sol5 - CO 350 Linear Optimization (Spring 2009) Solutions...

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
CO 350 Linear Optimization (Spring 2009) Solutions to Assignment 5 [10] 1. (a) The variables that may enter are x 4 and x 5 . For x 4 all coefficients in A 4 are positive. Since in each case b i = 0 any basis element may be leaving. Thus we have ( x 4 ,x 2 ),( x 4 ,x 3 ) and ( x 4 ,x 6 ). If x 5 is entering then x 2 is leaving, so we have ( x 5 ,x 2 ). (b) The current matrix is 2 1 0 2 - 4 0 β - 1 1 x 4 enters so we compute (after substituting β = 1) the lexicographic minimum of (0 , 1 , - 2 , 0) = min ± (0 , 2 , 1 , 0) 2 , (0 , 2 , - 4 , 0) 2 , (0 , 1 , - 1 , 1) 1 ² therefore x 3 is leaving. (c) We have the same matrix as in part (b). Note that (0 , 1 , 0 . 5 , 0) > L (0 , 1 , - 2 , 0), so we need to ensure that (0 , 1 , - 2 , 0) > L ± 0 , 1 , - 1 β , 1 β ² ⇒ - 2 > - 1 β 0 . 5 > β. To have x 6 leaving the basis, we need 0 . 5 > β > 0. [20] 2. A = ³ 1 1 2 0 2 2 3 3 1 2 ´ , b = ³ 3 7 ´ , c = [5 , 8 , 4 , 2 , 3] T . We use x 4 as basic variable. Phase 1 is the LP A = ³ 1 1 2 0 2 1 2 3 3 1 2 0 ´ , b = ³ 3 7 ´ , c = [0 , 0 , 0 , 0 , 0 , - 1] T . Iteration 1, B = { 6 , 4 } and x * B = [3 , 7] T 1. Solve A T B y = c B ³ 1 0 0 1 ´³ y 1 y 2 ´ = ³ - 1 0 ´ y = ³ - 1 0 ´ 2. Choose entering variable c 1 = c 1 - A T 1 y = 0 - [1 , 2][ - 1 , 0] T = 1 > 0 So x 1 enters the basis. 1
Background image of page 1

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

View Full DocumentRight Arrow Icon
3. Choose leaving variable - solve A B d = A 1 ± 1 0 0 1 ²± d 6 d 4 ² = ± 1 2 ² ± d 6 d 4 ² = ± 2 1 ² t = min ³ x * 6 d 6 , x * 4 d 4 ´ = min ³ 3 1 , 7 2 ´ = 3 r = 6 so x 6 is leaving. 4. x * 1 := t = 3 and x * 4 := x * 4 - td 4 = 7 - 3 × 2 = 1 5. B = { 1 , 4 } and all artificial variables are zero, hence the first phase is over. For Phase II we have B = { 1 , 4 } and corresponding x * = [3 , 1] T together with A = ± 1 1 2 0 2 2 3 3 1 2 ² , b = ± 3 7 ² , c = [5 , 8 , 4 , 2 , 3] T . Iteration 1, 1. Solve A T B y = c B ± 1 2 0 1 ²± y 1 y 2 ² = ± 5 2 ² y = ± 1 2 ² 2. Choose entering variable c 2 = c 2 - A T 2 y = 8 - [1 , 3][1 , 2] T = 1 > 0 So x 2 enters the basis. 3. Choose leaving variable - solve A B d = A 2 ± 1 0 2 1 ²± d 1 d 4 ² = ± 1 3 ² d = ± 1 1 ² t = min ³ x * 1 d 1 , x * 4 d 4 ´ = min ³ 3 1 , 1 1 ´ = 1 hence x 4 is leaving. 4. x * 2 := t = 1 and x * 1 := x * 1 - td 1 = 3 - 1 × 1 = 2 5. B = { 1 , 2 } Iteration 2, 1. Solve A T B y = c B ± 1 2 1 3 ²± y 1 y 2 ² = ± 5 8 ² y = ± - 1 3 ² 2
Background image of page 2
2. Choose entering variable c 3 = c 3 - A T 3 y = 4 - [2 , 3][ - 1 , 3] T = - 3 0 c 4 = c 4 - A T 4 y = 2 - [0 , 1][ - 1 , 3] T = - 1 0 c 5 = c 5 - A T 5 y = 3 - [2 , 2][ - 1 , 3] T = - 1 0 No entering variable can be selected. We have optimality, the optimal solution is x * = [2 , 1 , 0 , 0 , 0] T and the optimal value is 18. [20]
Background image of page 3

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

View Full DocumentRight Arrow Icon
Image of page 4
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 7

sol5 - CO 350 Linear Optimization (Spring 2009) Solutions...

This preview shows document pages 1 - 4. Sign up to view the full document.

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