I am faced with the following question
The logistic growth model given below is often used to describe the population
growth where r is the population growth rate ad K is the maximum population.
dP/dt = r*P((K-P)/K)
Use loop to iterate through possible values for parameters r and Kto fit the logistic growth
ODE within 15 seconds time limit. The code should automatically find parameters where
mean absolute error of the world population is at a minimum. The growth rate of world
population is typically 1%-3% and the maximum world population is typically less than 50
Some more background information
- The data was recorded from 1950:2010
- initial population is 2532229237
- Already defined in a vector 'Pop' which holds the world population for each year
- There is a 15second time limit for MATLAB to compute the answer using my code
My current code is quiet inefficient as it can not find the values of K and r within this 15 second window.
K = 10*10^9 - 2*10^6;
P_q2f(1:21,1) = 2532229237;
myx = 0;
%Begin a nested bunch of loops
while K <= 50*10^9
K = K + 2*10^6;
myz = 0;
r = 0.009;
while r <= 0.03
r = r + 0.001;
myz = myz + 1;
myx = myx + 1;
for i = 2:61
P_q2f(myz,i) = P_q2f(myz,i-1) + r*P_q2f(myz,i-1)*((K-P_q2f(myz,i-1))/K);
error_q2f(myx,1) = sum(abs(P_q2f(myz,1:end)-Pop));
[minVal, I] = min(error_q2f)
My code is a pretty ugly mess of nested loops. Can anyone think of a simpler and/or more efficient way of computing:
- The best fitting K and r values
- A vector of the population using these parameters
Any help is greatly appreciated I have spent so many hours on this one question. Thanks
"Jac " <email@example.com> wrote in message
> I am faced with the following question
> The logistic growth model given below is often used to describe the
> population growth where r is the population growth rate ad K is the
> maximum population.
> dP/dt = r*P((K-P)/K)
> Use loop to iterate through possible values for parameters r and Kto fit
> the logistic growth ODE within 15 seconds time limit. The code should
> automatically find parameters where mean absolute error of the world
> population is at a minimum. The growth rate of world population is
> typically 1%-3% and the maximum world population is typically less than 50
> Some more background information
> - The data was recorded from 1950:2010
> - initial population is 2532229237
> - Already defined in a vector 'Pop' which holds the world population for
> each year
> - There is a 15second time limit for MATLAB to compute the answer using my
> My current code is quiet inefficient as it can not find the values of K
> and r within this 15 second window.
> K = 10*10^9 - 2*10^6;
> P_q2f(1:21,1) = 2532229237;
> myx = 0;
> %Begin a nested bunch of loops
> while K <= 50*10^9
> K = K + 2*10^6;
So you want to go from K = 9998000000 to K = 50000000000 by increments of
2000000 -- how many increments will you need in the worst-case scenario?
K0 = 10*10^9 - 2*10^6
K1 = 50*10^9
d = 2*10^6
nInc = (K1-K0)/d
doWeHitTheUpperLimit = K0+(nInc*d) == K1
We perform nInc+1 = 20002 iterations of the WHILE loop in the worst case.
> myz = 0;
> r = 0.009;
> while r <= 0.03
> r = r + 0.001;
21 iterations here per outer loop iteration in the worst case scenario. That
could mean 20002*21 iterations if your upper limits for K and r are the
optimal values you seek.
> myz = myz + 1;
> myx = myx + 1;
> for i = 2:61
> P_q2f(myz,i) = P_q2f(myz,i-1) +
60 iterations of this FOR loop for each iteration of the inner and outer
loops, so 20002*21*60 assignments into P_q2f.
And preallocate this to be 61 rows long so you don't have to grow it each
iteration through the loop.
> error_q2f(myx,1) = sum(abs(P_q2f(myz,1:end)-Pop));
> [minVal, I] = min(error_q2f)
> My code is a pretty ugly mess of nested loops. Can anyone think of a
> simpler and/or more efficient way of computing:
> - The best fitting K and r values
> - A vector of the population using these parameters
If you have Optimization Toolbox, look at FMINUNC or FSOLVE with one of the
ODE solvers in MATLAB like ODE45.
To contact Technical Support use the Contact Us link on
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