Chapter 15

Simulation

Learning Objectives

1.Understand what simulation is and how it aids in the analysis of a problem.

2.Learn why simulation is a significant problem-solving tool.

3.Understand the difference between static and dynamic simulation.

4.Identify the important role probability distributions, random numbers, and the computer play in implementing simulation models.

6.Understand the following terms:

simulationMonte Carlo simulation

simulation modeldiscrete-event simulation

Solutions:

1.a.Profit=(249 - c1 - c2 ) x - 1,000,000

=(249 - 45 - 90) (20,000) - 1,000,000

=\$1,280,000 (Engineer's)

b.Profit=(249 - 45 - 100) (10,000) - 1,000,000

=\$40,000 (Financial Analyst)

c.Simulation will provide probability information about the various profit levels possible. What if scenarios show possible profit outcomes but do not provide probability information.

2.a.Letc=variable cost per unit

x=demand

Profit=50x - cx - 30,000

=(50 - c) x - 30,000

b.Base case:Profit=(50 - 20) 1200 - 30,000 = 6,000

Worst case:Profit=(50 - 24) 300 - 30,000 = -22,200

Best case:Profit=(50 - 16) 2100 - 30,000 = 41,400

c.The possibility of a \$41,400 profit is interesting, but the worst case loss of \$22,200 is risky. Risk analysis would be helpful in evaluating the probability of a loss.

3.

 Random Number Direct Labor Cost 0.3753 \$45 0.9218 \$47 0.0336 \$43 0.5145 \$45 0.7000 \$46

4.a.

 Sales Interval 0 .00 but less than .08 1 .08 but less than .20 2 .20 but less than .48 3 .48 but less than .72 4 .72 but less than .86 5 .86 but less than .96 6 .96 but less than 1.00

b.2, 5, 2, 3, 2, 4, 2, 1, 1, 2

c.Total Sales = 24 units

5.a.

 Stock Price Change Probability Interval -2 .05 .00 but less than .05 -1 .10 .05 but less than .15 0 .25 .15 but less than .40 +1 .20 .40 but less than .60 +2 .20 .60 but less than .80 +3 .10 .80 but less than .90 +4 .10 .90 but less than 1.00

b.

 Random Number Price Change Ending Price Per Share 0.1091 -1 \$38 0.9407 +4 \$42 0.1941 0 \$42 0.8083 +3 \$45

Ending price per share = \$45

6.a.

 Number of New Accounts Opened Probability Interval of Random Numbers 0 .01 .00 but less than .01 1 .04 .01 but less than .05 2 .10 .05 but less than .15 3 .25 .15 but less than .40 4 .40 .40 but less than .80 5 .15 .80 but less than .95 6 .05 .95 but less than 1.00

b.

 Trial Random Number Number of New Accounts Opened 1 0.7169 4 2 0.2186 3 3 0.2871 3 4 0.9155 5 5 0.1167 2 6 0.9800 6 7 0.5029 4 8 0.4154 4 9 0.7872 4 10 0.0702 2

c.For the 10 trials Gustin opened 37 new accounts. With an average first year commission of \$5000 per account, the total first year commission is \$185,000. The cost to run the 10 seminars is \$35,000, so the net contribution to profit for Gustin is \$150,000 or \$15,000 per seminar. Because the seminars are a very profitable way of generating new business, Gustin should continue running the seminars.

7.Time=a + r (b - a )

=10 + r (18 - 10)

=10 + 8r

 r Time 0.1567 11.25 minutes 0.9823 17.86 minutes 0.3419 12.74 minutes 0.5572 14.46 minutes 0.7758 16.21 minutes

8.a.The following table can be used to simulate a win for Atlanta

 Game Interval for Atlanta Win 1 .00 but less than .60 2 .00 but less than .55 3 .00 but less than .48 4 .00 but less than .45 5 .00 but less than .48 6 .00 but less than .55 7 .00 but less than .50

b.Using the random numbers in column 6 beginning with 0.3813, 0.2159 and so on, Atlanta wins games 1 and 2, loses game 3, wins game 4, loses game 5 and wins game 6. Thus, Atlanta wins the 6-game World Series 4 games to 2 games.

c.Repeat the simulation many times. In each case, record who wins the series and the number of games played, 4, 5, 6 or 7. Count the number of times Atlanta wins. Divide this number by the total number of simulation runs to estimate the probability that Atlanta will win the World Series. Count the number of times the series ends in 4 games and divide this number by the total number of simulation runs to estimate the probability of the World Series ending in 4 games. This can be repeated for 5-game, 6-game and 7-game series.

9.a.Base case using most likely completion times.

 A 6 B 5 C 14 D 8 33 weeks

Worst case:8 + 7 + 18 + 10 = 43 weeks

Best case:5 + 3 + 10 + 8 = 26 weeks

b.

 Activity Random Number Completion Time A 0.1778 5 B 0.9617 7 C 0.6849 14 D 0.4503 8 Total: 34 Weeks

c.Simulation will provide a distribution of project completion time values. Calculating the percentage of simulation trials with completion times of 35 weeks or less can be used to estimate the probability of meeting the completion time target of 35 weeks.

10.a.

 Hand Value Probability Interval 17 .1654 .0000 but less than .1654 18 .1063 .1654 but less than .2717 19 .1063 .2717 but less than .3780 20 .1017 .3780 but less than .4797 21 .0972 .4797 but less than .5769 Broke .4231 .5769 but less than 1.000

b/c.

 Hand Dealer Value Player Value Hand Dealer Value Player Value 1 Broke Broke 11 21 17 2 18 Broke 12 Broke Broke 3 21 17 13 17 Broke 4 17 Broke 14 Broke 20 5 21 21 15 18 20 6 17 17 16 Broke 18 7 18 17 17 19 Broke 8 18 Broke 18 Broke 20 9 Broke 17 19 20 Broke 10 Broke Broke 20 21 Broke

d.Dealer wins 13: 1-4, 7, 8, 10-13, 17, 19, 20

Pushes = 2: 5, 6

Player wins 5: 9, 14, 15, 16, 18

At a bet of \$10 per hand, the player loses \$80.

e.Player wins 7: 1, 9, 10, 12, 14, 16, 18

At a bet of \$10 per hand, the player loses \$60.

On the basis of these results, we would not recommend the player take a hit on 16 when the dealer is showing a 6.

11.a.Letr=random number

a=smallest value = -8

b=largest value = 12

Return % = a + r(b - a)

= -8 + r(12-(-8)) = -8 + r20

1st Quarterr = .52

Return % = -8 + .52(20) = 2.4%

For all quarters:

 Quarter r Return % 1 0.52 2.4% 2 0.99 11.8% 3 0.12 -5.6% 4 0.15 -5.0% 5 0.50 2.0% 6 0.77 7.4% 7 0.40 0.0% 8 0.52 2.4%

b.For each quarter,

Ending price = Beginning price + Change

For Quarter 1: Ending price=\$80.00 + .024(\$80.00)

=\$80.00 + \$1.92 = \$81.92

For Quarter 2:Ending price=\$81.92 + .118(\$81.92)

=\$81.92 + \$9.67 = \$91.59

 Quarter Starting Price/Share Return % Change \$ Ending Price/Share
 1 \$80.00 2.4% \$1.92 \$81.92 2 \$81.92 11.8% \$9.67 \$91.59 3 \$91.59 -5.6% -\$5.13 \$86.46 4 \$86.46 -5.0% -\$4.32 \$82.13 5 \$82.13 2.0% \$1.64 \$83.78 6 \$83.78 7.4% \$6.20 \$89.98 7 \$89.98 0.0% \$0.00 \$89.98 8 \$89.98 2.4% \$2.16 \$92.14

Price per share at the end of two years = \$92.14

c.Conducting a risk analysis would require multiple simulations of the eight-quarter, two-year period. For each simulation, the price per share at the end of two years would be recorded. The distribution of the ending price per share values would provide an indication of the maximum possible gain, the maximum possible loss and other possibilities in between.

12.a.Profit = Selling Price - Purchase Cost - Labor Cost - Transportation Cost

Base Case using most likely costs

Profit=45 - 11 - 24 - 3 = \$7/unit

Worst Case

Profit= 45 - 12 - 25 - 5 = \$3/unit

Best Case

Profit=45 - 10 - 20 - 3 = \$12/unit

b.

 Purchase Cost Interval Labor Cost Interval Transportation Cost Interval \$10 .00 but less than .25 \$20 .00 but less than .10 \$3 .00 but less than .75 11 .25 but less than .70 22 .10 but less than .35 5 .75 but less than 1.00 12 .70 but less than 1.00 24 .35 but less than .70 25 .70 but less than 1.00

c.Profit=45 - 11 - 24 - 5 = \$5/unit

d.Profit=45 - 10 - 25 - 3 = \$7/unit

e.Simulation will provide a distribution of the profit per unit values. Calculating the percentage of simulation trials providing a profit less than \$5 per unit would provide an estimate of the probability the profit per unit will be unacceptably low.

13.Use the PortaCom spreadsheet. Simulation results will vary, but a mean profit of approximately \$710,000 with a probability of a loss in the 0.07 to 0.10 range can be anticipated.

14.The Excel worksheet for this problem is as follows:

Selected cell formulas are as follows:

CellFormula

B13=\$C\$7+RAND()*(\$C\$8-\$C\$7)

C13=NORMINV(RAND(),\$G\$7,\$G\$8)

D13=(\$C\$3-B13)*C13-\$C\$4

a.The mean profit should be approximately \$6,000. Simulation results will vary with most simulations having a mean profit between \$5,500 and \$6,500.

b.120 to 150 of the 500 simulation trails should show a loss. Thus, the probability of a loss should be between 0.24 and 0.30.

c.This project appears too risky. The relatively high probability of a loss and only roughly \$6,000 as a mean profit indicate that the potential gain is not worth the risk of a loss. More precise estimates of the variable cost per unit and the demand could help determine a more precise profit estimate.

15.The Excel worksheet for this problem is as follows:

Selected cell formulas are as follows:

CellFormula

B15=VLOOKUP(RAND(),\$A\$6:\$C\$11,3)

C15=VLOOKUP(RAND(),\$A\$6:\$C\$11,3)

D15=B15+C15

H15=COUNTIF(D15:D1014,7)

H16=H15/COUNT(D15:D1014)

Simulation results will vary with most simulations showing between 155 and 180 7's. The probability of a 7 should be approximately 0.1667.

a.Simulation runs will vary. Generally, 340 to 380, or roughly 36% of the simulation runs will show \$130,000 to be the highest and winning bid.

b.\$150,000. Profit = \$160,000 = \$150,000 = \$10,000

c.Again, simulation results will vary. Simulation results should be consistent with the following:

 Amount Bid Win the Bid Profit per Win Average Profit \$130,000 340 to 380 times \$30,000 Approx. \$10,800 \$140,000 620 to 660 times \$20,000 Approx. \$12,800 \$150,000 1000 times \$10,000 \$10,000

Using an average profit criterion, both the \$130,000 and \$140,000 bids are preferred to the \$150,000 bid. Of the three alternatives, \$140,000 is the recommended bid.

17.The Excel worksheet for this problem is as follows:

Selected cell formulas are as follows:

CellFormula

B9=NORMINV(RAND(),\$C\$4,\$C\$5)

F10=COUNTIF(B9:B508,”>40000”)

a.Most simulations will provide between 105 and 130 tires exceeding 40,000 miles. The percentage should be roughly 24%.

b.

 Mileage In Most Simulations Number of Tires Approximate Percentage 32,000 80 to 100 18% 30,000 42 to 55 10% 28,000 18 to 30 4%

c.Of mileages considered, 30,000 miles should come closest to meeting the tire guarantee mileage guideline.

18.The Excel worksheet with data in thousands of dollars is as follows:

Selected cell formulas are as follows:

CellFormula

B11=\$C\$4+RAND()*(\$C\$5-\$C\$4)

C11=NORMINV(RAND(),\$H\$4,\$H\$5)

D11=MAX(B11:C11)

G11=COUNTIF(D11:D1010,”<750”)

H11=G11/COUNT(D11:D1010)

a.Cell G11 provides the number of times the contractor's bid of \$750,000 will beat the highest competitive bid shown in column D. Simulation results will vary but the bid of \$750,000 should win roughly 600 to 650 of the 1000 times. The probability of winning the bid should be between 0.60 and 0.65.

b.Cells G12 and G13 provide the number of times the bids of \$775,000 and \$785,000 win. Again, simulation results vary but the probability of \$750,000 winning should be roughly 0.82 and the probability of \$785,000 winning should be roughly 0.88. Given these results, a contractor's bid of \$775,000 is recommended.

19.Butler Inventory simulation spreadsheet. The shortage cost has been eliminated so \$0 can be entered in cell C5. Trial replenishment levels of 110, 115, 120 and 125 can be entered in cell C7.

Since the shortage cost has been eliminated, Butler can be expected to reduce the replenishment level. This will allow more shortages. However, since the cost of a stockout is only the lost profit and not the lost profit plus a goodwill shortage cost, Butler can permit more shortages and still show an improvement in profit.

A replenishment level of 115 should provide a mean profit of approximately \$4600. The replenishment levels of 110, 115 and 120 all provide near optimal results.

20.The Excel worksheet for this problem is as follows:

Selected cell formulas are as follows:

CellFormula

B14=NORMINV(RAND(),\$H\$4,\$H\$5)

C14=IF(B14<\$D\$8,B14,\$D\$8)

D14=\$D\$5*C14

E14=IF(C14<\$D\$8,(\$D\$8-C14),0)

F14=\$D\$6*E14

G14=\$D\$3+\$D\$4*\$D\$8

H14=D14+F14-G14

The number of stockouts can be computed by using the cell formula

=COUNTIF(E14:E513,”=0”)

a.The simulated mean profit with a production quantity of 60,000 units should be in the \$170,000 to \$210,000 range. The probability of a stockout is about 0.50.

b.The conservative 50,000 unit production quantity is recommended with a simulated mean profit of approximately \$230,000. The more aggressive 70,000 unit production quantity should show a simulated mean profit less than \$100,000.

c.When a 50,000 unit production quantity is used, the probability of a stockout should be approximately 0.75. This is a relative high probability indicating that Mandrell has a good chance of being able to sell all the dolls it produces for the holiday season. As a result, a shortage of dolls is likely to occur. However, this production strategy will enable the company to avoid the high cost associated with selling excess dolls for a loss after the first of the year.

21.The Excel worksheet for this problem is as follows:

Selected cell formulas are as follows:

CellFormula

B16=VLOOKUP(RAND(),\$A\$6:\$C\$10,3)

C16=IF(B16>\$G\$7,\$G\$7,B16)

D16=\$G\$4*C16

E16=B16-C16

F16=\$G\$5*E16

G16=D16-F16

a.Without overbooking, the problem states that South Central has a mean profit of \$2,800 per flight. The overbooking simulation model with a total of 32 reservations (2 overbookings) projects a mean profit of approximately \$2925. This is an increase in profit of \$125 per flight (4.5%). The overbooking strategy appears worthwhile. The simulation spreadsheet indicates a service level of approximately 99.2% for all passenger demand. This indicates that only 0.8% of the passengers would encounter an overbooking problem. The overbooking strategy up to a total of 32 reservations is recommended.

b.The same spreadsheet design can be used to simulate other overbooking strategies including accepting 31, 33 and 34 passenger reservations. In each case, South Central would need to obtain data on the passenger demand probabilities. Changing the passenger demand table and rerunning the simulation model would enable South Central to evaluate the other overbooking alternatives and arrive at the most beneficial overbooking policy.

22.Use the Hammondsport Savings Bank spreadsheet. Changing the interarrival times to a uniform distribution between 0 and 4 is the only change needed for each spreadsheet.

The mean time between arrivals is 2 minutes and the mean service time is 2 minutes. On the surface it appears that there is an even balance between the arrivals and the services. However, since both arrivals and services have variability, simulated system performance with 1 ATM will probably be surprisingly poor. Simulation results can be expected to show some waiting times of 30 minutes or more near the end of the simulation period. One ATM is clearly not acceptable.

23.Use the Hammondsport Savings Bank spreadsheet.

a.The interarrival times and service times section of the spreadsheet will need to be modified. Assume that the mean interarrival time of 0.75 is placed in cell B4 and that the mean service time of 1 is placed in cell B8. The following cell formulas would be required.

CellFormula

B16=(1/\$B\$4)*LN(RAND())

F16=(1/\$B\$8)*LN(RAND())

The simulation results will vary but most should show an average waiting time in a 2 to 4 minute range.

b.The service time mean and standard deviation would be entered in cells B8 and B9 as in the original Hammondsport 1 ATM spreadsheet. Cell F16 would have its original cell formula =NORMINV(RAND(),\$B\$8,\$B\$9).

Again simulation results will vary. The lower variability of the normal probability distribution should improve the performance of the waiting line by reducing the average waiting time. An average waiting time in the range 1.4 to 2 minutes should be observed for most simulation runs.

24.Use the Hammondsport 2 ATMs spreadsheet on the CD that accompanies the text. The interarrival times section of the spreadsheet will need to be modified. Assume that the mean interarrival time of 4 is placed in cell B4. The following cell formula would be placed in cell B16: =(1/\$B\$4)*LN(RAND())

a.Both the mean interarrival time and the mean service time should be approximately 4 minutes.

b.Simulation results should provide a mean waiting time of approximately .8 minutes (48 seconds).

c.Simulation results should predict approximately 150 to 170 customers had to wait. Generally, the percentage should be 30 to 35%.