Index exceeds matrix dimensions

1 Ansicht (letzte 30 Tage)
MD NASAR
MD NASAR am 8 Mai 2012
Dear All I am trying to solve equation but i am getting error code is:
M=100;
r=0.8;
alpha=0.45;
beta=0.75;
gama=0.1;
d=0.004;
h=0.4;
n=100;
c(1)=1580;
c0=1500;
lembda=150;
delT=1;
x(1)=2;
a(1)=1000;
p(1)=100;
q(1)=0.8;
for t=1:20
q(t)=0.8;
xx=t+1;
x(xx)=x(t)+delT*(alpha+beta*log(a(t))+gama*x(t))*(M-x(t))*exp(-d*p(t))*exp(h*q(t))
%disp(x(t));
x(xx)
lembda(xx)=lembda(t)*(r*lembda(t)-exp(-d*p(t))*exp(h*q(t)*(gama*M-alpha-beta*log(a(t))-2*gama*x(t))*(p(t)-c(t)+lembda(t))+n*q(t)^x(t)-1*log(q(t))...
*(alpha+beta*log(a(t))+gama*x(t))*(M-x(t))*exp(-d*p(t))*exp(h*q(t))))
% disp('lembda(t+1)');
lembda(xx)
c(t)=n*q(t)^x(t)-1+c0
% disp('c(t)');
c(t)
p(t)=c(t)-lembda(t)+1/alpha
% disp('p(t)');
p(t)
a(t)=(p(t)-c(t)+lembda(t))*beta*(M-x(t))*exp(-d*p(t))*exp(h*q(t))
%disp('a(t)');
a(t)
end
when i am running this but i am getting error "??? Index exceeds matrix dimensions." actually i want the value of x(1) to x(20), lembda(1) to lembda(20), p(1) to p(20), c(1) to c(20),a(1) to a(20) please help me Thank you.

Antworten (1)

Andreas Goser
Andreas Goser am 8 Mai 2012
When I run this code in this line
x(xx)=x(t)+delT*(alpha+beta*log(a(t))+gama*x(t))*(M-x(t))*exp(-d*p(t))*exp(h*q(t))
because in the second time the loop gets executed, it want to have a(2), but a is still 1x1.
Why THAT is it is up to you, as I don't know what you want to achive.

Kategorien

Mehr zu Mathematics and Optimization finden Sie in Help Center und File Exchange

Tags

Community Treasure Hunt

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

Start Hunting!

Translated by