HI, I think that the first approximation generate a smaller model that require lower memory. But i don't know how much memory will consume.
I don't know if there are any way to estimate how much memory the model needs
Best regard
Maximizing the difference in the order of binary variables

 User
 Posts: 44
 Joined: 9 months ago
 Location: Rosario  Argentina
Re: Maximizing the difference in the order of binary variables
Hi I'm back again
and I have a question regarding the BigM formulation:
The objective function looks like this :
How can I model the absolute value of the inner sum? Meaning:
I had a look at my other topic: Modeling the absolute value (viewtopic.php?f=9&t=9923&start=10)
There one answer was:
Thanky in advance for your suggestion. I really appreciate it.
and I have a question regarding the BigM formulation:
The objective function looks like this :
Code: Select all
sum(tp, sum (t,xpos(tp,t)*ord(tp))  sum(t,ypos(tp,t)*ord(tp)));
Code: Select all
sum(tp, ABS{sum (t,xpos(tp,t)*ord(tp))  sum(t,ypos(tp,t)*ord(tp))});
There one answer was:
How can I transfer this to this problem?Re: Modeling the absolute value
Report this post
Quote
Accept this answer
Unread postby Manassaldi » 3 weeks ago
Hi again PeterBe, I think this can works
x(t) and c(t) are continuous variables
p(t) are binary variables
M is a sufficiently large number (BigM parameter)
if "p=1", by "eq3" (xc) is greater than 0 so by "eq1" and "eq2" absvalue = xc (the rest of equation are also satisfied)
if "p=0", by "eq6" (xc) is lower than 0 so by "eq4" and "eq5" absvalue = cx (the rest of equation are also satisfied)
eq1(t).. absvalue(t) =l= x(t)  c(t) + (1p(t))*M;
eq2(t).. absvalue(t) =g= x(t)  c(t)  (1p(t))*M;
eq3(t).. x(t)  c(t) =g= (1p(t))*M
eq4(t).. absvalue(t) =l= c(t)  x(t) + p(t)*M;
eq5(t).. absvalue(t) =g= c(t)  x(t)  p(t)*M;
eq6(t).. x(t)  c(t) =l= p(t)*M
eq7.. z =e= sum(t,absvalue(t))
All this restrictions are for "z = sum(t,abs(x(t)c(t)))" with x and c continuous variables
Bye!
Thanky in advance for your suggestion. I really appreciate it.

 User
 Posts: 44
 Joined: 9 months ago
 Location: Rosario  Argentina
Re: Maximizing the difference in the order of binary variables
Hi, I think you can try with this...
eq01(tp).. x(tp) =e= sum(t,xpos(tp,t)*ord(tp))
eq02(tp).. c(tp) =e= sum(t,ypos(tp,t)*ord(tp))});
eq1(tp).. absvalue(tp) =l= x(tp)  c(tp) + (1p(tp))*M;
eq2(tp).. absvalue(tp) =g= x(tp)  c(tp)  (1p(tp))*M;
eq3(tp).. x(tp)  c(tp) =g= (1p(tp))*M
eq4(tp).. absvalue(tp) =l= c(tp)  x(tp) + p(tp)*M;
eq5(tp).. vabsvalue(tp) =g= c(tp)  x(tp)  p(tp)*M;
eq6(tp).. x(tp)  c(tp) =l= p(tp)*M
eq7.. z =e= sum(tp,absvalue(tp))
eq01(tp).. x(tp) =e= sum(t,xpos(tp,t)*ord(tp))
eq02(tp).. c(tp) =e= sum(t,ypos(tp,t)*ord(tp))});
eq1(tp).. absvalue(tp) =l= x(tp)  c(tp) + (1p(tp))*M;
eq2(tp).. absvalue(tp) =g= x(tp)  c(tp)  (1p(tp))*M;
eq3(tp).. x(tp)  c(tp) =g= (1p(tp))*M
eq4(tp).. absvalue(tp) =l= c(tp)  x(tp) + p(tp)*M;
eq5(tp).. vabsvalue(tp) =g= c(tp)  x(tp)  p(tp)*M;
eq6(tp).. x(tp)  c(tp) =l= p(tp)*M
eq7.. z =e= sum(tp,absvalue(tp))
Re: Maximizing the difference in the order of binary variables
Thanks for your answer Manassaldi
Is x(tp) a binary variable. Because as far as I understood x(tp) is a continious variable whereas in the previous models (BigM) x(tp) is a binary variable. I am talking about the model that you posted some weeks ago:
eq01(tp).. x(tp) =e= sum(t,xpos(tp,t)*ord(tp))
eq02(tp).. c(tp) =e= sum(t,ypos(tp,t)*ord(tp))});
eq1(tp).. absvalue(tp) =l= x(tp)  c(tp) + (1p(tp))*M;
eq2(tp).. absvalue(tp) =g= x(tp)  c(tp)  (1p(tp))*M;
eq3(tp).. x(tp)  c(tp) =g= (1p(tp))*M
eq4(tp).. absvalue(tp) =l= c(tp)  x(tp) + p(tp)*M;
eq5(tp).. vabsvalue(tp) =g= c(tp)  x(tp)  p(tp)*M;
eq6(tp).. x(tp)  c(tp) =l= p(tp)*M
eq7.. z =e= sum(tp,absvalue(tp))
Code: Select all
et
t /1*1440/
;
alias(t,tp,tpp);
binary variable
x(t)
y(t)
;
variable
z
;
equation
eq1,eq2,eq3,eq4,eq5,eq6,eqobj
;
eq1(t,tp).. sum(tpp$(ord(tpp) le ord(t)),x(tpp)) =l= ord(tp) + (1xpos(tp,t))*1440;
eq2(t,tp).. sum(tpp$(ord(tpp) le ord(t)),x(tpp)) =g= ord(tp)  (1xpos(tp,t))*1440;
eq3(tp).. sum(t,xpos(tp,t)) =l= 1;
*same for y
eq4(t,tp).. sum(tpp$(ord(tpp) le ord(t)),y(tpp)) =l= ord(tp) + (1ypos(tp,t))*1440;
eq5(t,tp).. sum(tpp$(ord(tpp) le ord(t)),y(tpp)) =g= ord(tp)  (1ypos(tp,t))*1440;
eq6(tp).. sum(t,ypos(tp,t)) =l= 1;
eqobj.. z =e= sum(tp, sum(t,xpos(tp,t)*ord(tp))  sum(t,ypos(tp,t)*ord(tp)));

 User
 Posts: 44
 Joined: 9 months ago
 Location: Rosario  Argentina
Re: Maximizing the difference in the order of binary variables
Hi, x(tp) is the sum of the product between a binary variable and its position, so I suppose that is an integer variable (not binary). Anyway, I think it's better to define it as a continuous variable.
Re: Maximizing the difference in the order of binary variables
Thanks Manassaldi for your answer and your effort
x(t) has to be a binary variable . Basically I do not want to change anything except "insert" the absolute value in the last equation:
So the question is: how can I "insert" this absolute function without changing the model itself. So there was this idea by Mansassaldi:
How can I combine those two formulation (if that is possible)? You simply cannot define x(t) to be a continious variable because then the first fomulation (BIgM) will not make sense
I think I am getting confused. Let's have a look at the BigM fomulation:Hi, x(tp) is the sum of the product between a binary variable and its position, so I suppose that is an integer variable (not binary). Anyway, I think it's better to define it as a continuous variable.
Code: Select all
set
t /1*1440/
;
alias(t,tp,tpp);
binary variable
x(t)
y(t)
;
variable
z
;
equation
eq1,eq2,eq3,eq4,eq5,eq6,eqobj
;
eq1(t,tp).. sum(tpp$(ord(tpp) le ord(t)),x(tpp)) =l= ord(tp) + (1xpos(tp,t))*1440;
eq2(t,tp).. sum(tpp$(ord(tpp) le ord(t)),x(tpp)) =g= ord(tp)  (1xpos(tp,t))*1440;
eq3(tp).. sum(t,xpos(tp,t)) =l= 1;
*same for y
eq4(t,tp).. sum(tpp$(ord(tpp) le ord(t)),y(tpp)) =l= ord(tp) + (1ypos(tp,t))*1440;
eq5(t,tp).. sum(tpp$(ord(tpp) le ord(t)),y(tpp)) =g= ord(tp)  (1ypos(tp,t))*1440;
eq6(tp).. sum(t,ypos(tp,t)) =l= 1;
eqobj.. z =e= sum(tp, sum(t,xpos(tp,t)*ord(tp))  sum(t,ypos(tp,t)*ord(tp)));
Code: Select all
eqobj.. z =e= sum(tp, ABS{sum(t,xpos(tp,t)*ord(tp))  sum(t,ypos(tp,t)*ord(tp))});
Code: Select all
eq01(tp).. x(tp) =e= sum(t,xpos(tp,t)*ord(tp))
eq02(tp).. c(tp) =e= sum(t,ypos(tp,t)*ord(tp))});
eq1(tp).. absvalue(tp) =l= x(tp)  c(tp) + (1p(tp))*M;
eq2(tp).. absvalue(tp) =g= x(tp)  c(tp)  (1p(tp))*M;
eq3(tp).. x(tp)  c(tp) =g= (1p(tp))*M
eq4(tp).. absvalue(tp) =l= c(tp)  x(tp) + p(tp)*M;
eq5(tp).. vabsvalue(tp) =g= c(tp)  x(tp)  p(tp)*M;
eq6(tp).. x(tp)  c(tp) =l= p(tp)*M
eq7.. z =e= sum(tp,absvalue(tp))
Re: Maximizing the difference in the order of binary variables
So you canot thinkg about a way how to combine the inital BigM formulation with the "absoulte value" in the objective function? It ried to combine the two formulations but did not really know how to do that

 User
 Posts: 44
 Joined: 9 months ago
 Location: Rosario  Argentina
Re: Maximizing the difference in the order of binary variables
Hi, I do not completely remember your model. I think this can work.
eq1(t,tp).. sum(tpp$(ord(tpp) le ord(t)),x(tpp)) =l= ord(tp) + (1xpos(tp,t))*1440;
eq2(t,tp).. sum(tpp$(ord(tpp) le ord(t)),x(tpp)) =g= ord(tp)  (1xpos(tp,t))*1440;
eq3(tp).. sum(t,xpos(tp,t)) =l= 1;
eq4(t,tp).. sum(tpp$(ord(tpp) le ord(t)),y(tpp)) =l= ord(tp) + (1ypos(tp,t))*1440;
eq5(t,tp).. sum(tpp$(ord(tpp) le ord(t)),y(tpp)) =g= ord(tp)  (1ypos(tp,t))*1440;
eq6(tp).. sum(t,ypos(tp,t)) =l= 1;
eq7(tp).. sumx(tp) =e= sum(t,xpos(tp,t)*ord(tp))
eq8(tp).. sumy(tp) =e= sum(t,ypos(tp,t)*ord(tp));
eq9(tp).. absvalue(tp) =l= sumx(tp)  sumy(tp) + (1p(tp))*M;
eq10(tp).. absvalue(tp) =g= sumx(tp)  sumy(tp)  (1p(tp))*M;
eq11(tp).. sumx(tp)  sumy(tp) =g= (1p(tp))*M
eq12(tp).. absvalue(tp) =l= sumy(tp)  sumx(tp) + p(tp)*M;
eq13(tp).. absvalue(tp) =g= sumy(tp)  sumx(tp)  p(tp)*M;
eq14(tp).. sumx(tp)  sumy(tp) =l= p(tp)*M;
eq15.. z =e= sum(tp,absvalue(tp));
Bye
eq1(t,tp).. sum(tpp$(ord(tpp) le ord(t)),x(tpp)) =l= ord(tp) + (1xpos(tp,t))*1440;
eq2(t,tp).. sum(tpp$(ord(tpp) le ord(t)),x(tpp)) =g= ord(tp)  (1xpos(tp,t))*1440;
eq3(tp).. sum(t,xpos(tp,t)) =l= 1;
eq4(t,tp).. sum(tpp$(ord(tpp) le ord(t)),y(tpp)) =l= ord(tp) + (1ypos(tp,t))*1440;
eq5(t,tp).. sum(tpp$(ord(tpp) le ord(t)),y(tpp)) =g= ord(tp)  (1ypos(tp,t))*1440;
eq6(tp).. sum(t,ypos(tp,t)) =l= 1;
eq7(tp).. sumx(tp) =e= sum(t,xpos(tp,t)*ord(tp))
eq8(tp).. sumy(tp) =e= sum(t,ypos(tp,t)*ord(tp));
eq9(tp).. absvalue(tp) =l= sumx(tp)  sumy(tp) + (1p(tp))*M;
eq10(tp).. absvalue(tp) =g= sumx(tp)  sumy(tp)  (1p(tp))*M;
eq11(tp).. sumx(tp)  sumy(tp) =g= (1p(tp))*M
eq12(tp).. absvalue(tp) =l= sumy(tp)  sumx(tp) + p(tp)*M;
eq13(tp).. absvalue(tp) =g= sumy(tp)  sumx(tp)  p(tp)*M;
eq14(tp).. sumx(tp)  sumy(tp) =l= p(tp)*M;
eq15.. z =e= sum(tp,absvalue(tp));
Bye
Re: Maximizing the difference in the order of binary variables
Thanks a lot Manassaldi for your effort once again
My model is on the first page of the thread. Here you can see it again:
I will try your suggestion.
My model is on the first page of the thread. Here you can see it again:
Code: Select all
set
t /1*1440/
;
alias(t,tp,tpp);
binary variable
x(t)
y(t)
;
variable
z
;
equation
eq1,eq2,eq3,eq4,eq5,eq6,eqobj
;
eq1(t,tp).. sum(tpp$(ord(tpp) le ord(t)),x(tpp)) =l= ord(tp) + (1xpos(tp,t))*1440;
eq2(t,tp).. sum(tpp$(ord(tpp) le ord(t)),x(tpp)) =g= ord(tp)  (1xpos(tp,t))*1440;
eq3(tp).. sum(t,xpos(tp,t)) =l= 1;
*same for y
eq4(t,tp).. sum(tpp$(ord(tpp) le ord(t)),y(tpp)) =l= ord(tp) + (1ypos(tp,t))*1440;
eq5(t,tp).. sum(tpp$(ord(tpp) le ord(t)),y(tpp)) =g= ord(tp)  (1ypos(tp,t))*1440;
eq6(tp).. sum(t,ypos(tp,t)) =l= 1;
eqobj.. z =e= sum(tp, sum(t,xpos(tp,t)*ord(tp))  sum(t,ypos(tp,t)*ord(tp)));
Who is online
Users browsing this forum: No registered users and 8 guests