Question

Problem 2: Modeling inseet population dynamies The use of arrays and matrix algebra are ideal for modeling population dynamics of organisms. If n) is computed from represents the number of organisms at time t, then the number at a later time, t+1, can be n(t+1) An(t). The array (vector, n) consists of the nurnber of females in each age class, and the array (matrix, A), called a Leslie matrix, is the survival- replacement matrix. Multiplying the number of females in each age class at time t, by the survival-replacement matrix, gives the number of females in each age class at the next time step. We need only model the female population because they carry the entire reproductive potential for the population. Assume a female insect population consists of only 2 age classes, juveniles and adults. Assume that 60% ofjuveniles survive to become adults, and that each adult female produces 3 female Juveniles within a time step. The survival- replacement matrix is then: 0 0.6 Suppose also that there are 100 adults and 10 juveniles at the start (first generation) 100 10 n(1) Using the rules of matrix multiplication, and applying eq. 1, we can easily compute the numbers of females in each age class in the next generation. In this case, the number of new adults would be 0 x 100 +0.6 x 10- 6, and the number of new juveniles would be 3 x 100 +0 x 10-300. For the numbers in the next time step, matrix Λ would operate on the new numbers, 6 and 300 to give 180 and 18. You can see already that the population fluctuations are pretty wild. A few more steps would convince you that the general population level will continue to increase without bound. Write a program to compute insect populations through 20 (or more) generations. Create a plot which shows the population of adults, juveniles, and the total population for at least 20 generations. Here are some suggestions: n should be an array of the number of females, where row 1 is adults and row 2 is juveniles. Each column should contain the number of females for one generation. You'll need to initialize the first column of n (n(:,1) with the initial population values. lambda (A) is a 2 x 2 array for the survival-replacement matrix . Check a few values with hand calculations to be sure the program is doing what you expect it to. Once this is all working, try changing the values in the survival-replacement matrix so that, say, 40% survive, or there are only 2 juvenile females produced per adult female. (Make any changes you want- just something different than what I just described). Describe what happens to the population when you make those change. Hand your deseription and plots. Is the change what you would expect?
Thanks ahead of time!!!

Answer 1

The program will be as follows:

import matplotlib.pyplot as plt
import numpy as np
# n is the number of generations we are running the simulation
n = 20
lamb = np.matrix([[0, 0.6], [3, 0]])

# Initial number of adults and juveniles
nT = np.matrix([100, 10]).T

# Set up the vectors for calculation and plotting
x = np.linspace(1, n, n)
y1 = np.zeros(n)
y2 = np.zeros(n)
y3 = np.zeros(n)

for i in range(0, n):
    # Set the number of adults
    y1[i] = nT[0]
    # Set the number of juveniles
    y2[i] = nT[1]
    # Add the previous cumulative total to the current generation
    # (if we aren't currently calculating the first generation itself)
    y3[i] = nT[0]+nT[1] + (y3[i-1] if i != 0 else 0)
    # Perform the matrix multiplication and store it as the new value of nT
    nT = lamb * nT

# Print a few test values
print("Number of adults in generation 2: " + str(y1[1]))
print("Number of juveniles in generation 2: " + str(y2[1]))
print("Number of adults in generation 3: " + str(y1[2]))
print("Number of juveniles in generation 3: " + str(y2[2]))

# Plot the different values
plt.plot(x, y1, label='Adults')
plt.plot(x, y2, label='Juveniles')
plt.plot(x, y3, label='Total Population')

# Show the legend
plt.legend()

# Show the plot
plt.show()




import matplotlib.pyplot as plt import numpy as np f n is the number of generations we are running the simulation n20 lamb= np . matrix ( [ [0, 0.6], [3, 011) Initial number of adults and juveniles nT= np.matrix ( [100, 10]). T # Set up the vectors for calculation and plotting x np.linspace (1, n, n) yl- np.zeros (n) y3 = np. zeros (n) for i in range (0, n): # Set the number of adults # Set the number of juveniles y2[1] = n1 [1] # Add the previous cumulative total to the current generation # (if we aren't currently calculating the first generation itself) # Perform the matrix multiplication and store it as the new value of nT # Print a few test values print ("Number of adults in generation 2: str (yl[1])) print ("Number of juveniles in generation 2: str (y2[1])) print ("Number of adults in generation 3: str (y1 [2])) print("Number of juveniles in generation 3: str (y2 [2)) # Plot the different values plt.plot(x, yl, label-'Adults') pít.plot (x, y2, label= ' Juveniles') plt.plot (x, y3, label-Total Population') # Show the legend plt.legend ) # Show the plot plt.show ()

Graph:

_Adults Juveniles 175000 Total Population 150000 125000 100000 75000 50000 25000 2.5 5.0 7.5 10.0 12.5 15.0 17.5 20.0

Output:

Number of adults in generation 2: 6.0 Number of adults in generation 3: 180.0 Number of juveniles in generation 3: 18.0

Checking for accuracy of the program:

$\begin{bmatrix} 0 & 0.6\\3 & 0 \end{bmatrix} \times \begin{bmatrix} 100\\10 \end{bmatrix} = \begin{bmatrix} 6\\300 \end{bmatrix}$, which is correctly output by the program

$\begin{bmatrix} 0 & 0.6\\3 & 0 \end{bmatrix} \times \begin{bmatrix} 6\\300 \end{bmatrix} = \begin{bmatrix} 180\\18 \end{bmatrix}$, which is correctly output by the program

On changing the values in the lamb matrix from 0.6 to 0.4 and 3 to 2, we get a graph as follows:

_ Adults 1400 Juveniles Total Population 1200 1000 800 600 400 200 2.5 5.0 7.5 10.0 12.5 15.0 17.5 20.0

Note that the expected value of a "new" child in every generation = probability of survival * number of new offspring. If this is greater than 1, the population is expected to grow exponentially (as in the first graph). If this is less than 1, the population is expected to stagnate after some time (as in the second graph).

Let me know if there are any questions or queries!