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Posted by Moby Dick
on 21/9/2009, 9:49 am, in reply to "Piggy back"
There are many possible solutions. The following one minimises the total sum of money:
rtot 215
atot 215
rpenny 25
rnickel 30
rdime 4
rquarter 0
apenny 0
anickel 23
adime 0
aquarter 4
Here's the LP_Solve model:
/* Objective function */
min: rtot;
/* Variable bounds */
rtot = atot;
rtot = rpenny + 5 rnickel + 10 rdime + 25 rquarter;
atot = apenny + 5 anickel + 10 adime + 25 aquarter;
aquarter = rquarter + 4;
rpenny >= apenny + 1;
rnickel >= anickel + 1;
rdime >= adime + 1;
anickel = 23;
rpenny + rnickel + rdime + rquarter - apenny - anickel - adime - aquarter = 32;
int rpenny, apenny, rnickel, anickel, rdime, adime, rquarter, aquarter;
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