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Unable to find explicit solution in Lagrangian optimization
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I am trying to find the analytical solution to the following problem:
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/1614056/image.png)
I tried solving it by coding the Lagrangian by hand and use solve, but Matlab prints the warning: "Unable to find explicit solution".
I used the following code:
syms e1 e2 p1 p2 rho gamma lambda
syms E H(e1,e2)
H(e1,e2) = (e1^rho +e2^rho)^(1/rho)
L(e1, e2, lambda) = p1*e1 +p2*e2 + lambda*(H(e1,e2)-E)
L_e1 = diff(L,e1) == 0
L_e2 = diff(L,e2) == 0
L_lambda = diff(L,lambda) == 0
system = [L_e1,L_e2,L_lambda]
[e1_s,e2_s,lambda_s]=solve(system,[e1 e2 lambda])
Do you know what I could do to solve this? Or is there a different and better way to find an analytical solution?
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채택된 답변
Catalytic
2024년 2월 11일
편집: Catalytic
2024년 2월 11일
An analytical solution for 0<rho<1 is -
A=[1 0;
0 1;
-1 0;
0 -1]*E;
[fval,i]=min(A*[p1;p2]);
e1=A(i,1);
e2=A(i,2);
댓글 수: 2
Catalytic
2024년 2월 11일
편집: Catalytic
2024년 2월 11일
You can see this graphically by plotting the constrained region. The region always has extreme points at (
), so that's where the optimum must lie.
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/1614541/image.png)
E=1;
for rho=[0.1:0.2:0.9]
fimplicit(@(e1,e2) abs(e1).^rho + abs(e2).^rho - E.^rho, [-1.5,1.5]); hold on
end
Matt J
2024년 2월 11일
I like it. And, in fact, because the extreme points lie at points where H(e1,e2) is not differentiable, it shows that you will never find the true solution with Lagrange multiplier analysis.
추가 답변 (1개)
Matt J
2024년 2월 11일
If you make rho explicit, it seems to be able to find solutions. I doubt there would be a closed-form solution for general rho.
rho=2;
syms e1 e2 p1 p2 gamma lambda
syms E H(e1,e2)
H(e1,e2) = (e1^rho +e2^rho)
L(e1, e2, lambda) = p1*e1 +p2*e2 + lambda*(H(e1,e2)-E^rho)
L_e1 = diff(L,e1) == 0
L_e2 = diff(L,e2) == 0
L_lambda = diff(L,lambda) == 0
system = [L_e1,L_e2,L_lambda]
[e1_s,e2_s,lambda_s]=solve(system,[e1 e2 lambda])
댓글 수: 4
Matt J
2024년 2월 11일
Even when it can be explicitly solved, the result isn't nice:
rho=sym(1/4);
syms e1 e2 p1 p2 gamma lambda
syms H(e1,e2)
H(e1,e2) = (e1^rho +e2^rho);
L(e1, e2, lambda) = p1*e1 +p2*e2 + lambda*(H(e1,e2)-1);
L_e1 = diff(L,e1) == 0;
L_e2 = diff(L,e2) == 0;
L_lambda = diff(L,lambda) == 0;
system = [L_e1,L_e2,L_lambda];
[e1_s,e2_s,lambda_s]=solve(system,[e1 e2 lambda])
Walter Roberson
2024년 2월 11일
You can eliminate the root() constructs, but the result is confusing.
rho=sym(1/4);
syms e1 e2 p1 p2 gamma lambda
syms H(e1,e2)
H(e1,e2) = (e1^rho +e2^rho);
L(e1, e2, lambda) = p1*e1 +p2*e2 + lambda*(H(e1,e2)-1);
L_e1 = diff(L,e1) == 0;
L_e2 = diff(L,e2) == 0;
L_lambda = diff(L,lambda) == 0;
system = [L_e1,L_e2,L_lambda];
[e1_s,e2_s,lambda_s]=solve(system,[e1 e2 lambda], 'maxdegree', 3)
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