Question

a) You have an exponentially-growing population of rodents, with an annual λ of 1.35.

Calculate the number of individuals added to the population in a given month when there are 250, 500, 1000, 5000, or 10,000 individuals to start with. (2pts)

b) Now impose density dependence on the population with a logistic model in which K = 6500 individuals. Again, calculate the number of individuals added to the population in a given month when there are 250, 500, 1000, 5000, and 10,000 individuals to start with. (2pts)

c) How does population growth compare between the two models (exponential vs. logistic) at the low and at the high starting population numbers? (2pts)

Present the numbers (rounded to two decimal places) for a), and b) in a table in the space below, along with the answer for c).

Answer 1

for exponential growth,

N_t = N₀ x lambda^t

where N₀ = initial population size

t = time steps into the future (in years).

when N₀ = 250, = 1.35

then, N_t= 250 * (1.35)^1/12 = 250 * (1.35)^0.0833 =  250 * 1.03046 = 257.615

for N₀ = 500,

N_t = 500 * 1.03046 = 515.23

for, N₀ = 1000,

N_t = 1000 * 1.03046 = 1030.46

for, N₀ =5000

N_t = 5000 * 1.03046 =5152.3

N₀ = 10,000

N_t = 10,000 * 1.03046 = 10304.6

b) for logistic growth

ln() = r, ln(1.35) = 0.300104, r = 0.300104

for N₀= 250, K =6500

N_t = 6500/1+[6500-250/250](2.718)^-0.3*1.03 = 6500/1+[25]0.7334 = 6500/19.33 = 336.265

for N₀ = 500,

N_t = 6500/1+[12]0.733 = 6500/9.8 = 663.265

for, N₀ =1000

N_t = 6500/5.031 = 6500/5 = 1300

for N₀ = 5000

N_t = 6500/1.22 =5327.87

for, N₀ = 10,000

N_t = 6500/0.74 = 8783.784

c)