I am working on a chance-constrained optimisation model that takes into account uncertainty. I am aware of how to convert constraints that are of a probabilistic nature into the equivalent deterministic one, but only when the parameter follows a normal distribution. The majority of my constraints do follow a normal distribution, but I have a constraint that follows a gamma distribution with parameters $k = 3.11$ and $\theta = 9.86$. How would I then convert this to the equivalent deterministic form?

The formulation I used to convert the probabilistic constraints is as follows:

There are three coal types, $i = 1,2,3$

Each coal type's ash percentage, denoted by $a_i$, follows a normal distribution, $a_i \sim N(\mu_i,\sigma_{i}^2)$, $\forall i$. Each coal's ash percentage is independent.

The blend has a maximum allowable ash percentage of $10$, and $\alpha = 0.025$.

Let $x_i$ be the proportion of coal type $i$ to be used in the blend.

$$ P\bigg\{\sum_{i=1}^{3} a_ix_i \leq 10\bigg\} \geq 1 -\alpha, \quad x_i \geq 0 \quad \forall i $$ $$ P\Bigg\{\dfrac{\sum_{i=1}^{3} a_ix_i - \sum_{i=1}^{3} \mu_ix_i}{\sqrt{\sum_{i=1}^{3} \sigma_i^2x_i^2}} \leq \dfrac{10 - \sum_{i=1}^{3} \mu_ix_i}{\sqrt{\sum_{i=1}^{3} \sigma_i^2x_i^2}}\Bigg\} \geq 1 -\alpha $$

Where $\quad$ $\dfrac{\sum_{i=1}^{3} a_ix_i - \sum_{i=1}^{3} \mu_ix_i}{\sqrt{\sum_{i=1}^{3} \sigma_i^2x_i^2}}$ $\quad$ represents the standard normal variant with mean $0$ and variance of $1$.

The following inequality is then formed:

$$ \phi\Bigg(\dfrac{10 - \sum_{i=1}^{3} \mu_ix_i}{\sqrt{\sum_{i=1}^{3} \sigma_i^2x_i^2}}\Bigg) \geq \phi(K_{1-a}) $$

Where $\phi(K_{1-a}) = 1 - \alpha$ and $\phi()$ represents the standard normal cumulative distribution function. This gives the following deterministic constraint:

$$ \sum_{i=1}^{3}\mu_ix_i + K_{1-\alpha} *\sqrt{\sum_{i=1}^{3}\sigma_i^2x_i^2} \leq 10 $$

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    $\begingroup$ Can you give an example of the constraint that you are trying to formulate, as well as your method for formulating it if the parameter is normally distributed? $\endgroup$ Commented Apr 7, 2020 at 20:46
  • $\begingroup$ @LarrySnyder610 I am developing a model to blend coal together for the coke making process. The coal types blend together and the blend’s qualities are calculated by multiplying each coal’s proportion in the blend by the quality, and summed over all the coal types. An example of a constraint for my formulation would be limiting the ash percentage to 10, and each of the coal type’s ash percentage is normally distributed. I would use the chance-constrained approach to convert the constraint from a probabilistic one to a deterministic one. $\endgroup$ Commented Apr 7, 2020 at 21:10
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    $\begingroup$ You should look at this article: link.springer.com/article/10.1186/1029-242X-2011-108 $\endgroup$
    – kur ag
    Commented Apr 8, 2020 at 16:22
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    $\begingroup$ Can you edit your question to include the actual formulation of your constraint, in algebraic notation? $\endgroup$ Commented Apr 9, 2020 at 1:12
  • $\begingroup$ @LarrySnyder610 I have uploaded the formulation $\endgroup$ Commented Apr 9, 2020 at 15:11

1 Answer 1


Your constraint says that the $(1-\alpha)$th fractile of the normal distribution must be $\le 10$. The particular normal distribution that you are bounding depends on the values of the decision variables. In general, the $(1-\alpha)$th fractile of a $N(\mu,\sigma^2)$ distribution is $\mu + z_{1-\alpha}\sigma$, hence the particular form in your constraint.

But any distribution has a $(1-\alpha)$th fractile. So your constraint just needs to say that the $(1-\alpha)$th fractile is $\le 10$. The fractile itself will depend on your distribution, as well as on the values of the decision variables.

Note that in general, the $(1-\alpha)$th fractile of a distribution is $F^{-1}(1-\alpha)$, where $F(\cdot)$ is the cdf.

  • $\begingroup$ What if we have a single variable, just x1. Could we use the same equivalent for a single variable? $\endgroup$
    – user13149
    Commented Jan 30 at 21:37

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