Hi Manassaldi,
thank you a lot for your answer. Apparently I didn't realize, that you already provided the solution I needed.
I'm sorry for that. Thank you a lot for you help!
Cheers
logical equations & binary variables

 User
 Posts: 86
 Joined: 3 years ago
 Location: Rosario  Argentina
Re: logical equations & binary variables
Hi, To avoid taking the first element, in my opinion, it is preferable to use this code:
e_T_pi(i)$(ord(i) gt 1) .. T_pi(i)*(m_rc_pb(i)+m_ts_pb(i)) =e= (m_rc_pb(i)*T_ro + m_ts_pb(i)*T_to(i));
Bye!
e_T_pi(i)$(ord(i) gt 1) .. T_pi(i)*(m_rc_pb(i)+m_ts_pb(i)) =e= (m_rc_pb(i)*T_ro + m_ts_pb(i)*T_to(i));
Bye!
Re: logical equations & binary variables
Hi,
I have one question for you Manassaldi.
In my objective function I have this member that is z = [...]TWC(j)*HS(j). Where HS(j)is a variable.
Now, TWC(j) should be a parameter that works like this:
TWC(j) = 0 when HS(j) < 1000
and
TWC(j) = 3.21 when HS(j) >=1000.
Any idea how to implement this in GAMS?
I tried to defined an equation called TWCup(j) that was:
TWCup(j)$(HS.l(j) >= 1000).. TWC(j) =e= 3.21;
but is not working
Thanks
I have one question for you Manassaldi.
In my objective function I have this member that is z = [...]TWC(j)*HS(j). Where HS(j)is a variable.
Now, TWC(j) should be a parameter that works like this:
TWC(j) = 0 when HS(j) < 1000
and
TWC(j) = 3.21 when HS(j) >=1000.
Any idea how to implement this in GAMS?
I tried to defined an equation called TWCup(j) that was:
TWCup(j)$(HS.l(j) >= 1000).. TWC(j) =e= 3.21;
but is not working
Thanks

 User
 Posts: 86
 Joined: 3 years ago
 Location: Rosario  Argentina
Re: logical equations & binary variables
Hi, I think you can try a simple bigM reformulation.
I assume that "HS(j)" is continuous and positive.
The new variable TWC_HS(j) correspond to TWC(j)*HS(j)
x(j) is a binary variable:
x(j)=1 if HS(j) >=1000.
x(j)=0 if HS(j) < 1000
eq1(j).. HS(j) =g= 1000*x(j);
eq2(j).. TWC_HS(j) =g= 3.21*HS(j)  3.21*HS.up(j)*(1x(j));
eq3(j).. TWC_HS(j) =l= 3.21*HS(j) + 3.21*HS.up(j)*(1x(j));
eq4(j).. HS(j) =l= (10001e10) + HS.up*x(j);
eq5(j).. TWC_HS(j) =l= 3.21*HS.up(j)*x(j);
So,
if x(j)=1
HS(j) =g= 1000;
TWC_HS(j) =g= 3.21*HS(j);
TWC_HS(j) =l= 3.21*HS(j); (so, TWC_HS(j) = 3.21*HS(j))
HS(j) =l= (10001e10) + HS.up; (always satisfied)
TWC_HS(j) =l= 3.21*HS.up(j); (always satisfied)
if x(j)=0
HS(j) =g= 0; (always satisfied)
TWC_HS(j) =g= 3.21*HS(j)  3.21*HS.up(j);(always satisfied)
TWC_HS(j) =l= 3.21*HS(j) + 3.21*HS.up(j);(always satisfied)
HS(j) =l= (10001e10); (1e10 is to avoid the value of 1000)
TWC_HS(j) =l= 0; (so, TWC_HS(j) = 0*HS(j))
There may be mistakes, but I think it should work.
Bye!
I assume that "HS(j)" is continuous and positive.
The new variable TWC_HS(j) correspond to TWC(j)*HS(j)
x(j) is a binary variable:
x(j)=1 if HS(j) >=1000.
x(j)=0 if HS(j) < 1000
eq1(j).. HS(j) =g= 1000*x(j);
eq2(j).. TWC_HS(j) =g= 3.21*HS(j)  3.21*HS.up(j)*(1x(j));
eq3(j).. TWC_HS(j) =l= 3.21*HS(j) + 3.21*HS.up(j)*(1x(j));
eq4(j).. HS(j) =l= (10001e10) + HS.up*x(j);
eq5(j).. TWC_HS(j) =l= 3.21*HS.up(j)*x(j);
So,
if x(j)=1
HS(j) =g= 1000;
TWC_HS(j) =g= 3.21*HS(j);
TWC_HS(j) =l= 3.21*HS(j); (so, TWC_HS(j) = 3.21*HS(j))
HS(j) =l= (10001e10) + HS.up; (always satisfied)
TWC_HS(j) =l= 3.21*HS.up(j); (always satisfied)
if x(j)=0
HS(j) =g= 0; (always satisfied)
TWC_HS(j) =g= 3.21*HS(j)  3.21*HS.up(j);(always satisfied)
TWC_HS(j) =l= 3.21*HS(j) + 3.21*HS.up(j);(always satisfied)
HS(j) =l= (10001e10); (1e10 is to avoid the value of 1000)
TWC_HS(j) =l= 0; (so, TWC_HS(j) = 0*HS(j))
There may be mistakes, but I think it should work.
Bye!
Re: logical equations & binary variables
Hi all,
I have a question regarding logical conditions applied to choice variables. I will be happy if someone, specially Manassaldi, can help me on that.
I have two choice variables, a(n) and th, and a parameter p(n). I need to code the following logic in gams.
if p(n)<= th then a(n) =0;
else if th< p(n) <= th * 10^.5 then a(n) = .2;
else if th * 10^.5 < p(n) <= th * 10^1 then a(n) = .4;
else if th * 10^1 < p(n) <= th * 10^1.5 then a(n) = .6;
else if th * 10^1.5 < p(n) <= th * 10^2 then a(n) = .8;
else if th * 10^2 < p(n) then a(n) = 1
I know how to use binary variable when I have only an if then case, but here I need to code many else if cases.
I hope my description of the problem is clear enough and am waiting to hear from you.
Best,
Mary
I have a question regarding logical conditions applied to choice variables. I will be happy if someone, specially Manassaldi, can help me on that.
I have two choice variables, a(n) and th, and a parameter p(n). I need to code the following logic in gams.
if p(n)<= th then a(n) =0;
else if th< p(n) <= th * 10^.5 then a(n) = .2;
else if th * 10^.5 < p(n) <= th * 10^1 then a(n) = .4;
else if th * 10^1 < p(n) <= th * 10^1.5 then a(n) = .6;
else if th * 10^1.5 < p(n) <= th * 10^2 then a(n) = .8;
else if th * 10^2 < p(n) then a(n) = 1
I know how to use binary variable when I have only an if then case, but here I need to code many else if cases.
I hope my description of the problem is clear enough and am waiting to hear from you.
Best,
Mary
Last edited by mbh 10 months ago, edited 1 time in total.

 User
 Posts: 86
 Joined: 3 years ago
 Location: Rosario  Argentina
Re: logical equations & binary variables
Hi, if p (n) is a parameter, I think it is better to reformulate the intervals definition:
if th >= p(n) then a(n) =0;
else if th < p(n) and th >= p(n) * 10^.5 then a(n) = .2;
else if th < p(n)*10^.5 and th >= p(n) * 10^1 then a(n) = .4;
else if th < p(n)*10^1 and th >= p(n) * 10^1.5 then a(n) = .6;
else if th < p(n)*10^1.5 and th >= p(n) * 10^2 then a(n) = .8;
else if th < p(n)*10^2 then a(n) = 1
Now, using binary variables, you must identify which is the interval where "th" is located:
[0 ; p(n)*10^2) or
[p(n) * 10^2 ; p(n)*10^1.5) or
[p(n) * 10^1.5 ; p(n)*10^1) or
[p(n) * 10^1 ; p(n)*10^.5) or
[p(n)*10^0.5 ; p(n) ) or
[p(n) ; Max_value]
Is this analysis ok?
Equations for selecting the intervals are easy to implement. Let me know if you need help.
Bye!
if th >= p(n) then a(n) =0;
else if th < p(n) and th >= p(n) * 10^.5 then a(n) = .2;
else if th < p(n)*10^.5 and th >= p(n) * 10^1 then a(n) = .4;
else if th < p(n)*10^1 and th >= p(n) * 10^1.5 then a(n) = .6;
else if th < p(n)*10^1.5 and th >= p(n) * 10^2 then a(n) = .8;
else if th < p(n)*10^2 then a(n) = 1
Now, using binary variables, you must identify which is the interval where "th" is located:
[0 ; p(n)*10^2) or
[p(n) * 10^2 ; p(n)*10^1.5) or
[p(n) * 10^1.5 ; p(n)*10^1) or
[p(n) * 10^1 ; p(n)*10^.5) or
[p(n)*10^0.5 ; p(n) ) or
[p(n) ; Max_value]
Is this analysis ok?
Equations for selecting the intervals are easy to implement. Let me know if you need help.
Bye!
Re: logical equations & binary variables
Hi Manassaldi.
Thank you so much for helping me out. Your analysis was really helpful.
Best,
Mary
Thank you so much for helping me out. Your analysis was really helpful.
Best,
Mary
Re: logical equations & binary variables
Hi again Manassaldi,
I have a new question for you and will really appreciate it if you can help me as before. I am still struggling with these kind of logical conditions on endogenous variable in my model. The problem is that p(n) includes a binary variable itself. I mean
p(n) = sum(g, t(g) * c(g,n)) where t(g) is a binary variable. With your help, I ended up with a solution to make the intervals using binary variables (uu(n), vv(n), ww(n) , xx(n), yy(n), zz(n), yy(n)). But after that I have a multiplication of binary variables as below:
eq1(n).. th =g= p(n) * uu(n) + p(n) * (10 ** (1/2)) * vv(n) + p(n) * (10 ** (2/2)) * ww(n) + p(n) * (10 ** (3/2)) * xx(n) + p(n) * (10 ** (4/2)) * yy(n);
eq2(n).. delay(n) =e= 0.2 * vv(n) + 0.4 * ww(n) + 0.6 * xx(n) + 0.8 * yy(n) + 1 * zz(n);
eq3(n).. th =l= max * uu(n) + (p(n)  exp(10)) * vv(n) + (p(n) * (10 ** (1/2))  exp(10)) * ww(n) + (p(n) * (10 ** (1))  exp(10)) * xx(n)
+ (p(n) * (10 ** (3/2))  exp(10)) * yy(n) + (p(n) * (10 ** (4/2))  exp(10)) * zz(n);
eq4(n).. uu(n) + vv(n) + ww(n) + xx(n) + yy(n) + zz(n) =e= 1;
const13(n).. p(n) =e= sum(g, t(g) * c(g,n) );
Where c(g,n) is a parameter.
I am getting some wired results and my guess is that multiplication of binary variables makes the model much more complicated. Is there any way to avoid the multiplication of binaries, because I already have some other highly nonlinear constraints in the model.
And actually since you seem to have a very good knowledge of GAMS, I have another question for you .
I have a constraint that includes multiplication of two continuous variables (const12 below):
const11(n).. phi(n) =g= 1  (1 + (d(n) / 229)) ** (.2) ;
const12(n).. i(n) =g= p(n) * y(n) ;
where d(n), phi(n), y(n), and i(n) are all choice variables. Is there anyway to linearize const12?
Many thanks in advance,
Mary
I have a new question for you and will really appreciate it if you can help me as before. I am still struggling with these kind of logical conditions on endogenous variable in my model. The problem is that p(n) includes a binary variable itself. I mean
p(n) = sum(g, t(g) * c(g,n)) where t(g) is a binary variable. With your help, I ended up with a solution to make the intervals using binary variables (uu(n), vv(n), ww(n) , xx(n), yy(n), zz(n), yy(n)). But after that I have a multiplication of binary variables as below:
eq1(n).. th =g= p(n) * uu(n) + p(n) * (10 ** (1/2)) * vv(n) + p(n) * (10 ** (2/2)) * ww(n) + p(n) * (10 ** (3/2)) * xx(n) + p(n) * (10 ** (4/2)) * yy(n);
eq2(n).. delay(n) =e= 0.2 * vv(n) + 0.4 * ww(n) + 0.6 * xx(n) + 0.8 * yy(n) + 1 * zz(n);
eq3(n).. th =l= max * uu(n) + (p(n)  exp(10)) * vv(n) + (p(n) * (10 ** (1/2))  exp(10)) * ww(n) + (p(n) * (10 ** (1))  exp(10)) * xx(n)
+ (p(n) * (10 ** (3/2))  exp(10)) * yy(n) + (p(n) * (10 ** (4/2))  exp(10)) * zz(n);
eq4(n).. uu(n) + vv(n) + ww(n) + xx(n) + yy(n) + zz(n) =e= 1;
const13(n).. p(n) =e= sum(g, t(g) * c(g,n) );
Where c(g,n) is a parameter.
I am getting some wired results and my guess is that multiplication of binary variables makes the model much more complicated. Is there any way to avoid the multiplication of binaries, because I already have some other highly nonlinear constraints in the model.
And actually since you seem to have a very good knowledge of GAMS, I have another question for you .
I have a constraint that includes multiplication of two continuous variables (const12 below):
const11(n).. phi(n) =g= 1  (1 + (d(n) / 229)) ** (.2) ;
const12(n).. i(n) =g= p(n) * y(n) ;
where d(n), phi(n), y(n), and i(n) are all choice variables. Is there anyway to linearize const12?
Many thanks in advance,
Mary

 User
 Posts: 86
 Joined: 3 years ago
 Location: Rosario  Argentina
Re: logical equations & binary variables
Hi, in the previous answer I wrote this code and I saved it as a draft.
I hope this can help you.
Bye!
I hope this can help you.
Bye!
Code: Select all
set
i intervals /i1*i6/
n your nset /n1*n2/
l limits /l,u/
;
parameter
p(n) /n1 1, n2 3/
intervals(n,i,l)
anvalue(i)
/i1 0
i2 0.2
i3 0.4
i4 0.6
i5 0.8
i6 1/
evalue /1e6/
;
intervals(n,'i1','l')=0;
intervals(n,'i2','l')=p(n)*(10**(2));
intervals(n,'i3','l')=p(n)*(10**(1.5));
intervals(n,'i4','l')=p(n)*(10**(1));
intervals(n,'i5','l')=p(n)*(10**(0.5));
intervals(n,'i6','l')=p(n);
intervals(n,'i1','u')=p(n)*(10**(2))  evalue;
intervals(n,'i2','u')=p(n)*(10**(1.5)) evalue;
intervals(n,'i3','u')=p(n)*(10**(1))  evalue;
intervals(n,'i4','u')=p(n)*(10**(.5))  evalue;
intervals(n,'i5','u')=p(n)  evalue;
intervals(n,'i6','u')=p(n)*2;
display intervals;
variable
th
thd(n,i)
a(n)
;
binary variable
bin(n,i)
;
equation
eq1,eq2,eq3,eq4,eq5
;
eq1(n).. th =e= sum(i,thd(n,i));
eq2(n,i).. thd(n,i) =g= intervals(n,i,'l')*bin(n,i);
eq3(n,i).. thd(n,i) =l= intervals(n,i,'u')*bin(n,i);
eq4(n).. sum(i,bin(n,i)) =e= 1;
eq5(n).. a(n) =e= sum(i,bin(n,i)*anvalue(i));
*this is only for testing:
th.lo=0.5;
model test /all/
solve test using mip minimizing th