Chapter Fourteen: Simulation
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1 TaylCh14ff.qxd 4/21/06 8:39 PM Page 213 Chapter Fourteen: Simulation PROBLEM SUMMARY 1. Rescue squad emergency calls PROBLEM SOLUTIONS Car arrivals at a service station 3. Machine breakdowns 4. Income analysis (12 19) 5. Decision analysis 6. Machine repair time (14 3) 7. Product demand and order receipt 8. Bank drive-in window arrivals and service a) 9. Loading dock arrivals and service 10. Product demand and order receipt 11. Football game 12. Student advising, arrival and service 13. Markov process 14. Inventory analysis 15. Rental car agency 16. CPM/PERT network analysis 17. Store robbery and getaway 18. Stock price movement 19. Hospital emergency room staffing 20. Break-even analysis 21. Rating dates 22. Model analysis (14 21) 23. Production capacity 24. Baseball game 25. Crystal Ball, maintenance cost 26. Crystal Ball, B-E analysis (14 20) 27. Crystal Ball, rating dates (14 21) 28. Crystal Ball, production capacity (14 23) 29. Crystal Ball, investment selection 30. Crystal Ball, inventory management Crystal Ball, hotel room rates 32. Crystal Ball, CPM/PERT network *Note: Many of these solutions may be different from the solution the instructor or student derives depending on the stream of random numbers used. The Excel solutions (provided on the Instructor s CD) will also be different because they use Excel-generated random numbers, whereas many of these solutions were developed using the random number table in the text. 213
2 TaylCh14ff.qxd 4/21/06 8:39 PM Page Snowfall Cumulative Financial Level (in) Probability RN Return > $120, ,000 < ,00 40,000 Use fifth column of random numbers in Table 14.3 Random Number Return 45 40, , , , , , , , , , , , , , ,
3 TaylCh14ff.qxd 4/21/06 8:39 PM Page 215 Random Number Return , , , , , ,000 Average return = 640,000 = 20 $32,000 The expected value is $40,000 so the simulated average for 20 weeks is significantly lower indicating the need for more trials
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12 TaylCh14ff.qxd 4/21/06 8:39 PM Page Cummulative Direction Probability Probability RN Ranges East (x = +1) West (x = 1) North (y = +1) South (y = 1) ,00 224
13 TaylCh14ff.qxd 4/21/06 8:39 PM Page 225 Monte Carlo Simulation (using 16th row of random numbers from Table 14.3) Considering the city as a grid with an x and y axis with the store at point (0,0) each random number selected indicates a movement of 1 unit (block) in either an x or y direction. Trial 1 Trial 2 Trial 3 Trial 4 Trial 5 End of Block r (x,y) r (x,y) r (x,y) r (x,y) r (x,y) 1 58 (0,1) 68 (0,1) 20 (1,0) 53 (0,1) 24 (1,0) 2 47 ( 1,1) 13 (1,1) 85 (1, 1) 45 ( 1,1) 83 (1, 1) 3 23 (0,1) 09 (2,1) 59 (1,0) 56 ( 1,2) 05 (2, 1) 4 69 (0,2) 20 (3,1) 72 (1,1) 22 (0,2) 81 (2, 2) 5 35 ( 1,2) 73 (3,2) 88 (1,0) 49 ( 1,2) 07 (3, 2) 6 21 (0,2) 77 (3,1) 11 (2,0) 08 (0,2) 78 (3, 3) 7 41 ( 1,2) 29 (2,1) 89 (2, 1) 82 (0,1) 92 (3, 4) 8 14 (0,2) 72 (2,2) 87 (2, 2) 55 (0,2) 36 (2, 4) 9 59 (0,3) 89 (2,1) 59 (2, 1) 27 ( 1,2) 53 (2, 3) ( 1,3) 81 (2,0) 66 (2,0) 49 ( 2,2) 04 (3, 3) Within 2 blocks? no yes yes no no 18. In 2 of the 5 trials the robber is within 2 blocks of the store. As an example, at the end of 10 blocks in trial 1, the robber is 1 block west and 3 blocks north. Stock Price Cummulative Movement Probability Probability RN Ranges Increase (+) Same (0) Decrease ( ) ,00 Stock Probability Cummulative RN Probability Cummulative RN Price Increase Probability Ranges Decrease Probability Ranges 1/ / / / / / / , ,00 225
14 TaylCh14ff.qxd 4/21/06 8:39 PM Page 226 Stock Price Price Day r Movement r Change (+, ) Stock Price /4 61 3/ / /2 62 1/ /8 62 3/ /2 61 7/ / / /8 61 5/ /8 61 1/ /4 61 3/ /4 61 1/ /8 61 5/ / /4 60 7/ / / /8 60 5/ /8 61 1/ /8 61 5/ /2 61 1/ / /8 61 1/ /4 61 1/ / /8 61 7/ /4 61 5/ / / /8 61 3/ / In order to expand the model for practical purposes the length of the simulation trial would be increased to one year. Then this simulation would need to be repeated for many trials, i.e., 1,000 trials. Cummulative Time Between Arrivals Probability RN Ranges ,00 Cummulative Service Probability RN Ranges Doctor (D) Nurse (N) Both (B) ,
15 TaylCh14ff.qxd 4/21/06 8:39 PM Page 227 Doctor Nurse Both Cummulative RN Cummulative RN Cummulative RN Time Probability Ranges Time Probability Ranges Time Probability Ranges , , ,00 Arrival Time of Service (mins.) Departure Clock Patient r Clock r Service r Doctor Nurse Wait? D N 1 31 D D B B B B D N N D D B D D D D N N B B P( wait) = 15 = Average waiting time = 440 mins. = 22 mins. 20 It would seem that this system is inadequate given that the probability of waiting is high and the average time is high. Also, observing the actual simulation three customers had to wait an hour and two others had to wait 35 minutes which seems excessive. Of course in order to make a fully informed decision this simulation experiment would need to be extended for more patients and then repeated several hundred times. 227
16 TaylCh14ff.qxd 4/21/06 8:39 PM Page Sales Volume RN1 Range Price RN2 Range Variable Cost RN3 Range ,00 $ $ $ $ $ $ ,00 $ $ $ $ $ ,00 c f = $9,000 Sales Variable Month RN1 Volume (V) RN2 Price (p) RN3 Cost (C v ) Z=VP-9,000-VCv $ $ $ , $ , $ , $ $ , $ , $ $ , $ $ , $ , $ , $ , $ , $ , $ , $ , $ Probability of at least breaking even = 10 = 20 Average monthly loss = $ Attractiveness RN1 Range Intelligence RN2 Range Personality RN3 Range , , ,00 228
17 TaylCh14ff.qxd 4/21/06 8:39 PM Page 229 Date RN1 Attractiveness RN2 Intelligence RN3 Personality Average Rating Average overall rating of Salem dates = There are several ways to access the accuracy of the results. First the student can determine the expected value for each characteristic and average them to see if this results in a value close to the simulated result. E (Attractiveness) = 2.50 E (Intelligence) = 3.05 E (Personality) = 2.77 Average rating = = This is relatively close to the simulated result of 2.92 which tends to verify that result. Confidence limits can also be developed for the average rating. However, this is best/easiest done with Excel. 23. Capacity Fx ( ) = x r 2 x 360, = x 360, 000 = 360, 000r Demand RN Range ,00 229
18 TaylCh14ff.qxd 4/21/06 8:40 PM Page 230 Week RN1 Capacity RN2 Demand Capacity>Demand? yes yes no yes yes yes yes no yes yes no yes yes no yes yes yes no yes yes Probability capacity > demand = 15 = Average capacity = Average demand =
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20 TaylCh14ff.qxd 4/21/06 8:40 PM Page average maintenance cost for the life of the car = $3, P(cost $3,000) = average profit = $ probability of breaking even = average rating = 2.91 probability of rating better than 3.0 (i.e., P(x 3.0)) = average surplus capacity = probability of sufficient capacity = (a) In Crystal Ball include the parameters (, ) for each fund return distribution in a spreadsheet cell. Select 6 other spreadsheet cells for the investment combination returns. For example, if the return parameters for investment 1 is in cell D6, the return parameters for investment 2 are in cell D7 and the total return for the investment combination (1,2) is in cell C12, the formula in C12 is =50,000* (1+D6)^3 + 50,000(1+D67)^3 The simulation results are as follows, Combination Return P (Return 120,000) 1,2 = 157,572, = 21,459 (.974) 1,3 = 166,739, = 28,599 (.959) 1,4 = 161,888, = 23,016 (.981) 2,3 = 148,692, = 22,693 (.911) 2,4 = 144,841, = 13,683 (.980) 3,4 = 129,429, = 20,886 (.653) (b) The highest probability of exceeding $120,000 is virtually tied at.98 between (1,4) and (2,4). 30. Q: = 1,860.83, = TC: = 1,465.2, = (a) Select an Excel spreadsheet cell to include the normal distribution parameters ( = 800, = 270) for conference rooms, for example, cell D4. 232
21 TaylCh14ff.qxd 4/21/06 8:40 PM Page 233 In another cell, for example D9, include the cost equation for the number of rooms actually reserved that is being tested. For example if 600 rooms are being tested, the cost formula in D9 would be, = 80*MAX(600-D4,0) + 40*MAX(D4-600,0). The simulation average value (for 1,000 trials) for 600 rooms is $12,283. The costs for each room reservation value are, Rooms Reserved Cost ($) , , , ,176 1,000 20,756 Thus, the minimum cost is with 700 rooms reserved. (b) Testing different values between 600 and 800 shows that a more exact value is approximately 690 rooms. 32. In Crystal Ball define a cell for each activity s triangular distribution parameters, then add the specific cells on each path to get the total path times. (a) Path A: = 10.96, = 1.60 Path B: = 11.98, = 1.44 Path C: = 9.84, = 1.53 Path D: = 10.85, = 1.35 The critical path is B. (b) Enter the critical path time of in the lower left-hand corner of the frequency chart window and hit enter. This results in a value of 25.4% which is the percentage of time (or probability) that this path will exceed 11.98, the critical path time. In CPM/PERT analysis the critical path is determined analytically which assumes only one path will always be critical. The simulation result shows that other paths might also be critical a percentage of the time. Since the analytical result reflects only a single critical path, the more accurate critical path results are provided with simulation. CASE SOLUTION: JET COPIES The probability function for time between repairs, f(x), is, f( x) = x, x The cummulative function, F(x) is, x 2 f( x)= x dx x = and, r x 36 = 2 x 2 = 36r x=6 r The cummulative distribution and random number ranges for the distribution of repair times are, Repair Time Cummulative y (days) P(y) Probability RN Ranges ,000 The probability function for daily demand is developed by determining the linear function for the uniform distribution, which is, fz ()= 1 b a or, fz ()= 1 = 1 b a 6 This function is integrated to develop the cummulative probability distribution, 1 = = = ( ) z z z z z Fz () dz z 1 =
22 TaylCh14ff.qxd 4/21/06 8:40 PM Page 234 Letting F(z) = r. r = z and solving for z, 6( ) z= r+ 1 3 z = 6r + 2 Monte Carlo Simulation: Time Between Copies Breakdowns Repair Time Lost Revenue r x weeks x r y days r z Lost , = 9.04 $ , = , , = , ,.71, , = , , = , = , ,.08, = , , = , , = , ,.39, = , , = ,396 $15,094 They need to purchase the back up copier. CASE SOLUTION: BENEFIT-COST ANALYSIS OF THE SPRADLIN BLUFF RIVER PROJECT mean B/C = =.145 P(B/C 1.0) =
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