11

# Population growth formula

Fri 9 Jun, 2006 02:02 pm
I found this formula about population growth, but I am not sure if it is correct. Anyone who has an idea?

The formula is:
Px = Po (1+y) x

Variables explanation :
Px: First/Current population
y: Average growth rate of population (derivative)
x: Years passed
Po: Population after x years

P.S. Deaths because of wars/genocides and deceases are excluded of course.
• Topic Stats
• Top Replies
Type: Discussion • Score: 11 • Views: 96,266 • Replies: 18

2
Fri 9 Jun, 2006 02:17 pm
I think you have the definitions for Po and Px reversed.
0 Replies

ebrown p

3
Fri 9 Jun, 2006 02:18 pm
The function you gave has two mistakes.

1. Your descriptions for Po and Px are reversed (Po is the initial population and Px is the population after x years).

2. The "x" should be an exponent (usually a superscript)
The function should read - Px = Po(1+y)^x

Now let's look at what this function describes. Let's look at what "y" means.

First if y = 0, then the population won't change. (1 + 0 = 1 and 1^x = 1).

If y = .5, then the population will increase by 50% each year (for example after one year (lets call this P1) P1 = Po*1.5 and after two years P2 = P1 * 1.5 = Po * 1.5 * 1.5.

So this function assumes that the population will grow by the same percentage (that is (100 * y)%) each year.

Of course this assumption is not realistic since there are all kinds of things that effect population growth (disease patterns, resources, level of education, current events, how many people already are around etc. etc. )

This equation is useful for lots of other situations including calculating how much money you earn (or pay) in interest.
Ellinas

1
Fri 9 Jun, 2006 02:33 pm
You are right I made a mistake. Po is the current population and PX the population after x years (damn hurry )

0 Replies

2
Sat 1 Nov, 2008 06:31 am
@Ellinas,
The equation for any cumulative increase even like in your bank account yearly interst is:
X2=X1((1+i)^n)
X2= The Future Value
X1= The Current Value
i= the yearly flat increase
n=No. Of Years
Example:
if you put 1000\$ in a bank account for 20 years on a yearly interst of 10%, the value of the money after 20 years will be=
X2= 1000((1+0.1)^20)= 6727.5
ERandall

1
Tue 28 Feb, 2012 01:45 pm
How to solve for 'n' is my question. For a programming class, not math.
ERandall
raprap

1
Wed 4 Apr, 2012 12:11 am
@ERandall,
It really doesn't matter if the class is math or programming--its the same process

Xn=X0(1+i)^n

use properties of powers to solve for n (logarithms)

Xn/X0=(1+i)^n

Rap
kswaby10

1
Sat 21 Apr, 2012 01:59 am
@raprap,
You're right. It is the same process.

newegg promo code vision direct coupons
0 Replies

uvosky

1
Wed 2 May, 2012 10:44 pm
@Ellinas,
" Ellinas " you wrote , " y : Average growth rate of population (derivative) "
In any study of the growth of a population ( whether human , animal , or bacterial ), the function which counts the number " x " of individuals present at time t necessarily takes on only integer values.Therefore the "derivative"
dx / dt is either zero ( when t lies in an open interval where x is constant ) , or else the "derivative" dx/dt does not exist ( when x jumps from one integer to another) . Nevertheless , useful information can often be obtained by assuming the population x as a continuous function of t with a continuous derivative
dx / dt at each instant. Then by postulating varius " laws of growth " so that the derivative can be represented as dx / dt = f ( x , t ) ( must be an explicit function of x and t ) we can obtain the desired population at any time t by integration when the initial population at initial time is known.
cicerone imposter

1
Wed 2 May, 2012 11:42 pm
@ebrown p,
You're a good teacher, ebrown. 0 Replies

dvlawyer

1
Wed 30 May, 2012 09:28 am
@uvosky,
ok, so if the eagle population was 500 breeding pairs in 1967, and 11040 breeding pairs in 2007, 40 years later, what is the rate of increase?
0 Replies

dvlawyer

1
Wed 30 May, 2012 09:30 am
@ebrown p,
so if there were 500 breeding pairs of eagles in 1967 in the lower 48 states, and there were 11040 breeding pairs in 2007, 40 years later, what is the rate of increase?
ebrown p

3
Wed 30 May, 2012 10:14 am
@dvlawyer,
First, let me point out that the calculation I am about to do is completely bogus (i.e. it will have absolutely nothing to do with the reality of eagles). The population growth formula only works in a set of ideal conditions, namely there has to be unlimited food and space for the eagles and there has to be no changes in conditions during the time period involved.

Of course with eagles in the US over the a period of 40 years this isn't the case at all. There has been human activity with loss of habitat. There have been changes in laws to provide protection. There are ups and downs in disease rates and prey availability.

This means that the population growth formula is completely unsuited for describing anything having to do with reality when it comes to actual eagles in the US.

If you were growing eagles in a petri dish (assuming you provided ample food and space) this formula would be correct. Other than that circumstance, this calculation is bogus... but let's do it anyway.

Starting with the formula Px = Po(1+y)^n Let's solve for y...

Px/Po = (1+y)^n
(Px/Po)^(1/n) = 1 + y
y = (Px/Po)^(1/n) -1

(11040/500)^(1/40) -1

If I haven't made a stupid mistake in my quick calculations, I get about 8.04% per year.

But did I make it clear enough that this calculation is bogus?

dvlawyer

0
Wed 30 May, 2012 07:44 pm
@ebrown p,
I think it would be bogus if we did not have the real starting and ending numbers, but those are real numbers, so the rate of increase would be a real number right? (and I got the same number using interest income calculator, compounding interest yearly....)
Ok, does the population formula change given that eagles take 5 years to reach sexual maturity, brood once per year, and average 2 eaglets? or is the formula based on human sexual maturity taking much longer? and incubation being 9 months? it seems if its the same formula as interest income then only the interval of when it compunds, in this case yearly since one brood per year, but I am not clear why the age of sexual maturity does not seem to effect the math for humans vs. eagles....

thanks so much for this!!! I am posting on the Decorah eagle cam site on Ustream which is a Raptor Research project

Melissa
ebrown p

2
Thu 31 May, 2012 08:56 pm
@dvlawyer,
No, this use of the formula is still bogus even if the starting and ending values are "real". The calculation is bogus for the reasons I listed above. The fact it had real starting and ending numbers has nothing to do with it. Read my previous post again if you don't understand why it is bogus.

The rate of growth of human population is slowing down. Human population is most definitely not following a typical population growth curve.

You can only use this formula under very certain circumstances. It doesn't apply for eagles and it doesn't apply for humans.

Since eagles are preditors, the dashed green line in this graph might be a more appropriate curve for eagles in the US. The equation you would use for this would be But this depends on a different set of assumptions (only slightly less bogus).

My real point is that you can't throw numbers (even "real" numbers) into formulas and expect results that aren't bogus if you haven't made sure the underlying assumptions match reality.
dvlawyer

1
Fri 1 Jun, 2012 06:55 am
@ebrown p,
that makes sense. thanks!
cicerone imposter

0
Fri 1 Jun, 2012 10:46 am
@dvlawyer,
Here's a graph on world population growth. 0 Replies

Abhishek Sijali

-1
Tue 10 Jul, 2012 05:58 am
Nonsense Idea and formula. Idiots.
0 Replies

shelton

-1
Tue 11 Sep, 2012 10:35 am
can you all figuar this one out the present population of a town is 4750, if the population is increasing by 350 people per year how long will it take the town to reach a population of 7200. can you help thanks
0 Replies

### Related Topics

Population growth - Question by yas2010

1. Forums
2. » Population growth formula