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dealing with negative indices error

Asked by Gaone Ramadubu on 16 Jul 2019
Latest activity Commented on by Star Strider
on 16 Jul 2019
please help, i keep getting this error: "Array indices must be positive integers or logical values."
Here is my code:
randn('state',100)
lambda = 2;
mu = 1;
Xnot = 1;
T = 1;
N = 2^8;
r = 0.5;
gamma = 0.5;
delta = 1-gamma;
dt = 1/N;
t = 0.5;
dW = sqrt(dt)*randn(1,N);
W = cumsum(dW);
R = 4;
Dt = R*dt;
L = N/R;
beta = 1.5;
eta = 2;
rho = 0.2;
v = r + (lambda-r)^2/2*delta*mu^2;
D = rho-(gamma*v);
G = delta*(1-exp((D/delta)*(t-T)));
H = (delta*eta)/beta
X_temp = Xnot;
for T = 1
t = -1:1
end
F = X_temp + (((delta*eta)/(beta*r))*(1-exp(r(t-T))));
C = ((D*F)/G)-H;
M = (lambda-r)/(delta*mu^2)
J = (eta*(lambda-r))/(beta*r*mu^2)*X_temp;
pi = M+J
X_EM = zeros(1,L);
for j = 1:L
Winc = sum(dW(R*(j-1)+1:R*j));
X_temp = X_temp + Dt*((pi*(lambda-r)+r)-C) + (mu*pi)*Winc;
X_EM(j) = X_temp;
end
plot([0:Dt:T],[Xnot,X_EM],'g-', 'LineWidth', 2)

  1 Comment

Rik
on 16 Jul 2019
Have you tried using breakpoints to find out where your code deviates from what you expect?

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2 Answers

Answer by Star Strider
on 16 Jul 2019

There turned out to be three problelms with your code that I corrected here.
The original one was due to your forgetting an operator, that I assume should be a multiplication:
F = X_temp + (((delta*eta)/(beta*r))*(1-exp(r*(t-T))));
The second is that ‘X_EM’ was not preallocated correctly, and not addressed correctly, since each ‘X_temp’ is a (1x3) vector:
X_EM = zeros(L,3);
for j = 1:L
Winc = sum(dW(R*(j-1)+1:R*j));
X_temp = X_temp + Dt*((pi*(lambda-r)+r)-C) + (mu*pi)*Winc;
X_EM(j,:) = X_temp;
end
The third was that the dependent variable in the plot call was not concatenated correctly, now that ‘X_EM’ is a matrix:
plot([0:Dt:T],[Xnot*ones(1,3);X_EM],'g-', 'LineWidth', 2)
With these changes, your code works.

  2 Comments

Thank you, it did work.
Great!
Since my Answer solved your problem, please Accept it!

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Answer by Walter Roberson
on 16 Jul 2019

for T = 1
t = -1:1
end
After that T is the scalar 1 and t is -1 0 1
F = X_temp + (((delta*eta)/(beta*r))*(1-exp(r(t-T))));
t-T is -1 0 1 minus 1, so -2 -1 0.
r is a scalar so r(t-T) is a request to index r at -2 -1 0

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