Question

Use the simplex algorithm to find all optimal solutions to the following LP.

max z=2x1+x2
s.t. 4x1 + 2x2 ≤ 4

−2x1 + x2 ≤ 2

x1 ≥1

x1,x2 ≥0

Answer 1

Solution:

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 S, 2. As the constraint-2 is of type'<' we should add slack variables, 3. As the constraint-3 is of type'>'we should subtract surplus variable S, and add artificial variable 4 After introducing slack,surplus,artificial variables Max z = 2x1 + x2 + 0 S1 + 0 S2 + 0 Sz - M41 subject to 4x1 + 2x2 + Si - 2x1 + x2 + S2 X1 - Sz + 41 = 1 and x1,x2. S1:52. 53,4120 Iteration-1 0 0 1 z= -M -M 0 -1 0 0 0 0 0 0 Z; - C; -M-21 м Negative minimum Z;-;- M-2 and its column index is 1. So, the entering variable is x1- Minimum ratio is 1 and its row index is 3. So, the leaving basis variable is A. .: The pivot element is 1.

Entering = x1, Departing = 41, Key Element = 1 + Rz(new) = R3(old) + R, (new) = R (old) - 4R3(new) + R (new) = Rz(old) + 2R;(new) Iteration 2 1 0 O MinRatio XB 0 0 + z = 2 2 2; 2;-C; -1 0 0 -21 Negative minimum Z;-C, is -2 and its column index is 5. So, the entering variable is Sz. Minimum ratio is 0 and its row index is 1. So, the leaving basis variable is S. : The pivot element is 4. Entering = S3, Departing = S1, Key Element = 4 + R (new) = R (old) = 4 + R (new) = R (old) + 2R (new)

+ Rz(new) = Rz(old) + R (new) Iteration-3 C2 0 XB MinRatio 0 0.25 0.5 0.5 0.25 z = 2 1 0.5 Z;-C; 0 0 0.5 Since all Z;-C;20 Hence, optimal solution is arrived with value of variables as : *1 = 1,x2 = 0 Max z = 2