Modeling the absolute value

Problems with modeling
Manassaldi
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Posts: 44
Joined: 10 months ago
Location: Rosario - Argentina

Re: Modeling the absolute value

Postby Manassaldi » 2 months ago

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
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Posts: 8
Joined: 7 months ago

Re: Modeling the absolute value

Postby gnaeidj » 2 months ago

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)).

I hope this can help you
Bye!

Hi
Thank you so much
it was such a great help


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