Question

Consider the following growth function for fish given by F( X ) = rX(1 − X/K) where the intrinsic growth rate r = 0.2 and the carrying capacity K = 100 tons of fish. Let the harvest function be given by H = qEX where the catchability coefficient be q = 0.01 and H is harvest in tons of fish.

Compute the following:

a. The maximum yield of fish at the steady state.

b. Effort and Harvest when the price of fish is $1 per ton and the unit cost of effort is $0.5.

c. Determine the supply of fish under Open Access and a Managed Fishery. Consider the price of fish as $0.5, $1 and $2 per ton.

d. Plot these functions on a graph.

Answer 1

The growth function for fish is F(X) = rX(1 − X/K), where intrinsic growth rate r = 0.2 and carrying capacity K = 100 tons of fish. Harvest function is H = qEX where q is catchability coefficient = 0.01.

a. The growth function F(X) = rX(1 - X/K) and in order to get maximum yield F''(X) < 0.

F(X) = rX - rX²/K

F'(X) = r - 2rX/K = 0

2rX/K = r

X* = Kr/2r = 100 * 0.2 / 2 * 0.2 = 50

Hence, The maximum yield of fish at the steady state is X* = 50

b. At steady state or Bio-Economic equilibrium,

F(X) = H

rX(1 - X/K) = qEX

1 - X/K = qE/r

X/K = 1 - qE/r

X = K(1 - qE/r) = 100(1 - 0.01E/0.2) = 100(1 - 0.05E) = 100 - 5E

Putting the value of X at Harvest function, H = 0.01*E*(100 - 5E) = E - 0.05E²

At Open access, TR = TC

P*H(E) = cE where P is price of fish per ton and c is the unit cost of effort

Given, P = $1 and c = $0.5

E - 0.05E² = 0.5E

E(1 - 0.05E) = 0.5E

1 - 0.05E = 0.5

0.05E = 0.5

E = 0.5/0.05 = 10

At a managed fishery, = TR - TC

d / dE = 1 - 0.1E - 0.5 = 0

0.1E = 0.5

E = 0.5/0.1 = 5

Thus, Harvest at open access is H = 10 - 5 = 5

Harvest at Managed fishery is H = 5 - 0.05*25 = 3.75

c. When P = $0.5,

  At open access the supply of fish is X = 100

At managed fishery the supply of fish is X = 100

When P = $1

At open access the supply of fish is X = 100 - 5*10 = 50

At managed fishery the supply of fish is X = 100 - 25 = 75

When P = $2

  At open access the supply of fish is X = 100 - 5*15 = 25

At managed fishery the supply of fish is X = 100 - 5*7.5 = 62.5

d. A is the Bio-Economic Equilibrium and X* = 50 is the maximum yield.