In this project, we are to model the population growth of kafit birds in the Greater Boomchalaka game preserve. The populations were tabulated in several years as shown in the table below.
year | 0 | 1 | 5 | 6 |
# birds | 470 | 940 | 25,220 | 93,590 |
1.
We are to find the cubic polynomial
C(x) = a x^{3} + b x^{2} + c
x + d that goes through all four of the data points, then use this to
estimate the population of kafit birds in years 2, 3, and 4.
Since the polynomial is required to pass through the given points,
each of the points must satisfy the functional relationship. Thus,
whatever a, b, c, and d are, the following equations must all be
true.
We need to solve these equations simultaneously. There are many different schemes for doing this, although at some level they are all the same thing. We will just solve an equation for one variable in terms of the others, and substitute that into the next equation. Other methods may be more efficient, but this way is completely straightforward.
From C(0), we see immediately that d=470. Substituting into
the next equation, we see that
If C(x) described the population of kafit birds exactly, then plugging in the year for x should give us the population in that year.
year | population | comment | |
0 | C(0) = 470 | 470 | Matches our data |
1 | C(1) = 1890 - 10220 + 8800 + 470 | 940 | Matches our data |
2 | C(2) = 15120 - 40880 + 17600+ 470 | -7690 | A negative population |
3 | C(3) = 51030 - 91980 + 26400 + 470 | -14080 | Also negative |
4 | C(4) = 120960 -163520 +35200 + 470 | -6890 | Still negative |
5 | C(5) = 236250 -255500 +44000 + 470 | 25220 | Matches again |
6 | C(6) = 408240 -367920 +52800 + 470 | 93590 | also OK |
Here is a graph of C(x), with the known populations marked by
circles. Note that it passes through each of them, but is negative
from early in year 1 until the beginning of year 4.
This seems a little suspect as a description of bird populations, but everything was done correctly. Note that all the populations were were given match the values of C(x), but the predicted populations for years 2, 3, and 4 are negative, which is not physically possible. So it seems very unlikely that C(x) is a good choice of model for the population of kafit birds.
We could try to find a more reasonable cubic polynomial by relaxing the requirement that our model polynomial pass exactly through all the points. This is not an unreasonable thing to do, since the model is only intended to approximate the given data, not match it exactly. Also, the population counts are probably inexact, as well.
However, in order to get a cubic polynomial which stays positive, I had to adjust the population counts by approximately 25%. For example, using the data points (0,470), (1,1175), (5,32000), and (6,70000), we obtain the cubic polynomial which gives us estimated populations of 22, 1669.5, and 10776 in years 2, 3, and 4. The graph is shown below.
Another possibility is to use a higher degree polynomial. If we estimate (somewhat arbitrarily) a value for the population in, say, year 3, we can perform the same procedure as before to find the unique fourth degree polynomial which passes through all 5 of the given points. On the left below are the graphs of the polynomials resulting from 10 such choices. The general shape seems almost reasonable, look at the magnification of the graphs for 0 < x < 4 on the right. Note that while the population estimates which result are all positive, the shape of the graph is less than fully satisfying. The functions drop dramatically in the first and third years.
We could, of course, use an even higher degree polynomial to try to get a shape that matches better with our intuition. However, the whole process is beginning to seem quite artificial. The problem is that the shape of a low-degree polynomial is just too restrictive to model this data well.
2.
For the second part, we try to find an exponential function
that models the data, and use it to estimate the population
in years 2, 3, and 4.
Because E(x) depends only on two numbers k and a, we can only force it to pass exactly through two of our data points. We can, however, try to make it come as close as possible to the others.
First, it seems reasonable to assume that the population for year 0 is exact, and the other years may contain counting errors, because we should know exactly how many birds were resettled, but counting birds in the wild can be a tricky, inexact process. If we make this assumption, we have , and so we must have k=470.
To find the value of a, we can pick any one of the other data points, and solve. Let's use (5, 25220) to illustrate the process.
Plugging in, we have
This gives as our function.
It should be pretty clear that if we had used one of the other points instead, we should get either or .
The graphs of all three are shown above. The best of the three seems to be E_{5} (the one in the middle). Using E_{5} to estimate the bird populations in years 2, 3, and 4, we obtain 2311, 5127, and 11371, respectively. Note that E_{5} gives a population of 1042 in year 1 and 55934 in year 6. These are relative errors of about 20% and 40%-- not so great.
Of course, there is no reason we need go through any of the
points exactly. We could just play around with the values of a and
k until we got something that looks better, or pick a pair of points
that seem ``sort of in the middle'' and repeat the process above, or we
could average the points or the exponents in some way. However, there is
another way. Note that if we take the logarithm of E(x), we get
Taking the natural logarithm of the populations, we get the data to
fit as
Using regression, the best fit line is found to be
We should remark that with so little information (only four points),
making a good model is nearly impossible. Even with sufficient data,
modelling can be a very tricky business.