This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: 1. Consider the following linear program in standard inequality form: maximize 2 x 1 + x 2 subject to x 1 + x 2 ≤ 1 2 x 1 + x 2 x 3 ≤ 4 3 x 1 2 x 2 ≤ 3 x 1 , x 2 , x 3 ≥ . (a) [2 marks] Prove algebraically that [1 0 2] T is an optimal solution for this linear program. Let [ x 1 x 2 x 3 ] T be any feasible solution. Then x 2 ≥ and x 1 + x 2 ≤ 1 , so 2 x 1 + x 2 ≤ 2 x 1 + 2 x 2 ≤ 2 · 1 = 2 . This bound on the objective value is achieved by [1 0 2] . Since this point is feasible, it must therefore be optimal. (b) [2 marks] Introduce slack variables and write down the resulting linear program in standard equality form. With slack variables x 4 , x 5 , x 6 we obtain maximize 2 x 1 + x 2 subject to x 1 + x 2 + x 4 = 1 2 x 1 + x 2 x 3 + x 5 = 4 3 x 1 2 x 2 + x 6 = 3 x 1 , x 2 , x 3 , x 4 , x 5 , x 6 ≥ . (c) [2 marks] Find an optimal solution to the linear program you wrote down in (b). The two linear programs are equivalent, in particular the optimal values are the same. Hence [1 0 2 0 4 0] T , which is feasible and achieves the optimal value 2, is optimal. (This vector was obtained form the vector in (a) by setting the slack variables appropriately.) [ Alternatively, one can use the solution to Question 2. There the simplex method is applied to the system in (b). ] 1 Question 1 continued: (d) [2 marks] Write down the definition of a basic feasible solution . A vector x is a basic feasible solution if it is basic (i.e., there exists a basis B such that A B x B = b and x N = 0 , employing the usual notation from class), and also feasible. (e) [2 marks] Is the vector you found in (c) a basic feasible solution for the linear program in (b)? Justify your answer carefully. It is feasible, so we need to check whether it is basic. According to Theorem 3.5 (recognizing basic solutions), this is the case if and only if A b B is invertible, where b B = { 1 , 3 , 5 } . We have A b B = 1 0 0 2 1 1 3 0 0 . This matrix is not invertible, since the columns are not linearly independent (consider for instance x = [0 1 1] T 6 = 0 , but for which A b B x = 0 .) [ Alternatively, if part (c) was solved using the simplex method, the resulting optimal so lution is basic feasible. ] (d) [2 marks] Is the vector given in (a) an extreme point of the feasible region of the original linear program? Justify your answer carefully. By Corollary 4.5 (basic feasible solutions and extreme points II), the vector in (a) is an extreme point if and only if the vector in (c) is a basic feasible solution for the linear program in (b). According to (e), this is not the case. The answer is therefore “no”. [ This approach does not work if (c) was solved using the simplex method, because in that case there is no relation to the vector given in (a). A different argument is therefore needed. For instance, one could use the definition of an extreme point: identify an open line segment that contains the vector in (a), and is contained in the feasible region. Oneline segment that contains the vector in (a), and is contained in the feasible region....
View
Full
Document
This note was uploaded on 10/31/2011 for the course OR&IE 3300 at Cornell.
 '10
 TODD

Click to edit the document details