Tenleyville is a town of 8 people. There are two occupations, basket weaving and

Question

Tenleyville is a town of 8 people. There are two occupations, basket weaving and
fishing, that are equally enjoyable to work in. Basket weavers earn $5 per day. The
income, I, from fishing is a function of the number of people fishing on a given day. This
is because the more fishing boats out on the lake, the less fish each boat will catch.
Specifically, I = 16 – 2*n , where n is the number of fishers. The lake is open-access, or
publicly owned, so that any of the 8 villagers is free to fish on the lake. [Hint: you might
want to make a table, but this is not necessary.]
a) If the villagers sequentially decide which occupation to choose, with full
information of the previous choices made, how many villagers will choose to fish?
b) Does your answer to (a) maximize village income? If not, what is the optimal
number of fishers?
c) What is the deadweight loss of this Tragedy of the Commons?

Explanation / Answer

Consider the given problem, here there are “8” people living in the town. There are 2 types of works, 1) basket waves earns $5 per day, 2) fishing, the income earned by fishing is function of numbers of peoples engaged in fishing.

Let’s, assume that only 1 people will go for fishing, so the fishing income is given by “I=16-2=14 > 5. Since the income from fishing is more than the other alternative, => further new people will switch to fishing.

Now let’s say ”n=5”, => all people will get “16 – 2*5 = 16 -10 =6 > 5”, now if further few want to switch then the fishing income will be < the other alternative (=> 4 < 5).

So, at the equilibrium “n=5” people will go to the fishing and each will earn “6”, other “3” will work under basket waves and each will earns “5” per day.

b).

So, under the above solution total income earns by all the peoples is given by “6*5 + 5*3 =30 + 15 = 45”.

But this not the optimum solution, here we need to maximize the income earned by all the fisher.

So, here I=16-2*n, total income be “16*n – 2*n^2”, => at the optimum “MI = 16 – 4*n = 0”,

=> “16 = 4*n, = n=16/4=4”, so at the optimum the no. of people should be n=4.

So, now each individual will get “8” in fishing, now the total income earned by all the individual is given by “8*4 + 5*4 = 32+20=52 > 45.

So, the optimum no. of fishers are n=4.

c).

Here, the dead weight loses is given by “(1/2)*(5-4)*(8-6)=1”, where 5 is the no. of people employed initially and 4 is the no. of people employed in the 2nd case. 6 be the income earned by each fishers in the 1st case and 8 be the income earned by all the people in fishing in the 2nd case.

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