# Public goods on cost of abatement

There is a manufacturer and a fishery. Let $$E$$ be the effluent of the manufacturer. The marginal cost of abatement borne by the manufacturer is $$M C A=8-E$$. The marginal external cost borne by the fishery is $$M E C=E$$. Assume that the manufacturer has the right to release effluent subject to government control. (a) What is the efficient pollution level? (b) How can an effluent fee be used to achieve the efficient pollution level? (c) How can bargaining between the two parties be used to achieve the efficient pollution level?

(a) The efficient level of pollution satisfies $$M C A=8-E=E=M E C$$, which gives $$E=4$$

(b) Suppose that the effluent fee is $$F$$ per unit of pollution. The manufacturer would choose emissions to satisfy $$M C A=8-E=F$$, which gives $$E=8-F$$. Hence, an effluent fee of $$F=4$$ would induce the efficient pollution level $$E=4$$.

(c) The increase in the total cost of abatement from $$E=8$$ to $$E=4$$ is $$T C A=$$ $$\frac{1}{2} \cdot 4 \cdot 4=8$$. The decrease in the total external cost from $$E=8$$ to $$E=4$$ is TEC $$=4 \cdot 4+\frac{1}{2} \cdot 4 \cdot 4=24$$. Hence, the fishery could offer the manufacturer a payment between 8 and 24 to reduce effluent from 8 to 4 .

Question: I don't understand the solution of (c). Can anyone help me explain part (c)?

• What exactly do you not understand about the solution to (c)? Is it the calculations leading to $TCA=8$ and $TEC=24$? Or is it the conclusion drawn from these numbers? Commented May 8 at 19:54
• @AdamBailey I don't know how to get TCA and TEC. Commented May 8 at 19:56

The change in effluent required to reach the efficient level is from $$E=8$$ to $$E=4$$, so we are interested in areas on the right hand side of the diagram. Thus the change in $$TCA$$ is triangle a whose area (1/2 x base x height) is $$\frac{1}{2} \cdot 4 \cdot 4=8$$. For the change in $$TEC$$ we need the rectangle consisting of triangles a and b (base x height) which is $$4 \cdot 4 = 16$$ and also triangle c whose area is $$\frac{1}{2} \cdot 4 \cdot 4=8$$ making a total of $$16+8=24$$.