Hello Guys,
I do have to solve a cost minimizing problem for a factory with several machines which can be turned on and off.
For turning on a machine you have to charge starting cost. Turning it off is free.
These are the decision variables we use:
dvar float+ MachineUse[p][t]; //These are the machines we use in period p of type t
dvar float NewMachines[p][t]; //This variable includes the differences between MachineUse of a certain type in period p and in period p1
dexpr float StartupCost = sum (p in Period, t in Type) StartCost[t] * NewMachines[p][t]; //This is the expression which directly goes into our objective function
The following are the constraints we successfully used for creating the values for NewMachines over all periods and types:
forall(p in Period, t in Type)
ctMaxMachineUse[p][t]: MachineUse[p][t] <= MaxMachine[t];
forall (p in Period, t in Type)
if (p==1) MachineUse[p][t] == NewMachines[p][t];
forall (p in 2..lastPeriod, t in Type) MachineUse[p][t]MachineUsep1 [t]==NewMachines[p][t];
Now the problem is that NewMachines[p][t] can also be negative when having turned of certain machines. The consequence is that these values (<0) also go into the objective negatively. Therefore we need to create a binary variable which is 0 if NewMachines<=0 and which is 1 if NewMachines>0...
We tried to reach this through an if constraint but it's not possible to implement a decision variable into the if constraint like if(MachineUse[p][t]<=0) binvar==0. We also tried to create an auxiliary variable which takes all the values of NewMachines[p][t] through a forall statement and then to insert this auxiliary variable into our if constraint in order to create the binary variable. This firstly appeared to be a right solution but in fact it manipulates all computed values and sets all machine used = 0.
I hope somebody has an idea how to solve this problem. Thanks for your help in advance.
Topic

Re: Binary variable for machine use
20121112T08:42:41ZThis is the accepted answer. This is the accepted answer.This is a duplicate of this thread.