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Unformatted text preview: Industrial Engineering Fall Integer Programming So utions to Homework Prob em . e move is feasible. Consider the integer program given in the solution to Problem of Home work . If we minimize the total number of boxes used, ∑ m i = y i , instead of simply minimizing , we nd that the minimum number of boxes we need to use is . Prob em . (a) First, we show F ⊆ F . Take some arbitrary x ∈ F . It is straightforward to show that if x = then x + x + x ≤ , and if x = then x + x + x ≤ . Putting this together, we have that x satis es x + x + x ≤ x . erefore, x ∈ F . Next, we show F ⊆ F . Take some arbitrary x ∈ F . erefore, x + x + x + x ≤ . If x = , then x + x + x ≤ , and so x + x + x ≤ x + x + x + x ≤ , x + x + x ≤ x + x + x + x ≤ , x + x + x ≤ x + x + x + x ≤ . If x = , then the three linear inequalities de ning F are trivially satis ed by x . erefore, x ∈ F . Finally, we show F ⊆ F . Take some arbitrary x ∈ F . Note that ( x + x + x ≤ ) + ( x + x + x ≤ ) + ( x + x + x ≤ ) + ( x ≤ ) + ( x ≤ ) = x + x + x + x ≤ . erefore x ∈ F . (b) Let P , P , P be the linear relaxations of F , F , F , respectively. Using similar techniques to those used in part (a), we can show that P ⊆ P and P ⊆ P . erefore, the linear relaxation of F is weaker than that of F and F . However, the linear relaxations of F and F are incomparable with respect to inclusion. We have that P ⊆ P , since ( , , , ) is in P but not in P , and we have that P ⊆ P , since ( , , , ) is in P but not in P . Prob em . Since ∑ e ∈ δ ( i ) x e = for all i ∈ V , we have that i ∈ S e ∈ δ ( i ) x e = S for all S ⊆ V ∶ S ≠ , V , or in other words, e ∈ E ( S ) x e + e ∈ δ ( S ) x e = S for all S ⊆ V ∶ S ≠ , V . erefore, for all S ⊆ V ∶ S ≠ , V , e ∈ δ ( S ) x e ≥ ⇔ e ∈ E ( S ) x e ≤ S . As a result, P tspcut = P tspsub . Prob em . Here’s one way to do this problem. By dividing the inequality in the given IP by and rounding, we obtain the valid inequality x + x + x + x + x ≤ . Li ing x in the above inequality ( eorem . in BW) yields x + x + x + x + x + x ≤ . Adding this inequality to the original IP and solving the LP relaxation yields an integral solution x * with x * = x * = x * = and x * = x * = x * = ....
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This note was uploaded on 02/14/2012 for the course IE 530 taught by Professor Ravindran during the Spring '97 term at Purdue.
 Spring '97
 RAVINDRAN
 Industrial Engineering

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