the number of gallons of gasoline sold by a gasoline distributor over the past 12 weeks. To use moving averages to forecast gasoline sales, we must first select the number of data values to be included in the moving average. For example, let us compute forecasts Using a 3 week moving average. The moving average calculation for the first 3 weeks of the gasoline sales time series is : 17 + 21 +19 57 Moving average (weeks 1-3) = ------------------ = ------ = 19 3 3 We then use this moving average value as the forecast for week 4. The actual value observed in week 4 is 23, so the forecast error in week 4 is 23-19. In general, the error associated with a forecast is the difference between the observed value of the time series and the forecast. The calculation for the second 3 week moving average is : 21 + 19 + 23 Moving average (weeks 2-4) - -------------------- = 21 3 Page 219 of 380 26/4/2008

Table 1 Week 1 2 3 4 5 5 7 8 9 10 11 12 Sales (1000's of Gallons) 17 21 19 23 18 16 20 18 22 20 15 22 Hence, the forecast for week 5 is 21, and the error associated with this forecast is 18-21 -3. Thus, the forecast error may be positive or negative, depending on whether the forecast is too low or too high. A complete summary of the 3 weeks moving average calculations for the gasoline sales time series is shown in Table 2. To forecast gasoline sales for week.13 using a 3 week moving average, we need to compute the average of sales for weeks 10, 11 and 12. The calculation for this moving average is : 20 + 15 + 22 Moving average (weeks 10-12) = ------------------- = 19 3 TABLE2 SUMMARY OF 3 WEEKS MOVING AVERAGE CALCULATION Week Time Series Values Moving Average Forecast 1 17-2 21-3 19-4 23 19 5 18 21 6 16 20 7 20 19 8 18 18 9 22 18 10 20 20 11 15 20 12 22 19 13-19 Hence, the forecast for week 13 is 19, or 19,000 gallons of gasoline. Page 220 of 380 26/4/2008

Problem 1. You are given the following information about demand of an item : Month 1 2 3 4 5 6 7 8 9 10 11 Demand 220 228 217 219 258 241 239 244 256 260 265 Calculate forecasted values using (i) three - monthly moving averages, (ii) five - monthly moving averages. Solution : The required values are given in third and fourth columns of Table 1. The three monthly values are obtained as (220 + 228 + 217) / 3 221.67, (228 + 217 + 219)/3 = 221.33 and so on. Similarly, five-monthly values are obtained by considering five monthly data. Table 1 Calculation of Forecasted Demand Month Demand Y 3 - Monthly Average 5 - Monthly Moving Average 1 220 - - 2 228 - - 3 217 - - 4 219 221.67-5 258 221.33-6 241 231.33 228.40 7 239 239.33 232.60 8 244 246.00 234.80 9 256 241.33 240.20 10 260 246.33 247.60 11 265 253.33 248.00 12-260.33 252.80 Page 221 of 380 26/4/2008

Problem 2. Obtain the profit forecasts using (i) four yearly moving averages and (ii) five yearly moving averages, from the following data relating to profits (in '000 Rs.) Solution : Year 1986 1987 1988 1989 1990 1991 1992 1993 Profit 48 53 55 56 58 63 68 60 Year 1994 1995 1996 1997 1998 1999 2000 2001 Profit 61 68 58 63 70 76 83 88 The given data and the required moving averages are shown in Table 2. Year Profit 4 - Yearly Moving Average Table 2 5 - Yearly Moving Average 1986 48 - - 1987 53 - - 1988 55 - - 1989 56 - - 1990 58 53.00-1991 63 55.50 54.00 1992 68 58.00 57.00 1993 60 61.25 60.00 1994 61 62.25 61.00 1995 68 63.00 62.00 1996 58 64.25 64.00 1997 63 61.75 63.00 1998 70 62.50 62.00 1999 76 64.75 64.00 2000 83 66.75 67.00 2001 88 73.00 70.00 2002-79.25 76.00 Forecast Accuracy An important consideration in selecting a forecasting method is the accuracy of the forecast. Page 222 of 380 26/4/2008

There are two methods to determine the accuracy of the forecast. (i) (ii) Mean Absolute Deviation (MAD) Mean Squared Error (MSE) Of these two methods the MSE is often used measure of the accuracy of a forecasting method. As we indicated previously, to use the moving averages method, we must first select the number of data values to be included in the moving average. Not surprisingly, for a particular time series, different lengths of moving averages will affect the accuracy of the forecast. One possible approach to choosing the number of values to be included is to use trial and error to identify the length that minimizes the MSE, Then, if we assume that the length that is best for the past will also be best for the future, we would forecast the next value in the time series using the number of data values that minimized the MSE for the historical time series. To understand this concept let us take an illustration. Continuing with the Gasoline example, let us find the 3-weekly moving averages and 5- seekly moving averages and compute both MAD and MSE for these averages. MAD and MSE computations for 3-weekly moving averages. Week Profit Moving Forecast Average Forecast Error Absolute Forecast Squared Forecast Error 1 17 - - - - 2 21 - - - - 3 19 - - - - 4 23 19 4 4 16 5 18 21-3 3 9 6 16 20-4 4 16 7 20 19 1 1 1 8 18 18 0 0 0 9 22 18 4 4 16 10 20 20 0 0 0 11 15 20-5 5 25 12 22 19 3 3 9 13-19 - - - Total 24 92 24 Mean Absolute Deviation (MAD) = --------- = 2.67 9 Page 223 of 380 26/4/2008

92 Mean Absolute Deviation (MSE) = --------- = 10.22 9 Where 9 is Observations. MAD and MSE computations for 5-weekly moving averages. Week Profit Moving Forecast Average Forecast Error Absolute Forecast Squared Forecast Error 1 17 - - - - 2 21 - - - - 3 19 - - - - 4 23 - - - - 5 18 - - - - 6 16 19.60-4 3.6 12.96 7 20 19.40 1 0.6 0.36 8 18 19.20-1 1.2 1.44 9 22 19.00 3 3 9 10 20 18.80 1 1.2 1.44 11 15 19.20-4 4.2 17.64 12 22 19.00 3 3 9 13-19.40 - - - Total 16.8 51.84 16.8 Mean Absolute Deviation (MAD) = ----------- = 2.4 7 51.84 Mean Absolute Deviation (MSE) = ------------ = 7.41 7 Based on MAD and MSE, 5-weekly moving averages are better. Therefore, forecast for the 13 th week is 19,400 gallons. Sometimes MAD and MSE may give contradicting results. in that case we take decision based on MSE. Problem 3. Following data is available about, actual sales quantities for the past 14 years : Page 224 of 380 26/4/2008

Find the "forecast' for year 15 using Two Years' as well as Three Years' Moving Average. Which of the two forecasts is more `reliable' on the basis of 'Mean Squared Error' (MSE) criterion? Year Actual Sales 2 Years Moving Average Error ( Actual Forecasted ) Error Square 3 Years Moving Average Error ( Actual Forecasted ) Error Square 1 2,300 - - - - - - 2 2,200 - - - - - - 3 2,000 2250-250 62,500 - - - 4 2,250 2100 150 22,500 2,167 83 6,889 5 2,600 2125 475 225,625 2,150 450 202,500 6 3,300 2425 875 765,625 2,283 1,017 1,034,289 7 3,500 2950 550 302,500 2,717 783 613,089 8 4,100 3400 700 490,000 3,133 967 935,089 9 3,800 3800 0-3,633 167 27,889 10 4,000 3950 50 2,500 3,800 200 40,000 11 4,300 3900 400 160,000 3,967 333 110,889 12 4,200 4150 50 2,500 4,033 167 27,889 13 4,800 4250 550 302,500 4,167 633 400,689 14 5,200 4500 700 490,000 4,433 767 588,289 15 5000-4,733 - - Answers: Sum Square of Errors Mean Square of Errors MSE 2,826,250 3,987,501 235,520.83 362,500.1 1) Forecast based on 2 years moving average = 5,000 2) Forecast based on 3 years moving average = 4,733 3) 3) Mean Squared Error being less for the two year moving average, it is more reliable ( Note: 3-4 marks could be allotted if only the moving average forecasts are correctly calculated without MSE ) Page 225 of 380 26/4/2008

Problem 4. Following data is available about "actual sales quantities" for the 10 years Year 1 2 3 4 5 6 7 8 9 10 Sales 230 220 200 240 230 260 300 240 280 320 Find the "forecast" for year 11 using "Two years" as well as "Three years "Moving Averages. Which of the two forecasts is more reliable on the basis of "Mean Squared Error" (MSE) criterion? Solution : For Homework Page 226 of 380 26/4/2008

Problem 5. Compute the three year and five year moving average for the following set of data. Compare the results. Comment. Period Value Period Value Period Value 1 69 6 112 11 178 2 84 7 120 12 196 3 90 8 129 13 220 4 94 9 147 14 236 5 93 10 162 15 260 Solution : For Homework Page 227 of 380 26/4/2008

Drawbacks of moving average forecast : Moving average forecast is quick, easy and fairly inexpensive to implement. It provides a reasonably good forecast for the immediate future (very short term). However, practicing managers must remember the drawbacks given in the following paragraph. Moving averages do not react well to seasonal variations. All observations considered in a time horizon are given the same weights. A large amount of historical data should be gathered and maintained to update forecast values. The choice of the period(n) is generally arbitrary. Page 228 of 380 26/4/2008

Method II ( Exponential Smoothing ) Exponential smoothing is a forecasting method that is easy to use and is handled efficiently by computers. Although it is a type of moving average technique, it involves little record keeping of past data. The basic exponential smoothing formula can be shown as follows : New forecast = last period's forecast + (last period's actual demand - last period's forecast)... (1) Where is a weight (or smoothing constant) that has a value between 0 and 1, inclusive. Equation 1 can also be written mathematically as F t = F t-1 + (A t-1 F t-1 ) Where F t = new forecast (for time period t) F t-1 = previous forecast (for time period t 1 ) = smoothing constant (0 1 ) A t-1 = previous period's actual demand The concept here is not complex. The latest estimate of demand is equal to our old estimate adjusted by a fraction of the error (last period's actual demand minus the old estimate). The smoothing constant, a, can be changed to give more weight to recent data when the value is high, or more weight to past data when it is low. For example, when a = 0.5, it can be shown mathematically that the new forecast is based almost entirely on demand in the past three periods. When a = 0.1, the forecast places little weight on recent demand and takes many periods (about 19) of historic values into account. Page 229 of 380 26/4/2008

Example In January, a demand for 142 of a certain car model for February was predicted by a dealer. Actual February demand was 153 cars. Using a smoothing constant of a = 0.20, we can forecast of March demand using the exponential smoothing model. Substituting into the formula, we obtain. New forecast (for March demand) = 142 + 0.2 (153 142) = 144.2 Thus, the demand forecast for the cars in March is 144. Suppose that actual demand for the cars in March was 136. A forecast for the demand in April, using the exponential smoothing model with a constant of a = 0.20, can be made. New forecast (for April demand) = 144.2 + 0.2 (136 144.2) = 142,6, or 143 cars. Selecting the Smoothing Constant The exponential smoothing approach is easy to use and has been applied successfully by banks, manufacturing companies, wholesalers, and other organizations. The appropriate value of the smoothing constant, a, however, can make the difference between an accurate forecast and an inaccurate forecast. In picking a value for the smoothing constant, the objective is to obtain the most accurate forecast. Several values of the smoothing constant may be tried, and the one with the lowest MAD (Mean Absolute Deviation) could be selected Page 230 of 380 26/4/2008

Example : Let us apply this concept with a trial and error testing of two values of a in the following example. The port of Baltimore has unloaded large quantities of grain from ships during the past eight quarters. The port's operations manager wants to test the use of exponential smoothing to see how well the technique works in predicting tonnage unloaded. He assumes that the forecast of grain unloaded in the first quarter was 175 tons. Two values of a are examined : a = 0.10 and a = 0.50. Table shows the detailed calculations for a = 0.10 only. Table - 1 Quarter Actual Tonnage Rounded Forecast Rounded Forecast With a = 0.10 Unloaded With a = 0.50* 1 180 175 175 2 168 176 = 175.00 + 0.10 (180-175) 178 3 159 175 = 175.50 + 0.10 (168-175.50) 173 4 175 173 = 174.75 + 0.10 (159-174.75) 166 5 190 173 = 173.18 + 0.10 (175-173.18) 170 6 205 175 = 173.36 + 0.10 (190-173.36) 180 7 180 178 = 175.02 + 0.10 (205-175.02) 193 8 182 178 = 178.02 + 0.10 (180-178.02) 186 9? 179 = 178.22 + 0.10 (182-178.22) 184 Note : * forecast rounded to the nearest one. To evaluate the accuracy of each smoothing constant, we can compute the absolute deviations and MADs (see table). Based on this analysis, a smoothing constant of a = 0.10 is preferred to a = 0.50 because it s MAD is smaller. Page 231 of 380 26/4/2008

Table - 2 Quarter Actual Tonnage Unloaded Rounded Forecast With = 0.10 Absolute Deviations for = 0.10 Rounded Forecast with = 0.50 Absolute Deviations for = 0.50 1 180 175 5 175 5 2 168 176 8 178 10 3 159 175 16 173 14 4 175 173 2 166 9 5 190 173 17 170 20 6 205 175 30 180 25 7 180 178 2 193 13 8 182 178 4 186 4 Sum of absolute deviations 84 100 deviations MAD = ---------------------- = 10.50 ( for = 0.1 ) n MAD = = 12.50 ( for = 0.5 ) Page 232 of 380 26/4/2008

Problem - 1 A company manufacturing Xerox machines experienced the following sales for theft Xerox machines during their busiest time of the year. Month Actual Demand May 120 June 180 July 150 Forecasted demand for May was 80 units. Using exponential something with a = 0.20, forecast the demand for August. Solution : For Homework Page 233 of 380 26/4/2008

4 th April, 2008. Lecture XII. DECISION TREE Introduction A decision tree is a schematic diagram of a sequence of alternative decisions and the results of those decisions. A decision tree is beneficial for several reasons : First, Second, Third, it provides a pictorial representation of the sequential decision process which facilitates understanding of the process. it makes the expected value computation easier as they can be performed directly on the tree diagram. the actions of more than one decision maker can be considered. Drawing a decision tree (the basic rules) Every decision tree starts from a decision point with the decision options that are currently being considered. (a) (b) There should be a fine, or branch, for each option or alternative. It helps to identify the decision point, and any subsequent decision points in the tree, with a symbol. Here, we shall use a square shape. It is conventional to draw decision trees from left to right, and so a decision tree will start as follows. A B D 1 C D Page 234 of 380 26/4/2008

The square is the decision point, and A, B, C and D represent four alternatives from which choice must be made. If the outcome from any choice is certain, the branch of the decision tree for that alternative is complete. If on the other hand, the outcome of a particular choice is uncertain, the various possible outcomes must be shown. We show this on a decision tree by inserting an outcome point on the branch of the tree. Each possible outcome is then shown as a subsidiary branch, coming out from the outcome point. The probability of each outcome occurring should be written on to the branch of the tree which represents that outcome. To distinguish decision points from outcome points, a circle will be used as the symbol for an outcome point. A D 1 B 1 High Sales 0.6 Low Sales 0.4 In the example above, there are two choices facing the decision maker, A and B. The outcome if A is chosen is known with certainty, but if 8 is chosen, there are two possible outcomes, high sales (0.6 probability) or low sales ( 0.4 probability ). When several outcomes are possible it is usually simpler to show two or more stages of outcome points on the decision tree. Page 235 of 380 26/4/2008

Example: A company can choose to launch a new product XYZ or not. If the product is launched, expected sales and expected unit costs might be as follows : Sales Units Probability Unit cost Probability 10,000 0.8 Rs. 6 0.7 15,000 0.2 Rs. 8 0.3 a) The decision tree could be drawn as follows : Cost Rs. 6 Launch 3 Sales 10,000 0.8 1 0.7 Cost Rs. 8 0.3 Sales 15,000 0.2 2 Cost Rs. 6 0.7 D 1 Cost Rs. 8 0.3 Do Not Lauch Page 236 of 380 26/4/2008

b) The layout shown above will usually be easier to use than the alternative way of drawing the tree which is shown below : Launch 1 Sales 10,000, Cost Rs. 6 0.56 Sales 10,000, Cost Rs. 8 0.24 Sales 15,000, Cost Rs. 6 0.14 D 1 Sales 15,000, Cost Rs. 8 0.06 Do Not Lauch Sometimes, a decision taken now will lead to other decisions to be taken in the future. When this situation arises, the decision tree can be drawn as a two-stage tree, as follows : Decision A Decision X 1 0.7 D 2 Decision B Decision C D 1 0.3 D 3 Decision D Decision Y Page 237 of 380 26/4/2008

In this tree, either a choice between A and B or else a choice between C and D will be made depending on the outcome which occurs after choosing X. The decision tree should be in chronological order from left to right. When there are twostage, decision trees, the first decision in time should be drawn on the left. Advantages Decision trees are a useful management tool. They allow manages to consider the options they are facing and the possible consequence of those options. Moreover, provided that the parameters included in the decision tree have been estimated correctly, managers, choose the best course of action and hopefully save money. Bear in mind, however, that the calculation of probabilities is highly subjective and on occasions, only a small error in the calculation of probabilities or expected values could lead to an incorrect decision being taken. Decision trees are useful in giving a simple diagrammatic representation of the various options open to management, and the possible outcomes of each options. They also draw attention to the immediate decisions to be made, and may help management to avoid wasting unnecessary time deliberating later decisions. A decision tree might also help management to eliminate decision options which are not worth considering further, and to focus attention on the more viable options. A decision tree is a simplified representation of reality, however, it may omit some possible decision options, or it may simplify the possible outcomes. For example, in this question, 'success' and 'failure' are two extreme outcomes, whereas a variety of outcomes between success and failure may be possible. The decision tree is therefore likely to be a simplification of reality. If management choose to base their decisions on the expected value criterion, choosing the option with the highest expected value of profit, then decision trees (along with other methods of calculating and presenting expected value figures) are a useful guide to decision making. Expected values reflect the weighted average of probable outcomes. For one-off decisions, a weighted average value may be an inappropriate basis for comparing options. For example, if there is a 0.8 probability of earning Rs.100,000 and a 0.2 probability of losing Rs.50,000,000, the expected value of profit would be -Rs.99,20,000. If the decision is non recurring, management may need to consider all possible outcomes, and especially the worst possible outcomes. If the company cannot afford to lose Rs.50,000,000, a 0.2 probability of such a loss might be unacceptable. Decision trees, or expected value calculations, are most useful where decisions are of a recurring nature, where the long-run average profit should be approximately the expected value of the profit for a single decision. Examples of such situations would be : Page 238 of 380 26/4/2008

i) Decisions on test drilling for oil, where a company carries out many tests each year, ii) Decisions on whether or not to launch a new product on the market, where a large number of new products are considered each year, iii) Decisions on whether or not to obtain market research information for particular project, where market research is in common use through out the organization. Page 239 of 380 26/4/2008

Example. A company has an investable surplus of Rs.100 Crores. Investing this amount in this existing business can give a certain return of 8.0%. Alternatively, there is an opportunity for diversification which, if successful, is estimated to bring a return of 17.0%. However if diversification is not successful, expected return will be only 2.0%. What must be the probabilities for two outcomes of diversification in order to make the diversification worthwhile? Let probability of success in diversification be l x'. Hence probability of failure be (1 - x). Evaluation of diversification should be at least be 8% i.e. Rs.8.0 crores if not more. Existing Business 8.0 Cr D 1 Success X % 17.0 Cr Failure ( 1 X ) % 2.0 Cr Hence, 8 = 17 X + 2 ( 1 X ) 8 = 17 X + 2 2X 8-2 = 15X 6 = 15X 6 X = ------ = 0.4 15 Probability of success in diversification should be at least 0.4 or 40.0% Page 240 of 380 26/4/2008

Problem 1. A manufacturing company has to select one of the two products A or B for manufacturing. Product A required investment of Rs.20,000 and product B, Rs.40,000. Market research survey shows high medium and low demands with corresponding probabilities and return from sales, in Rs. thousand, for the two products, in the following table. Market Probability Return from Sales ( Rs in 000 ) A B A B High 0.4 0.3 50 80 Medium 0.3 0.5 30 60 Low 0.3 0.2 10 50 Construct an appropriate Decision Tree. What decision the company should take? Solution : Product - A * Rs 32,000/- 1 High 0.4 Medium 0.3 Sale Rs 50,000/- Sale Rs 30,000/- D 1 Investment Rs 20,000 Product - B Low 0.3 High 0.3 Sale Rs 10,000/- Sale Rs 80,000/- Investment Rs 40,000 ** Rs 64,000/- 2 Medium 0.5 Sale Rs 60,000/- Roll Back Technique to evaluate Decision Tree. Low 0.2 Sale Rs 50,000/- We need to calculate the EMV ( Earnest Monitory Value ) for each alternate. EMV at 1 = 50,000 x 0.4 + 30,000 x 0.3 + 10,000 x 0.3 = 32,000 * Page 241 of 380 26/4/2008

EMV at 2 = 80,000 x 0.3 + 60,000 x 0.5 + 50,000 x 0.2 = 64,000 ** So, Net Profit from Product A Product B 32,000 64,000-20,000-40,000 ------------- -------------- 12,000 24,000 Optimum decision at D1 is Product B Page 242 of 380 26/4/2008

Problem 2. Alfa Industries has to decide whether to set up a large plant or a small plant for its new range of refrigerators. A large plant will cost the company Rs.25 lakhs while a small plant will cost Rs.12 lakhs. An extensive market survey and a cost profit volume analysis carried out by the company reveal the following. High demand Probability = 0.5 Moderate demand Probability = 0.3 Low demand Probability = 0.2 (a) A large plant with high demand will yield an annual profit of Rs.100 lakhs. (b) A large plant with moderate demand will yield an annual profit of Rs.60 lakhs. (c) A large plant with low demand will lose Rs.20 lakhs annually because of production inefficiencies. (d) A small plant with high demand would yield Rs.25 lakhs annually, taking into account the cost of lost sales due to inability to meet demand. (e) A small plant with moderate demand will yield Rs.35 lakhs, as the losses due to lost sales will be lower. (f) A small plant with low demand will yield Rs.45 lakhs annually, as the plant capacity and demand will match. Draw a Decision Tree and find optimum solution. Page 243 of 380 26/4/2008

Solution : ( All figures in Lakhs ) High 0.5 Sale Rs 100 Large Plant * Rs 72 1 Medium 0.3 Sale Rs 60 D 1 Investment Rs 25 Low 0.2 Sale Rs 20 Small Plant High 0.5 Sale Rs 25 Investment Rs 12 ** Rs 32 2 Medium 0.3 Sale Rs 35 Low 0.2 Sale Rs 45 EMV at 1 = 100 x 0.5 + 60 x 0.3 + 20 x 0.2 = 72 * EMV at 2 = 25 x 0.5 + 35 x 0.3 + 45 x 0.2 = 32 * So, Net Profit from Large Plant Small Plant 72 32-25 - 12 --------- --------- 47 20 Optimum decision at D1 is build the Large Plant. Page 244 of 380 26/4/2008

Problem 3. A financial advisor has recommended two possible mutual funds for investment: Fund A and Fund B. The return that will be achieved by each of these depends on whether the economy is good, fair, or poor. A payoff table has been constructed to illustrate this situation. Investment State of Nature Good Economy Fair Economy Poor Economy Fund A Fund B $10,000 $ 6,000 $ 2,000 $ 4,000 - $ 5,000 0 Probability 0.2 0.3 0.5 (a) (b) Draw the decision tree to represent this situation. Perform the necessary calculations to determine which of the two mutual funds is better. Which one should you choose to maximize the expected value? Solution : ( All figures in $ ) Fund A * 100 1 Good 0.2 Fair 0.3 10,000 2,000 D 1 Fund B Poor 0.5 Good 0.2 ( 5,000 ) 6,000 ** 2400 2 Fair 0.3 4,000 Poor 0.5 0 EMV at 1 = 10,000 x 0.2 + 2,000 x 0.3 + - 5,000 x 0.5 = 100 * Page 245 of 380 26/4/2008

EMV at 2 = 6,000 x 0.2 + 4,000 x 0.3 = 2400 ** It is advisable to Invest in the Mutual Fund B. Page 246 of 380 26/4/2008

Problem 4. Grow fast Company is evaluating four alternative single-period investment opportunities whose returns are based on the state of the economy. The possible states of the economy and the associated probability distribution are as follows : State Fair Good Grate Probability 0.2 0.5 0.3 The returns for each investment opportunity and each state of the economy are as follows : Alternative Fair Good Great W Rs.1,000 Rs.3,000 Rs.6,000 X Rs. 500 Rs.4,500 Rs.6,800 Y Rs.2,000 Rs.5,000 Rs.8,000 Z Rs.4,000 Rs.6.000 Rs.8,500 Using the decision tree approach, determine the expected return for each alternative. Which alternative investment proposal would you recommend if the expected monetary value criterion is to be employed? Page 247 of 380 26/4/2008

Solution : Fair 0.2 1,000 * 3,500 1 Good 0.5 3,000 Alternate W Grate 0.3 Fair 0.2 6,000 500 Alternate X ** 4,390 2 Good 0.5 4,500 Grate 0.3 6,800 D 1 Alternate Y Fair 0.2 2,000 *** 5,300 3 Good 0.5 5,000 Alternate Z Grate 0.3 8,000 Fair 0.2 4,000 **** 6,350 4 Good 0.5 6,000 Grate 0.3 8,500 EMV at 1 = 1,000 x 0.2 + 3,000 x 0.5 + 6,000 x 0.3 = 3,500 * Page 248 of 380 26/4/2008

EMV at 2 = 500 x 0.2 + 4,500 x 0.5 + 6,800 x 0.3 = 4,390 ** EMV at 3 = 2,000 x 0.2 + 5,000 x 0.5 + 8,000 x 0.3 = 5,300 *** EMV at 4 = 4,000 x 0.2 + 6,000 x 0.5 + 8,500 x 0.3 = 6,350 **** Alternate Z is the Better Proporals. Page 249 of 380 26/4/2008

Problem 5. Company, X has to decide whether to manufacture and market product A or product B. Profits will be the decision criterion. In case product A is manufactured and marketed, there is : i) A 40 percent chance or probability 0.4 that sales will be excellent and profit realized will be Rs. One lakh. ii) A 50 percent chance (or 0.5 probability ) of sales being merely fair, realizing a profit of Rs.60,000 and, iii) A 10 percent chance (or 0.1 probability ) of poor sales and loss of Rs.20,000. Similarly for Product B, a) A 70 percent chance or (0.7 probability ) of realizing a profit of Rs.1,50,000 and, b) A 30 percent chance (or 0.3 probability) of making a loss of Rs.30,000. Which product should be manufactured and marketed? Solution : ( All figures in Rs. ) Product A * Rs, 30,000 1 Excellent 0.4 Fair 0.5 1,00,000 60,000 D 1 Poor 0.1 ( 20,000 ) Product B Excellent 0.7 1.50,000 ** 2400 2 Poor 0.3 ( 30.000 ) Page 250 of 380 26/4/2008

EMV at 1 = 1,00,000 x 0.4 + 60,000 x 0.5 + - 20,000 x 0.2 = 30,000 * EMV at 2 = 1,50,000 x 0.7 + - 30,000 x 0.3 = 96,000 * Product B is Referred. Page 251 of 380 26/4/2008

Problem 6. The investment staff of TNC Bank is considering four investment proposals for a client Shares, Bonds, Real Estate and Savings Certificates. These investments will be held for one year. The past data regarding the four proposals are given below : Share : There is a 25 percent chance that shares will decline by 10 percent, a 30 percent chance that they will remain stable and a 45 percent chance that they will increase in value by 15 percent. Also the shares under consideration do not pay any dividends. Bonds : These bonds stand a 40 percent chance of increase in value by 5 percent and 60 percent chance of remaining stable and they yield 12 percent. Real Estate : This proposal has a 20 percent chance of increasing 30 percent in value, a 25 percent chance of increasing in 20 percent value, a 40 percent chance of increasing in 10 percent value, a 10 percent chance of remaining stable and a 5 percent chance of losing 5 percent of its value. Saving Certificate : These certificates yield 8 1/2 percent with certainty. Use a decision tree to structure the alternatives to the investment staff and using the expected value criterion, choose the alternative with the highest expected value. Page 252 of 380 26/4/2008

Solution : Assumption, Initial Investment is Rs. 100/- ( Figures in Rs. ) * 51.75 1 Decline 0.25 Stable 0.30 100 10 % = 90 100 Shares Bonds ** 114 2 Increase 0.45 Increase 0.40 Stable 0.60 100 + 15 % = 115 100 + 5 % = 105 + 12 = 117 100 + 12 = 112 D 1 30 % Increase 100 + 15 % = 120 0.20 Real Estate *** 113.75 3 20 % Increase 100 + 25 % = 125 0.25 10 % Increase 100 + 10 % = 110 0.40 Saving Certificate Stable 100 0.10 5 % Decrease 0.05 100 + 5 % = 90 **** 108 4 Increase 8.5 % 1.0 100 + 8.5 % = 108.50 EMV at 1 = 90 x 0.25 + 100 x 0.30 + 115 x 0.45 = 51.75 * Page 253 of 380 26/4/2008

EMV at 2 = 117 x 0.40 + 112 x 0.60 = 114 ** EMV at 3 = 120 x 0.20 + 125 x 0.25 + 110 x 0.40 + 100 x 0.10 + 90 x 0.05 = 113.75 *** EMV at 4 = 108 x 1.00 = 108 **** Bond is Referred option Page 254 of 380 26/4/2008

Problem 7. Suppose you have Rs.5,00,000 (Rs. Five Lakh) to invest in share market. Your broker has suggested to invest in either Company A or Company B. Shares in Company A are risky but could yield a 40% return on investment during next year if stock market conditions are favourable ("Bull" market). If stock market conditions are not favourable ("Bear" market), the stock many loose 20% of its value. Company B provides safe investment with 15% return in a "bull" market and only 5% in a "bear" market. The chance of the market being "bull" is 55%. Draw the decision tree and specify optimum course of action. Solution : Here important statement is The chance of the market being "bull" is 55%. ( All figures in Lakhs ) Bull Market 0.55 * Rs 5.65 Company A 1 5 + 40 % = 7 D 1 Investment Rs 5 L Bear Market 0.45 5-20 % = 4 Company B Investment Rs 5 L ** Rs 5.52 2 Bull Market 0.55 5 + 15 % = 5.75 EMV at 1 Bear Market 0.45 5 + 5 % = 5.25 = 7,00,000 x 0.55 + 4,00,000 x 0.45 = 5,65,000 * EMV at 2 = 5,65,000 x 0.55 + 5,25,000 x 0.45 = 5,52,500 ** Company A is a preferred option. Page 255 of 380 26/4/2008

Problem 8. A road transport company renews its fleet of luxury buses every 4 years. By replacing the entire fleet at one time favourable purchase prices are obtained. The size of the present fleet is 10 buses. The proprietor of the company could vary the size of the fleet depending on the economic climate. He estimates that if he buys 8 buses, he could earn a profit of Rs.10,000/- per bus over the next 4 years if the economic climate is upbeat, Rs.8000/- per bus if the climate is moderate and Rs.5000/- per bus if the economic climate is depressing. If he buys 12 buses, he could earn a profit of respectively Rs.13000/-, Rs.8000/- or Rs.1000/- per bus depending on the three types of economic climates. If he keeps the fleet at its present strength, the profits per bus are respectively Rs.12000/-, Rs.8000/- and Rs.3500/- over the next four years. He estimates that, the chance that the economic climate will be upbeat is 0.4 and that it will be moderate is 0.3. Use decision tree to get proprietor's optimal decision. Solution : Upbeat 0.4 Rs 10,000 * Rs 63,200 1 Moderate 0.3 Rs 8,000 D 1 8 Buses 10 Buses ** Rs 82,500 2 Depressing Rs 5,000 0.2 Upbeat 0.4 Moderate 0.3 Rs 12,000 Rs 8,000 12 Buses Depressing 0.2 Upbeat 0.4 Rs 3,500 Rs 13,000 *** Rs 94,800 3 Moderate 0.3 Rs 8,000 Depressing Rs 10,000 0.2 Page 256 of 380 26/4/2008

EMV at 1 = 10,000 x 0.40 + 8,000 x 0.30 + 5,000 x 0.30 = 7,900 As this fleet has 8 buses, 8 x 7,900 = 63,200/- * EMV at 2 = 12,000 x 0.40 + 8,000 x 0.30 + 3,500 x 0.30 = 8,250 As this fleet has 10 buses, 10 x 8,250 = 82,500/- ** EMV at 3 = 13,000 x 0.40 + 8,000 x 0.30 + 1,000 x 0.30 = 7,900 As this fleet has 12 buses, 12 x 7,900 = 94,800/- *** 12 Buses Fleet is Recomended. Page 257 of 380 26/4/2008

Problem 9. Suppose that you want to invest Rs.10,000 in the stock market by buying shares in one of two companies A and B. Share in company A are risky but could yield a 50% return on investment during the next year. If the stock market conditions are not favorable (i.e. "bear" market), the stock may loose 20% of its value. Company B provides safe investment with 15% return in a "bull" market and only 5% in a "bear" market. All the publications you have consulted are predicting a 60% chance for a 'bull" market and 40% for "bear" market. Draw the decision tree, and specify a course of action. Page 258 of 380 26/4/2008

Problem 10. 'Thanda Cool' Company has developed a new cold drink. A total of Rs.25 lakh is spent on the new product. One plan is to market it in small cans with other products of the company. The plan will cost Rs.8 lakhs and it might result in high, moderate or low market response with probabilities 0.3, 0.5 and 0.2 respectively, with revenues of Rs.55 lakhs, Rs.35 lakhs and Rs.15 lakhs for the corresponding market responses. The second marketing plan is to fully concentrate on T. V. advertisements with a cost of Rs.20 lakhs. The resulting market response can be either excellent or very good, with probabilities 0.4 and 0.6 respectively. The revenues in these cases will be Rs.50 lakhs and Rs.35 lakhs respectively. Draw a decision tree to determine the plan the company should follow, to maximize the profit. Page 259 of 380 26/4/2008

Problem 11. A company has the opportunity of marketing a new package of computer games. It has two possible courses of action : to test market on a limited scale or to give up the project completely. A test market would cost Rs.1,60,000 and current evidence suggests that consumer reaction is equally likely to be "Positive" or 'Negative". If the reaction to the test marketing were to be "Positive", the company could either market the computer games nationally or still give up the project completely. Research suggests that a national launch might result in the following sales : Sales High Average Low Contribution Rs. Million 1.2 0.3-0.24 Probability 0.25 0.50 0.25 If the test marketing were to yield "Negative" results, the company would give up the project. Giving up the project at any point would result in a contribution of Rs,60,000 from the sale of copyright etc. to another manufacturer. You are required to: (a) (b) Draw a decision tree to represent this situation including all relevant probabilities and financial values. Recommend a course of action for the company on the basis of expected values. Page 260 of 380 26/4/2008

Solution : Give Up 2 Cost Rs. 60,000/- 3 Positive 0.5 D 2 Market 1 High 0.25 Average 0.50 Rs 12,00,000 Rs 3,00,000 Limited Scale Cost Rs. 1,60,000/- Negative 0.5 Cost Rs. 60,000/- Low 0.25 Rs (-2,40,000) D 1 Give-up Cost Rs. 60,000/- 4 EMV at 1 = 12,00,000 x 0.25 + 3,00,000 x 0.50 + - 2,40,000 x 0.25 = 3,90,000 * EMV at 2 = 3,90,000 x 0.5 + 60,000 x 0.5 = 2,45,000 ** ( In Complete Problem ) Page 261 of 380 26/4/2008

Problem 12. Raman Industries Ltd. has a new product which they expect has a great potential. At present they have two courses of action open to them. To test market (S1) and to drop product (S2). Test market will cost Rs.50,000 and the response could be positive or negative with probabilities 0.70 and 0.30 respectively. If the result of the test marketing is positive they could either market it with full scale or drop the product. If they market with full scale then the result might be low, medium or high demand and the respective net pay offs would be Rs.1,00,000, Rs.2,00,000 or Rs.5,00,000. The outcomes have probabilities of 0.25, 0.55 and 0.20 respectively. If the result of the test marketing is negative they have decided to drop the product. If at any point they drop the product there is a net gain of Rs.25,000/- from the sale of the scrap. All financial values have been discounted to the present. Draw a decision tree for the problem and indicate the most preferred decision. Problem 13. A businessman has two independent investments A and B available to him but he lacks of capital to undertake both simultaneously. He can choose to take A first and then stop or if A s successful then take B or vice versa. The probability of success of A is 0.7 while for B it is 0.4. Both investments require an initial capital outlay of Rs.2,00,000 and both return nothing if unsuccessful. Successful completion of A will return Rs.30,000 over cost and the successful of B will return Rs.15,000 over cost. Draw the decision tree and determine the best strategy. Page 262 of 380 26/4/2008

Problem 14. An investment company is evaluating four alternative investment opportunities whose returns are based on the state of the economy. The possible states of economy and associated probabilities are as follows : State of Economy Fair Good Very Good Probability 0.2 0.5 0.3 The returns for each investment opportunity and each state of the economy are as follows : Investment State of Economy Fair Good Very Good Shares 1000 3000 6000 Bonds 500 4500 6800 Real Estate 0 5000 8000 Savings Certificates - 4000 6000 8500 Using decision tree approach determine the expected return for each alternative. Which investment proposal would you recommend? Problem 15. A company is currently working with a process which after paying for materials, labour etc., brings a profit of Rs.10 crores. The following alternatives are made available to the company. (i) (ii) The company can conduct research (R1) which is expected to cost Rs.10 crores having 90% chances of success. If it proves a success, the company gets a gross income of Rs.25 crores. The company can conduct research (R2) which is expected to cost Rs.5 crores having a probability of 60% success. If successful, the gross income will be Rs.25 crores. (iii) The company can pay Rs.6 crores as royalty of a new process which will bring a gross income of Rs.20 crores. (iv) The company can continue the current process. Because of limited resources, it is assumed that only one of the two types of research can be carried cut at a time. Draw a decision tree and evaluate the optimum decision. Page 263 of 380 26/4/2008

Problem 16. An agricultural company wants to decide which commodity it should stock to get maximum profit. It was supplied with the following information. The respective probabilities of monsoon will be excess, normal and deficient are being 0.40, 0.30 and 0.30. The estimated profit or loss of three commodities in respect of these different kinds of monsoon are : Monsoon Excess ( Rs. ) Normal ( Rs. ) Deficient ( Rs. ) Profit Per Ton Rice + 10,000-4,000 +15,000 Wheat + 4,000-3,000 + 8,000 Rice + 4,000 + 1,000 + 1,000 Problem 17. New India Telecom company has invented a new device which can be used for internet telephoning. The following options are available : (i) Manufacture the device. (ii) Sell the rights to other company. (iii) Be paid on royalty from other company. The probabilities associated with the level of sales in the market are 0.15 for 'High sales ; 0.25 for `Medium Sales' and 0.6 for 'Low Sales'. The profit (in lakhs of Rs.) which can be expected in each case is given in the following table. Draw the associated decision tree and specify course of action. (Profit in laksh of Rs.) Action Market Condition High Sales Medium Sales Low Sales Manufacture 85 25-20 Sell the rights 20 20-20 Get royalty 40 20-10 Page 264 of 380 26/4/2008

Problem 18. A manufacturer of ultimate coffee products has come to realize that its top brand coffee powder sale has started falling down. The board meets to discuss the alternative courses of action drafted out by the research cell of the company. They recommended :- Course (I) Course (ii) to step up the promotional efforts incurring an additional cost of Rs.3.5 lakhs. to modify the container slightly and introduce it as new ultimate coffee. Expenses for this change and promotion would be Rs.5 Lakhs. Course (III) to maintain statusquo. The estimated revenue in each of these courses of action is as follows : Course (I) Course (II) Course (III) Revenue (Rs.'000) 700 600 500 400 300 200 Probability: 0.15 0.25 0.25 0.20 0.10 0.05 Revenue( Rs.'000) 900 800 700 600 500 400 Probability: 0.05 0.15 0.30 0.20 0.15 0.15 Revenue( Rs.'000) 400 300 200 100 Probability: 0.25 0.30 0.25 0.20 Draw a decision tree to decide optimum course of action. Page 265 of 380 26/4/2008

Problem 19. A manufacturer of a popular brand of refrigerator intends to market its new fuel' economy model. It has 3 options namely (i) to go ahead and launch (ii) to run a test market (iii) to abandon the model. If the product is launched the demand may be poor, good or excellent. Market research conducted by a consulting firm indicates the probabilities of poor, good and excellent demand as 20%, 40%, 40% respectively. If a test market is used at a cost of Rs.2 Lakhs, it will give the indication of market being either favourable or unfavourable, there being 50 : 50 chance for the same. If the market response is favourable, the probabilities of poor, good and excellent market are 10%, 20% and 70 % respectively and if the response is unfavourable and the product is launched the corresponding probabilities are 80%, 10%, 10%. The pay offs are Rs.4,00,000, Rs.8,00,000 and Rs.14,00,000. Draw a decision tree and select an optimum strategy. Page 266 of 380 26/4/2008

Problem 20. Doily Toys Private Limited is considering the addition of a new toy to its existing product line. Three alternatives are available to the company. (i) Work overtime to meet the demand of the new toy. Overtime expenses are estimated at Rs.20,000 per month. (ii) Install a new equipment for which fixed expenses per month are expected at Rs.80,000. (iii) Lease (rent) a machine at the rate of Rs.35,000 per month. The expected demand for the toy are as given below : 10,000 pieces with a probability of 0.5 20,000 pieces with a probability of 0.3 50,000 pieces with a probability of 0.2 Monthly contribution from the sales of different units under different alternatives are calculated as under : Alternatives Demand ( in pieces) (Monthly Contribution) (in Rs.) 10,000 20,000 50,000 Overtime working 60,000 1,20,000 3,00,000 New equipment 80,000 1,60,000 4,00,000 Leasing 70,000 1,40,000 3,50,000 Draw the Decision Tree and recommend the optimum course of action to the company. Page 267 of 380 26/4/2008

10 th April, 2008. Lecture XIII. CAPITAL BUDGETING Introduction Capital budgeting is a process of planning capital expenditure which is to be made to maximize the long term profitability of the organization. The term capital budgeting refers to planning for capital assets. The capital budgeting decision means a decision as to whether or not money should be invested in long term projects such as installing a machinery or creating additional capacities to manufacture a part which is at present purchased from outside. Investment Appraisal Techniques The techniques available for appraisal of investment proposals are classified as follows : Traditional Techniques (a) (b) Payback period method Accounting rate of return method Discounted Cash flow Techniques (DCF Techniques) i) Net Present Value method ii) Internal Rate of Return method iii) Profitability Index Method Page 268 of 380 26/4/2008

Payback Period Method The payback period is usually expressed in years, which it takes the cash inflows from a capital investment project to equal the cash outflows. The method recognizes the recovery of original capital invested in a project. At payback period the cash inflows from a project will be equal to the project's cash outflows. This method specifies the recovery time, by accumulation of the cash inflows (inclusive of depreciation) year by year until the cash inflows equal to the amount of the original investment. When deciding between two or more competing projects the usual decision is to accept the one which has the shortest payback. In simple terms it can be defined as the number of years required to recover the cost of the investment. Merits It is simple to apply and easy to understand. This method is most suitable when the future is very uncertain. Shorter the payback period, the less risky is the project. Therefore, it can be considered as an indicator of risk. This method gives an indication to the prospective investors specifying when their funds are likely to be repaid. It does riot involve assumptions about future interest rates. Demerits It does not indicate whether an investment should be accepted or rejected, unless the payback period is compared with an arbitrary managerial target. The method ignores cash generation beyond the payback period and this can be seen more a measure of liquidity than of profitability. It fails to take into account the timing of returns and the cost of capital. It fails to consider the whole life time of a project. It is based on a negative approach and gives reduced importance to the going concern concept and stresses on the return of capital invested rather than on the profits occurring from the venture. The traditional payback approach does not consider the salvage value of an investment. It fails to determine the payback period required in order to recover the initial outlay if things go wrong. This method makes no attempt to measure a percentage return on the capital invested and is often used in conjunction with other methods. The projects with long payback periods are characteristically those involved in long term planning and which determine an enterprise's future. However, they may not yield their highest returns for a number of years and the result is that the payback method is biased against the very investments that are most important to long term. Page 269 of 380 26/4/2008

Examples : 1. When the Cash inflows are UNIFORM every year. Here the initial cost of investment is divided by the constant annual cash inflow as defined above to get the payback period. For example, an investment of Rs.40,000 in a machine is expected to produce a cash inflow of Rs.8,000 per annum then 40,000 Payback period = ---------------- = 5 years 8,000 2. When the projected cash inflows are NOT EQUAL every year. In such a situation payback period is calculated by a process of cumulating cash inflows till they equate the original investment outlay. e.g. Suppose initial cost is Rs.50 000/- Annual Inflows ( Rs ) Rs. Year I 10,000 Year II 15,000 Year III 20,000 Year IV 25,000 Year Inflows ( Rs ) Cumulative Inflows Rs. 1 10,000 10,000 2 15,000 25,000 3 20,000 45,000 4 25,000 70,000 The initial cost of Rs.50,000 will be recovered between years 3 and 4. Therefore, payback period will be 3 years plus a fraction of the 4th year. By the 3rd year, Rs.45,000 is recovered. The remaining Rs.5,000 would be recovered in the 4th year whose annual inflow in Rs.25,000. Therefore the fraction of the 4th Year needed to reach the cost would be : 5,000 25,000 = 1 5 Page 270 of 380 26/4/2008

1 Payback Period = 3 -------- Years 5 An alternative way of expressing the payback period is the payback period Reciprocal which is expressed as : 1 ----------------------- x 100 Payback period Higher the reciprocal more profitable will be the project. 3. Comparison between two projects. Project X ( 1,00,000 ) Project Y ( 1,00,000 ) Initial Cumulative Cumulative Investment Cash Inflows Cash Inflows Inflows Inflows Year I 20,000 20,000 25,000 25,000 Year II 20,000 40,000 25,000 50,000 Year III 30,000 70,000 50,000 1.00,000 Year IV 30,000 1,00,000 20,000 1.20,000 Year V 50,000 1,50,000 10,000 1,30,000 Analysis In this example, Project Y would be selected as its payback period of 3 years is shorter than the 4 years payback period of Project X. Page 271 of 380 26/4/2008

Accounting Rate of Return Method The accounting rate of return is also known as 'Return On Investment' or 'return on capital employed method' It employs the normal accounting technique to measure the increase in profit expected to result from an investment by expressing the net accounting profit arising from the investment as a percentage of the capital investment. In this method, most often the following formula is applied to arrive at the accounting rate of return. Average Annual Profit After Tax Accounting Rate of Return = ------------------------------------------ X 100 Average Investment Initial Investment + Salvage Value Average Investment = ---------------------------------------------- 2 Sometimes, initial investment is used in place of average investment. Of the various accounting rates of return on different alternative proposals, the one having highest rate of return is taken to be the best investment proposal. For example, in three alternative proposals A, B and C with expected accounting rates of return of 10%, 20% and 18% respectively, projects will be selected in order of B, C ad A. if he prevailing rates of interest is taken to be 15% p.a. then only proposals B and C will qualify for consideration and in that order. Merits It is easy to calculate because it makes use of readily available accounting information. It is not concerned with cash flows but rather based upon profits which are reported in annual accounts and sent to shareholders. Unlike payback period method, this method does take into consideration all the years involved in the life of a project. Where a number of capital investment proposals are being considered, a quick decision can be taken by use of ranking the investment proposals. If high profits are required, this is certainly a way of achieving them. Demerits It does not take into accounting time value of money. It fails to measure properly the rates of return on a project even if the cashflows are even over the project life. It uses the straight line method of depreciation. Once a change in method of depreciation takes place, the method will not be easy to use and will not work practically. Page 272 of 380 26/4/2008

This method fails to distinguish the size of investment required for individual projects. Competing investment proposals with the same accounting rate of return may require different amounts of investment. It is biased against short term projects in the same way that payback is biased against long term ones. Several concepts of investment are used for working out accounting rates of return. Thus there is no full agreement on the proper measure of the term investment. Thus different managers have different meanings when they refer to accounting rate of return. The accounting rates of return does not indicate whether an investment should be accepted or rejected, unless the rates of return is compared with the arbitrary management target. It measures the return in relation to the outlay and does not evaluate the absolute worth of the returns. Problems can arise in defining yearly profits, which will depend. to a certain extent, on the accounting policies adopted by the term which respect to such items as stock valuation, treatment of depreciation, research and development etc. Example The cost of the machine is Rs.80,000/-. It is expected to have a scrap value of Rs.10,000/- at the end of the Five year period. It is estimated to generate a profit over its life as follows : Year Net Profit ( Rs. ) 1 6,000 2 26,000 3 16,000 4 1,000 5 11,000 Total 60,000 Calculate the Accounting Rate of Return (ARR) Average Annual Profit ARR = ------------------------------- x 100 Average Investment 60,000 Average Annual Profit = ------------- = 12,000 5 Page 273 of 380 26/4/2008

( Original Investment + Salvage Value ) Average Investment = ---------------------------------------------------------- 2 ( 80,000 + 10,000 ) 90,000 = -------------------------- = ---------- = 45,000 2 2 12,000 ARR = ----------- x 100 = 26.76 % 45,000 Page 274 of 380 26/4/2008

Net Present Value Method The objective of the firm is to create wealth by using existing and future resources to produce goods and services. To create wealth, inflows must exceed the present value of all anticipated cash outflows. Net present value is obtained by discounting all cash outflows and inflows attributable to a capital investment project by a chosen percentage. The method discounts the net cash flows from the investment by the minimum required rate of return, and deducts the initial investment to give the yield from the funds invested. If yield is positive the project is acceptable. If it is negative the project in unable to pay for itself and is thus unacceptable. The exercise involved in calculating the present value is known as 'Discounting' and the factors by which we have multiplied the cash flows are known as the 'Discount Factors'. The discount factor is given by the following expression: Where 1 n (1 + r) r = Rate of interest per annum. n = number of years over which we are discounting. Merits It is based on the assumption that cash flows, and hence dividends, determine share holders wealth. Cash flows are subjective than profits. It considers the time value of money. It considers the total benefits arising out of proposals over its life-time. The future discount rate normally varies due to longer time span. This rate can be applied in calculating the NPV by altering the denominator. This method is particularly useful for the selection of mutually exclusive projects. ( In mutually exclusive projects acceptance of one project tantamount to rejection of the other project. ) This method of project selection is instrumental in achieving the financial objective. i.e., the maximization of the shareholders wealth. Demerits It is difficult to calculate as well as understand as compared to accounting rate of return method or payback period method. Calculation of the desired rates of return presents serious problems. Generally cost of capital is the basis of determining the desired rate. The calculation of cost of capital is itself complicated. Moreover, desired rates of return will vary from year to year. Page 275 of 380 26/4/2008

This method is an absolute measure. When two projects are being considered, this method will favour the project which has higher NPV. This method may not give satisfactory results when two projects having different effective lives are being compared. Normally, the project with shorter economic life is preferred, if other things are equal. This method emphasizes the comparison of net present value and disregards the initial investment involved. Thus, this method may not give dependable results. Example - 1 : A firm can invest Rs.10,000/- in a project with a life of Three years. The projected cash inflow are : Year-1 Rs. 4,000/-, Year-2 Rs.5,000/- and Year-3 Rs. 4,000/-. The cost of capital is 10 % p.a. Should the investment be made? Firstly the discount factors can be calculated based on Re.1 received in with `r' rate of interest in 3 years. 1 n (1 + r) 1 1 Year 1 = --------------------- = ---------- = 0.909 ( 1 + 10 / 100 ) ( 1.10 ) 1 1 Year 2 = --------------------- = ---------- = 0.826 ( 1 + 10 / 100 ) 2 ( 1.10 ) 2 1 1 Year 3 = --------------------- = ---------- = 0.751 ( 1 + 10 / 100 ) 3 ( 1.10 ) 3 Year Cash Flow ( Rs ) Discount Factor Present Value ( Rs ) 0 (10,000) 1.000 (10,000) 1 4,000 0.909 3,636 2 5,000 0.826 4.130 3 4,000 0.751 3,004 NPV = 770 Analysis : Since the net present value is positive, investment in the project can be made. Page 276 of 380 26/4/2008

Example - 2 : National Electronics Ltd., an electronic goods manufacturing company manufactures large range of electronic goods. It has under consideration two projects X and Y, each costing Rs.120 Lakhs. The projects are mutually exclusive and the company is considering the question of selecting one of the two projects. Cash flows have been worked out for both the projects and the details are given below. Project X has a life of 8 years and Project Y has a life of 6 years. Both will have zero salvage value at the end of their operational lives. The company is already making profits and its tax rate is 50%. The cost of Capital of the Company is 15%. At the end of Year Net Cash Inflow Present Value of Project X Project Y Rupee at 15 % 1 25 40 0.870 2 35 60 0.765 3 45 80 0.685 4 65 50 0.572 5 65 30 0.497 6 55 20 0.432 7 35-0.376 8 15-0.327 The company follows straight line method of depreciating assets. Advise the company regarding the selection of the project. Computation of Net Present Value of the Project X and Project Y. Page 277 of 380 26/4/2008

Project X End of year Cash flow Depreciation PBT Tax PAT Net C. F. = PAT + Depreciation Discount Factor P. V. 1 25 15 10 5 5 20 0.870 17.40 2 35 15 20 10 10 25 0.756 18.90 3 45 15 30 15 15 30 0.658 19.74 4 65 15 50 25 25 40 0.572 22.88 5 65 15 50 25 25 40 0.497 19.88 6 55 15 40 20 20 35 0.432 15.12 7 35 15 20 10 10 25 0.375 9.40 8 15 15 - - - 15 0.327 4.91 P V of cash inflows 128.23 Less: Initial Investment 120.00 Net Present Value 8.23 Project Y End of year Cash flow Depreciation PBT Tax PAT Net C. F. = PAT + Depreciation Discount Factor P. V. 1 40 20 20 10 10 30 0.870 26.10 2 60 20 40 20 20 40 0.756 30.24 3 80 20 60 30 30 50 0.658 32.90 4 50 20 30 15 15 35 0.572 20.02 5 30 20 10 5 5 25 0.497 12.43 6 20 20 - - - 20 0.432 8.64 P V of cash inflows 130.33 Less: Initial Investment 120.00 Net Present Value 10.33 Analysis: As Project Y has a higher net present value, it is advisable to take up Project Y. Page 278 of 380 26/4/2008

Internal Rate of Return Method Internal rate of return (IRR) is a percentage discount rate used in capital investment appraisals which brings the cost of a project and its future cash inflows into equality. It is the rate of return which equates the present value of anticipated net cash flows with the initial outlay. The IRR is also defined as the rate at which the net present value is zero. The rate for computing IRR depends as bank lending rate or opportunity cost of funds to invest which is often called as personal discounting rate or accounting rate. The test of profitability of a project is the relationship between the IRR (%) of the project and the minimum acceptable rate of return (%). The IRR can be stated in form of a ratio as shown below : Cash Inflows -------------------- = 1 Cash Outflows Or P. V. of Cash inflows P.V. of Cash Outflows = 0 The IRR is to be obtained by trial and error method to ascertain the discount rate at which the present values of total cash inflows will be equal to the present values of total cash out flows. If the cash inflow is not uniform, then IRR will have to be calculated by trial and error method. In order to have an approximate idea about such discounting rate, it will be better to find out the 'factor'. The factor reflects the same relationship of investment and cash inflows as in case of payback calculations. F = I/C F = Factor to be located. Where, I = Original Investment C = Average Cash Inflow per year. In appraising the investment proposals, IRR is compared with the desired rate of return or weighted average cost of capital, to ascertain whether the project can be accepted or not. IRR is also called as 'cut off rate' for accepting the investment proposals. The trial and error method for computing IRR when cash inflows are not even is summarized, step by step, as follows: 1. Compute NPV at the cost of capital, denoted here as r 1-2. See if NPV is positive or negative. 3. If NPV is positive, then pick another rate (r 2 ) much higher than r1. 4. Compute NPV using r 2. If NPV is positive then choose another value of r 2 which is higher than previous r 2. If NPV is negative at new r 2 then, the true IRR at which NPV = 0 must be somewhere between these two rates r 1 and r 2 (new). 5. Use interpolation for the exact rate. Page 279 of 380 26/4/2008

Interest Rate NPV ( X 1 ) r 1 NPV 1 > 0 ( Y 1 ) ( X 0 ) IRR NPV = 0 ( Y 0 ) ( X 2 ) r 2 NPV 2 < 0 ( Y 2 ) Merits : Then, ( X 2 X 1 ) ( Y 1 Y 0 ) X 0 = X 1 + --------------------------- ( Y 1 Y 2 ) It Considers the time value of Money It takes into account the total cash inflows and cash outflows. It is easy to understand for example if told that IRR of an investment is 20% against the desired rate of an investment is Rs.15,396/-. Demerits It does not use the concept of desired rate of return whereas it provides the rate of return which is indicative of the profitability of investment proposal. It involves tedious calculations, based on trial and error method. It produces multiple rates which can be confusing. Projects selected based on higher IRR may not be profitable. Unless the life of the project can be accurately estimated, assessment of cash flows cannot be correctly made. Single discount rate ignores the varying future interest rates. Page 280 of 380 26/4/2008

Example: A company has to select one of the following two projects : Particular Project A Project B Cost 11,000 10,000 Cash In Flows : Year 1 6,000 1,000 Year 2 2,000 1,000 Year 3 1,000 2,000 Year 4 5,000 10,000 Using the Internal rate of return method suggest which project is preferable. Factor in case of Project A = 11,000 / 3,500 = 3.14 Factor in case of Project B = 10,000 / 3,500 = 2.86 The factor thus calculated will be located in table given at the end of the handout on the line representing number of years corresponding to estimated useful life of the asset. This would give the expected rate of return to be applied for discounting the cash inflows the internal rate of return. In case of Project A, the rate comes to 10% while in case of Project B it comes to 15%. Project A Year Cash Inflows Discounting Present ( Rs. ) Factor at 10 % Value ( Rs. ) 1 6,000 0.909 5,454 2 2,000 0.826 1,652 3 1,000 0.751 751 4 5,000 0.683 3,415 Total Present Value 11,272 The present value at 10% comes to Rs.11,272. The initial investment is Rs.11,000. Internal rate of return may be taken approximately at 10%. Page 281 of 380 26/4/2008

In case more exactness is required another trial rate which is slightly higher than 10% (since at this rate the present value is more than initial investment ) may be taken. Taking a rate of 12%, the following results would emerge: Year Cash Inflows Discounting Present ( Rs. ) Factor at 12 % Value ( Rs. ) 1 6,000 0.893 5,358 2 2,000 0.797 1,594 3 1,000 0.712 712 4 5,000 0.636 3,180 Total Present Value 10,844 The internal rate of return is thus more than 10% but less than 12%. The exact rate may be calculated as follows : P. V. Required 11,000 P. V. at 10 % 11,272 (+) 272 P. V. at 12 % 10,844 (-) 156 272 Actual IRR = 10 + ---------------------- x 2 = 11.27% 272 (-156) Project B Year Cash Inflows Discounting Present ( Rs. ) Factor at 15 % Value ( Rs. ) 1 1,000 0.870 870 2 1,000 0.756 756 3 2,000 0.658 1,316 4 10,000 0.572 5,720 Total Present Value 8,662 Since present value at 15% comes only to Rs.8,662, a lower rate of discount should be taken. Taking a rate of 10% the following will be the result. Year Cash Inflows Discounting Present ( Rs. ) Factor at 10 % Value ( Rs. ) 1 1,000 0.909 909 2 1,000 0.826 826 3 2,000 0.751 1,502 4 15,000 0.683 6,830 Total Present Value 10,067 Page 282 of 380 26/4/2008

The present value at 10% comes to Rs.10,067 which is more or less equal to the initial investment Hence the internal rate of return may be taken as 10%. In order to have more exactness to Internal rate of return can be interpolated as done in case of Project 'A'. P. V. Required 10,000 P. V. at 10% 10,067 (+) 67 P. V. at 15% 8,662 (-) 1,338 67 Actual IRR = 10 + ---------------------- x 5 = 10.24% 67 (-1,338) Analysis : Thus, internal rate of return in case of Project A is higher as compared to Project B. Hence, Project A is preferable. Page 283 of 380 26/4/2008

Relative Ranking of Projects : IRR vs. NPV The relative ranking of projects, using the different DCF methods will be considered initially in simple accept / reject situations. This will be extended later to a detailed assessment of situations where a choice has to be made between two or more alternatives. In simple accept / reject situations a firm is able to implement all projects showing a return at or above the firms cost of capital. Both NPV and IRR would appear to be equally valid in the sense that they will both lead to accept or reject the same projects. Using NPV, all projects with a positive net present value, when discounted at the firm's cost of capital, will be accepted. Using IRR all projects which yield an internal rate of return in excess of the firms cost of capital will be chosen. Although IRR and NPV lead to the same conclusion regarding project acceptability, the ranking of a set of projects, obtained from IRR does not necessarily agree with that produced using NPV. Since, in the latter case, the ranking may vary according to particular discount rate used. The IRR measures only the quality of the investment while NPV takes into account both the quality and the scale. This is because the IRR provides a relative measure of value (% IRR) while the NPV provides an absolute measure (Rs. surplus). The IRR would rank, for example, a 100% return on an investment of Re.l.considerably higher than a 20% return on an investment of Rs.10 Lakhs, whereas the reverse would be true using NPV (as long as the cost of capital is below 20%). While one project may have a higher rate of profit per unit of capital invested than another, if it has fewer units of capital invested in it, it may make a smaller contribution to the wealth of the firm. Thus, if the objective is to maximize the firms wealth, then the ranking of project NPVs provides the correct measure. If the objective is to maximize, the rate of profitability per unit of capital invested, then RR would provide the correct ranking of projects, but this objective could be achieved by rejecting all but the most highly profitable projects. This is clearly unrealistic and therefore, one would conclude that NPV ranking is correct and!rr unsatisfactory as a measure of relative project value. When two investment proposals are mutually exclusive, both methods will give contradictory results. When two mutually exclusive projects are not expected to have the same life, NPV and IRR methods will give conflicting ranking. Page 284 of 380 26/4/2008

Example A company is considering which of the two mutually exclusive projects it should undertake. The Finance Director thinks that the project with the higher NPV should be chosen whereas the Managing Director thinks that the one with the higher irr should be undertaken especially as both projects have the same initial outlay and length of fife. The company anticipates a cost of capital of 10% and the net after tax cash flows of the projects are as follows : Year 0 1 2 3 4 5 Cash Flows Project X ( 200 ) 35 80 90 75 20 Project Y ( 200 ) 218 10 10 4 3 Required : (a) (b) (c) Calculate the NPV and IRR of each project. State, with reasons, which project you would recommend. Explain the inconsistency in the ranking of the two projects. The discount factors are as follows : Year 0 1 2 3 4 5 ( 10 % ) 1 0.91 0.83 0.75 0.68 0.62 Discount Factors ( 20 % ) 1 0.83 0.69 0.58 0.48 0.41 (a) Calculation of the NPV and!rr of each Project. NPV of Project X Year Cash Flow Discount Factors @ 10 % Discounted Values Discount Factors @ 20 % Discounted Values 0 (200) 1.00 (200) 1.00 (200) 1 35 0.91 31.85 0.83 29.05 2 80 0.83 66.40 0.69 55.20 3 90 0.75 67.50 0.58 52.20 4 75 0.68 51.00 0.48 36.00 5 20 0.62 12.40 0.41 8.20 NPV +29.15-19.35 Page 285 of 380 26/4/2008

IRR of Project X At 20% NPV is - 19.35 At 10% NPV is + 29.15 29.15 IRR = 10 + -------------------- x 10 = 16.01 % 29.15 + 19.35 NPV of Project Y Year Cash Flow Discount Factors @ 10 % Discounted Values Discount Factors @ 20 % Discounted Values 0 (200) 1.00 (200) 1.00 (200) 1 218 0.91 198.38 0.83 180.94 2 10 0.83 8.30 0.69 6.90 3 10 0.75 7.50 0.58 5.80 4 4 0.68 2.72 0.48 1.92 5 3 0.62 1.86 0.41 1.23 NPV +18.76-3.21 IRR of Project Y At 20% NPV is - 3.21 At 10% NPV is + 18.76 18.76 IRR = 10 + -------------------- x 10 = 18.54 % 18.76 + 3.21 (b) Both the projects are acceptable because they generate the positive NPV at the company's cost of capital at 10%. However, the company will have to select Project X because it has a higher NPV. If the company follows IRR method, then Project Y should be selected because of higher internal rate of return (IRR). But when NPV and IRR give contradictory results, a project with higher NPV is generally preferred because of higher return in absolute terms. Hence, Project X should be selected. (c) The inconsistency in the ranking of the projects arises because of the difference in the pattern of cash flows. Project X's major cash flows occur mainly in the middle three years, whereas Y generates the major cash flows in the first year itself. Page 286 of 380 26/4/2008

Profitability Index Method The profitability index (PI) is the present value of an anticipated future cash inflows divided by the initial outlay. The only difference between the net present value method and profitability index method is that when using the NPV technique the initial outlay is deducted from the present value of anticipated cash inflows, whereas with the profitability index approach the initial outlay is used as a divisor. in general terms, a project is acceptable if its profitability index value is greater than 1. Clearly, a project offering a profitability index greater than 1 must also offer a net present value which is positive. When more than one project proposals are evaluated, for selection of one among them, the project with higher profitability index will be selected. Mathematically. PI (profitability index) can be expressed as follows : - Present Value of Cash Inflows Profitability Index (PI) = ---------------------------------------- Present Value of Cash Outflow This method is also called 'cost benefit ratio' or 'desirability ratio' method. Example The following mutually exclusively projects can be considered : Particulars Project A Project B P V of Cash Inflows (i) 20,000 8,000 Initial Cash Outlay (ii) 15,000 5,000 Net Present Value 5,000 3,000 Profitability Index (i)/(ii) 33 1.60 Analysis : According to the NPV method, Project A would be preferred, whereas according to Profitability Index, Project B would be preferred. Although P1 method is based on NPV, it is a better evaluation technique than NPV in a situation of capital rationing. For example two projects may have the same NPV of Rs.10,000 but Project A requires initial outlay of Rs.1.00,000 whereas B only Rs.50,000. Then Project B would be preferred as per the yard stick of PI method. Limitations Profitability Index can not be used in capital rationing problems where projects are indivisible. Once a single large project with high NPV is selected, the possibility of accepting several small projects which together may have higher NPV than the single project, is excluded. Some times the project with lower profitability index may have to be selected if it generates cashflows in the earlier years, which can be used for setting up of another project to increase the overall NPV. Page 287 of 380 26/4/2008

Assumptions in Use of DCF Techniques In use of discounted cash flow techniques the following assumptions should be given consideration: The discount rate is constant over the life of the investment. All cash flows can be predicted with certainty so that a risk premium need not be added to the discount rate in order to compensate for risk. In project appraisal, managers work with uncertain future events and estimated cash flows expected to occur in future years. This involves a substantial amount of estimation which in practical terms, means that spurious accuracy is something which needs to be avoided The discount figures used can be calculated with great accuracy but when they are applied to these future estimated cash flows, the resultant calculation is only as accurate as the cash flows estimates. In many companies there is a tendency to produce discounted cash flow computation with several decimal places on each of the present values. This creates a totally fallacious appearance of accuracy in the evaluation process. To enable convenient calculation s to be performed, it is normal practice in capital project evaluations to assume that all cash flows take place at the end of the year. The initial cash outflows or investment in a project is assumed to take place now. The cash flows which go out now are taken to be at Year 0. The concept of Year C does not mean a year in general terms, but a point in time, i.e. today. Year 1 cash flows are assumed to take place at the end of the first year. The second year cash flows occurring at the end of the Year 2 and similarly for subsequent years. In appraising long projects. it is normal to use an arbitrary horizon period f 10 to 15 years. Firms do not consider cash flows beyond the horizon even if they expect the project to last longer. Page 288 of 380 26/4/2008

Capital Rationing Capital rationing is a situation where a constraint of budget ceiling is placed on the total size of capital expenditures during a particular period. Often firms draw up their capital budget under the assumption that the availability of financial resources is limited. Capital rationing refers to the selection of the investment proposals in a situation of constraint on availability of capital funds, to maximize the wealth of the company by selecting those projects which will maximize overall NPV of the concern. In capital rationing situation a company may have to forego some of the projects whose IRR is above the overall cost of the firm due to ceiling on budget allocation for the projects which are eligible for capital investment. Capital rationing refers to a situation where a company cannot undertake all positive NPV projects it has identified because of shortage of capital. Under this situation a decision maker is compelled to reject some of the viable projects having positive net present value because of shortage of funds. It is known as a situation involving capital rationing. In terms of financing investment projects, the following important questions are to be answered. What would be the requirement of funds for capital investment decisions in the forthcoming planning period? How much quantum of funds available for capital investment? How to assign the available funds to the acceptable proposals which require more funds than are available? The answer to the first and second questions are given with reference to the capital investment appraisal decisions made by the top management. The third question is answered with specific reference to the appraisal of investment decisions from the angle of capital rationing. Situations of Capital Rationing Capital rationing decisions can be studied under the following situations : Situation I Projects are Divisible and Constraint is a Single Period One The following are the steps to be adopted for solving the problem under this situation: (a) (b) (c) Calculate the profitability index of each project. Rank the projects on the basis of the profitability index calculated in (a) above. Choose the optimal combination of the projects. Page 289 of 380 26/4/2008

Example Project Required Initial Capital NPV The appropriate Cost of Capital A 1,00,000 20,000 B 3,00,000 35,000 C 50,000 16,000 D 2,00,000 25,000 E 1,00,000 30,000 Total fund available is Rs.3,00,000. Determine the optimal combination of projects assuming that the projects are divisible Project Required initial outlay (Rs.) NPV at the appropriate cost of capital (Rs.) Profitability index [ (3) / (2) ] Rank (1) (2) (3) (4) (5) A 1,00.000 20,000 0.2 3 B 3,00,000 35,000 0.117 5 C 50.000 16,000 0.32 1 D 2,00,000 25,000 0.125 4 E 1,00,000 30,000 0.3 2 Rank of Investment Project Required Initial (Rs.) 1 C 50,000 2 E 1,00,000 3 A 1,00,000 4 1 / 4 th of D 50,000 * Total 3,00,000 * (Rs.2,00,000 x 1 / 4) Therefore, the optimal combination of projects is C, E, A and 114t11 portion of D. Page 290 of 380 26/4/2008

Situation II Projects are Indivisible and Constraint is a Single Period one The following steps to be followed for solving the problem under this situation. (a) (b) Construct a table showing the feasible combinations of the project (whose aggregate of initial outlay does not exceed the fund available for investment). Choose the combination whose aggregate NPV is maximum and consider it as the optimal project mix. Example Using the same data as used in the previous illustration, determine the optimal project mix on the basis of the assumption that the projects are indivisible. Feasible Combinations Aggregate of NPVs A, C 36,000 A, D 45,000 A, E 50,000 C, D 41,000 C, E 46,000 D, E 55,000 A, C, E 66,000 By a careful inspection of the feasible combinations constructed in the above table, we can conclude that the optimal project mix is A, C and E because the aggregate of their NPVs is maximum. Page 291 of 380 26/4/2008

Problem 1. The initial investment for a project is Rs.15,000/- The annual cash flow due to the project are as given below : Year 1 2 3 4 5 6 7 Cash flow (in Rs.) 4000 2000 4000 3000 2000 1000 2000 If the cost of capital is 10% p.a., what is the net present value of the project? Solution : Year Cash Flow PV Factor PV of Cash ( Rs ) @ 10 % Flow 0 (15,000) 1.000 (15,000) 1 4,000 0.909 3,636 2 2,000 0.826 1,652 3 4,000 0.751 3,004 4 3,000 0.683 2,049 5 2,000 0.621 1,242 6 1,000 0.564 564 7 2,000 0.513 1,026 NPV = -1,827 Page 292 of 380 26/4/2008

Problem 2. The management of the Play Toy Factory, a manufacturer of toys in New Delhi, is considering the introduction of a new type of a toy-remote control motorbike. In the past, the management has been quite conservative in making investments in new product and consider this project quite a risky one. The management feels that the normally used required rate of return of 10% is not proper in this case and instead, a return of 16% is expected on this project. The project, requiring an outlay of Rs.1,50,000 has the following expected returns over its estimated life of 6 years. Year 1 2 3 4 5 6 Cash Flow ( 000 Rs.) 30 30 50 60 40 25 Should the project be undertaken? Solution : Year Cash Flow PV Factor ( Rs ) @ 16 % PV of Cash Flow 0 (1,50,000) 1.000 (1,50,000) 1 30,000 0.862 25,860 2 30,000 0.743 22,290 3 50,000 0.641 32,050 4 60,000 0.552 33,120 5 40,000 0.476 19,040 6 25,000 0.410 10,250 NPV = -7,390 Since NPV is Negative Do Not take this Project. Page 293 of 380 26/4/2008

Problem 3. A management wants to judge whether a project X is worth taking up or not The data with regarding the project (having life of 8 years' ) is given below : Year Net Cash Net Cash Year Flow Flow 1 6,000 5 9,700 2 8,500 6 7,600 3 10,000 7 5,700 4 11,800 8 4,000 The initial outlay of the project is Rs.45,000/- with a salvage value of Rs.10,000/-. If the cost of capital is 10%, find the Net Present Value (NPV) of the project and hence determine whether it is worth taking up the project or not. Solution : Salvage value means re-sale value Now since in the 8 th year i.e. last year of the project, there will be sale of capital asset which will generate Rs. 10,000/- at extra flow of cash. So net cash flow for 8 th year will be 4,000 + 10,000/- = 14,000/- Year Cash Flow PV Factor PV of Cash ( Rs ) @ 10 % Flow 0 (45,000) 1.000 (45,000) 1 6,000 0.909 5,454 2 8,500 0.826 7,021 3 10,000 0.751 7,510 4 11,800 0.683 8,059 5 9,700 0.621 6,024 6 7,600 0.564 4,286 7 5,700 0.513 2,924 8 14,000 0.467 6,538 NPV = 2816.60 Page 294 of 380 26/4/2008

Problem 4. Consider two projects A and B with the following cash flow streams. Year 0 1 2 3 4 Project A 1,00,000 28,000 35,000 41,000 45,000 Project B 1,80,000 65,000 65,000 65,000 65,000 Assuming discount rate as 10% per annum, and both projects are equally risky, which projects would you accept if you have to choose only one project between the two? Justify your answer. Solution : Year PV Factor @ 10 % Cash Flow ( Rs ) Project - A PV of Cash Flow Cash Flow ( Rs ) Project - B 0 1.000-100,000-100,000-180,000-180,000 1 0.909 28,000 25,452 65,000 59,085 2 0.826 35,000 28,910 65,000 53,690 3 0.751 41,000 30,791 65,000 48,815 4 0.683 45,000 30,735 65,000 44,395 NPV 15,888 NPV 25,985 Since NPV of Project B is > Project A, Project B is Better. PV of Cash Flow Justification Project A Project B PV 1,15,888 2,06,050 Investment 1,00,000 1,80,000 PV 1,15,888 2,06,050 Profitability = ---- = ------------ ------------ Index ( PI ) I 1,00,000 1,80,000 PI = 1.16 1.14 Now PI of Project A is > than Project B, Hence Project A is Better. Page 295 of 380 26/4/2008

Problem 5. A company is deciding to choose between two mutually exclusive projects A and B. Project A requires an initial investment of Rs.3,00,000 and is expected to generate cash flow of Rs.1,50,000 per annum for the 3 years of its life. Project B, on the other hand, requires Rs.3,40,000, and has a life of 6 years and would generate Rs.1,00,000 every year. Which proposal should be accepted? Assume a 10% rate of interest. Solution : Year PV Factor @ 10 % Cash Flow ( Rs ) Project - A PV of Cash Flow Cash Flow ( Rs ) Project - B PV of Cash Flow 0 1.000-300,000-300,000-340,000-340,000 1 0.909 150,000 136,350 100,000 90,900 2 0.826 150,000 123,900 100,000 82,600 3 0.751 150,000 112,650 100,000 75,100 4 0.683 100,000 68,300 5 0.621 100,000 62,100 6 0.564 100,000 56,400 NPV 72,900 NPV 95,400 Since NPV of Project B is > Project A, Project B is Better. Page 296 of 380 26/4/2008

Problem 6. A company has to decide which of the four projects W, X, Y, Z is a better investment. The initial investments along with each year's cash flows of these projects are given below : Project Life (in Years) Initial Investment (In Rs.) Annual Cash flow (in Rs.) W 8 16,000 4,000 X 3 6,000 3,500 Y 5 12,000 4,000 Z 6 8,000 3,000 Rank these projects on the basis of NPV, if the cost of capital s 15% per annum. Solution : Projects Life in Years Initial Investme nt ( I ) Annual Cash Flow PV Factor 15 % ( Cumulative ) PV of Cash Flow ( PV ) NPV = ( PV - I ) RANK PI = ( PV / I ) W 8 16,000 4,000 4.487 17,948 1,948 3 1.12 3 X 3 6,000 3,500 2.283 7,991 1,991 2 1.33 2 Y 5 12,000 4,000 3.352 13,408 1,408 4 1.12 4 Z 6 8,000 3,000 3.784 11,352 3,352 1 1.42 1 RANK Page 297 of 380 26/4/2008

Problem 7. One of the two machines A and B is to be purchase. From the following information, find out which of the two will be more profitable? Machine A ( Rs. ) Machine B ( Rs. ) Cost of Machine Life Earning before Tax Rs. 50,000 4 Years Rs. 80,000 6 Years 1 10,000 8,000 2 15,000 14,000 3 20,000 25,000 4 15,000 30,000 5-18,000 6-13,000 Answer ARR A 1.875 % B 2.33 % ( Machine B is Profitable ) Payback A 3 years and 9 months B 5 years and 2 months ( Machine A is Profitable ) Solution : Since PV factor is not given take it as 8%. Year PV Factor @ 8 % Project - A Cash Flow ( Rs ) PV of Cash Cash Flow ( Rs ) Project - B PV of Cash Flow Flow 0 1.000-50,000-50,000-80,000-80,000 1 0.926 10,000 9,260 8,000 7,408 2 0.857 15,000 12,855 14,000 11,998 3 0.794 20,000 15,880 25,000 19,850 4 0.735 15,000 11,025 30,000 22,050 5 0.681 18,000 12,258 6 0.630 13,000 8,190 NPV -980 NPV 1,754 Page 298 of 380 26/4/2008

Since NPV of Project B > NPV of Project A, Project B is profitable. Project A Project B PV 49,020 81,754 Investment 50,000 80,000 PV 49,020 81,754 Profitability = ---- = ------------ ------------ Index ( PI ) I 50,000 80,000 By PI index method Project B is better. PI = 0.98 1.02 Page 299 of 380 26/4/2008

Problem 8. A company whose cost of capital is 12% is considering two projects A and B. The following data are available. Project A Rs. Project B Rs. Investments 1,40,000 1,40,000 Cash Flows : Year 1 20,000 1,00,000 2 40,000 80,000 3 60,000 40,000 4 1,00,000 20,000 5 1,10,000 20,000 Select the most profitable project by using the following methods: (a) (b) (c) Payback period NPV profitability Index P. V. or Re.1 at 12% are : Year 1 2 3 4 5 P. V..9.8.7.6.55 Answer a) Payback : A B 1 3 ------ years 5 1 1 ------ years 2 Project B is Profitable. Page 300 of 380 26/4/2008

Solution : a) By Payback Period Method Project A ( 1,40,000 ) Project B ( 1,40,000 ) Initial Cumulative Cumulative Investment Cash Inflows Cash Inflows Inflows Inflows Year 1 20,000 20,000 1,00,000 1,00,000 Year 2 40,000 60,000 80,000 1,80,000 Year 3 60,000 1,20,000 40,000 2.20,000 Year 4 1,00,000 2,20,000 20,000 2.40,000 Year 5 1,10,000 3,30,000 20,000 2,60,000 For Project :. 20,000 1 1 A ------------- = ------ that is 3 ------ 1,00,000 5 5 40,000 1 1 B ------------- = ------ that is 1 ------ 80,000 2 2 Hence by Payback Period Method Project B is Profitable. b) By NPV Method Year PV Factor @ 12 % Project - A Cash Flow ( Rs ) PV of Cash Cash Flow ( Rs ) Project - B PV of Cash Flow Flow 0 1.000-140,000-140,000-140,000-140,000 1 0.900 20,000 18,000 100,000 90,000 2 0.800 40,000 32,000 80,000 64,000 3 0.700 60,000 42,000 40,000 28,000 4 0.600 100,000 60,000 20,000 12,000 5 0.550 110,000 60,500 20,000 11,000 NPV 72,500 NPV 65,000 Since NPV of Project A > NPV of Project B, Project A is profitable. Page 301 of 380 26/4/2008

c) By PI Method Project A Project B PV 2,12,500 2,05,000 Investment 1,40,000 1,40,000 PV 2,12,500 2,05,000 Profitability = ---- = --------------- --------------- Index ( PI ) I 1,40,000 1,40,000 By PI index method Project A is better. PI = 1.517 1.464 Page 302 of 380 26/4/2008

Problem 9. A Company has an investment opportunity costing Rs. 40,000 with the following expected net cash flow after tax but before depreciation. Year Net Cash Flow ( Rs ) 1 7,000 2 7,000 3 7,000 4 7,000 5 7,000 6 8,000 7 10,000 8 15,000 9 10,000 10 4,000 Using 10% as the cost of capital, determine the following : (a) (b) (c) (d) Payback period NPV at 10% discount factor Profitability Index at 10% discount factor IRR within 10% and 15% discount factor Answers : a) 5 years and 7.5 months b) Rs.8961 c) At 10% = 1.224 At 15% =.9855 d) 14,696% Page 303 of 380 26/4/2008

Solution : (a) Payback Method. Investment ( Rs. 40,000 ) Year Cash Inflows Cumulative Inflows 1 7,000 7,000 2 7,000 14,000 3 7,000 21,000 4 7,000 28,000 5 7,000 35,000 6 8,000 43,000 7 10,000 53,000 8 15,000 68,000 9 10,000 78,000 10 4,000 82,000 b) NPV Method. Cash Flow 5,000 ------------- = 0.625 8,000 So multiply this by 12, 0.625 x 12 = 7.5. So 5 years and 7.5 months. PV @ 10 % of Cash Flow PV @ 15 % of Cash Flow Year ( Rs ) PV Factor @ 10 % PV Factor @ 15 % 0-40,000 1.000-40,000 1.000-40,000 1 7,000 0.909 6,363 0.870 6,090 2 7,000 0.826 5,782 0.756 5,292 3 7,000 0.751 5,257 0.658 4,606 4 7,000 0.683 4,781 0.572 4,004 5 7,000 0.621 4,347 0.497 3,479 6 8,000 0.564 4,512 0.432 3,456 7 10,000 0.513 5,130 0.376 3,760 8 15,000 0.467 7,005 0.324 4,860 9 10,000 0.424 4,240 0.284 2,840 10 4,000 0.386 1,544 0.247 988 NPV 8,961 NPV -625 Page 304 of 380 26/4/2008

c) By PI Method @ 10 % @ 15 % PV 48,961 39,375 Investment 40,000 40,000 PV 48,961 39,375 Profitability = ---- = --------------- --------- Index ( PI ) I 40,000 40,000 d) IRR Method At 15% NPV is - 625 At 10% NPV is + 8961 PI = 1.224 0.984 8961 IRR = 10 + ------------------ x 5 = 14.67 % 8961 + 625 Page 305 of 380 26/4/2008

Problem 10. Consider an investment which has the following cash flows : Year 0 1 2 3 4 5 Cash Flows $ ( 31,000 ) 10,000 20,000 10,000 10,000 5,000 Compute the following : (a) (b) (c) Payback period Net present value (NPV) at 14 percent cost of capital Internal rate of return (IRR) Problem 11. A company is considering two mutually exclusive projects. Both require an initial cash out-lay of Rs.10,000 each, and have a life of five years. The company's required rate of return is 10 percent and pays tax at a 50 percent rate. The projects will be depreciated on a straight line basis. The before taxes cash flows expected to be generated by the projects are as follows : Year Before tax Cash Flow (Rs.) 1 2 3 4 5 Project A 4,000 4,000 4,000 4,000 4,000 Project B 6,000 3,000 2,000 5,000 5,000 Calculate for each project : (1) the payback. (2) the average rate of return, (3) the net present value and profitability index, and (4) the internal rate of return. Which project should be accepted and why? Problem 12. A company with a 10% cost of funds & limited investment of Rs.160 lakhs is evaluating the desirability of several investment proposals. Project Initial Investment Annual Life ( Rs in Lakhs ) Cash Flow A 120 5 30 B 80 3 32 C 80 4 25 D 40 7 8 E 120 9 15 Page 306 of 380 26/4/2008

(i) (ii) Rank the projects according to PI & NPV Determine optimal investment package. Page 307 of 380 26/4/2008

Problem 13. A company with 12% cost of funds and limited investment fund of Rs.4,00,000 is evaluating the desirability of several investment proposals. ( Cash flow in Rs. / Project life in years ) Total Project Life Project Investment Years Annual Cash Inflow A 3,00,000 2 1,87,600 B 2,00,000 5 66,000 C 2,00,000 3 1,00,000 D 1,00,000 9 20,000 E 3,00,000 10 66,000 (i) (ii) (iii) Rank the projects according to the profitability index and NPV method. Determine optimal investment package. Which projects should be chosen if investment limit is raised to Rs.5,00,000? Note : - The following is an extract from table for the 'Present Value Factor' for an annuity. Years 1 2 3 4 5 6 7 8 9 10 12% 0.893 1.690 2.402 3.037 3.605 4.111 4.564 4.968 5.328 5.650 Solution : Projects Life in Years Initial Investment ( I ) Annual Cash Flow PV Factor 12 % ( Cumulative ) PV of Cash Flow ( PV ) NPV = ( PV - I ) RANK PI = ( PV / I ) RANK A 2 300,000 187,600 1.690 317,044 17,044 4 1.06 5 B 5 200,000 66,000 3.605 237,930 37,930 3 1.19 3 C 3 200,000 100,000 2.402 240,200 40,200 2 1.20 2 D 9 100,000 20,000 5.328 106,560 6,560 5 1.07 4 E 10 300,000 66,000 5.650 372,900 72,900 1 1.24 1 Note : Always take PI Method for Ranking. Take project C & E, i.e. 2,00,000 + 3,00,000 = 5,00,000. If 4,00,000 would have been available then, we would have taken E & D projects Page 308 of 380 26/4/2008

16 th April, 2008. Lecture XIV. Cash Flow Evaluation In this the cash flow is not given hence cash flow needs to be generated. Before you start :- 1. Use Pencil for this exercise 2. Workout depreciation Schedule 3. Start Writing values Remember : 1. You must Remember the Rows & columns of the table. 2. Mentioned below are to be written with Negative Signs. a. Fixed Assets. b. Net working Capital c. Costs d. Depreciation e. Tax g. Initial Outlay h. Net Cash flow ( For First Year ) Page 309 of 380 26/4/2008

Problem 14. Futura Limited is considering a capital project about which the following information is available, (i) (ii) (iii) (iv) The investment outlay on the project will be Rs.200 million. This consists of Rs.150 million on the plant and machinery and Rs.50 million on net working capital. The entire outlay will be incurred in the beginning. The life of the project is expected to be 5 years. At the end of 5 years, fixed assets will fetch a net salvage of Rs.48 million whereas the net working capital will be liquidated at its book value. The project is expected to generate the revenue of the firm by Rs.250 million per year. The increase in costs on account of the project is expected to be Rs.100 million per year. (This includes all items on cost other than depreciation, interest and tax). The tax rate is 30%. Plant and machinery will be depreciated at the rate of 25% per year as per the written down method. Estimate the post-tax cash flows of the project, assuming cost of capital 12%. Solution : Working of Depreciation ( WDV Method ) Figures in millions Year Depreciation Book Value 1 25 % of 150 = 37.5 150 37.5 = 112.50 2 25 % of 112.5 = 28.12 112.5 28.12 = 84.37 3 25 % of 84.37 = 21.90 84.37 21.09 = 63.28 4 25 % of 63.28 = 15.82 63.28 15.82 = 47.46 5 25 % of 47.46 = 11.86 All computation with 2 decimal round-off. Page 310 of 380 26/4/2008

All Figure in Millions Sr No Particulars 0 1 2 3 4 5 1 Fixed Assets -150.00 2 Net working Captial -50.00 3 Revenues 250.00 250.00 250.00 250.00 250.00 4 Cost ( Other than Deprication & Tax ) -100.00-100.00-100.00-100.00-100.00 5 Depriciation -37.50-28.12-21.09-15.82-11.86 6 PBT (3) - (4) - (5) 112.50 121.88 128.91 134.18 138.14 7 Tax -33.75-36.56-38.67-40.25-41.44 8 PAT (6) - (7) 78.75 85.32 90.24 93.93 96.70 9 Net Salvage Value of FA 48.00 10 Recovery of Net working Capital 50.00 11 Initial Outlay -200.00 12 Operating (8) + (5) Cash Flows 116.25 113.44 111.33 109.75 108.56 13 Terminal Cash Flows (9) + (10) 98.00 14 Net Cash Flows (11) + (12) + (13) -200.00 116.25 113.44 111.33 109.75 206.56 15 PV Factor 1.000 0.893 0.797 0.712 0.636 0.567 PV of Cash Flow 16 (15) * (16) -200.00 103.81 90.41 79.26 69.80 117.12 17 NPV (14) - (17) 12.44 23.03 32.06 39.95 89.44 196.92 Page 311 of 380 26/4/2008

Problem 15. A company is considering a capital project about which the following information is available : (a) The investment outlay on the project will be Rs.400 lakhs. This consists of Rs.300 lakhs on the plant and machinery and Rs.100 lakhs on net working capital. The entire outlay will be incurred at the beginning of the project. (b) The life of the project is expected to be 5 years, fixed assets will fetch a net salvage value of Rs.96 lakhs where as net working capital will be liquidated at its book value. (c) The project is expected to increase the revenue of the firm by Rs.440 lakhs per year. The increase in costs on account of the project is expected to be Rs.250 lakhs per year (This includes all items of cost other than depreciation, interest and tax). The tax rate is 30%. (d) Plant and machinery will be depreciated at the rate of 20% per year as per the written down value method (e) Cost of capital 10% Using Net Present Value (NPV) method, determine whether the company should undertake the above proposal or not: Year 1 2 3 4 5 6 7 Present value at 10% 0.909 0.826 0.751 0.683 0.621 0.564 0.513 Page 312 of 380 26/4/2008

Solution : Working of Depreciation ( WDV Method ) Figures in Lakhs Year Depreciation Book Value 1 20 % of 300 = 60 300 60 = 240 2 20 % of 240 = 48 240 48 = 192 3 20 % of 192 = 38.4 192 38.4 = 153.6 4 20 % of 153.6 = 30.72 153.6 30.72 = 122.88 5 20 % of 122.88 = 24.58 All computation with 2 decimal round-off. All Figure in Lakhs Sr No Particulars 0 1 2 3 4 5 1 Fixed Assets -300.00 2 Net working Captial -100.00 3 Revenues 440.00 440.00 440.00 440.00 440.00 4 Cost ( Other than Deprication & Tax ) -250.00-250.00-250.00-250.00-250.00 5 Depriciation -60.00-48.00-38.40-30.72-24.58 6 PBT (3) - (4) - (5) 130.00 142.00 151.60 159.28 165.42 7 Tax -39.00-42.60-45.48-47.78-49.63 8 PAT (6) - (7) 91.00 99.40 106.12 111.50 115.79 9 Net Salvage Value of FA 96.00 10 Recovery of Net working Capital 100.00 11 Initial Outlay -400.00 12 Operating (8) + (5) Cash Flows 151.00 147.40 144.52 142.22 140.37 13 Terminal Cash Flows (9) + (10) 196.00 14 Net Cash Flows (11) + (12) + (13) -400.00 151.00 147.40 144.52 142.22 336.37 15 PV Factor 1.000 0.909 0.826 0.751 0.683 0.621 PV of Cash Flow 16 (15) * (16) -400.00 137.26 121.75 108.53 97.13 208.89 17 NPV (14) - (17) 13.74 25.65 35.99 45.08 127.49 247.94 Page 313 of 380 26/4/2008

Problem 16. A company is considering a capital project about which the following information is available. (i) The initial outlay of the project would be Rs.50 Lakhs with a salvage value of Rs.5 Lakhs. (ii) The cost of capital is 12%. (iii) The working capital required would be Rs.4 Lakhs which will be liquidated at the book value when the project is terminated. (iv) The life of the project is 6 years (v) The yearly cost is Rs.12 Lakhs which exclude depreciation and tax. (vi) The revenue generated in the first year is Rs.24 Lakhs which will increase by Rs.4 Lakhs every year. (vii) The depreciation will be charged as per the written down value method and the rate is 25%. (viii) The income tax rate is 40%. Solution : Working of Depreciation ( WDV Method ) Figures in Lakhs Year Depreciation Book Value 1 25 % of 50 = 12.50 50 12.50 = 37.50 2 25 % of 37.5 = 9.37 37.5 9.37 = 28.12 3 25 % of 28.12 = 7.03 28.12 7.03 = 21.09 4 25 % of 21.09 = 5.27 21.09 5.27 = 15.82 5 25 % of 15.82 = 3.95 15.82 3.95 = 11.86 6 25 % of 11.86 = 2.97 All computation with 2 decimal round-off. Page 314 of 380 26/4/2008

All Figure in Lakhs Sr No Particulars 0 1 2 3 4 5 6 1 Fixed Assets -50.00 2 Net working Captial -4.00 3 Revenues 24.00 28.00 32.00 36.00 40.00 44.00 Cost 4 ( Other than Deprication & Tax ) -12.00-12.00-12.00-12.00-12.00-12.00 5 Depriciation -12.50-9.38-7.03-5.27-3.96-2.97 6 PBT (3) - (4) - (5) -0.50 6.62 12.97 18.73 24.04 29.03 7 Tax -2.45-5.19-7.49-9.62-11.61 8 PAT (6) - (7) -0.50 4.17 7.78 11.24 14.42 17.42 9 Net Salvage Value of FA 5.00 10 Recovery of Net working Capital 4.00 11 Initial Outlay -54.00 Operating (8) + (5) 12 Cash Flows 12.00 13.55 14.81 16.51 18.38 20.39 Terminal 13 Cash Flows (9) + (10) 9.00 Net Cash Flows 14 (11) + (12) + (13) -54.00 12.00 13.55 14.81 16.51 18.38 29.39 15 PV Factor 1.000 0.893 0.797 0.712 0.636 0.567 0.507 PV of Cash Flow 16 (15) * (16) -54.00 10.72 10.80 10.55 10.50 10.42 14.90 NPV 17 (14) - (17) 1.28 2.75 4.27 6.01 7.96 14.49 35.47 Note : 1. Since there is a Loss in the First Year No Tax will be Paid. 2. In the Next year such loss of the last year will be adjusted before paying tax, So in the 2 nd year tax will be ( 6.62 0.5 ) x 40 % = 2.45 Page 315 of 380 26/4/2008

Problem 17. XYZ ENTER PRISE is contemplating a new investment project for which they are considering the following information. (a) (b) (c) Total cash outflow of the project will be Rs.10 crores, which consists of Rs.6 crores on plant and machinery and Rs.4 crores on gross working capital. The entire outflow will be incurred at the beginning of the project. Project has a life of 5 years at the end of 5 years, plant and equipment would fetch a salvage value of Rs.2 crores. Working capital will be liquidated at end of 5 years which will be equal to its book value (Rs.4 crores) The project will entail incremental revenues for the firm to the tune of Rs.8 crores per annum, the incremental expenses on account of the project will be Rs.4 crores per annum, which includes all items of expenses excluding depreciation and taxes. (d) The effective tax rate is 50%. (e) Cost of capital 14%. (f) (g) Depreciation is charged at 33.33% on the basis of Written Down Value Method. Help the enterprise, whether it should undertake the project or not on the basis of NPV criterion. Solution : Working of Depreciation ( WDV Method ) Figures in Lakhs Year Depreciation Book Value 1 33.33 % of 600 = 199.98 600 199.98 = 400.02 2 33.33 % of 400.02 = 133.32 400.02 133.32 = 266.66 3 33.33 % of 266.69 = 88.88 266.69 88.88 = 177.80 4 33.33 % of 177.80 = 59.26 177.80 59.26 = 118.54 5 33.33 % of 118.54 = 39.51 Page 316 of 380 26/4/2008

Solution : All Figure in Lakhs Sr No Particulars 0 1 2 3 4 5 1 Fixed Assets -600.00 2 Net working Captial -400.00 3 Revenues 800.00 800.00 800.00 800.00 800.00 Cost 4 ( Other than Deprication & Tax ) -400.00-400.00-400.00-400.00-400.00 5 Depriciation -199.98-133.32-88.88-59.26-39.51 6 PBT (3) - (4) - (5) 200.02 266.68 311.12 340.74 360.49 7 Tax -100.01-133.34-155.56-170.37-180.25 8 PAT (6) - (7) 100.01 133.34 155.56 170.37 180.25 9 Net Salvage Value of FA 200.00 10 Recovery of Net working Capital 400.00 11 Initial Outlay -1000.00 Operating (8) + (5) 12 Cash Flows 299.99 266.66 244.44 229.63 219.76 Terminal 13 Cash Flows (9) + (10) 600.00 Net Cash Flows 14 (11) + (12) + (13) -1000.00 299.99 266.66 244.44 229.63 819.76 15 PV Factor 1.000 0.877 0.769 0.675 0.592 0.519 PV of Cash Flow 16 (15) * (16) -1000.00 263.09 205.06 165.00 135.94 425.45 NPV 17 (14) - (17) 36.90 61.60 79.44 93.69 394.30 665.93 Page 317 of 380 26/4/2008

Problem 18. Naveen Enterprises is considering a capital project about which he following information is available: (i) (ii) (iii) (iv) (v) The investment outlay of the project will be Rs.100 million. This consists of Rs.80 million on plant and machinery and Rs.20 million on net working capital. The entire outlay will be incurred at the beginning of the project The project will be financed with Rs.45 million of equity capital, Rs.5 million of preference capital and Rs.50 million of debt capital. Preference capital will carry a dividend rate of 15%, debt capital will carry an interest 15%. The life of the project is expected to be 5 years. At the end of 5 years, fixed assets will fetch a net salvage value of Rs.30 million whereas net working capital will be liquidated at its book value. The project is expected to increase the revenue of the firm by Rs.120 million per year. The increase in cost on account of the project is expected to be Rs.80 million per year (This includes all items of cost other than depreciation, interest and tax). The effective tax rate will be 30%. Plant and machinery will be depreciated at the rate of 25% per year as per the written down value method. Hence the depreciation charges will be: First Year : Rs.20 million Second Year : Rs.15 million Third Year : Rs.11.25 million Fourth Year : Rs.8.44 million Fifth Year : Rs.6.33 million Given the above details develop the project cash flows. Page 318 of 380 26/4/2008

Problem 19. The following is available for a project. (i) (ii) (iii) (iv) (v) (vi) Initial project outlay is Rs.200 lakhs. The finance structure is Working Capital Loan Rs.20 lakhs, Term Loan Rs.100 lakhs and equity Rs.8 lakhs. The assets structure is land Rs.5 lakhs, building Rs.55 lakhs and machinery Rs.120 lakhs and current liabilities Rs.20 lakhs. Term loan is repaid in 20 equal quarterly instalments as principal repayment. Interest will be 20% p.a. The first instalment is due after 2 years. The three products P,S and K manufactured have a capacity of 30,000 units each per year. The selling prices are Rs.250, Rs.190 and Rs.130 respectively. Working Capital remains unchanged for all 3 years and carries rate of interest of 12% p.a. Operating cost is 55% of sales revenue. The capacity utilization is 80% first year, 90% second year and 100% third year. Depreciation is charged on fixed assets 20% p.a. on Written Down Value method. Evaluate project cash flows for first three years. Solution : Notes : 1. All fixed assets to be added, that is 5 + 55 +120 = 180 2. Evaluation is to be done for only 3 years as asked in the problem 3. Since Tax in not given 30 % is assumed 4. Since Salvage value is not given it is assumed to be Zeros 5. Recover of working capital is Zero 6. Interest to be ignored. Page 319 of 380 26/4/2008

Working of Depreciation ( WDV Method ) Figures in Lakhs Year Depreciation Book Value 1 20 % of 180 = 36 180 36 = 144.00 2 20 % of 144 = 28.80 144 28.80 = 115.20 3 20 % of 115.20 = 23.04 Calculation of Revenues Year 1 Product Max Qty Produced Net Qty Sale Rate Revenue P 30000 80% 24000 250 6,000,000 S 30000 80% 24000 190 4,560,000 K 30000 80% 24000 130 3,120,000 13,680,000 Year 2 Product Max Qty Produced Net Qty Sale Rate Revenue P 30000 90% 27000 250 6,750,000 S 30000 90% 27000 190 5,130,000 K 30000 90% 27000 130 3,510,000 15,390,000 Year 3 Product Max Qty Produced Net Qty Sale Rate Revenue P 30000 100% 30000 250 7,500,000 S 30000 100% 30000 190 5,700,000 K 30000 100% 30000 130 3,900,000 17,100,000 Page 320 of 380 26/4/2008

Sr No Particulars 0 1 2 3 1 Fixed Assets -180.00 2 Net working Captial -20.00 3 Revenues 136.80 153.90 171.00 Cost ( Other than Deprication & Tax ) -75.24-84.65-94.05 4 5 Depriciation -36.00-28.80-23.04 6 PBT (3) - (4) - (5) 25.56 40.45 53.91 7 Tax -7.67-12.14-16.17 8 PAT (6) - (7) 17.89 28.32 37.74 9 Net Salvage Value of FA 0.00 10 Recovery of Net working Capital 0.00 11 Initial Outlay -200.00 Operating (8) + (5) 12 Cash Flows 53.89 57.12 60.78 13 Terminal Cash Flows (9) + (10) 0.00 14 Net Cash Flows (11) + (12) + (13) -200.00 53.89 57.12 60.78 15 PV Factor 1.000 0.909 0.826 0.621 16 PV of Cash Flow (15) * (16) -200.00 48.99 47.18 37.74 17 NPV (14) - (17) 4.90 9.94 23.03 Page 321 of 380 26/4/2008

Problem 20. The details of a capital investment project are given below: Initial investment for plant and machinery : Rs. 100 Lakhs Additional investment in working capital : Rs. 40 Lakhs Sales (units) per years for years 1 to 5 : Rs. 1 Lakhs Selling price per unit : Rs. 120/- Variable cost per unit : Rs. 60/- Fixed overheads (excluding depreciation) per year for years 1 to 5 : Rs. 15 Lakhs Rate of depreciation on plant and machinery : 25 % by WDV Method Salvage value of plant and machinery (years) : Rs. 10 Lakhs Applicable tax rate : 40 % Time Horizon : 5 years Post-tax cut off rate : 12 % Calculate NPV of the project. Discount Years Factor 0 1 2 3 4 5 6 7 12% 1.000 0.8929 0.7972 0.7118 0.6355 0.5674 0.5066 0.4523 25% 1.000 0.8000 0.6400 0.5120 0.4096 0.3277 0.2621 0.2097 40% 1.000 0.7143 0.5102 0.3644 0.2603 0.1859 0.1328 0.0949 Solution : Notes : 1. Variable cost to be added in Cost, also it is there for 0 th year. 2. Since Time Horizon is 5 year Depreciation will be for 5 year. 3. Salvage & Recovery of working capital is not considered as the information about it is not given. 4. Since Post Tax cut off rate is given it is considered for PV Factor. Page 322 of 380 26/4/2008

Working of Depreciation ( WDV Method ) Figures in Lakhs Year Depreciation Book Value 1 25 % of 100 = 25 100 25 = 75 2 25 % of 75 = 18.75 75 18.75 = 56.25 3 25 % of 56.25 = 14.06 56.25 14.06 = 42.18 4 25 % of 42.18 = 10.55 42.18 10.55 = 31.63 5 25 % of 31.63 = 7.91 Sr No Particulars 0 1 2 3 4 5 1 Fixed Assets -100.00 2 Net working Captial -40.00 3 Revenues 120.00 120.00 120.00 120.00 120.00 Cost ( Other than Deprication & Tax ) -60.00-75.00-75.00-75.00-75.00-75.00 4 5 Depriciation -25.00-18.75-14.06-10.55-7.91 6 PBT (3) - (4) - (5) 20.00 26.25 30.94 34.45 37.09 7 Tax -8.00-10.50-12.38-13.78-14.84 8 PAT (6) - (7) 12.00 15.75 18.56 20.67 22.25 Net Salvage 9 Value of FA 10 Recovery of Net working Capital 11 Initial Outlay -140.00 Operating (8) + (5) 12 Cash Flows 37.00 34.50 32.62 31.22 30.16 13 Terminal Cash Flows (9) + (10) 0.00 0.00 14 Net Cash Flows (11) + (12) + (13) -140.00 37.00 34.50 32.62 31.22 30.16 15 PV Factor 1.0000 0.8929 0.7972 0.7118 0.6355 0.5674 16 PV of Cash Flow (15) * (16) -140.00 33.04 27.50 23.22 19.84 17.12 17 NPV (14) - (17) 3.96 7.00 9.40 11.38 13.05 Page 323 of 380 26/4/2008

25 th April, 2008. Lecture XV. ( Last Lecture ) Network Technique There are two techniques AOA ( Activity on Arrow ) by which activity can be shown as below : A 1 2 5 There is another Method AON ( Activity on Node ) where the activity can be shown as below : A/5 Where A is the Activity and 5 is the duration. This method is also known as Precedence Diagram. Below activity of AOA A 1 2 B 5 6 3 Can be shown in AON as A/5 B/6 Suppose B & C starts after A then with AON it can be shown as B A C Page 324 of 380 26/4/2008

Suppose Z starts after X & Y X Z Y Example -1 : A B C D E F G H - A A A B C D E,F,G B 3 E A C F 1 2 4 H 6 7 D G 5 B E A C F H D G Page 325 of 380 26/4/2008

Example - 2 A B C D E F G H - - A A B D.E C F,G C 4 G 1 A B 2 E D 5 F 6 H 7 3 C G A D F H End Start B E Page 326 of 380 26/4/2008

Example - 3 A B C D E F - A A B C D ( Dangler ) B 3 D 5 1 A 2 C F 4 E 6 B D C Start A End C E Page 327 of 380 26/4/2008

Example - 4 A B C D E F - A B B C D,E 3 5 4 8 7 6 12 12 0 0 8 3 8 8 C 4 1 A B 2 3 5 3 4 D E 7 19 19 25 25 4 5 6 E 6 0 EST Activity Duration EFT LST 0 LFT 8 12 C/4 12 19 E/7 3 0 3 8 A/3 B/5 0 0 3 3 0 8 8 0 12 12 0 19 19 25 F/6 5 13 19 25 D/8 11 3 19 Page 328 of 380 26/4/2008

Example 5. A B C D E F G - - B B B E A,D,C 4 5 6 3 8 11 4 ( Dummy ) 0 11 0 20 1 A 4 4 C B D1 5 6 4 G 24 24 5 5 2 8 E D 3 8 3 20 F 11 6 5 13 13 0 4 A/4 16 20 0 0 Start 5 11 C/6 14 20 11 15 G/4 0 0 0 5 5 8 20 24 B/5 0 5 D/3 17 20 5 13 13 24 26 24 END 24 24 E/8 F/11 5 13 13 24 Page 329 of 380 26/4/2008

Two Approaches to Network Drawing The two approaches used to develop project networks are known as activity on node (AON) and activity on arrow (AOA). Both methods use two building blocks the arrow and the node. Their names derive from the fact that the former uses a node to depict an activity, while the second uses an arrow to depict an activity. Basic Rule to Follow in Developing Project Networks The following eight rules apply in general when developing a project network: 1. Networks flow typically from left to right. 2. An activity cannot begin until all preceding connected activities have been completed. 3. Arrows on networks indicate precedence and flow. Arrows can cross over each other. 4. Each activity should have a unique identification number. 5. An activity identification number must be larger than that of any activities that precede it. 6. Looping is not allowed (in other words, recycling through a set of activities cannot take place). 7. Conditional statements are not allowed that is this type of statement should not appear. If successful, do something, if not, do nothing. 8. Experience suggests that when there are Qlultiple starts, a common start node can be used to indicate a clear project beginning on the network. Similarly a single project end node can be used to indicate a clear ending. Activity- on- Node (AON) Fundamentals The wide availability of personal computes and graphics programs has served as an impetus for use of the activity on node (AON) method (sometimes called the precedence diagram method). Figure 1 shows a few typical uses of building blocks for the AON network construction. An activity is represented by a node (box). The node can take many forms, but in recent years the node represented as a rectangle (box) has dominated. The dependencies among activities are depicted by arrows between the rectangles (boxes) on the AON network. The arrows indicate how the activities are related and the sequence in which things must be accomplished. The length and slope of the arrow are arbitrary and set for convenience of drawing the network. The letters in the boxes server here to identify the activities while you learn the fundamentals of network construction and analysis. In practice, activities have identification numbers and descriptions. Page 330 of 380 26/4/2008

There are three basic relationships that must be established for activities included in a project network. The relationships can be found by answering the following three questions for each activity. 1. Which activities must be completed immediately before this activity? These activities are called predecessor activities. 2. Which activities must immediately follow this activity? These activities are called successor activities. 3. Which activities can occur while this activity is taking place? This is known as a concurrent or parallel relationship. Activity - on - Node Network Fundamentals (i) A is preceded by nothing B is preceded by A C is preceded by B (ii) Y and Z are preceded by X Y and Z can begin at the same time, if you wish (iii) J, K & L can all begin at the same time, if you wish (they need not occur simultaneously) but All (J, K, L) must be completed before M can begin (iv) Z is preceded by X and Y AA is preceded by X ad Y Page 331 of 380 26/4/2008

Above (i) is analogous to a list of things to do where you complete the task at the top of the list first and then move to the second task, etc. This figure tells the projects manager that activity A must be completed before activity B can begin, and activity B must be completed before activity C can begin. Above (ii) tells us that activities Y and Z cannot begin until activity X is completed. This figure also indicates that activities Y and Z can occur concurrently or simultaneously if the project manager wishes: however, it is not a necessary condition. For example, pouring concrete driveway (activity Y) can take place while landscape planting (activityz) is being accomplished, but land clearing (activity X) must be completed before activities Y and Z can start. Activities Y and Z are considered parallel activities. Parallel paths allow concurrent effort, which may shorten time to do a series of activities. Activity X is sometimes referred to as a burst activity because more than one arrow bursts from the node. The number of arrows indicates how many activities immediately follow activity X. Above (iii) shows us activities, J, K and L can occur simultaneously if desired and activity M cannot begin until activities J, K and L are completed. Activities J, K and L are parallel activities. Activity M is called a merge activity because more than one activity must be completed before M can begin. Activity M could also be called a milestone. And in (iv), activities X and Yare parallel activities that can take place at the same time, activities Z and AA are also parallel activities. But activities Z and AA can not begin until activities X and Yare both completed. Given these fundamentals of AON, we can practice developing a simple network. Remember, the arrows can cross over each other (e.g. (iv above)), be bent, or be of any length or slope. Neatness is not a criterion for a valid, useful network - only accurate inclusion of all projects activities, their dependencies, and time estimates. Information for a simplified project network is given in Table 1. This project represents a new business center that is to be developed and the work and services the county engineering design department must provide as it coordinates with other groups '- such as the business center owners and contractors. Page 332 of 380 26/4/2008

Table 1 Network Information Activity Description Preceding A Application approval None B Construction plans A C Traffic study A D Service availability check A E Staff report B,C F Commission approval B,C,D G Wait for construction F H Occupancy E,G Koll Business Center - Complete Network Page 333 of 380 26/4/2008

AON Method Advantages : 1. No dummy activities are used. 2. Events are not used. 3. AON is easy to draw if dependencies are not intense. 4. Activity emphasis is easily understood by first-level managers. 5. The CPM approach uses deterministic times to construct networks. Disadvantages : 1. Path tracing by activity number is difficult. If the network is not available, computer outputs must list the predecessor and successor activities for each activity. 2. Network drawing and understanding are more difficult when dependencies are numerous. AOA Method Advantages : 1. Path tracing is simplified by activity event numbering scheme. 2. AOA is easier to draw if dependencies are intense. 3. Key events or milestones can easily be flagged. Disadvantages : 1. Use of dummy activities increases data requirements. 2. Emphasis on events can detract from activities. Activity delays cause events and projects to be late. Page 334 of 380 26/4/2008

Choice of method - AON or AOA In AOA networks, dummy activities meet two needs. First, when two parallel activities have the same start and end nodes, a dummy must be inserted to give each activity a unique identification number. Next, dummy activities can be used to clarify dependency relationships. Dummy activities are very useful when activity dependencies are far apart on the network. The major advantage of the AOA method is the avoidance of having to list all the predecessor and successor activities for each activity in the network so activity sequence and dependency can be traced when a network is not available or shows incomplete information. Computer output is reduced many fold. LAD AND LAG The method of showing relationships among activities discussed earlier is called the Finish -to Start relationship because it assumes all immediate preceding connected activities must be completed before the next activity can begin. In an effort to come closer to the realities of projects, some useful extensions have been added. Te use of laddering was the first obvious extension practitioners found very useful. Laddering The assumption that all immediate preceding activities must be 100 percent complete is too restrictive for some situations found in practice. This restriction occurs most frequently when one activity overlaps the start of another and has a long duration. Under the standard Finish - to- Start relationship, when an activity has a long duration and will delay the start of an activity immediately following it, the activity can be broken into segments and the network drawn using a Laddering approach so that the following activity can begin sooner and not delay the work. Page 335 of 380 26/4/2008

e.g. LAYING OF PIPE The trench must be dug, pipe laid and the trench refilled. If the pipeline is one mile long, it is not necessary to dig one mile of trench before the laying of pipe can begin or to lay one mile of pipe before refill can begin. Figure below shows how these overlapping might appear in an AON network using the standard Finish - to - Start approach Page 336 of 380 26/4/2008

Use of Lags : The use of lags has been developed to offer greater flexibility in network construction. A lag is the minimum amount of time a dependent activity must be delayed to begin or to end. The use of lags in project networks occurs for two primary reasons. (1) When activities of long duration delay the start or finish of successor activities, the network designer normally breaks the activity into smaller activities to avoid the long delay of the successor activity. Use of lags can avoid such delays and reduce network detail. (2) Lags can be used to constrain the start and finish of an activity. Examples: A lag is positive and indicates the number of time periods that must pass before the succeeding activity can start. E.g. a lag of +5 placed on a finish - to - start link means that activity B can start five time units after activity A has ended. A lag is positive and indicates the number of time periods that must pass before the succeeding activity can start. E.g. a lag of +5 placed on a finish - to - start link means that activity B can start five time units after activity A has ended. The most commonly used relationship extensions are Start - to - Start, Finish - to - Finish and the combination of these two. Page 337 of 380 26/4/2008

Finish - to - Start Relationship The arrowhead on the line indicates the direction of flow and thus the precedence. Activity B is A's successor. The logical link shown here is called a "Finish - to - Start link, as it connects the finish of activity A to the start of activity B. This means that activity B can not start unless activity A has ended. There are situations in which the next activity in a sequence must be delayed even when the preceding activity is complete. Lag - 2 Activity Y starts say 2 days after activity X is completed. Finish - to - start lags are frequently used when ordering materials. For example, it may take 1 day to place orders but take 19 days to receive the goods. The use of finish - to - start allows the activity duration to be 1 day and the lag 19 days. This approach ensures the activity cost is tied to placing the order only rather than charging the activity for 20 days of work. Page 338 of 380 26/4/2008

Start - to - Start Relationship Typical Start - to - Start relationships are shows in figure below. Figure - A shows the start - to start relation with zero lag, while figure - B shows the same relationship with a lag of 4 time units. It is important to note that the relationship may be used with or without lag. If time is assigned, it is usually shown on the dependency arrow of an AON network. Figure A below shows that activities M and N can start at the same time. In figure B, activity Q cannot begin until five units after activity P begins. This type of relationship typically depicts a situation in which you can perform a portion of one activity and begin a following activity before completing the first. This relationship can be used on the pipe-laying project. Figure C shows the project using an AON network. The Start - to - Start relationship reduces network detail and project delays by using lag relationship. Page 339 of 380 26/4/2008

It is possible to change finish to start relations to start to start relationships. e.g. in place of a finish - to start activity "design house, then build foundation", a start - to - start relationship could be used in which the foundation can be started, say, five days (lag) after design has started - the design of the foundation is the first part of the total design activity. Page 340 of 380 26/4/2008

EXAMPLE Draw the network using AON method. Activity Immediate Duration Description Predecessor (weeks) Start - 0 A Start 16 B Start 20 C Start 30 D B 15 E B 10 G 0 3 H D 16 J A 15 K J, G, E 12 Finish K, H,C 0 Calculating the critical path duration The network described in the activity sequencing table above and shown as the network diagram will be used to illustrate the network calculations. The network calculations will result in a number of values being determined for each activity. All the values need to be Page 341 of 380 26/4/2008

recorded on the diagram or in a table. If recorded in the network diagram, the values are attached to the activity as follows : 0 EST Activity Duration EFT LST Float 0 LFT Where, EST = Earliest Start Time EFT = Earliest Finish Time LST = Latest Start Time LFT = Latest Finish Time The early start time ( EST ) for an activity is the earliest time that an activity can start given that all its predecessor activities have been completed. The EST for an activity having no predecessor is always set to 0. The early finish time ( EFT ) of an activity is the earliest time that an activity can be completed. It is equal to the EST Plus its estimated duration. The EST time of an activity that have one predecessor is the EFT of the predecessor. The EST of an activity that has two or more predecessor is the greatest of the EFT of all its predecessors. To calculate these ties, work forward through the network performing a forward pass. The late start (LST) time and late Finish time ( LFT ) of an activity are the latest times at which an activity can start (LST) or be completed (LFT) without affecting the project duration. The LST of an activity is equal to its LFT minus its estimated duration. The LFT of an activity that has only one successor is the LST of the successor. The LFT of an activity having two or more successor is the least of the LS of all successor. To calculate these times, work backwards through the network performing a backward pass. Page 342 of 380 26/4/2008

The Forward Pass A forward pass is performed as follows : 1. Set the EST of the first activity in the network to 0 and its EFT equal to the activity duration. 2. Then for each other activity in the network calculate its EST as the greatest of the EFT of all its predecessor activities. Calculate each activity s EFT as its EST plus duration. After performing the forward pass, the network now looks as shown below : Page 343 of 380 26/4/2008

The Backward Pass A Backward pass is performed as follows : 1. First set the LFT of the last activity in the network to EFT of that activity. Calculate its LST as its LFT minus the activity duration. 2. Then for each other activity in the network calculate its LFT as the least of the LST of all the activities for which it is the predecessor. Calculate each activity s LST as its LFT minus duration. After performing the Backword pass, the network now looks as shown below : The resulting network diagram indicates the various paths through the project network logic, the relationship between the various activities and paths through the network and provides information on time duration all of which are necessary to determine the critical path. Page 344 of 380 26/4/2008

Exercise 1. Draw a project network from the following information. Activity Predecessor A None B None C A,B D A,B E A,B F C,D G E H F I F,G Exercise 2. Draw a project network from the following information. Activity Predecessor A None B None C None D A,B E B,C F D,E G F H F I G J H,I Exercise 3. Draw a project network from the following information. Activity Predecessor A None B A C A D A E B F B G C H D I F,G J E,I,H Page 345 of 380 26/4/2008

Definition of A Project : Theory ( Project Management ) A project is a Temporary Endeavour undertaken to create a unique product or Service. The amended definition is : A project is a Temporary Endeavour undertaken to Crate a unique outcome or results. Project Characteristics Project are Unique. Projects are Temporary in Nature and have definite beginning and ending dates. Projects are completed when the project goals are achieved ore it s determined that the project is no more viable. A successful project is one that meats or exceeds the expectations of the stakeholders. It is focused on the customer and customer expectations. It is made of collection of activities, that are linked together to achieve certain results. It is complex and involved different departments. It has to be flexible to accommodate the changes. It has many unknown factors & external influences. It has cost constraints. Involves risks. Comprises many sub-projects. Requires expertise in many fields. Challenges Traditional line of authority Provides opportunity for learning. Builds team spirit. It consumes large resources. It requires special controls It is tasks & performance oriented. It Requires a task & performance oriented leader. Stakeholders Stakeholder are those individuals ( or Organisations ) with a vested interest in your project. They are the people who are actively involved with the work of the project or have something to either gain or loose as a result of the project. Page 346 of 380 26/4/2008

Stages or Phases of a Project. Due to their uniqueness, projects involves a high degree of uncertainty. Organisations performing projects usually divide a project into phases to better provide management planning and control and identify appropriate links to the organisations ongoing operations. Collectively, the project phases are known as the Project Life Cycle. The various of Phases of Project are : 1. Project Conception Phase : i) Project Scouting and Establishment of Need ii) Identification of Project and Preliminary Screening iii) Feasibility Testing 2. Project Definition Phase : i) Visualization of the Project Features. ii) Preparation of Detailed Project Report. 3. Project Planning and Organization Phase. i) Appraisal of Project Examination of Technical, Organisational, Marketing, Financial, Economic and Social Aspects of the Project. ii) Final Selection and Investment Decision. 4. Implementation Phase. i) Implementation of Project Establishment of Project organisation for, Planning and Mangement. ii) Project follow-up : Monitoring and Control 5. Project Clearing Phase. i) Project Fruitarion : Production / Service, Distribution and Consumer Services. ii) Ex-Post Evaluation. Page 347 of 380 26/4/2008

Project Management Project Management is defined as : The application of knowledge, skills, tools and techniques to project activities in order to meet or exceed stakeholders needs and expectations from a Project. Meeting or exceeding stakeholders needs and expectations involves balancing competing demands among : Scope, time, cost and quality Stakeholders with differing needs and expectations Identified requirements ( needs ) and unidentified requirements ( Expectations ) Project Management Knowledge Areas : Modern project management recognises nine knowledge areas as described below : 1. Project Integration Management : relates to the process require to ensure that the various elements of the project co-ordinated. It includes plan development, execution and overall change control 2. Project Scope Management : means ensuring that the project includes all the work required and only the work required, to complete the project successfully. It covers scope initiation, planning, definition, verification and change control. 3. Project Time Management : includes the decision and required to ensure timely completion of the project, such as activity definition, sequencing, duration estimating, schedule development and control. 4. Project Cost management : refers to the process required to ensure that the project is completed within the approved budget resource planning, cost estimating, cost budgeting and control 5. Project Quality requirement : means the processes required to ensure that the project that the project will satisfy the needs for which it was undertaken. It covers quality planning, assurance and control. 6. Project Human Resource management : is making the most effective use of the people involved with the project. It includes organizational planning, staff acquisition and team development. 7. Project Communication Management : refers to the processes required to ensure timely and appropriate generation, collection, dissemination, storage and ultimate disposition of project information. It entails communications planning, information distribution, performance reporting and administrative closure. Page 348 of 380 26/4/2008

8. Project Risk management means identifying, analyzing and responding to project risk. It includes risk identification, risk quantification, risk response development and risk response control. 9. Project Procurement management covers the processes required to acquire goods and services from outside the performing organization - procurement planning, solicitation planning, solicitation, source selection, contract administration and contract close-out. SCOPE OF PROJECT MANAGEMENT The project management covers: (i) (ii) (iii) (iv) (v) (vi) (vii) (viii) (ix) Idea generation, analysis & finalization of one or more ideas for implementation, Preparation of feasibility reports for various projects ( ideas), working out facilities & finance requirements, benefits & long-term viability as well as profitability. Identification of partners needed. Technology requirements. Organisational requirements. Probable sites & building requirements. Commercial aspects. Environmental effects & action required. Govt. concessions available etc. After finally getting the Feasibility report approved by the concerned authorities- as the case may be, further actions required will be: - PREPARATION OF DETAILED PROJECT REPORT (DPR) This will deal with all the aspects of the project in details namely. Man power: How many managerial, supervisory & workmen? Machines: Types, quantity & suppliers. Materials: Detailed specifications of various materials & their procurement pattern- make or by decisions. Layout & space requirements. Funds requirements & budget. Pay back period / cost benefit analysis. Process & technology Planning & progress control. etc Page 349 of 380 26/4/2008

APPROVAL OF DPR. A Thorough study & scrutiny will be made of the DPR by the TOP management & finance experts. If it requires clearance by specific Govt. Agencies, then a copy will be sent to them for obtaining c1earance- Such as, from factory inspector, environment dept., fire fighting dept., industries dept., municipality etc. If the project is limited to a particular industry -such as introducing a new product, increasing production, replacement of old plant & machinery & like, approval of board of directors is necessary. Detailed Project Report (DPR) A Detailed Project Report (DPR) is a document containing detailed description of important aspects of a Project. This is prepared after initial hurdles in the process of getting the Project cleared, have been crossed and the need for Project has been established. A detailed Project Report generally contains the following details: 1. Background of the Project : The background of the project is described in terms of its basis. This generally include the type of project, the sector of economy it belongs to, its location, gestation period, brief description, technical details such as scope, process, scale, maps and designs, target beneficiaries, etc. 2. Objectives of the Projects : The objectives are specified in terms of general objectives, specific objectives, physical targets to be achieved etc. 3. Justification for the Project : The justification for the Project is based on technical organizational, marketing, financial, economic and environmental appraisal of the project. 4. Cost of Project and Sources of Funding : The cost of Project is determined by considering and including all the items of capital outlay. The arrangement made for meeting the requirement of funds for the project is specified in terms of capital structure, sources of funds and amount proposed to be raised from each source. 5. Salient Features of the Project : The salient features descried included, foreign exchange requirement and contribution, estimated sales revenue, estimated production cost, expected return on investment, social cost and benefits, extent of public participation, role of government, participation of non-governmental organizations, etc. Page 350 of 380 26/4/2008

6. Project Organization : Regarding the project organization and its personnel, the details regarding organization chart, line of control, authority-responsibility structure, extent of delegation, mechanism for monitoring and follow up, project control mechanism, etc. are given. 7. Implementation Details : The implementation details for a project are given in terms of sequence of tasks and activities, resource requirement, precautions, safety requirement, etc. The programme of activities is also given under it. This includes details regarding the activities covered, such as implementation requirement, time schedule, activities and events, critical activities, activity-wise resource requirement management control system, budgeting and budgetary control system, management information system, etc. Project Results : The project results are given in terms of quantitative as well as non-quantitative information both, such as year-wise targets to be achieved, area to be covered, persons benefiting etc. Page 351 of 380 26/4/2008

APPOINTMENT & ROLE OF CONSULTANT IN PROJECT. Introduction Most of the project owners use the services of a consultant - An individual consultant or a consulting firm or a captive consulting organization throughout the project duration. Even when the owner has a competent project team, a consultant may be appointed for the following reasons. a) The functional experts of various departments who are also the members of the project team will have many other parallel responsibilities which are bound to divert their attention from the project. b) A multi-disciplinary consultant's deep knowledge, rich experience and concentrated attention to all important aspects of a project will, no doubt, boost its efficiency, justifying the consultancy cost. c) An independent consultant will view all matters in an unbiased manner. General Issues Concerning Appointment Determination of consultant's role in the project. (a) (b) (c) (d) (e) (f) (g) Types of consultant The Role of Domestic consultant Selection & appointment Terms of reference Actual use of consultant Co-ordination procedure Professional - liability Consultant's Role A consultant can be used either in an "Advisory Role or in a Participatory Role". In the advisory role, after accomplishing the task of the study phase, he might continue as an advisor, without involving himself in the implementation. His assignments in this role will be : - Pre-Investment investigation Preparation of feasibility report Preparation of detailed project report Preparation of project specifications & tender documents Giving advice on problems. Page 352 of 380 26/4/2008

He shall play the roles of a planner, an organizer, and effective co-ordinator, an advisor in decision making, a counselor in overcoming the usual resistance to change and a multidisciplinary management guide, committed to the client's success and profitability. His role can further be increased to macro and micro- level planning of various activities such as defining project scope, organizing project team, assisting in procurement of critical equipments, scheduling & process chasing. Supply of technology or design is optional. Types of Consultants (i) (ii) (iii) Technical specialists. Functional experts. Multi-Disciplinary generalists The last one are the most sought- after Domestic Consultant : Whenever a foreign consultant is appointed, he has to work with a domestic consultant, as per central govt. guidelines Selection and Appointment : Following selection process is followed : a) Determination of consultant's role (term of reference) b) Determination of pre-qualification criteria. i) Cost ii) Short - Listing iii) Final selection iv) Negotiations on terms & conditions v) Appointment ( Decide type) Other points are self explanatory The selection criteria should also include: i) General & technical qualifications of personnel proposed to be employed on the project. ii) Suitability of their experience to the specific requirements of the tasks. iii) Local language ability, knowledge of the cultural background of project area. iv) Probable impact on the project of the consultancy assignment. Page 353 of 380 26/4/2008

ECONOMIC APPRAISAL OF A PROJECT Introduction In case of commercial projects, the real justification for the project comes from the economic viability. Accordingly, a project should be capable of producing an adequate return on the investment and the rate of return should be higher than the cost of funds or the required rate of return. A project is doomed to fail without economic viability. The object of economic appraisal of a project is to analyse the aspects of cost of funds and ROI over the life of the project and to determine whether the project shall be able to give adequate ROI or not. Scope (a) Initial Investment outlay. This specifies the requirement of funds for the project to begin with. It includes the expenditure on design, survey, consultancy, land, plant & machinery, building etc. & Minimum working capital, contingency etc. b) Subsequent Investment outlays. Some times the project is taken in phases. In such cases, subsequent investment has to be provided for c) Economic life of project: This is determined by the stage upto which the project is expected to produce positive cash flows and income flows. In other words, the project should produce adequate ROI to justify its continuance on economic consideration. (d) Salvage value: At the end of economic life, different assets, belonging to the project, when scrapped & disposed off, are expected to realize some value for the project. (e) Operating cash flows: These are generated from the sales revenue after deducting production cost and overheads. Page 354 of 380 26/4/2008

(f) Depreciation : Depreciation is charged with the object of distributing the cost of fixed assets over their entire useful life. However, since it does not involve cash out flow, it is added to the earning after interest and tax to determine the net cash inflow/ out flow. g) Rate of Tax : Tax is charged by the state on profit and also on other earnings of the business enterprise. This affects the surplus available for payment as dividend and also results in outflow of cash. h) Cost of Funds : Funds required for a project are generally loans from bank or other sources, which have got a cost. Cost of funds forms the basis for determining the present value of inflows and the net present value of the project. i) Opportunity cost : The Opportunity cost of funds invested in a project is the expected return from the alternative investment or the opportunity of investment foregone by the investors while making investment in the project. The expected ROI, should be higher than opportunity cost. Page 355 of 380 26/4/2008

PROJECT RISK What is Risk? A risk is any uncertain event, if it occurs, could prevent the project realizing the expectations of the stakeholders as stated in the agreed business case, project brief or agreed definition. A risk that becomes reality is trended as an "Issue". A Risk always has a cause and, if it occurs, a consequence, Risk can have negative or positive consequences, success is dependent on maintaining a high commitment to risk management procedures throughout the project. Two fundamental type of risks are always present. :- a) Project Risk Associated with the technical aspects of the work to achieve the required outcomes and b) Process Risk: - Associated with the project process, procedures, tools & technique employed, controls, communication, stakeholders and team performance, Risk V/s Uncertainty "Risk" can be defined as the variability of return from an investment & the possibilities of the effect are known., but in uncertainty, the outcome cannot be predicated. Kinds of Risks 1. Project Completion Risk : Completing a project in time and within the estimated cost itself is a major achievement. A project that is delayed will result in time over run which will consequently result in cost overrun. If the promoters are not able to fund the costover-run, project gets delayed. There can be technology failures or consultants non availability may cause delay. 2. Resource Risk : Manpower, raw materials, power, fuel, P & M etc. form resources. Delay in receipt of raw materials or P & M etc. will cause delay in project completion. Page 356 of 380 26/4/2008

3. Price Risk : Price fluctuations of both inputs and outputs will affect the project. 4. Competitor s Risk : The competitors may try to disturb the project by increasing or decreasing prices which will force us to review the project for calculation of ROI. 5. Technology Risk : The collaborator may supply us old technology or during the project time, new technology may have been developed, which may effect the profitability. We have to have a good consultant to guard against old technology supply. 6. Political Risk : Suddenly the govt. may impose some new tax or withdraw some facility earlier extended, such as lesser excise or octroi or no sales tax for 2 yrs or cheaper power etc. Also import duly changes may affect project. 7. Interest Rate Risk : Fluctuations in interest rate year after year may bring in adverse effect. Say project is funded by way of long -term borrowings at a particular rate of interest and if the interest rate falis down subsequently, there will be bad effect on project. If the interest rate increases in future, the working capital will be available at higher cost & will lower the profit. 8. Exchange Rate Risk : There are currency fluctuations & international currency rate may vary, resulting into more project cost. 9. Open Policy Risk or Risk from Global Competitors : The new risk emerged with the 'open door' policy of GOI, has thrown out number of industries as they cannot sustain the price-war. The project envisaged today may face such risks in coming days and the project will suffer 10 Risk Due to Trading Activities : Number of entrepreneurs are finding it easy to "assemble" products in producing in India rather than producing components & making the products ( e.g.. computers, DVDS etc. ) such risks in The project runs future times. Page 357 of 380 26/4/2008

11 Risk of Elimination of Product Due to fast changing technology & varying customer demand, some of the products ( now envisaged for project) may not be required., resulting into termination of the project. (example- TV screen for computer monitor, or plasma monitor in place of tube - technology) Techniques of Risk Analysis Though there are many mathematical techniques available for risk analysis, the following are the simple tools that come handy for analyzing small & medium sized projects: (1) Break - Even analysis (2) Sensitivity analysis (3) Decision - Tree analysis (4) Monte - carlo technique etc. (1) Break-Even Technique The costs are divided into fixed & variable costs as below : Fixed Assets Rent Insurance Depriciation Administrative Exp Variable Assets Raw Materials Consumables Power, Fuel Tools Etc. Fixed Cost BEP ( Qty ) = ---------------------------------------------------- Selling Price Unit Variable Cost / Unit Contribution = Sales realization Variable Cost Also, can be represented as Fixed Cost x Selling Price / Unit BEP (Rs) = ---------------------------------------------------- Contribution / Unit Page 358 of 380 26/4/2008

(2) Sensitivity Analysis. It is a technique that measures the change in the profitability of a project caused by changes in the factors that affect the cash inflows of a project. If a small change in one factor leads to a major change in the profitability of tile proposed investment, the project is considered more sensitive to that factor, in other words, the project is more risky. Other things being equal, a project that is less sensitive is preferable to projects that are more sensitive. Example Hind bulbs proposes to start a new venture for the manufacture of fluorescent bulbs. The estimates of the new venture are as under :- Output of bulbs per annum : 3 L Numbers Expected Sales revenue / annum : Rs. 1.50 cr Fixed Cost : Rs. 35 L Variable Costs : Rs. 65 L i) If the selling price comes down to Rs 40 per unit, find out its effect on BEP. ii) If the fixed costs increase to Rs 40 L find out its effect on BEP. iii) If the variable costs increase by 1 %, find out its effect on BEP. Solution Fixed cost Rs : 35,00,000 Sales Revenue Rs : 1,50,00,000 Selling price per unit = ( 150L / 3 L ) = Rs 50 Variable cost per unit = ( Rs. 66 L / 3 L ) Rs 22 Fixed Cost BEP = ---------------------------------------------- Selling Price / Unit V.C Per Unit 35,00,000 35,00,000 = -------------- = ------------- = 1.25,000 Units. 50 22 28 Selling Price / Unit comes down to Rs 40 L 35,00,000 35,00,000 BEP = -------------- = ------------- = 1.94,999 Units. 40 22 18 Page 359 of 380 26/4/2008

b) Fixed Cost increase to Rs 40 L 40,00,000 40,00,000 BEP = -------------- = ------------- = 1.42,875 Units. 50 22 28 c) Variable Cost increase by 10% Revised V.C. = / unit = ( 1.1 x 22 ) = Rs 24.20 35,00,000 40,00,000 BEP = -------------- = ------------- = 1.35,569 Units. 50 24.20 25.80 Results 1. Reduction in selling price by 20 %, increases BEP by 56 % 2. Increase in F.C. by 14.21 %, Increases BEP by 9 % 3. Increase in V.C. by 10 %, Increases BEP by 9 %. Hence BEP is more sensitive to selling price. Page 360 of 380 26/4/2008

Project Cost Estimation Cost of The Project Correct estimation of the capital cost of a project is the foundation over which the edifice of financial appraisal stands. Resources for the project are tied up after the project cost is estimated. Any under-estimation will put the project in problem. Components of Capital Cost of A Project. Following are the components that constitute the capital cost of any project : (i) (ii) (iii) (iv) (v) (vi) (vii) (viii) (ix) (x) (i) Land & land development Buildings Plant and machinery Electricals Transport and erection charges Know - how & consultancy fees Miscellaneous assets. Preliminary & pre- operative expenses Provision of contingencies Margin money for working capital Land & Land Development Before deciding upon the extent of land required for the project, re-assure that investment on land is absolutely essential, after considering the aspect of lease or hire etc. The proposed land has to be leveled in order to construct. The land may uneven - high low or bushy or hilly etc. Approach roads are also to be made simultaneously. Detailed estimates of cost for these work is required. (ii) Buildings Provision for different type of buildings such as : main office (admn.), factory sheds, stores, canteen, creche, dispensary, security, rest room, time office, toilets, facilities like dg set / ac. / compressor rooms etc. A detailed estimate of cost of each building with its drawings is required to take decision at appropriate time, for differing, if required. (iii) Plant and machinery At DPR stage, cost of each of the items of plant & machinery is required. Also inclusion of materials handling equipments, inspection & testing equipments etc. should be there. While, arriving at the cost estimates customs only, octroi, insurance charges in addition to Sales Tax etc. to be considered. Page 361 of 380 26/4/2008

(iv) Electricals These must include fees to electricity board, transformer, panel, cabling, voltage stabilizers, uninterrupted power supply etc. (v) Transport and erection charges Transport charges fill the P & M reach the factory site, including loading & unloading charges are to be accounted for. (vi) Know - how & consultancy fees Know how fees to technical consultant or the consulting firm and the fees for transported technology to the firm inclusive of training is to be provided. While working out fees, the extent of participation of consultant is also to be specified. (vii) Miscellaneous assets. These include provision for office equipments, furniture, fire- fighting equipments, water coolers, air conditioners etc. Also included in this are deposits with electricity board, advance for lease etc. (viii) Preliminary & pre- operative expenses These expenses include investigation fees, service charges of banks/ financing institutions etc. interest amount payable to the term loan landing institution during the period of implementation of the project is to be worked out ( Included project cost) (ix) Provision of contingencies While actually implementing the project, there may be some deviations like changes in the price for plant & machinery, changes in interest rate or parity rate of currency etc. Some un-planned equipments & items may be required. A provision of about 5 to 15% is therefore provided in the project cost estimates. (x) Margin money for working capital Normally bank & financial institutions extend loan to the extent of 75 % to 80 % on capital cost or for working capital. The balance amount is to be contributed by the promoter. This provision is essential. Page 362 of 380 26/4/2008

MONITORING AND CONTROL OF PROJECT Definition Monitoring is collecting, recording, and reporting information concerning any and all aspects of project performance that, the project manager or others in the organisation wish to know. It is important to remember that monitoring, as an activity, is distinct from "Controlling as well as from evaluation. Control is the act of reducing the difference between 'plan' & 'reality'. It involves the regular comparison of performance against targets, a search for the causes of deviation and a commitment to check adverse variances.. Project control Project Control No sooner is the project launched, control becomes the dominant concern of the project manager, indeed, once the launch phase is over, planning & control become closely intertwined in an integrated managerial process. Project control involves a regular comparison of performance against targets, a search for the causes of deviation, and a commitment to check adverse variances. It serves two major functions: (a) (b) It ensures regular monitoring of performance & Motivates project personnel to strive for achieving project objectives. Page 363 of 380 26/4/2008

A few things that can cause a project to require the control of performance cost or time are as follows: - A) Performance B) Cost i) Un-expected technical problems. ii) Insufficient resources when needed. iii) Quality & Reliability problems. iv) Client requires changes. v) Inter-functional complications. i) Tech. Difficulties require more resources. ii) Scope of work increases. iii) Initial bids/ estimates were low. iv) Budgeting was inadequate. v) Input price changes. vi) Corrective control was not exercised in time. C) Time: i) Long time to solve technical difficulties ii) Initial time estimates too optimistic iii) Task sequencing was incorrect. iv) Inputs delay v) Preceding talks delayed. vi) Customer requirements changed. Vii) Govt. Regulations were altered. Page 364 of 380 26/4/2008

PROJECT FINANCING Introduction Project financing may be defined as the raising of funds required to finance an economically separable capital investment proposal in which the lenders mainly rely on the estimated cash flow from the project to service their loans. Project financing differs from conventional financing in the following aspects. : A) In conventional financing, cash flow from different assets and business are comingled. A creditor makes an assessment of repayment of his loan by looking at all the cash flows and resources of the borrower. In project financing, cash flows from the project related assets alone are considered for assessing the re-paying capacity. b) In conventional financing, end use of the borrowed funds is not strictly monitored by the lenders. In project financing, the creditors ensure proper utilization of funds & creation of assets. As envisaged in the project proposal. Funds are also released in stages as & when assets are created. C) In conventional financing, the creditors are not interested in monitoring the performance of the enterprise and they are interested only in their money getting repaid in one way or the other. Project financiers are keen to watch the performance of the enterprise and suggest / take remedial measures as & when required to ensure that the project repays the debt out of its cash generations. They, at times, appoint their nominee in the board of directors, of their clients, in order to monitor the performance. Sources of Finance After the project cost is ascertained, the sources of finances available for meeting the project cost are to be analysed and a proper combination of different sources shall be chosen that is most suitable for the project. The various sources of finance can be broadly divided into 2 categories viz. 1) Equity capital & 2) Debt / Borrowed capital The combination of the above two should be judiciously chosen. Debt capital enforces upon the organisation an obligation for repayment of principal & payment of interest. Equity capital does not impose any such obligation. Equity capital serves as cushion at times when the business conditions are unfavourable leading to operational difficulties. Interest paid on the debt (term loan, debenture etc.) is a deductible item of expenditure from the profit earned by the organisation for the purpose of arriving at the tax payable by the organisation on its earnings). Page 365 of 380 26/4/2008

The contributors of equity capital anticipate a return on their investment by way of dividend. The following are the main sources of project finance. (i) (ii) (iii) (iv) (v) (vi) (vii) (viii) (ix) (x) (xi) (xii) (i) Ordinary shares Preference shares Debentures Bonds Term loans Deferred credits Capital investment subsidy lease financing unsecured loans lnternal accruals Bridge loans Public deposits Ordinary shares Ordinary shares ( or common shares or equity shares) are the source of permanent capital. The holders of ordinary shares are the legal owners of the company. For the capital contributed by the shareholders towards purchase of ordinary shares, they are entitled for dividends. Equity shareholders being the owners of the company, bear the risk of ownership. They are entitled to dividends on their capital invested, only after interest obligations and dividends to preference shareholders are paid. Since equity shareholders bear such risks associated with ownership of the company, they anticipate handsome return on their investment, by way of attractive dividends and price appreciation of their shares. (ii) Preference shares Preference shares bear a predetermined rate of dividend. They have priority of claim over equity shares in the matter of payment of dividend. They also have priority over equity shares on the assets of the company in the event of liquidation of the company. Though dividend on preference shares is payable out of profits. If the company incurs loss in a particular year, the dividend not paid during that year. is to be carried forward and is to be paid in subsequent year/years when the company earns profit. Dividends on non-cumulative preference shares do not add up for future payment. When the company earns profit, it has to pay arrears of dividends to preference shareholders before declaring any dividend on equity share. The Page 366 of 380 26/4/2008

dividend rate is fixed in the case of preference shares and the dividends pain on preference shares are not tax deductible. Preferences shares are of two types, viz, redeemable preference shares and irredeemable preference shares. While redeemable preference shares are redeemed after the stipulated period., Irredeemable preference shares do not have any maturity date. (iii) Debentures Debentures are instruments for raising long term debt capital. The debenture holders are the creditors of the company. The company that has borrowed money by way of debentures has the obligation to repay interest and debt on specified dates. (iv) Bonds A bond is more or less similar to a debenture and these tow terms of frequently used interchangeably. In India, there is a tendency to reserve the term 'bond' to public debt securities issued by the Government and public sector undertakings. (v) Term loans The term 'term loan' denotes long term loans offered for project financing. The period of principal repayment of such long term loans vary from 5 to 10 years depending upon the nature of the project (It will be more for infrastructure projects, say in order of 20 to 25 years) Initial moratorium (Holiday period) for the repayment of principal of one of two years is normally provided. The length of the repayment period depends upon the period of implementation of the project. (vi) Deferred Credits Some machinery suppliers provide the facility of deferred credit, provided the credit-taker offers a Bank guarantee. A project promoter when wants to avail the deferred credit facilities offered by a machinery supplier should approach a bank for offering guarantee for the repayment of deferred installments to the machinery supplier. Banks examine the viability of the project proposal before giving their guarantee. Normally, banks obtain mortgage of additional securities from the credit-takers to ensure that the bank does not stand to loose in the event of the guarantee being invoked by the machinery supplier due to non repayment of deferred installments. Page 367 of 380 26/4/2008

(vii) Capital investment subsidy Government provides subsidy for the setting up of industries. The subsidy offered is two types. Viz. a) Area subsidy b) Product subsidy a) Area subsidy : Area subsidy is available for projects (both new and expansion projects) set up in notified backward areas. Government notifies backward areas from time to time based on the industrial activity prevailing in different parts of the country. The objective of notifying certain areas as backward/most backward is to promote industrial development in those area. b) Product subsidy : Project subsidy is available for projects that manufacture specified products. These products that are eligible for subsidy are identified by the government by keeping in view the potential for the economic development of the country in such sectors of industries and notified by the Government. Projects that are set up for the manufacture of products that find place in the list of eligible products. can avail product subsidy, irrespective of the location of the project i.e. the project can avail su bsidy irrespective of whether the unit is set up in a notified backward area or not. The quantum of product subsidy is also fixed at a certain percentage on the investment on fixed assets ( in the range of 10 % to 20 % for different types of notified products) (viii) lease financing lease is a contract whereby the lessor (owner of an asset) give to the lessee ( the user of the asset) the right to use the asset, usually for an agreed period of time, in return for the consideration of periodical payments by the lessee to the lessor called, lease rentals. Lease, as a source of project finance is mainly suitable for expansion projects This is because of the reason that repayment of lease rental starts immediately. (ix) unsecured loans If there is some shortfall in the means of finance, the promoters/ directors can mobilize funds form their friends, relatives and well wishers in the form of loan to make good the shortfall. Such loans are always unsecured i.e. the lenders can not have any charge over the assets of the company. Unsecured loans can Page 368 of 380 26/4/2008

mobilized only based on the rapport that the project promoters have with their friends and relatives. Banks and financial institutions view unsecured loans with an eye of caution. They normally stipulate the following conditions if unsecured loan is to form part of the means - of finance. - The promoters shall not repay the unsecured loan till the term loan persists. - Interest, if any payable on unsecured loan shall be paid only after meeting the term loan repayment commitments (both repayment of principal and interest) - The rate of interest payable on unsecured loan shall not be higher that the rate of interest applicable for term loan. Banks/ Financial institutions also stipulate the maximum limit for unsecured loan. Normally unsecured loan component is expected not to exceed 50 % of the equity capital. (x) lnternal accruals Internal accruals from part of the means of finance in respect of expansion projects. An existing company that goes for an expansion (or diversification or modernization) project may opt to finance a portion of the capital investment out of internal cash accruals. Depreciation which is not a cash expenditure and profits retained after payment of dividends are the main sources of internally generated funds. Apart from the internal funds that have already been generated. The likely internal generation during the course of project implementation can also be used as a source for funding expansion projects. (xi) Bridge loans This is a temporary loan meant for tieing up the capital cost of a project. Bridge loans are sanctioned by banks and financial institutions in order to help speedy implementation of the project. In the absence of bridge loan, the project implementation may get delayed for want of sufficient funds. The necessity for bridge finance arises in situations where finance from a particular source is getting delayed. However, the availability of finance from that source is certain. In such situations, if the project is funded with bridge-loan to the extent needed, the implementation of the project will go on stream without getting held up for want of funds. The following example will explain the concept well. The project cost of a new project is estimated at Rs. 100.00 lakhs. The promoters are able to bring in a capital of Rs. 30.0 Lakhs. The project is eligible for an investment subsidy of Rs. 10.00 lakhs. The Financial institution that has Page 369 of 380 26/4/2008

appraised the project is ready to sanction a term loan of Rs. 60.00 lakhs. Thus, the means of finance for the project cost gets tied up as under: (xii) Promoter's contribution Investment subsidy Term loan Total Public deposits Rs. 30.00 Lakhs Rs. 10.001akhs Rs. 60.001akhs ---------------------- 100.00 Lakhs ---------------------- Public deposits are mobilized from the public in general and also from the share holders. Public deposits can be taken for a minimum period of six months and a maximum period of 36 months. Section 58 A of the Companies act regulates public deposits. The interest rate on Public deposits depends on the period of deposit. Government issues directions on the interest rate payable on public deposits from time to time. Raising of public deposit is a more simple and convenient form of mobilizing money as compared to raising loans from organized financial institutions. Page 370 of 380 26/4/2008

Question : Discuss Human Aspects of Project Management in relation to Authority, Personnel Orientation, Motivation and Team Building. Although the use of proper planning and budgetary control techniques help in successful project management, their use and effectiveness depends on the people who are involved and are responsible for various project activities at various levels. It is therefore very important to understand human nature and to achieve satisfactory human relations in project setting. Project manager has to handle problems and challenges relating to following issues. 1. Authority : Project manager very often does not have a direct authority over the project team. He has to co-ordinates efforts of various functional groups, experts, consultants external agencies etc. Team leadership and influencing professionals assumes importance than exercising authority. Project manager's authority therefore emanates from his ability to develop rapport with the team members, skillful resolution of conflicts, skill of communication and persuasion and ability to act as a buffer between technical engineering, financial and commercial personnel. 2. Personnel Orientation : Overemphasis on planning and control techniques that are mathematics and accounts oriented tends to adopt a structured and mechanical approach to project management. However, project being subject to many uncertainties, a more creative and adaptive approach is necessary to solve un-programmed and un-structured problems as the project progresses. Such orientation is required for all the project team. 3. Motivation : Project being a short team endeavour carried out by lossely bound professional team, it is very difficult to keep it motivated throughout the project term. The team members sometimes tend to get confused due to split authority and dual subordination structure of the organization. To keep a high level of motivation, the project manager has to ensure that the project goals are clearly defined and are visible to all involved. He has to encourage participative management with proper delegation of authority and responsibility that creates a sense of belonging to the project and to make individuals job sufficiently challenging to have greater personal commitment. 4. Team Building : Most of the project activities are inter-related and interdependent and most 'of the problems need inter-disciplinary solutions. Page 371 of 380 26/4/2008

Question When do you judge a project to be a failure? What are the possible causes of project failures? A project is considered to be a failure if it fails to complete in scheduled time or it exceeds the budgeted cost or is not able to fully accomplish the project objectives. While time overruns and cost overruns are the common causes of project failure, inability to meet the key performance indicators is also observed occasionally. Although the project is may not be a total failure as it is finally implemented and starts functioning, it tends to be uneconomical to run, does not create sufficient surplus for investing into future growth of the organization and eventually becomes a drain on organizations' funds. Common causes that result in time / cost overruns or failure to meet specifications are: Inadequate project formulation : Superficial field investigations, ill-defined project scope, wrong assessment of input requirements, inaccurate methods for estimation in of costs & benefits, deliberate underestimation of costs and overestimation of benefits etc. Unsuitable Project Organization : Incompetent project leader, inadequate authority to the project leader, confusing or improperly understood roles of project team members. Improper Implementation Planning : Insufficient breakdown of project activities, undue reliance on intuition and judgment rather than available information and data, improper definition of interlinkages between activities. Failure to take advance action : As project is subjected to a lot of risks during its course of implementation, it is necessary to be proactive at the signs of impending problems rather than react only when the problems are fully blown up. Non-ability of funds in time : Proper cash flow management is the key to success of the project. Non-availability of funds in time delays the project implementation as well as increases the project costs. Injudicious equipment tendering, procurement and contracts management : It is generally observed that equipment and materials form 65-70% of the project cost and procurement of the same covers almost 70% of the project implementation time. Similarly-, Proper selection of contractors ensures timely completion within budgeted costs. Ineffective project monitoring : As the projects are implemented under a dynamic, ever changing environment, effective monitoring and timely modifications in implementation strategy are required for the success of the project. Page 372 of 380 26/4/2008

Question What is a Gantt Chart? One of the oldest methods of presenting time schedule information is the Gantt chart, developed around 1917 by Henery Gantt. A Gantt chart is one of the most convenient, most commonly used, easy-to-grasp presentations of project activities. It is a two-dimensional graphical representation of the activities that make up the project. The vertical axis lists the project activities, one per tline, while the horizontal axis indicates time. Once the scheduled start and completion dates for every activity have been determined, the Gantt chart can be constructed. Figure shows a typical Gantt chart as planned and shows the same chart progressed. An added value of the Gantt chart is that the activities are time-scaled, which provides a perspective not possible with other project charts such as network diagrams. A time scaled network diagram can be developed which allows progress to be indicated similar to a Gantt chart. Networks are described in the next section. 'Critical path analysis'. The Gantt chart is a particularly effective and easy-to-read method of indicating the actual current status of activities compared to the planned progress. As a result, the Gantt chart can be helpful in expediting, sequencing and reallocating resources to activities, as well as keeping track of progress. In addition, the charts can contain a number of specialized symbols to designate or highlight items of special concern to the situation being charted. Advantages and disadvantages of the Gantt Chart The Gantt has the following general advantages over other planning tools such as network diagrams: Although they contain a great deal of information, they are easily understood. While they require frequent updating (as does any scheduling / control tool), they are easy to maintain. They provide a clear, simple picture of the state of the project. They are easy to construct and are not based on any mathematical model. The Gantt maybe constructed without a critical path analysis (CPA) being needed. They can, however, also graphically represent the output of a CPA. A close relationship exists between Gantt charts and CPA networks (PERT or CPM). Generally Gantt charts are derived from CPA networks by plotting the activity from its calculated earliest time for the length of its duration. If an activity has float, then this can be shown as differently patterned bar at the back of the plotted activity bar. Page 373 of 380 26/4/2008

The major disadvantage of the Gantt chart is that the relationships or dependencies between activities are not a explicit as in CPA networks. Unless connecting lines are drawn on the Gantt chart, it is not possible to determine the links between activities. Modern computer packages, however, can now show these connecting lines if required. The Gantt chart has survived for more than 80 years. Considering that in 1917 Henry Gantt could not have visualized the proliferation of sophisticated project management tools and techniques, the Gantt chart continues to prove its usefulness beyond any doubt. Page 374 of 380 26/4/2008

Table : Area Under the Standard Normal Curve From 0 to z Z 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09.0 0.0000 0.0040 0.0080 0.1200 0.0160 0.0199 0.0239 0.0279 0.3190 0.0359.1 0.0398 0.0438 0.0478 0.0517 0.0557 0.0596 0.0636 0.0675 0.0714 0.0753.2 0.0793 0.0832 0.0871 0.0910 0.0948 0.0987 0.1026 0.1064 0.1103 0.1141.3 0.1179 0.1217 0.1255 0.1293 0.1331 0.1368 0.1406 0.1443 0.1480 0.1517.4 0.1554 0.1591 0.1628 0.1664 0.1700 0.7360 0.1772 0.1808 0.1844 0.1879.5 0.1915 0.1950 0.1985 0.2019 0.2054 0.2088 0.2123 0.2157 0.2190 0.2224.6 0.2257 0.2291 0.2324 0.2357 0.2389 0.2422 0.2454 0.2486 0.2518 0.2549.7 0.2580 0.2612 0.2642 0.2673 0.2705 0.2734 0.2764 0.2794 0.2823 0.2852.8 0.2881 0.2910 0.2939 0.2967 0.2995 0.3023 0.3051 0.3078 0.3106 0.3133.9 0.3159 0.3186 0.3212 0.3238 0.3264 0.3289 0.3315 0.3340 0.3365 0.3389 1.0 0.3413 0.3438 0.3461 0.3485 0.3508 0.3531 0.3554 0.3577 0.3599 0.3621 1.1 0.3643 0.3665 0.3686 0.3708 0.3729 0.3749 0.3770 0.3790 0.3810 0.3830 1.2 0.3849 0.3869 0.3888 0.3907 0.3925 0.3944 0.3962 0.3980 0.3997 0.4015 1.3 0.4032 0.4049 0.4066 0.4082 0.4099 0.4115 0.4131 0.4147 0.4162 0.4177 1.4 0.4192 0.4207 0.4222 0.4236 0.4251 0.4265 0.4279 0.4292 0.4306 0.4319 1.5 0.4332 0.4345 0.4357 0.4370 0.4382 0.4394 0.4406 0.4418 0.4429 0.4441 1.6 0.4452 0.4463 0.4474 0.4484 0.4495 0.4505 0.4515 0.4525 0.4535 0.4545 1.7 0.4554 0.4564 0.4573 0.4582 0.4591 0.4599 0.4608 0.4616 0.4625 0.4633 1.8 0.4641 0.4649 0.4656 0.4664 0.4671 0.4678 0.4686 0.4693 0.4699 0.4706 1.9 0.4713 0.4719 0.4726 0.4732 0.4738 0.4744 0.4750 0.4756 0.4761 0.4767 2.0 0.4772 0.4778 0.4783 0.4788 0.4793 0.4798 0.4803 0.4808 0.4812 0.4817 2.1 0.4821 0.4826 0.4830 0.4838 0.4838 0.4842 0.4846 0.4850 0.4854 0.4857 2.2 0.4861 0.4864 0.4868 0.4871 0.4875 0.4878 0.4881 0.4884 0.4887 0.4890 2.3 0.4893 0.4896 0.4898 0.4901 0.4904 0.4906 0.4909 0.4911 0.4913 0.4916 2.4 0.4918 0.4920 0.4922 0.4925 0.4927 0.4931 0.4931 0.4932 0.4934 0.4936 2.5 0.4938 0.4940 0.4941 0.4943 0.4945 0.4946 0.4948 0.4949 0.4951 0.4952 2.6 0.4953 0.4955 0.5956 0.4958 0.4959 0.4960 0.4961 0.4962 0.4963 0.4964 2.7 0.4965 0.4966 0.4967 0.4968 0.4969 0.4970 0.4971 0.4972 0.4973 0.4974 2.8 0.4974 0.4975 0.4976 0.4977 0.4977 0.4978 0.4979 0.4979 0.4980 0.4981 2.9 0.4981 0.4982 0.4982 0.4983 0.4984 0.4984 0.4985 0.4985 0.4986 0.4986 3.0 0.49865 0.4987 0.4987 0.4988 0.4988 0.4989 0.4989 0.4989 0.4990 0.4990 4.0 0.4999684 Illustration for Z = 1.96, shaded area is 0.4750 out of total area of 1. Page 375 of 380 26/4/2008

Tables Present Value Factor (PVF) Table rate of interest r % Periods (n) 1% 2% 3% 4% 5% 6% 7% 8% 9% 10% 11% 12% 13% 0 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1 0.990 0.980 0.971 0.962 0.952 0.943 0.935 0.926 0.917 0.909 0.901 0.893 0.885 2 0.980 0.961 0.943 0.925 0.907 0.890 0.873 0.857 0.842 0.826 0.812 0.797 0.783 3 0.971 0.942 0.915 0.889 0.864 0.840 0.816 0.794 0.772 0.751 0.731 0.712 0.693 4 0.961 0.924 0.889 0.855 0.823 0.792 0.763 0.735 0.708 0.683 0.659 0.636 0.613 5 0.951 0.906 0.863 0.822 0.784 0.747 0.713 0.681 0.650 0.621 0.593 0.567 0.543 6 0.942 0.888 0.838 0.790 0.746 0.705 0.666 0.630 0.596 0.564 0.535 0.507 0.480 7 0.933 0.871 0.813 0.760 0.711 0.665 0.623 0.583 0.547 0.513 0.482 0.452 0.425 8 0.923 0.853 0.789 0.731 0.677 0.627 0.582 0.540 0.502 0.467 0.434 0.404 0.376 9 0.917 0.837 0.766 0.703 0.645 0.592 0.544 0.500 0.460 0.424 0.391 0.361 0.333 10 0.905 0.820 0.744 0.676 0.614 0.558 0.508 0.463 0.422 0.386 0.352 0.322 0.295 11 0.895 0.804 0.722 0.650 0.585 0.527 0.475 0.429 0.388 0.350 0.317 0.287 0.261 12 0.887 0.788 0.701 0.625 0.557 0.497 0.444 0.397 0.356 0.319 0.286 0.257 0.231 13 0.879 0.773 0.681 0.601 0.530 0.469 0.415 0.368 0.326 0.290 0.258 0.229 0.204 14 0.870 0.758 0.661 0.577 0.505 0.442 0.388 0.340 0.299 0.263 0.232 0.205 0.181 15 0.861 0.743 0.642 0.555 0.481 0.417 0.362 0.315 0.275 0.239 0.209 0.183 0.160 16 0.853 0.728 0.623 0.534 0.458 0.394 0.339 0.292 0.252 0.218 0.188 0.163 0.141 17 0.844 0.714 0.605 0.513 0.436 0.371 0.317 0.270 0.231 0.198 0.170 0.146 0.125 18 0.836 0.700 0.587 0.494 0.416 0.350 0.296 0.250 0.212 0.180 0.153 0.130 0.111 19 0.828 0.686 0.570 0.475 0.396 0.331 0.276 0.232 0.494 0.164 0.138 0.116 0.098 20 0.820 0.673 0.554 0.456 0.377 0.312 0.258 0.215 0.178 0.149 0.124 0.104 0.087 25 0.780 0.610 0.478 0.375 0.295 0.233 0.184 0.146 0.116 0.092 0.074 0.059 0.047 30 0.742 0,552 0.412 0.308 0.231 0.174 0.131 0.099 0.075 0.057 0.044 0.033 0.026 Page 376 of 380 26/4/2008

Tables Present Value Factor (PVF) Table rate of interest r % Periods (n) 14% 15% 16% 17% 18% 19% 20% 24% 28% 32% 36% 40% 0 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1 0.877 0.870 0.862 0.855 0.847 0.840 0.833 0.806 0.781 0.758 0.735 0.714 2 0.769 0.756 0.743 0.731 0.718 0.706 0.694 0.650 0.610 0.574 0.541 0.510 3 0.675 0.658 0.641 0.624 0.609 0.593 0.579 0.524 0.477 0.435 0.398 0.364 4 0.592 0.572 0.552 0.534 0.516 0.499 0.482 0.423 0.373 0.329 0.292 0.260 5 0.519 0.497 0.476 0.456 0.437 0.419 0.402 0.341 0.291 0.250 0.215 0.186 6 0.456 0.432 0.410 0.390 0.370 0.352 0.335 0.275 0.277 0.189 0.158 0.133 7 0.400 0.376 0.354 0.333 0.314 0.296 0.279 0.222 0.178 0.143 0.116 0.095 8 0.351 0.324 0.305 0.285 0.266 0.249 0.233 0.179 0.139 0.108 0.085 0.068 9 0.308 0.284 0.263 0.243 0.226 0.209 0.194 0.144 0.108 0.082 0.063 0.048 10 0.270 0.247 0.227 0.206 0.191 0.176 0.162 0 116 0.085 0.062 0.046 0.035 11 0.237 0.215 0.195 0.178 0.162 0.148 0.135 0.094 0.066 0.047 0.034 0.025 12 0.208 0.187 0.168 0.152 0.137 0.124 0.112 0.076 0.052 0.036 0.025 0.018 13 0.182 0.163 0.145 0.130 0.116 0.104 0.093 0.061 0.040 0.027 0.018 0.013 14 0.160 0.141 0.125 0.111 0.099 0.088 0.078 0.049 0.032 0.021 0.014 0.009 15 0.140 0.123 0.108 0.095 0.084 0.074 0.065 0.040 0.025 0.016 0.010 0.006 16 0.123 0.107 0.093 0.081 0.071 0.062 0.054 0.032 0.019 0.012 0.007 0.005 17 0.108 0.093 0.080 0.069 0.060 0.052 0.045 0.026 0.015 0.009 0.005 0.003 18 0.095 0.081 0.069 0.059 0.051 0.044 0.038 0.021 0.012 0.007 0.004 0.002 19 0.083 0.070 0.060 0.051 0.043 0.037 0.031 0.017 0.009 0.005 0.003 0.002 20 0.073 0.061 0.051 0.043 0.037 0.031 0.026 0.014 0.007 0.004 0.002 0.001 25 0.038 0.030 0.024 0.020 0.016 0.013 0.010 0.005 0.002 0.001 0.000 0.000 30 0.020 0.015 0.012 0.009 0.007 0.005 0.004 0.002 0.001 0.000 0.000 0.000 Page 377 of 380 26/4/2008

Tables Present Value Factor for an Annuity Factor for Annuity ( PVFA ) Table ( Cumulative ) Rate of Interest r % Periods (n) 1% 2% 3% 4% 5% 6% 7% 8% 9% 10% 11% 12% 13% 0 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1 0.990 0.980 0.971 0.962 0.952 0.943 0.935 0.926 0,917 0.909 0.901 0.893 0.885 2 1.970 1.942 1.913 1.886 1.859 1.833 1.181 1.783 1.759 1.735 1.713 1.690 1.668 3 2.941 2.884 2.829 2.775 2.723 2.673 2.624 2.577 2.531 2.487 2.444 2.402 2.361 4 3.902 3.808 3.717 3.630 3.546 3.465 3.387 3.312 3.240 3.170 3.102 3.037 2.974 5 4.853 4.713 4.580 4.452 4.329 4.212 4.100 3.993 3.890 3.791 3.696 3.605 3.517 6 5.795 5.601 5.417 5.242 5.076 4.917 4.767 4.623 4.486 4.355 4.231 4.111 3.998 7 6.728 6.472 6.230 6.002 5.786 5.582 5.389 5.206 5.033 4.868 4.712 4.564 4.423 8 7.652 7.325 7.020 6.733 6.463 6.210 5.971 5.747 5.535 5.335 5.146 4.968 4.799 9 8.566 8.162 7.786 7.435 7.108 6.802 6.515 6.247 5.995 5.759 5.537 5.328 5.132 10 9.471 8.983 9.530 8.111 7.722 7.360 7.024 6.710 5.418 6.145 5.889 5.650 5.426 11 10.368 9.787 9.253 8.760 8.306 7.887 7.499 7.139 5.805 6.495 6.207 5.938 5.687 12 11.255 10.575 9.954 9.385 8.863 8.384 7.943 7.536 7.161 6.814 6.492 6.194 5.918 13 12.134 11.348 10.635 9.986 9.394 8.853 8.358 7.904 7.487 7.103 6.750 6.424 6.122 14 13.004 12.106 11.296 10.563 9.899 9.295 8.745 8.244 7.786 7.367 6.982 6.628 6.302 15 13.865 12.849 11.938 11.118 10.380 9.712 9.108 8.559 8.061 7.606 7.191 6.811 6.462 16 14.718 13.578 12.561 11.652 10.838 10.106 9.447 8.851 8.313 7.824 7.379 6.974 6.604 17 15.562 14.292 13.166 12.166 11.274 10.477 9.763 9.122 8.544 8.022 7.549 7.120 6.729 18 16.398 14.992 13.754 12.659 11.690 10.828 10.059 9.372 8.756 8.201 7.702 7.250 6.840 19 17.226 15.678 14.324 13.134 12.085 11.158 10.336 9.604 8.950 8.365 7.839 7.366 6.938 20 18.046 16.351 14.877 13.590 12.462 11.470 10.594 9.818 8.129 8.514 7.963 7.469 7.025 25 22.023 19.523 17.413 15.622 14.094 12.783 11.654 10.675 8.823 9.077 8.422 7.843 7.330 30 25.808 22.396 19.600 17.292 15.372 13.765 12.409 11.258 10.274 9.427 8.694 8.055 7.496 Page 378 of 380 26/4/2008

Tables Present Value Factor for an Annuity Factor for Annuity ( PVFA ) Table ( Cumulative ) Rate of Interest r % Page 379 of 380 26/4/2008