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

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Muhammad Sulhi on 28 Mar 2011
This is my initial:
function V = fitnessFcn(Q)
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)
if (V <= 1.05 V >= 0.95) true else false end
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??

Aleksander on 29 Mar 2011
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 on 2 Apr 2011
It does not give different value of V...

Jan on 28 Mar 2011
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)
else
end
Or shorter:
reply = (0.95 <= V) && (V <= 1.05);
Muhammad Sulhi on 28 Mar 2011
I have try from yours..
1)
function V = fitnessFcn(Q)
X=0.202;
if (0.95 <= V) && (V <= 1.05)
else
end;
V = fitnessFcn(rand(1,30))
V =
0.8027
2)
function V = fitnessFcn(Q)
X=0.202;
reply = (0.95 <= V) && (V <= 1.05);
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....

Aleksander on 28 Mar 2011
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);
X=0.202;
while (0.95 > V) || (V > 1.05)
Q = rand(1,30);
X=0.202;
end
V
Muhammad Sulhi on 29 Mar 2011
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...??