Chapter 5 Scheduling the Project This chapter covers the topic of scheduling, probably the most extensively researched and published subject dealing with project management. In addition to the usual PERT and CPM networks, Gantt charts, etc., the subject of project uncertainty and risk management is also discussed. The use of computer simulation and Crystal Ball to generate the approximate distribution of project completion times to help better understand the implications associated with schedule uncertainty is also discussed. Cases and Readings A case appropriate to the subject of this chapter is: Harvard: 9-613-021 Arrow Diagramming Exercise This 3-page case describes the marketing campaign for a newly developed industrial hardware item. Over two-dozen activities are noted and described. The case asks for the network diagram and critical path. A reading appropriate to the subject of this chapter is: L.P. Leach. Critical Chain Project Management Improves Project Performance (Project Management Journal, June 1999, p. 39-51). This article explains the procedures developed by E. Goldratt in his Critical Chain approach to project management. Includes a discussion of project and feeder buffers. Projects using the critical chain often report significantly improved schedule, cost, and scope performance. Answers to Review Questions 1. What is a critical path? Answer: By definition, a critical path is the path with the longest duration. It is critical in the sense that if it is delayed the completion of the entire project will be delayed. 2. How is slack determined? Answer: Slack for a particular task is calculated by: • Subtracting the earliest time the task can start from the latest time the task can start, • Subtracting the earliest time the task can finish from the latest time the task can finish. Both calculations result in the same slack and indicate a window in which the task can be started and finished without delaying the entire project. The slack for a particular path is calculated by subtracting the path’s duration from the critical path’s duration and provides an indication of how much the path (i.e., all activities in total on that path) can be delayed without delaying the completion of the project. 3. How do you determine the ES for an activity with two predecessors? How do you determine the LF for an activity with two successors? Answer: The earliest start time for an activity with two predecessors is equal to the later (larger) earliest finish time of the two predecessors since both predecessors must be completed for the task to begin. The latest finish time for an activity with two successors is equal to the smaller (earlier) latest start time of the two successors. If the larger were used, then the preceding task would be permitted to finish after the latest allowable start time of the other successor. 4. Will all activities on a noncritical path have the same slack? Why or why not? Answer: No, all activities on the same non-critical path will not necessarily have the same slack. This is because a particular activity may be on multiple paths. When an activity is on more than one path, its slack is determined by the path with the least amount of slack. 5. For the following project, (a) List all predecessors of task 5. (b) List all predecessors of task 4. (c) List all predecessors of the network finish (F). Answer: (a) Task 4 is the only immediate predecessor of task 5; other predecessors are 1, 2, and 3. (b) Tasks 2 and 3 are both immediate predecessors of task 4 the other predecessor is 1. (c) Task 5 is the immediate predecessor of the network finish (F); the others are 1-4. 6. What is a “dummy” activity? Answer: This type of activity only is seen in activity-on-arrow diagrams. When two activity arrows have the same beginning and ending nodes they do not have a unique identity in the project network. To solve this exercise a new ending or starting node is created for one of the activities to provide them with a unique identity. Then, using a dashed arrow, a dummy activity with no duration is added to preserve the precedence relationship. 7. Consider Figure 5-14. Paths a-b-c and a-b-d converge at activity f, but we ignore this potential merge problem in the text. Why? Answer: Activities a and b are common to both paths and make the same impact on both paths, so do not need to be considered. Also, activity f is not on the critical path, so is less likely to delay the project. 8. What is meant by “project slack”? Answer: If the promised delivery date for a project is greater than the critical path time required to complete the project, the project is said to have “project slack.” The amount of the project slack is equal to the delivery time minus the expected project completion time. 9. When using AON networks, how does one indicate an event such as a project milestone? Answer: A milestone could be added as a node to the AON network with zero duration. 10. A probabilistic network has a critical path of 21 days and a .95 probability of completing this path in 24 days. Therefore, the project has a .95 chance of being finished by the end of the 24th day. True or False? Briefly explain your answer. Answer: False. Only the path claimed to be critical has a 95 percent chance of being completed within 24 days. However, there may be one or more other paths that also have a chance of taking longer than 24 days. If we are comfortable making the assumption that the paths are independent of one another, then the probability the project will be completed in 24 days or less can be calculated as the product of the probabilities that each path is finished on or before day 24. 11. “Not uncommonly, the Gantt chart is deceptive in its apparent simplicity.” Briefly explain. Answer: Because the Gantt chart is so easy to construct and read, people may use this tool with little project management training and no technical knowledge about the project. One danger is drawing conclusions and making decisions based on the relatively simple information displayed in the chart. 12. When activity times are known with certainty, the critical path is defined as the set of activities on a path from the project’s start event to its finish event that, if delayed, will delay the completion date of the project. Why must this definition be modified in situations where the activity times are not known with certainty? Are there dangers associated with not modifying the definition? Answer: In cases where the activity times are not known with certainty, it is not possible to determine the actual duration of each path. Therefore, it is not possible to determine the critical path a priori. A path that is determined to be critical at the beginning of the project based on expected activity durations may turn out not to be critical when the project is, say, half completed, perhaps as a result of the extra management attention this path received. Indeed, it is common for the status of various paths to alternate between being critical and not being critical as the project is completed. One danger is that what is thought to be the critical path at the beginning of the project consumes all of management’s attention only to have other paths fall behind and actually end up delaying the project. The implication is that all paths that have the potential to delay the project should be appropriately managed. Suggested Answers to Discussion Questions 1. Should a PM manage critical path tasks differently than noncritical path tasks? Answer: By definition, critical tasks are those tasks that if delayed will delay the completion of the entire project. Therefore, these tasks should be managed more closely than non-critical tasks. In cases where the activity times are not known with certainty, the tasks assumed to be critical at the beginning of the project may turn out not to be so critical. Therefore, when tasks times are uncertain, all tasks that may reasonably delay project completion must be carefully monitored. 2. How might you use the network approach to help prepare cost estimates? Answer: One way to use the network approach to prepare cost estimates would be to simply estimate the cost of each task in the diagram and then sum these costs up. The time estimates for the activities would likely be of significant help in estimating some of the costs (particularly when human labor is required), and then developing a cash flow schedule. To get a more accurate cost estimate of the project, you could develop three-point estimates for the costs, just like was done for the completion times, and then simulate the project taking both cost variations and schedule variations (when schedule changes would impact the activity cost) into consideration. 3. When would it be accurate to determine the probability of project completion by multiplying the probabilities of all the paths through the network together? When would it not be accurate? Answer: It would be accurate to multiply the probabilities together when the paths are independent of one another. In reality, the paths are not likely to be truly independent because the paths have activities in common and common resources are shared across paths. This latter point is particularly noteworthy as the network diagram only shows technological precedence relationships and most often does not include information about how the resources will be allocated. In many textbooks it is common to argue that while true independence across the paths is rarely met, statistical independence is achieved for large network diagrams with only a few violations. Of course, the typical homework-type problems assigned are not large enough to justify the independence assumption and the calculations required for realistically-sized network diagrams would be far too tedious for most managers. That is why simulation is the recommended approach in this text. However, it is important to note that understanding the statistical approach facilitates understanding the simulation approach. 4. Reconcile Question 5.3. If this approach is not accurate, would the probability of completion considering the critical path alone be more accurate? How might you estimate the correct probability without resorting to simulation? Answer: No, the probability would not be more accurate if only the critical path were considered unless this one path was much longer than all the other paths. In this case the other paths would have virtually no chance of delaying the entire project. When activity times are uncertain, properly calculating the probability that the project is completed by a certain date requires considering the probability that all paths are completed by the specified date. If the assumption of path independence is reasonable then the product of the probability of each path completing by the specified time can be calculated. Otherwise, simulation must be used. If the project is large, a reasonable estimate can be obtained by identifying those paths that appear to be potential possibilities of becoming the critical path at some point of the project due to the variability of both their time and variance. (That is, a time close to the critical time and a large variance.) 5. Why do you think most PMs use MSP’s Gantt chart format (see Figure 5-20) more commonly than the network format? Answer: It is much more intuitive, easier to read, and may contain much more information relevant to the project. 6. Which of the linkages in precedence diagramming do you think is most commonly used? Why? Answer: As noted in the chapter, the finish-to-start is the most commonly used linkage because typically certain activities must be completed before other activities can start. The start-to-start and finish-to-finish linkages are occasionally applicable. In these cases certain activities must either be started or finished at the same time. The start-to-finish linkage is probably used the least frequently. 7. In the calculation of variance for optimistic and pessimistic activity duration estimates made at the 95 or 90 percent level, the denominator of the fraction that approximated the standard deviation of the time distribution changes from the traditional (b - a)/6 to (b - a)/3.3 for 95 percent and to (b - a)/2.6 for 90%. Where did the 3.3 and the 2.6 come from? Answer: The 3.3 corresponding to 95 percent is based on a z-value of 1.645 which has 5 percent of the area in the upper tail. Doubling this yields the 3.3 given in the text. Note that the 3.3 is appropriately used when management specifies an optimistic time estimate that has a 95 percent chance of being achieved and a pessimistic time estimate that has only a 5 percent chance of being exceeded. The 2.6 corresponding to 90 percent is based on a z-value of 1.28 which has 10 percent of the area in the upper tail. Doubling 1.28 yields the 2.6 given in the book. Again, the 2.6 is appropriately used when management specifies an optimistic time estimate that has a 90 percent chance of being achieved and a pessimistic time estimate that has only a 10 percent chance of being exceeded. 8. Given all the estimating done to determine the duration of project activities, what does it mean to say that “only after the fact do we know which path was actually the critical path”? Answer: When activity times are uncertain, we can only estimate how long they will actually take. This means that we can only estimate the duration of the paths also. Since these are only estimates, we will not know for certain which path actually took the longest to complete until the project is complete. 9. It was noted that “the PM must manage the project team as well as the project.” Explain why. Answer: There are actually two sets of trade-offs project managers must make. Most commonly, we talk about making trade-offs between cost, schedule, and performance. However, there is often another set of trade-offs the project manager must deal with. Project managers must often make trade-offs between achieving the project goals and maintaining the health and viability of the project team. Managing this second set of trade-offs is what is meant by managing the project team while the first set of goals refers to managing the project. 10. Why do you think scheduling has been the major focus of effort throughout the history of project management rather than performance or budgeting? Answer: Performance … the performance of a project is measured by three criteria: is it on time? Is it on budget? And does it meet the agreed-upon specifications? These are outcomes which depend, to a large extent, on the effectiveness of the planning process which, in turn, depends on the accuracy of the scheduling. Budgeting … as pointed out at the beginning of the chapter, one cannot prepare a budget without knowing the specifics of each task and the time period(s) during which the task must be undertaken both for access to scarce resources and for resources whose cost depends on the duration of their use. The effectiveness of the budgeting process therefore depends on the accuracy of the scheduling. Thus, both performance and budgeting rely on the scheduling … and the more accurate and effective that is, the more likely the later stages (i.e., performance and budgeting) will be. Solutions to Exercises Note to instructor: All Crystal Ball solutions are simulations. As such, every answer will be different; although, the results will be similar to those shown. Note too that each time a student runs the simulation, they will get a different answer. 1. Refer to the network in Figure 5-14. What is the probability that path a-b-c-f will interfere with the promised project completion of 50 days? Recall that the critical path, a-b-d-g-h, had a probability of .86 for a 50-day completion. What is the probability that both paths will be complete in 50 days? Answer: The expected duration and variance of path a-b-c-f are 44.5 (10 4/6 + 12 1/6 + 12 2/6 + 9 2/6) and 6.47 (1.78 + .25 + 4.00 + .44) respectively. The probability that this path will take longer than 50 days and therefore interfere with the project completion can be calculated as follows: From Appendix A, the area in the upper tail for a z-value of 2.17 can be easily calculated as 1.5%. This means there is a 98.5% chance that this path will not interfere with the project being completed in 50 weeks. The probability that both paths finish by time 50 (assuming the paths are reasonably independent of one another) is .985 .86 = 84.71%. NB: Since both paths have activities a and b common to them, we are counting their variance twice, so our probability of 84.71% is more pessimistic than needed and the true probability is probably slightly greater than this, say 85%. 2. Refer to Table 5-4 and Figure 5-14. Recalculate the variance for each activity on the assumption that the optimistic and pessimistic estimates were made with a 95 percent probability. Recalculate the probability that the critical path will be completed in 50 days. Answer: Occasionally, students make the assumption that by 95% we mean that there is a 95% probability that the task will be completed within the range defined between the optimistic and pessimistic range. However, in these cases, (b-a) should be divided by 3.92, rather than 3.3. The spreadsheet below provides the solutions for the 99+ percent probability estimates, as well as the 95% estimates using both the 3.92 and 3.3 interpretations. Students may find the use of 3.92 more intuitive. Using this factor, the probability that path a-b-d-g-h finishes on or before time 50 is 76.5%. If students follow the text and make the assumption that there is a 5% chance of completing the tasks faster than the optimistic time estimates and there is a 5% chance that completing the tasks takes longer than the pessimistic time estimates, then the probability of path a-b-d-g-h finishing on or before time 50 is 72.8% 3. Refer to Table 5-4 and Figure 5-14. Recalculate the variance for each activity on the assumption that the optimistic and pessimistic estimates were made with a 90 percent probability. Again, recalculate the likelihood that the critical path will be finished in 50 days. Answer: If students assume that by 90% we mean that there is a 90% probability that the task will be completed with the range defined by the optimistic and pessimistic range then (b - a) should be divided by 3.29, rather than 2.6. The spreadsheet below provides the solutions for the 99+ percent probability estimates, as well as the 90% estimates using both the 3.29 and 2.6. Students may find the use of 3.29 more intuitive. Using this factor, the probability that path a-b-d-g-h finishes on or before time 50 is 72.7%. If students follow the text and make the assumption that there is a 10% chance of completing the tasks faster than the optimistic time estimates and there is a 10% chance that completing the tasks takes longer than the pessimistic time estimates, then the probability of path a-b-d-g-h finishing on or before time 50 is 68.4% 4. Given the information in the following table: Answer: a. Network diagram. b. and d. Each activity’s ES, EF, LS, and LF. Also slack. See following spreadsheet: Activity ES EF LS LF Slack A 0 4 0 4 0 B 4 10 4 10 0 C 4 8 4 8 0 D 8 10 8 10 0 E 10 14 15 19 5 F 10 15 10 15 0 G 8 11 12 15 4 H 15 19 15 19 0 I 19 21 19 21 0 c. Alternative paths … and the critical path: There are, therefore, two critical paths. d. Each activity’s slack. See spreadsheet above (section b). e. The length of time to complete the project. The project will take 21 (units of time) to complete (i.e., the length of time of the two critical paths). 5. In the following table, task durations are given in weeks. The estimates were made at the 95 percent level (see Section 5.2, Calculating Probabilistic Activity Times subsection). Answer: a. The spreadsheet below can be used to calculate the expected time and variance for each activity. (See our note on the use of 3.92 in Exercise 3.) b. The precedence diagram using an AOA (Activity on Arrow) network is as follows: The critical path is a-c-e-g since it has the longest expected duration of 19.5 weeks. Note: although this path has the longest expected duration, given the uncertainty associated with the activity times, it may not be the path with the longest actual duration. c. Referring to column H in the spreadsheet above, the probability of completing path a-c-e-g in 23 weeks or less is 93.4%. d. Both of the other two paths have virtually a 100% chance of being completed by week 23. e. As shown in the spreadsheet, there is a 93.1% probability of completing the entire project by week 23. 6. Given an auditing project with the following activities: Answer: For this problem, activities not on the critical path can be ignored (not enough information is given to draw a PERT chart). Those activities on the critical path are used to calculate TE and the project variance. Activity Duration Std. Dev. Variance A 2 2 4 C 4 0 0 E 1 1 1 G 4 2 4 H 2 0 0 Total 13 9 Therefore the critical path expected time is 13 weeks and path standard deviation is 3 weeks. Using this information gives the following probabilities for parts a-c: Completion Time Probability Less Than 12 0.37 NORM.DIST(12,13,3,TRUE) 13 0.50 NORM.DIST(13,13,3,TRUE) 16 0.84 NORM.DIST(16,13,3,TRUE) d. The number of weeks required to assure a 92.5 percent chance of completion, as guaranteed by the auditing firm. Or NORM.INV(0.925,13,3) 7. Resolve the previous exercise using computer simulation and compare your answers. Explain any differences you observe. Which methodology do you have more confidence in and why? Answer: A model using Crystal Ball was created to simulate the completion of the critical path as shown below: The expected duration and standard deviation were entered in rows 3 and 4, respectively. Assumption cells were entered in row 5 using the normal distribution since we have the mean and standard deviation (and not the optimistic, most likely, and pessimistic time estimates). Also note that activities c and f are known with certainty and therefore are constants in row 5. Finally, cell B6 sums up the simulated activity times generated by Crystal Ball in row 5 and was defined as the Forecast cell. The distribution of the completion time of the critical path is shown below based on 10,000 replications: A comparison of the results using the analytical method and simulation is shown below in the table. Question Analytical Approach Simulation a .37 .38 b .50 .50 c .84 .84 d 17.32 17.24 The results are very similar. Thus in the case of estimating probabilities for a single path, both the simulation and analytical approach yield similar results. In more realistic situations where the goal is to estimate the probability of the entire project being completed by a specified time, simulation will yield more accurate results due to the assumptions underlying the analytical approach. 8. Referring to the previous two questions, modify your simulation model to develop distributions for the slack time of each path. What do these distributions tell you? Answer: In Exercise 6a it was noted that the client desired a completion time of 12 weeks. The spreadsheet created for Exercise 7 was modified to calculate the path slack from 12 weeks: More specifically, cell B7 contains the formula =12-B6. The distribution of slack is shown below based on 10,000 replications of the simulation model. The average slack was 0.99 (standard deviation of 2.97) implying on average the project finished almost one week earlier than the desired 12 weeks. The minimum and maximum slack were -9.75 and 12.81, respectively. 9. Given the following information regarding a project involving an initial public offering (IPO): Answer: a. c. PROJECT COMPLETION TIME = 9 weeks d. he has two weeks of slack so a one or two week delay does not delay the project, but a two week delay will create another critical path. Delaying e three weeks creates a one week delay and replaces the previous critical path. 10. Enter the following information into an Excel spreadsheet. The time estimates were made at the 90 percent level (see Section 5.2, Calculating Probabilistic Activity Times subsection). All activity times are in days. Answer: a. Draw the network b. Calculate the expected time and variance for each activity. Recall that since the estimates were made at 90 percent, you must divide by 2.6 to get the standard deviation and then square that number to get the variance. Activity Optimistic Normal Pessimistic Expected Time Standard Deviation Variance a 5 6 9 6.33 1.5385 2.3669 b 4 4 6 4.33 0.7692 0.5917 c 7 9 15 9.67 3.0769 9.4675 d 6 6 6 6.00 0.0000 0.0000 e 4 5 7 5.17 1.1538 1.3314 f 12 16 17 15.50 1.9231 3.6982 g 8 12 20 12.67 4.6154 21.3018 h 7 9 16 9.83 3.4615 11.9822 i 10 14 18 14.00 3.0769 9.4675 j 6 12 20 12.33 5.3846 28.9941 k 7 9 14 9.50 2.6923 7.2485 c. Using the expected times, find the critical path and critical time. d. Find the probability that the critical path will be completed in 38 days or less. A Z value of 0.4308 yields a probability of 66.67 percent. e. Are there any serous sources of merge bias? There are a number of mergers, all at the last activity. Computing the expected completion time, standard deviation, and probability of exceeding 38 days for each path yields the following: Expected Time Path Variance Path Standard Deviation P= Max Path Duration, 1, 0) Thus when the path is the longest path the Forecast cell takes on a value of 1, otherwise it is assigned a 0. The average of the Forecast cell is the percent of times the path is the longest path or the critical path. Based on simulation the model 10,000 times, it is estimated that this path has a 74% chance of being the critical path. Given this, there is a 24% chance that other paths will be the critical path which reinforces the need to monitor all paths that can potentially delay the project’s completion. Launching E-Collar Question 1: The activities, paths, and times are shown below: Questions 2 and 3: The results will vary given the randomness inherent in simulation analysis. Below are the results from 3 different replications of the Crystal Ball model. Note that each replication below is based on simulating the project 1,000 times. Replication Mean and Std Dev of Completion Time Mean and Std Dev of Amount Client Billed Probability of Costs < $100,000 Probability Project will Take Between 350 to 450 Hours 1 412.46; 67.53 $91,034; $19,324 69.96% 51.71% 2 413.71; 66.99 $91,074; $18,939 70.22% 53.07% 3 414.32; 66.39 $91,434; $18,874 68.20% 53.46% Question 4: Providing an estimate for the project cost and duration depends on the certainty level desired. The table below summarizes the cost and duration such that you have the probability specified of not exceeding the estimates (estimates in the table are based on Replication 3 above). For example, there is a 90% probability that the costs and duration of the project will be less than or equal to $116,442 and 502.76 hours, respectively. Note that because the data is skewed, the 50% estimates in the table below differ from the means in the table above. Probability of Not Exceeding Costs and Duration Costs Duration (hours) 50% $90,730 410.88 60% $95,502 430.42 70% $100,756 449.70 80% $107,097 471.72 90% $116,442 502.76 95% $124,013 523.99 99% $139,818 576.73 Below is sample of the way the Crystal Ball model can be set up with assumption cells shaded green and the forecast cells shaded in blue. Solution Manual for Project Management in Practice Jack R. Meredith , Scott M. Shafer , Samuel J. Mantel 9781119385622
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