store solutions of a for loop

Hi, this is my M-file named nle.m

function F = nle(x,b1,b2)
F = [(1-x(1))*x(2)*30+(1-x(1))*(1-x(2))*150-b1;
    x(1)*189+(1-x(1))*x(2)*55.5+(1-x(1))*(1-x(2))*70-b2];

b1 and b2 are vectors (e.g. 5x1)

b1 = [24.1230 24.8000 25.4770 26.1540 26.8310]'

b2 = [67.4820 67.7000 67.9170 68.1340 68.3510]'

and the start point

x0 = [1 -1]

I would find the zero for each pair of b1 and b2, then I expect five pair of solutions. I write

for i=1:length(b1)
fun = @(x) nle(x, b1(i), b2(i));
x = fsolve(fun, x0);
end

Appears five time the statement: Optimization terminated: first-order optimality is less than options.TolFun

but in the workspace only the last solution is stored

x = [0.0965 1.0025]

If I write

for i=1:length(b1)
fun = @(x) nle(x, b1(i), b2(i));
x(i) = fsolve(fun, x0);
end

Optimization terminated: first-order optimality is less than options.TolFun.

??? In an assignment A(I) = B, the number of elements in B and I must be the same.

where is the mistake?

Answer 1

[Daniel Shub]

So close. The statement x(i) is expecting a single value, but you are giving it a row. The statement x(i, :) is expecting a row. Replace

x(i) = fsolve(fun, x0);

with

x(i, :) = fsolve(fun, x0);

You also might want to preallocate x by adding

x = zeros(length(b1), 2);

before your loop. This will allocate memory and can speed things up.

[gianluca]

perfect!

thanks for the help

Gianluca

[gianluca]

Can I constrain the solutions between two values (e.g. 0 and 1)?

[Daniel Shub]

You should ask that as a new question.