Growth Rates coupled to Survival Probabilities

For mammals to reproduce and the population grow or be stable, the following conditions have to be met:

• A Female mammal has to be born

• There is some probability that female will survive to fertility and have offspring

• Some of those offspring have to be females.

All of these conditions can be put together into a particular form for the growth rate.

Yes this is another scary looking and meaningless equation. We will use an iterative spreedsheet to solve it under various conditions

The terms in the above equation are the following:

• λ = the fractional growth rate per year for a species:
  • for a stable species λ = 1
  • values of λ less than 1 lead to population crashes. For instance a value of λ = 0.98 means the species population is declining at the rate of 2% per year (this is an exponential decline, of course).

• α = the age (in years) at which the species starts breeding
  

• s = the adult annual survival probability (in percent)

• b = adult average female fecundity - fecundity means the power of a species to multiply rapidly or its capacity to form reproductive elements. It is one measure of fitness. More explicitly b is average reproductive rate of female offspring per adult female in the overall population per year. So, if on average a female gives birth to another female once every 4 years, then b = 0.25.

• l_α represents the probability that a new born will survive to age α (the breeding age).
  note that l_α in turn = s_o * s₁ * s, where:
  • s_o = survival probability at birth and in fledgling stage of life
  • s₁ = sub-adult annual survival probability,
  • s is the adult annual survival probability
  

  Note that, of course, s, must be correlated at some level with s_o and s₁; it is also useful to seperate out s_o explicitly because some species might have a relatively large mortality rate at birth, but survival after birth as a higher probability.

Environmental alteration and/or habitat loss basically directly influence l_α as that would impact values for s_o and s₁ more than anything else. Generally speaking, in mammalian populations, if you manage to survive to adulthood your basically likely to survive long enough to breed.

Hence species survivability or stability depends on how well the growth equation shown above can be optimized. Things that help species are:

• low values for α
• high values for b (note that b and α are correlated; low values of α would result in higher b values)
• high survival probabilities

Conversely, environmental alterations that can trigger a population crash are

• stimuli that lower b (hard to do)
• habitat loss that lowers l_α (easy to do)

So this kind of growth equation (which is numerical in nature, that is, not analytic (solvable)) can be used to apply to populations with known or estimated survival rates, birth rates, and fertility ages. But something is missing from this framework. What is it?

The special case of λ = 1.

Our equation greatly simplifies for λ = 1.

In that case, we have the condition that

1 -s = l_αb

This mathematical condition is the same as the physical condition of an exact balance between female survival probability and females born per year in the ecosystem.

• For instance, for s = 0.9 and b = 0.25 then l_α must be .1/.25 = 0.40.
• Since l_α = s_o * s₁ * s and s = 0.9, then the stability condition occurs whenever s_o * s₁ = 0.44.
• This represents and example of the necessary "fine tuning" of a parameter set to achieve stability. This is why stability is rare and certainly difficult to maintain if you ever get there.
• As another example, consider the case when all 3 survival probabilities are the same (s= s_o=s₁ = 0.5).
• In this case l_α = .125 and 1 -s = 0.5. Therefore b would have to be 4 to achieve stability. Clearly this is an unphysical situation.

In this way, all of these various parameters are related at some level and are not completely independent and arbitrary. This is another way of saying that, since the species has survived, there has obviously been some balance achieved, over time, between these parameters.

However, at any given time in any system of mammals (except humans), λ is always less than 1 (sometimes only slightly less than 1), because of the difficulty in balancing the parameters to achieve stability. On long timescales, such balance can occur, but at any snapshot in time, there is no balance.

The importance of determining λ is that, once we know its value we can then estimate the characteristic timescale of the population crash as

where N is the number of known individuals in the population (8000 in this case).

As stated earlier, λ is fractional growth rate, so if you find that λ = 0.98 then that means 2% species reduction per year and the doubling time (or in this case the halving time) of the population is 70/2 = 35 years. After so many halving times, the species is essentially extinct. For instance, using 8000 and .98 produces a population crash time of 450 years, or 450/35 = 13 halving times. Note that 2¹³ = 8192.

So let's work out an example: ( also see iter spreadsheet )

• α = 4 years
• s = 0.90
• s_o = 0.15
• s₁ = 0.70
• b = 0.25

So we would have on the right hand side (RHS) of the equation:

l _α = 0.90 * 0.15 * 0.70 = .095

l _α b = .095 *.25 = 0.0236

so,

λ⁴(1 - .90/λ) = .024

as our equation to solve. The value of λ that is needed for the LHS to be equal to the RHS can not be solved analytically, but only by trial and error. Note, however, that λ can not be less than .90 else the left hand side of the equation becomes negative. So you only have to guess at λ in the range of 0.90 to 1.

The solution for this case is λ = 0.93 which leads a population crash time of 123 year for a starting N=8000. Note, if N would be 80000 then the crash time is 155 years, so again, N is not the driver, λ is.