• Join over 1.2 million students every month
• Accelerate your learning by 29%
• Unlimited access from just £6.99 per month
Page
1. 1
1
2. 2
2
3. 3
3
4. 4
4
5. 5
5
6. 6
6
7. 7
7

# IB MATH METHODS SL - FISH ESCAPEMENT

Extracts from this document...

Introduction

IB MATH METHODS SL FISH ESCAPEMENT Portfolio Assignment #3 Type III The Department of Fisheries and Oceans (DFO) sets targets for each river system for escapement (the number of fish that escape being caught by the ocean fishery and return to spawn). The salmon escapement in a river is currently estimated to be 2 million but has been declining by 8% per year. (That is, each year there are 8% fewer fish than the precious year.) a) i) If this decline continues, predict the population (of the escapement) in 8 years. The initial population of 2 million declines each year by 8%. Therefore we can write the population after 1 year as following: Let PO be the initial population of 2 million. Let PT be the population after T years. P1 = PO - (PO x 0.08) P1 = PO (1 - 0.08) P1 = PO (0.92) This will give us the answer for the population after 1 year, with an annual decline of 8%. Then to calculate the new population after 2 years, we calculate take the population after 1 year, while using the annual decline of 8% each year. P2 = P1 (0.92) P2 = PO (0.92)(0.92) This is beginning to show a pattern for the population after a set amount of years, after an annual decline of 8% each year. Therefore, observing this pattern, we can utilize the following formula to generate the population after 8 years with the continuing decline: PT = PO (0.92)T Therefore, the population after T years (PT) ...read more.

Middle

Equation 1: P8 = PO (r)8 2 500 000 = 2 000 000 (r)8 = r8 1.25 = r8 = r = 1.0283** ** We can throw out the negative number because the population is increasing and not decreasing, according to the question. Using equation 2 and the values given, we solve for r: Equation 2: P8 = PO (0.92)8 (r)8 2 500 000 = 2 000 000 (0.92)8 (r)8 = (0.92)8 (r)8 1.25 = (0.92)8 (r)8 = r8 = r = 1.1177** ** We can throw out the negative because there is only an increase in population. The rate of increase, for equation 1, is 1.0283 or 2.83%, while for equation 2, it requires a rate of increase of 1.1177 or 11.77%. This makes sense because in equation 2, we assumed that the population was decreasing by 8% while needing to increase to 2 500 000 million fish. Therefore a larger rate of increase was needed to ensure it reached a larger population. d) Because of recent court decisions, the DFO must now allocate an additional 100 000 fish to an in-river harvest (i.e. an additional 100 000 fish will be taken each year from the escapement). Assume that the current population of 2 million continues to decline by 8% per year and in addition, 100 000 fish are harvested from the river. Predict the population in 8 years. The initial population of the fish is 2 million, and it decreases each year by 8%. ...read more.

Conclusion

Using the TVM solver, we would input the information given as follows: N = 8 I% = 7 PV = 2 000 000 PMT = -100 000 FV = -2 410 392.103 P/Y = 1 C/Y = 1 END The TVM solver produces the same result as the derived formula. Therefore the population after 8 years of growth, increased to 2.4104 million fish. f) Assume that the current population is 2 million and that enhancement programs can produce a 7% annual growth in population. How many fish could be harvested each year in the in-river fishery and still have the population reach 2.5 million in 8 years? This question implies that the population is increasing, therefore we can assume that the 7% annual growth already accounts for the 8% annual loss per year due to escapement. Let x be the number of fish harvested each year. Let P8 be 2.5 million. P8 = PO (1.07)8 - x n 2 500 000 = 2 000 000 (1.07)8 - 10.25980257x 2 500 000 - 2 000 000 (1.07)8 = -10.25980257x = x x = 91 266.11875 x = 0.9127 million Each year, there can be 0.9127 million fish harvested and still have the fish population reach 2.5 million in 8 years. Using the TVM solver, we would input the following information: N = 8 I% = 7 PV = 2 000 000 PMT = -91 266.11875 FV = 2 500 000 P/Y = 1 C/Y = 1 END The TVM solver produces the same result as the derived formula. Therefore there can be 0.9127 million fish harvested each year while still allowing the population to reach 2.5 million in 8 years. TJ Dhaliwal Block F ...read more.

The above preview is unformatted text

This student written piece of work is one of many that can be found in our AS and A Level Population & Settlement section.

## Found what you're looking for?

• Start learning 29% faster today
• 150,000+ documents available
• Just £6.99 a month

Not the one? Search for your essay title...
• Join over 1.2 million students every month
• Accelerate your learning by 29%
• Unlimited access from just £6.99 per month

# Related AS and A Level Population & Settlement essays

1. ## Data Interpretation (assignment 1)

However in the winter months the crime rates hardly dropped with around 17,000 cases between October - December 2003 and around 18,000 cases between January - March 2004. The crime statistics for Norwich for Vehicle & other theft are nearly double that of the national average for each season of the year.

2. ## China Portfolio.

Forced abortions or sterilizations applied to all who had more than one child, this has caused much international concern, as it is considered 'inhumane' by many countries. Into the 80's and 90's, the one child policy did not apply strictly to farmers.

• Over 160,000 pieces
of student written work
• Annotated by
experienced teachers
• Ideas and feedback to