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Example 12.5 Operations Models. 12.112.1 | 12.2 | 12.3 | 12.4 | 12.6 | 12.7 |12.8 | 12.9 | 12.10 | 12.11 | 12.12 | 12.13 | 12.14 | 12.15 | 12.16 | 12.1712.212.312.412.612.712.812.912.10.

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Presentation on theme: "Example 12.5 Operations Models. 12.112.1 | 12.2 | 12.3 | 12.4 | 12.6 | 12.7 |12.8 | 12.9 | 12.10 | 12.11 | 12.12 | 12.13 | 12.14 | 12.15 | 12.16 | 12.1712.212.312.412.612.712.812.912.10."— Presentation transcript:

1 Example 12.5 Operations Models

2 | 12.2 | 12.3 | 12.4 | 12.6 | 12.7 |12.8 | 12.9 | | | | | | | | Background Information n Tom Lingley, an independent contractor, has agreed to build a new room on an existing house. n He plans to begin work on Monday morning June 1. n The main question is when will he complete his work, given that he works only weekdays. n The owner of the house is particularly hopeful that the room will be ready by Saturday, June 27, that is, in 20 or fewer working days. n The work proceeds in stages, labeled A through J, as summarized in the table on the next slide.

3 | 12.2 | 12.3 | 12.4 | 12.6 | 12.7 |12.8 | 12.9 | | | | | | | | Background Information -- continued n Three of the activities, E, F, and G, will be done by separate independent subcontractors. The expected durations of the activities are shown in the table. Activity Time Data DescriptionIndexPredecessorsExpected Duration Prepare foundationANone4 Put up frameBA4 Order custom windowsCNone11 Erect outside wallsDB3 Do electrical wiringED4 Do plumbingFD3 Put in duct workGD4 Hang dry wallHE, F, G3 Install windowsIB, C1 Paint and clean upJH2

4 | 12.2 | 12.3 | 12.4 | 12.6 | 12.7 |12.8 | 12.9 | | | | | | | | Background Information -- continued n However, these are only best guesses. n Lingley knows that the actual activity times can vary because of unexpected delays, worker illnesses, and so on. n He would like to use a computer simulation to see –How long the project is likely to take, and –How likely it is that the project will be completed by the deadline, and –Which activities are likely to be critical.

5 | 12.2 | 12.3 | 12.4 | 12.6 | 12.7 |12.8 | 12.9 | | | | | | | | Solution n We first need to choose distributions for the uncertain activity times. n Then, given any randomly generated activity times, we will illustrate a method for calculating the length of the project and identifying the activities on the critical path.

6 | 12.2 | 12.3 | 12.4 | 12.6 | 12.7 |12.8 | 12.9 | | | | | | | | The Pert Distribution n As always, there are several reasonable candidate probability distributions we could use for the random activity times. n Here we illustrate a distribution that has become popular in project scheduling, called the Pert distribution. As shown on the next slide, it is rounded version of the triangular distribution that is specified by three parameters: a minimum value, and a maximum value.

7 | 12.2 | 12.3 | 12.4 | 12.6 | 12.7 |12.8 | 12.9 | | | | | | | | The Pert Distribution -- continued

8 | 12.2 | 12.3 | 12.4 | 12.6 | 12.7 |12.8 | 12.9 | | | | | | | | The Pert Distribution -- continued n The distribution in the figure uses the values 7, 10, and 19 for these three values, which implies a mean of 11. We will use this distribution for activity C. n Similarly, for the other activities, we choose parameters for the Pert distribution that lead to the means in the table. n In reality, it would be done the other way around. The contractor would estimate the minimum, most likely, and maximum parameters for the various activities, and the means would follow from these.

9 | 12.2 | 12.3 | 12.4 | 12.6 | 12.7 |12.8 | 12.9 | | | | | | | | Developing the Simulation Model n The key to the model is representing the project network in activity-on-arc form, as in the diagram below, and then finding E j for each j, where E j is the earliest time we can get to node j. n We stated in Chapter 5 that when the nodes are numbered so that all arcs go from lower-numbered nodes to higher-numbered nodes, we can calculate the E j s iteratively, starting with E 1 =0, with the equation E j = max(E i + t ij ). n Here, the maximum is taken overall arcs leading into node j, and t ij is the activity time on such an arc.

10 | 12.2 | 12.3 | 12.4 | 12.6 | 12.7 |12.8 | 12.9 | | | | | | | | Developing the Simulation Model -- continued n Then En is the time to complete the project, where n is the index of the finish node. n This will make it very easy to calculate the project length. n We also need a methods for identifying the critical activities for any given activity times. n By definition, an activity is critical is a small increase in its activity time cause the project time to increase. Therefore, we will keep track of two sets of activity times, and associated project times.

11 | 12.2 | 12.3 | 12.4 | 12.6 | 12.7 |12.8 | 12.9 | | | | | | | | Developing the Simulation Model -- continued n The first uses the simulated activity times. The second adds a small amount, such as day, to a selected activitys time. By using the RISKSIMTABLE function with a list as long as the number of activities, we can make each activity the selected activity in this method.

12 | 12.2 | 12.3 | 12.4 | 12.6 | 12.7 |12.8 | 12.9 | | | | | | | | PROJECTSIM.XLS n The spreadsheet model appears on the next slide. n This file contains the model.

13 | 12.2 | 12.3 | 12.4 | 12.6 | 12.7 |12.8 | 12.9 | | | | | | | |

14 | 12.2 | 12.3 | 12.4 | 12.6 | 12.7 |12.8 | 12.9 | | | | | | | | Developing the Simulation Model -- continued n The details of the model are followed. –Inputs. Enter the parameters of the Pert activity time distributions in the shaded cells and the implied means next to them. As discussed above, we actually chose the minimum, most likely, and maximum values while Model window to achieve the means in the activity table. Note that some of these distributions are symmetric about the most likely value, whereas other are skewed. –Activity time. Generate random activity times in column I by entering the formula =RISKPERT(E5, F5, G5) in cell I5 and copying it down.

15 | 12.2 | 12.3 | 12.4 | 12.6 | 12.7 |12.8 | 12.9 | | | | | | | | Developing the Simulation Model -- continued –Augmented activity times. We want to successively add a small amount of each activitys time to determine whether it is on the critical path. To do this, enter the formula =RISKSIMTABLE({1, 2, 3, 4, 5, 6, 7, 8, 9, 10}) in cell B16. Then enter the formula =I5+IF(Index=C5,0.001,0) in cell J5 and copy it down. –Event times. We want to use the equation to calculate the node event times in the range B20:B27. There is no quick way to enter the required formulas. We need to use the project network as a guide for each node. Begin by entering 0 in cell B20. Then enter the appropriate formulas in the other cells. For example, the formulas in cells B22, B23, and B27 are =B21+I6, =MAX(B20+I7,B21+I6) and =RISKOUTPUT( )+MAX(B23+I13,B26+I14)

16 | 12.2 | 12.3 | 12.4 | 12.6 | 12.7 |12.8 | 12.9 | | | | | | | | Developing the Simulation Model -- continued –To understand these, note the node 3 has only one arc leading into it, and this arc originates at node 2. No MAX is required for this modes equation. In contrast, node 4 has two arcs leading into it, from nodes 1 and 2, so a MAX is required. Similarly, node 8 requires a MAX because it has two arcs leading into it. Also, it is the finish node, so we designate its event time cell as output cell – it contains the time to complete the project. –Augmented even times. Copy the formulas in the range B20:B27 to the range C20:C27 to calculate the even times when the selected activitys time is augmented by

17 | 12.2 | 12.3 | 12.4 | 12.6 | 12.7 |12.8 | 12.9 | | | | | | | | Developing the Simulation Model -- continued –Project time increase?. To check whether the selected activitys increased activity time increases the project time, enter the formula =RISKOUTPUT( )+IF(C27>B27,1,0) If this calculates to 1, then the selected activity is critical for these particular activity times. Otherwise, it is not. Note that this cell is also designated as output cell.

18 | 12.2 | 12.3 | 12.4 | 12.6 | 12.7 |12.8 | 12.9 | | | | | | | | n We set the number of iterations to 1000 and the number of simulations to 10. n After we request the histogram of project times shown on the next slide. n Recall from the example in Chapter 5 that when the activity times are not considered random, the project time is 20 days. Now it varies from a low of days to a high of days, with an average of days. n Although the 5 th and 95 th percentiles appear in the figure, it might be more interesting to Tom Lingley to see the probabilities of various project times being exceeded.

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20 | 12.2 | 12.3 | 12.4 | 12.6 | 12.7 |12.8 | 12.9 | | | | | | | | continued n For example, we entered 20 in the Left X box next to the histogram. The Left P value implies there is about a 59% chance that the project will not be completed within 20 days. n Similarly, the values in the Right X and Right P boxes imply that the chance of the project lasting longer/ than 23 days is slightly greater than 5%. n This is certainly not good news for Lingley, and he might have to resort to the crashing we discussed earlier.

21 | 12.2 | 12.3 | 12.4 | 12.6 | 12.7 |12.8 | 12.9 | | | | | | | | continued n The summary measures for the B29 output cell appears in this table.

22 | 12.2 | 12.3 | 12.4 | 12.6 | 12.7 |12.8 | 12.9 | | | | | | | | continued n Each simulation in this output represents one selected activity being increased slightly. n The Mean column indicates the fraction of iterations where the project time increases as a result of the selected activitys time increase. n Hence, it represents the probability that this activity is critical. n For example, the first activity (A) is always critical, the third activity (C) is never critical, and the fifth activity (E) is critical about 44% of the time.More specifically, we see that the critical path always includes activities A, B, D, H, J, and one of the three parallel activities E, F, and G.


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