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hmwk10sol

# hmwk10sol - ISyE 3232 Stochastic Manufacturing and Service...

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ISyE 3232 Stochastic Manufacturing and Service Systems Spring 2011 H. Ayhan and J. Dai Solutions to Homework 10 1. (a) P ( T 1) = 1 - e - 10 / 7 0 . 7603 and P ( T 2) = 1 - e - 20 / 7 0 . 9426. (b) The tardiness will be T 0 = max { ( T - 1) , 0 } . E ( T 0 ) = Z 1 ( t - 1) λe - λt dt = e - λ Z 0 sλe - λs dt = e 10 / 7 0 . 7 0 . 1678 (c) Let g 1 ( t ) , g 2 ( t ) , g 3 ( t ) denote the profit for the “\$1250”, “\$1000”, “\$750” contract with delivery time t respectively. Then they can be written as: g 1 ( t ) = 1250 0 t 1 / 2 2500(1 - t ) 1 / 2 < t 1 0 t > 1 g 2 ( t ) = 1000 0 t 1 1000(2 - t ) 1 < t 2 0 t > 2 g 3 ( t ) = 750 0 t 7 750(14 - t ) / 7 7 < t 14 0 t > 14 Let T be a exponential random variable with mean δ , let’s compute E [ g 1 ( T )]: f 1 ( δ ) = E [ g 1 ( T )] = Z 0 . 5 0 1250 1 δ e - t/δ dt + Z 1 0 . 5 2500(1 - t ) 1 δ e - t/δ dt = 1250 - 2500 δ ( e - 1 2 δ - e - 1 δ ) Similarly, f 2 ( δ ) = E [ g 2 ( T )] = 1000 - 1000 δ ( e - 1 δ - e - 2 δ ) f 3 ( δ ) = E [ g 3 ( T )] = 750 - 750 7 δ ( e - 7 δ - e - 14 δ ) Plug-in 0 . 7 into each of the expected profit function, f 1 (0 . 7) = 812 . 69, f 2 (0 . 7) = 872 . 45 and f 3 (0 . 7) = 749 . 99. Hence, choosing the “\$1000” contract will give us the max expected profit. (d) Using a software package like Maple or Mathematica, we can plot f i ( δ ) for i = 1 , 2 , 3. f 1 ( · ) achieves the largest value, which means that “\$1250 contract” is optimal, if 0 δ 0 . 5375; f 2 ( · ) achieves the largest value, which means that “\$1000 contract” is optimal, if 0 . 5375 δ 1 . 055; f 2 ( ·

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