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Unformatted text preview: IEOR 162, Fall 2011 Suggested Solution to Homework 10 Problem 1 (Problem 9.2.10) Let z be a binary variable such that z = ‰ 0 if x + y ≤ 3 is satisfied and 1 if 2 x + 5 y ≤ 12 is satisfied . Let M be a very large number, then the following two constraints x + y 3 ≤ Mz 2 x + 5 y 12 ≤ M (1 z ) ensures that at least one of x + y ≤ 3 and 2 x + 5 y ≤ 12 is satisfied. The condition that both x and y are integers is not important. Note that because there is no information regarding the possible values of x and y , there is no way for us to find a specific value for M . Problem 2 (Problem 9.2.11) First, note that “if x ≤ 2 then y ≤ 3” is equivalent to “ x > 2 or y ≤ 3”. Before we apply the technique of modeling “eitheror” requirements, note that it is not allowed to have strict inequalities in an LP or IP formulation. Therefore, we must apply the condition that x is an integer to convert x > 2 into a weak inequality. To do this, note that x > 2 is equivalent to x ≥ 3 if x is an integer. Therefore, all we need to do is to write constraints so that “ x ≥ 3 or y ≤ 3”. Let z be a binary variable such that z = ‰ 0 if x ≥ 3 is satisfied and 1 if y ≤ 3 is satisfied . Let M be a very large number, then the following two constraints 3 x ≤ Mz y 3 ≤ M (1 z ) ensures that at least one of x ≥ 3 and y ≤ 3 is satisfied, i.e., if x ≤ 2 then y ≤ 3. Note that because there is no information regarding the possible values of x and y , there is no way for us to find a specific value for M . Problem 3 (Problem 9.2.12) Let New York, Los Angeles, Chicago, and Atlanta be city 1, 2, 3, and 4. We then define x ij = units shipped from city i to region j , i = 1 ,..., 4, j = 1 ,..., 3 and y i = ‰ 1 if city i is chosen 0 otherwise , i = 1 ,..., 4 as our decision variables. We also define F = (400 , 500 , 300 , 150) as the fixed cost vector for cities, D = (80 , 70 , 40) as the demand vector for regions, and C = 20 40 50 48 15 26 26 35 18 24 50 35...
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 Fall '07
 Zhang
 Yi, optimal solution, Konrad Zuse

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