02.sol

02.sol - Industrial Engineering Fall Integer Programming So...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

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

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

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 * = ....
View Full Document

{[ snackBarMessage ]}

Page1 / 5

02.sol - Industrial Engineering Fall Integer Programming So...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online