hw5_solns

# hw5_solns - ORIE 4580/5580/5581 Solutions for Homework...

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ORIE 4580/5580/5581 Solutions for Homework Assignment 5 Question 1. a) See the attached spreadsheet. b) We have to pick M such that M max x [1 , 3] f ( x ) . From the ﬁgure in the accompanying spreadsheet, it is easy to see that picking M = 0 . 8 would do the job. However taking the “tightest” upper bound will decrease the number of “wasted” random numbers by our acceptance rejection method. In order to ﬁnd the tightest upper bound, we ﬁnd the maximum of f ( · ) over the interval [1 , 4]. Taking its derivative and setting it equal to 0, we get 3 32 [ 3 x 2 + 8 x + 1 ] = 0 = ⇒ − 3 x 2 + 8 x + 1 = 0 . The points at which this second-order polynomial takes value 0 are 8 ( 8) 2 4 × ( 3) × 1 2 × ( 3) . It turns out that - 8+ ( - 8) 2 - 4 × ( - 3) × 1 2 × ( - 3) is outside the interval [1 , 3]. - 8 - ( - 8) 2 - 4 × ( - 3) × 1 2 × ( - 3) is in the interval [1 , 3]. Furthermore, the second derivative of f ( · ) at this point is less than 0. (This means that

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## This note was uploaded on 03/08/2012 for the course ORIE 4580 at Cornell.

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hw5_solns - ORIE 4580/5580/5581 Solutions for Homework...

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