Let K be the new rate of decline.
0.92 = eK
ln(0.92) = ln (eK)
*ln(0.92) = **K ln(e)
K = ln(0.92)
* Using the laws of Logarithms, we can bring down the exponent of e and multiply by ln(e), as well as using ln on the other side of the equation.
** The figure “ln(e)” is equal to 1, therefore K multiplied by ln(e) is equal to K.
Using the new rate of decline, we can generate this new equation:
PT = PO (eln(0.92))T
The new rate of decline is ln(0.92), which is equal to -0.083381609 or -0.0834 or -8.34%. This new rate is not equal to the old rate of 8% because the new rate is a continuous rate of decline, which means that the fish population is decreasing at every moment in time. This is different from the geometric rate of decline modelled by the first equation, PT = PO (0.92)T, which assumes the fish population is decreasing by discreet amounts each year. Such as for exactly one year, it decreases by 8%, or 0.92.
b) Based on estimates of survival from egg to mature fish, a minimum spawning population of 600 000 is required to sustain an economically viable ocean fishery. Continuing to use an 8% rate of decline, when will the population in this river system fall below 600 000 salmon? (Assume a yearly count).
For this question, I will use the original equation, PT = PO (0.92)T, because it is assuming a yearly count with the 8% rate of decline.
Let Population be 600 000 and solve for Time:
PT = PO (0.92)T
600 000 = 2 000 000 (0.92)T
= 0.92T
ln(0.3) = T ln(0.92)
T =
T = 14.43930886
T 14.44
The fish population will fall below 600 000 after approximately 14.44 years.
c) The spawning grounds can sustain about 3 million spawners, so the current population of 2 million could be allowed to increase. What annual percent increase would be necessary to allow a population of 2 million to grow to 2.5 million in 8 years.
This question is vague because we can interpret it in two different ways. Firstly, we can just assume the fish population is not decreasing and so the equation would be:
Let r be the rate, and where 8 represents the number of years of growth.
P8 = PO (r)8
Secondly, we can assume there is still an 8% rate of decline in population, and thus, we need a higher rate of increase and use the formula:
P8 = PO (0.92)8 (r)8
Using equation 1 and the values given, we solve for r:
Let population be 2 500 000 and solve for r.
Let r be the rate, and where T = 8, represents the number of years of growth.
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%. In addition, after each year it decreases by 100 000 due to in-river harvesting.
Let PO be the initial population.
Let PT be the population at T years.
P1 = PO (0.92) - 100 000
P2 = P1 (0.92) - 100 000
= [PO (0.92) - 100 000] (0.92) - 100 000
= PO (0.92)2 - 100 000 (0.92)1 - 100 000
Here, there is a noticeable pattern developing with the exponents on 0.92. Therefore we can begin to implicate the foundations of a formula:
Let T be the number of years of growth.
PT = PT-1 (0.92) - 100 000
PT = PO (0.92)T - 100 000 (0.92)T-1 - 100 000 (0.92)T-2 - 100 000 (0.92)T-3 - … -
100 000 (0.92) - 100 000 (0.92)0
PT = PO (0.92)T - n
PT = PO (0.92)T - 100 000 n
This is a geometric series that can be easily summed as follows if T = 8:
n = 0.920+ 0.921+ 0.922+ 0.923+ 0.924+ 0.925+ 0.926+ 0.927
n = 1+ 0.92+ 0.922+ … 0.927
S8 =
S8 =
S8 = 6.084764086
Thus, in the equation:
PT = PO (0.92)T - 100 000(6.084764086)
The summation of the sigma notation multiplied by 100 000 provides the right side of the equation for the 8 year period.
PT = PO (0.92)T - 608476.4086
PT = 2 000 000 (0.92)8 - 608476.4086
PT = 417 961.3377
PT 0.4180 million
The population after 8 years of an 8% rate of decline, and after in-river harvesting of
100 000 fish each year, is approximately, 0.4180 million.
This question can also be solved using the Time Value Money (TVM) solver on the Texas Instruments (TI)-83 calculator. The TVM solver is used to obtain any of the following:
Figure 1: TVM Solver Symbols
In question d) the information given would be inputted as follows:
N = 8
I% = -8
PV = 200 000
PMT = -100 000
PV = -417 961.3377
P/Y = 1
C/Y = 1
END
Thus, the same result is reached using the TVM solver and the formula derived. The population after 8 years of decline and in-river harvesting is 417 961.3377 or
0.4180 million.
e) Enhancement programs can increase the proportion of eggs that survive to adult by enlarging the spawning grounds (so there is less competition among females for spawning sites) and protecting the stream beds from environmental insults. Assume that the current population is 2 million and the annual in-river harvest is 100 000 but that the enhancement program can produce a 7% annual growth in population. Predict the population in 8 years.
This question implies that the fish population is increasing, and therefore we can assume that the 7% annual growth is the overall growth in population, meaning that it has already accounted for the 8% annual loss per year due to escapement:
Let PO be the initial population.
Let PT be the population at T years.
P8 = PO (1.07)8 - 100 000 n
P8 = 2 000 000 (1.07)8 - 100 000 (10.25980257)
P8 = 2 410 392.1030
P8 = 2.4104 million
With this equation, the fish population would increase to 2.4104 million after 8 years.
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.