Ch05

5/Scheduling the Project

Chapter 5Scheduling the Project

This chapter covers the topic of scheduling, probably the most extensively covered 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 to generate the approximate distribution of project completion times is also discussed. Appendix C illustrates how Crystal Ball can facilitate this analysis and be used to help better understand the implications of schedule uncertainty.

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. 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 task times are uncertain, all tasks that may reasonably delay project completion must be carefully managed.)

2. Slack for a particular task is calculated by subtracting the earliest time the task can start from the latest time the task can start or by 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

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the path’s duration from the critical path’s duration and provides an indication of how much the path can be delayed without delaying the completion of the project.

3. The earliest start time for an activity with two predecessors is equal to the later 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 latest start time of the two successors. If the larger were used, then the preceding task would be permitted to finish after the latest start time of the other successor.

4. 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. (As noted in Section 3.3, p.64, only immediate predecessors should be listed.)a. Task 4 is the only immediate predecessor of task 5. b. b) Task 2 and 3 are both immediate predecessors of task 4. c. Task 5 is the immediate predecessor of the network finish (F).

6. When two activities have the same beginning and ending nodes they do not have a unique identity in the project network. To solve this problem a new ending or starting node is created for one of the activities to provide them with a unique identity. Then a dummy activity or an activity with no duration is added to preserve the precedence relationship.

7. Activities a and b are common to both paths and so do not need to be considered. They must take the same impact on both paths. We did consider the partial paths d-g-h and c-f.

8. If the promised delivery date for a project is greater than the 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 project completion time.

9. A milestone could be added as a node to the AON network with zero duration.

10. 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. 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

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danger is drawing conclusions and making decisions based on the relatively simple information displayed in the chart.

12. 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 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

13. 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 developing a cash flow schedule.

14. 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.

15. No, the probability would not be more accurate if only the critical path was 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.

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16. It is much more intuitive, easier to read, and may contain much more information relevant to the project.

17. 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.

18. On page 121 we begin this discussion the way project managers traditionally think about these probabilities, i.e., an estimate for a such that the actual duration will be a or less some specified percent of the time and an estimate for b such that the actual duration will be b or larger less than some percent of the time. From this perspective the ensuing discussion on page 122 is misleading since the probabilities are now defined in terms of the areas between a and b. Thus, 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 (consistent with the discussion on page 121). 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.

If the discussion on page 122 is used the percent refers to the probability of the project being completed within the range of the optimistic and pessimistic time estimates. In the case of 95 percent estimates the appropriate value would provide 0.025 percent of the area in each tail. For the standard normal distribution this corresponds to a z-value of 1.96. In other words, + 1.96 standard deviations (or 3.92 standard deviations) encompasses 95 percent of the area under the standard normal curve. Likewise 3.29 standard deviations encompass 90 percent of the area under the standard normal curve. Therefore, based on the discussion in the text, it is possible that students will interpret these probabilities in two different ways. Given this, we provide solutions for the end-of-chapter problems based on both interpretations.

19. 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.

20. 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.

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Namely, project managers must often make trade-offs between achieving the project goals and the project team viability. 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.

Solutions to Problems

21. The expected duration and variance of path a-b-c-f are 44.5 and 6.47, respectively. The probability that this path will take longer than 50 weeks 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%.

22. (Note to instructor. Refer to Discussion Question 18 for a discussion of two possible interpretations for the probabilities associated with these types of problems). If we assume that by 95% we mean that there is a 95% probability that the task will be completed within the range defined by the optimistic and pessimistic range then (b-a) should be divided by 3.92, rather than 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. 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%.

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23. (Note to instructor. Please refer to answer to Discussion Question 18.) If we 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%.

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12345678910111213141516171819

A B C D E F G HOpt. Norm. Pess. Var. Var. Var.

Activity a m b T E ((b-a)/6)2 ((b-a)/3.92)2 ((b-a)/3.3)2

a 8 10 16 10.67 1.78 4.16 5.88b 11 12 14 12.17 0.25 0.59 0.83c 7 12 19 12.33 4.00 9.37 13.22d 6 6 6 6.00 0.00 0.00 0.00e 10 14 20 14.33 2.78 6.51 9.18f 6 10 10 9.33 0.44 1.04 1.47g 5 10 17 10.33 4.00 9.37 13.22h 4 8 11 7.83 1.36 3.19 4.50

a-b-d-g-h 47.00 7.39 17.31 24.43Std Dev 2.72 4.16 4.94Prob 86.5% 76.5% 72.8%

Key FormulasCell E12 =E3+E4+E6+E9+E10 {copy to cells F12:H12}Cell F13 =SQRT(F12) {copy to cells G13:H13}Cell F 14 =NORMDIST(50,47,F13,TRUE) {copy to cells G14:H14}

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a 8 10 16 10.67 1.78 5.91 9.47b 11 12 14 12.17 0.25 0.83 1.33c 7 12 19 12.33 4.00 13.30 21.30d 6 6 6 6.00 0.00 0.00 0.00e 10 14 20 14.33 2.78 9.24 14.79f 6 10 10 9.33 0.44 1.48 2.37g 5 10 17 10.33 4.00 13.30 21.30h 4 8 11 7.83 1.36 4.53 7.25

a-b-d-g-h 47.00 7.39 24.57 39.35Std Dev 2.72 4.96 6.27Prob 86.5% 72.7% 68.4%

Key FormulasCell E12 =E3+E4+E6+E9+E10 {copy to cells F12:H12}Cell F13 =SQRT(F12) {copy to cells G13:H13}Cell F 14 =NORMDIST(50,47,F13,TRUE) {copy to cells G14:H14}


24. Note to instructor. In order to familiarize student with MSP’s “Help” facilities, we have not given precise instructions in the text for changing MSP’s project calendars, and for using MSP to find slack. If you wish to make these available, they follow:

The MSP calendar must be reset to a 7-day work week from its usual default of a 5-day work week. To reset the MSP calendar, click “Tools,” and then click “Change working time.” On the calendar you will note that Saturdays and Sundays for June and July are shaded, i.e., nonworking time. Click on each of these days and then click on “Working time.” Alternatively, you may highlight the entire months of June and July and then click on “Working time.”

To show slack or float in MSP, click on “View,” then on “More views,” on “Detail Gantt,” and “Apply.” Activity slack will be shown on the Gantt chart. Then click on “View,” “Table,” and then “Schedule.” Drag the divider bar to the right, and you will find “Total slack” and “Free slack” listed in the MSP table. (For these instructions to take effect, project activity data must have been entered into MSP.)

One way to find the critical path is to identify all paths and calculate their duration. As shown below, this project has six paths and path b-e-h-j-l is the critical path with a duration of 40.

Path Duration Critical?a-d-i-k 29 Noa-d-h-j-k 35 Nob-e-i-k 35 Nob-e-h-j-l 35 Yesb-f-j-l 35 Noc-g-j-l 35 No

The slack for each task is calculated as follows:

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1

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7g

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25. a. The spreadsheet below was used to calculate the expected time and

variance for each activity. b. The precedence diagram may be drawn as follows:

c. Path a-c-e-g 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.) Referring to column G in the spreadsheet below, we see the probability of completing path a-c-e-g in 23 weeks or less is 96.3%.

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Activity ES EF LS LF Slacka 0 5 5 10 5b 0 7 0 7 0c 0 4 14 18 14d 5 11 10 16 5e 7 16 7 16 0f 7 13 16 22 9g 4 8 18 22 14h 16 22 16 22 0I 16 24 22 30 6j 22 31 22 31 0k 24 34 30 40 6l 31 40 31 40 0

a

b

c

d

e

f

g


Spreadsheet for Problem 25

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 96.3% probability of completing the project by week 23.

26. a.

b. The expected time and variance for each activity are calculated in the spreadsheet below.

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1011121314151617181920212223242526272829



a 2 4 6 4.00 0.44 1.04 1.47b 3 5 9 5.33 1.00 2.34 3.31c 4 5 7 5.17 0.25 0.59 0.83d 4 6 10 6.33 1.00 2.34 3.31e 4 5 7 5.17 0.25 0.59 0.83f 3 4 8 4.50 0.69 1.63 2.30g 3 5 8 5.17 0.69 1.63 2.30

Expected Variance Variance VariancePaths Duration Column F Column G Column H

a-d-f 14.83 2.14 5.01 7.07a-c-e-g 19.50 1.64 3.84 5.42b-e-g 15.67 1.94 4.56 6.43

Prob Path Fin by 23a-d-f 100.0% 100.0% 99.9%a-c-e-g 99.7% 96.3% 93.4%b-e-g 100.0% 100.0% 99.8%

Prob Project Fin by 23 99.7% 96.3% 93.1%

Key Formulas:Cell B13 =E3+E6+E8 {copy to cells C13:E13}Cell B14 =E3+E5+E7+E9 {copy to cels C14:E14}Cell B15 =E4+E7+E9 {copy to cells C15:E15}Cell C18 =NORMDIST(23,$B13,SQRT(C13),TRUE) {copy to cells D18:E18 and C19:E20}Cell C22 =PRODUCT(C18:C20) {copy to cells D22:E22}

a

b

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d

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kg


c. As shown in the spreadsheet above, this project has five paths. Path c-g-i has the longest expected duration of 34.67 days although this path may turn out not to be the critical path given the uncertainty associated with the activity times.

d. As is calculated in the above spreadsheet, the probability that path c-g-i will be completed in 38 days or less is 70.7% assuming that the 90% refers to the probability that of each activity’s duration occurring between the optimistic and pessimistic time estimate (see answer to Discussion Question 18)

e. Assuming the paths are independent, for a 38 day delivery, paths a, d, i and b, e, i values greater than 3.0 and with associated probabilities almost 1.0, they can be ignored. The probabilities for paths b, f, i and c, h, k are .84 and .95 respectively. Given the probability that the nominal critical path is .667 (see Q. 25d), the chance that all three paths will be completed in 38 days is the product of their individual probabilities, .55.

f. Column F in the spreadsheet above corresponds to the 99+ percent level. According to the calculations shown, there is an 84% chance that the project will be completed in 38 days or less. The difference in probabilities is caused by the fact that the estimation of a and b at the 99+

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1011121314151617181920212223242526272829



a 5 6 9 6.33 0.44 1.48 2.37b 4 4 6 4.33 0.11 0.37 0.59c 7 9 15 9.67 1.78 5.91 9.47d 6 6 6 6.00 0.00 0.00 0.00e 4 5 7 5.17 0.25 0.83 1.33f 12 16 17 15.50 0.69 2.31 3.70g 8 12 20 12.67 4.00 13.30 21.30h 7 9 16 9.83 2.25 7.48 11.98i 10 14 18 14.00 1.78 5.91 9.47j 6 12 20 12.33 5.44 18.11 28.99k 7 9 14 9.50 1.36 4.53 7.25

ExpectedPaths Duration Variance Variance Variancea-d-i 26.33 2.22 7.39 11.83b-e-i 23.50 2.14 7.11 11.39b-f-j 32.17 6.25 20.79 33.28c-g-j 34.67 11.22 37.32 59.76c-h-k 29.00 5.39 17.92 28.70

Prob of Fin Path c-g-j < 38 84.0% 70.7% 66.7%

90% Chance of Fin path c-g-j 39.0 42.5 44.6

=NORMDIST(38,$E20,SQRT(F20),TRUE) copy to cells G25:H25

=NORMINV(0.9,$E20,SQRT(F20)) copy to cells G29:H29


percent level produces a distribution of activity times (and thus of path lengths) that is much smaller than the range produced by estimations at the 90 percent level. The greater uncertainty reduces that chance that any path will be completed in a specific time.

g. As shown in the spreadsheet, if the estimates were made at the 99+ percent level, the project would have a 90 percent chance of being completed by day 39. On the other hand, if the estimates were made at the 90 percent level (and assuming that 90% refers to the area between a and b), the project would have a 90 percent chance of being completed within 42.5 days.

27. A portion of the spreadsheet developed to simulate this project 150 times is shown below. (Note: the variance was based on the interpretation that there was a 95% chance that the task times would fall between a and b.)

a. Path a-d-f had the longest duration in 3.3% of the replications. Similarly, paths a-c-e-g and b-e-g had the longest durations in 81.3% and 15.3% of the simulation replications, respectively. This clearly demonstrates the difficulty in determining which path will be the critical path when activity times are uncertain.

b. In 140 of the 150 replications (or in 93.3% of the replications) the project was completed in 23 weeks or less. More formally, a histogram of the project completion times could be easily developed as shown below. The histogram appears to be approximately normally distributed and in fact almost always passes the chi-squared goodness of fit test confirming that the normal distribution provides a reasonable fit. Based on the assumptions that the distribution of project completion times follows a normal distribution, a mean project completion time of 19.97 weeks, and standard deviation of 2.07, it can be easily calculated that the probability of completing the project in 23 weeks or less is 92.86% which is quite close to the empirical estimate of 93.3%.

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A B C D E F G H I J K L M N OActivity Activity Activity Activity Activity Activity Activity Path Path Path Project a-d-f a-c-e-g b-e-g Less than

a b c d e f g a-d-f a-c-e-g b-e-g Duration Crirical? Crirical? Crirical? 23 Weeks?3.69 4.74 4.76 7.20 3.92 6.30 3.52 17.20 15.89 13.40 17.20 1 0 0 12.70 5.38 6.04 6.23 5.58 3.92 4.08 12.84 18.40 16.35 18.40 0 1 0 14.25 5.70 6.06 5.48 4.22 2.27 8.84 11.99 23.36 15.63 23.36 0 1 0 05.30 5.79 6.22 6.71 6.10 3.85 4.68 15.87 22.30 17.67 22.30 0 1 0 15.22 4.39 5.11 7.90 6.03 3.29 4.80 16.41 21.17 14.82 21.17 0 1 0 15.77 3.30 5.81 7.12 5.44 4.09 4.88 16.98 21.90 12.05 21.90 0 1 0 11.77 0.99 4.86 6.16 5.53 4.51 4.01 12.45 16.17 7.52 16.17 0 1 0 13.76 5.23 4.17 6.27 5.10 7.47 3.53 17.51 16.57 15.56 17.51 1 0 0 15.12 8.45 4.38 5.50 6.51 3.05 6.70 13.67 22.72 23.42 23.42 0 0 1 02.89 5.93 3.04 5.13 5.55 5.60 5.93 13.62 17.41 17.42 17.42 0 0 1 13.30 6.15 3.60 7.61 4.22 4.11 5.81 15.01 16.93 16.52 16.93 0 1 0 12.28 7.05 6.19 5.61 5.73 5.21 5.77 13.10 19.97 19.83 19.97 0 1 0 12.12 4.55 7.20 6.42 4.82 4.28 5.14 12.82 19.28 13.93 19.28 0 1 0 13.00 2.84 4.73 6.01 5.70 6.78 6.28 15.79 19.72 11.39 19.72 0 1 0 13.21 5.76 6.02 5.54 5.17 3.76 6.49 12.51 20.89 16.69 20.89 0 1 0 11.84 3.86 6.49 5.15 5.27 2.92 5.36 9.91 18.97 13.00 18.97 0 1 0 1


c. The 93.3% calculated here is slightly smaller than the 96.3% calculated in Problem 25. The difference could be due to a sampling error, not replicating the project enough times, or related to the assumption that the activity times are normally distributed.

28. Since the times for activities are now known for activities a - d, updating the simulation model simply requires replacing the randomly generated activity times for these tasks with the known times. After entering these known times in columns A – D, the revised probability of the project being completed by week 23 drops to 40%. With the updated information, it is clear that path b-e-g will be the critical path. Furthermore, task B’s actual duration was equal to its pessimistic time estimate. This and the fact that tasks E and G are skewed to the right helps explain the decrease in the probability of completing this project by week 23.

Suggested Solution to Discussion Problem

29. Note: If students use MSP to generate the AON network, and if they use a “Start” node to begin the project, the Start node will be numbered “1.” This will increase by one all the task numbers used in the problem statement. This is a common source of confusion in reality as well as in the classroom.

There appear to be three activities that have no logical predecessors: Organize the sales office, order stock from the manufacturer, and design the package. Organize the office has three logical successors: select distributors, hire sales personnel, and select advertising agency. Given these relationships, the other predecessor-successor relationships are obvious.

Besides the need to manage the critical path carefully, there are only two managerial problems or opportunities readily apparent. If resources could be made available to

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Distribution of Project Completion Times

0

5

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25

30

15 16 17 18 19 20 21 22 23 24Mor

e

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Fre

qu

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cy


speed up organization of the sales office, project slack would result. Second, selecting distributors is the only activity with low slack, 2 weeks. It bears watching.

Students may advise shifting resources from ordering stock from the manufacturer, but once the purchase order has been made out and sent, the 13 week duration is mostly waiting for delivery and there are no resource commitments involved.

Incident for Discussion Suggested Answer

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A

StartB

C

D

E

F

G

H

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Finish

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A B C D E F GTask Description ES EF LS LF Slack

A Design package 0 2 5 7 5B Order stock from manf. 0 13 4 17 4C Org. sales office 0 6 0 6 0D Set up packing facility 2 12 7 17 5E Select distributors 6 15 8 17 2F Hire sales personnel 6 10 6 10 0G Select adv. agency 6 8 13 15 7H Package initial stocks 13 19 17 23 4I Sell to distributors 17 23 17 23 0J Train sales personnel 10 17 10 17 0K Plan adv. campaign 8 12 15 19 7L Ship stock to distributors 23 29 23 29 0M Conduct adv. campaign 12 22 19 29 7


Springville Fire Department: The scheduling techniques mentioned are not mutually exclusive, they are complementary and could be used together. PERT/CPM serves well for planning and control. A Gantt chart would provide the ease of use, showing durations, monitoring and on-going analysis. It might also be noted that CPM has traditionally been favored by the construction industry. If only one technique must be used, the modified PERT/CPM method, plotted on a time scale, would probably be the best choice.

Suggested Case Analyses and Solutions

St. Dismas Assisted Living Facility Project Action Plan -- 3

Teaching Purpose: This installment of the St. Dismas case provides students with an opportunity to further develop their skills in creating and using Gantt charts.

1. Draw a Gantt chart for the construction phase of the project. What is the completion date if construction starts in March? What is the completion date of the project if construction is started in November?

The following is the Gantt chart using MSP with a March 1, 2000 start date. This was entered using the standard calendar defaults used by MSP of a Monday through Friday 8 am to 5 pm workday with an hour off for lunch. The project completion date is 7/30/01. (Please note: Start and finish milestones were added to the action plan for ease in identifying the project’s start and finish dates.)

The following is the Gantt chart using MSP with a November 9, 1999 start date. This start date was chosen because one of the constraints placed on the project was that it does not begin until after the elections in November. Elections are usually held the first Tuesday in the month of November. The case stated that one to two months was estimated as needing to be added to the project schedule to allow for bad weather conditions during the outside construction phases of the project (30 – 60 working days). Days of work need to be added to the duration of each of the steps where work takes place outside if it will happen during the winter months. The authors chose to add 30 days to each of the steps affected, step # 4 and #5. Step #6 is also work done outside, but with the changes made to #4 and #5, step #6 will not start until the month of April. Students may also change the calendars to let the workers off for the holidays of 1999 in addition to those allowed for in 2000. The new ending date is May 18, 01. The project

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will take a total of 30 additional days to complete. By adding 30 days of working time to each of the steps possibly affected by the weather, we only ended up adding 30 days total to the project. This is a good discussion point of how the tasks affected by the increased durations were not both on the critical path – only Step #4 extended the length of the project. However, the student’s must keep in mind resource availability and the increased cost of the project extension.

2. Why is it not possible to meet the scheduling constraints set by the Board? What is your recommendation to handle the scheduling problem?

The case outlined two specific constraints that the Board placed on the project. The first is that the project should not start until after the elections in November. The second constraint is that the building be ready for occupancy by July of the following year. The board wanted to target occupancy for the summer months.

The constraint of the building opening by July of the year following construction beginning can not be met. No matter when the project begins it takes longer than one year to complete. If the project begins immediately after the November 1999 elections it will be completed by May 18, 2001, if the project begins in March of 2000, as recommended by the construction manager it will be completed by July 30, 2001.

The constraint of construction beginning after the November elections can be met with out any affect on the project. Meeting the July complete occupancy constraint is possible only if the project is started in the winter months, this would add cost and time to the project. It would also make the first units available in April, which is before the targeted “summer” occupancy. Recall that the case stated that research showed that most people shopped for assisted living facilities during the summer months. A summer occupancy could be met by starting the project March 1, 2000, without additional time or cost added to the original estimated project action plan. 45 units would be available for occupancy as early as June 25, 2001 (see step #9 of Gantt chart with 3/1/00 start date).

3. When will the project be completed based on your recommendation?

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If students chose March 1, 2000, the entire project will be completed by July 31, 2001.

If students chose to start the project February 1, 2000, without any schedule changes due to weather conditions, the project would be completed by 6/29/01. By using MSP to change the project’s start date, students can easily choose various new start dates and see the associated project end date.

4. Draw a Gantt Chart of the Marketing Plan and Implementation Phase of the Project. Determine the start date of the Marketing Plan phase of the project in order to meet your recommended facility ready for occupancy date?

We assumed a March 1, 2000 start date for the project.

Below is the Gantt chart for the steps in the Marketing Plan and Implementation phase of the St. Dismas Assisted Living Project. The action steps were taken directly from the Marketing plan developed and implemented section of the broad marketing plan that was presented in the case.

To determine the project’s start date, you must first determine the start date of the final step in the project action plan, “Implementation of the Marketing Plan”. This must be started 5 months prior to the building being ready for occupancy so that marketing has time to find residents to move in when the facility is available. The marketing plan must be implemented based on the date that the first 45 units are ready for residents. This date is June 25, 2001.

In order to determine the start date of the Implementation task, Step #7, one can schedule this project backwards, we know the completion date, we do not know when to start. Once we enter the projects completion date, MSP will determine when each step of the project should take place. First enter all of the tasks names, precedences, and their durations, as shown below:

Next, we must enter this project’s overall finish date, using the project information dialog box, found in the “Project” menu on the tool bar.

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Enter an end date of 6/25/01, and select “Schedule from the project finish date”. MSP will automatically determine each step’s start and end date to meet the constraint you set. See below:

5. What is the next step the team members must take in order to complete their action plans? Each member of the project steering team needs to prepare final action plans, including dates and resources. The team must also determine the predecessors from outside their specific plan that link to their plans; for example if a step can not be completed on the marketing phase of the project until Legal has completed a step in their project plan, this must be noted on the action plan. This will enable a complete overall integrated project action plan to be tied to the project budget, monitored and controlled.

Nutristar

Teaching Purpose: The purpose of this case is to reinforce students’ skills in analyzing projects with probabilistic time estimates. The case also provides students with an opportunity to use spreadsheets to simulate the completion of the project and use the results of the simulation to perform standard probability calculations.

1. Draw a network diagram for this project. Identify all the paths through the network diagram.

The following abbreviations will be used for the activities.

Activity Description AbbreviationConcept Development APlan Development Define project scope B Develop broad schedule C Detailed cost estimates D Develop staffing plan EDesign and Construction Detailed engineering F Facility construction G Mobilization of construction Employees

H

Procurement of equipment IStart-up and Turnover Pre-startup inspection J Recruiting and training K

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Solving start-up problems L Centerlining M

Using these abbreviations and the information provided in the case, the following AOA network diagram can be constructed.

B F G A C H I J K D L M E

There are 32 paths through the network as follows

A, B, F,G, J, K, MA, B, F,G, J, L, MA, B, F, I, J, K, MA, B, F, I, J, L, MA, B, H,G, J, K, MA, B, H,G, J, L, MA, B, H, I, J, K, MA, B, H, I, J, L, M

A, C, F,G, J, K, MA, C, F,G, J, L, MA, C, F, I, J, K, MA, C, F, I, J, L, MA, C, H,G, J, K, MA, C, H,G, J, L, MA, C, H, I, J, K, MA, C, H, I, J, L, M

A, D, F,G, J, K, MA, D, F,G, J, L, MA, D, F, I, J, K, MA, D, F, I, J, L, MA, D, H,G, J, K, MA, D, H,G, J, L, MA, D, H, I, J, K, MA, D, H, I, J, L, M

A, E, F,G, J, K, M

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A, E, F,G, J, L, MA, E, F, I, J, K, MA, E, F, I, J, L, MA, E, H,G, J, K, MA, E, H,G, J, L, MA, E, H, I, J, K, MA, E, H, I, J, L, M

2. Simulate the completion of this project 100 times assuming that activity times follow a normal distribution. Estimate the mean and standard deviation of the project completion time.

Prior to simulating the project, the mean and standard deviation for each activity must be calculated. The spreadsheet shown below was developed to calculate these parameters.

Next, the actual spreadsheet to simulate the completion of the project can be developed. As demonstrated in the textbook, one approach is to dedicate a column to each activity and then generate random activity times. Thus, in the spreadsheet developed for this case, columns A - M were dedicated to activities A - M, respectively. Then random numbers were generated for one activity at a time using Excel’s random number generation capability. Using Excel’s Random Number Generator only requires the specification of the type of random variable, its mean, standard deviation, and the range for the random numbers. For example, to generate the random activity times for Activity A, normal was specified for the type of distribution, 12.5 for the mean, 3.5 for the standard deviation, and A3:A102 for the output range. A snapshot of the spreadsheet developed is shown below.

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1011121314151617181920

A B C D E F GOptimistic Most Pessimistic Expected Standard

Activity Time Likley Time Time Variance DeviationA 3 12 24 12.500 12.250 3.500B 1 2 12 3.500 3.361 1.833C 0.25 0.5 1 0.542 0.016 0.125D 0.2 0.3 0.5 0.317 0.003 0.050E 0.2 0.3 0.6 0.333 0.004 0.067F 2 3 6 3.333 0.444 0.667G 8 12 24 13.333 7.111 2.667H 0.5 2 4 2.083 0.340 0.583I 1 3 12 4.167 3.361 1.833J 0.25 0.5 1 0.542 0.016 0.125K 0.25 0.5 1 0.542 0.016 0.125L 0 1 2 1.000 0.111 0.333M 0 1 4 1.333 0.444 0.667

Key Formulas:Cell E3: =(B3+(C3*4)+D3)/6 {copy to cells E4:E15}Cell F3: =((D3-B3)/6)^2 {copy to cells F4:F15}Cell G3 =SQRT(F3) {copy to cells G4:G15}


After generating the random numbers, one column is dedicated to each path and formulas are entered to calculate the path completion time based on the random activity times generated. Again, a snapshot of the spreadsheet developed is shown below. Entering the formulas is very straightforward since the letters used to label the activities correspond directly to the column labels. To illustrate, the formula for path A - B - F- G - J - K - M in row three is: =A3+B3+F3+G3+J3+K3+M3.

To determine the time to complete the project, a final column was entered that calculates the maximum path completion time for each row in the spreadsheet. Then, summary statistics for this column were calculated. One simulation of 100 replications of the project yielded an average project completion time of 35.53 months with a standard deviation of 4.65 months. The maximum completion time was 46.17 months and the minimum completion time was 24.85 months.

3. Develop a histogram to summarize the results of your simulation.

Excel’s histogram function can be used to quickly calculate the frequency distribution of the project completion times. Since the shortest project completion time was just under 25 months and the longest just over 46, and the standard deviation was close to five, intervals of 0 - 25, 25 - 30, 30 - 35, 35 - 40, 40 - 45, and 45 - 50 were specified. Once the frequency distribution was generated, Excel’s ChartWizard was used to develop the actual histogram shown below. The distribution appears to be approximately normally distributed.

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A B C D EActivity Activity Activity Activity Activity

A B C D E11.44919 4.702878 0.535049 0.356053 0.4210748.028109 2.884924 0.694415 0.317893 0.48930413.3549 3.234122 0.338099 0.291584 0.241881

16.96766 2.398613 0.739816 0.279394 0.30374516.69423 3.21668 0.72604 0.293843 0.375742

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L M N O PActivity Activity Path Path Path

L M A B F G J K M A B F G J L M A B F I J K M0.441605 0.813718 38.36431331 38.30425754 26.175707420.974028 1.393954 28.75870589 28.94689201 20.957389341.115277 1.663059 33.8563155 34.61093393 28.589271291.13499 1.029766 31.93146016 32.6103619 29.56046329

0.502369 1.374516 36.15716899 36.18521673 31.82999899


4. Calculate the probability that the project can be completed within 30 months. What

is the probability that the project will take longer than 40 months? What is the probability that the project will take between 30 and 40 months?

Normally, the approach to calculate the probability that a project is completed by some specified time is to calculate the probability that all paths finish by the specified time. With 32 paths to consider, just determining which paths should be considered in the analysis is a tedious process. Further, this approach is based on the assumption that the paths are independent of one another. In the present case, this assumption is clearly violated due to the number of times several activities appear on alternate paths. Another approach is to use the information generated from simulating the completion of the project. Specifically, the results of the simulation indicated that the average completion time of the project was 35.33 months with a standard deviation of 4.65 months. Based on this and assuming a normal distribution, the probability that the project is completed within 30 months can be calculated as:

Thus, the probability of completing the project in 30 months or less (refer to normal table in Appendix A) is 12.51 percent. In a similar fashion, the probability that the project is not completed within 40 months can be calculated as:

Thus, the probability that the project is not completed within 40 months is 15.87 percent. The probability that the project is completed within 30 to 40 months is 71.62 (1.0 - .1251 - .1587) percent.

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A B C D E F G H I J K LBin Frequency

25 130 1535 3040 3845 1450 2

More 0

05

10152025303540

Number of Replications

25 30 35 40 45 50

Project Completion Time

15.165.4

33.3530

z

00.165.4

33.3540

z


Test Questions

True/False and Multiple Choice

F 1. ___When it was originally developed, PERT used certain (deterministic) methods to estimate activity duration.

T 2. ___ AON and AOA networks can both be used to depict any project network.

T 3. ___ The shortest time to complete a network is equal to the duration of the longest path through the network.

T 4. ___ Critical path tasks always have zero slack.

F 5. ___ To manage a project successfully, it is only necessary that the project manager pay close attention to tasks on the critical path.

F 6. ___ A project manager should use probabilities to determine project durations only on complex projects.

F 7. ___ If task duration estimates are carefully made, the project manager needs to only examine the critical path when conducting a risk analysis. T 8. ___The actual project duration will be known with certainty after the project is completed.

T 9. ___ Milestones on a Gantt chart are tasks with a duration of zero.

F 10. ___ It is easiest to see lead and lag time in a project task on a PERT/CPM chart.

F 11. ___ A start-to-finish linkage is the most common way of linking to successive task.

T 12. ___ The formula for the expected time of an activity in a network assumes that the optimistic, pessimistic and most likely time estimates have a Beta distribution.

T 13. ___ A project schedule is a project action plan converted into a timetable.

F 14. ___ A Gantt chart can not depict a critical path, only a PERT/CPM chart can.

T 15. ___ The difference between the LST and EST is called slack.

F 16. ___ A disadvantage of Gantt charts is that they are hard to draw.

F 17. ___ When discussing completion dates with senior management, a project manager should include a 10% safety factor.

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F 18. ___Lead and lags cannot be shown in an AOA network.

F 19. ___Leads and lags can only be shown on an AON network.

F 20. ___ The difference between LST and LFT is called slack.

F 21. ___ The difference between EST and LFT is called slack.

T 22. ___ The difference between LFT and EFT is called slack.

T 23. ___ A big advantage of AON networks is that they are easier to draw.

a 24. ___ What is it a milestone? a. a significant event in the projectb. a mark on a chart that depicts project progressc. an activity on the critical pathd. an activity with an uncertain completion timee. all of the above

b 25. ___What is project slack: a. the amount of time a non-critical task can be delayed without making the

project late.b. the amount of time the critical path of a project can be delayed without

making the project late.c. the amount of time an activity on the critical path can be delayed without

making the project lated. the difference between how long the project would take if all tasks were

completed based on their pessimistic versus optimistic time estimatese. none of the above

d 26. ___ For which purpose is simulation not used with regard to project scheduling: a. to overcome the limitations associated with statistical techniques used to

develop probability of completion time estimatesb. to investigate the range of project completion timesc. investigate the distribution of project completion timesd. to verify the accuracy of the optimistic, pessimistic and most likely time

estimatese. all of the above

c 27. ___ Technical dependencies on a project plan are easiest to see on a: a. Gantt chartb. GERT chartc. PERT/CPM chartd. Work Breakdown Structuree. Linear Responsibility Chart

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c 28. ___ PERT was originally used for what type of project? a. construction b. R & Dc. militaryd. computer software developmente. advertising

b 29. ___ Which of the following is not an element of the Gantt chart? a. actual progressb. variance of the critical pathc. the current dated. scheduled milestonese. all of the above are elements.

a 30. ___ Which of the following is typically used as the best estimate of task duration? a. expected timeb. pessimistic timec. optimistic timed. most likely timee. all of the above

Short Answer31. Define the term Critical Path.

The set of activities on a path from the project’s start to finish that, if delayed, will delay the completion of the project.

32. What is activity slack? The amount of time a non-critical activity can be delayed without delaying the project.

33. What are some of the drawbacks of letting the project team and management know how much slack is in the project?The team will tend to delay the start of a task with slack, thinking they have plenty of time to complete it—the student syndrome. Management will want the slack removed and the project duration shortened.

34. What is a Gantt chart?A graphical depiction of a project action plan. It displays project activities as a bar chart against a time scale.

35. What are the four methods of linking steps in a project using precedence diagramming?Finish to start, start to finish, start to start, finish to finish.

Problems

36. The following information has been compiled for a project that is about to begin.

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Activity Activity Duration Preceding ActivitiesA 3 NoneB 5 AC 3 AD 1 CE 3 BF 4 B, DG 2 CH 3 G, FI 1 E, H

a. Construct the network for the project.b. Determine the earliest and latest start times for each activity as well as the

earliest finish and latest finish times for each activity.c. Calculate the slack for each activity.d. Which activities are critical?e. How long will it take to complete this project?

Solution:

a.

b. and c.

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A

B

C

G

D

F

E

H

I

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A B C D E FActivity ES EF LS LF Slack

A 0 3 0 3 0B 3 8 3 8 0C 3 6 4 7 1D 6 7 7 8 1E 8 11 12 15 4F 8 12 8 12 0G 6 8 10 12 4H 12 15 12 15 0I 15 16 15 16 0


d. Activities A, B, F, H, and I have zero slack and are therefore critical.

e. The critical path is A-B-F-H-I and has a duration of 16.

37. The following information has been compiled for a project that is about to begin. Assume the time estimates were made at the 99+ percent level.

ActivityPreceding Activities

OptimisticTime

MostLikely

PessimisticTime

A None 6 7 14B None 8 10 12C A 2 3 4D A 6 7 8E B, C 5 5.5 9F B, C 5 7 9G D, E 4 6 8H F 2.5 3 3.5

a. Construct the network.b. What is the probability that the path with the longest expected duration will be

completed within 21 days?c. Assuming the paths are independent, what is the probability that the project

will be completed within 21 days?

Solution

a.

b. As shown in the spreadsheet below, path A-C-E-G has the longest duration of 23 and a probability of being completed by time 21 of 11.5%.

c. The probabilities of the other four paths being completed by time 21 can be calculated in a similar fashion. Based on the assumption that the paths are independent, the probability that the project is completed by time 21 can be calculated by taking the product of these respective probabilities. In this case, the probability that the entire project is completed by 21 days

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A

B

C

D

E

F

G

H


is .5 percent. In other words, there is only half a percent chance that all five paths will be completed by time 21.

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A B C D E FTask Opt. Most Likely Pess. t E Variance

A 6 7 14 8 1.78B 8 10 12 10 0.44C 2 3 4 3 0.11D 6 7 8 7 0.11E 5 5.5 9 6 0.44F 5 7 9 7 0.44G 4 6 8 6 0.44H 2.5 3 3.5 3 0.03

Prob ofExpected Path

Paths Duration Variance z Fin < 21A-D-G 21 2.3 0.00 50.0%A-C-E-G 23 2.8 -1.20 11.5%A-C-F-H 21 2.4 0.00 50.0%B-E-G 22 1.3 -0.87 19.3%B-F-H 20 0.9 1.04 85.2%

Prob of Proj < 21 0.5%

Ch05

Documents

project slack

project chapter

project network

entire project

discussion of project

project danger

subject of project uncertainty

latest time