# Linearizing min function Problem

How can I linearize $\min(x_1,x_2,x_3)$ in a maximization linear programming problem? Please help me. I've tried many things but I didn't solve.. My LP equations are as follows:

Objective function is: maximize $z=\min(x_1,x_2,x_3)$

Constraints:

$0.6 x_1 + 0.8 x_2 \leq 4500 \times 20$

$0.2 x_2 \leq 4500 \times 3$

$0.3 x_2 \leq 4500 \times 6$

$0.4 x_1 + 0.6 x_2 \leq 4500 \times 10$

$0.3 x_1 \leq 4500 \times 5$

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Did you draw the lines/area coming from those inequalities? – Sigur Dec 9 '12 at 0:56
No, but I don't need this since I will solve the problem in GAMS. I just need to linearize the objective function. Please help me – John Dec 9 '12 at 0:57
I don't know what is GAMS. – Sigur Dec 9 '12 at 1:00
Don't you know how to linearize z=min(x1,x2,x3) ? I do not need the solution of the problem but the linearization of z=min(x1,x2,x3) my friend ? Thank you for your care. – John Dec 9 '12 at 1:03

How you implement min($x_1, x_2, x_3$) in an LP solver depends on what you are trying to do with it. Since you are maximizing it you can do the following

$${\rm maximize } \ z$$ subject to $$z \le x_1 \\ z \le x_2 \\ z \le x_3$$ $z$ will take the value of min($x_1, x_2, x_3$) in any optimal solution.

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Shouldn't z consist of x1,x2 or x3 in an LP? I thought that situation but I didn't implement this solution's code. – John Dec 9 '12 at 2:00

assume $y=\min\{x_1,x_2,x_3\}$

Now we need to maximize $y$

since value of $y$ can either be $x1$, $x2$ or $x3$, we can safely add the following constraints:

$$x_1\geq y \qquad x_2 \geq y \qquad x_3\geq y$$

or

$$x_1-y\geq 0 \qquad x_2-y \geq 0 \qquad x_3-y\geq 0$$

now solve the problem.

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I wish I understood the purpose of giving a hint to a three-year-old question. But I guess I never will. – G. Sassatelli Nov 21 '15 at 20:28
Viraj, welcome to Math.SE. You will find that MathJax and $\LaTeX$ can be used to post mathematical expressions. – hardmath Nov 21 '15 at 20:35