Method other than for loop?

My code is quite long and outputs two values called beam1 and beam2. The values of these variables are dependant on another variable labeled length.

What I want my code to do is output a length value that allows beam1-beam2<0.0001. As of now I am using a for loop to cycle through a range of different length values to output the ones that fit my criteria. However this method takes an extremely long time to run.

Do you have any suggestions that I could use instead of a for loop that might help the program take less time to run?

[Sean de Wolski]

Don't name your variable 'length' as that is a very useful MATLAB function that you don't want to have overwritten.

[Sean de Wolski]

Other than that we're probably going to have to see the code (at least the relevant portions) to help.

Answer 1

[Walter Roberson]

You could possibly use something like fmincon to minimize on (beam1-beam2).^2 with a tolerance of 0.0001^2 . Ummm -- can your code come out with beam2 > beam1 ?

Does your computation have a lot of local minima? If so then a more powerful minimizer might be needed.

[Derek]

I was told to treat the majority of the code as a black box. the functions that determine beam1 and beam2 are,

Psb(N-x)=Psb(N-x+1)+gg1;

Psf(N-x)=Psf(N-x+1)+g1;

Psf(1) is beam1 in my output, and Psb(1) is beam2.

How would I go about using those functions with fmincon? Everything I have found online describing fmincon is very confusing. Thanks again!

[Andrew Newell]

By itself your code doesn't determine beam1 or beam2. It is a recursive definition for Psb and Psf. Normally such definitions start with f(i) = f0 for i=1 and then work upwards to (say) i=N. You seem to be working downwards from some value. What is that value?

[Walter Roberson]

If those are definitions for Psb and Psf, then Psb and Psf are linear. You can for example rewrite as

Psb(N-x+1) = Psb(N-x)-gg1

which makes it clear that Psb decreases by gg1 for each increase of 1 in its argument, and so Psb(s) = c(N) - gg1 * x for some constant c dependent upon N.

[Derek]

I am working downwards from N with Psb(N)=Psf(N)=100e-6

[Andrew Newell]

So what are you trying to vary to optimize |beam1-beam2|?

[Walter Roberson]

Are gg1 and g1 known constants? Are they variables to be determined but which are not not dependent on N or x ? Or are they a series of values indexed by (N-x) ? Or ... ?

[Andrew Newell]

To sum up the questions, you need a function that inputs x, or whatever your variable is, and outputs |beam1-beam2|.

[Derek]

Yes that is correct. I need to find a "length" value which causes |Psf(1)-Psb(1)|<0.0001.

Currently I am looping through different length values,

for length=1.7:0.01:1.9

and have an if statement telling the program to output the proper length that satisfys the condition as well as outputting Psb(1) and Psf(1).

My loop is outputting length=1.86 (which i already know is correct), I just need a faster way to do this.

Also, I keep getting the error,

"??? Attempted to access Pb1(1880); index must be a positive integer or logical.

Error in ==> cutandpaste2 at 143

Pb1(N)=5.6e-3;"

when I run the loop. It is not always at Pb1(1880) and changes depending on where I begin the loop. I have tried doing a long format (no help). I still get the optimal length with this error but no output for Psb and Psf.

finally for Walter,

g1=nSf*h*(Sesf*n2(N-x+1)-Sasf*n1(N-x+1)-alphaSf/(nSf*Nt))*Psf(N-x+1)*Nt;

gg1=nSb*h*(Sesb*n2(N-x+1)-Sasb*n1(N-x+1)-alphaSb/(nSb*Nt))*Psb(N-x+1)*Nt;

[Walter Roberson]

In Pb1(N)=5.6e-3; your N has been calculated as an expression involving non-integers. The calculated value is very close to 1880 (or whatever) but is not *exactly* an integer. You should avoid such calculations. Meanwhile, you can work-around using round()

Answer 2

[Andrew Newell]

Your problem sounds like an optimization problem. Using the Optimization Toolbox to find the length that minimizes beam1-beam2 would probably be the best way to speed up your code.

[Derek]

Sorry for the delay, it was the weekend and I just got back to the problem today. I am unable to use the optimization toolbox because I am student. I am looking into other ways of aquiring it.

[Walter Roberson]

The Student Edition includes the Optimization Toolbox.

http://www.mathworks.com/academia/student_version/details.html

[Derek]

whoops didn't notice that