Solutions - CSA 273 Solutions to Homework Set#9 Problem#1...

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CSA 273 Solutions to Homework Set #9 Problem #1 Adding artificial variables a 1 and a 2 to each of the constraints and a surplus variable e 2 to the second constraint, the "artificial" problem in standard form becomes: Minimize Z=3x 1 + 2x 2 + 4x 3 + Ma 1 + Ma 2 such that 2x 1 + x 2 + 3x 3 + a 1 = 10 3x 1 +3x 2 + 5x 3 -e 2 + a 2 = 20 x 1 ,x 2 ,x 3 ,a 1 ,a 2 ,e 2 0 Start out with the tableau: Z x 1 x 2 x 3 e 2 a 1 a 2 RHS Row 0 1 -3 -2 -4 0 -M -M 0 Row 1 0 2 1 3 0 1 0 10 Row 2 0 3 3 5 -1 0 1 20 To get an initial feasible solution with basic variables z,a 1 and a 2 , the coefficients of a 1 and a 2 must be cleared in row 0. This can be done by applying the eros: New Row 0 = Old Row 0 + M*Old Row1 + M*Old Row 2 The other rows would remain unchanged. Our Initial Simplex Tableau: Basic Z x 1 x 2 x 3 e 2 a 1 a 2 RHS Solution Ratio Row 0 1 5M-3 4M-2 8M-4 -M 0 0 30M Z=30M Row 1 0 2 1 3 0 1 0 10 a 1 =10 10/3=3.33 Row 2 0 3 3 5 -1 0 1 20 a 2 =20 20/5=4 has an initial basic feasible solution of a 1 =10, a 2 =20, x i =0 for i=1,2,3, e 2 =0 , and Z=30M to the artificial problem. The entering basic variable should be x 3 since it has the largest coefficient in Row 0. The leaving basic variable should be a 1 by the ratio test. Problem #2
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Solutions - CSA 273 Solutions to Homework Set#9 Problem#1...

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