Lagrange multiplier for sum of products with sum constraint
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Is it possible to use Lagrange multipliers (or another technique) to easily find a maximum of a function like
$$
f:
begincases
mathbbR^3_ge0&to mathbbR_ge0\
(x_1,x_2,x_3)&mapsto x_1x_2x_3+ 2x_1x_2+3x_1x_3+x_2
endcases
$$ with the constraint that the sum of the arguments $sum_i=1^3x_i=s$ for some $sinmathbbR_>0$. I know that the product $x_1x_2x_3$ is maximized when all variables are the same but here, we have a weighted sum of partial products. (The function is given only as an example, I actually want to find a maximum of a function with more variables but of the same kind, i.e. weighted sum of partial products where each variable has either power 0 or 1 in the exponent)
optimization lagrange-multiplier
asked Aug 19 at 10:29
phinz
1326
add a comment |Â
up vote
0
down vote
favorite
Is it possible to use Lagrange multipliers (or another technique) to easily find a maximum of a function like
$$
f:
begincases
mathbbR^3_ge0&to mathbbR_ge0\
(x_1,x_2,x_3)&mapsto x_1x_2x_3+ 2x_1x_2+3x_1x_3+x_2
endcases
$$ with the constraint that the sum of the arguments $sum_i=1^3x_i=s$ for some $sinmathbbR_>0$. I know that the product $x_1x_2x_3$ is maximized when all variables are the same but here, we have a weighted sum of partial products. (The function is given only as an example, I actually want to find a maximum of a function with more variables but of the same kind, i.e. weighted sum of partial products where each variable has either power 0 or 1 in the exponent)
optimization lagrange-multiplier
asked Aug 19 at 10:29
phinz
1326
Yes, it is possible to solve this constrained optimization problem using Lagrange multipliers.
– molarmass
Aug 19 at 10:47
@molarmass Can you please indicate how? Setting the differential of the Lagrangian to zero gives a non-linear problem where I do not see how it can be solved.
– phinz
Aug 19 at 11:14
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Is it possible to use Lagrange multipliers (or another technique) to easily find a maximum of a function like
$$
f:
begincases
mathbbR^3_ge0&to mathbbR_ge0\
(x_1,x_2,x_3)&mapsto x_1x_2x_3+ 2x_1x_2+3x_1x_3+x_2
endcases
$$ with the constraint that the sum of the arguments $sum_i=1^3x_i=s$ for some $sinmathbbR_>0$. I know that the product $x_1x_2x_3$ is maximized when all variables are the same but here, we have a weighted sum of partial products. (The function is given only as an example, I actually want to find a maximum of a function with more variables but of the same kind, i.e. weighted sum of partial products where each variable has either power 0 or 1 in the exponent)
optimization lagrange-multiplier
asked Aug 19 at 10:29
phinz
1326
Is it possible to use Lagrange multipliers (or another technique) to easily find a maximum of a function like
$$
f:
begincases
mathbbR^3_ge0&to mathbbR_ge0\
(x_1,x_2,x_3)&mapsto x_1x_2x_3+ 2x_1x_2+3x_1x_3+x_2
endcases
$$ with the constraint that the sum of the arguments $sum_i=1^3x_i=s$ for some $sinmathbbR_>0$. I know that the product $x_1x_2x_3$ is maximized when all variables are the same but here, we have a weighted sum of partial products. (The function is given only as an example, I actually want to find a maximum of a function with more variables but of the same kind, i.e. weighted sum of partial products where each variable has either power 0 or 1 in the exponent)
optimization lagrange-multiplier
asked Aug 19 at 10:29
phinz
1326
asked Aug 19 at 10:29
phinz
1326
asked Aug 19 at 10:29
phinz
1326
asked Aug 19 at 10:29
phinz
1326
1326
Yes, it is possible to solve this constrained optimization problem using Lagrange multipliers.
– molarmass
Aug 19 at 10:47
@molarmass Can you please indicate how? Setting the differential of the Lagrangian to zero gives a non-linear problem where I do not see how it can be solved.
– phinz
Aug 19 at 11:14
add a comment |Â
Yes, it is possible to solve this constrained optimization problem using Lagrange multipliers.
– molarmass
Aug 19 at 10:47
@molarmass Can you please indicate how? Setting the differential of the Lagrangian to zero gives a non-linear problem where I do not see how it can be solved.
– phinz
Aug 19 at 11:14
Yes, it is possible to solve this constrained optimization problem using Lagrange multipliers.
– molarmass
Aug 19 at 10:47
Yes, it is possible to solve this constrained optimization problem using Lagrange multipliers.
– molarmass
Aug 19 at 10:47
@molarmass Can you please indicate how? Setting the differential of the Lagrangian to zero gives a non-linear problem where I do not see how it can be solved.
– phinz
Aug 19 at 11:14
@molarmass Can you please indicate how? Setting the differential of the Lagrangian to zero gives a non-linear problem where I do not see how it can be solved.
– phinz
Aug 19 at 11:14
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
According to the Lagrange's Multipliers technique the formulation could be
Given
$$
f(x) = x_1 x_2 x_3 +2x_2 x_2+3x_1 x_3+x_2\
g_1(x) = x_1+x_2+x_3 - s = 0\
g_2(x,epsilon) = x_1 - epsilon_1^2 = 0\
g_3(x,epsilon) = x_2 - epsilon_2^2 = 0\
g_4(x,epsilon) = x_3 - epsilon_3^2 = 0\
$$
Determine the stationary points for
$$
L(x,lambda,epsilon) = f(x)+lambda _1g_1(x)+sum_k=1^3lambda_k+1g_k+1(x_k,epsilon_k)
$$
Here $lambda_k$ are the so called Lagrange multipliers and $epsilon_k$ are slack variables to transform the restrictions $x_k ge 0$ into equivalent equality restrictions.
Now the stationary points are the solutions for
$$
nabla L = left{
beginarrayrcl
lambda_1+lambda_2+2 x_2+x_2x_3+3 z=0 \
lambda_1+lambda_3+2 x_1+x_1 x_3+1=0 \
lambda_1+lambda_4+3 x_1+x_1 x_2=0 \
x_1+x_2+x_3-s=0 \
x_1-epsilon_1^2=0 \
x_2-epsilon_2^2=0 \
x_3-epsilon_3^2=0 \
-2 epsilon_1 lambda_2=0 \
-2 epsilon_2 lambda_3=0 \
-2 epsilon_3 lambda_4=0 \
endarray
right.
$$
After that, the solutions should be qualified as local minimum, local maximum or saddle point. This is done with the Hessian from
$$
(fcirc g_1)(x) = f(x_1,x_2,s-x_1-x_2)
$$
NOTE
For $s = 2$ we have
$$
left(
beginarrayccccccccccc
x_1 & x_ 2 & x_3 & lambda_1 & lambda_2 & lambda_3 & lambda_4 & epsilon_1 & epsilon_2 & epsilon_3 & f(x)\
0 & 0 & 2. & 0 & -6. & -1. & 0 & 0 & 0 & -1.41421 & 0. \
0 & 0 & 2. & 0 & -6. & -1. & 0 & 0 & 0 & 1.41421 & 0. \
0 & 2. & 0 & -1. & -3. & 0 & 1. & 0 & -1.41421 & 0 & 2. \
0 & 2. & 0 & -1. & -3. & 0 & 1. & 0 & 1.41421 & 0 & 2. \
0.75 & 1.25 & 0 & -2.5 & 0 & 0 & -0.6875 & -0.866025 & -1.11803 & 0 & 3.125 \
0.75 & 1.25 & 0 & -2.5 & 0 & 0 & -0.6875 & -0.866025 & 1.11803 & 0 & 3.125 \
0.75 & 1.25 & 0 & -2.5 & 0 & 0 & -0.6875 & 0.866025 & -1.11803 & 0 & 3.125 \
0.75 & 1.25 & 0 & -2.5 & 0 & 0 & -0.6875 & 0.866025 & 1.11803 & 0 & 3.125 \
1. & 0 & 1. & -3. & 0 & -1. & 0 & -1. & 0 & -1. & 3. \
1. & 0 & 1. & -3. & 0 & -1. & 0 & -1. & 0 & 1. & 3. \
1. & 0 & 1. & -3. & 0 & -1. & 0 & 1. & 0 & -1. & 3. \
1. & 0 & 1. & -3. & 0 & -1. & 0 & 1. & 0 & 1. & 3. \
2. & 0 & 0 & 0 & 0 & -5. & -6. & -1.41421 & 0 & 0 & 0. \
2. & 0 & 0 & 0 & 0 & -5. & -6. & 1.41421 & 0 & 0 & 0. \
endarray
right)
$$
The superabundant solutions are due to $epsilon_k^2$
A solution with $epsilon_k = 0$ means that the restriction $g_k+1(x,epsilon)$ is active.
answered Aug 19 at 11:55
Cesareo
5,9052412
Thank you very much! But can you also explain how one determines these solutions (since one cannot use Linear Algebra)? Is there a software (i.e. NumPy) that can determine these solutions (since my setting is bigger)
– phinz
Aug 19 at 14:13
Is it correct with in the definitions of $g_2,...,g_4$ without a square of the $epsilon$ variables?
– phinz
Aug 19 at 14:15
@phinz I use MATHEMATICA which handles well those equations. You can formulate also the problem without the slack $epsilon_k$ I did this way to show one of the variants when handling inequality restrictions. The slack variables are convenient in more complicated situations. MATHEMATICA is especially useful in those simple problems.
– Cesareo
Aug 19 at 15:48
@phinz Please verify the corrections introduced in the definition of $g_2,cdots, g_4$ The correct formulation is with $epsilon_k^2$ as it is now.
– Cesareo
Aug 19 at 15:52
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
According to the Lagrange's Multipliers technique the formulation could be
Given
$$
f(x) = x_1 x_2 x_3 +2x_2 x_2+3x_1 x_3+x_2\
g_1(x) = x_1+x_2+x_3 - s = 0\
g_2(x,epsilon) = x_1 - epsilon_1^2 = 0\
g_3(x,epsilon) = x_2 - epsilon_2^2 = 0\
g_4(x,epsilon) = x_3 - epsilon_3^2 = 0\
$$
Determine the stationary points for
$$
L(x,lambda,epsilon) = f(x)+lambda _1g_1(x)+sum_k=1^3lambda_k+1g_k+1(x_k,epsilon_k)
$$
Here $lambda_k$ are the so called Lagrange multipliers and $epsilon_k$ are slack variables to transform the restrictions $x_k ge 0$ into equivalent equality restrictions.
Now the stationary points are the solutions for
$$
nabla L = left{
beginarrayrcl
lambda_1+lambda_2+2 x_2+x_2x_3+3 z=0 \
lambda_1+lambda_3+2 x_1+x_1 x_3+1=0 \
lambda_1+lambda_4+3 x_1+x_1 x_2=0 \
x_1+x_2+x_3-s=0 \
x_1-epsilon_1^2=0 \
x_2-epsilon_2^2=0 \
x_3-epsilon_3^2=0 \
-2 epsilon_1 lambda_2=0 \
-2 epsilon_2 lambda_3=0 \
-2 epsilon_3 lambda_4=0 \
endarray
right.
$$
After that, the solutions should be qualified as local minimum, local maximum or saddle point. This is done with the Hessian from
$$
(fcirc g_1)(x) = f(x_1,x_2,s-x_1-x_2)
$$
NOTE
For $s = 2$ we have
$$
left(
beginarrayccccccccccc
x_1 & x_ 2 & x_3 & lambda_1 & lambda_2 & lambda_3 & lambda_4 & epsilon_1 & epsilon_2 & epsilon_3 & f(x)\
0 & 0 & 2. & 0 & -6. & -1. & 0 & 0 & 0 & -1.41421 & 0. \
0 & 0 & 2. & 0 & -6. & -1. & 0 & 0 & 0 & 1.41421 & 0. \
0 & 2. & 0 & -1. & -3. & 0 & 1. & 0 & -1.41421 & 0 & 2. \
0 & 2. & 0 & -1. & -3. & 0 & 1. & 0 & 1.41421 & 0 & 2. \
0.75 & 1.25 & 0 & -2.5 & 0 & 0 & -0.6875 & -0.866025 & -1.11803 & 0 & 3.125 \
0.75 & 1.25 & 0 & -2.5 & 0 & 0 & -0.6875 & -0.866025 & 1.11803 & 0 & 3.125 \
0.75 & 1.25 & 0 & -2.5 & 0 & 0 & -0.6875 & 0.866025 & -1.11803 & 0 & 3.125 \
0.75 & 1.25 & 0 & -2.5 & 0 & 0 & -0.6875 & 0.866025 & 1.11803 & 0 & 3.125 \
1. & 0 & 1. & -3. & 0 & -1. & 0 & -1. & 0 & -1. & 3. \
1. & 0 & 1. & -3. & 0 & -1. & 0 & -1. & 0 & 1. & 3. \
1. & 0 & 1. & -3. & 0 & -1. & 0 & 1. & 0 & -1. & 3. \
1. & 0 & 1. & -3. & 0 & -1. & 0 & 1. & 0 & 1. & 3. \
2. & 0 & 0 & 0 & 0 & -5. & -6. & -1.41421 & 0 & 0 & 0. \
2. & 0 & 0 & 0 & 0 & -5. & -6. & 1.41421 & 0 & 0 & 0. \
endarray
right)
$$
The superabundant solutions are due to $epsilon_k^2$
A solution with $epsilon_k = 0$ means that the restriction $g_k+1(x,epsilon)$ is active.
answered Aug 19 at 11:55
Cesareo
5,9052412
Thank you very much! But can you also explain how one determines these solutions (since one cannot use Linear Algebra)? Is there a software (i.e. NumPy) that can determine these solutions (since my setting is bigger)
– phinz
Aug 19 at 14:13
Is it correct with in the definitions of $g_2,...,g_4$ without a square of the $epsilon$ variables?
– phinz
Aug 19 at 14:15
@phinz I use MATHEMATICA which handles well those equations. You can formulate also the problem without the slack $epsilon_k$ I did this way to show one of the variants when handling inequality restrictions. The slack variables are convenient in more complicated situations. MATHEMATICA is especially useful in those simple problems.
– Cesareo
Aug 19 at 15:48
@phinz Please verify the corrections introduced in the definition of $g_2,cdots, g_4$ The correct formulation is with $epsilon_k^2$ as it is now.
– Cesareo
Aug 19 at 15:52
add a comment |Â
up vote
2
down vote
accepted
According to the Lagrange's Multipliers technique the formulation could be
Given
$$
f(x) = x_1 x_2 x_3 +2x_2 x_2+3x_1 x_3+x_2\
g_1(x) = x_1+x_2+x_3 - s = 0\
g_2(x,epsilon) = x_1 - epsilon_1^2 = 0\
g_3(x,epsilon) = x_2 - epsilon_2^2 = 0\
g_4(x,epsilon) = x_3 - epsilon_3^2 = 0\
$$
Determine the stationary points for
$$
L(x,lambda,epsilon) = f(x)+lambda _1g_1(x)+sum_k=1^3lambda_k+1g_k+1(x_k,epsilon_k)
$$
Here $lambda_k$ are the so called Lagrange multipliers and $epsilon_k$ are slack variables to transform the restrictions $x_k ge 0$ into equivalent equality restrictions.
Now the stationary points are the solutions for
$$
nabla L = left{
beginarrayrcl
lambda_1+lambda_2+2 x_2+x_2x_3+3 z=0 \
lambda_1+lambda_3+2 x_1+x_1 x_3+1=0 \
lambda_1+lambda_4+3 x_1+x_1 x_2=0 \
x_1+x_2+x_3-s=0 \
x_1-epsilon_1^2=0 \
x_2-epsilon_2^2=0 \
x_3-epsilon_3^2=0 \
-2 epsilon_1 lambda_2=0 \
-2 epsilon_2 lambda_3=0 \
-2 epsilon_3 lambda_4=0 \
endarray
right.
$$
After that, the solutions should be qualified as local minimum, local maximum or saddle point. This is done with the Hessian from
$$
(fcirc g_1)(x) = f(x_1,x_2,s-x_1-x_2)
$$
NOTE
For $s = 2$ we have
$$
left(
beginarrayccccccccccc
x_1 & x_ 2 & x_3 & lambda_1 & lambda_2 & lambda_3 & lambda_4 & epsilon_1 & epsilon_2 & epsilon_3 & f(x)\
0 & 0 & 2. & 0 & -6. & -1. & 0 & 0 & 0 & -1.41421 & 0. \
0 & 0 & 2. & 0 & -6. & -1. & 0 & 0 & 0 & 1.41421 & 0. \
0 & 2. & 0 & -1. & -3. & 0 & 1. & 0 & -1.41421 & 0 & 2. \
0 & 2. & 0 & -1. & -3. & 0 & 1. & 0 & 1.41421 & 0 & 2. \
0.75 & 1.25 & 0 & -2.5 & 0 & 0 & -0.6875 & -0.866025 & -1.11803 & 0 & 3.125 \
0.75 & 1.25 & 0 & -2.5 & 0 & 0 & -0.6875 & -0.866025 & 1.11803 & 0 & 3.125 \
0.75 & 1.25 & 0 & -2.5 & 0 & 0 & -0.6875 & 0.866025 & -1.11803 & 0 & 3.125 \
0.75 & 1.25 & 0 & -2.5 & 0 & 0 & -0.6875 & 0.866025 & 1.11803 & 0 & 3.125 \
1. & 0 & 1. & -3. & 0 & -1. & 0 & -1. & 0 & -1. & 3. \
1. & 0 & 1. & -3. & 0 & -1. & 0 & -1. & 0 & 1. & 3. \
1. & 0 & 1. & -3. & 0 & -1. & 0 & 1. & 0 & -1. & 3. \
1. & 0 & 1. & -3. & 0 & -1. & 0 & 1. & 0 & 1. & 3. \
2. & 0 & 0 & 0 & 0 & -5. & -6. & -1.41421 & 0 & 0 & 0. \
2. & 0 & 0 & 0 & 0 & -5. & -6. & 1.41421 & 0 & 0 & 0. \
endarray
right)
$$
The superabundant solutions are due to $epsilon_k^2$
A solution with $epsilon_k = 0$ means that the restriction $g_k+1(x,epsilon)$ is active.
answered Aug 19 at 11:55
Cesareo
5,9052412
Thank you very much! But can you also explain how one determines these solutions (since one cannot use Linear Algebra)? Is there a software (i.e. NumPy) that can determine these solutions (since my setting is bigger)
– phinz
Aug 19 at 14:13
Is it correct with in the definitions of $g_2,...,g_4$ without a square of the $epsilon$ variables?
– phinz
Aug 19 at 14:15
@phinz I use MATHEMATICA which handles well those equations. You can formulate also the problem without the slack $epsilon_k$ I did this way to show one of the variants when handling inequality restrictions. The slack variables are convenient in more complicated situations. MATHEMATICA is especially useful in those simple problems.
– Cesareo
Aug 19 at 15:48
@phinz Please verify the corrections introduced in the definition of $g_2,cdots, g_4$ The correct formulation is with $epsilon_k^2$ as it is now.
– Cesareo
Aug 19 at 15:52
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
According to the Lagrange's Multipliers technique the formulation could be
Given
$$
f(x) = x_1 x_2 x_3 +2x_2 x_2+3x_1 x_3+x_2\
g_1(x) = x_1+x_2+x_3 - s = 0\
g_2(x,epsilon) = x_1 - epsilon_1^2 = 0\
g_3(x,epsilon) = x_2 - epsilon_2^2 = 0\
g_4(x,epsilon) = x_3 - epsilon_3^2 = 0\
$$
Determine the stationary points for
$$
L(x,lambda,epsilon) = f(x)+lambda _1g_1(x)+sum_k=1^3lambda_k+1g_k+1(x_k,epsilon_k)
$$
Here $lambda_k$ are the so called Lagrange multipliers and $epsilon_k$ are slack variables to transform the restrictions $x_k ge 0$ into equivalent equality restrictions.
Now the stationary points are the solutions for
$$
nabla L = left{
beginarrayrcl
lambda_1+lambda_2+2 x_2+x_2x_3+3 z=0 \
lambda_1+lambda_3+2 x_1+x_1 x_3+1=0 \
lambda_1+lambda_4+3 x_1+x_1 x_2=0 \
x_1+x_2+x_3-s=0 \
x_1-epsilon_1^2=0 \
x_2-epsilon_2^2=0 \
x_3-epsilon_3^2=0 \
-2 epsilon_1 lambda_2=0 \
-2 epsilon_2 lambda_3=0 \
-2 epsilon_3 lambda_4=0 \
endarray
right.
$$
After that, the solutions should be qualified as local minimum, local maximum or saddle point. This is done with the Hessian from
$$
(fcirc g_1)(x) = f(x_1,x_2,s-x_1-x_2)
$$
NOTE
For $s = 2$ we have
$$
left(
beginarrayccccccccccc
x_1 & x_ 2 & x_3 & lambda_1 & lambda_2 & lambda_3 & lambda_4 & epsilon_1 & epsilon_2 & epsilon_3 & f(x)\
0 & 0 & 2. & 0 & -6. & -1. & 0 & 0 & 0 & -1.41421 & 0. \
0 & 0 & 2. & 0 & -6. & -1. & 0 & 0 & 0 & 1.41421 & 0. \
0 & 2. & 0 & -1. & -3. & 0 & 1. & 0 & -1.41421 & 0 & 2. \
0 & 2. & 0 & -1. & -3. & 0 & 1. & 0 & 1.41421 & 0 & 2. \
0.75 & 1.25 & 0 & -2.5 & 0 & 0 & -0.6875 & -0.866025 & -1.11803 & 0 & 3.125 \
0.75 & 1.25 & 0 & -2.5 & 0 & 0 & -0.6875 & -0.866025 & 1.11803 & 0 & 3.125 \
0.75 & 1.25 & 0 & -2.5 & 0 & 0 & -0.6875 & 0.866025 & -1.11803 & 0 & 3.125 \
0.75 & 1.25 & 0 & -2.5 & 0 & 0 & -0.6875 & 0.866025 & 1.11803 & 0 & 3.125 \
1. & 0 & 1. & -3. & 0 & -1. & 0 & -1. & 0 & -1. & 3. \
1. & 0 & 1. & -3. & 0 & -1. & 0 & -1. & 0 & 1. & 3. \
1. & 0 & 1. & -3. & 0 & -1. & 0 & 1. & 0 & -1. & 3. \
1. & 0 & 1. & -3. & 0 & -1. & 0 & 1. & 0 & 1. & 3. \
2. & 0 & 0 & 0 & 0 & -5. & -6. & -1.41421 & 0 & 0 & 0. \
2. & 0 & 0 & 0 & 0 & -5. & -6. & 1.41421 & 0 & 0 & 0. \
endarray
right)
$$
The superabundant solutions are due to $epsilon_k^2$
A solution with $epsilon_k = 0$ means that the restriction $g_k+1(x,epsilon)$ is active.
answered Aug 19 at 11:55
Cesareo
5,9052412
According to the Lagrange's Multipliers technique the formulation could be
Given
$$
f(x) = x_1 x_2 x_3 +2x_2 x_2+3x_1 x_3+x_2\
g_1(x) = x_1+x_2+x_3 - s = 0\
g_2(x,epsilon) = x_1 - epsilon_1^2 = 0\
g_3(x,epsilon) = x_2 - epsilon_2^2 = 0\
g_4(x,epsilon) = x_3 - epsilon_3^2 = 0\
$$
Determine the stationary points for
$$
L(x,lambda,epsilon) = f(x)+lambda _1g_1(x)+sum_k=1^3lambda_k+1g_k+1(x_k,epsilon_k)
$$
Here $lambda_k$ are the so called Lagrange multipliers and $epsilon_k$ are slack variables to transform the restrictions $x_k ge 0$ into equivalent equality restrictions.
Now the stationary points are the solutions for
$$
nabla L = left{
beginarrayrcl
lambda_1+lambda_2+2 x_2+x_2x_3+3 z=0 \
lambda_1+lambda_3+2 x_1+x_1 x_3+1=0 \
lambda_1+lambda_4+3 x_1+x_1 x_2=0 \
x_1+x_2+x_3-s=0 \
x_1-epsilon_1^2=0 \
x_2-epsilon_2^2=0 \
x_3-epsilon_3^2=0 \
-2 epsilon_1 lambda_2=0 \
-2 epsilon_2 lambda_3=0 \
-2 epsilon_3 lambda_4=0 \
endarray
right.
$$
After that, the solutions should be qualified as local minimum, local maximum or saddle point. This is done with the Hessian from
$$
(fcirc g_1)(x) = f(x_1,x_2,s-x_1-x_2)
$$
NOTE
For $s = 2$ we have
$$
left(
beginarrayccccccccccc
x_1 & x_ 2 & x_3 & lambda_1 & lambda_2 & lambda_3 & lambda_4 & epsilon_1 & epsilon_2 & epsilon_3 & f(x)\
0 & 0 & 2. & 0 & -6. & -1. & 0 & 0 & 0 & -1.41421 & 0. \
0 & 0 & 2. & 0 & -6. & -1. & 0 & 0 & 0 & 1.41421 & 0. \
0 & 2. & 0 & -1. & -3. & 0 & 1. & 0 & -1.41421 & 0 & 2. \
0 & 2. & 0 & -1. & -3. & 0 & 1. & 0 & 1.41421 & 0 & 2. \
0.75 & 1.25 & 0 & -2.5 & 0 & 0 & -0.6875 & -0.866025 & -1.11803 & 0 & 3.125 \
0.75 & 1.25 & 0 & -2.5 & 0 & 0 & -0.6875 & -0.866025 & 1.11803 & 0 & 3.125 \
0.75 & 1.25 & 0 & -2.5 & 0 & 0 & -0.6875 & 0.866025 & -1.11803 & 0 & 3.125 \
0.75 & 1.25 & 0 & -2.5 & 0 & 0 & -0.6875 & 0.866025 & 1.11803 & 0 & 3.125 \
1. & 0 & 1. & -3. & 0 & -1. & 0 & -1. & 0 & -1. & 3. \
1. & 0 & 1. & -3. & 0 & -1. & 0 & -1. & 0 & 1. & 3. \
1. & 0 & 1. & -3. & 0 & -1. & 0 & 1. & 0 & -1. & 3. \
1. & 0 & 1. & -3. & 0 & -1. & 0 & 1. & 0 & 1. & 3. \
2. & 0 & 0 & 0 & 0 & -5. & -6. & -1.41421 & 0 & 0 & 0. \
2. & 0 & 0 & 0 & 0 & -5. & -6. & 1.41421 & 0 & 0 & 0. \
endarray
right)
$$
The superabundant solutions are due to $epsilon_k^2$
A solution with $epsilon_k = 0$ means that the restriction $g_k+1(x,epsilon)$ is active.
answered Aug 19 at 11:55
Cesareo
5,9052412
edited Aug 19 at 15:49
answered Aug 19 at 11:55
Cesareo
5,9052412
answered Aug 19 at 11:55
Cesareo
5,9052412
answered Aug 19 at 11:55
Cesareo
5,9052412
5,9052412
Thank you very much! But can you also explain how one determines these solutions (since one cannot use Linear Algebra)? Is there a software (i.e. NumPy) that can determine these solutions (since my setting is bigger)
– phinz
Aug 19 at 14:13
Is it correct with in the definitions of $g_2,...,g_4$ without a square of the $epsilon$ variables?
– phinz
Aug 19 at 14:15
@phinz I use MATHEMATICA which handles well those equations. You can formulate also the problem without the slack $epsilon_k$ I did this way to show one of the variants when handling inequality restrictions. The slack variables are convenient in more complicated situations. MATHEMATICA is especially useful in those simple problems.
– Cesareo
Aug 19 at 15:48
@phinz Please verify the corrections introduced in the definition of $g_2,cdots, g_4$ The correct formulation is with $epsilon_k^2$ as it is now.
– Cesareo
Aug 19 at 15:52
add a comment |Â
Thank you very much! But can you also explain how one determines these solutions (since one cannot use Linear Algebra)? Is there a software (i.e. NumPy) that can determine these solutions (since my setting is bigger)
– phinz
Aug 19 at 14:13
Is it correct with in the definitions of $g_2,...,g_4$ without a square of the $epsilon$ variables?
– phinz
Aug 19 at 14:15
@phinz I use MATHEMATICA which handles well those equations. You can formulate also the problem without the slack $epsilon_k$ I did this way to show one of the variants when handling inequality restrictions. The slack variables are convenient in more complicated situations. MATHEMATICA is especially useful in those simple problems.
– Cesareo
Aug 19 at 15:48
@phinz Please verify the corrections introduced in the definition of $g_2,cdots, g_4$ The correct formulation is with $epsilon_k^2$ as it is now.
– Cesareo
Aug 19 at 15:52
Thank you very much! But can you also explain how one determines these solutions (since one cannot use Linear Algebra)? Is there a software (i.e. NumPy) that can determine these solutions (since my setting is bigger)
– phinz
Aug 19 at 14:13
Thank you very much! But can you also explain how one determines these solutions (since one cannot use Linear Algebra)? Is there a software (i.e. NumPy) that can determine these solutions (since my setting is bigger)
– phinz
Aug 19 at 14:13
Is it correct with in the definitions of $g_2,...,g_4$ without a square of the $epsilon$ variables?
– phinz
Aug 19 at 14:15
Is it correct with in the definitions of $g_2,...,g_4$ without a square of the $epsilon$ variables?
– phinz
Aug 19 at 14:15
@phinz I use MATHEMATICA which handles well those equations. You can formulate also the problem without the slack $epsilon_k$ I did this way to show one of the variants when handling inequality restrictions. The slack variables are convenient in more complicated situations. MATHEMATICA is especially useful in those simple problems.
– Cesareo
Aug 19 at 15:48
@phinz I use MATHEMATICA which handles well those equations. You can formulate also the problem without the slack $epsilon_k$ I did this way to show one of the variants when handling inequality restrictions. The slack variables are convenient in more complicated situations. MATHEMATICA is especially useful in those simple problems.
– Cesareo
Aug 19 at 15:48
@phinz Please verify the corrections introduced in the definition of $g_2,cdots, g_4$ The correct formulation is with $epsilon_k^2$ as it is now.
– Cesareo
Aug 19 at 15:52
@phinz Please verify the corrections introduced in the definition of $g_2,cdots, g_4$ The correct formulation is with $epsilon_k^2$ as it is now.
– Cesareo
Aug 19 at 15:52
add a comment |Â
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Yes, it is possible to solve this constrained optimization problem using Lagrange multipliers.
– molarmass
Aug 19 at 10:47
@molarmass Can you please indicate how? Setting the differential of the Lagrangian to zero gives a non-linear problem where I do not see how it can be solved.
– phinz
Aug 19 at 11:14