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2. Solve the following LP problem using the simplex method s.t. - 3Xl- X22-6 X1 +X224 and Xl 2 0,X2 u.r.s. HINT: Use the Big-M Method to find an initial bfs.

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Solution Find solution using Simplex(BigM) method subject to -3x1 X2-6 -x1 x2 4 and x1,x20 Solution Problem is subject to 3 X1 Here b-60 so multiply this constraint by -1 to makeb >0 3 x1 *2 2 and x1,x2 2 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 we should add slack variable S 2. As the constraint-2 is of type 2we should subtract surplus variable S2 and add artificial variable A

After introducing slack,surplus,artificial variables subject to and x,x2, S1, S2.A12 0 teration-1 MinRatio S, Z-4M Z. Z, - C M-2 -M-1 ↑ Negative minimum Z,- C, is -M-1 and its column index is 2. So, the entering variable is x2 Minimum ratio is 4 and its row index is 2. So, the leaving basis variable is A . The pivot element is 1

Entering-x2, Departing -A1. Key Element-1 R2(new) = R2(old) R1(new)- R1(old) R2(new) Iteration-2 C, MinRatio NS x, 0 40.5 Z-4 Negative minimum Z- C, is -3 and its column index is 1. So, the entering variable is x Minimum ratio is 0.5 and its row index is 1. So, the leaving basis variable is S1 The pivot element is4 Enteringx. DepartingSKey Element 4

Ri (new) = R1(old) +4 R, (new)-R2(old) + R1(new) MinRatio CB *1 S1 S, *1 =2

Negative minimum Z, - C, is - and its column index is 4. So, the entering variable is S Minimum ratio is 2 and its row index is 1. So, the leaving basis variable is x The pivot element is EnteringS2, Departing x1, Key Element- RI (new) = R1(old) × 4 > R2(new) 2(old)R,(new) Iteration-4 MinRatio 2

Since all 2-C,20 Hence, optimal solution is arrived with value of variables as x10,x2 6 Max Z-6

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