Illustrative Example of Using the Steepest Descent Method to Solve the Log Barrier Function

Solve the NLP problem

min f(x) = -x₁ x₂
s.t. g₁ = x₁ + x²₂ - 1 ≤ 0
g₂ = -x₁- x₂ ≤ 0

Use the method of barrier functions by employing the log function. In solving the sub-problems, carry three iterations of the steepest descent method.  Let the starting solution be . Carry two iterations of the barrier function.

Solution

The log barrier function B(x) for this problem is

The composite function becomes

The gradient of T(x) is given by

For every iteration of the barrier method, there are three iterations of the steepest descent method. We will use the following notation to differentiate between the outer and the inner iterations:

the solution computed in iteration t of the steepest descent method (inner loop)

the solution computed in iteration k of the barrier method (outer loop)

Iteration 1

Start with r₀ = 1 and .

Iteration 1 (Steepest Descent)

Set ,

To find α, solve min_α≥0 T(α)

The optimal solution is α = 0.053589839, and

Iteration 2 (Steepest Descent)

To find α, solve min_α≥0 T(α)

The optimal solution is α = 0.329838461

Iteration 3 (Steepest Descent)

α = 0.08878

This completes three iterations of the steepest descent method.

The solution after one iteration of the barrier method is

Iteration 2

We start with the previous optimal solution, which is

Set r₁ = 10

Iteration 1 (Steepest Descent)

Set

α = 0.572927856

Iteration 2 (Steepest Decent)

Iteration 3 (Steepest Descent)

The optimal solution after the second iteration of the barrier function is

Dr Muhammad Al-Salamah, Industrial Engineering, KFUPM, Dhahran.