On Wed, 11 Oct 2023, Leonardo Corato wrote:
About the upper bound, I realized the bins could't have been enough, so my
first code was a greedy solution: 1 bin per 1 item.
Two items per bin would also work.
Do you think I should write the full bpp with types, code? Maybe it could
help
Thank you Michael, I already did, I realized immediately it was a typing
error.
I have to double thank you because you made me understand a thing in which
I was stuck.
About the upper bound, I realized the bins could't have been enough, so my
first code was a greedy solution: 1 bin per 1 item.
I
On Fri, 6 Oct 2023, Michael Hennebry wrote:
YY{y in mouldTypes, b in 1..n, i in I: mo[i]==t} y[t, b] >= x[i, b] ;
Oops. A y where a t was needed. The corrected version:
YY{t in mouldTypes, b in 1..n, i in I: mo[i]==t} y[t, b] >= x[i, b] ;
--
Michael henne...@mail.cs.ndsu.nodak.edu
On Fri, 6 Oct 2023, Michael Hennebry wrote:
Your estimate of the number of bins necessary could be an underestimate.
It does not enforce the only two types in a bin requirement.
For larger problems, I think that that would be necessary.
I think that all that is needed is another else if:
else
Your estimate of the number of bins necessary could be an underestimate.
It does not enforce the only two types in a bin requirement.
For larger problems, I think that that would be necessary.
--
Michael henne...@mail.cs.ndsu.nodak.edu
"Occasionally irrational explanations are required" --
On Fri, 6 Oct 2023, Leonardo Corato wrote:
As you can see in bin 2 and 3 there are 3 mold types, so sadly s.t. fred is
not limiting to 2
I'd misunderstood.
I'd thought that you want no more than two of a given type.
Using an auxillary array is probably easiest:
y[t, b] == 1 iff bin b has an
Por favor no me envíen más correos, no pertenezco al grupo!!
El El vie, oct. 6, 2023 a la(s) 5:32 p.m., Leonardo Corato
escribió:
> Thank you very much Michael,
> and sorry for missing the maling list and sending the email directly to
> you.
>
> I applied
> s.t. fred{b in 1..n, t in
Thank you very much Michael,
and sorry for missing the maling list and sending the email directly to you.
I applied
s.t. fred{b in 1..n, t in mouldTypes}: sum{i in I: mo[i]==t} x[i, b] <= 2 ;
doesn't limit the types to max 2.
To check whether s.t. fred worked properly, I added some printf
I'm assuming the your e-mail was intended for the list.
I'm quoting most of it because the list does not have it.
In either case, forget about what I wrote.
GMPL is more powerful than I remembered.
I think the desired constraint is
fred{b in 1..n, t in mouldTypes} sum{i in I: mo[i]==t} x[i, b]
On Sat, 30 Sep 2023, Leonardo Corato wrote:
param m := 6;
param w := 1 50, 2 60, 3 30, 4 40, 5 40, 6 40;
--> param t := 1 A, 2 B , 3 B, 4 C, 5 C, 6 C;
param c := 100;
end;
I have to add a constraint so that the number of types for every bin is
limited to maximum 2.
Each bin can contain
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