Suppose the government wants to reduce the total pollution emitted by three loca

Question

Suppose the government wants to reduce the total pollution emitted by three local firms. Currently, each firm is creating 4 units of pollution in the area, for a total of 12 pollution units. If the government wants to reduce total pollution in the area to 6 units, it can choose between the following two methods: Available Methods to Reduce Pollution 1. The government sets pollution standards using command-and-control regulation 2. The government allocates tradable pollution permits Each firm faces different costs, so reducing pollution is more difficult for some firms than others. The following table shows the cost each firm faces to eliminate each unit of pollution. For each firm, assume that the cost of reducing pollution to zero (that is eliminating all 4 units of pollution) is prohibitively expensive First Unit of Pollution (Dollars) 130 600 90 Cost of Eliminating the.. . Second Unit of Pollution (Dollars) 165 750 115 Third Unit of Pollution Firm Firm X Firm Y Firm Z (Dollars) 220 1,200 140

Explanation / Answer

Ans:

1) Computation of cost of elimination of two units of pollution by each firm(In dollars)

X = 130 + 165

= 295

Y = 600 + 750

   = 1350

z = 90 + 115

   = 205

2) The cost of elimination of 3rd unit of pollution for firm X is $220 and firm Z is $140. Hence firm X and Z will trade a permit for a price which is above $220.

Following will apply - $244 , $345

3) If the government sets a price of $523 per permit both firm X and Z will sell one permit to firm Y.

4) Total cost of eliminating six units of pollution

Regulation = $295 + $1350 + $205

   = $1850

Tradable permits = $515 + $345

= $860

5) Less

Eliminating pollution by regulation is more costly to society than the pollution permits.

Firm Initial pollution permit allocation Action Final amount of pollution eliminated Cost of pollution reduction X 2 Sell one permit 3 units 130+165+220 = 515 Y 2 Buy two permits 0 0 Z 2 Sell one permit 3 units 90+115+140 = 345

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