## Modeling the absolute value

Problems with modeling
Manassaldi
User
Posts: 65
Joined: 1 year ago
Location: Rosario - Argentina

### Re: Modeling the absolute value

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)).

Bye!

gnaeidj
User
Posts: 8
Joined: 1 year ago

### Re: Modeling the absolute value

Manassaldi wrote: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)).