Answers to Exercise 8

Transcription

1 Answers to Exercise 8 Logistic Population Models 1. Inspect your graph of N t 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 N t increases. Per capita death rate increases as N t increases. This is the meaning of density dependence: per capita birth and death rates change as N t changes. Per capita birth and death rates become equal when N t reaches K. (This was demonstrated algebraically in the Introduction to this exercise.) 3. Inspect your graph of N t and N t /N t, or dn/dt and dn/dt/n, against N t. You should see the following: N t or dn/dt starts low, rises to a maximum when N t K/2, and then declines to 0 when N t reaches K. This confirms what you saw in the graph of N t against time: The population grows fastest when its size is half its carrying capacity. This property of logistic models will be important in our harvesting model. N t /N t or dn/dt/n starts high, declines linearly as N t increases, and reaches 0 when N t reaches K. These two features differ from the exponential model, in which N t increased linearly with N t and N t /N t was constant. 4. If you extrapolate the line for N t /N t 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 x- intercept 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 Change the initial population size to 100 by entering that value into cell B6. Look at your graph of N t 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. Exercise 8 Page 1 of 6

2 Try other values for initial population size and see what happens. If you make the initial population too large, N t 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 N t. It should resemble the graph below. Logistic Model, Explicit b and d Per capita birth and death rates Population size (Nt) Birth rate Death rate 6. Carrying capacity is a stable equilibrium. Two lines of evidence support this: Population size moves toward K from below (N 0 = 1) and from above (N 0 = 100). The lines for per capita births and deaths (graph above) cross at N t = K. When N t < K, per capita births are greater than per capita deaths, so the population will grow. When N t > K, per capita births are less than per capita deaths, so the population will shrink, returning toward K. At N t = 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 N 0 to 1.00, and set b = 2.00 and d = R will become Try making b = 0.04 and d = 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 N t should be a sigmoid curve stabilizing at N t = 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 N t = 100. Exercise 8 Page 2 of 6

3 Exercise 8 Page 3 of 6

4 Try making b' = 0.02 and d' = 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 N t should be a sigmoid curve stabilizing at N t = 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 N t = 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 N t is small. Per capita births < per capita deaths when N t 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 = Set b' = 0 and d' = 0 and examine your graphs and spreadsheet. Then set b' = and d' = 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 N t /N t 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 N t against time to a logarithmic scale. It should become a straight line. Also examine the graph of N t /N t 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 = Set b' = and d' = 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 N t. If your graph of N t 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 N t /N t 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 Exercise 8 Page 4 of 6

5 exponentially growing population, in which N t /N t 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' = If you built the model with explicit carrying capacity, set N 0 = 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 N t 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: N t shows smooth sigmoid growth, stabilizing at K b = 2.50, R = 1.50: N t 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 N t will reveal that these oscillations are also damped, but it will take much longer for N t 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! N t 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 N t values. Examine your graph of N t and N t /N t against N t. You should see that even in the most chaotic scenarios, N t still forms a smooth parabola with its peak at N t = K/2, and N t /N t still Exercise 8 Page 5 of 6

6 forms a straight line with its y-intercept at R. This shows the order underlying the chaos of N t. In other words, the values of N t 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(n t+1 /N t ). We can perform this calculation on human population estimates for 1963 to 2000 (available from the U.S. Census Bureau Web site, to estimate dn/dt/n for each time interval. This is one way to set up the spreadsheet: A B C D Date Year Nt dn/dt/n ,205,706, ,276,816, ,345,837, ,416,065, ,485,807, ,557,675, ,632,341, ,707,610, ,785,190, ,862,197, ,938,708, ,014,598, ,088,224, ,160,391, ,232,928, ,305,403, ,380,776, ,456,705, ,532,964, ,613,401, ,693,932, ,773,566, ,854,602, ,937,607, ,023,570, ,110,153, ,196,333, ,283,755, ,366,938, ,449,663, ,531,001, ,610,978, ,690,865, ,768,612, ,846,804, ,924,574, ,002,509, ,080,141,683 Exercise 8 Page 6 of 6

7 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 N t. 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 y = -3E-12x R 2 = Population size (billions) The R 2 here refers to the coefficient of determination, not to the geometric growth factor. Such a high value of R 2 indicates that the trendline fits the data very well. The equation of the line is y = (10 12 )x, where y = dn/dt/n and x = N t. From this we can see that the y-intercept is , 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 Exercise 8 Page 7 of 6

8 The Census Bureau projects a population of 9.1 billion in the year 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 Remember, however, that projections are not guarantees. Future human population may behave quite differently, and such prognostications are an active area of research. Exercise 8 Page 8 of 6