I have the following objective function:


Where $x$ is my decision variable. $prob_{n}$ is a piecewise function that can look like:

$prob_{n} = $

\begin{cases} 0.5, & x_n \leq 4 \\ 0.05, & x_n > 4 \end{cases}

The other constraints are:

$\sum_{n}(1-prob_{n}) > k, k \in R^{+}$

Seeing as $prob_n$ depends on the decision variable, does this mean that the objective function is no longer linear? I 'linearized' it using Big-M notation but when I try to solve it using glpk, I get an error saying that the objective is non-linear. The only reason I can think of is the one stated above or I have misspecified something in my code.

I am a little confused because I am not multiplying the decision variable (i.e. there is no $x^2$ term). So I don't see what type of non-linear problem this is.

  • 1
    $\begingroup$ Yes, $\text{prob}_n$ times $x_n$ is nonlinear. Also, you have a conflict when $x_n=4$. What type of decision variable is $x_n$? $\endgroup$
    – RobPratt
    Jul 24, 2023 at 21:32
  • $\begingroup$ LaTeX typo, apologies. I have left out a lot of constraints but x is a continuous variable. So I suppose I would need to use a MINLP solver for this (assuming it is convex) $\endgroup$
    – akkha
    Jul 24, 2023 at 22:02
  • 1
    $\begingroup$ I have added one constraint to the post. Note that this is the only constraint. What I have left out are definitions of variables. $\endgroup$
    – akkha
    Jul 25, 2023 at 6:07

1 Answer 1


This can be MILP using the escape-hatch that binary constraints can capture what would otherwise be a non-linear expression:

With $d_n \in \lbrace 0, 1 \rbrace $ as an auxiliary binary decision variable saying whether $prob_n$ is "low or high",

$$ \sum_{n}(1-prob_{n})(1+x_n) $$

decomposes to

$$ \sum_n (1 - 0.05 - (0.5 - 0.05) d_{n})(1+x_n) $$

$$ = \sum_n (0.95 - 0.45 d_{n})(1+x_n) $$

$$ = 0.95 n + \sum_n \left( 0.95 x_n - 0.45 d_n ( 1 + x_n ) \right) $$

$$ = 0.95 n + \sum_n \left( 0.95 x_n - 0.45 d_n -0.45 x'_n \right) $$

where $x'_n$, an auxiliary continuous (or semicontinuous) variable, is constrained to be either 0 or $x_n$ based on the value of $d_n$ with a big-$M$ constraint. In many problem constructions you can drop the first $0.95n$ term as it is not a function of your variables. You will need constraints smelling roughly like:

$$ x_n \le 4 + M_0 (1-d_n) $$

$$ x_n > 4 - M_1 d_n $$

$$ x'_n \le M_2 d_n $$ $$ x'_n \le x_n $$ $$ x'_n \ge x_n - M_3 (1-dn) $$

  • $\begingroup$ Very cool approach! I will have to read through it a couple of times before I can wrap my head around this. Just one quick question: would this also be applicable if my piecewise function had several segments instead of just two? Or is this the limiting factor since d_n has to be binary? $\endgroup$
    – akkha
    Jul 25, 2023 at 6:02
  • $\begingroup$ The approach would need modification and potentially more variables but will work for more segments $\endgroup$
    – Reinderien
    Jul 25, 2023 at 12:03
  • $\begingroup$ So, I've gone through this a few times now and it is such a nice solution! Is does 'trick' have some name so I can read about using it for more complex functions? $\endgroup$
    – akkha
    Jul 25, 2023 at 16:42
  • 1
    $\begingroup$ It's basically all in the theme of linearization using integral variables. Some piecewise functions with regular intervals may use integer (non-binary) variables. Too many cases to describe exhaustively. Generally look for expressions that may take a fixed number of values. $\endgroup$
    – Reinderien
    Jul 25, 2023 at 17:05

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