Answers to Exercise 8
Logistic Population Models
1. Inspect your graph of Nt against time. You should see the following:
• Population size increases slowly at first, then accelerates (the curve gets steeper), then
decelerates (the curve gets less steep), and finally stabilizes. A curve of this shape is said to
be sigmoid, and is typical of logistic population growth. A geometrically or exponentially
growing population grows at an ever-increasing rate and does not stabilize.
• The size of the stable population is the carrying capacity, K, which is 50 in this example.
2. Inspect your graph of per capita birth and death rates against population size (this applies
only if you used the version of the model with explicit birth and death rates). You should see
the following:
• Per capita birth rate decreases as Nt increases.
•Per capita death rate increases as Nt increases.
This is the meaning of density dependence: per capita birth and death rates change as Nt
changes. Per capita birth and death rates become equal when Nt reaches K. (This was
demonstrated algebraically in the Introduction to this exercise.)
3. Inspect your graph of ∆Nt and ∆Nt/Nt, or dN/dt and dN/dt/N, against Nt. You should see the
following:
• ∆Nt or dN/dt starts low, rises to a maximum when Nt ≈ K/2, and then declines to 0 when Nt
reaches K. This confirms what you saw in the graph of Nt against time: The population
grows fastest when its size is half its carrying capacity.
• ∆Nt/Nt or dN/dt/N starts high, declines linearly as Nt increases, and reaches 0 when Nt reaches
K.
These two features differ from the exponential model, in which ∆Nt increased linearly with Nt
and ∆Nt/Nt was constant.
4. If you extrapolate the line for ∆Nt/Nt until it crosses both axes of the graph, its y-intercept is
R and its x-intercept is K. If you extrapolate the line for dN/dt/N, its y-intercept is r and its xintercept is K. You can use this method of analysis to estimate R (or r) and K for real
populations from periodic population counts or estimates, even if you know nothing about
per capita birth and death rates. You will use this analysis to answer Question 11.
5. Change the initial population size to 100 by entering that value into cell B6. Look at your
graph of Nt against time. In the discrete-time models, you should see the population start out
at 100, fall below the carrying capacity, and then increase back to the carrying capacity of 50.
Answers: Exercise 8
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Try other values for initial population size and see what happens. If you make the initial
population too large, Nt may go negative in the discrete-time models, which makes no sense.
If you used the model with explicit birth and death rates, look at your graph of per capita
birth and death rates against Nt. It should resemble the graph below.
Per capita birth and death
rates
Logistic Model, Explicit
b and d
1.20
1.00
0.80
Birth rate
Death rate
0.60
0.40
0.20
0.00
0
50
100
150
Population size (Nt)
6. Carrying capacity is a stable equilibrium. Two lines of evidence support this:
• Population size moves toward K from below (N0 = 1) and from above (N0 = 100).
• The lines for per capita births and deaths (graph above) cross at Nt = K. When Nt < K, per
capita births are greater than per capita deaths, so the population will grow. When Nt > K,
per capita births are less than per capita deaths, so the population will shrink, returning
toward K. At Nt = K, per capita births and deaths are equal and the population stays at
equilibrium.
7. You can use the spreadsheet to answer this question only if you built the version with explicit
birth and death rates. To simplify matters, restore N0 to 1.00, and set b = 2.00 and d = 1.00.
R will become 1.00.
Try making b′ = 0.04 and d′ = 0.05. This represents a scenario in which per capita birth rate
increases with increasing population size, while per capita death rate increases because of
resource limitation.
Your graph of Nt should be a sigmoid curve stabilizing at Nt = 100 = K. Your graph of per
capita births and deaths should show both increasing, but because per capita deaths increase
more steeply, the two lines still cross, producing a stable equilibrium at Nt = 100.
Answers: Exercise 8
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Answers: Exercise 8
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Try making b' = –0.02 and d' = –0.01. This represents a scenario in which per capita birth
rate decreases with increasing population size because of resource limitation, while per capita
death rate decreases.
Your graph of Nt should be a sigmoid curve stabilizing at Nt = 100 = K. Your graph of per
capita births and deaths should show both decreasing, but because per capita births decrease
more steeply, the two lines still cross, producing a stable equilibrium at Nt = 100.
The important point here is that the population will have a stable equilibrium size if the
following conditions are met:
• Per capita births > per capita deaths when Nt is small.
• Per capita births < per capita deaths when Nt is large.
You can use the spreadsheet to answer Questions 8 and 9 only
if you built the version with explicit birth and death rates.
8. Set b = 1.50 and d = 1.00. Set b' = 0 and d' = 0 and examine your graphs and spreadsheet.
Then set b' = 0.001 and d' = 0.001 and examine your graphs and spreadsheet.
In both cases, you should see that K becomes undefined—that is, the spreadsheet will show
an error message in the cell for K. (If your graph of ∆Nt/Nt shows nonsensical ups and downs,
make sure the minimum of the left-hand y-axis is set to zero.)
You should also see that the population grows at an increasing rate. Does it grow
geometrically (exponentially)? You can test this by changing the y-axis of the graph of Nt
against time to a logarithmic scale. It should become a straight line. Also examine the graph of
∆Nt/Nt and its column in your spreadsheet. All this should convince you that the population
grows exponentially.
In both scenarios, even though per capita birth and death rates are changing, the difference
between them is constant, as you can see by examining the graph of per capita birth and death
rates (they will be parallel lines). The result is exponential population growth.
Try other values of b' and d', keeping them equal to each other.
9. Set b = 1.25 and d = 1.00. Set b' = 0.005 and d' = –0.001. Examine the graph of per capita
birth and death rates. You should see per capita births increasing linearly, and per capita
deaths decreasing linearly, with increasing Nt.
If your graph of Nt against time has a linear y-axis, you will see what looks like an exponential
curve. Try changing the y-axis to a logarithmic scale. You should see that the line still curves
upward. What does this mean?
It means the population is growing faster than exponentially.
Examine the graph of ∆Nt/Nt and its column in the spreadsheet. You should see that the per
capita rate of population growth increases as the population grows, in contrast to an
Answers: Exercise 8
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exponentially growing population, in which ∆Nt/Nt is constant. Something like this may have
occurred in human populations during demographic transitions.
10. You can use the spreadsheet to answer Question 10 only if you built one of the discrete-time
models.
• If you built the model with explicit birth and death rates, set b = 2.00, d = 1.00, b' = –0.001,
and d' = 0.001.
• If you built the model with explicit carrying capacity, set N0 = 1.00, R = 1.00, and K = 500.
• If you built the model with explicit birth and death rates, increase b by small increments,
keeping the other parameters unchanged.
•If you built the model with explicit carrying capacity, increase R by small increments,
keeping K unchanged.
Your graph of ∆Nt will be easier to read if you remove the line connecting the data points.
Double-click on one of the data points. In the resulting dialog box, select the Patterns tab. In
the left-hand box, for Line, choose None.
You should observe a variety of behaviors. The values given here are a few interesting
examples. Try these, and experiment with others as well. We give values of b for the model
with explicit birth and death rates, and values of R for the model with explicit carrying
capacity.
• b = 2.00, R = 1.00: Nt shows smooth sigmoid growth, stabilizing at K
• b = 2.50, R = 1.50: Nt overshoots K slightly, but soon stabilizes at K.
• b = 2.75, R = 1.75: overshoot, followed by damped oscillations (decreasing amplitude),
eventually stabilizing at K.
• b = 3.00, R = 2.00: overshoot, followed by persistent oscillations. Careful examination of
the spreadsheet column of Nt will reveal that these oscillations are also damped, but it will
take much longer for Nt to stabilize at K.
• b = 3.10, R = 2.10: oscillations of very slowly increasing amplitude.
• b = 3.25, R = 2.25: essentially stable oscillations around K.
• b = 3.50, R = 2.50: an interesting pattern, called a 2-point limit cycle, meaning oscillations
of two amplitudes. A large oscillation is followed by a small one, which is followed by a
large one, and so on. Not only 2-point cycles, but 4-, 8-, and 16-point cycles exist—see if
you can find them.
• b = 3.75, R = 2.75: Chaos! Nt changes radically, and with no apparent pattern, from each
time to the next. If you change the initial population size, you will see a completely
different sequence of Nt values.
Examine your graph of ∆Nt and ∆Nt/Nt against Nt. You should see that even in the most
chaotic scenarios, ∆Nt still forms a smooth parabola with its peak at Nt = K/2, and ∆Nt/Nt still
Answers: Exercise 8
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forms a straight line with its y-intercept at R. This shows the order underlying the chaos of
Nt. In other words, the values of Nt are not truly random, and we can recover an interpretable
pattern from them with the proper analysis.
11. Recall from Exercise 7 that we calculated dN/dt/N of a continuous-time exponential model
from population sizes by the formula dN/dt/N = r = ln(Nt+1 /Nt). We can perform this
calculation on human population estimates for 1963 to 2000 (available from the U.S. Census
Bureau Web site, http://www.census.gov/) to estimate dN/dt/N for each time interval. This is
one way to set up the spreadsheet:
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
Answers: Exercise 8
A
Date
1963
1964
1965
1966
1967
1968
1969
1970
1971
1972
1973
1974
1975
1976
1977
1978
1979
1980
1981
1982
1983
1984
1985
1986
1987
1988
1989
1990
1991
1992
1993
1994
1995
1996
1997
1998
1999
2000
B
Year
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
C
Nt
3,205,706,699
3,276,816,764
3,345,837,853
3,416,065,246
3,485,807,350
3,557,675,690
3,632,341,351
3,707,610,112
3,785,190,759
3,862,197,286
3,938,708,588
4,014,598,416
4,088,224,047
4,160,391,803
4,232,928,595
4,305,403,287
4,380,776,827
4,456,705,217
4,532,964,932
4,613,401,886
4,693,932,150
4,773,566,805
4,854,602,890
4,937,607,708
5,023,570,176
5,110,153,261
5,196,333,209
5,283,755,345
5,366,938,089
5,449,663,819
5,531,001,812
5,610,978,348
5,690,865,776
5,768,612,284
5,846,804,802
5,924,574,901
6,002,509,427
6,080,141,683
D
dN/dt/N
0.02194
0.020845
0.020772
0.02021
0.020408
0.02077
0.02051
0.020709
0.02014
0.019617
0.019084
0.018173
0.017499
0.017285
0.016977
0.017355
0.017184
0.016966
0.017589
0.017305
0.016823
0.016834
0.016954
0.01726
0.017089
0.016724
0.016684
0.01562
0.015296
0.014815
0.014356
0.014137
0.013569
0.013464
0.013214
0.013069
0.01285
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In this case, the formula in cell D6 is =LN(C7/C6), and this formula is pasted down the
column to cell D42.
As you did with your continuous-time logistic model, you can now graph dN/dt/N against Nt.
If you determine the linear trendline for the resulting graph, you can find the overall r as the
y-intercept and K as the x-intercept. To add a linear trendline to the graph, click on the graph,
then open Chart | Add Trendline . Choose a Linear trendline and click on the Options tab. From the
options, select “Display equation on chart” and “Display R-squared value on chart.” The
resulting graph should look like the one below.
Human Population 1963 to 2000: Logistic?
Per capita change in population size
0.025
0.02
0.015
0.01
y = -3E-12x + 0.0305
R2 = 0.9356
0.005
0
3.0
3.5
4.0
4.5
5.0
5.5
6.0
6.5
Population size (billions)
The R2 here refers to the coefficient of determination, not to the geometric growth factor.
Such a high value of R2 indicates that the trendline fits the data very well.
The equation of the line is y = 0.0305 – 3(10–12)x, where y = dN/dt/N and x = Nt. From this we
can see that the y-intercept is 0.0305, which we can take as an estimate of r. A little algebra
shows that the x-intercept is 10,166,666,000, which we can take as an estimate of K. If we
use these values in the continuous-time logistic model, with an initial population size of
3,205,706,699 (the population size in 1963), we get excellent agreement between predicted
and observed population sizes through the year 2000.
Answers: Exercise 8
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The Census Bureau projects a population of 9.1 billion in the year 2050. Extrapolating
beyond the year 2000, our model reaches that population about 9 years later—not bad for an
estimate based solely on population sizes. Our model predicts a population of 10 billion in
the year 2122, and reaches K at 10,166,666,000 in the year 2756. Remember, however, that
projections are not guarantees. Future human population may behave quite differently, and
such prognostications are an active area of research.
Answers: Exercise 8
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