Accumulator Error Feedback
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(Difference between revisions)
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% See also SUM | % See also SUM | ||
% | % | ||
- | % | + | % clear all |
- | % | + | % % v=sort(randn(13e6,1),'descend'); %better when sorted |
- | % | + | % v=randn(13e6,1); |
- | % | + | % rsumv = abs(sum(v) - sum(v(end:-1:1))); |
- | % | + | % disp(['rsumv = ' num2str(rsumv,'%18.16f')]); |
+ | % csumv = abs(csum(v) - csum(v(end:-1:1))); | ||
+ | % disp(['csumv = ' num2str(csumv,'%18.16f')]); | ||
+ | % % vsumv = sum(vpa(v)) - sum(vpa(v(end:-1:1))); %vpa toolbox 32GB RAM | ||
+ | % % disp(['vsumv = ' char(vsumv)]) | ||
s_hat=0; e=0; | s_hat=0; e=0; | ||
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=== links === | === links === | ||
- | [http://servidor.demec.ufpr.br/CFD/bibliografia/Higham_2002_Accuracy%20and%20Stability%20of%20Numerical%20Algorithms.pdf Accuracy and Stability of Numerical Algorithms, ch.4.3, Nicholas J. Higham, | + | [http://servidor.demec.ufpr.br/CFD/bibliografia/Higham_2002_Accuracy%20and%20Stability%20of%20Numerical%20Algorithms.pdf Accuracy and Stability of Numerical Algorithms 2e, ch.4.3, Nicholas J. Higham, 2002] |
For multiplier error feedback, see: | For multiplier error feedback, see: |
Revision as of 19:20, 25 September 2017
function s_hat=csum(x) % CSUM Sum of elements using a compensated summation algorithm. % % For large vectors, the native sum command in Matlab does % not appear to use a compensated summation algorithm which % can cause significant roundoff errors. % % This code implements a variant of Kahan's compensated % summation algorithm which often takes about twice as long, % but produces more accurate sums when the number of % elements is large. -David Gleich % % See also SUM % % clear all % % v=sort(randn(13e6,1),'descend'); %better when sorted % v=randn(13e6,1); % rsumv = abs(sum(v) - sum(v(end:-1:1))); % disp(['rsumv = ' num2str(rsumv,'%18.16f')]); % csumv = abs(csum(v) - csum(v(end:-1:1))); % disp(['csumv = ' num2str(csumv,'%18.16f')]); % % vsumv = sum(vpa(v)) - sum(vpa(v(end:-1:1))); %vpa toolbox 32GB RAM % % disp(['vsumv = ' char(vsumv)]) s_hat=0; e=0; for i=1:numel(x) s_hat_old = s_hat; y = x(i) + e; s_hat = s_hat_old + y; e = (s_hat_old - s_hat) + y; %calculate difference first (Higham) end
links
Accuracy and Stability of Numerical Algorithms 2e, ch.4.3, Nicholas J. Higham, 2002
For multiplier error feedback, see:
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