This post goes over an input/output decision choice, where the input is a common resource that may need to be managed sustainably. The production schedule exhibits diminishing returns so that the optimal input choice will be less than the maximum available.
Farmers in an arid region in Mexico draw their irrigation water from an
underground aquifer. The aquifer has a natural maximum recharge rate of 340
gallons per day (that is 340 gallons per day filter into the underground resource),
which could be regarded as the maximum sustainable withdrawal. The total
product schedule for well operations looks like this:
| Wells Operating (N) | Total Water Output (Gal/Day) |
| 1 | 100 |
| 2 | 200 |
| 3 | 280 |
| 4 | 340 |
| 5 | 380 |
| 6 | 400 |
| 7 | 400 |
| 8 | 380 |
| 9 | 340 |
A. Suppose that the cost of operating each well is 39 pesos per day and the value to the farmer, in terms of increased crop production and revenue, of each gallon of water is 1 peso. Calculate the total daily revenue (TR = output times value) for each number (N) of wells operating.
B. If each well is privately owned by a different farmer, how many wells will
operate? (To calculate this first calculate the average revenue, which is TR/N,
which is the expected revenue receive by an irrigator adding the Nth well).
Analyze this result in terms of economic efficiency and sustainability.
C. What would be the economically efficient number of wells if only the present time period is considered? (To calculate this you will need marginal revenue, which is given by ΔTR/ΔN.) Show that net social benefit in the present period is maximized at this number of wells.
D. In this case is the “short term” economically efficient level identified in Part C also ecologically sustainable?
E. Suppose now that the costs of operating each well are doubled (say through a tax on operating a well or a rise in input costs) to 78 pesos.
This question seems very complicated and tricky at first, but we just have to absorb all of the information given to us. Basically, there exists a water resource, that can supply up to 340 gallons of water per day. The farmers can choose to drill into that resource. The amount of holes they drill will produce more water at a diminishing rate (the 8th whole actually reduces water production).
For the first question, suppose the cost of each additional hole in the ground is 39 pesos, and the revenue is 1 peso for each gallon of water. To calculate the total revenue, you simply multiply total water output by 1 (revenue from each gallon). So the answer will be the same as your total water output (Gal/Day) column.
For the second question, what if each well is privately owned by a different farmer. In this case, 8 of the 9 wells will operate, because the average revenue when 8 wells are open is 47.5 which is higher than the cost to operate the well of 39. If the 9th well is opened, than average revenue drops to 37.78 which is below 39, so all of the farmers will be losing money, and at least one of them will close up shop, lowering the number of wells and increasing revenue. (see the table below for average revenue values).
For the third question, we would need to set marginal benefit (MB) equal to marginal cost (MC). In this case, MC is equal to 39, and MB is our marginal revenue (MR). MR will change depending on how many wells are opened. Since we want to get as close to MB=MC without MB<MC we would choose to open 5 wells, the MB/MR of the 5th well is 40, which is greater than our MC. If we were to open up the 6th well, then the MR would be 20 which is less than 39 which we don’t want. (see the table below for marginal revenue values).
Is this amount of wells ecologically feasible? The answer is no, because we would be producing 380 gallons of water per day, while the aquifer can only sustainably produce 340 per day. This type of shortsightedness is called being myopic. It means you care only for the present, without regard to future consequences. The way this problem is set up induces myopic behavior.
Now consider an operating cost change to 78 pesos per well. This would reduce the number of optimal wells be operated to only 3, because 80 (the marginal revenue from the 3rd well) is larger than 78 while 60 (the marginal revenue from the 4th well) is not. Under a each well is owned by a different farmer format, we would see 4 wells being operated because the average revenue at 4 wells is still larger than 78 while the average revenue of the 5th well puts us below that cost.
| Wells Operating (N) | Total Water Output (Gal/Day) | Average Revenue | Marginal Revenue |
| 1 | 100 | 100 | 100 |
| 2 | 200 | 100 | 100 |
| 3 | 280 | 93.33 | 80 |
| 4 | 340 | 85 | 60 |
| 5 | 380 | 76 | 40 |
| 6 | 400 | 66.67 | 20 |
| 7 | 400 | 57.14 | 0 |
| 8 | 380 | 47.5 | -20 |
| 9 | 340 | 37.78 | -40 |
