Homework 6 Solutions

# Homework 6 Solutions - let students know that data are...

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Homework 6 BUSQOM 1070, Spring 2011 Assigned: Wednesday, March 2, 2011 Due: Wednesday, March 16, 2011 From Module A (pages 682 – 685) in Heizer/Render 10 th Edition: Problem A.2 For part a – the decision table example (Table A.2) is on page 671. Build your table in the same format. b) Maximax decision: very large station (c) Maximin decision: small station (d) Equally likely decision: very large station (e)

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Problem A.3 A.3 (a) EMV (large stock) = 0.3(22) + 0.5(12) + 0.2(–2) = 12.2 EMV (average stock) 0.3(14) 0.5(10) 0.2(6) 10.4 EMV (small stock) 0.3(9) 0.5(8) 0.2(4) 7.5 Maximum EMV is large inventory \$12,200 (b) EVPI \$13,800 – 12,200 \$1,600 where: \$13,800 0.3(22) 0.5(12) 0.2(6) Problem A.4 A.4 Note: In the text, the states of nature appeared in the left column and the decision alternative across the top row. This is to

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Unformatted text preview: let students know that data are sometimes presented in alternative formats. (a) (b) using the maximin criterion; No floor space (N). Problem A.9 A.9 (a) Under conditions of risk, the company should choose batch processing, with an expectation of \$1,000,000, which is twice as high as the next best choice. (b) EVPI is \$170,000. Problem A.22 A.22 (a) A .25 (b) EMV (X) .45(18) .55(–18) –1.8 EMV (Y) .45( 9) .55(15) 4.2 Best EMV (X&Y) .45(9) .55( 3) 2.4 EMV (Nothing) 0 Y is best choice Problem A.23 A.23 The EMV of this game is \$0.59, as illustrated in the diagram below: Decision At X At Y At X At Y X 45 – 27 9 – 27 18 –18 Y 6–15 30–15 –9 15 X&Y (45 6)–(27 15) (30 9)–(27 15) 9 –3 Nothing 0 0 0 0 Probability .45 .55 .45 .55...
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Homework 6 Solutions - let students know that data are...

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