% I have got this simple program runnning a loop to satisfy the condition 'nvvni = 0' stop the program
PT= 50;
c = [0.20,0.10,0.10;0.20,0.20,0.20;0 0 0];
P= [190 72.2 51.6; 20.44 15.57 4.956;0 0 0];
K = P/PT;
A = c(1,1)*(K(1,1)-1)+ c(1,2)*(K(1,2)-1)+c(1,3)*(K(1,3)-1)+c(2,1)*(K(2,1)-1)+c(2,2)*(K(2,2)-1)+c(2,3)*(K(2,3)-1);
B = (c(1,1)*(K(1,1)-1)/K(1,1))+(c(1,2)*(K(1,2)-1)/K(1,2))+(c(1,3)*(K(1,3)-1)/K(1,3))+(c(2,1)*(K(2,1)-1)/K(2,1))+(c(2,2)*(K(2,2)-1)/K(2,2))+(c(2,3)*(K(2,3)-1)/K(2,3));
%disp(B)
for i=1:10
n=0;
nvni = (A )/(A-B);
n=n+1;
nvvni = (c(1,1)*(K(1,1)))/(nvni*(K(1,1)-1)+1)+(c(1,2)*(K(1,2)))/(nvni*(K(1,2)-1)+1)+(c(1,3)*(K(1,3)))/(nvni*(K(1,3)-1)+1)+(c(2,1)*(K(2,1)))/(nvni*(K(2,1)-1)+1)+(c(2,2)*(K(2,2)))/(nvni*(K(2,2)-1)+1)+(c(2,3)*(K(2,3)))/(nvni*(K(2,3)-1)+1);
if nvvni == 0;
disp(nvni);
break
else
nvvvi = (c(1,1)*(K(1,1)^2)/(nvni*(K(1,1) + 1)^2))+(c(1,2)*(K(1,2)^2)/(nvni*(K(1,2) + 1)^2))+(c(1,3)*(K(1,3)^2)/(nvni*(K(1,3) + 1)^2))+(c(2,1)*(K(2,1)^2)/(nvni*(K(2,1) + 1)^2))+(c(2,2)*(K(2,2)^2)/(nvni*(K(2,2) + 1)^2))+(c(2,3)*(K(2,3)^2)/(nvni*(K(2,3) + 1)^2));
nvvni = nvni - (nvvni/nvvvi);
nvni =nvvni;
end
end

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Jan
Jan 2012년 2월 19일
Please format your code properly. I've done this for you this time.
What is your question?

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답변 (1개)

Jan
Jan 2012년 2월 19일

0 개 추천

It is very unlikely that the result of such a complicated formula is exactly 0.0 due to rounding errors. Better check for a certain limit:
if abs(nvvni) < 1e-14
The size of the limit depends on the physical meaning of the formula and cannot be defined uniquely. An analysis of the sensitivity is recommended in addition: How sensitive is nvvni to tiny variations of the inputs, e.g. when using c+eps.

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Walter Roberson
Walter Roberson 2012년 2월 19일
http://matlab.wikia.com/wiki/FAQ#Why_is_0.3_-_0.2_-_0.1_.28or_similar.29_not_equal_to_zero.3F

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