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Creating a logistic model

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Introduction

Mathematics Higher Level

Assignment:                Mathematics Portfolio Type II:

Creating a Logistic Model

School:                Trinity Grammar School

School Code:

Name of Student:        Cassian Ho

Student Number:

Introduction

If a hydroelectric project is expected to cerate a large lake into which some fish are to be placed, a biologist estimates that if 10,000 fish were introduced into the lake, the population of fish would increase by 50% in the first year, but the long-term sustainable limit would be about 60,000.

In order to estimate the growth rate of the population of fish, it is best to find a linear growth factor for. We do this by finding two ordered pairs in the form (u0, r0), (un, rn).

rn is the growth rate when the population is n.

since we know from the information given that when there are 60000 fish in the lake, the growth rate is stable i.e. 1, this can be represented as one of our ordered pairs: (60000, 1). We also know that when 10000 fish are in the lake, the growth rate is 50% i.e. population is multiplied by 1.5, so the second ordered pair is (10000, 1.5).

from these pairs we have:

r10000 = 1.5

r60000 = 1

and since we are trying to search for a linear function of the growth rate, we can also denote rn as:

rn = mn + b        where n = population of fish. Substituting the two ordered pairs, we have:

1.5 = m(10000) + b

1 = m(60000) + b

Putting this into the GDC, we find that the solutions to the two unknowns are:

m = -0.00001

b = 1.6

from this, we can make the conjecture for the linear growth factor of fish in terms of Un:

rn = -0.00001 × un + 1.6

Since geometric population growth models take the form: un+1 = r × un

and we also have that rn = -0.00001 × un + 1.6, we find that the function for un+1 is:

un+1         = (-0.00001 × un + 1.6) × un

Middle

8

75211.05288

0.330713673

9

24873.32356

2.545573763

10

63316.87986

0.854057286

11

54076.24259

1.260645326

12

68170.96248

0.640477651

13

43661.97792

1.718872972

14

75049.39373

0.337826676

15

25353.68721

2.524437763

16

64003.80542

0.823832562

17

52728.41897

1.319949565

18

69598.8537

0.577650437

19

40203.80828

1.871032436

20

75222.62933

0.33020431

Although normally, I would look for a logistic function, however, seeing the fluctuation of results after seeing the graph (see next page), I have concluded that the fluctuations and oscillations of population of fish around the sustainable limit is far too great for the estimated logistic function to be precise, and the function obtained would have so much error that no conjectures or generalisations can be formed from it.

Following the same method as above, the graph for an initial growth rate of 3.2 is: During smaller initial growth rates (i.e. r = 1.5, 2, 2.3, 2.5), it can be said that over time, the population size of fish will settle down and converge at a population of 60000.

When we look at higher initial growth rates however, (i.e. r = 2.9, 3.2), the population of fish does not converge towards the sustainable limit, but fluctuates around it. Looking at the graph for r=2.9, we see that these fluctuations in fact, get larger, until it settles down at a certain limit. The same can be said for the graph for r=3.2, where the population of fish oscillates in a pattern over time.

Model for Growth Rate = 3.5

When we have a new initial growth rate, the ordered pairs i.e. (un, rn) have now changed yet again to:

(60000, 1)

(10000, 3.5)

To find the linear growth factor, we form the two equations:

3.5 = m(10000) + b

1 = m(60000) + b

Using the GDC to solve this, we have that m = -0.00006 and b = 4.6 so the linear growth factor is:

rn = -0.00006 × un + 4.6

and therefore the function for un+1 for is:

un+1 = (-0.00006 × un + 4.6) un

If we let 10000 fish with an initial growth rate of 3.5 cultivate for a while, the population of growth would be:

 Year Population of Fish Growth Rate t un+1 =(-0.00006 × un +4.6) × un rn =-0.00006 × un +4.6 0 10000 2.9 1 40000 2.2 2 88000 -0.68 3 -59840 N/A

Putting this in a graph, we have:

We see that by the third year, the population of fish would have exterminated to 0 (a negative number of fish is impossible of course). We can conclude from this that as the initial growth rate of fish increases, the fluctuations and oscillations of population of fish every year will become greater, so much to the point that the fluctuations may lead to a population of fish ‘below zero’.

Although we only have data for the three years that the fish existed, the graph shows hints of trying to take the ‘S-shape’ form as the other graphs did (i.e. r = 2.9, 3.2), however its growth rate was much too high in the beginning, causing the population of fish to soar much above the sustainable limit. Since when there are 60000 fish, growth rate is 1, any population of fish higher than this will result in a growth rate of less than 1 i.e. number of fish will diminish. In this case, the number of fish has diminished to such an extreme extent that there are no fish left.

So far we have been looking at examples where a population of fish has been left alone to cultivate. In real life situations however, it is most likely that governments or other organizations would like to harvest from this population of fish. Therefore, another area that is worthy of our investigation is how a harvest would affect the population of fish in the lake.

Let us assume that the population of fish in a lake follows the first model made in this portfolio. That is:

r10000 = 1.5, r60000 = 1

this also means that the population of fish follows the equations that I had calculated before:

rn = -0.00001 × un + 1.6

and

un+1         = (-0.00001 × un + 1.6) × un

Let us say that a government body wishes to harvest from a population of fish, AFTER its population has settled (i.e. reached its sustainable limit). This means that until there is 60000 fish in the lake, no harvesting will take place, so 60000 is our ‘initial number of fish’.

Harvest Size of 5000

Imagine that 5000 fish were taken from the lake at the end of each year. How would that affect the population of fish in the lake? It is best to first draw up a table of values:

 Year No. of Fish at beginning of year Rate t un+1 =un × rn - 5000 rn =-0.00001 × un +1.6 0 60000 1 1 55000 1.05 2 52750 1.0725 3 51574.375 1.08425625 4 50919.83843 1.090801616 5 50543.44203 1.09456558 6 50323.11193 1.096768881 7 50192.82314 1.098071769 8 50115.32208 1.098846779 9 50069.06026 1.099309397 10 50041.38846 1.099586115 11 50024.81595 1.099751841 12 50014.88341 1.099851166 13 50008.92783 1.099910722 14 50005.3559 1.099946441 15 50003.21325 1.099967867 16 50001.92785 1.099980722 17 50001.15667 1.099988433 18 50000.69399 1.09999306 19 50000.41639 1.099995836 20 50000.24983 1.099997502

Putting the data for the population of fish in the last 20 years (following this model) in a graph, we have: As we can see from this graph, the long term sustainable limit of fish has dropped from 60000 to approximately 50000. So, in a way, we can say that it is feasible for people to harvest 5000 fish per annum, as a small harvest such as this would not deplete the population of fish, even though it will lower the maximum sustainable limit of fish.

Since Rate = -0.00001 × un +1.6, as the population of fish decrease, the rate of growth increases so that at some no. of fish, there will be an increase in 5000 fish due to growth, and a decrease in 5000 fish due to harvest. This leads to a new stable sustainable limit of fish.

Next, we should explore other harvest sizes to see how they effect the sustainable limit of the fish population. I will now explore harvest sizes 3000, 4000, 5000 (already done), 6000, 7000, 8000 and 9000.

The general formula for finding population of fish when there is a harvest is:

un = [(-0.00001 × un + 1.6) × un] – H

where H is the harvest size per year.

Harvest Size of 3000

We first use Excel to tabulate the data for us:

 Year No. of Fish at beginning of year Growth Rate t un+1 =un × rn -3000 rn =-0.00001 × un +1.6 0 60000 1 1 57000 1.03 2 55710 1.0429 3 55099.959 1.04900041 4 54799.87958 1.052001204 5 54649.53931 1.053504607 6 54573.54143 1.054264586 7 54534.95204 1.05465048 8 54515.31333 1.054846867 9 54505.30745 1.054946925 10 54500.20652 1.054997935 11 54497.60532 1.055023947 12 54496.27866 1.055037213 13 54495.60198 1.05504398 14 54495.25681 1.055047432 15 54495.08075 1.055049192 16 54494.99094 1.055050091 17 54494.94513 1.055050549 18 54494.92176 1.055050782 19 54494.90984 1.055050902 20 54494.90376 1.055050962

Then, putting this into a graph, we have: The sustainable limit seems to have dropped to approximately 54494 fish.

Harvest Size of 4000

We first use Excel to tabulate the data for us:

 Year No. of Fish at beginning of year Growth Rate t un+1 =un × rn -4000 rn =-0.00001 × un +1.6 0 60000 1 1 56000 1.04 2 54240 1.0576 3 53364.224 1.06635776 4 52905.35437 1.070946456 5 52658.80178 1.073411982 6 52524.5888 1.074754112 7 52451.01779 1.075489822 8 52410.53579 1.075894642 9 52388.21465 1.076117854 10 52375.8931 1.076241069 11 52369.08718 1.076309128 12 52365.32657 1.076346734 13 52363.24824 1.076367518 14 52362.09952 1.076379005 15 52361.46457 1.076385354 16 52361.11359 1.076388864 17 52360.91958 1.076390804 18 52360.81234 1.076391877 19 52360.75305 1.076392469 20 52360.72028 1.076392797

Then, putting this into a graph, we have: The sustainable limit seems to have dropped to approximately 52360 fish.

Harvest Size of 6000

We first use Excel to tabulate the data for us:

 Year No. of Fish at beginning of year Growth Rate t un+1 =un × rn -6000 rn =-0.00001 × un +1.6 0 60000 1 1 54000 1.06 2 51240 1.0876 3 49728.624 1.102714 4 48836.43795 1.111636 5 48288.324 1.117117 6 47943.69606 1.120563 7 47723.93377 1.122761 8 47582.55549 1.124174 9 47491.09291 1.125089 10 47431.7096 1.125683 11 47393.0646 1.126069 12 47367.87764 1.126321 13 47351.4459 1.126486 14 47340.71915 1.126593 15 47333.71375 1.126663 16 47329.13742 1.126709 17 47326.14739 1.126739 18 47324.19355 1.126758 19 47322.91673 1.126771 20 47322.08229 1.126779

Then, putting this into a graph, we have: This time, the sustainable limit of fish has dropped to approximately 47322 fish.

Harvest Size of 7000

We first use Excel to tabulate the data for us:

 Year No. of Fish at beginning of year Growth Rate t un+1 =un × rn -7000 rn =-0.00001 × un +1.6 0 60000 1 1 53000 1.07 2 49710 1.1029 3 47825.159 1.12175 4 46647.79607 1.13352 5 45876.30493 1.14124 6 45355.73435 1.14644 7 44997.74857 1.15002 8 44748.42395 1.15252 9 44573.26386 1.15427 10 44449.46366 1.15551 ……… ………………………………….. ……………….. 25 44144.20251 1.15856 26 44143.61786 1.15856 27 44143.1986 1.15857 28 44142.89793 1.15857 29 44142.68231 1.15857 30 44142.52768 1.15857 31 44142.41679 1.15858 32 44142.33726 1.15858 33 44142.28023 1.15858 34 44142.23933 1.15858

Then, putting this into a graph, we have: The sustainable limit of fish has become somewhere close to 44142 fish.

Harvest Size of 8000

We first use Excel to tabulate the data for us:

 Year No. of Fish at beginning of year Growth Rate t un+1 =un × rn -8000 rn =-0.00001 × un +1.6 0 60000 1 1 52000 1.08 2 48160 1.1184 3 45862.144 1.141379 4 44346.06788 1.156539 5 43287.97124 1.16712 6 42522.26944 1.174777 7 41954.19712 1.180458 8 41525.16884 1.184748 9 41196.87367 1.188031 10 40943.17387 1.190568 ……… ……………………….. ……………. 35 40003.36435 1.199966 36 40002.69136 1.199973 37 40002.15302 1.199978 38 40001.72237 1.199983 39 40001.37787 1.199986 40 40001.10227 1.199989 41 40000.88181 1.199991 42 40000.70544 1.199993 43 40000.56434 1.199994 44 40000.45147 1.199995

Then, putting this into a graph, we have: The sustainable limit for a harvest size of 8000 is roughly 40000 fish.

Harvest Size of 9000

We first use Excel to tabulate the data for us:

Year

No. of Fish at beginning of year

Growth Rate

t

un+1 =un × rn -9000

rn =-0.00001 × un +1.6

0

60000

1

1

51000

1.09

2

46590

1.1341

3

43837.719

1.16162281

4

41922.89433

1.180771057

5

40501.34024

1.194986598

6

39398.55877

1.206014412

7

38515.2297

1.214847703

8

37790.13833

1.222098617

9

37183.27578

1.228167242

Conclusion

Using the data that I calculated about real increase in fish, we can now see that the point at which the increase in population size begins to slow down is after population reaches 30000. This coheres with my information about the real growth of fish. This also helps to explain why it is possible for there to be 2 sustainable limits for one harvest size. Looking back at our table for real increase of fish, we see that there are two points in which a real increase of 8000 fish can be reached. This is at growth rates 1.4 and 1.2; when population size is 20000 or 40000.

From this, we can also say that if there were to be an annual harvest of 5000 fish, the new possible sustainable limits would be 10000 and 50000, as they are the points at which there is an increase of 5000 fish per year, nullifying the harvest.

For a harvest of 9000 however, the only way to allow the real growth rate to reach 9000 is when population is 30000, there is only one solution. So, for a harvest size of 9000, there is only one sustainable limit.

For any harvest size, the way to find out when it is feasible to begin harvesting is by solving the quadratic equation model of the population of fish. The two (or one) values that we obtain for un are the sustainable limits. It is only possible to begin harvesting when the population has reached the lower sustainable limit i.e. the smaller solution. Any harvest starting before this population is reached will lead to a depletion of fish. Any harvest starting when population is above this will lead to a convergence of the population of fish and the higher sustainable limit i.e. the larger solution.

This student written piece of work is one of many that can be found in our International Baccalaureate Maths section.

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