# Why we need maths in economics 2(a) - The Oil Tank Model

**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".]*

\(Q =\) 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:

\begin{equation}

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

\end{equation}

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:

\begin{equation}

V = q \left[ -c + \frac{p \left(1-e^{-rT}\right)}{r} \right]

\end{equation}

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}

\end{equation}

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.

*"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.

\begin{equation}

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

\end{equation}

We can then use a general derivation rule,

\begin{equation}

\mathrm{d}f = \frac{\partial f}{\partial r} \mathrm{d}r + \frac{\partial f}{\partial q} \mathrm{d}q,

\end{equation}

to get our second last equation (with a little help from the chain, quotient and exponent rules):

\begin{align}

\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 .

\end{align}

Simplifying and setting equal to zero again for optimization yields,

\begin{equation}

\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.

\end{equation}

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}} .

\end{align}

(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} }.

\end{equation}

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:

**Impatience.**We want our money more quickly (and capital costs are high).**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.

[*] 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...

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