Kuhn-Tucker question with two inequality constraints
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I've been asked to solve the following problem.
max $10x-5x^2+2y-y^2+25$
subject to
$1-x-yge0$
$1-x^2-y^2ge0$
Is anyone able to solve this by hand? NB: I've been told that the KT assumptions are not valid for this problem, but I do not understand why
$L=10x-5x^2+2y-y^2+25 + lambda_1(1-x-y) + lambda_2(1-x^2-y^2) $
First condition~:
$partial L/partial x =10-x-lambda_1-2lambda_2$=0
$partial L/partial y =2-2y-lambda_1-2lambda_2$=0
Second condition:
$lambda_1^*ge0,g_1(x,^*y^*)ge0,lambda_2^*ge0,g_2(x,^*y^*)ge0$
Third condition:
$lambda_1^*g_1(x,^*y^*)=0$
and
$lambda_2^*g_2(x,^*y^*)=0$
I understand that with two constaints there are four cases:
both constraints binding
both constraints are not binding
constraint one is binding and constraint two is not binding
constraint two is binding and constraint one is not binding
but after this point i become lost.
optimization karush-kuhn-tucker
asked Aug 20 at 12:54
Atif Khan
61
add a comment |Â
up vote
0
down vote
favorite
I've been asked to solve the following problem.
max $10x-5x^2+2y-y^2+25$
subject to
$1-x-yge0$
$1-x^2-y^2ge0$
Is anyone able to solve this by hand? NB: I've been told that the KT assumptions are not valid for this problem, but I do not understand why
$L=10x-5x^2+2y-y^2+25 + lambda_1(1-x-y) + lambda_2(1-x^2-y^2) $
First condition~:
$partial L/partial x =10-x-lambda_1-2lambda_2$=0
$partial L/partial y =2-2y-lambda_1-2lambda_2$=0
Second condition:
$lambda_1^*ge0,g_1(x,^*y^*)ge0,lambda_2^*ge0,g_2(x,^*y^*)ge0$
Third condition:
$lambda_1^*g_1(x,^*y^*)=0$
and
$lambda_2^*g_2(x,^*y^*)=0$
I understand that with two constaints there are four cases:
both constraints binding
both constraints are not binding
constraint one is binding and constraint two is not binding
constraint two is binding and constraint one is not binding
but after this point i become lost.
optimization karush-kuhn-tucker
asked Aug 20 at 12:54
Atif Khan
61
The objective functions and both constraints are concave in $x$ and $y$
– Atif Khan
Aug 20 at 12:56
elcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
– José Carlos Santos
Aug 20 at 12:57
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I've been asked to solve the following problem.
max $10x-5x^2+2y-y^2+25$
subject to
$1-x-yge0$
$1-x^2-y^2ge0$
Is anyone able to solve this by hand? NB: I've been told that the KT assumptions are not valid for this problem, but I do not understand why
$L=10x-5x^2+2y-y^2+25 + lambda_1(1-x-y) + lambda_2(1-x^2-y^2) $
First condition~:
$partial L/partial x =10-x-lambda_1-2lambda_2$=0
$partial L/partial y =2-2y-lambda_1-2lambda_2$=0
Second condition:
$lambda_1^*ge0,g_1(x,^*y^*)ge0,lambda_2^*ge0,g_2(x,^*y^*)ge0$
Third condition:
$lambda_1^*g_1(x,^*y^*)=0$
and
$lambda_2^*g_2(x,^*y^*)=0$
I understand that with two constaints there are four cases:
both constraints binding
both constraints are not binding
constraint one is binding and constraint two is not binding
constraint two is binding and constraint one is not binding
but after this point i become lost.
optimization karush-kuhn-tucker
asked Aug 20 at 12:54
Atif Khan
61
I've been asked to solve the following problem.
max $10x-5x^2+2y-y^2+25$
subject to
$1-x-yge0$
$1-x^2-y^2ge0$
Is anyone able to solve this by hand? NB: I've been told that the KT assumptions are not valid for this problem, but I do not understand why
$L=10x-5x^2+2y-y^2+25 + lambda_1(1-x-y) + lambda_2(1-x^2-y^2) $
First condition~:
$partial L/partial x =10-x-lambda_1-2lambda_2$=0
$partial L/partial y =2-2y-lambda_1-2lambda_2$=0
Second condition:
$lambda_1^*ge0,g_1(x,^*y^*)ge0,lambda_2^*ge0,g_2(x,^*y^*)ge0$
Third condition:
$lambda_1^*g_1(x,^*y^*)=0$
and
$lambda_2^*g_2(x,^*y^*)=0$
I understand that with two constaints there are four cases:
both constraints binding
both constraints are not binding
constraint one is binding and constraint two is not binding
constraint two is binding and constraint one is not binding
but after this point i become lost.
optimization karush-kuhn-tucker
asked Aug 20 at 12:54
Atif Khan
61
edited Aug 20 at 13:09
asked Aug 20 at 12:54
Atif Khan
61
asked Aug 20 at 12:54
Atif Khan
61
asked Aug 20 at 12:54
Atif Khan
61
61
The objective functions and both constraints are concave in $x$ and $y$
– Atif Khan
Aug 20 at 12:56
elcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
– José Carlos Santos
Aug 20 at 12:57
add a comment |Â
The objective functions and both constraints are concave in $x$ and $y$
– Atif Khan
Aug 20 at 12:56
elcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
– José Carlos Santos
Aug 20 at 12:57
The objective functions and both constraints are concave in $x$ and $y$
– Atif Khan
Aug 20 at 12:56
The objective functions and both constraints are concave in $x$ and $y$
– Atif Khan
Aug 20 at 12:56
elcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
– José Carlos Santos
Aug 20 at 12:57
elcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
– José Carlos Santos
Aug 20 at 12:57
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
0
down vote
Hint.
Attached a plot showing the three stationary points $(a,b,c)$ and as we can observe the point $c$ obeys the KKT conditions.
answered Aug 20 at 13:14
Cesareo
5,9052412
Why does point b not obey the KKT conditions?
– Atif Khan
Aug 20 at 13:29
@AtifKhan Because at $b$ the gradient of the objective function (black) is not a positive combination of the restrictions gradient.(red)
– Cesareo
Aug 20 at 15:32
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Hint.
Attached a plot showing the three stationary points $(a,b,c)$ and as we can observe the point $c$ obeys the KKT conditions.
answered Aug 20 at 13:14
Cesareo
5,9052412
Why does point b not obey the KKT conditions?
– Atif Khan
Aug 20 at 13:29
@AtifKhan Because at $b$ the gradient of the objective function (black) is not a positive combination of the restrictions gradient.(red)
– Cesareo
Aug 20 at 15:32
add a comment |Â
up vote
0
down vote
Hint.
Attached a plot showing the three stationary points $(a,b,c)$ and as we can observe the point $c$ obeys the KKT conditions.
answered Aug 20 at 13:14
Cesareo
5,9052412
Why does point b not obey the KKT conditions?
– Atif Khan
Aug 20 at 13:29
@AtifKhan Because at $b$ the gradient of the objective function (black) is not a positive combination of the restrictions gradient.(red)
– Cesareo
Aug 20 at 15:32
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Hint.
Attached a plot showing the three stationary points $(a,b,c)$ and as we can observe the point $c$ obeys the KKT conditions.
answered Aug 20 at 13:14
Cesareo
5,9052412
Hint.
Attached a plot showing the three stationary points $(a,b,c)$ and as we can observe the point $c$ obeys the KKT conditions.
answered Aug 20 at 13:14
Cesareo
5,9052412
answered Aug 20 at 13:14
Cesareo
5,9052412
answered Aug 20 at 13:14
Cesareo
5,9052412
answered Aug 20 at 13:14
Cesareo
5,9052412
5,9052412
Why does point b not obey the KKT conditions?
– Atif Khan
Aug 20 at 13:29
@AtifKhan Because at $b$ the gradient of the objective function (black) is not a positive combination of the restrictions gradient.(red)
– Cesareo
Aug 20 at 15:32
add a comment |Â
Why does point b not obey the KKT conditions?
– Atif Khan
Aug 20 at 13:29
@AtifKhan Because at $b$ the gradient of the objective function (black) is not a positive combination of the restrictions gradient.(red)
– Cesareo
Aug 20 at 15:32
Why does point b not obey the KKT conditions?
– Atif Khan
Aug 20 at 13:29
Why does point b not obey the KKT conditions?
– Atif Khan
Aug 20 at 13:29
@AtifKhan Because at $b$ the gradient of the objective function (black) is not a positive combination of the restrictions gradient.(red)
– Cesareo
Aug 20 at 15:32
@AtifKhan Because at $b$ the gradient of the objective function (black) is not a positive combination of the restrictions gradient.(red)
– Cesareo
Aug 20 at 15:32
add a comment |Â
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The objective functions and both constraints are concave in $x$ and $y$
– Atif Khan
Aug 20 at 12:56
elcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
– José Carlos Santos
Aug 20 at 12:57