HW6_solution.pdf - IEOR E4101 Fall 2017 Assignment 6...

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IEOR E4101, Fall 2017, Assignment 6 Solution November 22, 2017 Problem 1. P ( X > n + m | X > m ) = P ( X > n + m ) /P ( X > m ) = (1 - p ) n + m / (1 - ρ ) m = (1 - ρ ) n = P ( X > n ) (1) Problem 2 Let X be the total demand during the lead time. We want to compute the expected value and standard deviation of max ( X - 1 . 5 , 0). For any x 0, P ( max ( X - 1 . 5 , 0) > x ) = P ( X > 1 . 5 + x ) = e - 1 . 5+ x 1 . 8 (2) E [ max ( X - 1 . 5 , 0)] = Z 0 P ( max ( X - 1 . 5 , 0) > x ) dx = Z 0 e - 1 . 5+ x 1 . 8 dx = 0 . 7823 (3) P ( max ( X - 1 . 5 , 0) 2 > x ) = P ( X > 1 . 5 + x ) = e - 1 . 5+ x 1 . 8 (4) E [ max ( X - 1 . 5 , 0) 2 ] = Z 0 P ( max ( X - 1 . 5 , 0) 2 > x ) dx = Z 0 e - 1 . 5+ x 1 . 8 dx = Z 0 e - 1 . 5+ u 1 . 8 2 udu = 2 . 816 (5) where we have substituted x = u . Finally p V ar ( max ( X - 1 . 5 , 0)) = 2 . 816 - 0 . 7823 2 = 1 . 485 (6) Problem 3. (a). Let E k be the event that exactly k bids exceed x . P ( E k ) = 10 k (1 - x ) k x 10 - k Let X be the second largest bid. P ( X > x ) = P ( 10 k =2 E k ) = 10 X k =2 P ( E k ) = 10 X k =2 10 k (1 - x ) k x 10 - k 1
(b). E [ X ] = Z 1 0 P ( X > x ) dx = Z 1 0 10 X k =2 10 k (1 - x ) k x 10 - k dx = 10 X k =2 10 k Z 1 0 (1 - x ) k x 10 - k dx = 10 X k =2 10 k (10 - k )! k ! 11! = 10 X k =2 1 / 11 = 9 / 11 Problem 4. Let X i be the lifetime of the i th component, for i = 1 , 2 , ..., 10. Let Y be the lifetime of the system. Then

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