Recipe

Solve a constrained problem with a log barrier

Use when: you have inequality constraints, a strictly feasible starting point is available, and you want every iterate to stay feasible along the way

Goal: a central path of strictly feasible points converging to the constrained optimum as μ shrinks

Quick recipe

1
For each inequality constraint $g_i(x)\le0$ , add $-\mu\ln(-g_i(x))$ to the objective.

Example

For $\min x^2+y^2$ s.t. $4-x-y\le0$ : $B(x,y;\mu)=x^2+y^2-\mu\ln(x+y-4)$ .

Why: the log term is finite only where every constraint is strictly satisfied, so it's automatically undefined outside the feasible region.
2
Locate any $x_0$ with $g_i(x_0)<0$ for every constraint, before running any solver at all.

Why: $B$ isn't even defined outside the feasible region, so there is no fallback if you start on the wrong side.
3
Run an unconstrained solver on $B(x;\mu_k)$ , warm-started from the previous $x(\mu_{k-1})$ .

Why: the barrier term keeps every step of this solve inside the feasible region automatically.
4
Set $\mu_{k+1}=\beta\mu_k$ for a shrink factor $\beta\in(0,1)$ (often around $0.1$ – $0.5$ ), and repeat until $\mu_k$ is small enough.
- $\mu_k$ is still large relative to your tolerance — repeat the previous step.
- $\mu_k$ is small enough — stop; $x(\mu_k)$ is your answer, and it is genuinely feasible.
Why: shrinking μ too aggressively can overshoot past where the warm start is still a good guess, the same stiffness tension as growing ρ too fast in Section 2's recipe.

Common mistakes

Starting from an infeasible or boundary point, where $B$ isn't even defined.
Shrinking μ so fast that each new subproblem's minimizer is far from the last one, defeating the warm start.
Forgetting that, unlike a penalty method's answer, a barrier iterate really is feasible at every μ, not just in the limit.