# Natural Growth and Logistic Models

1. May 4, 2005

### Beez

Hi, I have a question regarding natural growth and logistic models. According to my textbook, any exponential equations, P(t) = e^kt can be expressed logistically as P(t) = K/(1 + Ae^-kt) where A = (K-P0)/P0. When I applied this rule to P(t) = e ^0.0015t, and P(t) = 6000 / (1 + 5e^-0.0015t) it worked fine. The results were so close. But when I tried to do the same thing for P(t) = 400e^1.0986t and P(t) = 10000/(1 + 24e^-1.0986t), the results were so different from each other; 1199.98 and 1111.098 for P(1). Did I do something wrong here or they don't work if k>1?

I would appreciate it if I can get any help with this problem. Thank you in advance.

Naoko

2. May 4, 2005

### arildno

No, you have misuderstood the issue here:

The logistic model and the exponential model are nearly identical AT SMALL TIMES!!!
That is, it is only when time has gone a bit that the logistic growth restriction kicks in, and flattens the population level to K.

3. May 4, 2005

### Beez

Problem remains

I understood there is nothing to do with the value of k. Thank you.
But then how come the second logistic equation does not work for only t=1?

N0=400
N1=1200 (after a year the population tripled)
The lake can hold up to 10,000 fish
P(1)= 400e^(1.09861*1)=1199.99... =N1

P(1) = 10,000/(1+24e^(-1.09861*1))=1111.10... where 24 = (10000-400)/400

Am I doing something wrong here?

Thanks.

4. May 5, 2005

### arildno

Do you understand that the natural growth model is a DIFFERENT model than the logistic model; that is, there are some effects included in the logistic model that the natural growth model does not take into account?

5. May 5, 2005

### Beez

Yes I do

Yes I do. The logistic model takes the maximum population, in this case 10,000, and A, in this case (K-P0)/Po, in to account. But what I understood was while t is very small, they really don't have much effect and the exponential model and logistic model are almost identical. I used t=1 for the problem and the results were already so different. I used t=50 for the other problems that I mentioned earlier, but they still showed identical answers. That is why I thought the value of k or A might be the factors which determine if a size of the population satisfies both the exponential and logistic models or the only exponential model. But you told that it was nothing to do with them. Thus my question remained.

I hope I described my questions more clearly this time.

So, according to your explanation, the size of the population in question does not satisfy the logistic model. Am I right?

Thanks.

6. May 6, 2005

### arildno

First, wherever have I stated that some of the parameters don't matter???
I've never said anything to that effect!

Let's take it in full:
Let's call the exponential work model $$p_{E}(t)=P_{0}e^{kt}$$
And the logistic model $$p_{L}(t)=\frac{K}{1+\frac{K-P_{0}}{P_{0}}e^{-kt}}$$
Rewrite the logistic model as $$P_{L}(t)=\frac{P_{0}e^{kt}}{1+\frac{P_{0}(e^{kt}-1)}{K}}=\frac{P_{E}(t)}{1+\frac{P_{0}(e^{kt}-1)}{K}}$$
Thus, if we have: $$\frac{P_{0}(e^{kt}-1)}{K}<<1$$, then it follows $$P_{L}(t)\approx{P}_{E}(t)$$
Let us make this into a condition on "t":
We rewrite, and find:
$$t<<\frac{1}{k}ln(1+\frac{K}{P_{0}})$$
This condition determines what a "small time" is, in this particular context.

If you plug into your values of $$k,P_{0},K$$ here, you'll see why the approximation holds in one case but not in the other..
(that is, the "small times"-regions are different)

Note that a concept like "small time" is not really meaningful if you haven't specified a REFERENCE time.
That is, "small times" means :the time is small compared with the reference time for the problem.

I think it was this usage of mine of "small times" which confused you.
I hope that I have made it more explicit now.

Last edited: May 6, 2005
7. May 6, 2005

### Beez

Thank you

Thank you so much for providing me the detailed explanation.
Yes, I was taking "small amount" literally.
Now my sky is as blue as it could be.
Thank you for your help again.