How do I solve an equation numerically for a grid of values and then plot the relationship with the dependent variable?

Hi!

I would like to solve the following equation for pi = linspace(0.95,1.08,30) eqn = pi*(pi-1)' - beta*pi*(pi-1)' == (v/(alpha*gamma))*(c+g)^((1+eps)/alpha)+((1-v)/gamma)*(c+g)*c^(-sigma); Where all the parameters are specified and only c is unknown. How can I solve this relationship for all the different values of pi and how can I plot this relationship with c?

Thanks in advance!

Answer 1

[Torsten]

beta=...;
v=...;
alpha=...;
gamma=...;
g=...;
eps=...;
sigma=...;
c0=...;
pin=linspace(0.95,1.08,30);
for i=1:numel(pin)
  pi_actual=pin(i);
  fun=@(c) (1-beta)*pi_actual*(pi_actual-1)-(v/(alpha*gamma))*(c+g).^((1+eps)/alpha)+((1-v)/gamma).*(c+g).*c.^(-sigma); 
  c_sol(i)=fzero(fun,c0);
  c0=c_sol(i);
end
plot(pin,c_sol)

Best wishes

Torsten.

[Sean]

Thanks a lot. But could I insert pi*(pi-1)' - beta*pi*(pi-1)' == (v/(alpha*gamma))*(c+g)^((1+eps)/alpha)+((1-v)/gamma)*(c+g)*c^(-sigma) after the fun=@(c)? Or should I drop the accent ('), I mean pi is a vector right?

It should be looking something like the picture I added, but by following these lines of code I seem to get negative consumption (c).

[Torsten]

I don't know what pi*(pi-1)' should indicate, but I assume that you want to determine c for each value of the vector pi, and that this single value should be inserted as pi*(pi-1). At least this is what the code from above does.

To get an impression of your function, you should plot it first and see whether it may have multiple zeros.

Best wishes

Torsten.