Dtjytyk

Air travel with refrigerated insulin

Homology of the fiber

What is the tangent at a sharp point on a curve?

Jem'Hadar, something strange about their life expectancy

CLI: Get information Ubuntu releases

Is xar preinstalled on macOS?

Friend wants my recommendation but I don't want to

Should I be concerned about student access to a test bank?

Why does Surtur say that Thor is Asgard's doom?

How to test the sharpness of a knife?

How can an organ that provides biological immortality be unable to regenerate?

How do you justify more code being written by following clean code practices?

Turning a hard to access nut?

Output visual diagram of picture

Why is indicated airspeed rather than ground speed used during the takeoff roll?

How to find the largest number(s) in a list of elements, possibly non-unique?

Exit shell with shortcut (not typing exit) that closes session properly

How are passwords stolen from companies if they only store hashes?

How to determine the greatest d orbital splitting?

How can a new country break out from a developed country without war?

Did Nintendo change its mind about 68000 SNES?

What (if any) is the reason to buy in small local stores?

Unfrosted light bulb

Should a narrator ever describe things based on a characters view instead of fact?

Sufficient condition for a point on the boundary of a feasible set to be a minimizer

First-order condition for one kind of optimizationKKT point of a constrained optimization problemInequality optimization, KKT condition.KKT Conditions and ConvexityWhy are the KKT conditions sufficient in this case?Is there only one set of KKT conditions for a given optimization problem?First order sufficient condition for minimumminimize $x_1$ subject to $ f_1(x_1,x_2) le C_1 $ $f_2(x_1,x_2) le C_2$Optimality condition for convex QCLPConfusion about KKT points

0

$begingroup$

Consider the optimization problem

minimize $-2x^2-y^2$ subject to

$x+y=2$

$xgeq0$

$ygeq0$

The global minimum of this problem is when $y=0$ and $x=2$. How can we check the second order sufficiency condition at this point?

---

The Lagrangian is
$$L=-2x^2-y^2+lambda_1 (x+y-2)+lambda_2(-y)$$

The KKT conditions are
$$-4x+lambda_1 =0$$
$$-2y+lambda_1-lambda_2=0$$
$$x+y-2=0,~~y=0$$
Then we get $x=2, y=0$ as valid stationary point with $lambda_1=8$ and $lambda_2=8>0$.

For second order sufficiency condition we need
$$z^T nabla^2_{x,y} L(x^star,y^star,lambda^star)z>0~~~ forall zin{thetain R^2|[1~1]theta=0,~[0~~-1]theta=0}={(0,0)}$$

The question is since the valid variation space does not exists, how to we carry out our second order analysis.

optimization

share|cite|improve this question

edited Mar 12 at 4:45

Eric Brown

asked Mar 6 at 4:12

Eric BrownEric Brown

435211

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The optimal point has to be on the boundary by the maximum principle (minimum of a concave function on a convex set).
  $endgroup$
  – LinAlg
  Mar 6 at 17:46
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Yes the minimum point is in the boundary. But we assess the second order necessary condition that the Hessian of the Lagrangian over the valid variation space has to be positive definite, the valid variation space is just point zero and we cannot asses the second order sufficiency condition.
  $endgroup$
  – Eric Brown
  Mar 7 at 1:41
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I added the KKT conditions
  $endgroup$
  – Eric Brown
  Mar 12 at 4:45
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I do not understand why $y=0$ is a KKT condition. Should you not get $lambda_2y=0$, $xgeq 0$, $ygeq 0$? Since the KKT conditions are necessary, you can just enumerate all solutions to the KKT conditions and check whichs one is a minimum and which one is a maximum.
  $endgroup$
  – LinAlg
  Mar 12 at 13:55
  

  
  
  
  

add a comment | 

0

$begingroup$

Consider the optimization problem

minimize $-2x^2-y^2$ subject to

$x+y=2$

$xgeq0$

$ygeq0$

The global minimum of this problem is when $y=0$ and $x=2$. How can we check the second order sufficiency condition at this point?

---

The Lagrangian is
$$L=-2x^2-y^2+lambda_1 (x+y-2)+lambda_2(-y)$$

The KKT conditions are
$$-4x+lambda_1 =0$$
$$-2y+lambda_1-lambda_2=0$$
$$x+y-2=0,~~y=0$$
Then we get $x=2, y=0$ as valid stationary point with $lambda_1=8$ and $lambda_2=8>0$.

For second order sufficiency condition we need
$$z^T nabla^2_{x,y} L(x^star,y^star,lambda^star)z>0~~~ forall zin{thetain R^2|[1~1]theta=0,~[0~~-1]theta=0}={(0,0)}$$

The question is since the valid variation space does not exists, how to we carry out our second order analysis.

optimization

share|cite|improve this question

edited Mar 12 at 4:45

Eric Brown

asked Mar 6 at 4:12

Eric BrownEric Brown

435211

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The optimal point has to be on the boundary by the maximum principle (minimum of a concave function on a convex set).
  $endgroup$
  – LinAlg
  Mar 6 at 17:46
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Yes the minimum point is in the boundary. But we assess the second order necessary condition that the Hessian of the Lagrangian over the valid variation space has to be positive definite, the valid variation space is just point zero and we cannot asses the second order sufficiency condition.
  $endgroup$
  – Eric Brown
  Mar 7 at 1:41
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I added the KKT conditions
  $endgroup$
  – Eric Brown
  Mar 12 at 4:45
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I do not understand why $y=0$ is a KKT condition. Should you not get $lambda_2y=0$, $xgeq 0$, $ygeq 0$? Since the KKT conditions are necessary, you can just enumerate all solutions to the KKT conditions and check whichs one is a minimum and which one is a maximum.
  $endgroup$
  – LinAlg
  Mar 12 at 13:55
  

  
  
  
  

add a comment | 

$begingroup$

Consider the optimization problem

minimize $-2x^2-y^2$ subject to

$x+y=2$

$xgeq0$

$ygeq0$

The global minimum of this problem is when $y=0$ and $x=2$. How can we check the second order sufficiency condition at this point?

---

The Lagrangian is
$$L=-2x^2-y^2+lambda_1 (x+y-2)+lambda_2(-y)$$

The KKT conditions are
$$-4x+lambda_1 =0$$
$$-2y+lambda_1-lambda_2=0$$
$$x+y-2=0,~~y=0$$
Then we get $x=2, y=0$ as valid stationary point with $lambda_1=8$ and $lambda_2=8>0$.

For second order sufficiency condition we need
$$z^T nabla^2_{x,y} L(x^star,y^star,lambda^star)z>0~~~ forall zin{thetain R^2|[1~1]theta=0,~[0~~-1]theta=0}={(0,0)}$$

The question is since the valid variation space does not exists, how to we carry out our second order analysis.

optimization

share|cite|improve this question

edited Mar 12 at 4:45

Eric Brown

asked Mar 6 at 4:12

Eric BrownEric Brown

435211

$endgroup$

share|cite|improve this question

edited Mar 12 at 4:45

Eric Brown

asked Mar 6 at 4:12

Eric BrownEric Brown

435211

share|cite|improve this question

edited Mar 12 at 4:45

Eric Brown

asked Mar 6 at 4:12

Eric BrownEric Brown

435211

asked Mar 6 at 4:12

Eric BrownEric Brown

435211

asked Mar 6 at 4:12

Eric BrownEric Brown

435211