Maximizing the difference in the order of binary variables

Problems with modeling
Manassaldi
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Re: Maximizing the difference in the order of binary variables

Postby Manassaldi » 3 months ago

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

PeterBe
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Re: Maximizing the difference in the order of binary variables

Postby PeterBe » 1 month ago

Hi I'm back again :-)

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)));
How can I model the absolute value of the inner sum? Meaning:

Code: Select all

 sum(tp, ABS{sum (t,xpos(tp,t)*ord(tp)) - sum(t,ypos(tp,t)*ord(tp))});
I had a look at my other topic: Modeling the absolute value (viewtopic.php?f=9&t=9923&start=10)
There one answer was:

Re: Modeling the absolute value

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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" (x-c) is greater than 0 so by "eq1" and "eq2" absvalue = x-c (the rest of equation are also satisfied)
if "p=0", by "eq6" (x-c) is lower than 0 so by "eq4" and "eq5" absvalue = c-x (the rest of equation are also satisfied)

eq1(t).. absvalue(t) =l= x(t) - c(t) + (1-p(t))*M;
eq2(t).. absvalue(t) =g= x(t) - c(t) - (1-p(t))*M;
eq3(t).. x(t) - c(t) =g= -(1-p(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!
How can I transfer this to this problem?

Thanky in advance for your suggestion. I really appreciate it.

Manassaldi
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Re: Maximizing the difference in the order of binary variables

Postby Manassaldi » 1 month ago

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) + (1-p(tp))*M;
eq2(tp).. absvalue(tp) =g= x(tp) - c(tp) - (1-p(tp))*M;
eq3(tp).. x(tp) - c(tp) =g= -(1-p(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))

PeterBe
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Re: Maximizing the difference in the order of binary variables

Postby PeterBe » 1 month ago

Thanks for your answer Manassaldi

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) + (1-p(tp))*M;
eq2(tp).. absvalue(tp) =g= x(tp) - c(tp) - (1-p(tp))*M;
eq3(tp).. x(tp) - c(tp) =g= -(1-p(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))
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:

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$(ord(tpp) le ord(t)),x(tpp)) =l= ord(tp) + (1-xpos(tp,t))*1440;
eq2(t,tp).. sum(tpp$(ord(tpp) le ord(t)),x(tpp)) =g= ord(tp) - (1-xpos(tp,t))*1440;
eq3(tp)..   sum(t,xpos(tp,t)) =l= 1;
*same for y
eq4(t,tp).. sum(tpp$(ord(tpp) le ord(t)),y(tpp)) =l= ord(tp) + (1-ypos(tp,t))*1440;
eq5(t,tp).. sum(tpp$(ord(tpp) le ord(t)),y(tpp)) =g= ord(tp) - (1-ypos(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)));

Manassaldi
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Re: Maximizing the difference in the order of binary variables

Postby Manassaldi » 1 month ago

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.

PeterBe
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Re: Maximizing the difference in the order of binary variables

Postby PeterBe » 1 month ago

Thanks Manassaldi for your answer and your effort :-)
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.
I think I am getting confused. Let's have a look at the BigM fomulation:

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$(ord(tpp) le ord(t)),x(tpp)) =l= ord(tp) + (1-xpos(tp,t))*1440;
eq2(t,tp).. sum(tpp$(ord(tpp) le ord(t)),x(tpp)) =g= ord(tp) - (1-xpos(tp,t))*1440;
eq3(tp)..   sum(t,xpos(tp,t)) =l= 1;
*same for y
eq4(t,tp).. sum(tpp$(ord(tpp) le ord(t)),y(tpp)) =l= ord(tp) + (1-ypos(tp,t))*1440;
eq5(t,tp).. sum(tpp$(ord(tpp) le ord(t)),y(tpp)) =g= ord(tp) - (1-ypos(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)));
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:

Code: Select all

eqobj.. z =e=  sum(tp, ABS{sum(t,xpos(tp,t)*ord(tp)) - sum(t,ypos(tp,t)*ord(tp))});
So the question is: how can I "insert" this absolute function without changing the model itself. So there was this idea by Mansassaldi:

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) + (1-p(tp))*M;
eq2(tp).. absvalue(tp) =g= x(tp) - c(tp) - (1-p(tp))*M;
eq3(tp).. x(tp) - c(tp) =g= -(1-p(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))
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

PeterBe
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Re: Maximizing the difference in the order of binary variables

Postby PeterBe » 1 month ago

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 :-(

Manassaldi
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Location: Rosario - Argentina

Re: Maximizing the difference in the order of binary variables

Postby Manassaldi » 1 month ago

Hi, I do not completely remember your model. I think this can work.


eq1(t,tp).. sum(tpp$(ord(tpp) le ord(t)),x(tpp)) =l= ord(tp) + (1-xpos(tp,t))*1440;
eq2(t,tp).. sum(tpp$(ord(tpp) le ord(t)),x(tpp)) =g= ord(tp) - (1-xpos(tp,t))*1440;
eq3(tp).. sum(t,xpos(tp,t)) =l= 1;
eq4(t,tp).. sum(tpp$(ord(tpp) le ord(t)),y(tpp)) =l= ord(tp) + (1-ypos(tp,t))*1440;
eq5(t,tp).. sum(tpp$(ord(tpp) le ord(t)),y(tpp)) =g= ord(tp) - (1-ypos(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) + (1-p(tp))*M;
eq10(tp).. absvalue(tp) =g= sumx(tp) - sumy(tp) - (1-p(tp))*M;
eq11(tp).. sumx(tp) - sumy(tp) =g= -(1-p(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

PeterBe
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Re: Maximizing the difference in the order of binary variables

Postby PeterBe » 1 month ago

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:

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$(ord(tpp) le ord(t)),x(tpp)) =l= ord(tp) + (1-xpos(tp,t))*1440;
eq2(t,tp).. sum(tpp$(ord(tpp) le ord(t)),x(tpp)) =g= ord(tp) - (1-xpos(tp,t))*1440;
eq3(tp)..   sum(t,xpos(tp,t)) =l= 1;
*same for y
eq4(t,tp).. sum(tpp$(ord(tpp) le ord(t)),y(tpp)) =l= ord(tp) + (1-ypos(tp,t))*1440;
eq5(t,tp).. sum(tpp$(ord(tpp) le ord(t)),y(tpp)) =g= ord(tp) - (1-ypos(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)));
I will try your suggestion.


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