Optimization growth rate Question

In summary, the intrinsic growth rate of the fin whale population is 0.08, the maximum sustainable population is K=400,000, and the number of whales harvested per year is about 0.00001 Ex. It is estimated that the growth rate of the fin whale population will eventually stabilize at the level where harvest rate equals growth rate.
  • #1
lazystudent
4
0
First post on these forums, thanks all for your help!

Homework Statement


It is estimated that the growth rate of the fin whale population (per year) is rx(1 - x/K), where r = 0.08 is the intrinsic growth rate, K = 400,000 is the maximum sustainable population, and x is the current population, now around 70,000. It is further estimated that the number of whales harvested per year is about 0.00001 Ex, where E is the level of fishing effort in boat-days. Given a fixed level of effort, population will eventually stabalize at the level where growth rate equals harvest rate.

EDIT- FORGOT A, B, AND C.

a. What level of effort will maximize the sustained harvest rate? Model as a one-variable optimization problem using the five step method.

b. Examine the sensitivity to the intrinsic growth rate. Consider both the optimum level of effort and the resulting population level.

c. Examine the sensitivity to the maximum sustainable population. Consider both the optimum level of effort and the resulting population level.

Homework Equations



G(x) = rx(1 - x/K)
r = 0.08 is the intrinsic growth rate
K = 400,000 is the maximum sustainable population
x is the current population, now around 70,000

H(x) = 0.00001 Ex

F(x) = G(x) - H(x) <------ this F(x) isn't given explicitly in the original problem, but I think it's necessary to solve the problem.

The Attempt at a Solution



Intuitively, I had thought you would differentiate F(x) with respect to E, and then solving for 0 (first derivative test to see when the slope = 0, therefore E would be maximized). However, that gives you dF/dE = .00001x which is basically useless in maxing H(x)

My next attempt will be to differentiate it with respect to x, and then perhaps run the first derivative test to see where population can be maxed? Because H(x) will be maxed when x is maxed (but E has to remain constant).

Whatever insight the reader of this post can provide to the writer will be greatly appreciated!

P.S. We are allowed to use Wolfram Mathematica to solve these problems.[/B][/B]
 
Last edited:
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  • #2
Hi lazystudent ! honest name :)
Yeah, I agree that your ##F(x)## function is important to this problem. But I don't think you should be differentiating it. First think of what that function means for this problem. ##G(x)## is a growth rate and ##H(x)## is the rate at which they are harvested, so this should tell you what ##F(x)## is, intuitively.
 
  • #3
Thank you for your response BruceW.

Well, the (nominal) growth rate minus the harvest rate will equal the REAL growth rate. But I believe the problem is asking for what level of constant E will max H(x).

I emailed my professor. I think your response just confused me more than I already was :( if we aren't supposed to use differentiation to optimize the harvest rate, I literally am stumped.
 
  • #4
lazystudent said:
Thank you for your response BruceW.

Well, the (nominal) growth rate minus the harvest rate will equal the REAL growth rate. But I believe the problem is asking for what level of constant E will max H(x).

I emailed my professor. I think your response just confused me more than I already was :( if we aren't supposed to use differentiation to optimize the harvest rate, I literally am stumped.

lazystudent said:
First post on these forums, thanks all for your help!

Homework Statement


It is estimated that the growth rate of the fin whale population (per year) is rx(1 - x/K), where r = 0.08 is the intrinsic growth rate, K = 400,000 is the maximum sustainable population, and x is the current population, now around 70,000. It is further estimated that the number of whales harvested per year is about 0.00001 Ex, where E is the level of fishing effort in boat-days. Given a fixed level of effort, population will eventually stabalize at the level where growth rate equals harvest rate.

Homework Equations



G(x) = rx(1 - x/K)
r = 0.08 is the intrinsic growth rate
K = 400,000 is the maximum sustainable population
x is the current population, now around 70,000

H(x) = 0.00001 Ex

F(x) = G(x) - H(x) <------ this F(x) isn't given explicitly in the original problem, but I think it's necessary to solve the problem.

The Attempt at a Solution



Intuitively, I had thought you would differentiate F(x) with respect to E, and then solving for 0 (first derivative test to see when the slope = 0, therefore E would be maximized). However, that gives you dF/dE = .00001x which is basically useless.

My next attempt will be to differentiate it with respect to x, and then perhaps run the first derivative test to see where population can be maxed? Because H(x) will be maxed when x is maxed (but E has to remain constant).

Whatever insight the reader of this post can provide to the writer will be greatly appreciated!

P.S. We are allowed to use Wolfram Mathematica to solve these problems.[/B][/B]

Given a value of ##E##, you have a recursion connecting the population ##x_{n+1}## at time ##t = n+1## with the population ##x_n## at time ##t = n##. This has the form
[tex] x_{n+1} = f(x_n), [/tex]
where
[tex] f(x) =x + r x \left( 1 - \frac{x}{k} \right) - c E x. [/tex]
Here, r = 0.08, k = 400,000, c = 0.00001 = 1.0 e(-5).
To find the "stable" (limiting value) ##x_{\infty} = \lim_{n \to \infty} x_n##, you can apply the "cobweb plot" method; see
https://en.wikipedia.org/wiki/Cobweb_plot or
http://www.ms.unimelb.edu.au/~andrewr/620341/pdfs/ga_sum.pdf , for example.

The actual level of ##x_{\infty}##---or if it exists at all---will depend on the value of ##E## that you choose. I guess that you are supposed to choose ##E## in such a way that something desirable happens to the value of ##x_{\infty}##, but your problem description does not say what that is.

I don't think differentiation has much to do with the problem, at least not until you have determined the value of ##x_{\infty}## as a function of ##E##.
 
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  • #5
Edited first post - totally freakin forgot A, B, C, and D!
 
  • #6
In rereading my attempt at a solution for A, I realized that the problems wants me to find at what constant E can we max H(x) given that F(x) = G(x) - H(x).

I don't know what exactly I was doing differentiating H(x) with respect to E.

EDIT

I differentiated F = rx(1-[x/k]) -0.00001Ex with respect to x, giving me

dF/dx= r(1-[x/k]) - (rx)/k - .00001E. I plugged in r=.08, K = 400000, and x = 70000. I set dF/dx = 0 and solved for E. I got

E = 5200.

Is this the correct max sustained harvest rate? Am I even on the right track?
 
Last edited:
  • #7
lazystudent said:
In rereading my attempt at a solution for A, I realized that the problems wants me to find at what constant E can we max H(x) given that F(x) = G(x) - H(x).

I don't know what exactly I was doing differentiating H(x) with respect to E.

EDIT

I differentiated F = rx(1-[x/k]) -0.00001Ex with respect to x, giving me

dF/dx= r(1-[x/k]) - (rx)/k - .00001E. I plugged in r=.08, K = 400000, and x = 70000. I set dF/dx = 0 and solved for E. I got

E = 5200.

Is this the correct max sustained harvest rate? Am I even on the right track?

The question stated clearly that the population will settle down to a sustainable value of x for which the growth rate = the harvesting rate. For any value of E, that condition gives you an equation in x, whose solution is the sustainable population. It will be a function of E, call it X(E). What is the harvesting rate corresponding to population x = X(E)? How can you maximize it?
 
  • #8
I agree with Ray :) You used x=70,000, which is the current population. But you should be finding the long-term 'sustainable population'. Also, you were differentiating F(x) but that means you were trying to find the max value of rate of population change? that is not a relevant quantity for this problem.
 

1. What is optimization growth rate?

Optimization growth rate refers to the process of finding the most efficient or effective way to increase a certain metric or achieve a desired outcome. This can be applied to various fields, such as economics, business, and computer science.

2. Why is optimization growth rate important?

Optimization growth rate is important because it allows for better resource allocation and decision-making. By finding the most efficient way to achieve a goal, businesses and organizations can save time, money, and resources. It also leads to improved productivity and competitiveness.

3. How do you calculate optimization growth rate?

The calculation of optimization growth rate depends on the specific scenario and goal. Generally, it involves analyzing data and identifying the factors that contribute to growth. Mathematical models, such as regression analysis and differential equations, can also be used to determine the optimal growth rate.

4. What are some strategies for optimizing growth rate?

Some strategies for optimizing growth rate include conducting market research, identifying areas for improvement, implementing new technologies or processes, and continuously monitoring and analyzing data. Collaboration and innovation are also key factors in achieving optimal growth.

5. What are the potential challenges in optimizing growth rate?

One potential challenge in optimizing growth rate is balancing short-term gains with long-term sustainability. Additionally, external factors such as economic conditions and competition can also impact optimization efforts. It is important to continuously adapt and adjust strategies in order to maintain and improve growth rate over time.

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