getting actual numbers instead of symbolic results when using solve

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Dear all,
I am stuck with a problem where I am trying to find a minimum-variance-portfolio, with the budget constraint, hence using the Lagrangian.
I have defined my variance-covariance-matrix as follows. The four unknown variables are the three weights and lambda. The Lagrangian is defined, and I want to use "solve" on the four equations below to get the four results. Since the variance-covariance matrix contains actual numbers, I should get actual numbers (i.e. asset weights) after putting the four equations with four unknowns into "solve", but it does not work at all and I don't know why.
Perhaps one can help.
Sigma = [0.04,0.023,0.045;0.023,0.03,0.02;0.045,0.02,0.06];
syms w1 w2 w3 lambda ;
risk = [w1,w2,w3]*Sigma*[w1;w2;w3];
Lagrangian = risk + lambda*(w1+w2+w3-1);
vals = solve(diff(Lagrangian,w1)==0,diff(Lagrangian,w2)==0,diff(Lagrangian,w3)==0,diff(Lagrangian,lambda)==0);

Respuesta aceptada

A Jenkins
A Jenkins el 6 de Abr. de 2015
Why do you say it doesn't work? I see your answers returned as a structure:
>> vals
vals =
lambda: [1x1 sym]
w1: [1x1 sym]
w2: [1x1 sym]
w3: [1x1 sym]
If you want to access each one to see the value, you can do the following:
>> vals.w1
ans =
15/88
If you don't like structures, have it return all your answers separately:
>> [lambda_star, w1_star, w2_star, w3_star]=solve(diff(Lagrangian,w1)==0,diff(Lagrangian,w2)==0,diff(Lagrangian,w3)==0,diff(Lagrangian,lambda)==0)
lambda_star =
-491/8800
w1_star =
15/88
w2_star =
65/88
w3_star =
1/11
(Also useful is the vpa() command if you want to evaluate the fraction to a decimal representation.)
  4 comentarios
Max
Max el 7 de Abr. de 2015
now I am ashamed to open another question for this issue, but if I want to plot the portfolio risk "risk" (as above) as a function of w1 and w2 (and setting w3=1-w1-w2), what do I need to do? I.e. I want the two weights on the lower axis and the results total risk on the third axis, to get a nice colored surface with the minimum visible.
Do I need to define the "risk" function differently?
plot (w1,w2,risk) and plot3(w1,w2,risk) do not work!
tuan nguyen minh
tuan nguyen minh el 27 de Sept. de 2021
thanks, you help me solve the question

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