Question

Use the simplex method and the Duality Principle to solve the following minimum problem: Minimize: C = 2001 +622 Subject to the constraints: - 201 +3.02 > 0 31 +3:02 > 9 C1 0 22 > 0 and using your final tableau answer the questions below by entering the correct answer in each blank box. Please enter fractions as 3/5, -4/7, and so on. 11 12 C=

Answer 1

After introducing slack variables Max Z 2X1 - 6X2 + 0 S1 + 0 S2 subject to 2 X1 - 3x2 + S1 = 0 X1 - 3X2 + S2 = -9 and X1, X2, S1, S220 Iteration-1 Cj -2 B CB XB X1 S1 0 0 2 -3 -6 OO X2 S1 S2 10 0 1 0 0 0 60 0 0 S2 0 -9 (-1) -3 Z = 0 Z; 0 2 Z;-C; 2;-9 Ratio S2j and S2, j < 0 -21 -2 Minimum negative Xo is -9 and its row index is 2. So, the leaving basis variable is S2. Maximum negative ratio is -2 and its column index is 1. So, the entering variable is X7. :: The pivot element is -1. Entering = Xq, Departing = S2, Key Element -1 + R2(new) = Rz(old) = (-1) + Ry(new) = R1(old) - 2R2(new)

Iteration-2 Cj -2 0 B CB X1 X2 S1 S2 S1 0 -18 0 (-9) 2 X1 -2 9 1 3 0 -1 Z = - 18 -2 -6 0 2 0 0 0 2 Z; Z; -C; Z; - C; Ratio Sj and S1,j< 0 = 0 1 Minimum negative Xp is -18 and its row index is 1. So, the leaving basis variable is Sz. Maximum negative ratio is 0 and its column index is 2. So, the entering variable is X2. :. The pivot element is -9. Entering = X2, Departing = S, Key Element -9 + R (new) = R1 (old) = (-9) + R (new) = R (old) - 3R_(new)

Iteration-3 Cj -2 0 -60 x2 si B Св X1 S2 2 X2 -6 2 0 1 1 -9 X1 -2 3 1 0 win Z = - 18 Z; -2 -6 0 2 Z;-C; 0 0 0 2 Ratio --- Since all 2; -C; 2 0 and all X Bi 2 0 thus the current solution is the optimal solution. Hence, optimal solution is arrived with value of variables as : X1 = 3, x2 = 2 Max Z = - 18 :. Min Z = 18

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