哦。
å‘自我的 iPad
在 2012-11-2,15:12,Renger van Nieuwkoop 写é“:
Hi Maria
I do not quite understand what Z(n,a) could be. Usually you have the flow on an arc, defined as f(a), but Z(n,a) refers to a node and some arc (but not necessarily the arc connecting to n).
I think you really want to have Z(a), but that is up to you.
Anyway, this would work in your setting (you cannot define Z as Z(a,n), because a is an assigned set, so you have to define it over all I,j. This is not a problem as the equation can be declared over a (e.g. f(n,a(I,j)).
set n nodes /n1*n10/
a(n,n) arcs;
table arcs(n,n)
n1 n2 n3 n4 n5
n1 1 1 1
n2 1 1 1
n3 1 1 1 1
n4 1 1 1
n5 1 1 1
;
alias(n,i,j);
a(i,j)= YES$arcs(i,j);
alias(n,i,j);
variables
Z(n,i,j),
y(n);
equation f(n,i,j);
f(n,a(i,j))…
Z(n,a)+ y(j) =G= 0 ;
Cheers
Renger
Von: gamsworld@googlegroups.com [mailto:gamsworld@googlegroups.com] Im Auftrag von Maria a. Rad
Gesendet: Donnerstag, 1. November 2012 19:49
An: gamsworld@googlegroups.com
Betreff: Re: exponents of a dynamic set
Hi Renger
thanks for ur time. Since I do not piont to the exponents of the dynamic set in my constraint, this way does not work. This is the exact constraint:
set n nodes /1*10/
a(n,n) arcs;
- here a(n,n) is evaluated as:
n1 n2 n3 n4 n5
n1 YES YES YES
n2 YES YES YES
n3 YES YES YES YES
n4 YES YES YES
n5 YES YES YES
f(n,a) . . Z(n,a)+ y(?) =e=0;
In which ‘?’ means the second exponent of a something like ord(a(*,2)) (which is not correct!)
Cheers
Maria
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