03q1sol - Massachusetts Institute of Technology Logistical...

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1 Massachusetts Institute of Technology Logistical and Transportation Planning Methods Fall 2003 Solutions to quiz 1 Prepared by Margr´ et Vilborg Bjarnad´ ottir (Bjarnad´ ottir, 2003, (Outline Kang, 2001)) X 1 , X 2 are uniformly distributed between 0 and a. Let G ( a ) E [ max ( x 1 , x 2 ) 3 ] and consider G ( a + ) when X 1 , X 2 are uniformly distributed between 0 and a + , where is very small. Suppose a < X 2 a + and 0 X 1 a . Then we know that max ( x 1 , x 2 ) = x 2 . There- fore E [ max ( x 1 , x 2 ) 3 ] E [ x 2 3 ]. Since X 1 and X 2 are independent, G ( a + ) for this case can be computed as follows: a + G ( a + ) = E [ max ( x 1 , x 2 ) 3 ] = E [ x 2 3 ] = ( x 2 ) 3 f X 2 ( x 2 ) dx 2 , a where f X 2 ( x 2 ) is the probability density function of X 2 . Because X 2 is uniformly distributed over ( a, a + ], f X 2 ( x 2 ) = a 1 . Thus, 1 a + G ( a + ) = ( x 2 ) 3 dx 2 a ± ² a + 1 1 4 = x 4 2 a = 1 · 1 ³ ( a + ) 4 a 4 ´ 4 = 1 · 1 ³ 4 a 3 + 6 a 2 2 + 4 a 3 + 4 ´ 4 1 1 ³ 3 ´ = · (4 a + o ( ) , 4 where o ( ) represents higher order terms of satisfying lim 0 o ( ) = 0 (“pathetic terms”). There- fore, G ( a + ) a 3 as ± 0. By symmetry we have G ( a + ) a 3 as ± 0 when 0 X 2 a and a < X 1 a + . Finally, we do not have to compute G ( a + ) for the case where a < X 1 a + and a < X 2 a + because the associated probability is negligible. Page 1 of 5
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± ² ± ² Massachusetts Institute of Technology Logistical and Transportation Planning Methods Fall 2003 The following table summarizes G ( a + )’s. Case
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This note was uploaded on 11/08/2011 for the course AERO 16.72 taught by Professor Hansman during the Fall '06 term at MIT.

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03q1sol - Massachusetts Institute of Technology Logistical...

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