ps1sol - 1.203J 6.281J 13.665J 15.073J 16.76J ESD.216J...

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Unformatted text preview: 1.203J / 6.281J / 13.665J / 15.073J / 16.76J / ESD.216J Logistical and Transportation Planning Methods Solution Problem Set #1 Problem 1 Two-horse race (a). The conditional pdf of U given that V = v is: ( u v )= f ( u , v ) Equation 1 f U V f V ( ) v The marginal pdf of V is given by: ∞ Equation 2 V ( ) = ∫ f ( u , v ) du f v −∞ (b). There are two random variables: • X , which is the finishing time for A, with x ∈[ 3,8 ] in minutes; ⎡ x ⎤ • Y , which is the finishing time for B, with y ∈ ⎢ ;2 x ⎥ in minutes. ⎣ 4 ⎦ The joint sample space is therefore: y 3 8 The probability law of the sample space is: f X , Y ( x , y ) = f Y ( y x )⋅ f X ( ) = 4 1 4 Equation 1 x ⋅ = X 7 x 5 35 x Then from question a): ⎡ 3 ⎤ 8 Equation 2 ∀ y ∈ ⎢ 4 ;16 ⎥ Y ( ) X , Y ( ) f y = ∫ f x , y dx ⎣ ⎦ 3 x 0 3/4 6 16 y=x/4 y=2x 3 2 1 Problem Set #1 1/11 1.203J / 6.281J / 13.665J / 15.073J / 16.76J / ESD.216J Logistical and Transportation Planning Methods 35 4 x From the joint sample space, we can see that there are 3 different cases for the boundaries of the integration: Case #1: ⎡ 3 ⎤ 4 y 4 y 4 4 4 y ⎢ ;2 ⎥ f Y ( ) = ∫ f X Y ( x , y ) dx = ∫ dx = ln ∀ y ∈ y , ⎣ 4 ⎦ 3 3 35 x 35 3 Case #2 ∀ y ∈[ 2;6 ] f Y ( ) = ∫ 8 f X , Y ( x , y dx = 4 ln 8 y ) 35 3 6 8 Case #3 ∀ y ∈[ 6;1 ] f Y ( y ) = ∫ f X , Y ( x , y ) dx = 4 ln 16 35 y 2 y (c). f X , Y ( x , y ) ⎡ 3 ⎤ Equation 1 f X Y ( x , y ) = ∀ y ∈ ⎢ ;16 ⎥ f ( y ⎣ 4 ⎦ Y There are three different cases: Case #1: ⎡ 3 ⎤ 1 ( x , y ) = = ∀ y ∈ ⎢ ;2 ⎥ f X Y ⎣ 4 ⎦ 4 ln 4 y x ⋅ ln 4 y 35 3 3 Case #2 ∀ y ∈[ 2;6 ] f ( x , y ) = 1 X Y 8 x ⋅ ln 3 Case #3 ∀ y ∈[ 6;1 ] f ( x , y ) = 1 X Y 16 x ⋅ ln y (d). A wins the race if and only if x p y . 3 ⎛ 3 ⎞ If y = , then y p 3 ≤ x and P ⎜ ⎜ S y = ⎟ ⎟ = 0 , whereas obviously, P ( S ) ≠ 0 . Thus, the event 4 ⎝ 4 ⎠ S and r.v. Y are not independent. (e). The winner will win the by less than 1 min if and only if x − y ≤ 1. The corresponding area on the joint sample space is: Problem Set #1 2/11 1.203J / 6.281J / 13.665J / 15.073J / 16.76J / ESD.216J Logistical and Transportation Planning Methods The probability that the winner will win by less than 1 minute is then: P = ∫∫ f , ( x , y ) dxdy X Y Area 4 P = ∫∫ dxdy 35 x Area P = − P ≈ 0.22 (f). The winner’s time is less than 6 minutes if and only if min ( x , y ) ≤ 6 . On the joint sample x y=x/4 y=x+1 y=x y=x-1 8 y 0 3 1 6 16 y=2x 1 ( ) 35 8 3 ln 8 16 6 3/4 space, this corresponds to: 8 y = 2 x 4 Thus : P = 1 − ∫∫ f X , Y ( x , y ) dxdy = 1 − ∫ ∫ dxdy = 0.77 35 x unshaded x = 6 y = x area 8 x y 0 3 y=x/4 y=2x y=x Problem Set #1 3/11 1.203J / 6.281J / 13.665J / 15.073J / 16.76J / ESD.216J Logistical and Transportation Planning Methods Problem 2 Cell Phones (a). Let i be the day of the month: i = 1..30 ....
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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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ps1sol - 1.203J 6.281J 13.665J 15.073J 16.76J ESD.216J...

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