"Greater than" condition in integer linear program with a binary variable

Question

How can one model the following condition in an integer linear program?

$$A = \begin{cases} 1 & \text{if } B > C\\ 0 & \text{otherwise}\end{cases}$$

where $A \in \{0,1\}$ and $B, C \in \mathbb N$. We have upper and lower bounds on both $B$ and $C$.

Possible duplicate of Converting If-else condition to Linear Programming — adrianN, Jan 30 '17 at 11:08
It's a completely different question. I've already searched for linked questions before posting. — Salah, Jan 30 '17 at 11:16
There is no "conversion to binary". An equation A=B-C means just that - that A equals B-C. — Yuval Filmus, Jan 30 '17 at 13:18
You say that B and C are positive integers. In that case B > C is equivalent to B ≥ C + 1, which makes it a lot easier. — gnasher729, Jan 30 '17 at 20:20

JobS · Answer 1 · 2021-05-07T15:32:07.090

The second inequality of the answer of Salah doesn't hold if $B=C$. I think the correct answer is the following two inequalities:

$$ B \geq C + 1 - M (1-A) $$ $$ B \leq C + M A $$

where $M$ is an upperbound on $|B-C| + 1$.

The first inequality restricts $A$ to $0$ if $B \leq C$.

For example, if $B=0$ and $C=0$, then

$$ 0 \geq 0 + 1 - M(1-A) \rightarrow A = 0 $$

The second inequality restricts $A$ to $1$ if $B > C$.

For example, if $B=1$ and $C=0$, then

$$ 1 \leq 0 + M A \rightarrow A = 1 $$

score 1 · Answer 2 · answered Jan 30 '17 at 15:03

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I think I found the answer here! It is sufficient to use the big-M method by introducing the following constraints:

B >= C + 1 - M*(1-A);

C >= B + 1 - M*A

answered Jan 30 '17 at 15:03

Salah

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This only works if we know an upper bound on $|B-C|$, and pick $M$ to be larger than that bound. If $B,C$ are unbounded, then this doesn't work -- there's no finite $M$ that will necessarily work. – D.W. Jan 30 '17 at 17:30
Absolutely, there are upper and lower bounds on B and C and thus on |B - C|. – Salah Jan 31 '17 at 08:51
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Please edit the question to state that fact. Thank you! – D.W. Jan 31 '17 at 14:35
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@D.W. I think the second inequality does not hold if B=C. Am I missing something (because you said it works?) You would end up with B>=B+1 , since A would be 0. – AlexGuevara Sep 23 '19 at 08:22
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@AlexGuevara, good point. I don't think you're missing anything -- it looks to me like it doesn't work when $B=C$. – D.W. Sep 23 '19 at 15:20

score 0 · Answer 3 · answered Aug 28 '21 at 16:23

This is another possibility for the "greater or equal" and "lower or equal" version. For the pure greater and pure lower version you must remove the +1 resp. -1 from first constraint. You must also add a +1 to the second constraint "... >= yM+1".

score 0 · Answer 4 · answered Jan 30 '17 at 12:08

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Is it possible to enforce this through your objective function? For instance, if your objective function was $$\max A,$$ the constraint $$A\leq (B-C) M,$$ where $M$ is a "big-$M$" (large value) will work provided B is larger than C by some tolerance. The tolerance will be $\frac{1}{M}$, i.e. $B-C \geq \frac{1}{M}$ before the constraint is properly enforced.

answered Jan 30 '17 at 12:08

Dylan Black

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Maybe I need to work around my formulation to incorporate your solution but as for now I don't see how this could be done given that variable A doesn't figure directly in the objective and also I have many constraints of this form. – Salah Jan 30 '17 at 12:37
It would be interesting to see more of the formulation. – Dylan Black Jan 30 '17 at 13:10
I have the variables Xi, Si, Ei and constants Ci where i=1, ...,n and a binary constant matrix A(i,j) where i,j=1, ...,n. Now Xi are computed by a set of constraints but the important for my question are following: Si>=Ej if A(j,i)==1 AND Xi>Xj (this is the one I am stuck with) Ei=Si+Ci y=max(Ei) The objective is min (y) – Salah Jan 30 '17 at 13:37
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This only works if we know an upper bound on $|B-C|$, and pick $M$ to be larger than that bound. If $B,C$ are unbounded, then this doesn't work -- there's no finite $M$ that will necessarily work. – D.W. Jan 30 '17 at 17:31

score 0 · Answer 5 · answered Aug 31 '17 at 12:36

If you are in a MILP setting, then I think the following should work. Assume $\underline{B},\underline{C}$ and $\overline{B},\overline{C}$ are your respective lower/upper bounds. Then the constraint $$A \geq \frac{B-C}{\overline{B}-\underline{C}}$$ ensures that $A=1$ if $B-C$ is positive, since the RHS of the equation is between 0 and 1 in this case. When $B-C$ is negative, then the constraint is useless. The constraint $$\frac{\underline{C}-\overline{B}}{C-B}\geq A$$ ensures that $A=0$ if $B-C$ is negative, for similar reasons.

"Greater than" condition in integer linear program with a binary variable

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