- 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.

The terms in the above horrible equation are the following:

- lambda = the fractional growth rate per year
for a species; for a stable species lambda = 1;
values of lambda less than 1 lead to population crashes. For instance
a value of lambda = 0.98 means the species population is declining
at the rate of 2% per year (this is an exponential decline, of course).
- alpha = the age at which the species starts breeding
- s = the adult annual survival probability
- 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, the b '
value would be 0.25.
- l
_{alpha}represents the probability that a new born will survive to age alpha (the breeding age).- note that l

_{alpha}in turn = s_{o}* s_{1}* s where s_{o}= survival probability at birth and in fledgling stage of life, s_{1}= sub-adult annual survival probability, and s is as given above. Note that, of course, s, must be correlated at some level with s_{o}and s_{1}.

Environmental alteration and/or habitat loss basically directly influence
l_{alpha} as that would impact values for s_{o} and
s_{1} more than anything else. Generally speaking, in
mammalian populations, if you manage to survive to adulthood your basically
okay (unless you have tenure ...).

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

- low alpha
- high b (note that b and alpha are correlated; low values of alpha promote high 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
_{alpha}

The special case of lambda = 1.

Our equation greatly simplifies for lambda = 1.

In that case, we have the condition that

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_{a}
must be .1/.25 = 0.40.

Since l_{a} = s_{o} * s_{1} * s
and s = 0.9, then the stability condition occurs whenever
s_{o} * s_{1} = 0.44.

As another example, consider the case when all 3 survival probabilities
are the same (s= s_{o}=s_{1} = 0.5).

In this case l_{a} = .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), lambda is always less than 1 (sometimes only slightly less than 1), because of the difficulty of balance 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 lambda 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, lambda is fractional growth rate, so if you find
that lambda = 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 2^{13} = 8192.

So lets work out the case for the first data set I gave you for the assignment. Data Set 1 (from university biologists):

- alpha = 3 years
- s = 0.95
- s
_{o}= 0.15 - s
_{1}= 0.72 - b = 0.24

So we would have

as our equation to solve. This can not be solved analytically, but only by trial and error (which is why you program this). Note, however, that l can not be less than .95 else the right hand side of the equation becomes negative. So you only have to guess at l in the range of 0.95 to 1.

The solution for this case is l = 0.98 which leads to the 450 year population crash timescale.

In principle, the formula that we have been using would allow the
female adults to live forever (and reproduce forever). This is
obviously unphysical and so one needs to introduce a correction
factor, * w *, which represents the maximum reproductive age
of a species (something that's actually hard to measure).

The corrected expression looks even more horrible than the first one.

Before we collectively freak out, let's examine this new equation to see if it makes sense in some physical limits.

In the first case, let w approach infinity. In that case,
s/l is less than 1 and (s/l)^{w} would be 0 and we recover our initial equation.

In the case where w=a (i.e. the female dies when they reach reproductive age) we would have the condition

Which is exactly what you get if you formally make s =0. And we just stated this condition: when the female reaches their reproductive age, a, they die and hence have no adult female survival probability.

As you will determine in your exercise, and as we will see below,
the addition of * w * makes a very big difference. Lets take
the case of * w * = 15 years for the previous data set. We
then have the same right hand side as before (0.024) and the
same numerator as before:

l^{3}(1 - .95/ l)

But this is now corrected by the term in the denominator which is

(1 - .95/ l)^{13}

Putting this into excel and incrementing lambda until we get an approximate solution yields lambda = 0.85 or a fractional decrease in the population of 15% per year - this is a lot bigger than the previous decrease of 2% a year and we get a population crash time of 56 years, instead of 450 years.

So, * w * makes a big damn difference!

Equilibrium Occupancy:

Finally, let's consider an ecosystem which is composed of a bunch of individual cells or areas. Within each cell or areas one can identify 3 possible conditions

- the cell is occupied by females; we will call this
condition p .
- the cell is habitable but is currently not occupied;
we will call this
condition h .
- The cell is unfit for habitation.

In general, a species will not inhabit all of
the available habitat as that could lead to extinction (as in the
classic overgrazing scenario). A working model is that
habitable and uninhabitable patches of land are randomly dispersed
throughout the ecosystem. This model makes an important prediction
for * p *
which is the equilibrium occupancy of suitable habitat
by females (the model assumes the males will find them ...) and it has
two parameters *h* is the proportion of the region which is habitable
and *k* is the demographic potential of the species.

Demographic
potential is the equilibrium proportion of total territory that would
be occupied by females in a completely suitable region. * k* is not
something easily measured but *p* and *h* can be.
These are related by the following:

For simplicity we will use now use **p** for **p _{o}**

Note that in the unique case of *h* = 1 (i.e. 100% of the
ecosystem is habitable), then * p * (the observed occupancy)
will equal * k * and the species is at is demographic potential.

More generally, lets consider a case of *h* = 0.5 for
a species with * k * = 0.7. In that case, we should observe
a value of

* p * = 1 - 0.3/0.5 = 0.4

So 40% of my ecosystem should be occupied by adult females.

If **h** is known exactly (which we will pretend that it is) then the
variance in k can be determined as follows:

Below is a data set each with N_{p} = 10.

Data Set 1 (from university biologists):

- p = 0.4
- h = 0.38

What do we get for this.

Putting in the values for p and h and solving for k shows that k = =0.77.

The standard deviation of k (which is the square root of the variance)
would be 0.06 in this case of N_{p} = 10. And therefore
k is determined to an accuracy of 6% k = 0.77 +/- 0.06.

If we assume, for the sake of argument, that before development, *
h * was 0.65 then that would lead to * p * = 1 - .23/.65 = 0.64
compared to the current value of * p * = 0.4. On this basis,
would could therefore argue that development has lead to a substantial
reduction of the species throughout the ecosystem.

However, this is now where you have to consider the error bars on k.
For instance a k value of 0.71 is perfectly consistent with the data
and that leads to * p * = 0.54 which is less of a difference.

Therefore, in the real world, its important to be able to determine
* k * to pretty high accuracy.

Finally we consider the critical case of * h * = 1 - * k *.

If * h * in the real world falls below this critical value,
then the species will no longer have enough habitat to support them.

Formally, for * k * = 0.77 we would get *h _{critical } * =
1 - 0.77 = 0.23 which is less than the present

That means you could argue that 15% more of the total ecosystem could
be developed without killing the species. However, once again, errors
on * k * are important. If you use a * k * = 0.71 this
gets you to *h _{critical } * = 0.29. So unless you know