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Case Problem - The following case problem gives an example of how PERT/CPM analysis can be used in project management.
This case problem is taken from page 855 in the 2000 version of Operations Management by Russell and Taylor. The case involves a group of college students seeking to have a band called the Bloodless Coup perform at their college. However, they only have 18 days to organize the activity. The case asks for PERT analysis to be used to determine the probability of the concert preparations being completed in 18 days or less.
1. The first step in solving this problem, is to create a chart listing each activity, the description, the activity's predecessors and the three time estimates. Doing this we come up with a chart like the following:
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2. Next we need to use POM for Windows (or some other type of Project Management Software) to get a PERT/CPM analysis of the chart. Make sure the project management module is selected and create a new triple time estimate chart. From our chart, we can see that the number of tasks will be twelve. Also, we want precedence list selected as our table structure.
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3. Once we have clicked ok on the "Create data set" dialog box, we can now enter the data into POM. The data should entered as follows:
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4. After pressing the solve button, you come up with the following results page:
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The chart shows the activity times for each activity along with the standard deviation from each time. From this, one can see that the total project activity time is under the 18 days that the students need to complete the project in time. Thus, it seems highly probable that the activity will be completed in time. We can also see that activities a, d, e, and h have no slack. This indicates that they are part of the critical path. The key values we will need to compute the probability of completing the tasks in 18 days are circled in black.
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5. In order to figure the probability of the project being finished in time, we must compute the z score. The formula is as follows:
Z = (the sample mean - the overall mean) / (the overall standard deviation)
Z = (18 - 15.1667) / (1.7873)
Z = 1.5852
Now one can look in a normal table to find the corresponding probability. In this case, the probability is 94.3 %. Thus it is highly likely that the students will finish the project in 18 days or less.
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