adjusting value in the function........

This is my initial:

function V = fitnessFcn(Q)

deltad = -24.268; X=0.202;

V = sqrt ((Q(30)*X)/(2*(1-cos(deltad/2))));

Answer: V = fitnessFcn(rand(1,30))

V =

    0.4335

>> V = fitnessFcn(rand(1,30))

V =

    0.4956

>> V = fitnessFcn(rand(1,30))

V =

    0.7894

>> V = fitnessFcn(rand(1,30))

V =

    0.6970

>> V = fitnessFcn(rand(1,30))

V =

    0.5132

>> V = fitnessFcn(rand(1,30))

V =

    0.7035

>> V = fitnessFcn(rand(1,30))

V =

    0.8027

>> V = fitnessFcn(rand(1,30))

V =

    0.7466

>> V = fitnessFcn(rand(1,30))

V =

    0.8897

>> V = fitnessFcn(rand(1,30))

V =

    0.7190

>> V = fitnessFcn(rand(1,30))

V =

    0.8302

>> V = fitnessFcn(rand(1,30))

V =

    0.4334

>> V = fitnessFcn(rand(1,30))

V =

    0.3439

>> V = fitnessFcn(rand(1,30))

V =

    0.8189

>> V = fitnessFcn(rand(1,30))

V =

    0.6515

>> V = fitnessFcn(rand(1,30))

V =

    0.8410

>> V = fitnessFcn(rand(1,30))

V =

    0.8377

>> V = fitnessFcn(rand(1,30))

V =

    0.7555

>> V = fitnessFcn(rand(1,30))

V =

    0.4869

>> V = fitnessFcn(rand(1,30))

V =

    0.2456

>> V = fitnessFcn(rand(1,30))

V =

    1.0383

So when I run the program, the answer for V will get a random number until I get the value in range 0.95<V<1.05.

After editing:

function V = fitnessFcn(Q)

deltad = -24.268; X=0.202;

V = sqrt ((Q(30)*X)/(2*(1-cos(deltad/2))));

if (V <= 1.05 V >= 0.95) true else false end

Answer:

V = fitnessFcn(rand(1,30))

ans =

     1

V =

    0.6990 (this value still not in range of 0.95 until 1.05)

How about if I want the answer between 0.95<V<1.05 automatically means that it will give the answer correct based on the range (0.95<V<1.05). Can anyone help me on this??

Answer 1

[Aleksander]

just put a for loop around your while statement for example (and this is quick and dirty - might want to make it more efficient and 'nice':
b=[];
for i=1:30
while
statement
end
V=[V b]
end

[Muhammad Sulhi]

thank you from your responding...I've try this...it seem to be works..but the value for all 30 values are the same...am i right..??how can I make it become another value not just fix for one value but still in the range of 0.95 until 1.05....????

[Aleksander]

I get all different values so that sounds strange - after all you are generating random numbers and there is no reason they should produce the same value. What version of Matlab do you have? It might be a problem with your random seed, that is, how Matlab generates a random number.

In older versions (before R2008b I think), if you generate some random numbers in a program, and I run the same program as you, we would get identical results due to the manner in which Matlab generated the random numbers. Newer versions use a different algorithm to avoid this (although, for training purposes it might be useful to use the old algorithm if you want to compare results when solving a problem).

Anyway, try this:

Create a number stream:

1)

s = RandStream('mt19937ar')

Make it the default stream:

2)

RandStream.setDefaultStream(s)

I don't remember off the top of my head how to see what your current seed is but I'm guessing someting like getDefaultStrem or RandStream.getDefaultstream

[Aleksander]

Just to be clear, simply copy the lines directly and run them.

You don't need to 'create' anything yourself per se. That is, don't let the 'mt19937ar' statement confuse you. It is just the name Matlab uses for that particular random number generator.

[Muhammad Sulhi]

sorry...I use R2009a version...this is my coding to get that value...

function V = fitnessFcn(Q)

deltad = -24.268;

X = 0.202;

V = sqrt ((Q(30)*X)/(2*(1-cos(deltad/2))));

b=[];

for i=1:30

while (0.95 > V) || (V > 1.05)

Q = rand(1,30);

deltad = -24.268;

X=0.202;

V = sqrt ((Q(30)*X)/(2*(1-cos(deltad/2))));

end

V=[V b]

end

end

Is it right..??Can you check it for me...from this coding, I get 30 value but give the same answer...can you help me on this matter..??

[Muhammad Sulhi]

It does not give different value of V...

Answer 2

[Jan]

This is a strange line:

if (V <= 1.05 V >= 0.95) true else false end

What does it mean? Perhaps you want this:

if (0.95 <= V) && (V <= 1.05)
  reply = true;
else
  reply = false;
end

Or shorter:

reply = (0.95 <= V) && (V <= 1.05);

[Muhammad Sulhi]

I have try from yours..

1)

function V = fitnessFcn(Q)

deltad = -24.268;

X=0.202;

V = sqrt ((Q(30)*X)/(2*(1-cos(deltad/2))));

if (0.95 <= V) && (V <= 1.05)

reply = true;

else

reply = false;

end;

answer:

V = fitnessFcn(rand(1,30))

V =

0.8027

2)

function V = fitnessFcn(Q)

deltad = -24.268;

X=0.202;

V = sqrt ((Q(30)*X)/(2*(1-cos(deltad/2))));

reply = (0.95 <= V) && (V <= 1.05);

answer:

V = fitnessFcn(rand(1,30))

V =

0.7466

both method I've try by replacing it..but the answer still the same..what i meant is..when i run the program, it will give the value exactly in the range of 0.95 until 1.05...I've try from your answer...but the value still not change..it still read the value out of the range..help me please....

Answer 3

[Aleksander]

From what I can tell, there's nothing in your statement that tells your program to continue executing until the condition 0.95<=V<=1.05 is true. I'm not sure what you're trying to do, but look copy and run this code for example. It will always produce a number within your range. you can "beautify" it but the point is that there is something in there that changes the value until the condition is true:

       Q = rand(1,30);
deltad = -24.268; 
X=0.202;

V = sqrt ((Q(30)*X)/(2*(1-cos(deltad/2))));
while (0.95 > V) || (V > 1.05)

      Q = rand(1,30);
deltad = -24.268; 
X=0.202;

V = sqrt ((Q(30)*X)/(2*(1-cos(deltad/2))));

end
V

[Muhammad Sulhi]

thank you from your responding..I had try edit from my coding by combination from yours and me and it works...from the coding,i just produce one value of V..my next question is...how can I produce many value of V..For example I want to produce V1 until V30...but the value V1 until V30 still in the range of 0.95 until 1.05...can anyone help me on this...??