# how to identify the mistakes in loop in the given following code?

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M.Rameswari Sudha on 10 Oct 2022
Commented: Jan on 10 Oct 2022
I have to find q, n value and TC. we need to find the these three values in iteration way. How to solve it?
clc
clear all
pr=7;
p=2500;
D=1600;
z=4;
zr=4;
m=10;
hr=5;
h1=12;
h2=11;
h3=13;
cL=4;
c=20;
c0=2;
c1=3;
c2=2;
ce=15;
k=0.1;
alpha=0.1;
beta=0.9;
u=1-alpha+alpha.*beta;
s=250;
A=110;
g0=15;
g1=20;
theta1=1;
theta2=0.6;
w0=0.3;
w1=0.2;
TC1=inf
TC1 = Inf
q=1:10:191
q = 1×20
1 11 21 31 41 51 61 71 81 91 101 111 121 131 141 151 161 171 181 191
n=1:1:20
n = 1×20
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
j=1;
l=1;
for i=1:length(q)
n=sqrt((D.*zr.*m+s+c.*g0+(A.*((q(i).*h2)/2))))/(q(i).*(((hr+cL.*k).*((D.*q(i))/(m.*p)))+((h1.*q(i)).*((1/2)-(((2.*u-1).*D)/(2.*p.*u))-D))+(D.*ce.*g1.*theta1)+(ce.*((w1.*(1-(((2.*u-1).*D)./(2.*D.*u))).*q(i))-(q(i)./2)).*theta2)));
q=sqrt((2.*D.*(zr.*m+s+g0.*(c+n(i).*ce.*theta1)+A+n(i).*h3))./(n(i).^2.*(((hr+cL.*k).*(D./m.*p))+(h1.*(1-(((2.*u-1).*D)./(2.*p.*u)))+(ce.*(w1.*(2-(((2.*u-1).*D)./(p.*u))-1).*theta2))))));
N(i+1)=n;
Q(i+1)=q;
j=j+1;
end
Index exceeds the number of array elements. Index must not exceed 1.
N
Q
z1=1;
for z=2:1:20
if ((N(z)~=N(z-1)) | Q(z)~=Q(z-1))
z1=z1+1;
else
break
end
end
z1
j=1;
for i=1:20
Q=Q(i);
N=N(i);
TC1=(pr.*D)+(z.*(m.*D./N(i).*Q(i)))+((hr+cL.*k).*(D.*N(i).*Q(i)./2.*m.*p))+(h1.*(1-(((2.*u-1).*D)./(2.*p.*u))).*N(i).*Q(i))-(h1.*N(i).*Q(i)/2)+(s.*D./N(i).*Q(i))+(D.*A./N(i).*Q(i))+(D.*h3./Q(i))+(D.*h2./2.*N(i))+((D./u).*(c0+(c1.*alpha)+(c0.*alpha)+(c2.*(1-beta).*alpha)))+((c.*D./N(i).*Q(i)).*(g0+g1.*N(i).*Q(i)))+(((D.*ce.*N(i).*theta1)./(N(i).*Q(i))).*(g0+g1.*N(i).*Q(i)))+((w0+w1.*(((1-(((2.*u-1).*D)./(2.*p.*u))).*N(i).*Q(i))-(N(i).*Q(i)/2))).*ce.*theta2);
TC(i)=TC1;
j=j+1;
end
round(TC)
for m=2:5
if(TC1 >=TC)
TC1=TC;
else
ansn = m+1;
break
end
end
round(TC)
Jan on 10 Oct 2022
@M.Rameswari Sudha: Codes like this must fail:
for i=1:20
Q=Q(i);
N=N(i);
...
end
In the first iteration the arrays Q and N are set to the scalar value of their first elements. Then there is no 2nd element in Q and N in the next iteration. The same problem here:
for i=1:length(q)
n = sqrt((D.*zr.*m+s+c.*g0+(A.*((q(i) ...
q = ... n(i) ...
% Now q and n are scalars and in the 2nd iteration "q(i)" and "n(i)" will fail.
end
By the way, the code can be made much more readable. Comapre
j = 1;
for i=1:length(q)
n=sqrt((D.*zr.*m+s+c.*g0+(A.*((q(i).*h2)/2))))/(q(i).*(((hr+cL.*k).*((D.*q(i))/(m.*p)))+((h1.*q(i)).*((1/2)-(((2.*u-1).*D)/(2.*p.*u))-D))+(D.*ce.*g1.*theta1)+(ce.*((w1.*(1-(((2.*u-1).*D)./(2.*D.*u))).*q(i))-(q(i)./2)).*theta2)));
q=sqrt((2.*D.*(zr.*m+s+g0.*(c+n(i).*ce.*theta1)+A+n(i).*h3))./(n(i).^2.*(((hr+cL.*k).*(D./m.*p))+(h1.*(1-(((2.*u-1).*D)./(2.*p.*u)))+(ce.*(w1.*(2-(((2.*u-1).*D)./(p.*u))-1).*theta2))))));
N(i+1)=n;
Q(i+1)=q;
j=j+1;
end
with
R = (2 * u - 1) * D;
S = R / (2 * p * u);
for i = 1:length(q)
N(i+1) = sqrt(D * zr * m + s + c * g0 + A * q(i) * h2 / 2) / ...
(q(i) * ((hr + cL * k) * D * q(i) / (m * p) + ...
h1 * q(i) * (0.5 - S - D) + ...
D * ce * g1 * theta1 + (ce * q(i) * (w1 * (1 - R / (2 * D * u)) - ...
0.5) * theta2)));
Q(i+1) = sqrt((2 * D * (zr * m + s + g0 * (c + n(i) * ce * theta1) + A + ...
n(i) * h3)) / (n(i)^2 * ((hr + cL * k) * D / m * p + ...
h1 * (1 - S) + (ce * w1 * (1 - 2 * S) * theta2))));
end
j = 1 + length(q);

Adam Danz on 10 Oct 2022
for i=1:length(q)
n=sqrt((D.*zr.*m+s+c.*g0+(A.*((q(i).*h2)/2))))/(q(i).*(((hr+cL.*k).*((D.*q(i))/(m.*p)))+((h1.*q(i)).*((1/2)-(((2.*u-1).*D)/(2.*p.*u))-D))+(D.*ce.*g1.*theta1)+(ce.*((w1.*(1-(((2.*u-1).*D)./(2.*D.*u))).*q(i))-(q(i)./2)).*theta2)));
% ^ ^ .... (and more)
q=sqrt((2.*D.*(zr.*m+s+g0.*(c+n(i).*ce.*theta1)+A+n(i).*h3))./(n(i).^2.*(((hr+cL.*k).*(D./m.*p))+(h1.*(1-(((2.*u-1).*D)./(2.*p.*u)))+(ce.*(w1.*(2-(((2.*u-1).*D)./(p.*u))-1).*theta2))))));
% ^ oops!
N(i+1)=n;
Q(i+1)=q;
j=j+1;
end
You're indexing q(i) assuming q is a vector which is the correct assumption on the first iteration of the loop. But then you're overwriting q and replacing the vector with a scalar. So when you index q(i) and i>1, you get an indexing error.

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