help in this plz

I have Values like this
C2 = 0.10;
C3 = 0.10;
C4 = 0.20;
C5 = 0.20;
C6 = 0.20;
P1=190;
P2=72.2;
P3=51.6;
P4=20.44;
P5=15.57;
P6= 4.956;
K1=P1/P;
K2=P2/P;
K3=P3/P;
K4=P4/P;
K5=P5/P;
K6=P6/P;
note :- this is very lengthy and I want to make it shorter and calculating like this
for c = [0.20,0.10,0.10,0.20,0.20,0.20]
for P = [190,72.2,51.6,20.44,15.57,4.956]
PG = 50;
K = (P/PG);
A = c*(K-1);
B = c*(K-1)/K;
nv = A/(A-B);
disp(nv)
end
end
Can I use these code to calculate it in a simpler way or I am need to change the code , suggest plz
thanks

1 comentario

Image Analyst
Image Analyst el 19 de Feb. de 2012
Inside your loop, the c's cancel out in the nv division, so nv, K, and PG don't depend on c at all so why are they in a loop over c? Also, in the first chunk of code, I assume P1 = P(1), so what is P? You can't do K(1) = P(1) ./ P because the dimensions don't match so I have no idea what you want to do.

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Respuestas (2)

Walter Roberson
Walter Roberson el 19 de Feb. de 2012

0 votos

Yes, you can use loops such as you show. Your "for" syntax is valid.
You might want to store the answers after you calculate them, though.

3 comentarios

Nasir Qazi
Nasir Qazi el 19 de Feb. de 2012
but it wont let me to calculate 'nv', try it yourself plz
Walter Roberson
Walter Roberson el 19 de Feb. de 2012
Unfortunately I do not have access to a MATLAB license on weekends.
Jan
Jan el 19 de Feb. de 2012
The code runs fine and "nv" is calculated - but only the last value is stored. Because "nv" does not depend on "c" at all, the outer loop can be omitted.

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Jan
Jan el 19 de Feb. de 2012

0 votos

You nv does not depend on c, such that the calculation of nv can be simplified to:
nv = K / (K-1);
or:
nv = 1 / (1 - PG/P);

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