Question

max Xi + 3x2 subject to: -X1 – x2 < -3 -21 + x2 = -1 X1 + 2x2 = 2 X1, x2 > 0

Answer 1

The solution too long, so i wrote it down in my word and pasted it here, You may download the image and view it s it will be clearly visible. Thank you!

Solution: Problem is Max Z = x + 3 x2 subject to - i - 22 -3 Here by = -3<0. so multiply this constraint by -1 to make bı > 0. & + 22 23 - 2 + 22 S -1 Here b2 = -1<0. so multiply this constraint by -1 to make b2 > 0. 21 - 22 1 ti + 2 x2 < 2 and X1,42 > 0; -->Phase-15- 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 subtract surplus variable S, and add artificial variable A1 2. As the constraint-2 is of type'>'we should subtract surplus variable S, and add artificial variable A2 3. As the constraint-3 is of type' < 'we should add slack variable S3

After introducing slack,surplus,artificial variables Max Z = - A1 - A2 subject to x1 + x2 - Si + A = 3 21 - 22 - S2 + A2 = 1 *1 + 2 x2 + S3 and 21, #2, S1, S2, S3, A1, A2 > 0 NACO Iteration-1 MinRatio ХВ -1 = 3 -1 1/2 = 1³ 0 0 2 = - 4 Z; Z; -; 0 -2 -2 oi 0 1 -1 0 -1 0 1

Negative minimum Z - C; is – 2 and its column index is 1. So, the entering variable is 21. Minimum ratio is 1 and its row index is 2. So, the leaving basis variable is A2. : The pivot element is 1. Entering = 31. Departing = A2, Key Element = 1 - R2 (new) = R (old) - R₂ (old) = 1 1 -1 0-100 Rz (new) = RZ (old) 1 1 -1 0 -1 0 0 - R1 (new) = R1 (old) - Rz (new) R1 (old) = 3 1 1 -1 0 0 1 R2 (new) = 1 1 -1 0-1 0 0 R1 (new) = R1 (old) - R2(new) 2 0 2 -1 1 0 1

- Rz (new) = R3 (old) - R2 (new) Rz(old) = 2 1 2 0 R2 (new) = 1 1 -1 0 R3 (new) = R3 (old) - R2(new) 1 0 3 0 0 1 -1 0 1 Iteration-2 0 MinRatio Хв XB 12 NIN = 0.3333 → 2= -2 1 20 Z; – C; - TT - 10

Negative minimum 2 - Cis -2 and its column index is 2. So, the entering variable is c.. Minimum ratio is 0.3333 and its row index is 3. So, the leaving basis variable is Sg .. The pivot element is 3. Entering = x2. Departing = S3, Key Element = 3 - R3 (new) = R3 (old) : 3 R3(old) = 10 3 0 1 1 0 R3(new) = R3 (old) = 3 0.3333 0 1 0 0.3333 0.3333 0 - Ri (new) = R1 (old) - 2R3 (new) R1 (old) = 20/2 – 1 1 R3 (new) = 0.3333 0 1 0 0.3333 0.3333 0 2 x R3 (new) = 0.6667 0 2 0 0.6667 0.6667 0 Ri (new) = R1 (old) - 2R3 (new) 1.3333 0 0 -1 0.3333 -0.6667 1

- Rz (new) = R2 (old) + R3 (new) R2 (old) = 11 - -1 0 0 R: (new) = 0.3333 0 1 0 0.3333 0.3333 0 Ry(new) = R (old) + R$ (new) 1.3333 1 0 0 -0.6667 0.3333 0 Iteration- 3 C 0 XB MinRatio Хв - S2 S3 S2 1.3333 -1 1.3333 0.3333 -0.6667 0.3333 0 1.3333 -0.6667 0.3333 0.333 0.3333 (0.3333) 0.3333 = 1 Z= - 1.3333 Z; -0.3333 0.6667 Z; – C; -0.3333 + 0.6667 0

Negative minimum Z; -C; is -0.3333 and its column index is 4. So, the entering variable is S2 Minimum ratio is 1 and its row index is 3. So, the leaving basis variable is 22. .. The pivot element is 0.3333. Entering = S2. Departing = x2, Key Element = 0.3333 - R3 (new) = R3 (old) : 0.3333 R3 (old) = 0.3333 0 1 0 0.3333 0.3333 Rs (new) = R3(old) = 0.3333 10 3 0 1 1 0 - R1 (new) = R1 (old) - 0.3333R3 (new) R1 (old) = 1.3333 0 0 -1 0.3333 -0.6667 1 R: (new) = 1 0 3 0 1 1 0 0.3333 x R3 (new) = 0.3333 0 1 0 0.3333 0.3333 0 R1 (new) = R1 (old) - 0.3333R3 (new) 1 0 -1 -1 0 -1 1

- Rz (new) = R2 (old) +0.6667R3 (new) R2 (old) = 1.3333 1 0 0 -0.6667 0.3333 R: (new) = 10 30 0.6667 x R3 (new) = 0.6667 0 2 0 0.6667 0.6667 0 R2 (new) = R2 (old) + 0.6667R3 (new) 1 2 1 2 0 Ci 0 S2 MinRatio Iteration-4 B CB A -1 210 S20 Z= -1 i 2 1 0 1 0 1 1 0 Z; – C; 0 1 0

Since all Z - C; > 0 Hence, optimal solution is arrived with value of variables as: 21 = 2, *2 = 0 Max Z=0 But this solution is not feasible because the final solution violates the 1° constraint & + 22 23 and the artificial variable Aj appears in the basis with positive value 1 So phase-2 is not possbile.