$x_1+x_2+x_3+x_4 = 100$ with

$1 \le x_1\le10$

$2\le x_2\le15$


$0\le x_4\le10$

Apparently this is the same as

$y_1 + y_2 + y_3 + y_4 = 92$ with

$y_1 \le 9$

$y_2 \le13$

$y_4 \le10$

I understand the $3$ conditions, but what has happened to $x_3$? I can see that it has been subtracted but what has happened to the conditions?

  • $\begingroup$ x_3 >= 5 and Z+x_3 =100 will have the same number of solutions as x_3 >= 0 and Z +x_3 = 95 has. We don't need to write x_3 >= 0 as that's presumed. $\endgroup$
    – fleablood
    Mar 4, 2016 at 6:01

1 Answer 1


We are defining $y_1=x_1-1, y_2=x_2-2,y_3=x_3-5,y_4=x_4$ and demanding that all the $y$'s be $\ge 0$. Once we defined $y_3=x_3-5$, there is no constraint on $y_3$ except that it be nonnegative, so it is not listed in the constraints.

  • $\begingroup$ Thank you. Just needed clarification. $\endgroup$
    – Aaron
    Mar 4, 2016 at 4:42

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