Question

Question 1: 1. Figure out what the following MATLAB code does (Notice that R n = 500 for jay - 1:4 ifjoy > 1;oldtime = time; end M rand(n); t cputime; forrep -1:5 R chol(A): en time = cputime-t; if jay > 1; ratio - time/oldtime, end en If you put these instructions into a file named, say, zap.m, you can run the program by typing zap from the MATLAB command line. The functions randn, cputime, and chol are built-in MATLAB functions, so you can learn about them by typing help randn, etc. You might find it useful to type more on before typing help topic. Alternatively you can read about theme in MATLAB's help browser. 2. Does the code produce reasonable values of ratio when you run it? 3. What value would you expect in theory? Depending on the speed of the machine on which you are running MATLAB, you may want to adjust the initial value of n or the number of times the rep loop is executed

Answer 1

1. This code runs four times. But most of the calculations happen from the second run, i.e. 3 times. In each run, it creates an nxn matrix M. Then performs multiplication between M and its transpose, and stores the value in A. Then it creates Cholesky factorization for 5 times. Then calculates how much time it took the cpu to perform the calculation. Then on the next run, it doubles the dimension of the matrices performs the entire calculation again, and compares how much more time it took for that iteration. And returns the ratio of new time to old time.

2.

The code provides different values each time, and it seems the difference between between 1st ratio and second ratio is not that much, however, in the last ratio, when n is 8 times it original value, difference is more.

Screenshot: i

Editor DAPrograms MATLAB zap.m Command Window > zap n 500; for jay-1 : 4 ratio - if jay>l:oldtime time:end M = randn (n) ; 4·6667 for rep 1:5 ratio = R chol (A) ; 4。5000 end matrixsize = n; timecputime - t 10 - if jay>l:ratio time/oldtime,end 12 13- ratio - end 7.8571 15 兵

Screenshot: ii

zap.m X > zap n = 500; for jay-1 : 4 ratio = if jay>l:oldtime time:end M = randn (n) ; 4 . 6667 t cputime for rep 1:5 ratio- R chol (A) ; 4.5000 end matrixsize = n; timecputime - t 10 - if jay>l:ratio time/oldtime,end 12 13- ratio = end 7.8571 15 > zap ratio- ratio = 5.5833 ratio = 7.4030 兵

3. I expected the ratio to increase gradually for each iteration, as each time the dimension of the matrices get doubled, that means double or squared the operations, so it should take more than double the time of that of previous time, hence the ratio should be increasing. So the result is somewhat reasonable, but not consistent enough for the first two ratio.

If we increase n (in the following case, n is set to 800, instead of 500), the result is more prominent:

or if number of times the loop rep is executed is increased (in the following case, rep is executing 8 times, instead of 5):

Editor D:Programs MATLAB zap.m Command Window zap.m > zap n = 500; for jay-1:4 ratio = if jay>l:oldtime time:end M randn (n); 5.1429 for rep 1:8 ratio- R chol (A) ; end matrixsize- n; timecputime - t if jay>1;ratio time/oldtime, end 7.3333 10 - 12 13- ratio = end 8.4356 15 > zap ratio = ratio = 6, 4375 ratio = 7.6796