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In Boyd's Convex Optimization, pp. 243,

for any optimization problem ... for which strong duality obtains, any pair of primal and dual optimal points must satisfy the KKT conditions

i.e. $\mathrm{strong ~ duality} \implies \mathrm{KKT ~ is ~ necessary ~ condition ~ for ~ optimal ~ solution}$

and in pp. 244,

(When the primal problem is convex) if $\tilde{x}, \tilde{\lambda}, \tilde{\mu}$ are any points that satisfy the KKT conditions, then $\tilde{x}$ and $(\tilde{\lambda}, \tilde{\mu})$ are primal and dual optimal, with zero duality gap.

If duality gap = 0, the problem satisfies strong duality, and in the 3rd paragraph:

If a convex optimization problem ... satisfies Slater’s condition, then the KKT conditions provide necessary and sufficient conditions for optimality

For me it means: (for any convex problems KKT is already sufficient for optimal)

$$\mathrm{KKT} \implies \mathrm{optimal ~ with ~ zero ~ duality ~ gap} \implies \mathrm{strong ~ duality} \implies \mathrm{KKT ~ is ~ also ~ necessary}$$

so KKT is necessary and sufficient for any convex problems? (Because Slater's condition can be automatically satisfied for the zero duality gap)

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  • $\begingroup$ slater condition is not automatic in convex programing. $\endgroup$ – Red shoes Nov 19 '17 at 19:14
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The KKT conditions are not necessary for optimality even for convex problems. Consider $$ \min x $$ subject to $$ x^2\le 0. $$ The constraint is convex. The only feasible point, thus the global minimum, is given by $x=0$. The gradient of the objective is $1$ at $x=0$, while the gradient of the constraint is zero. Thus, the KKT system cannot be satisfied.

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  • $\begingroup$ slater or any other constraint qualification does hold in your example $\endgroup$ – Red shoes Nov 19 '17 at 19:17
  • $\begingroup$ @Redshoes ? simply no. please check again. the constraint says $x^2$ LESS than zero. $\endgroup$ – daw Nov 20 '17 at 7:03
  • $\begingroup$ in fact, no constraint qualification holds: Please calculate the a KKT point $\endgroup$ – daw Nov 20 '17 at 7:04
  • $\begingroup$ Sorry I wanted to say ,, does NOT hold.. it was a typo. That's why KKT is not satisfied in minimum point $\endgroup$ – Red shoes Nov 21 '17 at 7:43
  • $\begingroup$ @daw I'm trying to solve the dual problem in this question: math.stackexchange.com/questions/2571031/… could you please help with it? Thank you! :) $\endgroup$ – ALannister Dec 18 '17 at 0:38
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Boyd and Vandenberghe considers convex optimization problems of the form \begin{align} \text{minimize} &\quad f_0(x) \\ \text{subject to} & \quad f_i(x) \leq 0 \quad \text{for } i = 1,\ldots, m \\ &\quad a_i^T x = b_i \quad \text{for } i = 1,\ldots, p, \end{align} where $f_0,\ldots, f_m$ are convex functions. The optimization variable is $x \in \mathbb R^n$. (See equation (4.15), p. 136 in Boyd and Vandenberghe.)

Let $x \in \mathbb R^n$, $\lambda \in \mathbb R^m$, and $\nu \in \mathbb R^p$. Then the following two statements are equivalent:

  1. $x$ and $(\lambda,\nu)$ together satisfy the KKT conditions.
  2. $x$ and $(\lambda,\nu)$ are primal and dual optimal, and strong duality holds.

If Slater's condition is satisfied, then strong duality is guaranteed to hold, and so we can make a simpler and more useful statement. In this case, the following are equivalent:

  1. $x$ and $(\lambda,\nu)$ together satisfy the KKT conditions.
  2. $x$ and $(\lambda,\nu)$ are primal and dual optimal.

Warning: If strong duality does not hold, then it is possible for $x$ and $(\lambda,\nu)$ to be primal and dual optimal without satisfying the KKT conditions.


By the way, if Slater's condition holds, then dual optimal variables $(\lambda,\nu)$ are guaranteed to exist. So if $x$ is primal optimal, then $x$ and $(\lambda,\nu)$ together satisfy the KKT conditions.

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  • $\begingroup$ What's an example of a problem where strong duality holds but Slater's conditions do not hold? $\endgroup$ – Y. S. Jul 27 '18 at 2:23
  • $\begingroup$ @Y.S. I think an example is minimize $x$ subject to $x^2 \leq 0$. The Lagrangian is $L(x,z)= x+ z x^2$ and the dual function is $g(z) = -1/4z$ if $z \neq 0$ and $g(z) = -\infty$ otherwise. The dual optimal value is $0$, but the dual optimal value is not attained. Maybe you wanted an example where the dual optimal value is attained, I'll have to think about that. (Is that impossible?) $\endgroup$ – littleO Jul 27 '18 at 20:24
  • $\begingroup$ Cool example! But then is slater's condition satisfied? The primal constraint has no interior. $\endgroup$ – Y. S. Jul 28 '18 at 19:50
  • $\begingroup$ @Y.S. I think Slater's condition for an inequality constraint $h(x) \leq 0$ would say that there exists a point $\tilde x$ such that $h(\tilde x) < 0$. $\endgroup$ – littleO Jul 28 '18 at 20:04
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    $\begingroup$ Btw, I realized I can simplify the above example. The primal problem could be to minimize $0$ subject to $x^2 \leq 0$. Then the Lagrangian is $L(x,z) = a x^2$ and the dual function is $g(z) = 0$ if $z \geq 0$ and $g(z) = -\infty$ otherwise. The dual optimal value is $0$, and in this case the dual optimal value is attained (for example by $z = 0$). $\endgroup$ – littleO Jul 28 '18 at 21:17
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A very good explanation of your question can be found here.

A table at the end of the explanation summarizes when the KKT conditions are necessary and sufficient.

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  • $\begingroup$ I think the sentence from your link "The KKT conditions are necessary to find an optimum" is just wrong. $\endgroup$ – LKS Mar 19 '18 at 3:55

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