1. Advanced Simulation Methods

2. © The McGraw-Hill Companies, Inc., 2004 Operations -- Prof. Juran2 Overview Advanced Simulation Applications Beta Distribution Operations –Project Management (PERT) “Textbook” method Crystal Ball Marketing –New Product Development decision

3. © The McGraw-Hill Companies, Inc., 2004 Operations -- Prof. Juran3 Beta Distribution Parameter Description Characteristics Min Minimum Value Any number-∞ to ∞ Max Maximum Value Any number-∞ to ∞ Alpha (α) Shape Factor Must be > 0 Beta (β) Shape Factor Must be > 0 The Beta distribution is a continuous probability distribution defined by four parameters:

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5. © The McGraw-Hill Companies, Inc., 2004 Operations -- Prof. Juran5 The Beta distribution is popular among simulation modelers because it can take on a wide variety of shapes, as shown in the graphs above. The Beta can look similar to almost any of the important continuous distributions, including Triangular, Uniform, Exponential, Normal, Lognormal, and Gamma. For this reason, the Beta distribution is used extensively in PERT, CPM and other project planning/control systems to describe the time to completion of a task.

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7. © The McGraw-Hill Companies, Inc., 2004 Operations -- Prof. Juran7 The project management community has evolved approximations for the Beta distribution which allow it to be handled with three parameters, rather than four. The three parameters are the minimum, mode, and maximum activity times (usually referred to as the optimistic, most-likely, and pessimistic activity times). This doesn’t give exactly the same results as the mathematically- correct version, but has important practical advantages. Most real-life managers are not comfortable talking about things like probability functions and Greek-letter parameters, but they are comfortable talking in terms of optimistic, most-likely, and pessimistic. PERT Approximations

8. © The McGraw-Hill Companies, Inc., 2004 Operations -- Prof. Juran8 3-step Procedure

9. © The McGraw-Hill Companies, Inc., 2004 Operations -- Prof. Juran9 Beta Distributions in Crystal Ball The Crystal Ball distribution gallery includes the Beta distribution, but in a form slightly different from the description above. Specifically, Crystal Ball assumes the minimum is zero. Instead of “maximum” or “pessimistic”, it asks for a “Scale” parameter.

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11. © The McGraw-Hill Companies, Inc., 2004 Operations -- Prof. Juran11 Assume we are given optimistic, most-likely, and pessimistic times of 1, 2, and 3 time units, respectively. We first use these parameters to calculate the mean (formula (iii)), standard deviation (formula (iv)), alpha (formula (v)), beta (formula (vi)), and the difference between the maximum and minimum, as shown here: Example

12. © The McGraw-Hill Companies, Inc., 2004 Operations -- Prof. Juran12 Next, we create a Crystal Ball assumption cell in A2, using the parameters shown: We make a cell next to the assumption cell, adding the random number to the minimum. Cell B3 will now be a Beta-distributed random variable with the optimistic, most-likely, and pessimistic activity times we specified.

13. © The McGraw-Hill Companies, Inc., 2004 Operations -- Prof. Juran13 Operations Example: Project Management (PERT) 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.

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16. © The McGraw-Hill Companies, Inc., 2004 Operations -- Prof. Juran16 Here’s an activity-on-arc diagram of the problem:

17. © The McGraw-Hill Companies, Inc., 2004 Operations -- Prof. Juran17 We start a spreadsheet model like this, calculating the mean and standard deviation using the PERT formulas:

18. © The McGraw-Hill Companies, Inc., 2004 Operations -- Prof. Juran18 Now we calculate shape and scale parameters:

19. © The McGraw-Hill Companies, Inc., 2004 Operations -- Prof. Juran19 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 A section for simulating the times of the activities Model Overview

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22. © The McGraw-Hill Companies, Inc., 2004 Operations -- Prof. Juran22 Example: Activity C

23. © The McGraw-Hill Companies, Inc., 2004 Operations -- Prof. Juran23 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. Nodes with only one preceding activity are easier (see Nodes 4 and 8 below).

24. © The McGraw-Hill Companies, Inc., 2004 Operations -- Prof. Juran24 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:

25. © The McGraw-Hill Companies, Inc., 2004 Operations -- Prof. Juran25 Now, for each path, and for each activity, we can set up an IF statement to say whether the path (or activity) was critical for any particular realization of the model:

26. © The McGraw-Hill Companies, Inc., 2004 Operations -- Prof. Juran26 Here’s a cell to tell whether the project was completed by week 43: 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.

27. © The McGraw-Hill Companies, Inc., 2004 Operations -- Prof. Juran27 Crystal Ball For each of the random activities, we create an assumption cell, as shown here for Activity A:

28. © The McGraw-Hill Companies, Inc., 2004 Operations -- Prof. Juran28 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):

29. © The McGraw-Hill Companies, Inc., 2004 Operations -- Prof. Juran29 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.

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35. © The McGraw-Hill Companies, Inc., 2004 Operations -- Prof. Juran35 Question 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. CPM analysis gives a completion time of 42 weeks. The critical path is A-B-D-E-F-G-J-K-L-M

36. © The McGraw-Hill Companies, Inc., 2004 Operations -- Prof. Juran36 “Textbook” 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.

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42. © The McGraw-Hill Companies, Inc., 2004 Operations -- Prof. Juran42 Marketing Example: New Product Development decision Cavanaugh Pharmaceutical Company (CPC) has enjoyed a monopoly on sales of its popular antibiotic product, Cyclinol, for several years. Unfortunately, the patent on Cyclinol is due to expire. CPC is considering whether to develop a new version of the product in anticipation that one of CPC’s competitors will enter the market with their own offering. The decision as to whether or not to develop the new antibiotic (tentatively called Minothol) depends on several assumptions about the behavior of customers and potential competitors. CPC would like to make the decision that is expected to maximize its profits over a ten-year period, assuming a 15% cost of capital.

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44. © The McGraw-Hill Companies, Inc., 2004 Operations -- Prof. Juran44 Customer Demand Analysts estimate that the average annual demand over the next ten years will be normally distributed with a mean of 40 million doses and a standard deviation of 10 million doses, as shown below. This demand is believed to be independent of whether CPC introduces Minothol or whether Cyclinol/Minothol has a competitor.

45. © The McGraw-Hill Companies, Inc., 2004 Operations -- Prof. Juran45 CPC’s market share is expected to be 100% of demand, as long as there is no competition from AMI. In the event of competition, CPC will still enjoy a dominant market position because of its superior brand recognition. However, AMI is likely to price its product lower than CPC’s in an effort to gain market share. CPC’s best analysis indicates that its share of total sales, in the event of competition, will be a function of the price it chooses to charge per dose, as shown below. The Cyclinol product at $7.50 would only retain a 38.1% market share, whereas the Minothol product at $6.00 would have a 55.0% market share.

46. © The McGraw-Hill Companies, Inc., 2004 Operations -- Prof. Juran46 Questions What is the best decision for CPC, in terms of maximizing the expected value of its profits over then next ten years? What is the least risky decision, using the standard deviation of the ten-year profit as a measure of risk? What is the probability that introducing Minothol will turn out to be the best decision?

47. © The McGraw-Hill Companies, Inc., 2004 Operations -- Prof. Juran47 Income statement-like calculations for each of four scenarios 3 Forecasts: NPV in $millions for each decision Yes/No New Product Better U~(0, 1) (whether or not AMI enters market) N~(40, 10) (Total market demand)

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49. © The McGraw-Hill Companies, Inc., 2004 Operations -- Prof. Juran49 Summary Advanced Simulation Applications Beta Distribution Operations –Project Management (PERT) “Textbook” method Crystal Ball Marketing –New Product Development decision