PERT 12 Quantitative Tools (1)

Transcription

1 PERT 12 Quantitative Tools (1) Proses keputusan dalam operasi Fundamental Decisin Making, Tabel keputusan. Konsep Linear Programming Problem Formulasi Linear Programming Problem Penyelesaian Metode Grafis Analisis Sensitivitas Masalah Minimasi Metode Simplek

2 The Decision-Making Process Problem Quantitative Analysis Logic Historic Data Marketing Research Scientific Analysis Modeling Decision? Qualitative Analysis Weather State and federal legislation New technological breakthroughs Election outcome

3 Learning Objectives When you complete this chapter, you should be able to: Describe or Explain: Decision making under risk Decision making under uncertainty Decision making under certainty

4 The Decision Process in Operations 1. Clearly define the problems and the factors that influence it 2. Develop specific and measurable objectives 3. Develop a model 4. Evaluate each alternative solution 5. Select the best alternative 6. Implement the decision and set a timetable for completion

5 Fundamentals of Decision Making 1. Terms: a. Alternative a a course of action or strategy that may be chosen by the decision maker b. State of nature an an occurrence or a situation over which the decision maker has little or no control

6 Fundamentals of Decision Making (Lanj) 2. Symbols used in a decision tree: a. decision node from which one of several alternatives may be selected b. a a state-of of-nature node out of which one state of nature will occur

7 Decision Tree Example A decision node A state of nature node Favorable market Construct large plant Construct small plant Do nothing Unfavorable market Favorable market Unfavorable market

8 Decision Table Example Alternatives State of Nature Favorable Market Unfavorable Market Construct large plant $200,000 $180,000 Construct small plant $100,000 $ $ 20,000 Do nothing $ 0 $ 0

9 Decision-Making Environments Decision making under uncertainty Complete uncertainty as to which state of nature may occur Decision making under risk Several states of nature may occur Each has a probability of occurring Decision making under certainty State of nature is known

10 Uncertainty 1. Maximax Find the alternative that maximizes the maximum outcome for every alternative Pick the outcome with the maximum number Highest possible gain

11 Uncertainty 2. Maximin Find the alternative that maximizes the minimum outcome for every alternative Pick the outcome with the minimum number Least possible loss

12 Uncertainty 3. Equally likely Find the alternative with the highest average outcome Pick the outcome with the maximum number Assumes each state of nature is equally likely to occur

13 Uncertainty Example Alternatives States of Nature Favorable Market Construct large plant $200,000 Construct small plant $100,000 Unfavorable Market Maximum in Row Minimum in Row Row Average $200,000 -$180,000 $200,000 -$180,000 $10,000 $100,000 -$20,000 $100,000 -$20,000 $40,000 Do nothing $0 $0 $0 $0 $0 Maximax Maximin 1. Maximax choice is to construct a large plant 2. Maximin choice is to do nothing 3. Equally likely choice is to construct a small plant Equally likely

14 Risk Each possible state of nature has an assumed probability States of nature are mutually exclusive Probabilities must sum to 1 Determine the expected monetary value (EMV) for each alternative

15 Expected Monetary Value EMV (Alternative i) = (Payoff of 1 st state of nature) x (Probability of 1 st state of nature) (Payoff of 2 nd state of nature) x (Probability of 2 nd state of nature) (Payoff of last state of nature) x (Probability of last state of nature)

16 EMV Example Alternatives Favorable Market States of Nature Unfavorable Market Construct large plant (A1) $200,000 -$180,000 Construct small plant (A2) $100,000 -$20,000 Do nothing (A3) $0 $0 Probabilities EMV(A 1 ) = (.5)($200,000) + (.5)(-$180,000) = $10, EMV(A 2 ) = (.5)($100,000) + (.5)(-$20,000) = $40, EMV(A 3 ) = (.5)($0) + (.5)($0) = $0

17 EMV Example Alternatives Favorable Market States of Nature Unfavorable Market Construct large plant (A1) $200,000 -$180,000 Construct small plant (A2) $100,000 -$20,000 Do nothing (A3) $0 $0 Probabilities EMV(A 1 ) = (.5)($200,000) + (.5)(-$180,000) = $10, EMV(A 2 ) = (.5)($100,000) + (.5)(-$20,000) = $40, EMV(A 3 ) = (.5)($0) + (.5)($0) = $0 Best Option

18 Certainty Is the cost of perfect information worth it? Determine the expected value of perfect information (EVPI)

19 Expected Value of Perfect Information EVPI is the difference between the payoff under certainty and the payoff under risk Expected value EVPI = under certainty Maximum EMV Expected value. under certainty = (Best outcome or consequence for 1 st state of nature) x (Probability of 1 st state of nature) Best outcome for 2 nd state of nature) x (Probability of 2 nd state of nature) Best outcome for last state of nature) x (Probability of last state of nature)

20 EVPI Example 1. The best outcome for the state of nature favorable market is build a large facility with a payoff of $200,000.. The best outcome for unfavorable is do nothing with a payoff of $0. Expected value under certainty = ($200,000)(.50) + ($0)(.50) = $100,000

21 EVPI Example 2. The maximum EMV is $40,000,, which is the expected outcome without perfect information. Thus: Expected value EVPI = under certainty Maximum EMV = $100,000 $40,000 = $60,000 The most the company should pay for perfect information is $60,000

22 Decision Trees Information in decision tables can be displayed as decision trees A decision tree is a graphic display of the decision process that indicates decision alternatives, states of nature and their respective probabilities, and payoffs for each combination of decision alternative and state of nature Appropriate for showing sequential decisions

23 Decision Trees

24 Decision Trees 1. Define the problem 2. Structure or draw the decision tree 3. Assign probabilities to the states of nature 4. Estimate payoffs for each possible combination of decision alternatives and states of nature 5. Solve the problem by working backward through the tree computing the EMV for each state-of of-nature node

25 Decision Tree Example EMV for node 1 = $10,000 = (.5)($200,000) + (.5)(-$180,000) Favorable market (.5) Payoffs $200,000 Construct large plant Do Do nothing nothing Construct small plant 1 2 Unfavorable market (.5) Favorable market (.5) Unfavorable market (.5) -$180,000 $100,000 -$20,000 EMV for node 2 = $40,000 = (.5)($100,000) + (.5)(-$20,000) Figure A.2 $0

26 Complex Decision Tree Example

27 Complex Example 1. Given favorable survey results EMV(2) = (.78)($190,000) + (.22)(-$190,000) = $106,400 EMV(3) = (.78)($90,000) + (.22)(-$30,000) = $63,600 The EMV for no plant = -$10,000 so, if the survey results are favorable, build the large plant

28 Complex Example 2. Given negative survey results EMV(4) = (.27)($190,000) + (.73)(-$190,000) = -$87,400 EMV(5) = (.27)($90,000) + (.73)(-$30,000) = $2,400 The EMV for no plant = -$10,000 so, if the survey results are negative, build the small plant

29 Complex Example 3. Compute the expected value of the market survey EMV(1) = (.45)($106,400) + (.55)($2,400) = $49, If the market survey is not conducted EMV(6) = (.5)($200,000) + (.5)(-$180,000) = $10,000 EMV(7) = (.5)($100,000) + (.5)(-$20,000) = $40,000 The EMV for no plant = $0 so, given no survey, build the small plant

30 Decision Trees in Ethical Decision Making Maximize shareholder value and behave ethically Technique can be applied to any action a company contemplates

31 Decision Trees in Ethical Decision Making Is actio n legal? Yes No Does action maximize company returns? Yes No Is it ethical? (Weigh the affect on employees, customers, suppliers, community against shareholder benefit) Is it ethical not to take action? (Weigh the harm to shareholders vs. the benefits to other stakeholders) Yes No Yes No Do it Don t do it Don t do it Do it, but notify appropriate parties Don t do it

32 Linear Programming A mathematical technique to help plan and make decisions relative to the trade-offs necessary to allocate resources Will find the minimum or maximum value of the objective Guarantees the optimal solution to the model formulated

33 LP Applications 1. Scheduling school buses to minimize total distance traveled 2. Allocating police patrol units to high crime areas in order to minimize response time to 911 calls 3. Scheduling tellers at banks so that needs are met during each hour of the day while minimizing the total cost of labor

34 LP Applications 4. Selecting the product mix in a factory to make best use of machine- and labor-hours available while maximizing the firm s s profit 5. Picking blends of raw materials in feed mills to produce finished feed combinations at minimum costs 6. Determining the distribution system that will minimize total shipping cost

35 LP Applications 7. Developing a production schedule that will satisfy future demands for a firm s product and at the same time minimize total production and inventory costs 8. Allocating space for a tenant mix in a new shopping mall so as to maximize revenues to the leasing company

36 Requirements of an LP Problem 1. LP problems seek to maximize or minimize some quantity (usually profit or cost) expressed as an objective function 2. The presence of restrictions, or constraints, limits the degree to which we can pursue our objective

37 Requirements of an LP Problem 3. There must be alternative courses of action to choose from 4. The objective and constraints in linear programming problems must be expressed in terms of linear equations or inequalities

38 Formulating LP Problems The product-mix problem at Shader Electronics Two products 1. Shader Walkman, a portable CD/DVD player 2. Shader Watch-TV, a wristwatch-size size Internet-connected color TV Determine the mix of products that will produce the maximum profit

39 Formulating LP Problems Hours Required to Produce 1 Unit Walkman Watch-TVs Available Hours Department (X 1 ) (X 2 ) This Week Electronic Assembly Profit per unit $7 $5 Decision Variables: X 1 = number of Walkmans to be produced = number of Watch-TVs to be produced X 2 Table B.1

40 Formulating LP Problems Objective Function: Maximize Profit = $7X 1 + $5X 2 There are three types of constraints Upper limits where the amount used is the amount of a resource Lower limits where the amount used is the amount of the resource Equalities where the amount used is = the amount of the resource

41 Formulating LP Problems First Constraint: Electronic time used is Electronic time available 4X 1 + 3X (hours of electronic time) Second Constraint: Assembly is time used Assembly time available 2X 1 + 1X (hours of assembly time)

42 Graphical Solution Can be used when there are two decision variables 1. Plot the constraint equations at their limits by converting each equation to an equality 2. Identify the feasible solution space 3. Create an iso-profit line based on the objective function 4. Move this line outwards until the optimal point is identified

43 Graphical Solution X 2 Number of Watch-TVs Feasible region Assembly (constraint B) Electronics (constraint A) Number of Walkmans X 1

44 Graphical Solution Choose 100 a possible value for the objective function Number of Watch TVs Iso-Profit Line Solution Method X Solve for the axis intercepts of the function and plot the line Feasible region Assembly (constraint B) $210 = 7X 1 + 5X 2 X 2 = 42 X 1 = 30 Electronics (constraint A) Number of Walkmans X 1

45 Graphical Solution X 2 Figure B.4 Number of Watch-TVs (0, 42) $210 = $7X 1 + $5X 2 (30, 0) Number of Walkmans X 1

46 Graphical Solution X 2 Figure B.5 Number of Watch-TVs $350 = $7X 1 + $5X 2 $280 = $7X 1 + $5X 2 $210 = $7X 1 + $5X Number of Walkmans $420 = $7X 1 + $5X 2 X 1

47 Graphical Solution X 2 Figure B.6 Number of Watch-TVs Maximum profit line Optimal solution point (X 1 = 30, X 2 = 40) Number of Walkmans $410 = $7X 1 + $5X 2 X 1

48 Corner-Point Method X 2 Number of Watch-TVs Number of Walkmans X 1

49 Corner-Point Method The optimal value will always be at a corner point Find the objective function value at each corner point and choose the one with the highest profit Point 1 : (X 1 = 0, X 2 = 0) Profit $7(0) + $5(0) = $0 Point 2 : (X 1 = 0, X 2 = 80) Profit $7(0) + $5(80) = $400 Point 4 : (X 1 = 50, X 2 = 0) Profit $7(50) + $5(0) = $350

50 Corner-Point Method Solve for the intersection of two constraints 4X 1 + 3X (electronics time) 2X 1 + 1X (assembly time) 4X 1 + 3X 2 = 240-4X 1-2X 2 = X 2 = 40 4X 1 + 3(40) = 240 4X = 240 X 1 = 30

51 Corner-Point Method The optimal value will always be at a corner point Find the objective function value at each corner point and choose the one with the highest profit Point 1 : (X 1 = 0, X 2 = 0) Profit $7(0) + $5(0) = $0 Point 2 : (X 1 = 0, X 2 = 80) Profit $7(0) + $5(80) = $400 Point 4 : (X 1 = 50, X 2 = 0) Profit $7(50) + $5(0) = $350 Point 3 : (X 1 = 30, X 2 = 40) Profit $7(30) + $5(40) = $410

52 Sensitivity Analysis How sensitive the results are to parameter changes Change in the value of coefficients Change in a right-hand hand-side value of a constraint Trial-and and-error approach Analytic postoptimality method

53 Sensitivity Report

54 Changes in Resources The right-hand hand-side values of constraint equations may change as resource availability changes The shadow price of a constraint is the change in the value of the objective function resulting from a one-unit change in the right- hand-side value of the constraint

55 Changes in Resources Shadow prices are often explained as answering the question How much would you pay for one additional unit of a resource? Shadow prices are only valid over a particular range of changes in right-hand hand-side values Sensitivity reports provide the upper and lower limits of this range

56 Sensitivity Analysis X Changed assembly constraint from 2X 1 + 1X1 2 = 100 to 2X 1 + 1X1 2 = 110 Corner point 3 is still optimal, but values at this point are now X 1 = 45, X 2 = 20,, with a profit = $415 Electronics constraint is unchanged X 1 Figure B.8 (a)

57 Sensitivity Analysis X Changed assembly constraint from 2X 1 + 1X1 2 = 100 to 2X 1 + 1X1 2 = 90 Corner point 3 is still optimal, but values at this point are now X 1 = 15, X 2 = 60,, with a profit = $405 Electronics constraint is unchanged X 1 Figure B.8 (b)

58 Changes in the Objective Function A change in the coefficients in the objective function may cause a different corner point to become the optimal solution The sensitivity report shows how much objective function coefficients may change without changing the optimal solution point

59 Solving Minimization Problems Formulated and solved in much the same way as maximization problems In the graphical approach an iso- cost line is used The objective is to move the iso- cost line inwards until it reaches the lowest cost corner point

60 Minimization Example X 1 = number of tons of black-and and-white chemical produced X 2 = number of tons of color picture chemical produced Minimize total cost = 2,500X 1 + 3,000X 2 Subject to: X 1 30 tons of black-and and-white chemical X 2 20 tons of color chemical X 1 + X 2 60 tons total X 1, X 2 $0 nonnegativity requirements

61 Minimization Example X 2 60 X 1 + X 2 = Feasible region 30 b X 1 = 30 a X 2 = X 1

62 Minimization Example Total cost at a = 2,500X 1 + 3,000X 2 = 2,500 (40) + 3,000(20) = $160,000 Total cost at b = 2,500X 1 + 3,000X 2 = 2,500 (30) + 3,000(30) = $165,000 Lowest total cost is at point a

63 LP Applications Production-Mix Example Product Wiring Drilling Department Assembly Inspection Unit Profit XJ $ 9 XM $12 TR $15 BR $11 Department Capacity (in hours) Product Minimum Production Level Wiring 1,500 XJ Drilling 2,350 XM Assembly 2,600 TR Inspection 1,200 BR

64 LP Applications X 1 = number of units of XJ201 produced X 2 = number of units of XM897 produced X 3 = number of units of TR29 produced X 4 = number of units of BR788 produced Maximize profit = 9X X X X 4 subject to.5x X X 3 + 1X 4 1,500 hours of wiring 3X 1 + 1X 2 + 2X 3 + 3X 4 2,350 hours of drilling 2X 1 + 4X 2 + 1X 3 + 2X 4 2,600 hours of assembly.5x 1 + 1X 2 +.5X 3 +.5X 4 1,200 hours of inspection X units of XJ201 X units of XM897 X units of TR29 X units of BR788

65 LP Applications Diet Problem Example Feed Product Stock X Stock Y Stock Z A B C D 3 oz 2 oz 1 oz 6 oz 2 oz 3 oz 0 oz 8 oz 4 oz 1 oz 2 oz 4 oz

66 LP Applications X 1 = number of pounds of stock X purchased per cow each month X 2 = number of pounds of stock Y purchased per cow each month X 3 = number of pounds of stock Z purchased per cow each month Minimize cost =.02X X X 3 Ingredient A requirement: 3X 1 + 2X 2 + 4X 3 64 Ingredient B requirement: 2X 1 + 3X 2 + 1X 3 80 Ingredient C requirement: 1X 1 + 0X 2 + 2X 3 16 Ingredient D requirement: 6X 1 + 8X 2 + 4X Stock Z limitation: X 3 80 X 1, X 2, X 3 0 Cheapest solution is to purchase 40 pounds of grain X at a cost of $0.80 per cow

67 LP Applications Production Scheduling Example Month Manufacturing Cost Selling Price (during month) July $60 August $60 $80 September $50 $60 October $60 $70 November $70 $80 December $90 X 1, X 2, X 3, X 4, X 5, X 6 = number of units manufactured during July (first month), August (second month), etc. Y 1, Y 2, Y 3, Y 4, Y 5, Y 6 = number of units sold during July, August, etc.

68 LP Applications Maximize profit = 80Y Y Y Y Y 6 - (60X X X X X 5 ) July: I 1 = X 1 August: I 2 = I 1 + X 2 - Y 2 September: I 3 = I 2 + X 3 - Y 3 October: I 4 = I 3 + X 4 - Y 4 November: I 5 = I 4 + X 5 - Y 5 December: I 6 = I 5 + X 6 - Y 6 New decision variables: I 1, I 2, I 3, I 4, I 5, I 6 Inventory at end of this month Inventory at end of previous month Current month s production = + This month s sales

69 LP Applications Maximize profit = 80Y Y Y Y Y 6 - (60X X X X X 5 ) July: I 1 = X 1 August: I 2 = I 1 + X 2 - Y 2 September: I 3 = I 2 + X 3 - Y 3 October: I 4 = I 3 + X 4 - Y 4 November: I 5 = I 4 + X 5 - Y 5 December: I 6 = I 5 + X 6 - Y 6 I 1 100, I 2 100, I 3 100, I 4 100, I 5 100, I 6 = 0 for all Y i 300

70 LP Applications Maximize profit = 80Y Y Y Y Y 6 - (60X X X X X 5 ) July: I 1 = X 1 August: Final Solution I 2 = I 1 + X 2 - Y 2 September: I 3 = I 2 + X 3 - Y 3 October: Profit = $19,000 I 4 = I 3 + X 4 - Y 4 November: X I 5 = I 4 + X 5 - Y 1 = 100, X 2 = 200, X 3 = 400, 5 December: I 6 = I 5 + X 6 - Y 6 X 4 = 300, X 5 = 300, X 6 = 0 Y 1 = 100, Y 2 = 300, Y 3 = 300, Y 4 = 300, Y 5 = 300, Y 6 = 100 I for all Y i = 100, I 2 = 0, I 3 = 100, I 4 = 100, I 5 = 100, I 6 = 0 I 1 100, I 2 100, I 3 100, I 4 100, I 5 100, I 6 = 0

71 LP Applications Labor Scheduling Example 9 AM 10 AM 11 AM Time Period Number of Tellers Required Time Period AM - 10 AM 10 1 PM - 2 PM AM - 11 AM 12 2 PM - 3 PM 11 AM - Noon 14 3 PM - 4 PM Noon - 1 PM 16 4 PM - 5 PM Number of Tellers Required 18 PM 18 PM 17 PM 15 PM 10 F P 1 P 2 P 3 P 4 P 5 = Full-time tellers = Part-time time tellers starting at 9 AM (leaving at 1 PM) = Part-time time tellers starting at 10 AM (leaving at 2 PM = Part-time time tellers starting at 11 AM (leaving at 3 PM = Part-time time tellers starting at noon (leaving at 4 PM) = Part-time time tellers starting at 1 PM (leaving at 5 PM) PM) PM)

72 LP Applications Minimize total daily manpower cost = $75F + $24(P 1 + P 2 + P 3 + P 4 + P 5 ) F + P 1 10 (9 AM - 10 AM needs) F + P 1 + P 2 12 (10 AM - 11 AM needs) 1/2 F + P 1 + P 2 + P 3 14 (11 AM - 11 AM needs) 1/2 F + P 1 + P 2 + P 3 + P 4 16 (noon - 1 PM needs) F + P 2 + P 3 + P 4 + P 5 18 (1 PM - 2 PM needs) F + P 3 + P 4 + P 5 17 (2 PM - 3 PM needs) F + P 4 + P 5 15 (3 PM - 7 PM needs) F + P 5 10 (4 PM - 5 PM needs) F 12 4(P 1 + P 2 + P 3 + P 4 + P 5 ).50( )

73 Minimize total daily manpower cost = $75F + $24(P 1 + P 2 + P 3 + P 4 + P 5 ) s F + P 1 10 (9 AM - 10 AM needs) F + P 1 + P 2 12 (10 AM - 11 AM needs) 1/2 F + P 1 + P 2 + P 3 14 (11 AM - 11 AM needs) 1/2 F + P 1 + P 2 + P 3 + P 4 16 (noon - 1 PM needs) F + P 2 + P 3 + P 4 + P 5 18 (1 PM - 2 PM needs) F + P 3 + P 4 + P 5 17 (2 PM - 3 PM needs) F + P 4 + P 5 15 (3 PM - 7 PM needs) F + P 5 10 (4 PM - 5 PM needs) F 12 4(P 1 + P 2 + P 3 + P 4 + P 5 ).50(112) F, P 1, P 2, P 3, P 4, P 5 0

74 LP Applications Minimize total daily manpower cost = $75F + $24(P 1 + P 2 + P 3 + P 4 + P 5 ) There are two alternate optimal solutions to this problem but both will cost $1,086 per day F + P 1 10 (9 AM - 10 AM needs) F + P 1 + P 2 12 (10 AM - 11 AM needs) 1/2 F + P 1 + P 2 First + P 3 14 Second (11 AM - 11 AM needs) 1/2 F + P 1 + Solution P 2 + P 3 + P 4 16 Solution (noon - 1 PM needs) F + F P 2 + = P P 4 + P 5 18 F (1 = PM 10-2 PM needs) F + P 3 + P 4 + P 5 17 (2 PM - 3 PM P needs) F 1 = 0 P + P 4 + P = 6 (3 PM - 7 PM needs) F P 2 = 7 + P 5 10 P 2 (4 = 1 PM - 5 PM needs) F P 3 = 2 12 P 3 = 2 P 4 = 2 P 4 = 2 P 5 = 3 P 5 = 3 4(P 1 + P 2 + P 3 + P 4 + P 5 ).50(112) F, P 1, P 2, P 3, P 4, P 5 0

75 The Simplex Method Real world problems are too complex to be solved using the graphical method The simplex method is an algorithm for solving more complex problems Developed by George Dantzig in the late 1940s Most computer-based LP packages use the simplex method

76 Flair Furniture Company Hours Required to Produce One Unit Department X 1 Tables X 2 Chairs Available Hours This Week Carpentry Painting/Varnishing Profit/unit Constraints: Objective: $7 $5 4 X1 + 3 X X1 + 1 X2 100 Maximize: (carpentry) (painting & varnishing) X + X

77 Flair Furniture Company's Feasible Region & Corner Points Number of Chairs X B = (0,80) Feasible Region 4 X + 3 X C = (30,40) 2X1 + 1X1 D = (50,0) X Number of Tables

78 Flair Furniture - Adding Constraints: 4 X + 3 X2 1 2 X + 1 X2 1 Slack Variables 240 (carpentry) 100 (painting & varnishing) Constraints with Slack Variables 4 X 2 X X + 1 X S 7 X + X S Objective Function 2 = 240 (carpentry ) = 100 (painting &varnishing Objective Function with Slack Variables X + S 1 X2 S1 1 2 )

79 Flair Furniture s s Initial Simplex Tableau Profit per Production Unit Mix Column Column C j $0 $0 Real Variables Columns Slack Variables Columns Constant Column $7 $5 $0 $0 Solution Mix X 1 X 2 S 1 S 2 Quantity S 1 S 2 Z j C j - Z j $0 $0 $0 $0 $7 $5 $0 $ $0 $0 Profit per unit row Constraint equation rows Gross profit row Net profit row

80 Identified in the Initial Simplex Tableau C j $0 $0 $7 $5 $0 $0 Solution Mix X 1 X 2 S 1 S 2 Quantity S 1 S 2 Z j C j - Z j Pivot number $0 $0 $0 $0 $7 $5 $0 $0 Pivot column $0 $0 Pivot row

81 Completed Second Simplex Tableau for Flair Furniture C j $7 $0 $7 $5 $0 $0 Solution Mix X 1 X 2 S 1 S 2 Quantity X 1 S 2 1 1/2 1/ Z j C j - Z j $7 $7/2 $7/2 $0 $0 $3/2 -$7/2 $0 $350

82 Identified in Second Simplex Tableau C j $7 $0 $7 $5 $0 $0 Solution Mix X 1 X 2 S 1 S 2 Quantity X 1 S 2 Z j C j - Z j 1 1/2 1/ Pivot number $7 $7/2 $7/2 $0 $0 $3/2 -$7/2 $0 Pivot column $350 (Total Profit) Pivot row

83 Calculating the New X 1 Row for Flair s s Third Tableau ( Number in new X 1 row) Number in old X 1 row 1 0 3/2-1/2 30 = ( ) ( Number ) Corresponding - x ( ) number in above pivot number = 1 - (1/2) x = 1/2 - (1/2) x = 1/2 - (1/2) x = 0 - (1/2) x = 50 - (1/2) x new X 2 row (0) (1) (-2) (1) (40)

84 Final Simplex Tableau for the Flair Furniture Problem C j $7 $5 $7 $5 $0 $0 Solution Mix X 1 X 2 S 1 S 2 Quantity X 1 X /2-1/ Z j C j - Z j $7 5 $1/2 $3/2 $0 $0 -$1/2 -$3/2 $410

85 Simplex Steps for Maximization 1. Choose the variable with the greatest positive C j - Z j to enter the solution. 2. Determine the row to be replaced by selecting that one with the smallest (non-negative) quantity-topivot-column ratio. 3. Calculate the new values for the pivot row. 4. Calculate the new values for the other row(s). 5. Calculate the C j and C j - Z j values for this tableau. If there are any C j - Z j values greater than zero, return to Step 1.

86 Surplus & Artificial Variables Constraints 5 X 25 X X 25 X X + 30 X X + 30 X X + 8 X = 900 Constraints-Surplus & Artificial Variables S A A 1 2 = 210 = 900 Objective Function Min: 5 X X2 + 7 X3 Objective Function-Surplus & Artificial Variables Min : 5 X S MA + MA 1 X2 X

87 Simplex Steps for Minimization 1. Choose the variable with the greatest negative C j - Z j to enter the solution. 2. Determine the row to be replaced by selecting that one with the smallest (non-negative) quantity-topivot-column ratio. 3. Calculate the new values for the pivot row. 4. Calculate the new values for the other row(s). 5. Calculate the C j and C j - Z j values for this tableau. If there are any C j - Z j values less than zero, return to Step 1.

88 Special Cases Infeasibility C j M M Solution Mix X 1 X 2 S 1 S 2 A 1 A 2 Qty 5 X X M A Z j M -21-M M M C j -Z j M-31 2M+21 0

89 Special Cases Unboundedness C j Solution Mix X 1 X 2 S 1 S 2 Qty 9 X S Z j C j -Z j Pivot Column

90 Special Cases Degeneracy C j Solution Mix X 1 X 2 X 3 S 1 S 2 S 3 Qty 8 X 2 1/ S / S / Z j C j -Z j Pivot Column

91 Special Cases Multiple Optima C j Solution Mix X 1 X 2 S 1 S 2 Qty 2 X 2 3/ S /2 1 3 Z j C j -Z j

92 Sensitivity Analysis High Note Sound Company Max : 50 X 2 X 3 X Subject to : X + 4 X + 1 X

93 Sensitivity Analysis High Note Sound Company

94 Simplex Solution High Note Sound Company C j Solution Mix X 1 X 2 S 1 S 2 Qty 120 X 2 1/2 1 ¼ S 2 5/2 0-1/ Z j C j -Z j

95 Simplex Solution High Note Sound Company C j Solution Mix X 1 X 2 S 1 S 2 Qty 120 X 2 1/2 1 ¼ S 2 5/2 0-1/ Z j C j -Z j

96 Nonbasic Objective Function Coefficients C j Solution Mix X 1 X 2 S 1 S 2 Qty 120 X 2 1/2 1 ¼ S 2 5/2 0-1/ Z j C j -Z j

97 Basic Objective Function Coefficients C j Solution Mix X 1 X 2 S 1 S 2 Qty 120+ X 2 1/2 1 ¼ S 2 5/2 0-1/ Z j 60+1/ / C j -Z j -10-1/ /4 0

98 Simplex Solution High Note Sound Company C j Solution Mix X 1 X 2 S 1 S 2 Qty 120 X 2 1/2 1 ¼ S 2 5/2 0-1/ Z j C j -Z j Objective increases by 30 if 1 additional hour of electricians time is available.

99 Shadow Prices Shadow Price: Value of One Additional Unit of a Scarce Resource Found in Final Simplex Tableau in C-Z Row Negatives of Numbers in Slack Variable Column

100 Steps to Form the Dual To form the Dual: If the primal is max., the dual is min., and vice versa. The right-hand-side values of the primal constraints become the objective coefficients of the dual. The primal objective function coefficients become the right-hand-side of the dual constraints. The transpose of the primal constraint coefficients become the dual constraint coefficients. Constraint inequality signs are reversed.

101 Primal & Dual Max : Primal: 50 X 1 Subject to : 2 X 3 X X + 4 X + 1 X Dual Min : 80 U Subject 1 to : 2U 4 U U + + 3U 1U

102 Primal s Optimal Solution C j $7 $5 Comparison of the Primal and Dual Optimal Tableaus Solution Mix X 2 S 2 Z j C j - Z j Quantity $2,400 $50 $120 $0 $0 X 1 X 2 S 1 S 2 1/2 1 1/4 0 5/2 0-1/ Dual s Optimal Solution C j $7 $5 Solution Mix U 1 S 1 Z j C j - Z j Quantity $2, $0 $0 X 1 X 2 S 1 S 2 1 1/4 0-1/4 0-5/2 1-1/ $ M M A 1 A 2 0 1/2-1 1/ M M-40