How can I determine the time and corresponding values of two functions when the value one function is 20% larger than the other one with a FOR oder WHILE loop?

Hi guys,

I'm new to Matlab and there is a task I don't know how to solve. So I have this code:

Pumax=75E3; % factor in front of exp-term
ku=0.045; % the decay rate
Pumin=1E5;
Psmax=3E5;
Po=1E4;
ks=0.08;
t = [0:1:200]
Ps=Psmax./(1+((Psmax/Po)-1)*exp(-ks*t)); %suburbs
Pu= Pumax*exp(-ku*t) + Pumin ; %city

Now i shall determine the time and corresponding values of Pu(t) and Ps(t) when the population of the suburbs are 20% larger than the city with a for or a while loop.

[Stephen23]

Note that the square brackets are not needed when creating a vector, as you are not concatenating anything. All you need is:

t = 0:1:200;

You will also notice that the MATLAB editor shows a warning saying that the square brackets are not required. See:

https://www.mathworks.com/matlabcentral/answers/35676-why-not-use-square-brackets

Answer 1

[Reza Bonyadi]

I would do the following: First, you want x such that Ps(x)=Pu(x)+0.2*Pu(x). So, define an anonymous function:

myf = @(x)(((Psmax./(1+((Psmax/Po)-1)*exp(-ks*x))))-1.2*(Pumax*exp(-ku*x) + Pumin));

This essentially defines the function myf(x)=Ps(x)-(Pu(x)+0.2*Pu(x))

You are then after an x in a way that myf(x) is 0. You can find such x using many methods, such as fzero:

fzero(myf,35)

35 is just a guess for the solution and the function returns 39.6068.

Indeed if you do

a=39.6068;Psmax./(1+((Psmax/Po)-1)*exp(-ks*a))-1.2*(Pumax*exp(-ku*a) + Pumin)

you will get close to zero.

You can see the corresponding values by:

a=39.6068;disp([(Psmax./(1+((Psmax/Po)-1)*exp(-ks*a))) Pumax*exp(-ku*a) + Pumin]);

Does that work?

[Stephen23]

Edit: changed incorrect reference to "inline function" to correct reference to "anonymous function", with hyperlink.