hw07solutions

# hw07solutions - IEOR 162 Linear Programming Spring 2007...

This preview shows pages 1–2. Sign up to view the full content.

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

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: IEOR 162 Linear Programming Spring 2007 Homework 7 Solutions 4.7.2 (4 points) To find multiple optimal solutions, we perform the simplex algorithm and instead of stopping when all reduced costs are nonpositive, we continue by entering a variable with 0 reduced cost. ↓ z x 1 x 2 x 3 x 4 RHS ratio 1 3- 6 5 7 1 35 5- 1 2 1 2 1 ↓ z x 1 x 2 x 3 x 4 RHS ratio 1 3 6 8.5 1- 7 2 28 56 17- 1 2 1 1 2 1- We would normally stop here because all reduced costs are nonpositive, but since we want to find more optimal solutions we enter x 1 into the basis. z x 1 x 2 x 3 x 4 RHS ratio 1 3 6 1 2 17- 7 17 56 17 1 1 17 5 17 45 17 So we have found 2 optimal solutions, x 1 = ( 1 28 ) and x 2 = ( 56 17 45 17 ) . We know that all points that lie on the line between x 1 and x 2 must also be optimal, so to find a third optimal solution, we can simply take the average of x 1 and x 2 , x 3 = x 1 + x 2 2 = ( 28 17 31 17 14 ) ....
View Full Document

## This homework help was uploaded on 04/02/2008 for the course IEOR 162 taught by Professor Zhang during the Spring '07 term at Berkeley.

### Page1 / 3

hw07solutions - IEOR 162 Linear Programming Spring 2007...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online