Transform from constrained to unconstrained optimization

Question

In a constrained optimization problem, I found in a paper a way to define new variables such that the constraints disappear. They only give the new variable definitions, and I would like to understand how they did.

The context is thermodynamics. A mixture of $c$ components can phase separates into $p$ phases. $z_i$ is the total mole number of component $i$ and $n_{ij}$ is the number of moles of component $i$ in the phase $j$. The free energy $G(\{n_{ij}\}_{ij})$ is known. The problem is to find the composition $\{n_{ij}\}_{ij}$ that minimizes the free energy $G$.

The mass conservation gives the constraints

$\sum_{j} n_{ij} = z_i$

$0 \leq n_{ij} \leq z_i$

From this, the paper defines the new variables $\beta_{ij} $

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$n_{i1} = \beta_{i1} z_i ~~~~~~~~~~~~i=1,...,c$

$n_{ij} = \beta_{ij} (z_i - \sum_{m=1}^{j-1} n_{im}) ~~~~~~~~~ i=1,...,c ~~~~~~~~ j=2,...,p-1$

$n_{ip}=z_i - \sum_{m=1}^{p-1}n_{im} ~~~~~~~ i=1,...,c$

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with $\beta_{ij} \in [0:1]$ and the problem is know unconstrained

Any help, direction, that will get me to understand how the authors achieve this variable change would be greatly appreciated.

Thank you.

Problem source: Srinivas, Rangaiah, "Differential Evolution with Tabu List for Global Optimization and Its Application to Phase Equilibrium and Parameter Estimation Problems." Ind. Eng. Chem. Res. 2007, 46, 3410-3421

Answer 1

Ok I got it.

Let's take a simple case, with only 3 components and 2 phases.

$n_{ij}$ is the number of moles of components i in phase j.

$z_i$ is the total number of moles of component i.

We must satisfy the mass conservation

$n_{11}+n_{12}+n_{13} = z_1$

$n_{21}+n_{22}+n_{23} = z_2$

$n_{31}+n_{32}+n_{33} = z_3$

Let's fill the phases, component by component. We start with the component 1 in the phase 1. The minimum quantity we can put is obviously 0, and the maximum quantity is the total number of mole of component 1 : $z_1$. Therefore one can define the variable $\beta_{11}$

$n_{11}=\beta_{11} z_1$

And $\beta_{11}$ is free to range between 0 and 1.

Now, we still use the component 1, and we want to fill the phase 2. The minimum quantity 0, and the maximum quantity is the total number of mole of component 1, minus the amount we already put in phase 1 : ($z_1-n_{11}$). So we define $\beta_{12}$

$n_{12}=\beta_{12} (z_1-n_{11})$

and $\beta_{12}$ is also free to range between 0 and 1.

Now we want to fill the 3rd phase with component 1, and there is only one possibility : the remaining quantity of component 1 ($z_1-n_{11}-n_{12}$) and

$n_{13}=z_1-n_{11}-n_{12}$

And so on.....

We then arrive to the general equations.