mt.s09.sol

mt.s09.sol - CO 350 Linear Optimization Midterm Examination...

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1 CO 350 Linear Optimization Midterm Examination - Spring 2009 June 15, 2009, 7–9 p.m. Instructors: L. Tun¸cel (Section 1), E. Teske (Section 2) SOLUTIONS 1. [10 points] (a) Convert the following problem to standard inequality form, while doing so, if there are [5] free variables you must eliminate them by using a suitable equation: maximize 10 x 1 - 6 x 2 subject to x 1 + 3 x 2 ≥ - 8 2 x 1 - x 2 = 12 x 1 0 . Solution. Upon elimination of the free variable, using the equation x 2 = 2 x 1 - 12, we obtain maximize - 2 x 1 (+72) subject to - 7 x 1 ≤ - 28 x 1 0 . (b) Convert the following problem to standard equality form: [5] minimize 18 x 1 + 9 x 2 subject to 2 x 1 - 7 x 2 ≥ - 20 3 x 1 + 5 x 2 1 x 1 , x 2 0 . Solution. maximize - 18 x 1 - 9 x 2 subject to - 2 x 1 + 7 x 2 + x 3 = 20 3 x 1 + 5 x 2 + x 4 = 1 x 1 , x 2 , x 3 , x 4 0 . 2. [20 points] Let A := 4 - 1 3 6 5 1 2 8 8 , b := 5 10 2 , c := 10 20 30 describe the data for an LP problem in standard inequality form.
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2 (a) Introduce slack variables x 4 , x 5 , x 6 0 and transform the given problem to an LP problem [5] in standard equality form. Solution. maximize 10 x 1 + 20 x 2 + 30 x 3 subject to 4 x 1 - x 2 + 3 x 3 + x 4 = 5 6 x 1 + 5 x 2 + x 3 + x 5 = 10 2 x 1 + 8 x 2 + 8 x 3 + x 6 = 2 x 1 , x 2 , x 3 , x 4 , x 5 , x 6 0 . (b) Using the simplex method with the smallest subscript rule solve the LP problem you [15] obtained in part (a). Show all your work. Remember to give, after each simplex iteration, the new basis and the new basic feasible solution. Solution. z - 10 x 1 - 20 x 2 - 30 x 3 = 0 4 x 1 - x 2 + 3 x 3 + x 4 = 5 6 x 1 + 5 x 2 + x 3 + x 5 = 10 2 x 1 + 8 x 2 + 8 x 3 + x 6 = 2 x 1 enters, min { 5 4 , 10 6 , 2 2 } = 1, x 6 leaves the basis. We pivot on (6 , 1) and obtain the tableau: z + 20 x 2 + 10 x 3 + 5 x 6 = 10 - 17 x 2 - 13 x 3 + x 4 - 2 x 6 = 1 - 19 x 2 - 23 x 3 + x 5 - 3 x 6 = 4 x 1 + 4 x 2 + 4 x 3 + 1 2 x 6 = 1 which is optimal. The new basis is B := { 1 , 4 , 5 } , the corresponding basic feasible solution is ¯ x := [1 , 0 , 0 , 1 , 4 , 0] T with the objective value 10. Thus, an optimal solution of the original LP is x * := [1 , 0 , 0] T , with objective value z * = 10. (Actually, it is the only optimal solution of the LP.) 3. [10 points] Let F be the set of x satisfying x 1 - 4849845 x 3 + x 4 = 3 2 x 2 + 1048576 x 3 = 2 x 1 , x 2 , x 3 , x 4 0 (a) Find two extreme points of F . Justify your answer. [8] Solution. Note that F = { x : Ax = b, x 0 } where A = ± 1 0 - 4849845 1 0 2 1048576 0 ² , b = [3 , 2] T . Let x 0 := [3 , 1 , 0 , 0] T and x 00 := [0 , 1 , 0 , 3] T . Clearly, x 0 ,x 00 F .
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3 A 1 and A 2 are linearly independent. So are A 2 and A 4 . By Theorem 5.1 ( Let A be m by n with rank m , and let ¯ x be a solution of Ax = b . Then ¯ x is a basic solution of Ax = b iff { A j : ¯ x j 6 = 0 } is a linearly independent set. ), both x 0 and x 00 are basic solutions of Ax = b .
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This note was uploaded on 05/28/2011 for the course CO 250 taught by Professor Guenin during the Fall '10 term at Waterloo.

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mt.s09.sol - CO 350 Linear Optimization Midterm Examination...

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