1. Consider the following polyhedron of an LP problem: 2x 1 x 2 + 5x 3 = 1 (1) 3x 2 + x 4 5 (2) 7x 1 4x 3 + x 4 4 (3) x 1, x 2, x 4 0

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1 MA Linear Programming Tutorial 3 Solution. Consider the following polyhedron of an LP problem: x x + x 3 = ( 3x + x 4 ( 7x 4x 3 + x 4 4 (3 x, x, x 4 Identify all active constraints at each of the following points x = (x, x, x 3, x 4 T and determine whether it is a basic feasible solution. (a ; (b 3 ; (c ; (d. [Solution] NOTE: When checking for active constraints, students may overlook constraints: x, x, x 4. Active constraints rank Conclusion (a x, x 4, (3. 3 Not feasible & not basic (not satisfied Not basic Feasible solution (b only(active. Not feasible, not basic (not satisfied Not basic Feasible solution (c (, ( feasible but not basic Not basic Feasible solution (d (, (, x, x 4 feasible, basic A basic Feasible solution

2 . Suppose the feasible set P in R is described by the following constraints: x + 3x, 3x x, 3x x, x + 3x 3, x, x. (a Sketch the feasible set P and find all extreme points of P. (b For each of the following objective, (i min x + x (ii max 4x x (iii max 3x + x compute the corresponding objective value at each extreme point of P found in Part (a; and determine all optimal solutions if the LP is not unbounded. (Answers ( ( ( 4 (a 4,, (iii Unbounded. (b (i (, (ii λ ( 4 4 +( λ (, λ [, ] [Solution] A sketch of the polyhedral set shows that the polyhedral set does not contain a line. Thus, for an LP with feasible set P, if there is an optimal solution, then an optimal solution occurs at some extreme point. (a Graphically, all extreme points of P can be located and computed. (b Objective Extreme points ( ( 4 4 ( (i min x + x 9 8 (ii max 4x x 3 λ ( 4 4 Optimal Solution ( ( + ( λ, λ [, ] 4 (iii max 3x + x 4 unbounded objective value For Part (iii, note that the objective function 3x + x is not bounded from above (see graph. Alternatively, note that x and x are unbounded from above, thus, 3x + x is unbounded from above.

3 3. Consider the polyhedron P = {(x, x, x 3 T R 3 x + x + x 3, x, x }. Does P contain a line? [Solution] P contains a line if and only if P does not have an extreme point. The point x = is a basic feasible solution. (Check! Hence it is an extreme point. Thus P does not contain a line. Alternatively, suppose P contains a line say x + λd where x P and d R 3 {}. That is, x + λd P for all λ R. x Let x = x and d = d. x 3 Then, for all λ R, (x + λ + (x + λd + (x 3 + λ (x + λ (x + λd The last two inequalities imply that = and d =. From the first inequality, we have Thus, =, and hence d =. x + x + x 3 + λ λ This contradicts the choice of d. Therefore, P does not contain a line. 3

4 4. A vector d R n is said to be a feasible direction at x in a polyhedron P if there exists a θ > such that x + θd P. (a Let P = {x R n Ax = b, x } and x P. Prove that a vector d R n is a feasible direction at x if and only if Ad = and d i for every i such that x i =. (b Let P = {x R 3 x + x + x 3 =, x } and consider the vector x = (,, T. Find the set of feasible directions at x. [Solution] (a Let d be feasible at x. Then there exists θ > suth that x + θd P. This implies Ax + θad = b. Because Ax = b and θ >, we have Ad =. Because x + θd and θ >, x i = implies d i. Suppose Ad = and d i if x i =. From Ax = b and Ad =, it follows A(x + θd = b for all θ R. Now we are left to show that there exists a θ > such that x + θ d. Let I = {i d i < } and let θ = min{ x i d i i I}. Because d i if x i =, we have x i > for every i I. Thus, θ >. For any i I, d i implies x i + θ d i x i. For any i I, θ x i d i implies θ d i x i. Thus, x i + θ d i x i x i =. The above shows that x + θ d P with θ >. Therefore, d is a feasible direction at x. (b Using (a, the set of feasible directions at x = (,, T can be described as {d R 3 + d + =,, d }. 4

5 . Consider the following LP: (a Maximize 4x + 3x x 3 Subject to x + x + x 3 = x + x + 4x 3 + x 4 = x, x, x 3, x 4 (i Verify that columns A and A 4 are linearly independent. (ii Determine the inverse A B of the associated basis matrix A B = (A, A 4. Identify all basic variables and all nonbasic variables. (iii Construct the associated basic solution. Is this solution feasible, i.e. x B? (b Determine all basic feasible solutions of the LP problem and their corresponding objective values. ( (Answers(a (ii A B = [Solution] (a (i [ A A 4 ] = [ ] [ Thus, A and A 4 are linearly independent. ( (ii B =. ( ( ( ( x (iii x B = = B x b = 4 =. 4 Thus the basic solution is x =. Since x 4 <, x is not a feasible 4 solution. (b All basic feasible solutions: ]

6 B B x B c ( ( ( T Bx B ( A A = 8 8 ( ( ( ( A A 3 = ( ( ( ( A A 4 = 3 ( ( ( ( A3 A 4 = 4 Note that columns A & A 3 are linearly dependent. Thus they cannot form a basis matrix.

7 . Consider the following linear programming problem, where b and A i are 3 column matrices for i =,, 3, 4. minimize c x + c x + c 3 x 3 + c 4 x 4 Subject to A x + A x + A 3 x 3 + A 4 x 4 = b x, x, x 3, x 4 Suppose x = (x,, x 3, x 4 T is a basic feasible solution, where A B = [A, A 3, A 4 ] is the basis matrix. Let d = (,,, T such that x + d is a feasible solution. Prove that = A B A. [Solution] Note that x + d is feasible if and only if A(x + d = b & x + θd A(x + d = b Ad =, since Ax = b. (A, A, A 3, A 4 = A + A + A 3 + A 4 = A + A 3 + A 4 = A (A, A 3, A 4 = A A B = A 3 = A B A 3 since A B is nonsingular. 7

8 7. Refer to Question. Prove that the change in the objective value from x to x + d is ( c (c, c 3, c 4 A B A. [Solution] Change in the objective value = c T (x + d c T (x = c T d = (c, c, c 3, c 4 d 4 = c + (c, c 3, c 4 = ( c (c, c 3, c 4 A since B A = A B A 3 from Q. 8