advanced simulation methods. © the mcgraw-hill companies, inc., 2004 operations -- prof. juran2...

Advanced Simulation Methods

© The McGraw-Hill Companies, Inc., 2004

Operations -- Prof. Juran 2

Overview

Advanced Simulation Applications • Beta Distribution• Operations

– Project Management (PERT)• “Textbook” method• Crystal Ball

• Marketing– New Product Development decision



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:



Here are sixteen different Beta distributions, all with a minimum of 0 and a maximum of 100.

β

0.5 1.0 2.0 4.0

0.5

0.0000

0.0050

0.0100

0.0150

0.0200

0.0250

0.0300

0.0350

0.0400

0.0450

0.0500

0 25 50 75 0.0000

0.0050

0.0100

0.0150

0.0200

0.0250

0.0300

0.0350

0.0400

0.0450

0.0500

0 25 50 75 0.0000

0.0050

0.0100

0.0150

0.0200

0.0250

0.0300

0.0350

0.0400

0.0450

0.0500

0 25 50 75 0.0000

0.0050

0.0100

0.0150

0.0200

0.0250

0.0300

0.0350

0.0400

0.0450

0.0500

0 25 50 75

1.0

0.0000

0.0050

0.0100

0.0150

0.0200

0.0250

0.0300

0.0350

0.0400

0.0450

0.0500

0 25 50 75 0.0000

0.0050

0.0100

0.0150

0.0200

0.0250

0.0300

0.0350

0.0400

0.0450

0.0500

0 25 50 75 0.0000

0.0050

0.0100

0.0150

0.0200

0.0250

0.0300

0.0350

0.0400

0.0450

0.0500

0 25 50 75 0.0000

0.0050

0.0100

0.0150

0.0200

0.0250

0.0300

0.0350

0.0400

0 25 50 75

2.0

0.0000

0.0050

0.0100

0.0150

0.0200

0.0250

0.0300

0.0350

0.0400

0.0450

0.0500

0 25 50 75 0.0000

0.0050

0.0100

0.0150

0.0200

0.0250

0.0300

0.0350

0.0400

0.0450

0.0500

0 25 50 75 0.0000

0.0050

0.0100

0.0150

0.0200

0.0250

0.0300

0.0350

0.0400

0.0450

0.0500

0 25 50 75 0.0000

0.0050

0.0100

0.0150

0.0200

0.0250

0.0300

0.0350

0.0400

0.0450

0.0500

0 25 50 75

α

4.0

0.0000

0.0050

0.0100

0.0150

0.0200

0.0250

0.0300

0.0350

0.0400

0.0450

0.0500

0 25 50 75 0.0000

0.0050

0.0100

0.0150

0.0200

0.0250

0.0300

0.0350

0.0400

0.0450

0.0500

0 25 50 75 0.0000

0.0050

0.0100

0.0150

0.0200

0.0250

0.0300

0.0350

0.0400

0.0450

0.0500

0 25 50 75 0.0000

0.0050

0.0100

0.0150

0.0200

0.0250

0.0300

0.0350

0.0400

0.0450

0.0500

0 25 50 75



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.



Mean: min - max min

(i)

Standard Deviation:

2

2 min - max1

(ii)



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



1. Get estimates for the minimum optimistic, most-likely, and pessimistic time to completion for the activity.

2. Estimate the mean and standard deviation using equations (iii) and (iv):

6

max mode 4min (iii)

6

min - max (iv)

3. Use equations (v) and (vi) to calculate shape factors that are consistent with the mean and standard deviation:

1

mean - maxmin -mean min - maxmin -mean

2 (v)

min -mean mean - max

(vi)

3-step Procedure



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.



We can still use Crystal Ball to simulate Betas, but it requires following these steps.

1. Make an assumption cell with the alpha and beta calculated with formulas (v) and (vi) above.

2. For the “Scale” parameter, enter the difference between the maximum and minimum.

3. In the spreadsheet model, create a separate cell for the activity time. This cell should be the minimum plus the random number generated by Crystal Ball in the assumption cell.



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

123456789

10

A B C D E F G H I J K L MOptimistic (o) Most Likely (m) Pessimistic (p) Mean StDev Alpha Beta Max - Min

1 2 3 2 0.333 4.000 4.000 2

=(D2+4*E2+F2)/6

=(F2-D2)/6

=((G2-D2)/(F2-D2))*((((G2-D2)*(F2-G2))/(H2^2))-1)

=((F2-G2)/(G2-D2))*I2

=F2-D2



12

A B C D E F G H I J KOptimistic (o) Most Likely (m) Pessimistic (p) Mean StDev Alpha Beta Max - Min

2 3 1 2 3 2 0.333 4.000 4.000 2=A2+D2

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.



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.



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



Create a PERT model of this project and use it to answer these questions: 1. What is the expected completion time of this project? 2. What completion time should Sharon use, if she wants to be 90%

confident? 3. What is the probability of completion 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? 6. Which activities are most likely to be on the critical path? 7. Compare the PERT results to those you would have found using (a) basic

CPM using the most-likely times, and (b) the “by-hand” PERT method from the textbook.



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:

123456789101112131415

A B C D E F G H I J K L MActivity Description Optimistic Pessimistic Most-likely Predecessors Start Node End Node Mean StDev

A Survey Site 2 4 3 None 0 1 3.00 0.33B Excavation 9 15 12 A 1 2 12.00 1.00C Prepare Drawings 4 18 9 None 0 3 9.67 2.33D Soil Study 1 1 1 B 2 3 1.00 0.00E Prelim. Report 1 3 2 C, D 3 4 2.00 0.33F Approve Plans 1 1 1 E 4 5 1.00 0.00G Concrete Forms 5 9 6 F 5 7 6.33 0.67H Procure Steel 2 10 5 F 5 6 5.33 1.33I Order Cement 1 1 1 F 5 8 1.00 0.00

Dummy 0 0 0 H 6 8 0.00 0.00J Deliver Gravel 2 5 3 G 7 8 3.17 0.50K Pour Concrete 8 14 10 H, I, J 8 9 10.33 1.00L Cure Concrete 2 2 2 K 9 10 2.00 0.00M Strength Test 2 2 2 L 10 11 2.00 0.00

=(C3+4*E3+D3)/6

=(D5-C5)/6



Now we calculate shape and scale parameters:

123456789101112131415

A B C D E F G H I J K L M N O P Q R SActivity Description Optimistic Pessimistic Most-likely Predecessors Start Node End Node Mean StDev Alpha Beta Scale

A Survey Site 2 4 3 None 0 1 3.00 0.33 4.00 4.00 2B Excavation 9 15 12 A 1 2 12.00 1.00 4.00 4.00 6C Prepare Drawings 4 18 9 None 0 3 9.67 2.33 3.11 4.57 14D Soil Study 1 1 1 B 2 3 1.00 0.00 0E Prelim. Report 1 3 2 C, D 3 4 2.00 0.33 4.00 4.00 2F Approve Plans 1 1 1 E 4 5 1.00 0.00 0G Concrete Forms 5 9 6 F 5 7 6.33 0.67 2.33 4.67 4H Procure Steel 2 10 5 F 5 6 5.33 1.33 3.23 4.52 8I Order Cement 1 1 1 F 5 8 1.00 0.00 0

Dummy 0 0 0 H 6 8 0.00 0.00 0J Deliver Gravel 2 5 3 G 7 8 3.17 0.50 2.94 4.62 3K Pour Concrete 8 14 10 H, I, J 8 9 10.33 1.00 2.94 4.62 6L Cure Concrete 2 2 2 K 9 10 2.00 0.00 0M Strength Test 2 2 2 L 10 11 2.00 0.00 0

=((I3-C3)/(D3-C3))*((((I3-C3)*(D3-I3))/(J3^2))-1)=((D4-I4)/(I4-C4))*K4

=D5-C5



123456789101112131415161718192021222324252627282930

A B C D E F G H I J K L M NActivity Description Predecessors Start Node End Node Min Alpha Beta Scale CB Time Simulated Time Start Time End Time Critical?

A Survey Site None 0 1 2 4.00 4.00 2 3.00 5.00 0.00 5.00 1B Excavation A 1 2 9 4.00 4.00 6 12.00 21.00 5.00 26.00 1C Prepare Drawings None 0 3 4 3.11 4.57 14 9.67 13.67 0.00 13.67 0D Soil Study B 2 3 1 0 1.00 1.00 26.00 27.00 1E Prelim. Report C, D 3 4 1 4.00 4.00 2 2.00 3.00 27.00 30.00 1F Approve Plans E 4 5 1 0 1.00 1.00 30.00 31.00 1G Concrete Forms F 5 7 5 2.33 4.67 4 6.33 11.33 31.00 42.33 1H Procure Steel F 5 6 2 3.23 4.52 8 5.33 7.33 31.00 38.33 0I Order Cement F 5 8 1 0 1.00 1.00 31.00 32.00 0

Dummy H 6 8 0 0 0.00 0.00 38.33 38.33J Deliver Gravel G 7 8 2 2.94 4.62 3 3.17 5.17 42.33 47.50 1K Pour Concrete H, I, J 8 9 8 2.94 4.62 6 10.33 18.33 47.50 65.83 1L Cure Concrete K 9 10 2 0 2.00 2.00 65.83 67.83 1M Strength Test L 10 11 2 0 2.00 2.00 67.83 69.83 1

Node Time Path Total Critical?0 0 A-B-D-E-F-H-K-L-M 60.67 01 5.00 A-C-E-F-H-K-L-M 52.33 02 26.00 A-B-D-E-F-I-K-L-M 54.33 03 27.00 A-C-E-F-I-K-L-M 46.00 04 30.00 A-B-D-E-F-G-J-K-L-M 69.83 15 31.00 A-C-E-F-G-J-K-L-M 61.50 06 38.33 69.837 42.338 47.509 65.8310 67.83 <= 43? Penalty11 69.83 0 268,333$

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



Here’s the section keeping track of the activity times (we’ll fill in the stuff in column J with Crystal Ball assumption cells later). Because of the way Crystal Ball handles Beta distributions, we need to add the random numbers (column J) to the minimums (column F) to get the simulated activity times (column K).

123456789

101112131415

A B C D E F G H I J K LActivity Description Predecessors Start Node End Node Min Alpha Beta Scale CB Time Simulated Time Start Time

A Survey Site None 0 1 2 4.00 4.00 2 3.00 5.00B Excavation A 1 2 9 4.00 4.00 6 12.00 21.00C Prepare Drawings None 0 3 4 3.11 4.57 14 9.67 13.67D Soil Study B 2 3 1 0 1.00 1.00E Prelim. Report C, D 3 4 1 4.00 4.00 2 2.00 3.00F Approve Plans E 4 5 1 0 1.00 1.00G Concrete Forms F 5 7 5 2.33 4.67 4 6.33 11.33H Procure Steel F 5 6 2 3.23 4.52 8 5.33 7.33I Order Cement F 5 8 1 0 1.00 1.00

Dummy H 6 8 0 0 0.00 0.00J Deliver Gravel G 7 8 2 2.94 4.62 3 3.17 5.17K Pour Concrete H, I, J 8 9 8 2.94 4.62 6 10.33 18.33L Cure Concrete K 9 10 2 0 2.00 2.00M Strength Test L 10 11 2 0 2.00 2.00

=F4+J4=F5



Now we set up an area in the spreadsheet to keep track of the nodes and their times:

18192021222324252627282930

A BNode Time

0 0123456789

1011

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.



Example: Activity C

123456789

101112131415161718192021222324252627282930

A B C D E F G H I J K L M NActivity Description Predecessors Start Node End Node Min Alpha Beta Scale CB Time Simulated Time Start Time End Time

A Survey Site None 0 1 2 4.00 4.00 2 3.00 5.00 0 5.00B Excavation A 1 2 9 4.00 4.00 6 12.00 21.00 0 21.00C Prepare Drawings None 0 3 4 3.11 4.57 14 9.67 13.67 0 13.67D Soil Study B 2 3 1 0 1.00 1.00 0 1.00E Prelim. Report C, D 3 4 1 4.00 4.00 2 2.00 3.00 0 3.00F Approve Plans E 4 5 1 0 1.00 1.00 0 1.00G Concrete Forms F 5 7 5 2.33 4.67 4 6.33 11.33 0 11.33H Procure Steel F 5 6 2 3.23 4.52 8 5.33 7.33 0 7.33I Order Cement F 5 8 1 0 1.00 1.00 0 1.00

Dummy H 6 8 0 0 0.00 0.00 0 0.00J Deliver Gravel G 7 8 2 2.94 4.62 3 3.17 5.17 0 5.17K Pour Concrete H, I, J 8 9 8 2.94 4.62 6 10.33 18.33 0 18.33L Cure Concrete K 9 10 2 0 2.00 2.00 0 2.00M Strength Test L 10 11 2 0 2.00 2.00 0 2.00

Node Time0 0123456789

1011

0

1

2

3

4

5

7

8

9

10

11

A

C

B

D

E

F

H

G J K

I

6

Dummy

L

M

0

1

2

3

4

5

7

8

9

10

11

A

C

B

D

E

F

H

G J K

I

6

Dummy

L

M

=VLOOKUP(D4,$A$19:$B$30,2,0)

=L4+K4



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).

123456789101112131415161718192021222324252627282930

A B C D E F G H I J K L MActivity Description Predecessors Start Node End Node Min Alpha Beta Scale CB Time Simulated Time Start Time End Time

A Survey Site None 0 1 2 4.00 4.00 2 3.00 5.00 0.00 5.00B Excavation A 1 2 9 4.00 4.00 6 12.00 21.00 5.00 26.00C Prepare Drawings None 0 3 4 3.11 4.57 14 9.67 13.67 0.00 13.67D Soil Study B 2 3 1 0 1.00 1.00 26.00 27.00E Prelim. Report C, D 3 4 1 4.00 4.00 2 2.00 3.00 27.00 30.00F Approve Plans E 4 5 1 0 1.00 1.00 30.00 31.00G Concrete Forms F 5 7 5 2.33 4.67 4 6.33 11.33 31.00 42.33H Procure Steel F 5 6 2 3.23 4.52 8 5.33 7.33 31.00 38.33I Order Cement F 5 8 1 0 1.00 1.00 31.00 32.00

Dummy H 6 8 0 0 0.00 0.00 38.33 38.33J Deliver Gravel G 7 8 2 2.94 4.62 3 3.17 5.17 42.33 47.50K Pour Concrete H, I, J 8 9 8 2.94 4.62 6 10.33 18.33 47.50 65.83L Cure Concrete K 9 10 2 0 2.00 2.00 65.83 67.83M Strength Test L 10 11 2 0 2.00 2.00 67.83 69.83

Node Time0 01 5.002 26.003 27.004 30.005 31.006 38.337 42.338 47.509 65.8310 67.8311 69.83

=MAX(M10:M12)

=M6



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:

12345678910111213141516171819202122232425

A B C D E F G H I J K L MActivity Description Predecessors Start Node End Node Min Alpha Beta Scale CB Time Simulated Time Start Time End Time

A Survey Site None 0 1 2 4.00 4.00 2 3.00 5.00 0.00 5.00B Excavation A 1 2 9 4.00 4.00 6 12.00 21.00 5.00 26.00C Prepare Drawings None 0 3 4 3.11 4.57 14 9.67 13.67 0.00 13.67D Soil Study B 2 3 1 0 1.00 1.00 26.00 27.00E Prelim. Report C, D 3 4 1 4.00 4.00 2 2.00 3.00 27.00 30.00F Approve Plans E 4 5 1 0 1.00 1.00 30.00 31.00G Concrete Forms F 5 7 5 2.33 4.67 4 6.33 11.33 31.00 42.33H Procure Steel F 5 6 2 3.23 4.52 8 5.33 7.33 31.00 38.33I Order Cement F 5 8 1 0 1.00 1.00 31.00 32.00

Dummy H 6 8 0 0 0.00 0.00 38.33 38.33J Deliver Gravel G 7 8 2 2.94 4.62 3 3.17 5.17 42.33 47.50K Pour Concrete H, I, J 8 9 8 2.94 4.62 6 10.33 18.33 47.50 65.83L Cure Concrete K 9 10 2 0 2.00 2.00 65.83 67.83M Strength Test L 10 11 2 0 2.00 2.00 67.83 69.83

Node Time Path Total0 0 A-B-D-E-F-H-K-L-M 60.671 5.00 A-C-E-F-H-K-L-M 52.332 26.00 A-B-D-E-F-I-K-L-M 54.333 27.00 A-C-E-F-I-K-L-M 46.004 30.00 A-B-D-E-F-G-J-K-L-M 69.835 31.00 A-C-E-F-G-J-K-L-M 61.506 38.33 69.83

=SUM(K2,K3,K5,K6,K7,K9,K13,K14,K15)

=MAX(G19:G24)



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:

12345678910111213141516171819202122232425

A B C D E F G H I J K L M N O P QActivity Description Predecessors Start Node End Node Min Alpha Beta Scale CB Time Simulated Time Start Time End Time Critical?



Node Time Path Total Critical?0 0 A-B-D-E-F-H-K-L-M 60.67 01 5.00 A-C-E-F-H-K-L-M 52.33 02 26.00 A-B-D-E-F-I-K-L-M 54.33 03 27.00 A-C-E-F-I-K-L-M 46.00 04 30.00 A-B-D-E-F-G-J-K-L-M 69.83 15 31.00 A-C-E-F-G-J-K-L-M 61.50 06 38.33 69.83

=IF(G23=$G$25,1,0)

=SUM(H19,H20)



27282930

A B C D E8 47.509 65.83

10 67.83 <= 43?11 69.83 0

=IF(B30<43,1,0)

282930

A B C D E F G H I9 65.83

10 67.83 <= 43? Penalty11 69.83 0 268,333$

=IF(B30>43,10000*(B30-43),0)

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.



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):

123456789101112131415






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.

123456789101112131415161718192021222324252627282930




Node Time Path Total Critical?0 0 A-B-D-E-F-H-K-L-M 60.67 01 5.00 A-C-E-F-H-K-L-M 52.33 02 26.00 A-B-D-E-F-I-K-L-M 54.33 03 27.00 A-C-E-F-I-K-L-M 46.00 04 30.00 A-B-D-E-F-G-J-K-L-M 69.83 15 31.00 A-C-E-F-G-J-K-L-M 61.50 06 38.33 69.837 42.338 47.509 65.8310 67.83 <= 43? Penalty11 69.83 0 268,333$



Question 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:



Question 3: What is the probability of completion by week 43?

We can answer that using the statistics from the “<= 43?” forecast cell:



Question 4: Give an estimated probability distribution for the amount of penalties Sharon will have to pay.

Question 5: What is the expected value of the penalty?

Here’s the frequency chart and the summary statistics:



Question 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 0.0010 A-C-E-F-H-K-L-M 0.0000 A-B-D-E-F-I-K-L-M 0.0000 A-C-E-F-I-K-L-M 0.0000 A-B-D-E-F-G-J-K-L-M 0.8940 A-C-E-F-G-J-K-L-M 0.1050



Here are results for the various activities, sorted in descending order of criticality:

Activity Estimated Probability of being Critical A 1.0000 E 1.0000 F 1.0000 K 1.0000 L 1.0000 M 1.0000 G 0.9990 J 0.9990 B 0.8950 D 0.8950 C 0.1050 H 0.0010 I 0.0000



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

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 7 16 7 D 15 16 15 16 0 E 16 18 16 18 0 F 18 19 18 19 0 G 19 25 19 25 0 H 19 24 23 28 4 I 19 20 27 28 8 J 25 28 25 28 0 K 28 38 28 38 0 L 38 40 38 40 0 M 40 42 40 42 0



“Textbook” MethodThe 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.

1234567891011121314151617181920212223242526

A B C D E F G H I JActivity Description Optimistic Pessimistic Most-likely Predecessors Start Node End Node Mean StDev

A Survey Site 2 4 3 None 0 1 3.00 0.33B Excavation 9 15 12 A 1 2 12.00 1.00C Prepare Drawings 4 18 9 None 0 3 9.67 2.33D Soil Study 1 1 1 B 2 3 1.00 0.00E Prelim. Report 1 3 2 C, D 3 4 2.00 0.33F Approve Plans 1 1 1 E 4 5 1.00 0.00G Concrete Forms 5 9 6 F 5 7 6.33 0.67H Procure Steel 2 10 5 F 5 6 5.33 1.33I Order Cement 1 1 1 F 5 8 1.00 0.00

Dummy 0 0 0 H 6 8 0.00 0.00J Deliver Gravel 2 5 3 G 7 8 3.17 0.50K Pour Concrete 8 14 10 H, I, J 8 9 10.33 1.00L Cure Concrete 2 2 2 K 9 10 2.00 0.00M Strength Test 2 2 2 L 10 11 2.00 0.00

Path Sum of Means Sum of VariancesA-B-D-E-F-H-K-L-M 38.67A-C-E-F-H-K-L-M 35.33A-B-D-E-F-I-K-L-M 34.33A-C-E-F-I-K-L-M 31.00A-B-D-E-F-G-J-K-L-M 42.83 2.92A-C-E-F-G-J-K-L-M 39.50

=SUM(I2,I3,I5,I6,I7,I8,I12:I15)

=SUM(J2^2,J3^2,J5^2,J6^2,J7^2,J8^2,J12^2, J13^2,J14^2,J15^2)



O nce w e hav e th e m ean and v ar iance, w e can estim ate the p robabi l i ty o f fi n ish ing by w eek 43, assu m ing that the total tim e is norm al ly d istr ibu ted :

1 81 92 02 12 22 32 42 52 62 7

D E F G H I J K L M NP a th S u m o f M e a n s S u m o f V a r ia n c e sA -B -D -E -F -H -K -L -M 3 8 .6 7A -C -E -F -H -K -L -M 3 5 .3 3A -B -D -E -F - I-K -L -M 3 4 .3 3A -C -E -F - I-K -L -M 3 1 .0 0A -B -D -E -F -G -J -K -L -M 4 2 .8 3 2 .9 2 1 .7 1A -C -E -F -G -J -K -L -M 3 9 .5 0

9 0 % c o m p le t io n t im e 4 5 .0 2 2P ro b (X < 4 3 ) 0 .5 3 8 9

= S Q R T (H 2 3 )

= N O R M IN V (0 .9 ,G 2 3 ,I2 3 )= N O R M D IS T (4 3 ,G 2 3 ,I2 3 ,1 )



90% Completion Time: Crystal Ball Textbook

45.11 46.57

Discussion: These results are consistent with each other; the estimates are both within a narrow range. 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 Textbook

0.553 0.5389

Discussion: Again, the estimates are consistent with each other. The “Textbook” method, as before, is based on a normal distribution for the total project time.



Expected Penalty Crystal Ball Textbook

$6,257 N/ A

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 “Textbook” method cannot be used to answer this question without employing some difficult calculus on the normal distribution.



Criticality Paths Crystal Ball

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

Activities Crystal Ball Textbook

A 0.995 1.000 B 0.995 1.000 C 0.005 0.000 D 0.995 1.000 E 1.000 1.000 F 1.000 1.000 G 0.999 1.000 H 0.001 0.000 I 0.000 0.000 J 0.999 1.000 K 1.000 1.000 L 1.000 1.000 M 1.000 1.000

Discussion: The textbook method assumes that there is only one path that could be critical (the one with the longest expect total time). With this model, any discussion of criticality is not very interesting. To the extent that several paths have the potential to be critical, the textbook method may underestimate the total time of the project and/ or the variability of the project time.



Marketing Example: New Product Development

decisionCavanaugh 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.



Costs and Revenues

Cyclinol costs $1.00 per dose to manufacture, and sells for $7.50 per dose. The proposed Minothol product would cost $0.90 per dose and sell for $6.00, allowing CPC to protect its market share against lower-priced competition. This would, however, require a one-time investment of $140 million.

Cyclinol Minothol Fixed Cost None $140 million Variable Cost $1.00 $0.90 Selling Price $7.50 $6.00

Competition

There is really only one other company with the potential to enter the market, Ahrens MethLabs, Inc. (AMI). Competitive analysis indicates that AMI is 20% likely to introduce a competing product if CPC stays with the higher-priced Cyclinol product, but only 5% likely to enter the market if CPC introduces Minothol.



Customer DemandAnalysts 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.

Average Antibiotic Demand

0.000

0.005

0.010

0.015

0.020

0.025

0.030

10 25 40 55

Millions of Doses

Pro

ba

bil

ity



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.

Market Share vs. Price

0%

20%

40%

60%

80%

100%

$- $2 $4 $6 $8 $10 $12 $14

Price

Mar

ket

Sh

are



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?



123456789

10111213141516171819

A B C D E0.70928

No Competition Competition No Competition CompetitionPrice 7.50$ 7.50$ 6.00$ 6.00$

Total Demand 40.2P(Competition) 20% 5%Competition? No NoMarket Share 100.0% 38.1% 100.0% 55.0%Cyclinol Units Sold 40.2 15.3 40.2 22.1Revenue 301.28$ 114.86$ 241.02$ 132.56$ Fixed Cost -$ -$ 140.00$ 140.00$ Variable Cost 1.00$ 1.00$ 0.90$ 0.90$ Annual profit 261.11$ 99.55$ 204.87$ 112.68$ Discount Rate 15%10-year PV $1,310.45 $499.61 $888.20 $425.51

ResultsCyclinol 1,310.45$ Minothol 888.20$ Minothol Better? 0

Cyclinol Minothol

Income statement-like calculations for each of four

scenarios

3 Forecasts:NPV in $millions for each

decisionYes/No New Product Better

U~(0, 1)(whether or not

AMI enters market)

N~(40, 10)(Total market

demand)



Summary

Advanced Simulation Applications • Beta Distribution• Operations

– Project Management (PERT)• “Textbook” method• Crystal Ball

• Marketing– New Product Development decision

advanced simulation methods. © the mcgraw-hill companies, inc., 2004 operations -- prof. juran2...

Documents

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