Download Midterm Exam Solutions - Optimization | MATH 0164 and more Exams Optimization Techniques in Engineering in PDF only on Docsity!
MATH 164/2 Optimization, Winter 2006, Midterm Exam Solutions Instructor: Luminita Vese Teaching Assistant: Ricardo Salazar
[1] (10 points)
(a) Consider the following feasible set S =
{ x ∈ Rn^ : Ax ≥ b, x ≥ ~ 0
}
. Show
that if the vector d satisfies d 6 = ~0, d ≥ ~0 and Ad ≥ ~0, then d is a direction of unboundedness for the set S.
(b) Consider the following linear programming problem: Minimize z = x 1 − 5 x 2 + x 3 − 3 x 4 subject to
3 x 1 − x 2 + 2 x 4 ≥ 2 − 2 x 1 + 3 x 4 ≥ − 1 x 2 − x 3 ≥ 2 5 x 1 − 3 x 3 ≥ 2 x 1 , x 2 , x 3 , x 4 ≥ 0
(b1) Show that x = (2, 4 , 2 , 1)T^ is a feasible point to the problem and label each of the constraints as active or inactive. (b2) Show that the vector d = (1, 2 , 1 , 1)T^ is a direction of unboundedness (note that this problem, as given, is not in standard form!) Solution: (a) We know by definition that d is a direction of unboundedness for S if d 6 = ~ 0 and if x + γd ∈ S, for any x ∈ S and any γ ≥ 0. Let x ∈ S and γ ≥ 0 arbitrary. Then Ax ≥ b, and using the assumptions on d, we have: A(x + γd) = Ax + A(γd) = Ax + γAd ≥ b + γ~0 = b. Therefore A(x + γd) ≥ b (1) Also, because x ≥ ~0, γ ≥ 0 and d ≥ ~0, we have x + γd ≥ ~0 + γ~0 = ~ 0 (2) From (1) and (2), we deduce that x + γd ∈ S for any x ∈ S and any γ ≥ 0, therefore d 6 = 0 is a direction of unboundedness for the set S.
(b1) We verify that x = (2, 4 , 2 , 1)T^ satisfies all eight constraints given. 3 · 2 − 4 + 2 · 1 = 4 > 2 (1st inactive) − 2 · 2 + 3 · 1 = −1 = −1 (2nd active) 4 − 2 = 2 = 2 (3rd active) 5 · 2 − 3 · 2 = 4 > 2 (4th inactive) x 1 = 2 > 0 (inactive), x 2 = 4 > 0 (inactive), x 3 = 2 > 0 (inactive), x 4 = 1 > 0 (inactive). Therefore the point x = (2, 4 , 2 , 1)T^ satisfies all eight constraints of the problem, so it is a feasible point, x ∈ S.
(b2) Notice that d = (1, 2 , 1 , 1)T^6 = ~0 and d ≥ ~0. Therefore, using the proof for (a), it is sufficient to show that Ad ≥ ~0, where A is the matrix corresponding to the first four constraints:
A =
.
Indeed,
Ad =
=
> ~ 0 ≥ ~ 0 ,
therefore d = (1, 2 , 1 , 1)T^ is a direction of unboundedness for S given in (b).
[2] (12 points) Consider the linear programming problem
Minimize z = x 1 − x 2 + 3x 3 subject to
x 1 + 2 x 3 ≥ 4 x 1 − x 2 ≥ 0 − 2 x 1 + x 2 + x 3 ≥ 1 x 1 , x 2 , x 3 ≥ 0
(a) Show that x = (1, 1 , 2)T^ is a feasible point to the problem. (b) Show that p = (− 2 , − 3 , 0)T^ is a feasible direction at the feasible point x = (1, 1 , 2)T^. (c) Determine the maximal step length α such that x + αp remains feasible, with x and p as in part (b). (d) Find the minimum value of p 3 , such that (− 2 , − 3 , p 3 )T^ is a feasible direction at x = (1, 1 , 2)T^. Solution: We will also label all six constraints, since this is needed for (b)-(d). (a) 1 + 2 · 2 = 5 > 4 (inactive) 1 − 1 = 0 = 0 (active) (−2) · 1 + 1 + 2 = 1 = 1 (active) x 1 = 1 > 0 (inactive), x 2 = 1 > 0 (inactive), x 3 = 2 > 0 (inactive). Therefore x = (1, 1 , 2)T^ is a feasible point.
(b) By the property from the course, it is sufficient to show that Apˆ ≥ ~0, where Aˆ is the submatrix of A corresponding to active constraints only:
Apˆ =
[ 1 − 1 0 − 2 1 1
]
=
[ 1 1
] ≥ ~ 0 ,
therefore p = (− 2 , − 3 , 0)T^ is a feasible direction at x = (1, 1 , 2)T^.
x 1 = 3x 2 − 5 x 3 − 6 x 5 + 1, x 4 = −x 2 + 2x 3 + x 5 + 1 and the new z function of non-basic variables is z = 4x 3 − 2 x 4 + 3x 5 + 2 = 4x 3 − 2(−x 2 + 2x 3 + x 5 + 1) + 3x 5 + 2 = +2x 2 + x 5. Notice now all variables inside z have postive or zero coefficients, therefore the basic feasible solution (1, 0 , 0 , 1 , 0)T^ is optimal, with optimal value
min(z) = z(1, 0 , 0 , 1 , 0) = 0.
[4] (7 points) Suppose that a linear program has l optimal extreme points {v 1 , v 2 , ..., vl}. Prove that if a feasible point x can be expressed as a convex combination of vi, then x is optimal. Solution: Let x be a convex combination of v 1 , v 2 , ..., vl, with x =
∑l i=1 αivi, for some αi ≥ 0 and
∑l i=1 αi^ = 1. Let M := minS z, with z(x) = cT^ x. Then z(vi) = cT^ vi = M, for all i = 1, 2 , ..., l. We have: z(x) = cT^ x = cT^
( ∑ l i=1 αivi
)
∑l i=1 c
T (^) (αivi) = ∑l i=1 αi(c
T (^) vi) = ∑l i=1 αiM^ =^ M^
∑l i=1 αi^ =^ M^ ·^ 1 =^ M^ =^ Miny∈S^ z(y), therefore^ x^ is also an optimal solution of the linear programming problem. (we use the fact that the function z is linear, the linearity was shown in class).
[5] (11 points) (a) Recall the definitions of a convex set S and of a convex function g on S. (b) Let g be a convex function on Rn. Prove that the set S =
{ x : g(x) ≤ 0
} is convex. (c) Let g 1 , g 2 be two convex functions on the real line, and let r > 0 be a fixed real number. Show that the function f (x) = x + g 1 (x) + rg 2 (x) is also convex on the real line. Solution: (a) The set S is convex if for any x, y ∈ S and any 0 ≤ α ≤ 1, we have αx + (1 − α)y ∈ S. The function g : S → R is convex on the convex set S if for any x, y ∈ S and any 0 ≤ α ≤ 1, we have g(αx + (1 − α)y) ≤ αg(x) + (1 − α)g(y).
(b) Let x, y ∈ S and let 0 ≤ α ≤ 1 be arbitrary. Then g(x) ≤ 0 and g(y) ≤ 0. We also have, by convexity of g: g(αx + (1 − α)y) ≤ αg(x) + (1 − α)g(y) ≤ α · 0 + (1 − α) · 0 = 0 (because α ≥ 0 and 1 − α ≥ 0), therefore g(αx + (1 − α)y) ≤ 0, or αx + (1 − α)y ∈ S. In conclusion, the set S is convex.
(c) Let x, y ∈ R and 0 ≤ α ≤ 1 be arbitrary. We have f (αx + (1 − α)y) = αx + (1 − α)y + g 1 (αx + (1 − α)y) + rg 2 (αx + (1 − α)y) ≤ by convexity of g 1 and g 2 and r > 0 ≤ αx + (1 − α)y + αg 1 (x) + (1 − α)g 1 (y) + r(αg 2 (x) + (1 − α)g 2 (y)) = αx + αg 1 (x) + αrg 2 (x) + (1 − α)y + (1 − α)g 1 (y) + (1 − α)rg 2 (y) = α(x + g 1 (x) + rg 2 (x)) + (1 − α)(y + g 1 (y) + rg 2 (y)) = αf (x) + (1 − α)f (y). In conclusion, f is a convex function.
