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parag_patil
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Problem with decision variable and cost function

Post by parag_patil » 2 months ago

Dear all,

I have a binary decision variable, y(X,t) as follows:

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sets
t    /t1*t5/
X  /X1*X2/ ;
binary variable
y(X,t);
As it can be seen, the variable y(X,t) can take 1 or 0 value for all X set. In other words:

y(X,t) can take as follows (an example):

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             X1         X2
t1          0           0
t2          1           0
t3          0           1
t4          1           1
t5          1          1
I want to define a variable, Y(t) ( for equation ):

such that if for any t,

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sum(X,y(X,t)) > 1 
Y(t) should be 1, else Y(t) should be 0.

Please note that, I do not want to have conditional-statement, because I cannot use it while writing equations. Also, there should be no dollar condition for variables.

Thanks

abhosekar
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Re: Problem with decision variable and cost function

Post by abhosekar » 2 months ago

Do you want
sum(X,y(X,t)) > 1 (means 2 or more) or sum(X,y(X,t)) >= 1 (means one or more)?

If it is the latter, it is very easy (first of all do not rename both variables y. I will use y(x, t) and y1(t) in the following equations)

eq1(t).. y1(t) =l= sum(X,y(X,t))
eq2(x,t).. y(x,t) =l= y1(t);

eq1 ensures that if sum(x, y(x, t)) is 1 or more, y1(t) has to be 1. M is a big-M constant (not too big I would say it should be same as card(x) in your case).
eq2 ensures that if all y(x,t) are 0, y1(t) is 0.

Hope this helps.

- Atharv

parag_patil
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Re: Problem with decision variable and cost function

Post by parag_patil » 2 months ago

Hi Atharv,

Thanks for your answer. Your answer is absolutely what I needed.

Yes, I wrote it by mistake (repeared Y), thanks that you have corrected it.

Meanwhile, before your answer was being posted , I thought the following :

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y(t) =e= [{[sum(X,Y(X,t)) = 0 ] AND 0} AND 0 ]

Can you please see if it makes sense ?

Thanks alot for your answer !

abhosekar
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Re: Problem with decision variable and cost function

Post by abhosekar » 2 months ago

No it does not make sense. "sum(X,Y(X,t)) = 0" is a condition that might keep changing during solve. This is no different than using $ values on variables. It is not allowed while defining equations.
Please use the constraints that I mentioned.

- Atharv

parag_patil
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Re: Problem with decision variable and cost function

Post by parag_patil » 2 months ago

Yes. You are correct.

I tried with it. And indeed it is like using $ over variables.

Thanks for your advice. I have to use the condition , which you have provided.

Thanks alot again.

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