Same code but different results

조회 수: 4 (최근 30일)
Alessandro Gottardi
Alessandro Gottardi 2020년 12월 24일
답변: Walter Roberson 2020년 12월 24일
Using the same code, I get different results than those obtained on a different pc and version of Matlab (the versions involved are R2020b and R2012b). Does anyone have any idea why this is happening?
  댓글 수: 2
KALYAN ACHARJYA
KALYAN ACHARJYA 2020년 12월 24일
Code?
Alessandro Gottardi
Alessandro Gottardi 2020년 12월 24일
편집: Alessandro Gottardi 2020년 12월 24일
clear all
close all
n = 20000;
maxiter = 100000;
epsilon = 1.0e-10;
A = sparse(-2*eye(n)+diag(ones(n-1,1),1)+diag(ones(n-1,1),-1));
y = linspace(1,5,n)';
b = A*y;
tcpu = cputime;
[x, z, k] = coniugato(-A,-b,epsilon,maxiter);
tcpu = cputime - tcpu;
if k==maxiter
disp('Attenzione: raggiunto il numero massimo di iterazioni');
end
if isempty(x); return; end
r = A*x-b;
fprintf('Dimensione %d, Iter %d, tcpu = %f, err = %e \n', n,k,tcpu,z(k))
and the function coniugato
function [x, z, k] = coniugato(A,b,epsilon,maxiter,x)
n = length(b);
[k,l] = size(A);
if k~=l || n~=k
x = [];
fprintf('Dimensioni di A e/o b non corrette \n');
return;
end
z = zeros(maxiter,1);
if nargin<5; x=zeros(n,1); end;
r = A*x-b;
p = r;
rr = r'*r;
for k=1:maxiter;
Ap = A*p;
pAp = p'*Ap;
alfa = rr/pAp;
x = x - alfa * p;
r = r - alfa * Ap;
rrn = r'*r;
z(k) = sqrt(rrn);
if z(k)<epsilon; break; end
beta = rrn/rr;
p = r + beta * p;
rr = rrn;
end
fprintf('Gradiente coniugato: ');
end

댓글을 달려면 로그인하십시오.

채택된 답변

Walter Roberson
Walter Roberson 2020년 12월 24일
It looks to me as if you are using mrdivide ( / ) a couple of times in your code. The / operator has been improved to be able to select more efficient algorithms in some cases, and the underlying high performance math libraries have been upgraded at least twice since 2012 (my memory is saying three times for Windows, but that would have to be confirmed.)
A few years ago, the upgrades to the underlying math libraries started causing error messages in code that had run before. What had happened was that more accurate algorithms were used, but the more accurate algorithms were also more likely to detect that the problem was nearly singular — that really the old libraries should have warned about singularity too but had been a little sloppy and had permitted calculations that were not numerically meaningful.

추가 답변 (0개)

카테고리

Help CenterFile Exchange에서 Signal Integrity Kits for Industry Standards에 대해 자세히 알아보기

태그

Community Treasure Hunt

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

Start Hunting!

Translated by