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When reading an example in "Fourier-Motzkin elimination extension to integer programming problems-H.P.Williams" :

"Suppose we wished lo eliminate x between the following two inequalities \begin{align*} 3x+5y \leq 19\quad(1)\\ 2x+y \geq 3\quad (2) \end{align*} 2(1) and 3(2) we have $-3y+9 \leq 6x \leq -10y +38 \quad (3)$

if $x$ is taken from continuum of real or rational numbers (3) is equivalent to $-3y+9 \leq -10y +38 \quad (3)$ giving $7y \leq 29$.

But $x$ is an integer (3) is equivalent to

$7y\leq 29$, y odd or $7y \leq 26$ , y even"

I cant' understand why "$7y\leq 29$, y odd or $7y \leq 26$ , y even", I think it better is "$7y\leq 29$, y odd or $7y \leq 28$,y even" please help me. Thanks a lot!

Thu Le
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1 Answers1

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If $y$ is even, $-3y+9$ is odd. The number $6x$ is even, so there can't be equality, but a strict inequality. Hence $-3y+9 \leq 6x - 1$, in other words

$$6x + 3y \geq 10.$$

But the equation $6x+3y = 10$ doesn't have solutions in integers, since $\gcd(6,3)= 3$ doesn't divide $10$. The smallest above $10$ the quantity $6x+3y$ can be is $12$. Hence we have the inequality $6x+3y \geq 12$. From this we get the inequality

$$12-3y \leq 6x$$

Continue this with the other inequality of $(3)$ to have

$$12-3y \leq 6x \leq -10y+38$$

And finally

$$7y\leq 26.$$

The other inequlity (when $y$ odd) is as is.

ploosu2
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