Answer to Question #211852 in Operations Research for almanya

Question

Solve the following LPP using dual simplex method: Max Z= -3x1- x2,

Subject to: x1+x2 ≥ 1, 2x1+3x2 ≥ 2, x1,x2 ≥ 0.

Answer 1

Solution.

$-3x_1-x_2\to max,$

$x_1+x_2\geq 1,\newline 2x_1+3x_2\geq 2.$

We replace signs in restrictions and we enter additional variables x3, x4.

Initial simplex table, where x3 and x4 are basic variables.

So, as $|b|_{max}=|-2|$ is in the 2nd line, and the maximum modulo element in the 2nd line is -3, x2 is a new basic variable.

Updated table

$|b|_{max}=|-\frac{1}{3}|$

is in 1 row, and the largest element modulo $-\frac{1}{3}$ is in 1 column, then x1 is a new basic variable.

Updated table.

Let's calculate and create a simplex table with deltas.

$∆_i=C_1•a_{1i}+C_2•a_{2i}-C_i$

$∆_4=-2<0,$ therefore, the plan is suboptimal.

Take the new base variable x4 instead of x1.

We will have the following table, as well as a table with deltas.

$∆_i>0,$ therefore, the plan (0,1,0,1) is optimal.

So,

$$F_{max}=-3•0+(-1)•1+0•0+0•1=-1.$$

Answer. -1.