# hw1-answers - IE 495 – Stochastic Programming Problem...

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Unformatted text preview: IE 495 – Stochastic Programming Problem Set #1 — Solutions 1 Random Linear Programs and the Distribution Problem Recall the random linear program that we saw in class: minimize x 1 + x 2 subject to ω 1 x 1 + x 2 ≥ 7 ω 2 x 1 + x 2 ≥ 4 x 1 ≥ x 2 ≥ with ω 1 ∼ U [1 , 4] and ω 2 ∼ U [1 / 3 , 1]. Let • ( x * 1 ( ω ) ,x * 2 ( ω )) be the optimal solution for a given value of ω = ( ω 1 ,ω 2 ). • v * ( ω ) = x * 1 ( ω ) + x * 2 ( ω ) be the optimal objective function value. 1.1 Problem Calculate x * 1 ( ω ) ,x * 2 ( ω ) , and v * ( ω ) for all ω ∈ Ω = [1 , 4] × [1 / 3 , 1]. Answer: There are six basic solutions. (0 , 0) and (0 , 4) are not basic feasible solutions. By inspection, we can exclude the points (0 , 7) and (4 /ω 1 , 0) as optimal solutions. There are therefore two cases: • The optimal solution occurs at the intersection of x 2 = 0 and ω 1 x 1 + x 2 = 7. This occurs when 7 /ω 1 > = 4 /ω 2 . • The optimal solution occurs at the intersection of the inequalities ω 1 x 1 + x 2 = 7 and ω 2 x 1 + x 2 = 4. IE495 Problem Set #1 Solutions Prof Jeff Linderoth Putting these two facts together, we can conclude that x * 1 = ‰ 3 ω 1- ω 2 if 7 ω 1 ≤ 4 ω 2 7 ω 1 Otherwise x * 2 = ‰ 4 ω 1- 7 ω 2 ω 1- ω 2 if 7 ω 1 ≤ 4 ω 2 Otherwise v * = ‰ 4 ω 1- 7 ω 2 +3 ω 1- ω 2 if 7 ω 1 ≤ 4 ω 2 7 ω 1 Otherwise q.e.d. 1.2 Problem Calculate the distribution function F v * of the random variable v * ( ω ) . Answer: This one was hard . In general, determining the distribution function of a sum of random variables is an exercise in messy integration, which is all this was. Let A I x = { ( ω 1 ,ω 2 ) | 7 /ω 1 ≤ 4 /ω 2 , 4 ω 1- 7 ω 2 + 3 ≤ x ( ω 1- ω 2 ) Let A II x = { ( ω 1 ,ω 2 ) | 7 /ω 1 ≥ 4 /ω 2 , 7 /ω 1 ≤ x We must compute F v * ( x ) = P ( v * ≤ x ) = Z A I x f ( ω 1 ) f ( ω 2 ) dω 1 dω 2 + Z A II x f ( ω 1 ) f ( ω 2 ) dω 1 dω 2 Note that f ( ω 1 ) = 1 / 3 ,f ( ω 2 ) = 3 / 2, so f ( ω 1 ) f ( ω 2 ) = 1 / 2. Therefore, the integral is just a matter of simple algebra and geometry. Consider the picture in Figure 1. In ω-space, as x increases, the line swings from top to bottom. First we’ll worry about A I x . Consider the picture drawn in Figure 2. The points specified by A,B,C are Point ω 1 coord. ω 2 coord A 7 /x 4 /x B (2 + x ) / 3( x- 4) 1/3 C 7/4 1 The picture is only valid when the intersection of the lines ( x- 4) ω 1 +(7- x ) ω 2 = 3 and ω 2 = 1 / 3 is ≤ 4. ω 1 = 2 + x 3( x- 4) ≤ 4 ⇔ x ≥ 50 / 11 Problem 1 Page 2 IE495 Problem Set #1 Solutions Prof Jeff Linderoth Figure 1: Integration Area I II Figure 2: Integration Area A I x A C B I I II III IV (x-4)w1 + (7-x)w2 = 3 Problem 1 Page 3 IE495 Problem Set #1 Solutions Prof Jeff Linderoth So the following formula we will derive for A I x is valid only when x ≥ 50 / 11....
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hw1-answers - IE 495 – Stochastic Programming Problem...

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