A sensitivity analysis was performed on the converter (d), where 10 values were

Question

A sensitivity analysis was performed on the converter (d), where 10 values were compared ranging from 0.10 to 0.40. Below are scatter plots of N at year 250 and P at year 250 for the different values of (d) (assume the system has reached a steady-state), with a table of these data given below 50 45 35 25 150 100 So 15 a. 10 0.1 02 0.3 0.4 0.5 0.3 0.4 0.5 0.1 0.2 Does the converter (d) have higher leverage for N or for P? d N at year 25 P at year 25a Here is a table of values for N and P at year 250 for different values of the converter (d). Assuming that the- system is in a steady-state, fill in the column for the necessary values for converter (a) given the value for -0.2-t-200 converter (d) and the steady value of N. that the0.1 100 45 40 35 30 0.3 300 0.4 400

Explanation / Answer

According to your data I have figured out that the Lotka-Volterra model is being used with following parameters:

N - > Prey population, P - > Predator population

dN/dt = bN - aNP, Where, b - > per capita reproduction rate of prey, a - > per capita rate of predation (fraction of prey population eaten per predator)
dP/dt = caNP - dP, Where, c -> no. of prey required to convert into one predator, d - > per capita mortality rate of predators

(Note: in the given graph, mortality rate (d) of predators is increasing, hence predator population is decreasing and prey population is in turn increasing.)

In the table, N and P at equilibrium is given along with d.
To find 'a' we do the following,
At equilibrium, dP/dt = 0
-> 0 = caNP - dP
-> a = d/cN   (eq . A)
We know d and N values (given in the table) but we don't know 'c'.

To find c we do the following:
1) Look at Prey population vs d graph, its slope will give you increase in prey population per unit increase in d.
Lets call it slope1, using the data in the table we get
slope1 = (200 - 100) / (0.2 - 0.1) = 100/0.1 = 1000, slope1 = 1000

2) Look at the Predators vs d graph, its slope will give you the decrease in prey population per unit increase in d.
Lets call it slope2, using the data in the table we get
slope2 = (40 - 45) / (0.2 - 0.1) = -5/0.1 = -50 (negative sign because the population is decreasing, we already know this so lets ignore the sign)
slope2 = 50

3) now,
no. of prey required to covert it into 1 predator (c) =
(decrease in predators per unit increase in d) / (increase in prey population per unit increase in d)
i.e. c = slope2/slope1
c = 50/1000 = 20, c = 1/20

Now to get a, just substitute the values in (eq. A) with given d and N, put c = 1/20

a = 0.02

d N P a = (d/cN) 0.1 100 45 (0.1*20)/100 = 2/100 = 0.02 0.2 200 40 (0.2*20)/200 = 2/100 = 0.02 0.3 300 35 (0.3*20)/300 = 2/100 = 0.02 0.4 400 30 (0.4*20)/400 = 2/100 = 0.02

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