hw3soln - (b) B-1 = 1 . 2500-1 . 0000-2 . 0000-. 8333 ....

Info iconThis preview shows pages 1–7. Sign up to view the full content.

View Full Document Right Arrow Icon
Background image of page 1

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

View Full DocumentRight Arrow Icon
Background image of page 2
Background image of page 3

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

View Full DocumentRight Arrow Icon
Background image of page 4
Homework 3 Additional Problems: Solutions AP1. The relevant matrices are given by £ A I / = 1 4 2 - 1 0 1 0 3 2 0 0 - 1 0 1 c = £ - 2 - 3 - 1 0 0 - M - M / b = 8 6 The simplex method required three iterations in problem 4.6-3(b). The basic variables in these iterations were { ¯ x 6 , ¯ x 7 } , { x 2 , ¯ x 7 } , and { x 1 ,x 2 } . (0) If x B = { ¯ x 6 , ¯ x 7 } , then B = 1 0 0 1 = B - 1 = 1 0 0 1 c B = £ - M - M / Therefore x B = B - 1 b = 1 0 0 1 ‚• 8 6 = 8 6 Z = c B B - 1 b = £ - M - M / 8 6 = - 14 M. (1) If x B = { x 2 , ¯ x 7 } , then B = 4 0 2 1 = B - 1 = 1 4 0 - 1 2 1 c B = £ - 3 - M / Therefore x B = B - 1 b = 1 4 0 - 1 2 1 ‚• 8 6 = 2 2 Z = c B B - 1 b = £ - 3 - M / 2 2 = - 6 - 2 M. (2) If x B = { x 1 ,x 2 } , then B = 1 4 3 2 = B - 1 = - 1 5 2 5 3 10 - 1 10 c B = £ - 2 - 3 /
Background image of page 5

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

View Full DocumentRight Arrow Icon
Therefore x B = B - 1 b = - 1 5 2 5 3 10 - 1 10 ‚• 8 6 = 4 5 9 5 Z = c B B - 1 b = £ - 2 - 3 / 4 5 9 5 = - 7 . These solutions agree with the values of x B and Z found in 4.6.3(b). (The objective function values would have to be multipled by - 1 to determine the objective for the original, minimization problem. But since we’re dealing with the revised, maximization, problem, it’s OK to leave them as negatives.) AP2. (a) A - 1 = 0 . 0714 0 . 0714 0 . 0714 - 0 . 5000 0 . 1667 0 . 5000 0 . 1071 - 0 . 0595 0 . 1071
Background image of page 6
Background image of page 7
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: (b) B-1 = 1 . 2500-1 . 0000-2 . 0000-. 8333 . 6667 1 . 0000 . 2500 . 0000 . 0000 (c) A-1 B = . 0714 . 2143 . 6429-. 5000-. 5000-1 . 1667-. 3929-. 6786 . 1310 MATLAB printout: >> A = [4 -1 2; 9 0 -6; 1 1 4] A = 4-1 2 9-6 1 1 4 >> inv(A) ans = 0.0714 0.0714 0.0714-0.5000 0.1667 0.5000 0.1071-0.0595 0.1071 >> B = [0 0 4; 3 6 5; -2 -3 0] B = 4 3 6 5-2-3 >> inv(B) ans = 1.2500-1.0000-2.0000-0.8333 0.6667 1.0000 0.2500 >> inv(A) * B ans = 0.0714 0.2143 0.6429-0.5000-0.5000-1.1667-0.3929-0.6786 0.1310 AP3. Rows 1 and 3 are linearly independent, and row 2 is a linear combination of rows 1 and 3 (row 2 equals row 1 minus row 3). Therefore the rank is 2....
View Full Document

This homework help was uploaded on 04/07/2008 for the course IE 220 taught by Professor Storer during the Spring '07 term at Lehigh University .

Page1 / 7

hw3soln - (b) B-1 = 1 . 2500-1 . 0000-2 . 0000-. 8333 ....

This preview shows document pages 1 - 7. Sign up to view the full document.

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