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10SimHw11Solution

# 10SimHw11Solution - IEOR E4404.001 SIMULATION Prof Mariana...

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Unformatted text preview: IEOR E4404.001 SIMULATION Prof. Mariana Olvera-Cravioto Assignment #11 Solutions 1. (a) Let X = T 1 + ... + T 10 (the raw estimator for θ ). For n = 1000 simulation runs the following values were obtained X ( x ) = 35 . 2562 and S 2 ( n ) = 318 . 1027 (b) For the antithetic variables approach we simulate X 1 = T 1 + ... + T 10 with one set of uniform(0,1) random variables ( U 1 ,...,U 20 ) and X 2 = T ′ 1 + ... + T ′ 10 , where T ′ i were generated using (1- U 1 ,..., 1- U 20 ) instead. The antithetic variables estimator is Z = ( X 1 + X 2 ) / 2. For m = 500 simulation runs plus 500 antithetic runs we obtained the values Z ( m ) = 35 . 5177 and S 2 ( m ) = 72 . 9137 The variance reduction over the raw estimator is approximately 77%. (c) Let Y = ∑ 10 i =1 S i , then E [ Y ] = 10 and var ( Y ) = 10 (since the service times are iid with an exponential distribution having mean 1). The control variable estimator is Z = X + c * ( Y- E ( Y )), where X is defined as in part (a) and c * =- cov ( X,Y ) /var ( Y ). We first run the simulation to obtain an estimator for cov ( X,Y ). For n = 1000 simulations we obtain a value of ˆ c * =- 1 var ( Y )( n- 1) n ∑ i =1 ( X i- X ( n ))( Y i- Y ( n )) =- 4 . 7921 We then use this value of ˆ c * to obtain the control variables estimator Z = X + ˆ c * ( Y- E [ Y ]). For a different set of n = 1000 simulation runs the following values were obtained Z ( n ) = 35 . 1607 and S 2 ( n ) = 90 . 15 The variance reduction over the raw estimator is approximately 71%....
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10SimHw11Solution - IEOR E4404.001 SIMULATION Prof Mariana...

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