I'm working on an optimization problem using Docplex and wonder if it's possible to incorporate a logical expression into the objective function of my model.

Specifically, I have three variables x, y, and z.

$$x_{ijk} \quad : \quad binary \quad variable$$ $$y_{ik} \quad and \quad z_{ik} \quad : \quad are \quad continuous \quad variables $$ And I want to model this objective function:

\begin{aligned} \min w_{1} \sum_{k \in K} \sum_{(i, j) \in V} c_{ijk}*x_{ijk} + w_{2}\sum_{k \in K} \sum_{(i, j) \in V} [(y_{jk} - y_{ik})*x_{ijk} \quad if \quad z_{ik} == 0] \end{aligned}

Is it feasible to include such a logical expression directly within the objective function in CPLEX? If so, how can I go about implementing it effectively?

  • 1
    $\begingroup$ So you want to minimize $\sum_k\sum_{i,j} (y_{jk}-y_{ik})x_{ijk}(1-\delta_{ik})$ where $\delta_{ik}\in\{0,1\}$ and $z_{ik} \ge \epsilon \delta_{ik}$ ($z_{ik}=0 \implies \delta_{ik}=0$). This needs to be linearized. $\endgroup$
    – Kuifje
    Commented Aug 5, 2023 at 14:29
  • $\begingroup$ What should the objective term be if $z_{ik}\neq 0?$ $\endgroup$
    – prubin
    Commented Aug 5, 2023 at 15:23
  • $\begingroup$ I have edited the question, it is just the second term of the objective function. In the model, the variable 'z' takes on discrete values from the set {0, 1, 2, 3}. When the value of 'z' is not equal to 0, the second term evaluates to 0. $\endgroup$
    – Nada.S
    Commented Aug 5, 2023 at 16:04

1 Answer 1


We can use a "big M" approach, assuming that you can find a positive constant $M$ such that $\vert y_{jk}-y_{ik} \vert \le M$ for all $k\in K$ and $(i,j)\in V.$

Introduce new binary variables $u_{ik}$ and continuous variables $v_{ijk}.$ The second objective summand (the mess in brackets) becomes $v_{ijk}.$ Now add the following constraints: $$3u_{ik} + z_{ik} \le 3\, (1)$$ $$u_{ik} + z_{ik} \ge 1 \, (2)$$ $$v_{ijk} \le M u_{ik}\, (3)$$ $$v_{ijk} \ge -M u_{ik}\, (3')$$ $$v_{ijk} \le M x_{ijk}\, (4)$$ $$v_{ijk} \ge -M x_{ijk}\, (4')$$ $$v_{ijk} \le y_{jk} - y_{ik} + M(2 - u_{ik} - x_{ijk}) \, (5)$$ and $$v_{ijk} \ge y_{jk} - y_{ik} - M(2 - u_{ik} - x_{ijk}) \, (5')$$

Constraints (1) and (2) combine to set $u_{ik} =1$ if and only if $z_{ik} = 0.$ Constraints (3) and (3') say that the $(i,j,k)$ objective term is 0 if $u_{ik}=0$ (meaning $z_{ik} \in \lbrace 1,2,3 \rbrace$). Constraints (4) and (4') say that the objective term is 0 if $x_{ijk} =0.$ Finally, constraints (5) and (5') say that if $u_{ik}=1$ (meaning $z_{ik}=0$) and $x_{ijk}=1,$ then the objective term equals $y_{jk}-y_{ik}.$

Addendum: If the solver supports "if-then" (implication) constraints (CPLEX does), then the "big M" constraints can be replaced with implications, as follows. $$u_{ik}=0 \implies v_{ijk}=0$$ $$x_{ijk}=0 \implies v_{ijk}=0$$ $$u_{ik} + x_{ijk} = 2 \implies v_{ijk} = y_{jk}-y_{ik}$$

As a general rule, you are probably better off with the "big M" constraints if you can find a reasonably tight value of $M.$

  • $\begingroup$ Thanks, the reasoning seems to be correct, Just finding the appropriate value of "M" isn't feasible. I just wondering taking into account that the second term will be calculated just in the case if there is an arc between (i,j) $$x_{ijk} =1 \quad and \quad z_{ik}=0$$ it s necessary to deal with the other cases. $\endgroup$
    – Nada.S
    Commented Aug 7, 2023 at 8:33
  • $\begingroup$ I'm not sure what you mean by "necessary to deal with the other cases". $\endgroup$
    – prubin
    Commented Aug 7, 2023 at 15:10
  • $\begingroup$ I extended the answer to describe use of implication constraints if you can't find a good value for $M.$ $\endgroup$
    – prubin
    Commented Aug 7, 2023 at 15:24

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