IEOR150F10_SampleFinal_Solution

# Because d 1 and d 2 are independent and no consumer

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Because D 1 and D 2 are independent and no consumer goes to retailer A after she tries retailer B, E and D 2 are also independent. Therefore, we have P ( R = 0) = P ( D 2 = 0) P ( E = 0) = 2 9 , P ( R = 1) = P ( D 2 = 0) P ( E = 1) + P ( D 2 = 1) P ( E = 0) = 1 3 , P ( R = 2) = P ( D 2 = 1) P ( E = 1) + P ( D 2 = 2) P ( E = 0) = 1 3 , and P ( R = 3) = P ( D 2 = 2) P ( E = 1) = 1 9 . Let F 2 ( x ) = P ( R x ) be the cdf of R , then F 2 (0) = 0 . 222, F 2 (1) = 0 . 556, F 2 (2) = 0 . 889, and F 2 (3) = 1. Now, because the critical ratio is again 0.6, which is between F 2 (1) and F 2 (2), we round up and conclude that the optimal inventory level is q * 2 = 2. With two units on hand, the expected sales quantity is h min { R, 2 } i = 0 × 2 9 + 1 × 1 3 + 2 × 1 3 + 2 × 1 9 = 11 9 = 1 . 222 . 4

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The expected sales revenue is thus 1 . 222 × \$10 = \$12 . 22. As the purchasing cost is \$4, the expected profit is \$12 . 22 - \$4 × 2 = \$4 . 22. 7. (a) Given w and T , the retailer chooses an nonnegative inventory level q to maximize his expected profit, which is the expected sales revenue r £ min { X, q } / minus the variable payment wq and the fixed payment T . Therefore, his problem is R I = max q 0 r h min { X, q } i - wq - T. (b) Note that R I = - T + max q 0 r £ min { X, q } / - wq because T has nothing to do with the decision variable q . Therefore, the optimal inventory level is q I = F - 1 r - w r · , which does not depend on T . (c) The manufacturer’s problem is M I = max w,T ( w - c ) q I + T s.t. r h min { X, q I } i - wq I - T 0 , q I = F - 1 r - w r · . The first constraint guarantees a nonnegative expected profit for the retailer. The second constraint set q I optimally. The objective function is to maximize the sales margin ( w - c ) q I plus the fixed payment T . (d) We know that q D is the optimal solution of M D = max q 0 r h min { X, q } i - cq. Therefore, q D = F - 1 ( r - c r ). Since q I = F - 1 ( r - w r ), we have q I = q D if w = c . (e) If w = c , the manufacturer’s problem becomes M I = max T T s.t. r h min { X, q I } i - cq I - T 0 , q I = F - 1 r - c r · . The first constraint requires T r £ min { X, q I } / - cq I . This allows the retailer to receive a nonnegative expected profit. The manufacturer also needs to choose T 0 so that he can make money. Therefore, the following requirement 0 T r h min { X, q I } i - cq I induces the participation of both players. Note that because w = c , we have q I = q D and thus r £ min { X, q I } / - cq I = M D , the manufacturer’s maximized expected profit under direct sales.
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