ps1sol

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

This preview shows pages 1–5. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

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 ....
View Full Document

{[ snackBarMessage ]}

### Page1 / 11

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

This preview shows document pages 1 - 5. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online