# hw5sol - IEOR 4404 Simulation Prof. Mariana Olvera-Cravioto...

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IEOR 4404 Assignment #5 Solutions Simulation 16th October 2005 Prof. Mariana Olvera-Cravioto Page 1 of ?? Assignment #5 Solutions 1. We use the Inverse-Transform Method. First, compute the c.d.f of X, given by: F ( x ) = 1 e - 1 R x t =0 e t dt = 1 e - 1 ( e x - 1) when 0 x 1. This gives: F - 1 ( x ) = ln (( e - 1) x + 1) when 0 x 1. The way to generate X is thus: STEP 1. Generate U Unif (0 , 1) STEP 2. Return F - 1 ( U ) 2. We see that: f ( x ) = 1 / 2 x 2 e - x , h ( x ) = λe - λx . Thus, we have the following inequality: 1 / 2 x 2 e - x cλe - λx which is equivalent to: 1 / 2 λx 2 e ( λ - 1) x c . Now, let r ( x ) = 1 / 2 λx 2 e ( λ - 1) x . Then, r ( x ) is maximized at x 0 = 2 1 - λ . Note that we need have λ < 1 since x > 0. Thus, the maximum of r ( x ) is achieved at r ( x 0 ) = 2 e - 2 λ (1 - λ ) 2 . Finally, to ﬁnd the value of λ that minimizes c , simply ﬁnd the λ that minimizes r ( x 0 ). This value of λ is 1 3 . 3. Let

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## This note was uploaded on 11/17/2010 for the course IEOR 4404 taught by Professor C during the Spring '10 term at Columbia.

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hw5sol - IEOR 4404 Simulation Prof. Mariana Olvera-Cravioto...

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