Question

4. Uranium Mine Extraction (10 points)

The owner of a uranium mine hires you as an economist and asks you to determine the optimal number of uranium ore which should be extracted from the mine this year (q0) and next year (q1), after which time no mining will be possible. The owner would therefore like to extract all the uranium ore by the end of the second year. There are 60 tons of uranium ore in the ground. The price this year is $30 a ton and the price next year is known to be $35 a ton. The cost of mining is given by the following function: c(qi) = 25 + 4.5qi + .1q2. The owner’s discount rate is .10.

(a) Write down the formula for determining the present value of mining all of the tons of uranium ore between the two periods.

(b) What three first order conditions hold at the optimum where the present value of the uranium mine is maximized? (hint: use the Lagrange multiplier approach)

(c) IN THREE SENTENCES OR LESS give a brief economic rationale for each of the three first order conditions. Solve for q0 and q1.

(d) How would you calculate the value to the owner of discovering an extra ton of uranium ore? What is this number?

(e) How would the allocation of the optimal amount of uranium ore to be mined in each of the two time periods change in a qualitative sense (e.g. q0 increases and q1 decreases) under each of the following situations? For each question assume that each of the other parameters remains fixed at values given at beginning of problem.

• Second period price increases to $40

• Fixed cost part of cost function increases from 25 to 35

• Parameters in second period cost function decrease with linear term going from 4.5 to 3.5 and quadratic term going from 0.1 to 0.09. (First period cost function remains unchanged)

• Price in the first period increases from $30 to $35

• Discount rate decreases from .10 to .06

Answer 1

This is a two-period economy, call those time t=0 and t=1. The single good is Uranium. We have the following details about the economy:

• Total Uranium= 60 tons. Let the quantity of Uranium mined at period t=0 be q₀ and q₁ . Therefore q₀+q₁=60 as all the Uranium needs to be mined by the end of period 1
• Price at time t=0, p₀= $30 and price at time t=1, t=1, p₁= $35 per ton.
• c(q_i) = 25 + 4.5q_i + .1q²_i
• Discount rate, denote it by d=0.10

Solution: a) Present value of mining all the Uranium between the two periods would be the discounted value of total money we receive by selling the mined uranium minus the cost of mining the uranium. Let us denote the present value by V₀ , then

V0(q₀, q₁) = p₀ .q₀ - c(q₀) + d( p₁ .q₁- c(q₁))

and q₀+q₁=60, i.e., q₁=60-q₀

This gives V₀= p₀ .q₀ - c(q₀) + d( p₁ .q₁- c(q₁ ))

V₀(q₀)= d p₀ .p₁ .q₁.q₀ - (c(q₀) + d.c(q₁))

=0.10*30*35 (q₁.q₀) - (25 + 4.5q₀ + .1q²₀) - 0.10( 25 + 4.5q₁ + 0.1q₁²)

b) Objective: max_q0,q1 L where

L = (V₀(q₀, q₁) + k(60-q₀-q₁) and k is the Lagrange multiplier

In order to maximize the present value obtained from Uranium, the following first-order condition holds:

- dL/dq₀=0

- dL/dq1=0

- dL/dk=0

c) The above three conditions imply that our Lagrangian is maximized such that increasing quantity mined in any period would only result in lowering our value. The conditions ensure that we maximize the revenue and minize the costs involved in both the periods while taking into account that the total quantity if Uranium available is limited.

Solving for q₀ and q₁

The solution for q at t=0 and t=1 is as shown in the images

d) If we now assume that given these optimum quantities, the owner discovers additional unit of Uranium, the additional value can be found by differentiating Value function with respect to Uranium quantity. If we want to know the present (discounted) value of the additional utility, there are two cases possible:

Case 1) The extra unit of Uranium appears in period 0: The change in value would be dV₀(q₀ ,q₁)/dq₀= p₀-dc(q₀)/q₀=31.5

2) The extra unit appears in period 1: The discounted value at time t=0 is (dV₀(q₀ ,q₁)/dq₁)*(0.1)=3.15