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Three decision variables M, O and Z,

forall (k in 1..Cells, i in 1..nbMachines, j in 1..nbComps)
{(U[k][i][j] == 1) && ( A[k][i] <= -1 +((D[k][j]*S[k][j])*52*5))
=> M[k][i][j]==1 ;  } 

forall (k in 1..Cells, i in 1..nbMachines, j in 1..nbComps)
{(V[k][i][j] == 1) && (A[k][i] >=  -1 + ((D[k][j]*S[k][j])*52*5)) 
=> Z[k][i][j]==1 ;} 

forall (k in 1..Cells, i in 1..nbMachines, j in 1..nbComps)
{(V[k][i][j] == 1) && ( A[k][i]  >= -1 + ((D[k][j]*S[k][j])*52*5)) 
=> O[k][i][j]==1;}
  • If U is 1, A is less than or equal to D*S*52*5, than M will be 1.
  • If V is 1, A is more than D*S*52*5, than Z will be 1.
  • If V is 1, A is more than D*S*52*5, than O will be 1.

In the run, M, Z and O are getting the same values. The sum of M and Z as well as sum of M and O will be equal to U and V. (U=V) Please help me to write the correct expressions for finding out M, O and Z. U and V are extracted in previous steps. A, D and S are input data.

Pavel
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Ram
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1 Answers1

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Please provide the decision expression for a less wild guess by my side. I am assuming that all three lines are inside a subject to {} block. As your constraints 2 and 3 are literally the same (besides a missing space and a different letter O/Z) I guess Cplex comes up with the easiest solution by assigning the same values to all Decision variables.

larsl
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  • The Objective function is Minimizing sum (k in 1..Cells, i in 1..nbMachines, j in 1..nbComps) (M[k][i][j] * A [k][i] + Z[k][i][j] * I [k][j] * D[k][j] + O[k][i][j] * S [k][j] * D [k][j] + W[k][i][j] * IA [k][j] * D[k][j]); The solution for decision variable W is obtained and the other variables M, Z and O have to find out. – Ram Dec 20 '17 at 14:33