Question

1) Capital stocks and resource maintenance

A fishery contains 5,000 tons of fish, and has a 3% annual replenishment rate (that is, 3% is its population growth rate). Assume that 200 tons of fish are harvested at the end of each year.

e) How many years will it take to fully deplete the fishery? Clearly show how you calculated this.
f) Now assume that the annual harvest is only 100 tons. How many tons will be in the fishery after 10 years? Compare to 10 years of harvesting 200/year and discuss.
g) For the fish harvest to be sustainable, what is the largest number of tons of fish that can be harvested each year (starting with the first year)? Explain why this amount does not change.

Answer 1

5000(0.03)-200=-50 then after 1 year fishery has 4950 tons of fish

now,

4950(0.03)-200=-51.5 after 2 years fishery has 4950-51.5 = 4898.5 tons of fish

That means it is depleting by 1.03 factor each year starting with -50 tons per year

50+50(1.03)+50(1.03)^2+...+50(1.03)^n =5000

1.03^m= 3

m(log(1.03))=log(3)

m=37.2 years

Ans f)

If we harvest 100 tons each year then after 10 years fishery will be

5000(1.03)-100 then it is a gain of 50 tons of fisheries

hence after 10 years we have [50((1.03)^10-1/0.03)] +5000 = 5573.2tons

If we harvest 200 tons each year then after 10 years fishery will be

5000(1.03)-200 then it is a gain of 50 tons of fisheries

hence after 10 years we have -50((1.03)^10-1)/0.03) +5000 = 4426.8 tons

As increment in fish each year is less than 200 tons & greater than 100 tons therefore we see increment in fishery in 100 tons case and decrement in 200 tons case

Ans g)

To sustain the level of present fishery level we need to harvest 150 tons of fish each year that is so evident from above

becoase 5000(1.03)-150 =5000 tons therefore level of 5000 tons will never change