# What Stopping Criteria to Use in Projected Gradient Descent

Suppose we want to solve a convex constrained minimization problem. We want to use projected gradient descent. If there was no constraint the stopping condition for a gradient descent algorithm would be that the gradient of function is close to zero.

But for a constrained problem the gradient at the optimal point is not necessarily (close to) zero. Now what stopping condition should we use for a projected gradient descent algorithm?

• I thought your question was about projected gradient descent. But the answer you accepted says nothing about projected gradient descent. – Rahul Nov 13 '12 at 0:20
• @RahulNarain Could you complete the answer? – user25004 Nov 13 '12 at 0:43
• I don't know the answer, or I would have posted it. I just didn't understand your acceptance criteria. Usually when one accepts an answer, it means that one has found out what they wanted to know and are not looking for any more answers. – Rahul Nov 13 '12 at 17:21

## 2 Answers

Just like in the gradient descent method, you want to stop when the norm of the gradient is sufficiently small, in the projected gradient method, you want to stop when the norm of the projected gradient is sufficiently small. Suppose the projected gradient is zero. Geometrically, that means that the negative gradient lies in the normal cone to the feasible set. If you had linear equality constraints only, it would mean that the gradient vector is orthogonal to the feasible set. In other words, it's locally impossible to find a descent direction, and you have first-order optimality.

• Thanks. Do you have a formal proof for this? (Do you know where I can find it?) – user25004 May 9 '14 at 2:22
• Look for example in Numerical Optimization (Nocedal & Wright, Springer), Section 12.3 "First-Order Optimality Conditions". For any optimization problem, the first-order conditions are that the negative gradient lie in the normal cone. – Dominique May 9 '14 at 8:16
• @ Dominique, what do you mean projected gradient in "projected gradient is sufficiently small"? Can you explain it more clearly? – Richkent Jan 19 '16 at 15:55
• @Richkent the concept of projected gradient depends on the type of constraints. If you have equality constraints, the projected gradient is the orthogonal projection of the usual gradient into the subspace tangent to the constraints. If you have inequality constraints, the projection is into the tangent cone to the active inequalities. That's what the Kuhn-Tucker conditions express when your objective and constraints are smooth and the Slater qualification holds. The section of Nocedal & Wright that I recommended above has all the details. I hope this helps. – Dominique Jan 27 '16 at 23:39

there are actually several methods for dealing with constraint optimization problems. See Penalty Method or Augmented Lagrangian Method as two examples.

To solve the problem \begin{align} \min f( x) \\ \text{ subject to: } c_i( x) \ge 0 ~\forall i \in I \end{align} We reformulate the problem for the Penalty Method as follows \begin{align} \min F(x)_k = f(x) + {\sigma_k}_i p(c_i(x)) \end{align} With $p(c) = \min(0,c)^2$.

Then inside each of your optimization iteration, you iterate over k, increasing the penalty coefficients ${\sigma_k}_i$.Thus you hope to obtain a solution to your minimization problem $F$ which also minimizes $f$, while fullfilling the constraints.

Edit:

Note that also the constraints have to be smooth as you evaluate the gradient.

• How is this related to the question??? – Dominique May 8 '14 at 22:35