Dear all:
What is a suggested syntax to accomplish this:
sets e,a(e),b(e),c(e);
a(e)=(ord(efm)…)$YES
b(e)=(ord(efm)=…)$YES;
(This syntax is not accepted, but shows what I want to do:)
c(e)=e$yes+a(e)$no+b(e)$no;
Basically, c(e) needs to have those elements in set e that are not members of subsets a(e) or b(e).
Thanks in advance !
Regards
Claudio
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Claudio,
Is this what you’re trying to do?
-Steve
sets
e / e1 * e10 /,
a(e), b(e)
c(e) ‘e - a - b’
;
scalars lo /2/, up /6/;
a(e) = (ord(e) up);
b(e) = (ord(e) = lo) or (ord(e) = up);
c(e) = not(a(e)) and not(b(e));
parameter chk(e);
chk(e) = 1 - 1$a(e) - 1$b(e) - 1$c(e);
abort$[card(chk)] ‘partition iff chk is zero here’, chk, a, b, c;
- this is only a necessary condition, and not so useful for pinpointing the violation
abort$[card(e) (card(a)+card(b)+card(c))] ‘bad partition’;
On Mon, Feb 1, 2016 at 8:41 AM, Claudio Delpino wrote:
Dear all:
What is a suggested syntax to accomplish this:
sets e,a(e),b(e),c(e);
a(e)=(ord(efm)…)$YES
b(e)=(ord(efm)=…)$YES;
(This syntax is not accepted, but shows what I want to do:)
c(e)=e$yes+a(e)$no+b(e)$no;
Basically, c(e) needs to have those elements in set e that are not members of subsets a(e) or b(e).
Thanks in advance !
Regards
Claudio
Steven Dirkse, Ph.D.
GAMS Development Corp., Washington DC
Voice: (202)342-0180 Fax: (202)342-0181
sdirkse@gams.com
http://www.gams.com
Steve: This was precisely what I was trying to do. I wasn’t aware you could just declare subset membership like that. Thanks!!!
On Mon, Feb 1, 2016 at 10:41 AM, Claudio Delpino wrote:
Dear all:
What is a suggested syntax to accomplish this:
sets e,a(e),b(e),c(e);
a(e)=(ord(efm)…)$YES
b(e)=(ord(efm)=…)$YES;
(This syntax is not accepted, but shows what I want to do:)
c(e)=e$yes+a(e)$no+b(e)$no;
Basically, c(e) needs to have those elements in set e that are not members of subsets a(e) or b(e).
Thanks in advance !
Regards
Claudio
\
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