getting actual numbers instead of symbolic results when using solve

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);

 Accepted Answer

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 Comments

Also
double(w3_star)
Will return the numeric value a double
thank you so much, I was sure I was close, but I couldn't figure out how to get the actual results from there.
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!
thanks, you help me solve the question

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Asked:

Max
on 6 Apr 2015

Commented:

on 27 Sep 2021

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