# Pcg method is imprecise

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Fedor2023-3-7

I must be wrong in my code, so feel free to note where I've mistaken. I'm testing pcg method's precision through following scheme: generate random positive definitive matrix, and let it be linear system with the desired solution as (1, 1, 1, 1...) - to simplify testing. We can solve that linear system with division or simply state that solution should be (1, 1, 1, 1...). Then we solve system again with pcg method and test that all xi are nearby 1. And it doesn't work - pcg's solution does not meet needed tolerance(precision). Maybe I understand the meaning of tolerance incorrectly?
Code:
tol = 1e-8;
iterations = 150;
matrix_1_size = 50;
matrix_1 = sprand(matrix_1_size,matrix_1_size,.5);
matrix_1 = matrix_1'*matrix_1;
B_1 = sum(matrix_1, 2);
expected_solution_1 = matrix_1 \ B_1;
pcg_solution_1 = pcg(matrix_1, B_1, tol, iterations);
diff = abs(pcg_solution_1 - expected_solution_1);
tolerance = zeros(matrix_1_size, 1) + tol;
assert(all(diff <= tolerance));

### 采纳的回答

Chunru 2023-3-8
The pcg method use tol as a relative tolerance such that norm(residual) < tol*norm(rhs). So you need a different assert.
tol = 1e-8;
iterations = 150;
matrix_1_size = 50;
matrix_1 = sprand(matrix_1_size,matrix_1_size,.5);
matrix_1 = matrix_1'*matrix_1;
B_1 = sum(matrix_1, 2);
expected_solution_1 = matrix_1 \ B_1;
pcg_solution_1 = pcg(matrix_1, B_1, tol, iterations);
pcg converged at iteration 74 to a solution with relative residual 9.8e-09.
diff = abs(pcg_solution_1 - expected_solution_1);
norm(diff)
ans = 1.3296e-06
norm(pcg_solution_1-expected_solution_1)
ans = 1.3296e-06
norm(B_1)
ans = 772.9681
norm(pcg_solution_1-expected_solution_1)/norm(B_1) < tol
ans = logical
1
assert(norm(pcg_solution_1-expected_solution_1)/norm(B_1) < tol)
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Fedor 2023-3-11
Thanks, that is the solution!

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