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Course: Differential equations>Unit 1

Lesson 6: Logistic models

Worked example: logistic model equations

The general logistic function is N(t)=(N₀K)/(N₀+(K-N₀)e⁻ʳᵗ). In this video, we solve a real-world word problem about logistic growth.

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• I wonder what is the logic to calculate to get "r", anual populatin growth rate, when N becomes 1.5N in 20 years.
I calculate in this way.
N(o)*exp(r20)= 1.5N(o),
exp(20r)= 1.5
20r = In(1.5)
r = In(1.5)/20
then i got 0.020273... I wonder why these two answer is different.
thx for ur time ^^
• OK this confused me too, but I think I figured out where the confusion comes in.

I think that at when Sal says that in 20 years, the population grows by 50%, he's not talking about the imaginary island he just made up. He's talking about unconstrained growth, growth with an infinite K, the non-Malthusian way of estimating population growth. Because remember r is the unconstrained growth constant.

In a world with no K, no cap on the population, the rate of growth would remain the same every year, which is why he can find r by just taking the 20th root of 1.5, resulting in the number 1.0205. If you multiply 1.0205 by your current population, you get the next year's population in an UNCONSTRAINED environment.

THEN Sal plugs in his r to the logistic differential equation to get the actual population growth of the island, with it's environmental constraint K. Notice at that the population of the island after 20 years is almost, but not quite, 150.
• When I solve for r using N(20)=150 I get 0.02313.
• This is because N(t) takes into account the population cap K, which stunts growth from the outset. Without K, a yearly growth of 2.05% would bring the population up 50% over 20 years. With K, the function actually requires a higher yearly growth rate to increase by 50% over 20 years, as you have calculated.
• For everyone confused about his r, I have it figured out. The formula for Compound Annual Growth rate (CAGR) is = [(Ending value/Beginning value)^(1/# of years)] - 1. In his example the ending value would be the population after 20 years and the beginning value is the initial population. Since it has grown by 50% we know this ratio will be 1.5, He sets this to the power of (1/20) since we are concerned with the number of years being 20, and then to get 0.0205 he subtracts 1 from 1.02048 and rounds. This gives the ANNUAL growth rate (which is fine in this example because the units for t are years), but if you are concerned with instantaneous growth rate, then there is a simple conversion from your annual growth rate, A, to r (instantaneous). r = ln(1+A). Hope this helped clear some stuff up. Here is the webpage I used for the CAGR formula: http://www.investopedia.com/terms/c/cagr.asp
• I'm not following what Sal did at when he got out the calculator and got a number for r. Did he just rearrange the whole logistic equation in his head and solve for r, or what?
• this is the equation he used:
future value / present value = (1+i)^n (growth rate equation google it)
i= growth rate
n=number of periods.
150/100=(1+i)^20---> i=[(1.5)^(1/20)] - 1
• The clarification at said that r=ln(1+A). How did that come about?
• At 1.02 changes to 0.020 which changes the final equation. Also, I calculated r by using the given equation with the given n(t)=150, t=20, n(0)=100, and k=1000 to get that r=0.023. What is r=ln(1+A)? I've never seen that before. Can someone explain?

N(t)=(100)(1000)/100+900e^-rt
N(t)=(1000)/1+9e^-rt
150=(1000)/1+9e^-r20
1+9e^-r20=1000/150
1+9e^-r20=20/3
9e^-r20=17/3
e^-r20=17/27
ln(e^-r20)=ln(17/27)
-20r=ln(17/27)
r=-1/20ln(17/27)
r=0.0231

Thank you
• Can we find this limit K? Assuming that our No is 2 (lets say Adam and Eve) and that humans have been on earth for approximately 200,000 years an now the population is approximately 7 billion people. Or are we still constrained by r?
• If you assume that a population has grown following the Logistinc function (human population does not), you can simply solve for K, but you would need to know the value of the other variables, namely `Nₒ`, `r` and a point in time `N(t)`.

In your example you have `Nₒ=2`, and a point in time `t=2⨉10^5` and `N(2⨉10^5) = 7⨉10^9`, but you still lack the value of `r` to get the value of `K`.

The equation solved for `K` is:
``    Nₒ·e^(-rt) - N(t)K = ――――――――――――――――――      e^(-rt) - Nₒ``
• Don't understand why he subtracts 1 from his calculation on the calculator to get r. Why do we subtract 1?
• What exactly is the significance of r in this kind of situation? What is its meaning? I don't get why it's "annual growth". Is it a preassumed term?