Hi Gustavo,
Assuming the set t is defined over / 1, …, T /, then writing the equation
eq1(i,t)… H(i,f) =e= H(i,t-1) + Y(i,t) ;
will result in generating the equations:
H(i,‘1’) =e= Y(i,‘1’) ;
H(i,‘2’) =e= H(i,‘1’) + Y(i,‘2’) ;
H(i,‘3’) =e= H(i,‘2’) + Y(i,‘3’) ;
…
H(i,‘T’) =e= H(i,‘T-1’) + Y(i,‘T’) ;
this may easily be checked in the *.LST file, under the “equation listing” section :
---- eq1 =E=
eq1(i0,1)… H(i0,1) - Y(i0,1) =E= 0 ; (LHS = 0)
eq1(i0,2)… - H(i0,1) + H(i0,2) - Y(i0,2) =E= 0 ; (LHS = 0)
eq1(i0,3)… - H(i0,2) + H(i0,3) - Y(i0,3) =E= 0 ; (LHS = 0)
…
Note that you might also make use of the “cycling” index such as in
eq2(i,t) … H(i,t) =e= H(i,t–1) + Y(i,t) ;
which would map t–1 on the last element of the set when t is the first.
Give it a try and see for yourself what the *.LST file provides you with.
Hope this helps.
cheers
dax
Le mardi 18 février 2014 19:19:19 UTC+1, Gustavo Gallardo a écrit :
I have a multiperiod problem that I need to solve.
I have a set t /t1*t10/ and I have the equation:
H(i,t)=e=H(i,t-1)+Y(i,t);
How can I define that H(i,‘0’)=0, or say that im going to start with cero, in the stage t-1.
Regards,
Gustavo
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