Exponential transformation

Example of Convexification in MINLP

The following MINLP problem can take a Covexification approach using exponential transformation:

min $ Z = 5{x_1^2}{x_2^8} + 2{x_1} + \frac{x_2^3} + 5{y_1} + 2 {y_2^2} $

s.t.

$ {x_1} \leq 7{x_2^{0.2}} $

$ 2{x_1^3} - y_1^2 \leq 1 $

$ x_1 \geq 0 $

$ x_2 \leq 4 $

$ y_1 = 0,1 $ $ y_2 = 0,1 $

Using the exponential transformation to continuous variables $ x_1, x_2, x_3 $ by substituting $ x_1 = e^{u_1} and x_2 = e^{u_2} $ described the problem becomes the following:

$ min Z = 5{e^{2{u_1}}}{e^{8{u_2}}} + 2{e^{u_1}}{e^{2u_2}} + \frac{e^{{u_2}{3}}} + 5{y_1} + 2 {y_2^2} $

$ s.t $

$ {e^{u_1}} \leq 7{e^{0.2{u_2}}} $

$ 2{e^{3{u_1}}} - y_1^2 \leq 1 $

$ e^{u_1} \geq 0 $

$ e^{u_2} \leq 4 $

$ y_1 = 0,1 $ $ y_2 = 0,1 $

With additional logarithmic simplification through properties of natural logarithm:

min $ Z = 5{e^{2{u_1}}}{e^{8{u_2}}} + 2{e^{u_1}}{e^{2u_2}} + \frac{e^{{u_2}{3}}} + 5{y_1} + 2 {y_2^2} $

s.t

$ u_1 \leq \ln 7 + 0.2{u_2} $

$ 2{e^{3{u_1}}} - y_1^2 \leq 1 $

$ {u_2} \leq \ln 4 $

$ y_1 = 0,1 $ $ y_2 = 0,1 $

Where $ u_1 $ is unbounded due to logarithmic of 0 being indefinite.

The transformed objective function can be show to be convex through the positive-definite test of the Hessian, for the example above the Hessian is as follows :

In order to prove the convexity of the transformed functions the positive definite test of Hessian is used as defined in "Optimization of Chemical Processes" can be used. This tests the Hessian defined as:

$ H(x) = H = \nabla^2f(x) $

to test that

$ Q(x)\geq 0 $

where

$ Q(x) = x^THx $

for all $ x \neq 0 $

for functions the Hessian is defined as:

$ \begin{bmatrix} X & X \\ X & Y \end{bmatrix} $

In the example above the hessian is defined as:

$ \begin{bmatrix} X & X \\ X & Y \end{bmatrix} $

Therefore H(x) is positive-definite and strictly convex.