Results for 3 52 f start ending branch profit use

This preview shows page 36 - 41 out of 46 pages.

Results for 3-52. f. Start Ending Branch Profit Use Node Node Node Node Prob. (End Node) Branch? Type Value Start 0 1 0 0 Dec 0.80 Branch 1 1 2 0 0 Ch 0.726 Branch 2 1 3 0 0 Yes Dec 0.80 Branch 3 2 4 0.6 0 Dec 0.81 Branch 4 2 5 0.4 0 Dec 0.60 Branch 5 3 8 0 0 Yes Ch 0.76 Branch 6 3 17 0 0.8 Fin 0.80 Branch 7 4 6 0 0 Yes Ch 0.81 Branch 8 4 11 0 0.6 Fin 0.60 Branch 9 6 9 0.9 0.9 Fin 0.90 Branch 10 6 10 0.1 0 Fin 0.00 Branch 11 5 7 0 0 Ch 0.18 Branch 12 5 14 0 0.6 Yes Fin 0.60 Branch 13 7 12 0.2 0.9 Fin 0.90 Branch 14 7 13 0.8 0 Fin 0.00 Branch 15 8 15 0.6 1 Fin 1.00 Branch 16 8 16 0.4 0.4 Fin 0.40 3-53. a. The decision table for Chris Dunphy along with the expected profits or expected mone- tary values (EMVs) for each alternative are shown on the next page.
Table for Problem 3-53a Return in $1,000 N O . OF W ATCHES E VENT 1 E VENT 2 E VENT 3 E VENT 4 E VENT 5 Probability 0.10 0.20 0.50 0.10 0.10 Expected Profit 100,000 100 110 120 135 140 119.5 150,000 90 120 140 155 170 135.5 200,000 85 110 135 160 175 131.5 250,000 80 120 155 170 180 144.5 300,000 65 100 155 180 195 141.5 350,000 50 100 160 190 210 145 400,000 45 95 170 200 230 151.5 450,000 30 90 165 230 245 151 500,000 20 85 160 270 295 155.5 b. For this decision problem, Alternative 9, stocking 500,000, gives the highest expected profit of $155,500. c. The expected value with perfect information is $175,500, and the expected value of perfect information (EVPI) is $20,000. d. The new probability estimates will give more emphasis to event 2 and less to event 5. The overall impact is shown below. As you can see, stocking 400,000 watches is now the best decision with an expected value of $140,700.
Return in $1,000: N O . OF W ATCHES E VENT 1 E VENT 2 E VENT 3 E VENT 4 E VENT 5 Probability 0.100 0.280 0.500 0.100 0.020 Expected Profit 100,000 100 110 120 135 140 117.1 150,000 90 120 140 155 170 131.5 200,000 85 110 135 160 175 126.3 250,000 80 120 155 170 180 139.7 300,000 65 100 155 180 195 133.9 350,000 50 100 160 190 210 136.2 400,000 45 95 170 200 230 140.7 450,000 30 90 165 230 245 138.6 500,000 20 85 160 270 295 138.7 3-54. a. Decision under uncertainty. b. Population Population Row Same Grows Average Large wing –85,000 150,000 32,500 Small wing –45,000 60,000 7,500 No wing 0 0 0 c. Best alternative: large wing. 3-55. a. Weighted Population Population Average with Same Grows α = 0.75 Large wing –85,000 150,000 91,250 Small wing –45,000 60,000 33,750 No wing 0 0 0 b. Best decision: large wing. c. No.
3-56. a. No Mild Severe Expected Congestion Congestion Congestion Time Tennessee 15 30 45 25 Back roads 20 25 35 24.17 Expressway 30 30 30 30 Probabilities (30 days)/(60 days) = 1/2 (20 days)/(60 days) = 1/3 (10 days)/(60 days) = 1/6 b. Back roads (minimum time used). c. Expected time with perfect information: 15 × 1/2 + 25 × 1/3 + 30 × 1/6 = 20.83 minutes Time saved is 3 1 3 ; minutes. 3-57. a. EMV can be used to determine the best strategy to minimize costs. The QM for Windows solution is provided. The best decision is to go with the partial service (maintenance) agreement. Solution to 3-57a Probabilities 0.2 0.8 Maint. No Maint. Expected Row Row Cost ($) Cost ($) Value Minimum Maximum ($) ($) ($) No Service Agreement 3,000 0 600 0 3,000 Partial Service Agreement 1,500 300 540 0 1,500 Complete Service Agreement 500 500 500 500 500 Column best 500 0 500 The minimum expected monetary value is $500 given by Complete Service Agreement b. The new probability estimates dramatically change Sim’s expected values (costs). The best decision given this new information is to still go with the complete service or maintenance policy with an expected cost of $500. The results are shown in the table.
Solution to 3-57b Probabilities 0.8 0.2 Does Not Expected Needs Repair Need Repair Value ($) ($) ($)

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture