Store result from for-loop and plot

6 Ansichten (letzte 30 Tage)
charlotte88
charlotte88 am 7 Mär. 2016
Bearbeitet: Torsten am 7 Mär. 2016
Hi,
I have a maximization problem which I want to solve in matlab. The problem consists of several parameters, but I am mostly interested in what happens to x when alpha increases from 0 to 0.5 with increments of 0.05. I then want to plot the value of x in a figure with x on the y-axis and alpha on the x-axis
I have defined the function in one file:
function b = maximizationpb2(x, alpha)
beta1 = 6;
beta2 = 6;
s1 = 1/3;
s2 = 1/3;
s3 = 1/3;
alpha = 0.5;
b = -(x(1) - beta1*((x(2)-s2)^2+(1-x(1)-x(2)-s3)^2))^(1-alpha) * (x(2) - beta2*((x(1)-s1)^2+(1-x(1)-x(2)-s3)^2))^alpha;
And then I have used fmincon to solve the problem:
lb = [0,0];
ub = [1,1];
A = [1,1];
b = [1];
Aeq = [];
beq = [];
x0 = [1/3, 1/3];
a = fmincon(@maximizationpb2,x0,A,b,Aeq,beq,lb,ub);
x3 = 1 - a(1) - a(2);
My first difficulty is to make a for loop which goes through all values of alpha = 0:0.05:0.5. I have to admit I have no clue where to start (I was so happy last week when I finally managed to solve the problem!!). Any help would be very much appreicated!
  1 Kommentar
Stephen23
Stephen23 am 7 Mär. 2016
You have alpha as a function input, but never use it. YOu need to change your function and learn to test your code:
function out = test
lb = [0,0];
ub = [1,1];
A = [1,1];
b = [1];
Aeq = [];
beq = [];
x0 = [1,1]/3;
s = [1,1,1]/3;
beta = [6,6];
alpha = 0:0.05:0.5;
out = NaN(numel(alpha),1);
for k = 1:numel(alpha)
a = fmincon(@(x)fun(x,alpha(k),beta,s),x0,A,b,Aeq,beq,lb,ub);
out(k) = 1 - a(1) - a(2);
end
end
%--------------------------------------------------------------------------
function b = fun(x, alpha, beta, s)
b = -(x(1) - beta(1)*((x(2)-s(2))^2+(1-x(1)-x(2)-s(3))^2))^(1-alpha) * (x(2) - beta(2)*((x(1)-s(1))^2+(1-x(1)-x(2)-s(3))^2))^alpha;
end
%--------------------------------------------------------------------------

Melden Sie sich an, um zu kommentieren.

Akzeptierte Antwort

Torsten
Torsten am 7 Mär. 2016
Bearbeitet: Torsten am 7 Mär. 2016
function driver
lb = [0,0];
ub = [1,1];
A = [1,1];
b = [1];
Aeq = [];
beq = [];
x0 = [1/3, 1/3];
for k=1:11
alpha(k)=0.05*(k-1);
a = fmincon(@(x)maximizationpb2(x,alpha(k)),x0,A,b,Aeq,beq,lb,ub);
M(k,1)=a(1);
M(k,2)=a(2);
M(k,3)= 1 - a(1) - a(2);
end
plot(alpha,M(:,1)) % plot a(1) over alpha
function b = maximizationpb2(x, alpha)
beta1 = 6;
beta2 = 6;
s1 = 1/3;
s2 = 1/3;
s3 = 1/3;
b = -(x(1) - beta1*((x(2)-s2)^2+(1-x(1)-x(2)-s3)^2))^(1-alpha) * (x(2) - beta2*((x(1)-s1)^2+(1-x(1)-x(2)-s3)^2))^alpha;
Best wishes
Torsten.

Weitere Antworten (0)

Kategorien

Mehr zu Numerical Integration and Differential Equations finden Sie in Help Center und File Exchange

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by