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Solutions to the midterm exam of math 164 - linear programming for the summer 2009 semester. It includes explanations for six problems covering topics such as extreme points, feasible directions, linear programming in standard form, and convex functions.
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MATH 164 - Lecture 1 - Summer 2009 Midterm Solutions - July 13, 2009
This is a closed-book and closed-note examination. Calculators are not allowed. Please show all your work. Use only the paper provided. You may write on the back if you need more space, but clearly indicate this on the front. There are 6 problems for a total of 100 points.
POINTS:
1
Solution. Since x is an extreme point , x ∈ S = {x : Ax = b, x ≥ 0 } and thus feasible. We need to show that x is a basic solution. Suppose x has k nonzero components. We can reorder the variables so that zero components of x are the last n − k. Denote the first k components of x by xB (xB ∈ Rk), and the first k columns of A by B (B is m×k matrix). Then Ax = BxB. If the columns of B are linearly independent then we are done. Suppose that it is not the case. Then there exist a vector p ∈ Rk, p 6 = 0 such that Bp = 0. Since xB > 0, we can find a small ≤ such that xB + ≤p > 0 and xB − ≤p > 0. Denote by y a vector in Rn^ with the first k coordinates xB + ≤p and the rest zero and by z a vector in Rn^ with the first k coordinates xB − ≤p and the rest zero. Then y, z ≥ 0 and
Ay = B(xB + ≤p) = BxB + ≤Bp = b, Az = B(xB − ≤p) = BxB − ≤Bp = b
So, y, z ∈ S. Since x = 12 y + 12 z contradicts the fact that x is an extreme point, we conclude that columns of B must be linearly independent. There- fore, x is a basic solution. §
in a standard form.
Solution. We make the substitutions x 1 = x′ 1 + 1, x 2 = x′ 2 + 2, x 3 = x′ 3 − x′′ 3 and z′^ = z − 1. Next, we take the right half of 0 ≤ x′ 1 ≤ 5 and impose it as a general constraint. Adding the appropriate slack/excess variables, we get: minimize z′^ = 3x′ 1 − x′ 2 + 5x′ 35 x′′ subject to
2 x′ 1 + 2x′ 2 − x′ 3 + x′′ 3 + s 1 = 14 −x′ 1 + 3x′ 2 − x′ 3 + x′′ 3 = 1 x′ 1 + x′ 2 + x′ 3 − x′′ 3 − e 1 = 2 x′ 1 + s 2 = 5 x′ 1 , x′ 2 , x′ 3 , x′′ 3 , s 1 , s 2 , e 1 ≥ 0 §
T = {x ∈ S : f (x) ≤ r}
is also convex.
Solution. Let y, z ∈ T. Then, f (y), f (z) ≤ r. Let 0 < α < 1 be arbitrary. Then, we have
f (αy + (1 − α)z) ≤ αf (y) + (1 − α)f (z) ≤ αr + (1 − α)r = r. Hence, αy + (1 − α)z ∈ T for all α ∈ (0, 1), and so, T is convex. §
Prove that a nonzero vector d is a direction of unboundedness if and only if Ad = 0 and d ≥ 0.
Solution. First, suppose that d is a direction of unboundedness. i.e. for all x ∈ S, x + γd ∈ S for all γ > 0. Hence, if x ≥ 0 and Ax = b, then
b = A(x + γd) = Ax + γAd = b + γAd, and so, γAd = 0, which in turn implies that Ad = 0. Now, for the sake of contradiction, suppose that di < 0 for some i, and fix x ∈ S. Let γ = − 2 x di+1i , which is nonnegative since di < 0 and xi ≥ 0. Then, (x+γd)i = −xi − 1 < 0, and so, x + γd /∈ S, which is a contradiction. Hence, d ≥ 0. Conversely, suppose that Ad = 0 and d ≥ 0. Let x ∈ S. Then, Ax = b and x ≥ 0. Let γ > 0 be arbitrary. Clearly, x + γd ≥ 0, and A(x + γd) = Ax + γAd = b + γ · 0 = b. Thus, x + γd ∈ S, and so, since γ was arbitrary, it follows that d is a direction of unboundedness. §
