Question

Find the minimum value. Minimize subject to C = 10xı + 31x2 - 5x3 X1 + 3x2 s6 4x2 + x3 s2 X1, X2, X3 20 Select the correct choice below and, if necessary, fill in the answer boxes to complete your choice. O A. The minimum value C= occurs at X1 - X2 X3 = OB. There is no solution.

Answer 1

Min Z 10X1 + 31 x2 - 5x3 subject to X1 + 3 x2 <6 4x2 + x3 = 2 and X1, X2, X3 = 0; The problem is converted to canonical form by adding slack, surplus and artificial variables as appropiate 1. As the constraint-1 is of type 's' we should add slack variable Si 2. As the constraint-2 is of type'<' we should add slack variable S2 After introducing slack variables Min Z = 10x1 + 31 x2 - 5x3 + 0 S1 + 0 S2 subject to X1 + 3x2 + Si = 6 4x2 + x3 + S2 = 2 and x1, x2, X3, S1, S2 = 0 Iteration-1 C; 10 31 -5 0 0 MinRatio Хв B CB Хв X1 X2 X3 Si S2 X3 Si 0 6 1 3 0 1 0 S2 0 2 0 4 (1) 0 * =2 – 1 Z= 0 Z; 0 0 0 0 0 Z; - C; -10 -31 51 0 0 Positive maximum Z; - C; is 5 and its column index is 3. So, the entering variable is xz. Minimum ratio is 2 and its row index is 2. So, the leaving basis variable is S2. :. The pivot element is 1. Entering X3, Departing = S2, Key Element = 1

+ R2(new) = R2(old) + Ri(new) = R (old) Iteration-2 C; 10 31 -5 0 0 B CB X B X1 X2 X3 Si S2 MinRatio Si 0 6 1 3 0 1 0 X3 -5 2 0 4 1 0 Z= -10 Z; 0 -20 -5 0 -5 Z; - C; -10 -51 0 0 -5 Since all Z; - C; = 0 Hence, optimal solution is arrived with value of variables as : x 1 = 0, x2 = 0, x3 = 2 Min Z= - 10

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