I went through the answers to this question: Modeling floor function exactly, but I still do not get how to model y = floor(x). Is that question answered and I just do not see it?
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1$\begingroup$ You are essentially re-asking the same question, so I'd suggest that you ask a more specific version of your question, otherwise this one will probably get closed. $\endgroup$ – LarrySnyder610♦ Sep 21 '20 at 23:52
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1$\begingroup$ The short answer is that you cannot model a floor function exactly; it fails at the integer values of $x$. $\endgroup$ – LarrySnyder610♦ Sep 21 '20 at 23:54
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$\begingroup$ Ok, now I get it. The whole discussion was about proving the statement "you cannot model a floor function exactly." $\endgroup$ – Clement Sep 22 '20 at 8:22
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$\begingroup$ Hi, this is why I answered. I hope this helps $\endgroup$ – Alex Fleischer Sep 22 '20 at 12:15
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in https://github.com/AlexFleischerParis/howtowithopl/blob/master/ceil.mod
range r=1..4;
float x[r]=[1.5,4.0,2.0001,5.9999];
dvar int y[r];
dvar float f[r] in 0..0.9999999;
subject to
{
forall(i in r) y[i]==x[i]+f[i];
}
execute
{
writeln(x," ==> ",y);
}
assert forall(i in r) y[i]==ceil(x[i]);
//which gives
// [1.5 4 2.0001 5.9999] ==> [2 4 3 6]
I gave an OPL CPLEX example about how to model ceil. Floor is not very different.
range r=1..4; float x[r]=[1.5,4.0,2.0001,5.9999]; dvar int y[r]; dvar float f[r] in 0..0.9999; subject to { forall(i in r) y[i]==x[i]-f[i]; } execute { writeln(x," ==> ",y); } assert forall(i in r) y[i]==floor(x[i]);
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I tried finding floor value together with ceil value. c = ceil(x) and f=floor(x)
$$x \leq c \leq x+1 $$ $$x \geq f \geq x-1 $$
$$ c-f \leq 1 $$ $$ x \leq f + M(c-x) $$ $$ x \geq c - M(x-f) $$ $$ x\in R^+, c,f \in Z^+ $$
Possible cases: Case 1: x=3.9 $$3.9 \leq c \leq 4.9 $$ $$3.9 \geq f \geq 2.9 $$
$$ c-f \leq 1 $$ $$ 3.9 \leq f + M(c-3.9) $$ $$ 3.9 \geq c - M(3.9-f) $$ Result c=4, f=3
Case 2: x=3.1 $$3.1 \leq c \leq 4.1 $$ $$3.1 \geq f \geq 2.1 $$
$$ c-f \leq 1 $$ $$ 3.1 \leq f + M(c-3.1) $$ $$ 3.1 \geq c - M(3.1-f) $$ Result c=4, f=3
Case 3: x=3 $$3 \leq c \leq 4 $$ $$3 \geq f \geq 2 $$
$$ c-f \leq 1 $$ $$ 3 \leq f + M(c-3) $$ $$ 3 \geq c - M(3-f) $$ 3.a. c=4, f=2 is possible from const 1 and 2 but const 3 inf.
3.b. c=4, f=3 is possible from const 1 and 2 but const 5 inf.
3.c. c=3, f=2 is possible from const 1 and 2 but const 4 inf. Therefore c=3, f=3.
M is big number and determines the sensivity of x. For example, x=3.9 , M must be equal or greater than 10. Likewise, if x=3.99, $M \geq 100$
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$\begingroup$ @RobPratt I gave a new answer, I do not know there is something missed. $\endgroup$ – kur ag Sep 22 '20 at 22:50
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$\begingroup$ Your third constraint is implied by your first two constraints and can be omitted. Also, Case 3 should have 3 instead of 3.1. $\endgroup$ – RobPratt Sep 22 '20 at 23:48
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$\begingroup$ I think you will find that $1/M$ plays the same role as $\epsilon$ in the linked question. $\endgroup$ – RobPratt Sep 23 '20 at 1:32
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$\begingroup$ You still need to replace 3,1 with 3 in Case 3. And $(c,f)=(4,3)$ is feasible. $\endgroup$ – RobPratt Sep 23 '20 at 5:54
