issue with polyval and polyfit measuring time
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Hi!
So I'm working on a code that evaluates how fast a loop can evaluate a varying range of numbers. I put these numbers into an array called x and then the times taken into an array called y. Then I used polyfit and polyval to try to see how long it would take to evaluate numbers between 1 and 1000000000, but when I get my answer, it is very small. I'd expect it to increase like all the others so I think I did something wrong but I am not sure what.
figure(6)
x = [ 1000, 10000, 100000, 1000000]; % represents the tested arrays
y = [ 0.076243, 0.251277, 3.053055, 64.876395]; % represents the time taken
loglog(x,y,'bo--');
title('Time Needed to Calculate the Prime Numbers in an Array');
xlabel('Array length [ - ]');
ylabel('Time taken [ in seconds ]');
grid off;
% d.
xlog = log(x);
tlog = log(y);
coefficients = polyfit(xlog,tlog,1) % which yields [ 0.9874 -9.8979 ]
% Given these coefficients with the polyfit function, the line of best
% fit can be represented by the equation --
% y = 0.9874x - 9.8979 where x is the logarithm of the array length
% and y is the logarithm of the time taken
% e. Polyval may also further be used to find fitted data when given the
% coefficients and the logarithm of 1000000000, which equals 9, as an
% input.
seconds = 10.^polyval(coefficients, 9); % which yields the total seconds
% needed to calculate the logarithm of 1000000000, though this must
% be expressed in hours.
time = seconds./3600 % yields a very small time that doesn't make sense!
3 Comments
Rik
on 11 Mar 2020
Why do you delete your question? It is very rude to do that. And as you see, it isn't very effective either. Please don't give people more work trying to restore the original text.
If you want private consultation: hire a consultant.
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