4
$\begingroup$
line1 line2 line3 line4
A 2.3 0 3.1 0
B 0 4 2.2 0
C 1.1 0 0 4.6

Let's say after optimization with certain constraints, my model will generate an optimal production allocation table similar to above. A, B, C are the three product types and the matrix displays the number of hours needed for each production line in order to finish the demand.

I need to add another constraint which allows only one non-integer hour amount for each of the products. what is the easiest way or logic to do this? Please show by model.addVars() and model.addConstrs() if possible.

Update: This is what I wrote according to the answer by Rob:

y = edm.addVars(lines, products, vtype=GRB.INTEGER)
z = edm.addVars(lines, products, vtype=GRB.BINARY)
for line in lines:
    for product in products:
        constraint1_1 = edm.addConstr(
            (
              #blocks[line, product] - y[line, product] >= - z[line, product]
              y[line, product] - blocks[line, product] <= z[line, product]
            ), 
            name = 'constraint 1.1'
        )
        
        constraint1_2 = edm.addConstr(
            (
              blocks[line, product] - y[line, product] <= z[line, product]
            ), 
            name = 'constraint 1.2'
        )

constraint2 = edm.addConstrs(
    (
      quicksum(z[line, product] for line in lines) <= 1 for product in products
      ),
    name = 'constraint 2'
    )   

But my model optimization keeps running and has not given a solution for 2 hours. Any ideas what could be the issue? Thanks

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1
  • $\begingroup$ It might help to add explicit bounds on $y$. $\endgroup$
    – RobPratt
    Sep 3 at 13:16
11
$\begingroup$

Let $x_{p,\ell}$ be the continuous variables in your table. Introduce integer variables $y_{p,\ell}$ and binary variables $z_{p,\ell}$, and impose linear constraints \begin{align} -z_{p,\ell} \le x_{p,\ell} - y_{p,\ell} &\le z_{p,\ell} &&\text{for all $p$ and $\ell$} \tag1 \\ \sum_\ell z_{p,\ell} &\le 1 &&\text{for all $p$} \tag2 \end{align} Constraint $(1)$ enforces $x_{p,\ell} \not= y_{p,\ell} \implies z_{p,\ell} = 1$. Constraint $(2)$ allows this to happen at most once per $p$.

If your solver supports indicator constraints, you can replace $(1)$ with $z_{p,\ell} = 0 \implies x_{p,\ell} = y_{p,\ell}$.

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2
  • $\begingroup$ Hi thanks for your answer. If you don't mind could you check my code above $\endgroup$
    – Tracy
    Sep 3 at 4:21
  • $\begingroup$ With respect to solve time, can you post log or any details on solver stats? $\endgroup$
    – anjikum
    Sep 3 at 12:16

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