Bio 481 - Problem Set 1

General Note: a^b stands for a raised to the b power.

1. If every individual produces one baby per day, how many individuals will be in the population after four days if N(0) = 1? If N(0) = 3? (Assume that no deaths occur and that a baby produced today does not reproduce until tomorrow.)
   
   This follows the exponential map of population growth: N(t+1)=R*N(t). Remember that R is supposed to contains birth, death rates, and survivorship. Here we have no death, 100% survivorship, and 1 offspring. Because of this R = 1 (for survivorship of the current individual) + 1 (for the new offspring). Therefore R =2. N(4) = N(0)*(R^4). When N(0) is equal to 1, N(4) = 1*(2^4) = 16. When N(0) = 3, N(4) = 3*(2^4) = 48.
   
   
2. If every individual produces four babies per day and N(0)=5, what will the values of N(1), N(2), N(3), and N(4)? Plot N(t) against t.
   
   This is essentially the same problem as number one. Here R = 1 (survivorship) + 4 (new births). N(1) = (5^1)*5 = 25. N(2) = (5^2)*5=125. N(3)=(5^3)*5=625. N(4)=(5^4)*5=3125. NOTE: quite a few people said that R was 4 here, rather than 5. To prove that it is 5, look at the case when there are no babies. If there are no deaths, N(t+1) = N(t). Here R=1.
   
   
   Exponential with N(0)=5, R=5.
   
   
3. Using the assumptions of exercise 1, what is the parameter of the exponential equation? (obviously, here as elsewhere, you should be able to explain how you get the answer, not just write it down)
   
   R = the rate of population growth in an exponential equation. The difference equations form is N(t) = N(0)*(R^t). In this case, R=2 since the population doubles every day.
   
   
4. Using the exponential differential equation, let r=0.83 and begin with a population size of 2. Plot N(t) against t for t=0,1,...,10.
   
   The differential exponential equation is: dN/dt=r*N. Here, r=0.83 and N(0) =2. The integrated form of this equation is N(t)=N(0)*(e^rt).
   
   
   
   
5. Repeat 4, but plot lnN(t) versus t. How would you estimate r, if you had only the data points and did not know the parameter?
   
   The plot is shown below. You can figure out r, since the slope of lnN(t) versus t is equal to r.
   
   
   
   
6. What is the doubling time (how long will it take for the population to double in size -- assuming an exponential process) if r=0.993 and N(0)=10? If r=0.993 and N(0)=20?
   
   It is easiest to do this analytically. N(t)=N(0)*(e^rt). For the population to double, we require N(t)=2*N(0). Therefore 2*N(0)=N(0)*(e^rt). This implies that e^rt=2. We are solving for t, so t= ln(2)/r. Thus t=.698. As you can see, if does not matter what the initial population size was. It will always take the same amount of time to double in size (for the same r value).
   
   
7. Derive a general equation for doubling time; for tripling time.
   
   We essentially derived the equation for doubling time in problem 8. Time to double = ln(2)/r. To triple, N(t)=3*N(0). Therefore e^rt=3. The time to triple is ln(3)/r.
   
   
8. Let N(0)=.1, K=1.0, and r=1. Plot the solution to the logistic differential equation. How long does it take to get to K? How long does it take to get to 90% of K?
   
   The plot to the logistic differential equation is shown below. If the population does not start at K, it asymptotically approaches K. Therefore it reaches K only when t approaches infinity. Note that the computer cannot keep track of very small differences so at some point, it rounds the population to K. N(t)=.9*K when t is approximately 4.4.
   
   
   
   Continuous logistic equation with r=1, K=1, and N(0)=.1.
   
   
9. Let K=100, N(0)=.1, and r=1.  Plot the solution to the logistic differential equation.  How long does it take to get to K?  How long does it take to get to 90% of K?  Repeat for r=2.0.
   
   Again, remember that the population only approaches K when t approaches infinity.   For r=1.0, N(t)=.9K when t is approximately 9.1.  For r=2.0, N(t)=.9K when t is approximately 4.55. The takehome message is that how fast the population reaches K is more a factor of r than what K actually is.
   
   
   
   Continuous logistic equation with r=1, K=100, N(0)=.1.
   
   
10. Begin with a population of 2 individuals: r=0.83, K=302.  Use the logistic equation to compute N(t) for t=0,1,2,.. 15.  Plot N(t) against t.  Do the same for the exponential equation and r=0.83.  Compare the two graphs.
    
    Here we can using the continuous time logistic and exponential equations.  The main point you should notice is that the logistic equation acts very much like an exponential equation with the same value of r at low population levels.  Only when the population increases, does density dependence begin to play a major factor in limiting population growth.
    
    
    
    Continuous logistic and exponential equations with r=0.83, K=302, and N(0)=2.
    
    
11. For the data generated in exercise 10, plot ln{[1/{K-N(t)}]*N(t)} versus time.  Draw a line through the points.  What is the slope of the line?
    
    The plot to the above function is a line.  If the computer doesn't have any rounding errors, you would be able to see this forever. But it does have some problems after a long enough time. It is a straight line with a slope of r.
    
    
    
    Plot of ln{[1/(K-N(t))]*N(t)} vs. time for the continuous logistic equation where r=.83, K=302, and N(0)=2.
    
    
12. Use Maple or a similar application for graphing functions and print out graphs for the logistic map where r=1.8, 2.2, 3.1, 3.8, and 4.1 (i.e. five different graphs).  By hand, stair step the population for each of these graphs.
    
    Note that to stairstep, you must have a graph of N(t+1) vs. N(t).  This curve gives you all possible projection values.  When you use the 45 degree line, you are performing one time increment of the projection.  A time series plot is equivalent to a single stairstepping where N(0) represents the point where you chose to start stairstepping the population. Plots are shown with problem 15.
    
    
13. Using the exponential map with migrations, let r = .9 and m=.2. What is the equilibrium point? Is it an attractor or repellor? Project the population beginning at various points to convince yourself that the equilibrium is what you say it is. (Vanderplot might be useful here).
    
    I would only use Vanderplot for projecting the population, not for solving for the equilibrium point. Instead an analytical solution is easiest. At equilibrium, we know that N(t+1) = N(t). The general form for the exponential map with constant migration is N(t+1) = r*N(t) + m. Here, N(t+1)=.9*N(t)+.2. Now we can solve for N(equilbrium). N(eq)=.9*N(eq) + 0.2. Then, 0.1*N(eq) = 0.2. This yields N(eq)=2.0. This point is an attractor. To prove this start with values slightly above and below N(eq). If they approach N(eq) with an increase in time, it is an attractor. Let N(t)=1.9. Then N(t+1)=1.91. Let N(t+1)=2.1. Then N(t+1)=2.09. In both cases, the system is attracted to N(eq). The following Vanderplot graphs also illustrate this.
    
    
    
    Shown above: approaching the equilibrium from N(0)=1.9 and N(0)=2.1. r=.9, m=.2.
    
    
14. Using the exponential map with a constant predation factor, let r=1.5 and p=.2. What is the equilibrium point? Is it an attractor or repellor? Project the population beginning at various points to convince yourself that the equilibrium is what you say it is.
    
    The exponential map with predation is N(t+1)=r*N(t) - p. Here, N(t+1) = 1.5*N(t) - .2. Since we know that N(t+1)=N(t) at equilibriun, N(eq) = 1.5*N(eq) -.2. Then, 0 = .5*N(eq) -.2. Therefore, N(eq)=.4. This is a repellor. When N(t)=.3, N(t+1)=.25. When N(t)=.5, N(t+1)=.55. In both cases, the system was repelled from .4. The following Vanderplot graphs show the same thing using population projections.
    
    
    
    Shown above: repelling away from the equilibrium when N(0)=.45 and N(0)=.35. r=1.5, p=.2.
    
    
15. Use Vanderplot or Maple to solve the logistic map for the above values of r (i.e. do five different projections of the population over time).  Qualitatively describe the behaviour of each.
    
    Here I have included plots for both 14 and 15.  The plots are grouped together by r value.  An r of 1.8 reaches equilibrium with a simple increasing trajectory.  A slightly larger r value (in this case 2.2) can overshoot the equilibrium value, oscillate around that equilibrium value, and eventually converge on the equilibrium point.  In our case, it only overshot the equilibrium once and didn't oscillate before converging on the equilibrium.  An r of 3.1 is within the range where we have a two-point oscillation.  An r of 3.8 describes a strange attractor.  Finally, an r of 4.1 is large enough that the population overshoots too far and density dependence causes the population to crash.  This can be seen by looking at the simple logistic map: N(t+1)=R*(1-N(t)).  If you look at the graph of N(t+1) vs. N(t) for r=4.1 there is a region of the curve where N(t+1) is greater than one.  This represents where the population will increase to greater than 1.0.  If you look to the equation, N(t+1) is negative for N(t) > 1.  This is what causes the population to go extinct (in a relatively short amount of time in our example).
    
    
    Shown above: logistic map with r=1.8 (simple equilibrium/point attractor).
    
    
    Shown above: logistic map with r=2.2 (point attractor, but overshot slightly).
    
    
    Shown above: logistic map with r=3.1 (two-point attractor).
    
    
    Shown above: logistic map with r=3.8 (strange attractor).
    
    
    Shown above: logistic map with r=4.1 (extinction).
    
    
16. Construct the bifurcation diagram for the logistic map.
    
    
    
    Shown above: logistic map bifurcation diagram for values of r between 1.0 and 4.5.
    
    
17. Construct the bifurcation diagram for the logistic map with constant predation, using r as the bifurcation parameter.
    
    Note that the graph is identical to problem 16, except shifted to the right. Also the degree of shift is not equal to p, but is greater.
    
    
    
    Shown above: logistic map with constant predation (p=.2).
    
    
18. Construct the bifurcation diagram for the logistic map with constant predation, using the predation coefficient as the bifurcation parameter.
    
    Here you can see that increased predation causes the population to go from a two-point oscillatory attractor to a stable point attractor, and eventually to a saddle node bifurcation point where the population becomes extinct.
    
    
    
    Shown above: logistic map with constant predation. K=1.0, r=3.2.
    
    

Extra: a more in-depth look at how predation can affect a logistic curve

• a) Given the general logistic equation (differential), how would you modify it to include a constant predation rate?  What does this do the graph of the logistic equation?
  
  A good form for this equation is: dn/dt = r*N((K-N)/K) - P
  A constant predation rate has the effect of moving the whole graph of the logistic equation down by the value of P. Note that you often start both equations at the same N(0). Since some of the logistic equation with predation has a negative trajectory, the plot is moved to the down and to the right.
  
  
  
  Shown above: logistic equations with K=100, r=2.5. Npred has a constant predation term of p=2.0.
  
  
• b) Again consider the general logistic equation.  This time develop a modified equation that includes a per-capita predation rate that decreases with population density (saturates at high density).  Plot one example of this using any values for r, K, N(0), and your predation parameters.  What would you expect to happen if predation saturates at a point below K, if it saturates above K?
  
  One form of density-dependent predation is: dn/dt = r*N((K1-N)/K1) - P*N((K2-N)/K2). If K1=K2 (they saturate at the same value), this can be rewritten as dn/dt = (r-P)*N((K-N)/K). Note that there is the problem that predation has a positive effect on the population when it is above carrying capacity. This can be explained in the following way: suppose you had a population with a carrying capacity of 100. If it was at a population of 150, density dependence would have a negative effect on the population, causing it to decrease dramatically. On the other hand, if predation ate 50 individuals, the "effective population size" for density dependence was only 100. Here predation would counteract the negative effect of the density dependence.
  
  
  Here the above model is represented with r=2.5, K=100, N(0)=1, and p=2.0. If the two K's are equal, predation merely has the effect of slowing the intrinsic rate of increase.
  
  
  Here the above model is represented with r=2.5, K1=100, p=2.0, N(0) =1, and K2=75. Here predation has a positive effect at smaller population levels (some below K1). Although it slows down the rate of increase, the population increases to reach a stable point attractor above K1.
  
  
  Here the above model is represented with r=2.5, K1=100, p=2.0, N(0) =1, and K2=150. Here predation has a positive effect only at large population levels (above K1). It slows down the rate of increase, as well as causing the population to reach a stable point attractor below K1.
  
  There are several problems with above model of density-dependent predation as described in the above model. Instead, I like a form where the amount of predation saturates quickly to reach a constant level (the graph of number eaten as a function of N would look like an increasing curve which asymptotes at a maximum value). Here the predation term has the following form: - Pmax*(N/(alpha + N)). Here Pmax is the maximum number that the predator can consume. Alpha is a measure of how fast the number eaten asymptotes (faster for a smaller alpha). Note that we have to start the population at a larger N(0) here or else it will crash from the predation pressure.
  
  
  Logistic model with K=100, r=2.5, N(0)=15, Pmax=10, alpha=1. Pred shows the numbers eaten at a given time (notice that it saturates almost instantly). Here the model increases slower and reaches a stable point attractor at a value lower than K (the difference between the attractors is less than Pmax).
  
  
  Logistic model with K=100, r=2.5, N(0)=15, Pmax=25, alpha=10. Pred shows the numbers eaten at a given time (notice that it saturates slower than before due to the larger alpha). Here the model increases slower and reaches a stable point attractor at a value lower than K (the difference between the attractors also is less than Pmax).
  
  
• c) Describe 5 examples of mechanisms that might provide density-dependent feedback in actual biological systems.
  
  Here I was expecting a little bit of an explanation about how the feedback mechanism would provide density-dependence. For example, with predation, an explanation like: if the per-capita predation level increased as a function of population density, then we would expect predation to have a greater effect at high population levels (negative density-dependence). One reason this might occur is if at higher population levels of the prey, the predator switched from another food source to eating our target prey species more frequently.
  
  So, I'm not going to write this reasoning out for all the mechanisms (instead why don't you go ahead and think it out for yourselves), but here are just a few mechanisms (negative density-dependence only):
  
  competition for a resource that exists at a constant level (or constant supply rate)
  increased chance of disease transmission (or increased mortality) at higher population densities
  a fixed amount of habitat space (it becomes harder to find suitable habitat at higher densities)
  our target species mimics a poisonous species (eg. viceroy) as it increases in density, predators encouter the mimic more frequently are more likely to predate on both species with increasing proportions of the mimic
  increased cannabilism (or aggressive interactions) in the population with increasing density
  bacterial system where at increasing density, waste toxin concentration increases
  

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