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# s7 - CO350 Assignment 7 Solutions Exercise 1 For the...

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CO350 Assignment 7 Solutions. Exercise 1: For the following pairs ( A, b ) solve the feasilbility problem ( Ax = b, x 0) by first formulating the auxiliary problem ( P ) and then solving ( P ) by the simplex method. If ( Ax = b, x 0) is feasible, give a feasible solution, otherwise find a vector y satisfying ( A T y 0 , b T y < 0). (a) A = - 1 1 2 1 - 2 1 1 - 1 0 1 1 1 - 1 1 0 , and b = 1 1 4 . Solution: The auxiliary problem is: ( P ) maximize - u 1 - u 2 - u 3 subject to - x 1 + x 2 +2 x 3 + x 4 - 2 x 5 + u 1 = 1 x 1 + x 2 - x 3 + x 5 + u 2 = 1 x 1 + x 2 - x 3 + x 4 + u 3 = 4 x 1 , x 2 , x 3 , x 4 , x 5 , u 1 , u 2 , u 3 0 Now let z = - u 1 - u 2 - u 3 = ( - x 1 + x 2 + 2 x 3 + x 4 - 2 x 5 - 1) + ( x 1 + x 2 - x 3 + x 5 - 1) + ( x 1 + x 2 - x 3 + x 4 - 4) = x 1 + 3 x 2 + 2 x 4 - x 5 - 6 . So the initial tableau is: z - x 1 - 3 x 2 - 2 x 4 + x 5 = - 6 - x 1 + x 2 +2 x 3 + x 4 - 2 x 5 + u 1 = 1 x 1 + x 2 - x 3 + x 5 + u 2 = 1 x 1 + x 2 - x 3 + x 4 + u 3 = 4 After solving ( P ) via the simplex method we arrive at the final tableau: z +2 x 2 + x 3 + u 1 +3 u 2 = - 1 x 1 +3 x 1 + x 4 + u 1 +2 u 2 = 3 x 1 + x 2 - x 3 + x 5 + u 2 = 1 - 2 x 2 - x 3 - u 1 - 2 u 2 + u 3 = 1 Since the optimal value of ( P ) is negative, the system ( Ax = b, x 0) is infeasible. The optimal basic dual solution for ( P ) satisfies 1 0 1 - 2 1 0 0 0 1 y = 0 0 - 1 , so y = [1 , 2 , - 1] T .

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