Scalar division and Subtraction ?!!
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I am trying to use some artificial data to see if my code is working.. but there is a error for the division and subtraction part.. See below the Code...
function Sa = trial(lambdaMax,lambda,T)
t = 0;
I = 0;
Sa = [];
u = {10,2,11,4,5,6};
t = t - log(u)/lambdaMax;
while t < T
u = {10,2,11,4,5,6};
if (u <= lambda(t)/lambdaMax)
I = I+1;
Sa(I) = t;
end
u = {10,2,11,4,5,6};
t = t - log(u)/lambdaMax;
end
lambdaMax=50;
T=20;
lambda =@(Sa) lambdaMax*cos(Sa);
Sa = trial(lambdaMax,lambda,T);
figure
hold on
%plot(Sa,lambda(Sa),'*')
xlabel('t')
ylabel ('cos(x)')
X = linspace(min(Sa),max(Sa),10);
Y = pchip(Sa,lambda(Sa),X);
plot(X,Y)
line(repmat(Sa,2,1),repmat([0;1],1,length(Sa)),'color','r' )
Thanks all in advance :)
1 Kommentar
Akzeptierte Antwort
Fangjun Jiang
am 13 Jul. 2011
Why do you use cell for your variable u? change it to be data array.
u = {10,2,11,4,5,6}
u = [10,2,11,4,5,6]
lambdaMax=50;
T=20;
lambda =@(Sa) lambdaMax*(cos(Sa)+1.1);
Sa = trial(lambdaMax,lambda,T);
figure;
hold on;
plot(Sa,lambda(Sa),'*')
xlabel('t')
ylabel ('cos(x)')
X = linspace(min(Sa),max(Sa),100);
Y = pchip(Sa,lambda(Sa),X);
plot(X,Y)
line(repmat(Sa,2,1),repmat([0;1],1,length(Sa)),'color','r' )
function Sa = trial(lambdaMax,lambda,T)
t = 0;
I = 0;
Sa = [];
u=rand;
t = t - log(u)/lambdaMax;
while t < T
u=rand;
if (u <= lambda(t)/lambdaMax)
I = I+1;
Sa(I) = t;
end
u=rand;
t = t - log(u)/lambdaMax;
end
6 Kommentare
Sean de Wolski
am 13 Jul. 2011
no it is not. What is lambdamac, lambda (a function handle we presume by looking at the recursive nature of your function, and T?
Weitere Antworten (3)
Sean de Wolski
am 13 Jul. 2011
t converges to:
-20888 -6288 -21753 -12576 -14600 -16254
All of those are less than T. The while loop never exits. Perhaps you want while t>T?
13 Kommentare
Sean de Wolski
am 13 Jul. 2011
The easiest way would just be to pull
I = I+1;
outside of the if statement. 'I' will get bigger every time and then all of the non-zero values in SA are places to be filled in.
Susan
am 13 Jul. 2011
12 Kommentare
Fangjun Jiang
am 13 Jul. 2011
Does it require the lambda function be positive? I modified your lambda function to make it always positive. See the code in my answer section.
Susan
am 13 Jul. 2011
1 Kommentar
Sean de Wolski
am 13 Jul. 2011
t = 0;
I = 0;
Sa = [];
u = rand;
t = t - log(u)/lambdaMax;
while t <= T
if (u <= lambda(t)/lambdaMax)
I = I+1;
Sa(I) = t;
end
u = rand;
t = t - log(u)/lambdaMax;
u = rand;
end
Is how I interpret that last page.
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