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 fÃ©vrier 2014 19:19:19 UTC+1, Gustavo Gallardo a Ã©crit :

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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