mip error

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Rofice
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Posts: 14
Joined: 6 months ago

mip error

Postby Rofice » 1 month ago

Hi all,
Can anybody help me in finding the problem of following model

binary variable
b1
b2
;
positive variable

p pressure 1
t temperature 1
t2 temperature 2
p2 pressure 2
;
variable
obj
;

equation
a1
a2
a3
a4
a6
a7

;

a1..
t =e= 2*p ;

a2..
p =l= 1000*b1 ;

a3..
t2 =l= 8*p2 ;


a4..
p2=e= 1000*b2;

a6..
b1+b2 =e= 1;

a7..
obj =e=10*(t+t2) ;
p.lo=1;
p.up=2;
p2.lo=2.1;
p2.up = 3;




model m /all/;
solve m minimizing obj using mip;

User avatar
Renger
Posts: 54
Joined: 1 year ago

Re: mip error

Postby Renger » 1 month ago

Hi

What do you mean by "problem"?
The problem is infeasible (no solution found). If I remove the bounds, I get a solution.

Cheers
Renger

Rofice
User
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Posts: 14
Joined: 6 months ago

Re: mip error

Postby Rofice » 1 month ago

Thanks Renger for your response. I figured out that problem but now I have other problem. I want to transform below equation

x = fc('1','H2O')/f('1')
where x = composition of water in stream 1
fc('1','H2O') = component flow of water in stream 1
f('1') = total flow of stream 1

Also all x,fc and f are variables and i want to transform above equation into linear one. So if you can help me in this regard I will be extremely thankful for your help.

I try to find this in literature, the best I could find is below
x * f('1') = fc('1','H2O')
but above trick is valid only if x is not a variable, however in my case it is variable.

I would also like to mention bounds, if some one needed for solving this problem as x is a ratio so value of x lies between 0 to 1, for f('1) upper bound is 6200
Any suggestions for reformulation?


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