# Why do we prefer Dual Problem over Primal Problem in convex optimization

In some applications of convex optimization (e.g. SVM) the way to go is via the Lagrangian Dual Problem.

I'm clear about the following statements (please correct me if I'm wrong):

• The dual problem provides a lower bound for the primal problem.
• The dual problem is always concave. If the primal problem is hard to optimise (not convex, high-dimensional) it's easier to find a lower bound via the dual and use it as an approximation for the primal problem.
• Under KKT-Conditions the optimal solutions of the primal and the dual problem are the same. Therefore the solution can be obtained by both, the primal and the dual problem.

Assuming KKT-conditions, WHEN and WHY is the dual problem the better (or the only) way to go?

• Another advantage: the dual problem has as many variables as constraints are there in the primal. So you can reduce the dimensionality, sometimes by much. – Anna SdTC Mar 21 '17 at 17:39
• @Anna SdTC is right: this is one of the main reasons. Sometimes also, the variables in the dual problem can be interpreted in such a way that it improves the unerstanding of the problem. – Jean Marie Mar 21 '17 at 22:21
• Strong duality (equal primal and dual optimal values) doesn't generally hold for non-convex problems or even for convex problems unless there is a suitable constraint qualification. Thus your third statement is incorrect. – Brian Borchers Mar 22 '17 at 1:22
• Many optimization methods (so-called primal-dual methods) simultaneously solve the primal and dual problems. – Brian Borchers Mar 22 '17 at 1:23