How can I go back and resolve failed attempts?

Hi everyone, so I have the following code:

clc;
clear;

% Import data
% PD: probability of default
% CM: covariance matrix
% DT: default threshold

PD = xlsread('Data_CIMDO.xlsx','PD');
CM = xlsread('Data_CIMDO.xlsx','COV');
DT = xlsread('Data_CIMDO.xlsx','DT');

Original_PD = PD;                       %Store original PD
LM_rows = 11;                           %Expected LM rows
LM_columns = length(PD)                 %Expected LM columns
LM_FINAL = zeros(LM_rows,LM_columns);    %Dimensions of LM_FINAL

for i = 1:length(PD)
    PD = Original_PD(:,i);
    options = optimset('Display','iter');
    x0 = rand(size(PD,1)+1,1);
    [LM,fval,exitflag] = fsolve(@(x)ConstLM(x,PD,CM,DT), x0, options);
    LM_FINAL(:,i) = LM;
end

Now since the code depends on the initial value (x0) when solving for LM, after one run of the code there are many unsolved values for LM as the initial x0 was incorrectly guessed. So how can I adjust the code such that it keeps running until all LM's have been solved?

Thanks.

Answer 1

[Jan]

...
for k = 1:100
  [LM, fval, exitflag] = fsolve(@(x)ConstLM(x,PD,CM,DT), x0, options);
  if exitflag == 1
    break;
  end
end
if exitflag ~= 1
  warning('FSOLVE did not find a solution.');
end
...

I'd prefer such a loop with a maximum loop counter to guarantee that the function stops in finite time.

[sittmo]

Thanks Jan Simon for your assistance.

As I am still quite a beginner with MATLAB, please excuse any silly questions.

For the "for k = 1:100" loop, is that basically running the following line of code:

"[LM, fval, exitflag] = fsolve(@(x)ConstLM(x,PD,CM,DT), x0, options)"

...100 times and stops if exitflag is equal to 1?

If after 100 cycles and exitflag is not 1, then it exits the loop and goes to the next equation to solve?

Also does the code you provided ensure that each time we attempt a re-solve, x0 changes to a new value?

Thank you again.

[Walter Roberson]

You can put the

x0 = rand(size(PD,1)+1,1);

before the fsolve() call to use a new starting point each time.