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Grant McDermott

Data. Economics. Environment.

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As promised, here is the first example of a highly stylised model, which \(-\) despite its unrealistic simplicity \(-\) still produces valid (and unexpected) results for real-life decisions. 

Topic: The optimal depletion of an oil resource.
Aim: To show that changes in the interest rate don't always have clear implications for our rate of withdrawal.

[NOTE: Those of you with no desire to go through all the equations (although that kinda defeats the purpose) can just scroll down to "THOUGHT FOR THE DAY".]

The model that I am going to refer to here is commonly known as the "oil tank" model. As the name suggests, imagine we have an oil resource that can be exploited exactly as if it were in a tank or drum. A key assumption here is that rate of production is proportional to the number of wells; you drill two holes, the oil comes out twice as fast. This is grossly unrealistic. As any geologist would probably tell you: drilling holes affects well pressure around the borehole, geological structure and reserves are not uniform, etc, etc. There are also other economic considerations that we ignore here, such as the fact that we shall assume a constant oil price[*], uniform costs of production, and an oil field that has been unitized. (Unitization essentially means that the resource is controlled by one supplier. This is not completely correct, but just go with it...) Bottom line: lots of simplifications. 

So, let's specify the model and then solve to see what we get.

\(Q =\) Amount of oil that can be recovered
\(q_t = q = \) The (constant) rate of production
\(T = Q/q =\) Extraction time (remember that extraction is proportional to the number holes you drill)
\(c =\) The cost to develop each well
\(p =\) Price of oil
\(r =\) Interest rate

The present value of the well to us as producers can thus be written as:

V = -cq + \int_0^T \! pqe^{-rt} \, \mathrm{d}t

Where the first term on the RHS equals the cost of our investment. (Note: We assume upfront payments for simplicity[**], and the second term under the integral represents the total revenue flow from production until we exhaust the resource at time, \(T\).)

Solving for the above gives:
V = q \left[ -c + \frac{p \left(1-e^{-rT}\right)}{r} \right]
Now that we have a PV equation for our oil field, we can used the standard optimization procedure to find the optimal production rate... simply derive w.r.t. \(q\) and set equal to zero. Recalling that \(T = Q/q\):

\begin{align} \frac{\mathrm{d}V}{\mathrm{d}q} = 0 &= - c + \frac{p \left(1-e^{-rT}\right)}{r} + pq \frac{\mathrm{d}T}{\mathrm{d}q}e^{-rT} \\ &= - c + \frac{p \left(1-e^{-rQ/q}\right)}{r} - \frac{pQ}{dq}e^{-rQ/q} \end{align}
Since it usually makes sense to think in terms of normalised costs (i.e. how much it costs to develop a field relative to the price of oil), we can just divide through by p and solve to get:

\frac{c}{p} = \frac{1-e^{-rQ/q}}{r} - \frac{Qe^{-rQ/q}}{q}
Okay, we're getting there. The above equation is important because we can use it to analyse how changes in two "given" variables \(-\) investment cost per well \(\left( \frac{c}{p} \right)\) and interest rate \(\left( r \right)\) \(-\) affect our "choice" variable, namely the production rate \(\left( q \right)\). I want to focus on the latter in this blog post. So let's think about what our intuition would tell us about the relationship between production rates and interest.

If an oil producer asks, "What should I do if the interest rate goes up?", the instinctive response is "produce faster". After all, this seems perfectly intuitive; higher rates of interest signal increasing impatience and we can put the money that we receive from production into the bank at the higher interest. So, let's see if this is how things work out.

For this last part, it will make things easier to follow if we define the previous equation as being equivalent to a function, which we'll call \(f\):

f \equiv \frac{c}{p} = \frac{1-e^{-rQ/q}}{r} - \frac{Qe^{-rQ/q}}{q}

We can then use a general derivation rule,

\mathrm{d}f = \frac{\partial f}{\partial r} \mathrm{d}r + \frac{\partial f}{\partial q} \mathrm{d}q,
to get our second last equation (with a little help from the chain, quotient and exponent rules):

\mathrm{d}f = &\left[ \frac{ \left(\frac{Q}{q}e^{-rQ/q} \right)r - \left( 1 - e^{-rQ/q} \right)}{r^2} - \frac{-Q}{q}\frac{Qe^{-rQ/q}}{q} \right] \mathrm{d}r \\ &+ \left[ \frac{-r \frac{Q}{q^2}e^{-rQ/q} }{r} - \frac{\left( r \frac{Q}{q^2}Qe^{-rQ/q}\right)q - Qe^{-rQ/q} }{q^2} \right] \mathrm{d}q .

Simplifying and setting equal to zero again for optimization yields,

\mathrm{d}f = 0 = \left[ \frac{ \left( r \frac{Q}{q} \right) e^{-rQ/q} - \left( 1 - e^{-rQ/q} \right)}{r^2} + \frac{Q^2}{{q}^2}e^{-rQ/q} \right] \mathrm{d}r - \left[ \frac{rQ^2}{q^3}e^{-rQ/q} \right] \mathrm{d}q.
And now \(-\) if you've made it this far! \(-\) we are at last ready to produce our final equation... i.e. The one that describes the relationship between the production rate \((q)\) and the interest rate \((r)\). All we need to do is rearrange the above equation and then simplify this bad boy as follows,

\frac{\mathrm{d}q}{\mathrm{d}r} &= \frac{\frac{ \left( r \frac{Q}{q}\right)e^{-rQ/q} - \left( 1 - e^{-rQ / q} \right) }{r^2} + \frac{Q^2}{q^2}e^{-rQ/q}}{\frac{rQ^2}{q^3} e^{-rQ/q}} \\ &= \frac{ -\frac{1}{r} \left[ \frac{ \left( 1 - e^{-rQ / q} \right) }{r} - \frac{Q}{q}e^{-rQ/q} \right] + \frac{Q^2}{q^2}e^{-rQ/q}}{\frac{rQ^2}{q^3} e^{-rQ/q}} \\ &= \frac{ -\frac{1}{r} \frac{c}{p} + \frac{Q^2}{q^2}e^{-rQ/q} }{\frac{rQ^2}{q^3} e^{-rQ/q}} .

(For this last step, just recall our equation for \(\frac{c}{p}\) above.)

Looking at our final equation above, it should be clear that we cannot directly see how the rate of production \((q)\) should change. That's because we are left with an equation of ambiguous sign:

\frac{\mathrm{d}q}{\mathrm{d}r} =\frac{ \text{something negative} + \text{something positive} }{\text{something positive} }.
In other words, it is not certain whether \(\frac{\mathrm{d}q}{\mathrm{d}r}\) will be negative (i.e. we should decrease our production rate) or positive (i.e. we should increase production). It all depends on how high our costs of investment \((\frac{c}{p})\) are! If investment costs are "high", then we should produce less, since \(\frac{\mathrm{d}q}{\mathrm{d}r} < 0\). On the other hand, if they are "low", then we should produce more, since \(\frac{\mathrm{d}q}{\mathrm{d}r} > 0\).

THOUGHT FOR THE DAY: The "oil tank" model is crude and dramatically simplified. Yet, it still provides very useful insights. We have isolated key variables to see how they might affect each other on a ceteris paribus basis. What we focused on here is how changing the rate of interest actually has an ambiguous effect on the optimal depletion of an oil resource. Accordingly, we can say that a higher rate of interest (r) plays two roles:
  1. Impatience. We want our money more quickly (and capital costs are high).
  2. Opportunity cost of capital. To get more money, we need to drill more wells. However, this is expensive in the face of high interest rates.
The first of these is the one that comes completely naturally to us. As such, we're almost always predisposed to think that we should just do something "quicker" (in this case producing oil) when faced with a higher rate of interest. However, having worked through the analysis, we see that the interest rate has another important impact that is easily glossed over; namely it represents the opportunity cost of capital. If we relied solely on our intuition, then it's easy to make the error of ignoring this latter effect. Indeed, as I argued in my previous post, working through the maths has helped us to make an intuitive argument that only becomes "intuitive" upon reflection!

[*] Although, we can compare the fact that oil companies typically make forecasts based on "set" prices to analyse profitability thresholds and so forth.
[**] We could take PVs for investment costs over time if we really wanted...