Solution to Brook Museum Problem 1

Solution to Brook Museum Problem1

Sharon Katz is project manager in charge of laying the foundation for the new Brook Museum of Art in New Haven, Connecticut. Liya Brook, the benefactor and namesake of the museum, wants to have the work done within 41 weeks, but Sharon wants to quote a completion time that she is 90% confident of achieving. The contract specifies a penalty of $10,000 per week for each week the completion of the project extends beyond week 43. Activity Description Optimistic Pessimistic Most-likely Predecessors A Survey Site 2 4 3 None B Excavation 9 15 12 A C Prepare Drawings 4 18 9 None D Soil Study 1 1 1 B E Prelim. Report 1 3 2 C, D F Approve Plans 1 1 1 E G Concrete Forms 5 9 6 F H Procure Steel 2 10 5 F I Order Cement 1 1 1 F J Deliver Gravel 2 5 3 G K Pour Concrete 8 14 10 H, I, J L Cure Concrete 2 2 2 K M Strength Test 2 2 2 L

1 David Juran. Adapted from Levin, Rubin, Stinson, and Gardner (1989) “Quantitative Approaches to Management”, 7th ed., p. 605. Create a PERT model of this project and use it to answer these questions: 1. What is the expected completion time of this project? Here’s an activity-on-arc diagram of the problem:

We start a spreadsheet model like this, calculating the mean and standard deviation using the PERT formulas: A B C D E F G H I J K L M 1 Activity Description Optimistic Pessimistic Most-likely Predecessors Start Node End Node Mean StDev 2 A Survey Site 2 4 3 None 0 1 3.00 0.33=(C3+4*E3+D3)/6 3 B Excavation 9 15 12 A 1 2 12.00 1.00 4 C Prepare Drawings 4 18 9 None 0 3 9.67 2.33 =(D5-C5)/6 5 D Soil Study 1 1 1 B 2 3 1.00 0.00 6 E Prelim. Report 1 3 2 C, D 3 4 2.00 0.33 7 F Approve Plans 1 1 1 E 4 5 1.00 0.00 8 G Concrete Forms 5 9 6 F 5 7 6.33 0.67 9 H Procure Steel 2 10 5 F 5 6 5.33 1.33 10 I Order Cement 1 1 1 F 5 8 1.00 0.00 11 Dummy 0 0 0 H 6 8 0.00 0.00 12 J Deliver Gravel 2 5 3 G 7 8 3.17 0.50 13 K Pour Concrete 8 14 10 H, I, J 8 9 10.33 1.00 14 L Cure Concrete 2 2 2 K 9 10 2.00 0.00 15 M Strength Test 2 2 2 L 10 11 2.00 0.00

B60.2350 2 Prof. Juran Now we calculate shape and scale parameters:

A B C D E F G H I J K L M N O P 1 Activity Description Predecessors Start Node End Node Min Mode Max Mean StDev Alpha Beta 2 A Survey Site None 0 1 2 3 4 3 0.333 4.000 =((I3-F3)/(H3-F3))*((((I3-F3)*(H3-I3))/(J3^2))-1)4.000 3 B Excavation A 1 2 9 12 15 12 1.000 4.000 4.000 4 C Prepare Drawings None 0 3 4 9 18 9.667 2.333 3.106 4.568 5 D Soil Study B 2 3 1 1 1 1 0.000 =((H6-I6)/(I6-F6))*K6 6 E Prelim. Report C, D 3 4 1 2 3 2 0.333 4.000 4.000 7 F Approve Plans E 4 5 1 1 1 1 0.000 8 G Concrete Forms F 5 7 5 6 9 6.333 0.667 2.333 4.667 9 H Procure Steel F 5 6 2 5 10 5.333 1.333 3.229 4.521 10 I Order Cement F 5 8 1 1 1 1 0.000 11 Dummy H 6 8 0 0 0 0 0.000 12 J Deliver Gravel G 7 8 2 3 5 3.167 0.500 2.938 4.617 13 K Pour Concrete H, I, J 8 9 8 10 14 10.33 1.000 2.938 4.617 14 L Cure Concrete K 9 10 2 2 2 2 0.000 15 M Strength Test L 10 11 2 2 2 2 0.000 On a separate sheet we set up the model, which will have three sections:  A section for simulating the times of the activities  A section to keep track of each node and when it occurs  A section to keep track of each path through the network, to identify the critical path in each simulated project completion.

B60.2350 3 Prof. Juran Activity Times Here’s the section keeping track of the activity times. The numbers in column K will be Crystal Ball assumption cells. A B C D E F G H I J K L 1 Activity Description Predecessors Start Node End Node Min Alpha Beta Scale CB Time Simulated Time Start Time 2 A Survey Site None 0 1 2 4.00 4.00 2 3.00 5.00 3 B Excavation A 1 2 9 4.00 4.00 6 12.00 21.00 =F4+J4 4 C Prepare Drawings None 0 3 4 3.11 4.57 14 9.67 13.67 =F5 5 D Soil Study B 2 3 1 0 1.00 1.00 6 E Prelim. Report C, D 3 4 1 4.00 4.00 2 2.00 3.00 7 F Approve Plans E 4 5 1 0 1.00 1.00 8 G Concrete Forms F 5 7 5 2.33 4.67 4 6.33 11.33 9 H Procure Steel F 5 6 2 3.23 4.52 8 5.33 7.33 10 I Order Cement F 5 8 1 0 1.00 1.00 11 Dummy H 6 8 0 0 0.00 0.00 12 J Deliver Gravel G 7 8 2 2.94 4.62 3 3.17 5.17 13 K Pour Concrete H, I, J 8 9 8 2.94 4.62 6 10.33 18.33 14 L Cure Concrete K 9 10 2 0 2.00 2.00 15 M Strength Test L 10 11 2 0 2.00 2.00

Nodes Now we set up an area in the spreadsheet to keep track of the nodes and their times: A B 18 Node Time 19 0 0 20 1 21 2 22 3 23 4 24 5 25 6 26 7 27 8 28 9 29 10 30 11 We need to link the node times to the starting and ending times for the activities. The start time for any activity is the time at which its beginning node occurs. The end time for any activity is the start time plus the activity time.

B60.2350 4 Prof. Juran Here’s an example using Activity C:

A B C D E F G H I J K L M N O P Q 1 Activity Description Predecessors Start Node End Node Min Mode Max Mean StDev Alpha Beta Simulated Time Start Time End Time 2 A Survey Site None 0 1 2 3 4 3 0.333 4.000 4.000 5.00 0.00 5.00 3 B Excavation A 1 2 9 12 15 12 1.000 4.000 4.000 21.00 5.00 26.00 =N4+M4 4 C Prepare Drawings None 0 3 4 9 18 9.667 2.333 3.106 4.568 13.67 0.00 13.67 5 D Soil Study B 2 3 1 1 1 1 0.000 1.00 26.00 27.00 =VLOOKUP(D4,$A$19:$B$30,2,0) 6 E Prelim. Report C, D 3 4 1 2 3 2 0.333 4.000 4.000 3.00 27.00 30.00 7 F Approve Plans E 4 5 1 1 1 1 0.000 1.00 30.00 31.00 8 G Concrete Forms F 5 7 5 6 9 6.333 0.667 2.333 4.667 11.33 31.00 42.33 9 H Procure Steel F 5 6 2 5 10 5.333 1.333 3.229 4.521 7.33 31.00 38.33 10 I Order Cement F 5 8 1 1 1 1 0.000 1.00 31.00 32.00 11 Dummy H 6 8 0 0 0 0 0.000 0.00 38.33 38.33 12 J Deliver Gravel G 7 8 2 3 5 3.167 0.500 2.938 4.617 5.17 42.33 47.50 13 K Pour Concrete H, I, J 8 9 8 10 14 10.33 1.000 2.938 4.617 18.33 47.50 65.83 14 L Cure Concrete K 9 10 2 2 2 2 0.000 2.00 65.83 67.83 15 M Strength Test L 10 11 2 2 2 2 0.000 2.00 67.83 69.83 16 17 18 Node Time 19 0 0 20 1 5.00 21 2 26.00 =MAX(O4,O5) 22 3 27.00 23 4 30.00 24 5 31.00 25 6 38.33 26 7 42.33 The story being told here:  The start time for Activity C is the time of Node 0, which comes from using the VLOOKUP formula in N4 to find the node time in the range A19:B30.  The duration of Activity C is the (random) value in M4.  The ending time of Activity C is its start time plus its duration, added up in cell O4.  The time of Node 3 is either the ending time of Activity C, the ending time of Activity D, whichever is greater. We get this maximum in cell B22. Any activities that start at Node 3 will use this value for their start time. It’s important to be careful with the nodes that have multiple activities leading into them (in this model, Nodes 3 and 8). The times for those nodes must be the maximum ending time for the set of activities leading in.

B60.2350 5 Prof. Juran Criticality Now we set up an area in the spreadsheet to track each of the paths through the network, to see which one is critical. This network happens to have six paths, so we set up a cell to add up all of the activity times for each of these paths: D E F G H I J K L M 1 Start Node End Node Min Mode Max Mean StDev Alpha Beta Simulated Time 2 0 1 2 3 4 3 0.333 4.000 4.000 5.00 3 1 2 9 12 15 12 1.000 4.000 4.000 21.00 4 0 3 4 9 18 9.67 2.333 3.106 4.568 13.67 5 2 3 1 1 1 1 0.000 1.00 6 3 4 1 2 3 2 0.333 4.000 4.000 3.00 7 4 5 1 1 1 1 0.000 1.00 8 5 7 5 6 9 6.33 0.667 2.333 4.667 11.33 9 5 6 2 5 10 5.33 1.333 3.229 4.521 7.33 10 5 8 1 1 1 1 0.000 1.00 11 6 8 0 0 0 0 0.000 0.00 12 7 8 2 3 5 3.17 0.500 2.938 4.617 5.17 13 8 9 8 10 14 10.3 1.000 2.938 4.617 18.33 14 9 10 2 2 2 2 0.000 2.00 15 10 11 2 2 2 2 0.000 2.00 16 17 18 Path Total Critical? 19 A-B-D-E-F-H-K-L-M 60.67 0 =SUM(M4,M6,M7,M9,M13,M14,M15) 20 C-E-F-H-K-L-M 47.33 0 21 A-B-D-E-F-I-K-L-M 54.33 0 =IF(G22=$G$25,1,0) 22 C-E-F-I-K-L-M 41.00 0 23 A-B-D-E-F-G-J-K-L-M 69.83 1 24 C-E-F-G-J-K-L-M 56.50 0 =MAX(G19:G24) 25 Max Path 69.83 26

Now, for each activity, we can set up an IF statement to say whether the activity was critical for any particular realization of the model. Note that Activity H (Procure Steel, in row 9) is part of two paths (A-B-D-E-F-H-K-L-M, in row 19, and C-E-F-H-K-L-M, in row 20). In this example, neither of those was the critical path, so Activity H is non-critical.

B60.2350 6 Prof. Juran D E F G H I J K L M N O P Q R 1 Start Node End Node Min Mode Max Mean StDev Alpha Beta Simulated Time Start Time End Time Critical? 2 0 1 2 3 4 3 0.333 4.000 4.000 5.00 0.00 5.00 1 3 1 2 9 12 15 12 1.000 4.000 4.000 21.00 5.00 26.00 1 4 0 3 4 9 18 9.67 2.333 3.106 4.568 13.67 0.00 13.67 0 5 2 3 1 1 1 1 0.000 1.00 26.00 27.00 1 6 3 4 1 2 3 2 0.333 4.000 4.000 3.00 27.00 30.00 1 7 4 5 1 1 1 1 0.000 1.00 30.00 31.00 1 8 5 7 5 6 9 6.33 0.667 2.333 4.667 11.33 31.00 42.33 1 =SUM(H19,H20) 9 5 6 2 5 10 5.33 1.333 3.229 4.521 7.33 31.00 38.33 0 10 5 8 1 1 1 1 0.000 1.00 31.00 32.00 0 11 6 8 0 0 0 0 0.000 0.00 38.33 38.33 12 7 8 2 3 5 3.17 0.500 2.938 4.617 5.17 42.33 47.50 1 13 8 9 8 10 14 10.3 1.000 2.938 4.617 18.33 47.50 65.83 1 14 9 10 2 2 2 2 0.000 2.00 65.83 67.83 1 15 10 11 2 2 2 2 0.000 2.00 67.83 69.83 1 16 17 18 Path Total Critical? 19 A-B-D-E-F-H-K-L-M 60.67 0 20 C-E-F-H-K-L-M 47.33 0 21 A-B-D-E-F-I-K-L-M 54.33 0 22 C-E-F-I-K-L-M 41.00 0 23 A-B-D-E-F-G-J-K-L-M 69.83 1 24 C-E-F-G-J-K-L-M 56.50 0 25 Max Path 69.83

Here’s a cell to tell whether the project was completed by week 43: A B C D E 27 8 47.50 28 9 65.83 29 10 67.83 <= 43? =IF(B30<43,1,0) 30 11 69.83 0 Here’s a cell to keep track of the penalty (if any) Sharon will have to pay. Note that we have assumed that the penalty applies continuously to any part of a week. A B C D E F G H I 28 9 65.83 29 10 67.83 <= 43? Penalty =IF(B30>43,10000*(B30-43),0) 30 11 69.83 0 $ 268,333

B60.2350 7 Prof. Juran Crystal Ball For each of the random activities, we create an assumption cell, as shown here for Activity A:

Here’s the model after doing this for every random activity time (Activities D, F, I, L, M, and the Dummy activity have no variability): A B C D E F G H I J K L M N O 1 Activity Description Predecessors Start Node End Node Min Mode Max Mean StDev Alpha Beta Simulated Time Start Time End Time 2 A Survey Site None 0 1 2 3 4 3 0.333 4.000 4.000 5.00 0.00 5.00 3 B Excavation A 1 2 9 12 15 12 1.000 4.000 4.000 21.00 5.00 26.00 4 C Prepare Drawings None 0 3 4 9 18 9.667 2.333 3.106 4.568 13.67 0.00 13.67 5 D Soil Study B 2 3 1 1 1 1 0.000 1.00 26.00 27.00 6 E Prelim. Report C, D 3 4 1 2 3 2 0.333 4.000 4.000 3.00 27.00 30.00 7 F Approve Plans E 4 5 1 1 1 1 0.000 1.00 30.00 31.00 8 G Concrete Forms F 5 7 5 6 9 6.333 0.667 2.333 4.667 11.33 31.00 42.33 9 H Procure Steel F 5 6 2 5 10 5.333 1.333 3.229 4.521 7.33 31.00 38.33 10 I Order Cement F 5 8 1 1 1 1 0.000 1.00 31.00 32.00 11 Dummy H 6 8 0 0 0 0 0.000 0.00 38.33 38.33 12 J Deliver Gravel G 7 8 2 3 5 3.167 0.500 2.938 4.617 5.17 42.33 47.50 13 K Pour Concrete H, I, J 8 9 8 10 14 10.33 1.000 2.938 4.617 18.33 47.50 65.83 14 L Cure Concrete K 9 10 2 2 2 2 0.000 2.00 65.83 67.83 15 M Strength Test L 10 11 2 2 2 2 0.000 2.00 67.83 69.83 Now we create forecast cells to track the completion time of the whole project (B30) as well as the criticalities of the various paths (H19:H24) and activities (N2:N15). We also make forecast cells to track whether the project took longer than 43 weeks, and what the penalty was.

B60.2350 8 Prof. Juran A B C D E F G H I J K L M N O P 1 Activity Description Predecessors Start Node End Node Min Mode Max Mean StDev Alpha Beta Simulated Time Start Time End Time Critical? 2 A Survey Site None 0 1 2 3 4 3 0.333 4.000 4.000 5.00 0.00 5.00 1 3 B Excavation A 1 2 9 12 15 12 1.000 4.000 4.000 21.00 5.00 26.00 1 4 C Prepare Drawings None 0 3 4 9 18 9.67 2.333 3.106 4.568 13.67 0.00 13.67 0 5 D Soil Study B 2 3 1 1 1 1 0.000 1.00 26.00 27.00 1 6 E Prelim. Report C, D 3 4 1 2 3 2 0.333 4.000 4.000 3.00 27.00 30.00 1 7 F Approve Plans E 4 5 1 1 1 1 0.000 1.00 30.00 31.00 1 8 G Concrete Forms F 5 7 5 6 9 6.33 0.667 2.333 4.667 11.33 31.00 42.33 1 9 H Procure Steel F 5 6 2 5 10 5.33 1.333 3.229 4.521 7.33 31.00 38.33 0 10 I Order Cement F 5 8 1 1 1 1 0.000 1.00 31.00 32.00 0 11 Dummy H 6 8 0 0 0 0 0.000 0.00 38.33 38.33 12 J Deliver Gravel G 7 8 2 3 5 3.17 0.500 2.938 4.617 5.17 42.33 47.50 1 13 K Pour Concrete H, I, J 8 9 8 10 14 10.3 1.000 2.938 4.617 18.33 47.50 65.83 1 14 L Cure Concrete K 9 10 2 2 2 2 0.000 2.00 65.83 67.83 1 15 M Strength Test L 10 11 2 2 2 2 0.000 2.00 67.83 69.83 1 16 17 18 Node Time Path Total Critical? 19 0 0 A-B-D-E-F-H-K-L-M 60.67 0 20 1 5.00 C-E-F-H-K-L-M 47.33 0 21 2 26.00 A-B-D-E-F-I-K-L-M 54.33 0 22 3 27.00 C-E-F-I-K-L-M 41.00 0 23 4 30.00 A-B-D-E-F-G-J-K-L-M 69.83 1 24 5 31.00 C-E-F-G-J-K-L-M 56.50 0 25 6 38.33 Max Path 69.83 26 7 42.33 27 8 47.50 28 9 65.83 29 10 67.83 <= 43? Penalty 30 11 69.83 0 $ 268,333 We run the model for 1,000 iterations, and then try to answer the questions.

2. What completion time should Sharon use, if she wants to be 90% confident? The best way to answer that is to look at the percentiles for the Project Time forecast cell:

We would interpret these results to indicate that the project will be completed by week 45.11 with 90% probability.

B60.2350 9 Prof. Juran 3. What is the probability of completion by week 43? We can answer that using the statistics from the “<= 43?” forecast cell, or by using the mean line on the frequency chart:

Another way would be to use the grabber on the frequency chart for the completion time:

This indicates that there is approximately 55% probability that the project will be completed by week 43.

4. Give an estimated probability distribution for the amount of penalties Sharon will have to pay. 5. What is the expected value of the penalty? Here’s the frequency chart and the summary statistics:

B60.2350 10 Prof. Juran The expected penalty is approximately $6,257, although the distribution is strongly right-skewed. While Sharon has some small probability of paying a large penalty ($30,000 or more) there is also a very good chance she will pay little or no penalty at all. With a skewed distribution, it is useful to look at the median as well as the mean. The median of $0 suggests that more than half of the time Sharon will pay no penalty.

B60.2350 11 Prof. Juran 6. Which activities are most likely to be on the critical path? Here are results for the various paths: Path Estimated Probability of being Critical A-B-D-E-F-H-K-L-M 0.001 C-E-F-H-K-L-M 0.000 A-B-D-E-F-I-K-L-M 0.000 C-E-F-I-K-L-M 0.000 A-B-D-E-F-G-J-K-L-M 0.994 C-E-F-G-J-K-L-M 0.005 It looks like almost all of the time the A-B-D-E-F-G-J-K-L-M path is critical, with C-E-F-G-J-K-L-M critical 0.5% of the time. One time out of the 1,000 iterations, the A-B-D-E-F-H-K-L-M path was critical. Here are results for the various activities, sorted in descending order of criticality: Activity Estimated Probability of being Critical E 1.000 F 1.000 K 1.000 L 1.000 M 1.000 G 0.999 J 0.999 A 0.995 B 0.995 D 0.995 C 0.005 H 0.001 I 0.000

B60.2350 12 Prof. Juran 7. Compare the PERT results to those you would have found using (a) basic CPM using the most-likely times, (b) the “by-hand” PERT method from the textbook, and (c) HOM. Details about the construction and operation of the “Textbook” and HOM methods appear as appendices to this document. CPM analysis gives a completion time of 42 weeks. The critical path is A-B-D-E- F-G-J-K-L-M Here are summaries of the results obtained from the four probabilistic methods, with some discussion as to their relative strengths and weaknesses. 90% Completion Time: Crystal Ball HOM (simulation) Textbook HOM (no simulation) 45.11 44.62 - 45.67 46.57 44.88 - 45.22 Discussion: All of these results are consistent with each other; the estimates are all within a narrow range. The HOM output provides somewhat arbitrary intervals, which makes it difficult to specify the completion time associated with a particular probability. The “Textbook” method is based on the assumption that the probability distribution of the total project time is normal. Probability of Completion by Week 43 Crystal Ball HOM (simulation) Textbook HOM (no simulation) 0.553 0.4190 – 0.6640* 0.5389 0.5000 - 0.5793* Discussion: Again, the estimates are all consistent with each other. In this case the HOM intervals are perhaps too wide to be useful. The “Textbook” method, as above, is based on a normal distribution for the total project time. Expected Penalty Crystal Ball HOM (simulation) Textbook HOM (no simulation) $6,257 $6,284 $0 $6,251 Discussion: Crystal Ball has a distinct advantage in answering this question; not only does it provide a precise estimate of the expected penalty, but it also provides a standard error for this estimate, which would be necessary if we were interested in constructing a confidence interval around the estimate. The HOM simulation method can be used to come up with an estimate, but it is complicated. For the estimate shown above, we took the midpoints of the bins in the completion time frequency histogram, calculated the penalties associated with the midpoints, and calculated the weighted average penalty using HOM’s estimated probabilities as weights. (Strangely, the frequencies added up to 999,

B60.2350 13 Prof. Juran not 1000.) A similar approach was used to create the HOM non-simulation estimate. The “Textbook” method cannot be used to answer this question without employing some difficult calculus on the normal distribution.

B60.2350 14 Prof. Juran Criticality Paths Crystal Ball HOM (sim.) Textbook HOM (no sim.) A-B-D-E-F-H-K-L-M 0.001 0.000 0.000 0.002 C-E-F-H-K-L-M 0.000 0.000 0.000 A-B-D-E-F-I-K-L-M 0.000 0.000 0.000 0.000 C-E-F-I-K-L-M 0.000 0.000 0.000 0.000 A-B-D-E-F-G-J-K-L-M 0.994 1.000 1.000 0.998 C-E-F-G-J-K-L-M 0.005 0.000 0.000

Activities Crystal Ball HOM (simulation) Textbook HOM (no simulation) A 0.995 0.990 1.000 1.000 B 0.995 0.990 1.000 1.000 C 0.005 0.010 0.000 0.000 D 0.995 0.990 1.000 1.000 E 1.000 1.000 1.000 1.000 F 1.000 1.000 1.000 1.000 G 0.999 0.998 1.000 1.000 H 0.001 0.002 0.000 0.000 I 0.000 0.000 0.000 0.000 J 0.999 0.998 1.000 1.000 K 1.000 1.000 1.000 1.000 L 1.000 1.000 1.000 1.000 M 1.000 1.000 1.000 1.000 Discussion: The non-simulation methods assume that there is only one path that could be critical (the one with the longest expect total time). With these models, any discussion of criticality is not very interesting. To the extent that several paths have the potential to be critical, these methods may underestimate the total time of the project and/or the variability of the project time. The HOM and Crystal Ball simulation models come up with very similar estimated probabilities, although HOM is not set up to look at the paths. Note that we don’t have enough information to completely dissect the frequencies that each path was critical in the HOM output. General Observations: HOM is the easiest of these methods to set up and run. There are a number of questions here where Crystal Ball provides easier/better answers, but it requires more work to set up the model to begin with. This HOM module was created specifically to deal with project management problems, whereas Crystal Ball is a more general simulation program. The choice of which to use would depend on one’s need for flexibility (Crystal Ball having more inherent flexibility), along with other factors.

B60.2350 15 Prof. Juran With respect to the non-simulation methods, they are clearly inferior in terms of answering some of the probabilistic questions in this case (highlighting the advantages of learning how to do simulations), but they do provide reasonable estimates for the other questions.

B60.2350 16 Prof. Juran Appendix: The “Textbook” PERT Method

The textbook method involves (a) finding the means and standard deviations for each path, (b) determining which path has the longest expected total time, and (c) summing the variances of the activities on that path to get the variance of the path. In our case, the longest path would be A-B-D-E-F-G-J-K-L-M, with a mean of 42.83 weeks and a variance of 2.92 weeks. A B C D E F G H I J K L M 1 Activity Description Predecessors Start Node End Node Min Mode Max Mean StDev Var 2 A Survey Site None 0 1 2 3 4 3 0.333 0.111 3 B Excavation A 1 2 9 12 15 12 1.000 1.000 4 C Prepare Drawings None 0 3 4 9 18 9.67 2.333 5.444 5 D Soil Study B 2 3 1 1 1 1 0.000 0.000 6 E Prelim. Report C, D 3 4 1 2 3 2 0.333 0.111 7 F Approve Plans E 4 5 1 1 1 1 0.000 0.000 8 G Concrete Forms F 5 7 5 6 9 6.33 0.667 0.444 9 H Procure Steel F 5 6 2 5 10 5.33 1.333 1.778 10 I Order Cement F 5 8 1 1 1 1 0.000 0.000 11 Dummy H 6 8 0 0 0 0 0.000 0.000 12 J Deliver Gravel G 7 8 2 3 5 3.17 0.500 0.250 13 K Pour Concrete H, I, J 8 9 8 10 14 10.3 1.000 1.000 14 L Cure Concrete K 9 10 2 2 2 2 0.000 0.000 15 M Strength Test L 10 11 2 2 2 2 0.000 0.000 16 17 18 Path Sum of Means 19 A-B-D-E-F-H-K-L-M 38.67 20 C-E-F-H-K-L-M 32.33 21 A-B-D-E-F-I-K-L-M 34.33 22 C-E-F-I-K-L-M 28.00 =SUM(K2,K3,K5,K6,K7,K8,K12,K13,K14,K15) 23 A-B-D-E-F-G-J-K-L-M 42.83 2.92 24 C-E-F-G-J-K-L-M 36.50 Once we have the mean and variance, we can estimate the probability of finishing by week 43, assuming that the total time is normally distributed: D E F G H I J K L M 18 Path Sum of Means Sum of Variances 19 A-B-D-E-F-H-K-L-M 38.67 20 C-E-F-H-K-L-M 32.33 21 A-B-D-E-F-I-K-L-M 34.33 22 C-E-F-I-K-L-M 28.00 =SQRT(H23) 23 A-B-D-E-F-G-J-K-L-M 42.83 2.92 1.708 24 C-E-F-G-J-K-L-M 36.50 25 =NORMINV(0.9,G23,I23) 26 90% completion time 45.02 =NORMDIST(43,G23,I23,1) 27 Prob(X < 43) 0.5389 28

B60.2350 17 Prof. Juran Appendix: HOM Model

Here is the HOM data sheet:

The HOM program allows for two possibilities, depending on whether the “Run Simulation” box is checked in the HOM parameters. HOM without Simulation

B60.2350 18 Prof. Juran When the box is not checked, HOM replicates the “textbook” method, and gives the following output:

Activity Early Early Late Late Name Start Finish Start Finish Slack ======A 0 3 0 3 0 B 3 15 3 15 0 C 0 9.66667 6.33333 16 6.33333 D 15 16 15 16 0 E 16 18 16 18 0 F 18 19 18 19 0 G 19 25.3333 19 25.3333 0 H 19 24.3333 23.1667 28.5 4.16667 I 19 20 27.5 28.5 8.5 J 25.3333 28.5 25.3333 28.5 0 K 28.5 38.8333 28.5 38.8333 0 L 38.8333 40.8333 38.8333 40.8333 0 M 40.8333 42.8333 40.8333 42.8333 0

Expected Completion Time : 42.8333 Standard Deviation on CP : 1.70783

Critical Path: A B D E F G J K L M

B60.2350 19 Prof. Juran Probability of Project Completion.

Time to Probablity Complete of Project Completion ======37.7099 0.001354 38.0514 0.002562 38.3930 0.004670 38.7346 0.008208 39.0761 0.013914 39.4177 0.022758 39.7592 0.035934 40.1008 0.054797 40.4424 0.080748 40.7839 0.115059 41.1255 0.158649 41.4671 0.211858 41.8086 0.274264 42.1502 0.344587 42.4918 0.420734 42.8333 0.500000 43.1749 0.579266 43.5165 0.655413 43.8580 0.725736 44.1996 0.788142 44.5412 0.841351 44.8827 0.884941 45.2243 0.919251 45.5659 0.945203 45.9074 0.964066 46.2490 0.977241 46.5905 0.986086 46.9321 0.991792 47.2737 0.995330 47.6152 0.997438 47.9568 0.998645

B60.2350 20 Prof. Juran HOM with Simulation

When the “Run Simulation” box is checked, HOM yields the following output:

PERT Simulation (Criticality) Calculations Results. Activity Criticality Name Indices Description ======A 99.0% B 99.0% C 1.0% D 99.0% E 100.0% F 100.0% G 99.8% H 0.2% I 0.0% J 99.8% K 100.0%

B60.2350 21 Prof. Juran L 100.0% M 100.0%

B60.2350 22 Prof. Juran Project completion time: Minimum : 38.3251 Maximum : 48.8158 Mean : 42.893 Standard Deviation : 1.74677

Completion frequency distribution (1000 runs performed): Interval Interval Frequency Start End (%) (Counts) ======38.33 39.37 1.6 16 39.37 40.42 6.3 63 40.42 41.47 12.9 129 41.47 42.52 21.1 211 42.52 43.57 24.5 245 43.57 44.62 17.1 171 44.62 45.67 10.7 107 45.67 46.72 4.2 42 46.72 47.77 1.3 13 47.77 48.82 0.2 2

B60.2350 23 Prof. Juran