Mathematical Models of Population Growth


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Exponential Growth

Under simplified conditions, such as a constant environment (and with no migration), it can be shown that change in population size (N) through time (t) will depend on the difference between individual birth rate (b0) and death rate (d0), and given by:

(equation 1)

dN/dt = (b0 - d0) / N0

where:

The difference between birth and death rates (b0 - d0) is also called r, the intrinsic rate of natural increase, or the Malthusian parameter. It is the theoretical maximum number of individuals added to the population per individual per time. By solving the differential equation 1 we get a formula to estimate a population size at any time:

(equation 2)

N = N0ert

where e = 2.718... (base of natural logs).

This equation shows us that if birth and death rates are constant, population size will grow exponentially. If you transform the equation to natural logarithms (ln), the exponential curve becomes linear, and the slope of that line can be shown to be r:

(equation 3)

ln(N) = ln(N0) + ln(e)rt      and

(equation 4)

r = [ln(N) - ln(N0)] / t

where ln(e) = 1. The population growth rate, r, is a basic measure in population studies, and it can be used as a basis of comparison for different populations and species.


Logistic Growth

(equation 5)

dN/dt = N [(b0 - kbN) - (d0 + kdN)]

We need to modify the basic equation 1 so that birth and death rates are no longer constant through time, but decrease and increase respectively as population size increases:

where kb and kd are the density-dependent birth and death rate constants respectively. This equation predicts that a population will stop growing (zero population growth) when birth rate equals death rate, or:

(equation 6)

b0 - kbN = d0 + kdN

This can be converted into an equation showing the size at which the population reaches a steady state:

(equation 7)

N = (b0 - d0) / (kb + kd)

The value of N when the population is at steady state is the carrying capacity of the environment, or K. This can be simplified:

(equation 8)

K = r / (kb + kd)

since b0 - d0 = r. If we combine this new form of the carrying capacity equation with equation 5 we get the familiar form of the logistic growth equation:

(equation 9)

dN/dt = rN [(K - N) / K]


Interspecific Competition

Mathematically, competition between species is treated as another density-dependent effect on the carrying capacity for each species. In the logistic equation 9, you have already been introduced to the density-dependent birth and intraspecific competition: as the number of individuals in one population increase, birth and death rates change because resources become limiting.

To incorporate the effects of interspecific competition (that is the interaction between individuals of different species) we simply include a competition term in the logistic equation 9. For the purposes of the model, competition now becomes defined as "the amount by which Species 1 lowers the K of Species 2, and vice versa," so that there are two competition coefficients for a two-species system.

Recall the basic logistic growth equation 9. For Species 1, we can incorporate a competition coefficient, , into this:

(equation 10)

dN1/dt = r1N1 [(K1 - N1 - N2) / K1]

where is the number of individuals of competing Species 2 which are equivalent to Species 1. Similarly, for Species 2, is the strength of competition it experiences from Species A, incorporated as:

(equation 11)

dN2/dt = r2N2 [(K2 - N2 - N1) / K2]

where:

The competition coefficient determines how much of an effect the population of another species will have on the growth rate of a species. For example, if Species 1 has a competition coefficient, = 1.0, then this means that it is experiencing the same intensity of competition from an individual of Species 2, as it is from an individual of its own species. And if Species 2 has a = 2.0, then it is experiencing much higher interspecific competition relative to intraspecific competition. What would be the outcome in a mixed culture under these conditions? Intuitively, it would seem that Species 1 would win since it is experiencing less competition overall. However, the outcome also depends on the relative intensity of intraspecific competition (or the ratio of the Ks). In this case there is not necessarily just one outcome.

There are four possible outcomes of a two species competitive interaction, as follows:

As an example, in Gause's competition experiment with P. caudatum and S. mytilus, he reported the competition coefficients were = 5.5 and = 0.12 and the carrying capacities were K1 = 75 and K2 = 15.6, respectively. These coefficients and the carrying capacities suggest that P. caudatum (Species 1) should be eliminated by S. mytilus (Species 2) since:

5.5 > 75 / 15.6 = 4.8     and    0.12 < 15.6 / 75 = 0.21


Predation

The simplest model for predation is the Volterra predator-prey equation which has two parts. The change in prey population through time is described by:

(equation 12)

dV/dt = rV - (aV)P

and the change in predator population size through time is:

(equation 13)

dP/dt = b(aV)P - d0P

where:

The efficiency factor, b, measures how well the predator converts food into offspring. A less efficient predator will expend more of its energy on maintenance than on reproduction compared to an efficient predator. This model is simplistic because it assumes exponential growth of prey, and a predator can eat as many prey as it can catch -- that is, it never gets full!

The basic model gives several important results:

  1. Predator and prey populations will oscillate through time. These so-called cycles have been shown in some situations, such as the now classic example of lynx and snowshoe hares in Canada.

    If the predator and prey populations are roughly in balance (populations rise and fall in the short-term, but the long-term average size of the populations does not change) then several more results accrue:

  2. The predator abundance depends more on the prey growth rate (r) than on its own death rate (d0), whereas prey abundance is controlled more by predator death rate than by its own growth rate. This is the basis of the Volterra Principle.

  3. A lower feeding rate (a) for each predator will result in a higher prey population but also more predators too -- because more predators will be recruited to maintain the overall predation rate on the prey population.

  4. If the predators increase their efficiency (b) and produce more new predators from captured prey, prey abundance decreases, but predator abundance is unaffected.

A More Realistic Model of Predation

The following model is more realistic because it incorporates logistic growth for the prey populations, and rather than assume that predation rate is simply related to prey density, it incorporates a predator satiation factor. When the prey are very abundant, a predator can no longer continue to increase its feeding rate because it becomes full. The model is:

(equations 14 and 15)

dV/dt = r [1 - (V/K)]V - c[1 - e(-aV/c)]P


dP/dt = bc [1 - e(-aV/c))P] - d0P

where:

With this model, oscillations are possible, but they are not the only behaviour predicted. We also get stable coexistence.

  1. The prey carrying capacity, K, is important in controlling the type and stability of the predator-prey interaction. At low prey carrying capacity (K), both predator and prey populations remain fairly steady with only small fluctuations in each. Oscillations occur if K is large enough, and is partly due to the destabilising effects of satiation which allow the prey to "get away" from the predators for a while until they approach their own carrying capacity. Predator numbers, on the other hand, still increase until they begin to regulate the prey again and prey abundance decreases. As a result, predator numbers fall, and the cycle begins again. At higher K, the oscillations can get so large that one or other population goes extinct.

  2. The Volterra Principle still holds in general. In addition, changing the environmental quality for the prey (e.g., increase prey food) influences the predator's abundance, and not the prey's.

  3. There is an optimum feeding rate which maximizes the predator population size. Too high a predation rate, and the prey are overexploited and the resource will become depleted. Predator population size is maximized when the prey population equals half its carrying capacity level (0.5 K).


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