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# Ch3hwsolutions - UTA Chapter 3 Decision Making PROBLEMS 1...

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Unformatted text preview: UTA Chapter 3 Decision Making PROBLEMS 1. Williams Products a. Break-even quantity [Q] 2 Fixed costar-"(Unit price — Unit variable costs} 2 3360.00031318 — 86) = 5.000 units The graphic approach is shown on the following illustration. using Break-Even Aimh'sis Solver of OM Explorer. \$200,000 -- \$ 80,000 -- D (10,000, 100,000) \$ 50.000 -- \$ 40,000 -- \$ 20000 ~ (10,000, 120,000) \$ 00 ,000 -- - -Revenues an .000 —- —Costs 50 ,000 40 ,000 W 20 ,000 -- \$0 4' . . . . . . 0 2 .000 4 ,000 E .000 8 .000 10 .000 12 .000 Quantity {0} Two lines must be drawn: Revenue: : 18;} Total cost: = 60.000 + 6Q b. Proﬁt=Revenue — Total cost = pQ — (F + (Q) = (\$14.00)10. 000 — [\$60. 000 + (\$6)10. 000] = \$140. 000 — \$120. 000 2 \$20. 000 c. Proﬁt=Revenue — Total cost =pQ—(F+CQ) =(\$12.50)15.000—[360.000+[36]15.000] = \$18?.500—\$150.000 = 337.500 12. Forsite Company a. Say that each criterion (arbitrarily) receives 20 points: Product Calculation Total Score A 201061—2010?)——20[0.41—20(1.01+20(0.2) = 58 B 20(0.81—20{0.3t—20t0.?t+20t0.4t—20t1.01 = 64 C 20t0.3t—20t0.9}—-20t0.51+20{O.61—20(0.5) = 56 The best alternative is service B and the worst is service C. This relationship holds as long as any arbitrary weight is equally applied to all performance criteria. b. Let x 2 point allocation to criteria 1. 3. 4. and 5 2x 2 point allocation to criteria 2 {ROI} 3: + 2x + .r + x + x = 100 points 6x = 100 points x = 16.? points Product Calculation Total Score A 16.?(061—33.3(0.7’t—16.?(0.4t——16.?{1.01——16.2102} =60.0 B 16.?(081—33.3(0.31—16.?t0.?)——16.?(0.4)+16.?[10] =58.4 C‘ 16210.3)—33.3[0.9)—16.?{0.5)——16.?{0.6} -—16.7(0.5j: = 61-? The rank order of the services has chrnged to C. A. B. 15. Build-Rite Construction a. Maximin Criterion—Best Decision: Subcontract Payoff: S100.000 b. Maximax Criterion—Best Decision: Hire Payoff: \$623000 c. Laplace Criterion—Best Decision: Subcontract Weighted Payoff: \$221.66? Alternative W’eighted Payoff Hire [—S250.000 —100.000 + \$62 5.00011? 2 \$158.33 3 Subcontract _ [\$ 100.000 —150.000 + S415.00_0]..=’3 = \$221.66? —IDo—-notlriu Er} : : 33 — n‘e:11}0:ﬁ[}:5[1 15111—80271] ﬁmzﬁatﬁei: 16. Decision Tree ‘ \$15 0.5 \$30 ° I ' 2 ' 20 Alternative] I \$60 0 a \$ . '“ \$18 ‘ \$24 \$25 Alternative2 m . 0.2 O ' \$20 em - - \$30 Work from right to left. Here we begin with Decision Node 2. although Decision Node 3 would be an equally good starting point. The key concept is that we carniot begin analysis of Decision Node 1 until we know the expected payoffs for Decision Nodes 2 and 3. Decision Node 2 1. Its first alternative (in the upper right portion of the tree) leads to an event node with an expected payoff of 822.50 [or 0.5(15) + 0.580)]. 2. Its second alternative leading downward reaches an event node with an expected payoff of 320.60 [or 0.4(20) + 0.3(18)+ 0.3(24)]. 3. Thus the expected payoff for decision node 2 is 522.50. because the ﬁrst alternative has the better expected payoff. Plune the second alternative. Decision Node 3 4. Its second alternative leads to an event node has an expected payoff of S24 [or 0.6(20) + 0.4(30)]. 5. Thus the payoff for decision node 3 is 825. because the fn‘st alternative (\$25) is better than the expected payoff for the second alternative (\$24). Pnuie the second alternative. Decision Node 1 6. The second alternative leads to an event node has an expected payoff of \$24 [or 0.2(25) + 0.5(26)+ 0.390)]. 7. Thus the expected payoff for decision node 1 is 524. because the second alternative {\$24) is better than the expected payoff for the second alternative (522.50). Prune the first alternative. Thus the best initial choice (Decision 1) is to select the lower branch. Alternative 2. If the top branch of the subsequent event occurs (a 20% probability). then Decision 3 must be made. Select its first alternative. Conclusion: Select the lower branch. with an expected payoff of \$24. 17. One machine or two. a. Decision Tree Subcontract 3160000 High demand (0.80 Do nothing S 120.000 S 140.000 Two . I-Iih demand 0.80 3180000 Low demand 0.20 390300 b. Working from right to left: Decision Node 2 1. The best choice is to subcontact (\$160,000). which becomes the expected payoff for Decision Node 2. Piune the “Do nothing“ and Buy second“ altematives. Decision Node 1 2. The alternative to buy one machine has an expected value of \$152,000 [01‘ 0.8(160000) -- 0.2(120.000]. The altemative to bu}r two machines has an expected value of \$162,000 [or 0.8(180000) -- 0.2(90.000]. 4. Thus the best choice is to buy two machines because it has a higher expected payoff (\$161000 versus \$152000). Piune the one machine alternative. Conclusion: Buy two machines. with an expected payoff of \$162,000. in ...
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