Question.
Fishing of a severely depleted population of fish was banned in the early 1990's. On 1 January 1995 it was estimated that the mass of the fish population was 5000 tonnes and increasing at a rate of 8.5% per year.
On 1 January a few years later, it was estimated that the mass of the fish population was 10000 tonnes and increasing at a rate of 7.3% per year.
Write down the annual proportionate growth rates for population masses of 5000 and 10000 tonnes.
Can you help?
Thanks
Hi
Sorry it's taken so long to respond.
What is being asked for is "What is the annual proportionate growth rates for population masses of 5000 and 10000 tonnes." Then you are asked to work out proportionate growth rate ,r, for low population levels and then ,E, the equilibrium population level. Finally, you are being asked to specify a logistic recurrence system for this model.
The first part of the question which read "Fishing of a severely depleted population of fish was banned in the early 1990's. On 1 January 1995 it was estimated that the mass of the fish population was 5000 tonnes and increasing at a rate of 8.5% per year." I managed ok as this was referring to an exponential model. Its the second part that I'm not sure about. I think the annual proportionate growth rate is 0.085 for population mass of 5000 and 0.073 for population mass of 10000 but I'm not sure.
Essentially they want us to compare the exponential model with the logistic model and comment on which one seems the better predictor.
Hope this helps. I've not actually got the question in front of me just now so I'm recalling from memory. Any help much appreciated.
Regards
Thanks. I already have referenced this and I'm happy with answering the rest of the question but the correct answers, i.e. r and E and the logistic recurrence relation depends upon the correct proportionate growth rates for population masses of 5000 tonnes and 10000 tonnes. I am assuming that 0.085 and 0.073 are the correct growth rates to use as 8.5% and 7.3% are the rate of increase for these population masses.
Not quite.
Don't forget these growth rates are an increase, therefore, just for example, an increase of 5% would mean a growth rate of 1.05. That is 1 + 5% or 1+0.05
If there was a decrease of 5%, the growth rate would be 1 - 5% or 1-0.05 = 0.95
I hope this helps you.
Hi
Now I'm really confused about growth rates and proportionate growth rates.
If it's 1.085 for population mass 5000 and 1.073 for population mass 10000 then when I work out r and E, the numbers don't seem right.
For example, proportionate growth rate has the form r(1-P/E) for a population size Pn. Hence
1.085 = r(1-5000/E) and 1.073 = r(1-10000/E)
Solving these simultaneous equations gives,
1.085/r + 5000/E = 1 and 1.073/r + 10000/E = 1
therefore subtracting the second equation from 2 times the first equation gives
2.17/r + 10000/E - (1.073/r + 10000/E) = 2 - 1
which gives r = 1.097
Substituting r into the first equation gives
1.085/1.097 + 5000/E = 1
= 5000/E = 1 - 1.085/1.097 = 0.0109 therefore E = 5000/0.0109 = 457083
which seems very high. Have I missed something?
Hi, I have gone throught this question and found the mistake made.
Sorry Oututorlm you are wrong.
the rate is not 1.085 its 0.085. It is proportionate of the population using the formula:
(Pn+1 - Pn)/Pn
which is:
Pn+1= 5000*1.085=5425
5425-5000=425
425/5000=
0.085
The other rate also starts with a 0.
0.073
Student999 if you use these values then you should be fine with the method you used above.