# Steepest Descent Method with Stopping Criteria using MATLAB

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Sayed Naseem Sadaat on 2 Apr 2022
I have the following quadratic function.
f(x1 ,x2 ) = 5x1^2 + x2^2 + 4x1x2 −14x1 − 6x2 + 20
I want to minimize it starting points X(0) = [x1^0, x2^0] = [0,10] and the stopping criteria (tolerance) = 10^-6 using steepest descent method. I need help.

Riccardo Scorretti on 2 Apr 2022
Edited: Riccardo Scorretti on 2 Apr 2022
Hi. I don't know it steepest descend is part of any of the Matlab toolbox (anyway, you can find a version here: https://fr.mathworks.com/matlabcentral/fileexchange/22532-steepest-descent-algorithm), but you can use the standard method fminsearch, which in this case works quite well, because the problem is very simple:
% Setup the problem
X0 = [0 10];
fun = @(X) 5*X(:,1).^2 + X(:,2).^2 + 4*X(:,1).*X(:,2) - 14*X(:,1) - 6*X(:,2) + 20;
% Perform the optimization (indeed not by using the steepest descent algorithm)
options = optimset('Display','iter', 'TolX', 1E-6);
[X,FVAL,EXITFLAG] = fminsearch(fun, X0)
X = 1×2
1.0000 1.0000
FVAL = 10.0000
EXITFLAG = 1
Let's check if the result is reasonable:
[x, y] = meshgrid(-2:0.1:3, -2:0.1:3);
f = reshape(fun([x(:) y(:)]), size(x));
figure
mesh(x, y, f) ; hold on ; plot3(X(1), X(2), FVAL, 'r*') ; view(30, 60); To be 100% sure, let's do some analytical computing:
syms x1 x2
f = 5*x1^2 + x2^2 + 4*x1*x2 - 14*x1 - 6*x2 + 20;
jac = jacobian(f)
jac = det([10 4 ; 4 2])
ans = 4.0000
The jacobian matrix is constant, and it is not singular: hence it exists only one solution:
solve(jac)
ans = struct with fields:
x1: 1 x2: 1

### More Answers (1)

Sayed Naseem Sadaat on 2 Apr 2022
Edited: Sayed Naseem Sadaat on 2 Apr 2022
Hello Ricardo,
Thank you very much for the useful information and sharing the relevant thread.