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Consider the following minimization problem,

$$\min_{x\in \mathbb{R}^2} (x_1-x_2^2) \text{ subject to } -x_1^2-x_2^2+2x_1\geq 0.$$

I am asked to find "feasible directions" from the point $(x_1,x_2)=(0,0).$ That is, directions along which the objective decreases.

My notes say the set of feasible directions at a point $x$ is given by:

$$\{ s | <s,\nabla c_i(x)> = 0 \text{ for each equality constraint } c_i, (<s, \nabla c_i(x)>) \geq 0 \text { for each active inequality constraint at } x\} .$$

Are these directions enough to ensure a decrease in the objective? Or do I need to add the condition that $(<s,\nabla f(x)>) < 0$ to ensure that $s$ is descent?

Dman
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  • Try drawing the constraint surface and some contours of the objective to guess. – copper.hat Jun 13 '19 at 19:02
  • The two conditions above are contradictory, if I want $s = (s_1,s_2)$ to be descent, I need that $s_1<0$ but if I want $s$ to be feasible, I need $s_1\geq 0$. So I am guessing that there are no feasible descent directions? – Dman Jun 13 '19 at 19:05
  • @copper.hat is this correct? – Dman Jun 13 '19 at 19:31

1 Answers1

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We see that $\{x | f(x) < f(0) \} = \{x | x_1 < x_2^2 \}$, and we see from this (consider $f(0+th)$) that if $h$ is a feasible direction then $h_1 \le 0$.

If we examine feasibility of such $h$, then we need $th$ to be feasible for small $t>0$, or in other words, $-t^2h_1^2-t^2h_2^2+2th_1 \ge 0 $. From this we see that we must have $h_1 \ge 0$ and so $h_1=0$. For $th$ to be feasible, we need $h_2 = 0$ and so there are no (first order) feasible direction.

copper.hat
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