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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.

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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.

This problem has been solved

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