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;

## mip error

### Re: mip error

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

What do you mean by "problem"?

The problem is infeasible (no solution found). If I remove the bounds, I get a solution.

Cheers

Renger

### Re: mip error

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?

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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