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1 1 of 14 4/27/2009 7:45 AM Chapter 7 - Network Models in Project Management INTRODUCTION Most realistic projects that organizations like Microsoft, General Motors, or the U.S. Defense Department undertake are large and complex. A builder putting up an office building, for example, must complete thousands of activities costing millions of dollars. NASA must inspect countless components before it launches a rocket. Avondale shipyards in New Orleans requires tens of thousands of steps in constructing an ocean-going tugboat. Almost every industries worries about how to manage similar large-scale, complicated projects effectively. It is a difficult problem, and the stakes are high. Millions of dollars in coast overruns have been wasted due to poor planning of projects. Unnecessary delays have occurred due to poor scheduling. How can such problems be solved? The first step in planning and scheduling a project is to develop the work breakdown structure. This involves identifying the activities that must be performed in the project. There may be varying levels of detail, and each activity may be broken in to its most basic components. The time, cost, resource requirements, predecessors and person(s) responsible are identified for each activity. When this has been done, a schedule for the project can be developed. The program evaluation and review technique (PERT) and the critical path method (CPM) are two popular quantitative analysis techniques that help managers plan, schedule, monitor, and control large and complex projects. They were developed because there was a critical need for a better way to manage. FRAMEWORK OF PERT AND CPM There are six steps common to both PERT and CPM. The procedure follows: Six Steps of PERT and CPM 1. Define the project and all of its significant activities or tasks. 2. Develop the relationships among the activities. Decide which activities must precede others Draw the network connecting all of the activities. Assign time and /or cost estimates to each activity. 5. Compute the longest time path through the network; this is called the critical path. 6. Use the network to help plan, schedule, monitor, and control the project. Finding the critical path is a major part of controlling a project. The activities on the critical path represent tasks that will delay the entire project if they are delayed. Managers drive flexibility by identifying non critical activities and replanning, rescheduling, and reallocating resources such as personnel and finances. Although PERT and CPM are similar in their basic approach, they do differ in the way activity times are estimated. For every PERT activity, three times estimates are combined to determine the expected activity completion time and its variance. Thus, PERT is a probabilistic technique; it allows us to find the probability that the entire project will be completed by any given date. On the other hand, CPM is called a deterministic approach. It uses to times estimates,

2 2 of 14 4/27/2009 7:45 AM PERT other hand, CPM is called a deterministic approach. It uses to times estimates, the normal time and the crash time, for each activity. The normal completion time is the time we estimate it will take under normal conditions to complete the activity. The crash completion time is the shortest time it would take to finish an activity if additional funds and resources were allocated to the task. In this chapter we investigate not only PERT and CPM, but also the technique called PERT/cost that combines the benefits of both PERT and CPM. Almost any large project can be sub divided into a series of smaller activities or tasks that can be analyzed with PERT. When you recognize that projects can have thousands of specific activities, you see why it is important to be able to answer such questions as the following: 1. When will the entire project be completed? 2. What are the critical activities or tasks in the project, that is, the ones that will delay the entire project if they are late? 3. Which are the noncritical activities, that is, the ones that can run late without delaying the entire project s completion? 4. What is the probability that the project will be completed by a specific date? 5. At any particular date, is the project on schedule, behind schedule, or ahead of schedule? 6. On any given date, is the money spent equal to, less than, or greater than the budgeted amount? 7. Are there enough resources available to finish the project on time? 8. If the project is to be finished in a shorter amount of time, what is the best way to accomplish this at the least cost? PERT (or PERT/Cost) can help answer each of these questions. The first step is to define the project and all project activities. When the project begins, the building of the internal components for the device (activity A) and the modifications that are necessary for the floor and roof (activity B) can be started. The construction of the collection stack (activity C) can begin once the internal components are completed, and pouring of the new concrete floor and installation of the frame (activity D) can be completed as soon as the roof and floor have been modified. After the collection stack has been constructed, the high-temperature burner can be built (activity E), and the installation of the pollution control system (activity F) can begin. The air pollution device can be installed (activity G) after the high-temperature burner has been built, the concrete floor has been poured, and the frame has been installed. Finally, after the control system and pollution device have been installed, the system can be inspected and tested (activity H). All of these activities seem rather confusing and complex until they are placed in a network. First, all of the activities must be listed. This information is shown in Table 7.1. We see in the table that before the collection stack can be constructed (activity C), the internal components must be built (activity A). Thus, activity A is the immediate predecessor of activity C. Similarly, both

3 3 of 14 4/27/2009 7:45 AM activities D and E must be performed just prior to installation of the air pollution device (activity G). Table 7.1 Activities and Immediate Predecessors for General Foundry, Inc. Activity Times The next step in the PERT procedure is to assign estimates of the time required to complete each activity. Time is usually given in units of weeks. For one-of-a-kind projects or for new jobs, providing activity time estimates is not always an easy task. Without solid historical data, managers are often uncertain as to activity times. For this reason, the developers of PERT employed a probability distribution based on three time estimates for each activity: Optimistic time (a) = time an activity will take if everything goes as well as possible. There should be only a small probability (say, 1/100) of this occurring. Pessimistic time (b) = time an activity would take assuming very unfavorable conditions. There should also be only a small probability that the activity will really take this long. Most likely time (m) = most realistic time estimate to complete the activity. PERT often assumes that time estimates follow the beta probability distribution (see Figure 7.1). This continuous distribution has been found to be appropriate, in many cases, for determining an expected value and variance for activity completion times. To find the expected activity time (t), the beta distribution weights the estimates as follows: t = To compute the dispersion or variance of activity completion time, we use this formula1: Variance =

standard deviations from the mean). Because (b-a) is 6 standard deviations, one standard deviation is (b a)/6. Thus the variance is [(b a)/6] 2. Table 7.

4 4 of 14 4/27/2009 7:45 AM Figure 7.1 Beta Probability Distribution with Three Time Estimates This formula is based on the statistical concept that from one end of the beta distribution to the other is 6 standard deviations (±3 standard deviations from the mean). Because (b-a) is 6 standard deviations, one standard deviation is (b a)/6. Thus the variance is [(b a)/6] 2. Table 7.2 Time Estimates (Weeks) for General Foundry, Inc Table 7.2 shows General Foundry s optimistic, most likely, and pessimistic time estimates for each activity. It also reveals the expected time (t) and variance for each of the activities, as computed with Equations 7.1 and 7.2. HOW TO FIND THE CRITICAL PATH Once the expected completion time for each activity has been determined, we accept it as the actual time of that task. Although Table 7.2 indicates that the total expected time for all eight of

5 of 14 4/27/2009 7:45 AM Although Table 7.2 indicates that the total expected time for all eight of General Foundry s activities is 25 weeks, it is obvious in Figure 13.

5 5 of 14 4/27/2009 7:45 AM Although Table 7.2 indicates that the total expected time for all eight of General Foundry s activities is 25 weeks, it is obvious in Figure 13.3 that several of the tasks can be taking place simultaneously. To find out just how long the project will take, we perform the critical path analysis for the network. The critical path is the longest time path route through the network. If Lester Harky wants to reduce the total project time for General Foundry, he will have to reduce the length of some activity on the critical path. Conversely, and delay of an activity on the critical path will delay completion of the entire project. Figure 7.2 General Foundry s Network with Excepted Activity Times To find the critical path, we need to determine the following quantities for each activity in the network: 1. Earliest start time (ES): the earliest time an activity can begin without violation of immediate predecessor requirements. 2. Earliest finish time (EF): the earliest time at which an activity can end. 3. Latest start time (LS): the latest time an activity can begin without delaying the entire project. 4. Latest finish time (LF): the latest time an activity can end without delaying the entire project. In the network, we represent these times as well as the activity times (t) in the nodes, as seen here: We first show how to determine the earliest times. When we find these, the latest times can be computed. Earliest Times There are two basic rules to follow when computing ES and EF times. The first rule is for the earliest finish time, which is computed as follows: activity time Earliest finish time = earliest start time + expected

The second rule is for the earliest start time, which is computed as follows: Earliest start = largest of the earliest finish times of immediate predecessors ES = largest EF of immediate predecessors.

6 6 of 14 4/27/2009 7:45 AM EF = ES + t Also, before any activity can be started, all of its predecessor activities must be completed. In other words, we search for the largest EF for all of the immediate predecessors in determining ES. The second rule is for the earliest start time, which is computed as follows: Earliest start = largest of the earliest finish times of immediate predecessors ES = largest EF of immediate predecessors. The start of the whole project will be set at time zero. Therefore, any activity that has no predecessors will have an earliest start time of zero. So ES = 0 for both A and B in the General Foundry problem, as seen here: The rest of the earliest times for General Foundry are shown in Figure 7.4. These are found using a forward pass through the network. At each step, EF = ES + t, and ES is the largest EF of the predecessors. Notice that activity G has an earliest start time of 8, since both D (with EF = 7) and E (with EF = 8) are immediate predecessors. Activity G cannot start until both predecessors are finished, and so we choose the larger of the earliest finish times for these. Thus, G has ES = 8. The finish time for the project will be 15 weeks, which is the EF for activity H. Figure 7.3 General Foundry s Earliest Start (ES) and Earliest Finish (EF) Times Latest Times The next step in finding the critical path is to compute the latest start time (LS) and the latest finish time (LF) for each activity. We do this by making a

7 7 of 14 4/27/2009 7:45 AM backward pass through the network, that is, starting at the finish and working backward. There are two basic rules to follow when computing the latest times. The first rule involves the latest start time, which is computed as Latest start time = latest finish time activity time LS = LF t Also, since all immediate predecessors must be finished before an activity can begin, the latest start time for an activity determines the latest finish time for its immediate predecessors. If an activity is the immediate predecessor for two or more activities, it must be finished so that all following activities can begin by their latest start times. Thus, the second rule involves the latest finish time, which is computed as Latest finish time = smallest of latest start times for following activities, or LF = smallest LS of following activities To compute the latest times, we start at the finish and work backwards. Since the finish time for the General Foundry project is 15, activity H has LF = 15. The latest start for activity H is LS = LF t = 15 2 = 13 weeks Continuing to work backward, this latest start time of 13 becomes the latest finish time for immediate predecessors F and G. All of the latest times are shown in Figure Notice that for activity C, which is the immediate predecessor for two activities (E and F), the latest finish time is the smaller of the latest start times (4 and 10) for activities E and F. Concept of Slack in Critical Path Computations When ES, LS, EF, and LF have been determined, it is a simple matter to find the amount of slack time, or free time, that each activity has. Slack is the length of time an activity can be delayed without delaying the whole project. Mathematically, Slack = LS ES or slack = LF EF Figure 7.4 General Foundry s Latest Start (LS) and Latest Finish (LF) Times

8 of 14 4/27/2009 7:45 AM Table 7.3 summarizes the ES, EF, LS, LF, and slack times for all of General Foundry s activities.

This means that it can be delayed up to 1 week without causing the project to run any longer than expected. Table 7.

Because of this, they are called critical activities and are said to be on the critical path. Lester Harky s critical path is shown in network form in Figure 7.5.

8 8 of 14 4/27/2009 7:45 AM Table 7.3 summarizes the ES, EF, LS, LF, and slack times for all of General Foundry s activities. Activity B, for example, has 1 week of slack time since LS ES = 1 0 = 1 (or, similarly, LF EF = 4 3 = 1). This means that it can be delayed up to 1 week without causing the project to run any longer than expected. Table 7.3 General Foundry s Schedule and Slack Times On the other hand, activities A, C, E, G, and H have no slack time; this means that none of them can be delayed without delaying the entire project. Because of this, they are called critical activities and are said to be on the critical path. Lester Harky s critical path is shown in network form in Figure 7.5. The total project completion time, 15 weeks, is seen as the largest number in the EF or LF columns of Table 7.3. Industrial managers call this a boundary timetable. Figure 7.5 General Foundry s Critical Path (A-C-E-G-H) PROBABILITY OF PROJECT COMPLETION The critical path analysis helped us determine that the foundry s expected project completion time is 15 weeks. Harky knows, however, that if the project is not completed in 16 weeks, General Foundry will be forced to close by

Variation in activities that are on the critical path can affect overall project completion possibly delaying it. This is one occurrence that worries Harky considerably.

9 9 of 14 4/27/2009 7:45 AM is not completed in 16 weeks, General Foundry will be forced to close by environmental controllers. He is also aware that there is significant variation in the time estimates for several activities. Variation in activities that are on the critical path can affect overall project completion possibly delaying it. This is one occurrence that worries Harky considerably. PERT uses the variance of critical path activities to help determine the variance of the overall project. If the activity times are statistically independent, the project variance is computed by summing the variances of the critical activities. Project variance = variances of activities on the critical path From Table 7.2 we know that Hence, the project variance is Project variance = 4/36 + 4/ / /36 + 4/36 = 112/36 = Figure 7.6 Probability Distribution for Project Completion Times We know that the standard deviation is just the square root of the variance, so Project standard deviation = = = 1.76 weeks How can this information be used to help answer questions regarding the probability of finishing the project on time? In addition to assuming that the activity times are independent, we also assume that total project completion time follows a normal probability distribution. With these assumptions, the bell-shaped curve shown in Figure 7.7 can be used to represent project

10 10 of 14 4/27/2009 7:45 AM bell-shaped curve shown in Figure 7.7 can be used to represent project completion dates. It also means that there is a 50% chance that the entire project will be completed in less than the expected 15 weeks and a 50% chance that it will exceed 15 weeks.2 For Harky to find the probability that his project will be finished on or before the 16-week deadline, he needs to determine the appropriate area under the normal curve. The standard normal equation can be applied as follows: Z = Where = = 0.57 Z is the number of standard deviations the due date or target date lies from the mean or expected date. Referring to the normal table in Appendix A, we find a probability of Thus, there is a 71.6% chance that the pollution control equipment can be put in place in 16 weeks or less. This is shown in Figure You should be aware that non-critical activities also have variability (as seen in Table 7.2). In fact, a different critical path can evolve because of the probabilistic situation. Figure 7.7 Probability of general Foundry s Meeting the 16-Week Deadlines WHAT PERT WAS ABLE TO PROVIDE PERT has thus far been able to provide Lester Harky with several valuable pieces of management information: 1. The project s expected completion data is 15 weeks. 2. There is a 71.6% chance that the equipment will be in place within the 16-week deadline. PERT can easily find the probability of finishing by any date Harky is interested in. 3. Five activities (A, C, E, G, H) are on the critical path. If any one of them is delayed for any reason, the entire project will be delayed. 4. Three activities (B, D, F) are not critical but have some slack time built in. This means that Harky can borrow from their resources, if needed, possibly to speed up the entire project.

11 11 of 14 4/27/2009 7:45 AM 5. A detailed schedule of activity starting and ending dates has been made available (see Table 7.3). SHORTEST-ROUTE TECHNIQUE The shortest-route technique finds how a person or item can travel from one location to another while minimizing the total distance traveled. In other words, it finds the shortest route to a series of destinations. Every day, Ray Design, Inc., must transport beds, chairs, and other furniture items from the factory to the warehouse. This involves going through several cities. Ray would like to find the route with the shortest distance. The road network is shown in figure 7.8 Figure 7.8 Roads from Ray s Plant to Warehouse STEPS OF THE SHORTEST-ROUTE TECHNIQUE Find the nearest node to the origin (plant). Put the distance in a box by the node. Find the next-nearest node to the origin (plant), and put the distance in a box by the node. In some cases, several paths will have to be checked to find the nearest node. Repeat this process until you have gone through the entire network. The last distance at the ending node will be the distance of the shortest route. You should note that the distance placed in the box by the each node is the shortest route to this node. These distances are used as intermediate results in finding the next-nearest node. Looking Figure 7.8, we can see that the nearest node to the plant is node 2, with a distance of 100 miles. Thus we will connect these two nodes. This first iteration is shown in Figure 7.9 Figure 7.9 Roads from Ray s Plant to Warehouse

12 12 of 14 4/27/2009 7:45 AM Now we look for the next-nearest node to the origin. We check nodes 3, 4 and 5. Node 3 is the nearest, but there are two possible paths. Path is nearest to the origin, with a total distance of 150 miles (See Figure 7.10) Figure 7.10 Roads from Ray s Plant to Warehouse

13 of 14 4/27/2009 7:45 AM We repeat the process. The next-nearest node is either node 4 or node 5. Node 4 is 200 miles from node 2, and node 2 is 100 miles from node 1.

13 13 of 14 4/27/2009 7:45 AM We repeat the process. The next-nearest node is either node 4 or node 5. Node 4 is 200 miles from node 2, and node 2 is 100 miles from node 1. Thus, node 4 is 300 miles from the origin. There are two paths for node 5, 2-5 and 3-5, to the origin. Note that we don t have to go all the way back to the origin because we already know the shortest route from node 2 and node 3 to the origin. The minimum distances are placed in boxes by these nodes. Path 2-5 is 100 miles, and node 2 is 100 miles from the origin. Thus the total distance is 200 miles. In a similar fashion, we can determine that the path from node 5 to the origin through node 3 is 190 (40 miles between node 5 and 3 plus 150 miles from node 3 to the origin). Thus, we pick node 5 going through node 3 to the origin (See Figure 7.11). Figure 7.11 Roads from Ray s Plant to Warehouse The next-nearest node will be either node 4 or node 6, as the last remaining nodes. Node 4 is 300 miles from the origin (300=200 from node 4 to node 2 plus100 from node 2 to the origin). Node 6 is 290 miles from the origin (290= ). Node 6 has the minimum distance, and because it is the ending node, we are done (refer to Figure 7.12). The shortest route is path , with a minimum distance of 290 miles. Figure 7.12 Roads from Ray s Plant to Warehouse

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Project Management Chapter 13

Lecture 12 Project Management Chapter 13 Introduction n Managing large-scale, complicated projects effectively is a difficult problem and the stakes are high. n The first step in planning and scheduling