Modeling the absolute value

[Manassaldi]

Hi, the dollar command is used to manipulate the sets.
For example, if the restrictions eq1, eq2, eq3 and eq4 does not apply to the last “t”:

eq1(n,m,t)(ord(t) ne card(t)).. 1 - z(n,m,t) + z(n,m,t+1) + absvalue(n,m,t) =g= 1; eq2(n,m,t)(ord(t) ne card(t))… 1 - z(n,m,t) + 1-z(n,m,t+1) + 1-absvalue(n,m,t) =g= 1;
eq3(n,m,t)(ord(t) ne card(t)).. z(n,m,t) + 1-z(n,m,t+1) + absvalue(n,m,t) =g= 1; eq4(n,m,t)(ord(t) ne card(t))… z(n,m,t) + z(n,m,t+1) + 1-absvalue(n,m,t) =g= 1;
eq5… sum((n,m,t),absvalue(n,m,t)) =l= 2*j;

“$(ord(t) ne card(t))” is mean that the position of the set “t” (card(t)) is not equal (ne) to the last position (card(t)).

I hope this can help you
Bye!

[gnaeidj]

@Manassaldi

Hi, the dollar command is used to manipulate the sets.
For example, if the restrictions eq1, eq2, eq3 and eq4 does not apply to the last “t”:

eq1(n,m,t)(ord(t) ne card(t)).. 1 - z(n,m,t) + z(n,m,t+1) + absvalue(n,m,t) =g= 1; eq2(n,m,t)(ord(t) ne card(t))… 1 - z(n,m,t) + 1-z(n,m,t+1) + 1-absvalue(n,m,t) =g= 1;
eq3(n,m,t)(ord(t) ne card(t)).. z(n,m,t) + 1-z(n,m,t+1) + absvalue(n,m,t) =g= 1; eq4(n,m,t)(ord(t) ne card(t))… z(n,m,t) + z(n,m,t+1) + 1-absvalue(n,m,t) =g= 1;
eq5… sum((n,m,t),absvalue(n,m,t)) =l= 2*j;

“$(ord(t) ne card(t))” is mean that the position of the set “t” (card(t)) is not equal (ne) to the last position (card(t)).

I hope this can help you
Bye!

Hi
Thank you so much
it was such a great help

[Monsef]

Does someone know how to linearize and optimize this absolute term |A - B|, where both A and B are free integers. And what if the term changed to |A-1|, what will the minimization look like?.

Here is my Linearization for |A-1|
X>=A-1
X>=1-A
X>=0.

Min X

Is this correct?

[mbussieck]

Look at GAMS Model Library model absmip (https://www.gams.com/latest/gamslib_ml/libhtml/gamslib_absmip.html) This shows how to linearly (with discrete variables) model abs, min, max, and sign.

-Michael