Solve the following LP by using the dual simplex method:min 3x1 + 2x2 + 10s.t.3x1 + x2 ≥ 3,4x1 + 3x2 ≥ 6,x1 ≥ 1,x2 free.2
Question
Solve the following LP by using the dual simplex method:
min 3x1 + 2x2 + 10\
s.t.
3x1 + x2 ≥ 3,
4x1 + 3x2 ≥ 6,
x1 ≥ 1,
x2 free.
Solution
The dual simplex method is a technique used in linear programming to solve problems that are not feasible at the start. Here's how you can solve the given problem using the dual simplex method:
Step 1: Convert the inequalities to equalities by introducing slack variables. The problem becomes:
min 3x1 + 2x2 + 10 s.t. 3x1 + x2 - s1 = 3, 4x1 + 3x2 - s2 = 6, x1 - s3 = 1, x2 free, s1, s2, s3 ≥ 0.
Step 2: Write the initial simplex tableau:
x1 | x2 | s1 | s2 | s3 | RHS | |
---|---|---|---|---|---|---|
s1 | 3 | 1 | -1 | 0 | 0 | 3 |
s2 | 4 | 3 | 0 | -1 | 0 | 6 |
s3 | 1 | 0 | 0 | 0 | -1 | 1 |
Z | -3 | -2 | 0 | 0 | 0 | 0 |
Step 3: Identify the pivot row. The pivot row is the one with the most negative right-hand side (RHS). In this case, it's the third row.
Step 4: Identify the pivot column. The pivot column is the one with the most negative coefficient in the pivot row. In this case, it's the fifth column.
Step 5: Perform row operations to make the pivot element 1 and all other elements in the pivot column 0. The new tableau becomes:
x1 | x2 | s1 | s2 | s3 | RHS | |
---|---|---|---|---|---|---|
s1 | 3 | 1 | -1 | 0 | 0 | 3 |
s2 | 4 | 3 | 0 | -1 | 0 | 6 |
x1 | 1 | 0 | 0 | 0 | -1 | 1 |
Z | -3 | -2 | 0 | 0 | 0 | 0 |
Step 6: Repeat steps 3-5 until all RHS values are non-negative. In this case, the tableau is already optimal.
So, the optimal solution is x1 = 1, x2 = 0, with an objective value of 3.
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