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According to this link binary variables - absolute values the formula of some binary variables like absolute value is :

y = | x1 - x2| for two variables x1, x2 with 0 ≤ xi ≤ U

Introduce binary variables d1, d2 to mean

d1 : 1 if x1 - x2 is the positive value d2 : 1 if x2 - x1 is the positive value

MIP formulation

0 ≤ xi ≤ U [1.i]

0 ≤ y - (x1-x2) ≤ 2 · U · d2 [2]

0 ≤ y - (x2-x1) ≤ 2 · U · d1 [3]

d1 + d2 = 1 [4]

Notice the bolded numbers, what do these 2s represent? Do they represent the value range? Aren't the value range is only from 0 to 1?

Sorry, I'm totally new to this.

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That's just basic inequalities, there is no trick.

  • if $d_1 = 1$, then $d_2 = 0$, and $y = x_1 - x_2$. That means that: $$y - (x_1 - x_2) = 0 \leqslant 2 \times U \times d_2 = 0$$
  • if $d_1 = 0$, then $d_2 = 0$, and $y = x_2 - x_1$. That means that: $$y - (x_1 - x_2) = 2x_2 - 2x_1 \leqslant 2\times U - 2\times 0 = 2\times U \times d_2$$

Same thing for the other inequality.

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  • $\begingroup$ According to the link, it said "Binary decision variables take value 0 or 1". The values can only be from 0 to 1. What do these 2s represent? I'm sorry, I have a very limited background in maths $\endgroup$ Commented Mar 17, 2024 at 11:46
  • $\begingroup$ $d_1$ and $d_2$ take values in $\{0,1\}$. Those are the binary decision variables. $\endgroup$ Commented Mar 17, 2024 at 11:50
  • $\begingroup$ Ok, thanks. So, those 2s in y−(x1−x2)=0⩽2×U×d2=0 and y−(x1−x2)=2x2−2x1⩽2×U−2×0=2×U×d2 represent 1? $\endgroup$ Commented Mar 17, 2024 at 15:19
  • $\begingroup$ What are you saying? Those 2's are representing 2, not 1… This is $2 = 1 + 1$, because there is two occurrences of $x_1$ and $x_2$. Honestly, I don't even understand what is the problem. $x + x = 2x$, and this $2$ is not $1$. $\endgroup$ Commented Mar 17, 2024 at 15:46
  • $\begingroup$ Oh sorry for that, so the value range is still 0 to 1? Not 0 to 2? My problem is that because it is binary variables which strictly involved only 0 and 1, so my initial thought is that there cannot be number 2 in there. But only 0 and 1 $\endgroup$ Commented Mar 17, 2024 at 16:33

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