1998examfinal-soln

1998examfinal-soln - IE 221 OPERATIONS RESEARCH /...

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IE 221 OPERATIONS RESEARCH / PROBABILISTIC MODELS FINAL EXAM SOLUTIONS , FALL 1998 1 . .. (a) production cost per dozen = (0.15)(12) = 1.8 price per dozen = (0.35)(12) = 4.2 “salvage” price per dozen = (0.05)(12) = 0.6 q = # dozens cookies baked daily d = $ dozens cookies demanded daily d < q : cost = -4.2d + 1.8q – 0.6(q – d) = 1.2q – 3.6d c 0 = 1.2 d > q : cost = -4.2d + 1.8q = -2.4q c u = 1.2 F(q*) c u / (c o + c u ) = 2.4 / 3.6 = 0.667 q* = 40 dozens (b) P ( z (q* – 50) / 20 ) = 0.667 z* = 0.43 q* = 50 + (0.43)(20) = 58.6 dozen 2 . .. (a) EOQ = (2KE(D) / h) = (2(100)(5000) / 2) = 707.1 P ( X 80 ) = 1 – P ( z (r – E(X)) / σ X ) = 1 – P ( z (80 – 20) / 30 ) = 1 – 0.9772 = 0.0228 P ( X r* ) = hq* / (c B E(D)) 0.0221 = (2)(707.1) / (c B 5000) c B = 12.41 (shortage cost) [ Also, “shortage cost” = (D/Q) c B NL(z*) = (5000 / 707.1)(12.41)(0.0085) = 0.75 / year ] (b) 0.0228 = (2)(707.1) / [ 2(707.1) + c LS
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1998examfinal-soln - IE 221 OPERATIONS RESEARCH /...

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