iteration with biscetion method

3 次查看（过去 30 天）
Claus Madsen
Claus Madsen 2013-3-12
Hello i need som help with program this code. I have the following parameters in a matrix[1566,1] R, I, Q_min, Q_max, A and d. And i want to solv the eq with the biscetion methode. i want the result to come out in a matriv?
%%%%%%%%%%%%%%%%%%%%%%%%%%Parameter %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
data=xlsread('ledning.xls');
Q_min=data(:,2);
R=data(:,5)/4; % Hydraulisk radius for fuldløbende ledning
I=data(:,4); %I=I_0 dvs. slope
A=(data(:,5)/2).^2*3.14;
M=75;
d=data(:,5);
for i=1:length(data)
Q_max(i,1)=abs(A(i,1)*M*(I(i,1)/100)^(1/2)*R(i,1)^(2/3));
end
for i=1:length(data)
p(i,1)=Q_min(i,1)/Q_max(i,1);
end
%%%%%%%%%%%%%%%%%%%%%%%%Ligning der skal løses %%%%%%%%%%%%%%%%%%%%%%%%%%%
%Ligningen løses via biscetion metoden
f=@(y) (0.46-0.5*cos(3.14*(y/d))+0.004*cos(2*3.14*(y/d)))-(p); %delfyldningsformel
format long
%%%%%%%%%%%%%%%%%%%%%%%%Betingelser til biscetion %%%%%%%%%%%%%%%%%%%%%%%%
eps_abs = 1e-5;
eps_step = 1e-5; %%%%abs og step angiver hvor tæt resultaterne skal være på hinanden
a = 0; % Initial Guess to your function such that f(a)>0.
b = 0.2; % Initial Guess to your function such that f(b)<0.
%%%%%%%%%%%%%%%%%%%%%%%%Påbegynd loop %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
while (b - a >= eps_step || ( abs( f(a) ) >= eps_abs && abs( f(b) ) >= eps_abs ) )
c = (a + b)/2;
if ( f(c) == 0 )
break;
elseif ( f(a)*f(c) < 0 )
b = c;
else
a = c;
end
end
  1 个评论
Jan
Jan 2013-3-12
What does the posted code do in opposite to your demands?

请先登录，再进行评论。

请先登录，再回答此问题。

回答（0 个）

类别

在 Help Center 和 File Exchange 中查找有关 Simulink 的更多信息

标签

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by