Removing variable from LP problem?

397 Views Asked by At

max $4x_1 + 3x_2 - 2x_3$

Subject to $x_1 + x_2 + 2x_3 \le 3$

$x_1 + 2x_2 + x_3 \le 6$

$ x_1 \le2$

$x_1, x_2, x_3 \ge 0$

I was told to remove one of the variables(variable fixing) by looking at the coefficients in the object function and contraints, but I don’t know how to do that.

2

There are 2 best solutions below

0
On

Guide:

From the objective funciton, you would want $x_1$ and $x_2$ to be large and $x_3$ to be small.

Minimizing $x_3$ doesn't stop $x_1$ and $x_2$ from being big, in fact it allows $x_1$ and $x_2$ to be big.

0
On

The coefficient $4$ of $x_1$ in objective function is the greatest compared with others ($3$ and $-2$) and the coefficient $1$ of $x_1$ in the constraints are less or equal to those of others ($1$ and $2$). So $x_1$ should be set equal to maximum value $2$. The adapted linear programming problem is: $$\text{Max} \ \ 8+3x_2-2x_3 \ \ \text{subject to}$$ $$x_2+2x_3\le 1 \\ 2x_2+x_3\le 4 \\ x_2,x_3\ge 0.$$