# Math Portfolio Type II

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Introduction

Mathematics Portfolio

Type II

Creating a Logistic Model

Description

A geometric population model takes the form where r is the growth factor and un is the population at year n. For example, if the population were to increase annually by 20%, the growth factor is r = 1.2, and this would lead to an exponential growth. If r = 1 the population is stable. A logistic model takes a similar form to the geometric, but the growth factor depends on the size of the population and is variable. The growth factor is often estimated as a linear function by taking estimates of the projected initial growth rate and the eventual population.

Part 1

Information which has been given in Part 1: -

- 10,000 fishes are introduced into a lake
- The population increase if 10,000 fishes are introduced into the lake would be by 50% in the first year.
- The long term sustainable limit in this case would be 60,000

It has been given that the geometric population growth model takes the form . Now, if we have to find out the ordered pair (u0,r0): -

- It is mentioned in the description that unis the population at year n. Therefore, u0 will be the population at year 0, or the initial population of the lake which is 10,000.

Therefore, u0 = 10,000 ------------------- (1)

- r is the growth factor as mentioned in the description. As the population would increase by 50% in the first year, so r0 = 1 + 1 x

= 1 + 0.5

= 1.5 ------------------- (2)

So the first ordered pair (u0,r0) as shown in (1) and (2) would be (10000,1.5).

Now, if we have to find out the ordered pair (un, rn): -

- It is given in the question that un = 60,000.

Therefore, un = 60,000 ---------------------- (3)

- r is the growth factor as mentioned in the description. As found out in (3) that un = 60,000 which shows that the population has reached its long term sustainable limit where population is stable.

Therefore, rn = 1 ---------------------- (4)

Middle

60323

6

57375

16

59737

7

61892

17

60207

8

58378

18

59382

9

61218

19

60133

20

59893

The above tabulated data can be represented by the following logistic function graph: -

The table and graph confirm how higher growth rates above 2 lead to less stability in population. Here, the variance from the long term sustainable limit (which is 60000) is much more (64703 in the 3rd year) compared to when the initial growth rate was 2.3 or 2. The successive year sees a similar larger fall to 55573 in population compared to the other two cases. Consequently, the table confirms how a high initial growth rate leads to longer time for the population to stabilize at the long term sustainable limit. In this case, it is not attained in the first 20 years. Rather, upon further calculations we can find out that the population becomes stable at the long term sustainable limit in the 41st year compared to the 17th year when the initial growth rate is 2.3 .

Part 5

Let us see what happens when the initial growth rate is 2.9

The ordered pair (u0, r0) for the first pair when the predicted growth rate ‘r’ for the year is 2 will be (10000, 2.9).

(un, rn) = (60000, 1) according to the description.

We can get the equation of the linear growth factor by entering these sets of ordered pairs in the STAT mode of the GDC Casio CFX-9850GC PLUS.

Thus, we obtain a linear graph which can be modeled by the following function: -

y = (-3.8 x 10-5)x + 3.28

Here, the variable y can be replaced by rn which represents the growth factor at some year n and the variable x can be replaced by un which represents the population at some year n. Therefore, the equation for the linear growth factor in terms of un will be: -

rn = (-3.8 x 10-5)un + 3.28

Now, to find the recursive logistic function, we shall substitute the value of rn calculated above in the following form un+1 = rn.un

un+1 = [(-3.8 x 10-5)un + 3.28]un

= (-3.8 x 10-5)un2 + 3.28un

To find out the changes in the pattern, let us see the following table which shows the first 20 values of the population obtained according to the logistic function un+1= (-3 x 10-5)un2 + 2.8un just calculated: -

n+1 | un+1 | n+1 | un+1 |

0 | 10000 | 10 | 70738 |

1 | 29000 | 11 | 41872 |

2 | 63162 | 12 | 70716 |

3 | 55572 | 13 | 41919 |

4 | 64922 | 14 | 70720 |

5 | 52779 | 15 | 41910 |

6 | 67261 | 16 | 70719 |

7 | 48701 | 17 | 41911 |

8 | 69611 | 18 | 70719 |

9 | 44187 | 19 | 41911 |

20 | 70719 |

The above tabulated data can be represented by the following logistic function graph: -

We can infer from the data in the table and the graph how a high initial growth rate of 2.9 can lead to fluctuating populations every year. The long term sustainable limit of 60,000 is crossed at the end of the 2nd year itself, becoming 63162. But, in such a case, the environment (in this case the lake) cannot withhold such a population and thus there are a lot of fishes either dying out or migrating in the subsequent year which reduces the population to 55572. Yet, again in the next year, due to high growth rate, the population reaches well above even 63162 to 64922. Such a fluctuating trend is observed in the succeeding years until the end of the 16th and 17th year during and after which the population fluctuates between 70719 and 41911.

Part 6

To initiate an annual harvest of 5000 fishes, let us first find out at what point the fish population stabilizes when the initial growth rate, r = 1.5.

The following table shows the growth in the population over the first 20 years taking r = 1.5 and thus making use of the following recursive relation as found out in Part 2:-

un+1 = (-1 x 10-5) un2 + 1.6un

In this relation, the starting population is the initial population u0 = 10000.

n+1 | un+1 | n+1 | un+1 |

0 | 10000 | 10 | 59772 |

1 | 15000 | 11 | 59908 |

2 | 21750 | 12 | 59963 |

3 | 30069 | 13 | 59985 |

4 | 39069 | 14 | 59994 |

5 | 47246 | 15 | 59997 |

6 | 53272 | 16 | 59999 |

7 | 56856 | 17 | 59999 |

8 | 58643 | 18 | 59999 |

9 | 59439 | 19 | 59999 |

20 | 59999 |

Here, we see that the population reaches a stable point, when r =1 at the end of the 16th year. According to the question, we shall initiate the harvest of 5000 fishes every year from this point onwards. The following changes are made in the original recursive relation:-

- The starting population is taken to be 59999
- In the relation un+1 = (-1 x 10-5) un2 + 1.6un , we subtract 5000 as this is the number which is removed every year. Thus, the new relation becomes

un+1 = (-1 x 10-5) un2 + 1.6un – 5000

It should be noted that although the start population or u0 is taken to be 59999, the starting year is the 16th year.

Entering these values in the RECUR mode of the GDC Casio CFX 9850GC Plus, we get the following table:-

n+1 | un+1 | n+1 | un+1 |

16 | 59999 | 26 | 50041 |

17 | 54999 | 27 | 50024 |

18 | 52749 | 28 | 50014 |

19 | 51574 | 29 | 50008 |

20 | 50919 | 30 | 50005 |

21 | 50543 | 31 | 50003 |

22 | 50323 | 32 | 50001 |

23 | 50192 | 33 | 50001 |

24 | 50115 | 34 | 50000 |

25 | 50069 | 35 | 50000 |

We see that there is a declining trend in the fish population with an annual harvest of 5000 fishes when the growth rate is 1.5. However, it should be noted that with this annual harvest, the population finally stabilizes in the 34th year when it becomes 50000.

The graph is as follows: -

Thus, we can conclude that it is feasible to initiate an annual harvest of 5000 fishes once the stable population of 59999 is reached in the 16th year with the growth rate being 1.5.

Initially, a declining trend in the fish population is observed as 5000 fishes are harvested every year. Yet, from the 34th year onwards, we observe that the new stable fish population becomes 50,000 with an annual feasible harvest of 5000 fishes.

Part 7

Taking the same model in which the growth rate r = 1.5, we shall investigate other harvest sizes.

n+1 | un+1 | n+1 | un+1 |

0 | 10000 | 10 | 59772 |

1 | 15000 | 11 | 59908 |

2 | 21750 | 12 | 59963 |

3 | 30069 | 13 | 59985 |

4 | 39069 | 14 | 59994 |

5 | 47246 | 15 | 59997 |

6 | 53272 | 16 | 59999 |

7 | 56856 | 17 | 59999 |

8 | 58643 | 18 | 59999 |

9 | 59439 | 19 | 59999 |

20 | 59999 |

- When the harvest size = 2500 fishes

We shall initiate the harvest of 2500 fishes after the 16th year when the population stabilizes at 59999. The following changes are made in the original recursive relation:-

- The starting population is taken to be 59999
- In the relation un+1 = (-1 x 10-5) un2 + 1.6un , we subtract 2500 as this is the number which is removed every year. Thus, the new relation becomes

un+1 = (-1 x 10-5) un2 + 1.6un – 2500

Entering these values in the RECUR mode of the GDC Casio CFX 9850GC Plus, we get the following table:-

n+1 | un+1 | n+1 | un+1 |

16 | 59999 | 26 | 55498 |

17 | 57499 | 27 | 55496 |

18 | 56437 | 28 | 55495 |

19 | 55948 | 29 | 55495 |

20 | 55715 | 30 | 55495 |

21 | 55602 | 31 | 55495 |

22 | 55547 | 32 | 55495 |

23 | 55520 | 33 | 55495 |

24 | 55507 | 34 | 55495 |

25 | 55501 | 35 | 55495 |

The graph is as follows: -

We see that when the annual harvest is 2500, there is a declining trend which stabilizes quite early, that is from the 28th year onwards compared to when the harvest is 5000 fishes. Also, the new stable population, 55495 is higher compared to when the harvest if 5000 fishes.

Thus, we can conclude that it is also feasible to initiate an annual harvest of 2500 fishes and the new stable population would be achieved from the 28th year onwards as 55495.

- When the harvest size = 7500

We shall initiate the harvest of 7500 fishes after the 16th year when the population stabilizes at 59999. The following changes are made in the original recursive relation:-

- The starting population is taken to be 59999
- In the relation un+1 = (-1 x 10-5) un2 + 1.6un , we subtract 7500 as this is the number which is removed every year. Thus, the new relation becomes

un+1 = (-1 x 10-5) un2 + 1.6un – 7500

The table settings are as follows: -

Entering these values in the RECUR mode of the GDC Casio CFX 9850GC Plus, we get the following table:-

n+1 | un+1 | n+1 | un+1 | n+1 | un+1 | n+1 | un+1 |

16 | 59999 | 26 | 42774 | 36 | 42278 | 46 | 42249 |

17 | 52499 | 27 | 42642 | 37 | 42270 | 47 | 42248 |

18 | 48937 | 28 | 42544 | 38 | 42265 | 48 | 42248 |

19 | 46851 | 29 | 42470 | 39 | 42260 | 49 | 42248 |

20 | 45511 | 30 | 42415 | 40 | 42257 | 50 | 42248 |

21 | 44605 | 31 | 42374 | 41 | 42255 | 51 | 42248 |

22 | 43972 | 32 | 42342 | 42 | 42253 | 52 | 42247 |

23 | 43520 | 33 | 42319 | 43 | 42251 | 53 | 42247 |

24 | 43192 | 34 | 42301 | 44 | 42250 | 54 | 42247 |

25 | 42951 | 35 | 42288 | 45 | 42249 | 55 | 42247 |

Conclusion

In this case also, we find that the population dies put during the 7th year which means that such a combination also is not sustainable.

- Let us try with an initial population size of 19000 fishes and initiate a harvest of 8000 fishes from the 1st year itself.

The new recursive function becomes un+1 = (-1 x 10-5) un2 + 1.6un – 8000. Also, the starting population, u0 = 19000. This shall be entered in the GDC Casio CFX-9850GC Plus.

The graph and GDC show that the population dies out during the 14th year which means that this combination is not sustainable.

- When the initial population size is 20000 fishes and a harvest of 8000 fishes is initiated from the 1st year itself.

The new recursive function becomes un+1 = (-1 x 10-5) un2 + 1.6un – 8000. Also, the starting population, u0 = 20000. This shall be entered in the GDC Casio CFX-9850GC Plus.

Using this combination, we have found out using the GDC that the population remains constant and stable starting from the 1st year itself. It neither depletes nor increases. Thus, this introductory fish population size of 20000 is the most sustainable for harvesting 8000 fishes.

- To prove that an initial population of 20000 fishes is the most suitable and sustainable for a harvest of 8000 fishes every year, we shall investigate using u0 = 19999.

The new recursive function becomes un+1 = (-1 x 10-5) un2 + 1.6un – 8000. Also, the starting population, u0 = 19999. This shall be entered in the GDC Casio CFX-9850GC Plus.

The GDC shows that in such a case, the population has a decreasing trend which slowly increases with increasing number of years and finally, dies out during the 52nd year. Hence, an initial population size of 19999 is not sustainable.

These results help us to infer that a minimum initial population size of 20000 fishes is required to sustain a harvest of 8000 fishes every year from the 1st year.

Candidate Name: - Sanchit Ladha

Candidate Session No.: -1070-006

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

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