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

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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

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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
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]

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

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