L29_Dynamic Programming (continued)

L29_Dynamic Programming (continued) - Problem 10 Problem...

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Problem 10 Problem Set 10.3A Page 414 Maximize z = y 1 y 2 …y n subject to y 1 +y 2 +…+y n = c, y i 0
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Thus there are n stages to this problem. At stage i, we have to choose the variable y i . The state of the problem at stage i is defined by the variable x i , which represents the sum of the variables to be decided at stages i, i+1, …, n. Thus n n n n y x y y y x y y y x = + + + = + + + = . . ... ... 3 2 2 2 1 1
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Let be the optimal return for ( ) i i F x stages i, i+1, …, n. Stage n: Thus x n = y n Optimal return for this stage = ( ) max{ } n n n n n y F x y x = =
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1 1 1 1 ( ) max{ ( )} n n n n n n y F x y F x - - - - = × Stage n-1: Optimal return for this stage = occurs when y n-1 = (x n-1 )/2 1 1 1 1 max{ ( )} n n n n n y y F x y - - - - = × - 1 1 1 1 max{ ( )} n n n n y y x y - - - - = × - = 2 1 2 n x -
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2 2 2 2 2 1 2 2 3 2 2 2 2 2 ( ) max{ ( )} ( ) max{ } 4 3 n n n n n n n n y n n n n y F x y F x y x y x y - - - - - - - - - - - - = × - - = × = Stage n-2: Optimal return for this stage = occurs when y n-2 = (x n-2 )/3
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1 ( ) 1 n i i i i x F x n i - + = - + Stage i: Optimal return for this stage (by induction on i) = occurs when y i = (x i )/(n-i+1)
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1 1 1 ( ) n n x c F x n n = = Stage 1: Optimal return for this stage = occurs when y 1 = x 1 / n = c/ n y 2 = x 2 /(n-1) = (x 1 -y 1 )/(n-1) = (c- c/n)/(n-1) = c/n Similarly, y i = c/n for all i.
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Problem 12 Problem Set 10.3A Page 415 Maximize z = (y 1 +2) 2 +y 2 y 3 +(y 4 -5) 2 subject to y 1 +y 2 +y 3 +y 4 5, y i 0, integers
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1 +2) 2 +(y 2 -5) 2 +y 3 y 4 subject to y 1 +y 2 +y 3 +y 4 5, y i 0, integers In order to get a proper decomposition of the objective function , we rewrite it as Thus there are 4 stages to this problem. At stage i, we have to choose the variable y i . The state of the problem at stage i is defined by
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L29_Dynamic Programming (continued) - Problem 10 Problem...

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