Adding some changes and code rewriting

[EDIT: 20110628 00:13 CDT - reformat - WDR]

i have code and i would like to make some changes and add 2 conditions for these 3 random variables (c, gama, fi). i want to limit each of them in a specific zone and check the random number which is generated in every loop to not be duplicated also not be out of range.

these are conditions:

• 2.7< fi <16.3
• 450< c <800
• 1.92< gama <1.98

and each of them must be check to not be repeating in every loop.

this is my code :

clear;
clc
B=1000;L=2000;Sc=1.1;Sq=1.1;Sgama=.8;
nsamples=10000;
   for i=1:nsamples
       C=normrnd(620,147.64);
       gama=normrnd(1.96,0.02);
       fi=normrnd(3.76,1.1034);
       Nq=tan((pi/4)+(pi*fi/360))*tan((pi/4)+(pi*fi/360))*2.718^(pi*tan(fi*pi/180));
       Nc=(Nq-1)*cot(fi*pi/180);
       Ngama=2*(Nq+1)*tan(fi*pi/180);
       qult(i)=(C*Nc*Sc)+(384*Nq*Sq)+(980*Ngama*Sgama);
   end

Answer 1

[Walter Roberson]

clc
B=1000; L=2000; Sc=1.1; Sq=1.1; Sgama=.8;
nsamples=10000;

rmem = zeros(3,nsamples);

for i=1:nsamples
  while true
    C = normrnd(620,147.64);
    if C<=450 || C >= 800 || ismember(C,rmem(1,1:K-1)); continue; end
    gama = normrnd(1.96,0.02);
    if gama <= 1.92 || gama >= 1.98 || ismember(gama,rmem(2,1:K-1)); continue; end
    fi = normrnd(3.76,1.1034);
    if fi <= 2.7 || fi >= 16.3 || ismember(fi,rmem(3,1:K-1)); continue; end
    rmem(:,K) = [C; gama; fi];
    break
  end

    Nq = tan((pi/4)+(pi*fi/360)) * tan((pi/4)+(pi*f/360)) * exp(pi*tan(fi*pi/180));
    Nc = (Nq-1)*cot(fi*pi/180);
    Ngama = 2*(Nq+1)*tan(fi*pi/180);
    qult(K) = (C*Nc*Sc)+(384*Nq*Sq)+(980*Ngama*Sgama);
  end

I am not clear as to why you want to check for duplicates: it isn't that duplicates are impossible, but they are relatively rare at 53 bit precision. In the 500000 normrnd(620,147.6) that I generated as a test, the two closest values were a good 2.9899638320785e-11 apart, which was 7 clear bits greater resolution than the largest value generated. I calculate that a 50% chance of a duplicate does not arise until slightly under 112 million samples.

If you eliminate the checks for duplicates or generate a few extra points and discard the duplicates and any extras afterwards, then the code can be made considerably more efficient.

[milad babaei]

Thank you so much.i will appreciate

[milad babaei]

i have run above code and make qult distribution but i dont understand why in workspace window there is only one value for C. SO i couldnt get 1e4 values for C to do sencitivity analysis with C and qult???how can i get distribution for C values??/

[Walter Roberson]

Looking back at the code, it appears I renamed a variable but inconsistently. Please change the line

for i=1:nsamples

to

for K=1:nsamples

At the end of the loops, rmem(1,:) will be all of the C values.

[milad babaei]

THANK yoU SO MUCH....before running the script i did that correction..appreciate your help.

Answer 2

[Krishna Kumar]

To get the random numbers within some limits, you can use randn( since you seem to want random distributed values. x= lowerlimit+(lower limit-upperlimit)*randn(size(array)); For the second question, you must weigh the utility of it against the computational burden it imposes. I suppose randn function does not return exactly equal values so frequently. If at all you want to check the same, you need to store all previous values in an array and check the new value with this each time, which I think does not add value to code. Or you can try some funny ideas like this. random_number= mean(randn(1)+rand(1)). Since both of the random number may not be repeated, chance of repetition are still lower. Or form the random numbers in a single matrix initially,(nsamples,1) sizes vector and check for repetition before entering the loop.

Hope you can fix these things easily.

[Walter Roberson]

The mean() of two random values with the same distribution will have a smaller standard deviation than a single random value of that distribution would have: the values will cluster more to the center.

For example, with six-sided dice, the probability of a "1" being thrown on a single dice is 1/6, but the probability of 1 being the result of the mean of two six-sided dice is only 1 in 36 (both would have to be thrown as 1's for the mean to be 1.)

[Krishna Kumar]

thanks WR for the insight.