logical equations & binary variables

GAMS
1

Hey guys,

I got another question on logical equations or rather binary variables. This is again a beginner question.

So right now I’m trying to understand how to set the value of binary variables with conditions.

Consider the binary variable x and variable m.
If i would write the code in other language synthex, what I’m trying to do would look like:

if m > 0
x = True
else
x = False
end

So I’m trying to set the value of x to 1, when the variable m is greater than zero.

How do I model this in GAMS? Do I need to use logical equations for it?
I can’t find a solution for this.

Thanks again for your help.

Cheers
Max

2

Just include the following constraint:

m-Mx <=0

where M is an upper bound for m.

3

Hi, try these equations

eq1… m =g= m.lox;
eq2… m =l= m.upx;

m.lo (lower bound)
m.up (upper bound)

if x=1
m greater than m.lo1;
m lower than m.up1
if x=0
m greater than m.lo0;
m lower than m.up0;
so, m = 0

Bye!

4

Thanks for your fast reply again, Manassaldi.
You are really helping me with my project.

I got another question:
I actually explained wrong, what I want to have. I need the binary variable x to equal 1, when the variable m equals zero.
So basically the opposite of the problem I stated before. Is there any good way to do it?
Unfortunately with your solution of the problem I can’t think of a similar solution for this topic.

Thanks in advance!

Cheers

5

Hi, It’s similar

eq1… m =g= m.lo*(1-x);
eq2… m =l= m.up*(1-x)

Bye!

6

Thanks a lot for your fast reply.
I could have got to this myself. I apologize for not thinking enough, before asking the question.

7

Hi guys,

I got another problem, that’s based in this here.

So coming back to my model, I still want to set the binary variable nrc to 1, when the mass flows m_rc_ts and m_rc_pb both equal to zero.
That’s why I implemented the following equations:

e_nrc_lo(i+1) … m_rc_ts(i+1)+m_rc_pb(i+1) =g= (1-nrc(i+1));
e_nrc_up(i+1) … m_rc_ts(i+1)+m_rc_pb(i+1) =l= m_rcmax*(1-nrc(i+1));

So if I get this right mathemetically, if the mass fows equal to zero, it should look like this:

0 + 0 >= 1 - nrc
0 + 0 <= 282 * (1-nrc)

which leads to:

nrc >= 1 and nrc <= 1, which leads to the only solution nrc = 1.

I implemented the variable, so these are the only equations were it is in. The inital value is 1.

If I run it now, I get the following from the equation list:

e_nrc_lo(1)..  m_rc_ts(1) + m_rc_pb(1) + nrc(1) =G= 1 ; (LHS = 1)
     
e_nrc_lo(2)..  m_rc_ts(2) + m_rc_pb(2) + nrc(2) =G= 1 ; (LHS = 0
     , INFES = 1 ****)
     
e_nrc_lo(3)..  m_rc_ts(3) + m_rc_pb(3) + nrc(3) =G= 1 ; (LHS = 0
     , INFES = 1 ****)

e_nrc_up(1)..  m_rc_ts(1) + m_rc_pb(1) + 282*nrc(1) =L= 282 ; (LHS = 282)
     
e_nrc_up(2)..  m_rc_ts(2) + m_rc_pb(2) + 282*nrc(2) =L= 282 ; (LHS = 0)
     
e_nrc_up(3)..  m_rc_ts(3) + m_rc_pb(3) + 282*nrc(3) =L= 282 ; (LHS = 0)

*The column list gives the following:

nrc(1)
                (.LO, .L, .UP, .M = 0, 1, 1, 0)
        1       e_nrc_lo(1)
      282       e_nrc_up(1)

nrc(2)
                (.LO, .L, .UP, .M = 0, 0, 1, 0)
        1       e_nrc_lo(2)
      282       e_nrc_up(2)

nrc(3)
                (.LO, .L, .UP, .M = 0, 0, 1, 0)
        1       e_nrc_lo(3)
      282       e_nrc_up(3)

m_rc_ts(1)
                (.LO, .L, .UP, .M = 0, 0, 282, 0)
        1       e_nrc_lo(1)
        1       e_nrc_up(1)
       (0)      e_nts(1)
        1       e_m_rc_same(1)
        1       e_ts_rc_same(1)
       (0)      e_charge(1)
       -5.8864  e_Re(1)
       -0.0001  e_U(1)
     (-12.5642) e_h(1)

m_rc_ts(2)
                (.LO, .L, .UP, .M = 0, 0, 282, 0)
        1       e_nrc_lo(2)
        1       e_nrc_up(2)
       (0)      e_nts(2)
        1       e_m_rc_same(2)
        1       e_ts_rc_same(2)
       (0)      e_charge(2)
       -5.8864  e_Re(2)
       -0.0001  e_U(2)

m_rc_ts(3)
                (.LO, .L, .UP, .M = 0, 0, 282, 0)
        1       e_nrc_lo(3)
        1       e_nrc_up(3)
       (0)      e_nts(3)
        1       e_m_rc_same(3)
        1       e_ts_rc_same(3)
       (0)      e_charge(3)
       -5.8864  e_Re(3)
       -0.0001  e_U(3)

m_rc_pb(1)
                (.LO, .L, .UP, .M = 0, 0, 112, 0)
        1       e_nrc_lo(1)
        1       e_nrc_up(1)
        1       e_m_pbmax(1)
    (-838)      e_T_pi(1)
        1       e_m_pb_same(1)
       (0)      e_T_po(1)
       (0)      e_P_out(1)
        1       e_m_rc_same(1)

m_rc_pb(2)
                (.LO, .L, .UP, .M = 0, 0, 112, 0)
        1       e_nrc_lo(2)
        1       e_nrc_up(2)
        1       e_m_pbmax(2)
    (-839)      e_T_pi(2)
        1       e_m_pb_same(2)
      (-0.0597) e_T_po(2)
   (30340)      e_P_out(2)
        1       e_m_rc_same(2)

m_rc_pb(3)
                (.LO, .L, .UP, .M = 0, 0, 112, 0)
        1       e_nrc_lo(3)
        1       e_nrc_up(3)
        1       e_m_pbmax(3)
    (-839)      e_T_pi(3)
        1       e_m_pb_same(3)
      (-0.0597) e_T_po(3)
   (30340)      e_P_out(3)
        1       e_m_rc_same(3)

I don’t know why, the equation e_nrc_lo does not set the binary variable nrc to 1, but just says, that the equation is infeasable. I mean the whole purpose of the equation is to set the variable to 1. Is there something obvious, which I can’t see? Or is it maybe more complicated?

Thanks in advance, guys!

Cheers,
Max

8

Hi, the equation listing and column Listing shows the value on the initial point, not in the solution. The equations are fine.
You should review the SolEQU and SolVAR section.
Bye!

9

Hi Manassaldi,

thanks again for your answer. I see, that I missunderstood the columen and the equation listing.
However I wanted to set the binary variable nrc to prevent the division by zero.

e_T_ri(i+1) … T_ri(i+1) =e= (m_ts_rc(i+1)*T_to(i+1) + m_pb_rc(i+1)*T_po(i+1)) / (m_ts_rc(i+1) + m_pb_rc(i+1) + nrc(i+1)) + nrc(i+1)*T_ri(i);

The binary variable should set the value to one when the mass flows m_ts_rc and m_pb_rc are zero, so that a division by zero is prevented. Because of this, I can’t review the solequ and solvar yet, because the model gets an execution error. Is there anyway to prevent this correctly? Or do I just have to add small numbers into the denominator to prevent the division by zero?

Thanks in advance!

Cheers

10

Hi, In another post I suggested this equation

e_T_pi(i+1) … T_pi(i+1)*(m_rc_pb(i+1)+m_ts_pb(i+1)) =e= (m_rc_pb(i+1)*T_ro + m_ts_pb(i+1)*T_to(i+1));

11

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

12

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!

13

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 :wink:

14

Hi, I think you can try a simple big-M 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= 1000x(j);
eq2(j)… TWC_HS(j) =g= 3.21HS(j) - 3.21HS.up(j)(1-x(j));
eq3(j)… TWC_HS(j) =l= 3.21HS(j) + 3.21HS.up(j)(1-x(j));
eq4(j)… HS(j) =l= (1000-1e-10) + HS.upx(j);
eq5(j)… TWC_HS(j) =l= 3.21HS.up(j)x(j);


So,
if x(j)=1
HS(j) =g= 1000;
TWC_HS(j) =g= 3.21HS(j);
TWC_HS(j) =l= 3.21HS(j); (so, TWC_HS(j) = 3.21HS(j))
HS(j) =l= (1000-1e-10) + HS.up; (always satisfied)
TWC_HS(j) =l= 3.21HS.up(j); (always satisfied)

if x(j)=0
HS(j) =g= 0; (always satisfied)
TWC_HS(j) =g= 3.21HS(j) - 3.21HS.up(j);(always satisfied)
TWC_HS(j) =l= 3.21HS(j) + 3.21HS.up(j);(always satisfied)
HS(j) =l= (1000-1e-10); (1e-10 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!

15

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!

16

Hi, in the previous answer I wrote this code and I saved it as a draft.
I hope this can help you.
Bye!

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 /1e-6/
;
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