Optimal strategy between extraction and storage of crude oil

Abstract

This paper deals with the optimal choice between extraction and storage of crude oil over time. An oil producer should decide on the proportion of oil extracted to sell and the proportion to store. This optimal operational strategy should be conducted on a daily basis while taking into consideration physical, operational and financial constraints such as: storage capacity, crude oil spot price, total quantity available for possible extraction or the maximum amount which could be invested at time t for the extraction choice. Our main results show that when the stock is close to be full, it is better to sell a higher part of the extracted quantity. Therefore, if the stock is empty, the best strategy is to secure the reserves against oil prices fluctuations. In the case of full stock, it is useless to put more quantity in reserves and the best strategy is to sell more output. But, if the stock and the available reserves for extraction are half full, the optimal strategy is to consider both selling and storing output.

Appendix

Proof of Proposition 3.1

From (4.2) and (3.5), we have by integration by parts that

$$\begin{aligned} -\int _0^T \lambda Q_u^sdu= & {} \lambda \left( -TQ_0^s-\int _0^T(T-u)(q_u^s-q_u^{v,s})du\right) \end{aligned}$$

(6.1)

The second term $Q_T^sP_T^v$ gives

$$\begin{aligned} Q_T^sp_T^v= & {} Q_0^sp_0^v+\int _0^Tp_u^vdQ_u^s+\int _0^TQ_u^sdp_u^v\nonumber \\= & {} \int _0^T\left( q_u^s-q_u^{v,s}\right) p_u^vdu+\int _0^TQ^s_up^v_u\mu du+\int _0^T \sigma Q^s_up^v_u dW_u \end{aligned}$$

(6.2)

Taking we expectation of (6.2), removing the martingale term, gives

$$\begin{aligned} \mathbb {E}\left[ Q_T^sp_T^v\right]= & {} Q_0^sp_0^v+\mathbb {E}\left[ \int _0^T\left( q_u^s-q_u^{v,s}\right) p_u^vdu\right] +\mathbb {E}\left[ \int _0^TQ^s_up^v_u\mu du\right] \end{aligned}$$

(6.3)

We study now the last term of (6.3),

$$\begin{aligned} \int _0^TQ^s_up^v_u\mu du= & {} \mu \int _0^TQ^s_up^v_u du\nonumber \\= & {} \mu \left( \int _0^T \left\{ Q^s_0+\int _0^u (q_s^s-q_s^{v,s})ds\right\} p^v_udu\right) \nonumber \\= & {} \mu \left( Q^s_0\int _0^Tp^v_udu+\int _0^T \int _0^T (q_s^s-q_s^{v,s})p^v_s1_{\{s\le u\}}dsdu\right) \nonumber \\= & {} \mu \left( Q^s_0\int _0^Tp^v_udu+\int _0^T \int _0^T p^v_udu 1_{\{s\le u\}}(q_s^s-q_s^{v,s})ds\right) \nonumber \\= & {} \mu \left( Q^s_0\int _0^Tp^v_udu+\int _0^T \left( \int _s^T p^u_tdu\right) (q_s^s-q_s^{v,s})ds\right) \end{aligned}$$

(6.4)

Replacing (6.4) in (6.3), we obtain

$$\begin{aligned} \mathbb {E}\left[ Q_T^sp_T^v\right]= & {} Q_0^sp_0^v+\mathbb {E}\left[ \int _0^T\left( q_u^s-q_u^{v,s}\right) p_u^vdu\right] +\mu Q^s_0 \mathbb {E}\left[ \int _0^Tp^v_udu\right] \nonumber \\&+\,\mu \int _0^T \mathbb {E}\left[ \mathbb {E}\left[ \int _s^T p^u_tdu\vert \mathcal{F}_s\right] (q_s^s-q_s^{v,s})ds\right] \end{aligned}$$

(6.5)

Since

$$\begin{aligned} p^v_u=p^v_s\exp \left( \mu (u-s)\right) \exp \left( \sigma (W_u-W_s)-\frac{\sigma ^2}{2}(u-s)\right) \end{aligned}$$

we get

$$\begin{aligned} \mathbb {E}\left[ p^v_u\vert \mathcal{F}_s\right] =p^v_s\exp \left( \mu (u-s)\right) \end{aligned}$$

Thus we get in (6.5),

$$\begin{aligned} \mathbb {E}\left[ Q_T^sp_T^v\right]= & {} Q_0^sp_0^v+\mathbb {E}\left[ \int _0^T\left( q_u^s-q_u^{v,s}\right) p_u^vdu\right] +\mu Q^s_0 \mathbb {E}\left[ \int _0^Tp^v_udu\right] \nonumber \\&+\,\mu \mathbb {E}\left[ \int _0^T\left( \int _s^Tp^v_s\exp \left( \mu (u-s)\right) du\right) (q_s^s-q_s^{v,s})ds\right] \nonumber \\= & {} Q_0^sp_0^v+\mathbb {E}\left[ \int _0^T\left( q_u^s-q_u^{v,s}\right) p_u^vdu\right] +\mu Q^s_0 \mathbb {E}\left[ \int _0^Tp^v_udu\right] \nonumber \\&+\, \mathbb {E}\left[ \int _0^T \left( e^{\mu (T-s)}-1\right) p^v_s (q_s^s-q_s^{v,s})ds\right] \end{aligned}$$

(6.6)

Thus $\mathbb {E}\left[ \int _0^T \left\{ \left( q_u^{v,s}-q_u^s\right) p_u^v-\lambda Q_u^s\right\} du+Q_T^sp_T^v\right]$ is equal to

$$\begin{aligned}= & {} \mathbb {E}\left[ \int _0^T \left( q_u^{v,s}-q_u^s\right) p_u^vdu-\lambda TQ_0^s-\int _0^T\lambda (T-u)(q_u^s-q_u^{v,s})du\right] \\&+\,Q_0^sp_0^v+\mu y \mathbb {E}\left[ \int _0^Tp^v_udu\right] +\mathbb {E}\left[ \int _0^T\left( q_u^s-q_u^{v,s}\right) p_u^vdu\right] \\&+\,\mathbb {E}\left[ \int _0^T \left( e^{\mu (T-s)}-1\right) p^v_s (q_s^s-q_s^{v,s})ds\right] \end{aligned}$$

After a simplification, we finally obtain

$$\begin{aligned}= & {} -\lambda TQ_0^s+Q_0^sp_0^v+\mu Q_0^s\int _0^T p^v_0\exp \left( \mu u\right) du\\&+\int _0^T \mathbb {E}\left[ (q_u^s-q_u^{v,s})\left\{ p^v_u+\left( e^{\mu (T-u)}-1\right) p^v_u-\lambda (T-u)-p^v_u\right\} \right] du \end{aligned}$$

That the expected result. $\square$

Proof of Theorem 4.2

(i) Let $w \in C^2(\mathbb {R}_+ \times [0, Q^D] \times [0, Q^D])$ and $q \in \mathcal{A}(x)$. By Itô’s forumla we have for any stopping times $\tau _n$

$$\begin{aligned} e^{-\rho \left( T \wedge \tau _n\right) } w\left( X^x_{T \wedge \tau _n}\right)= & {} w(x) + \int _0^{T \wedge \tau _n} e^{- \rho u} \left[ \mathcal{L}^{q_u}w\left( X^x_u\right) - \rho w\left( X^x_u\right) \right] du\\&+ \int _0^{T \wedge \tau _n} e^{- \rho u} \partial _p w\left( X^x_u\right) \sigma P_u d B_u . \end{aligned}$$

We consider the sequence of stopping times $(\tau _n)_{n \ge 1}$ defined by

$$\begin{aligned} \tau _n:= & {} \inf \left\{ t \ge 0 : \int _0^t \left| \partial _p w\left( X^x_u\right) P_u\right| ^2 \ge n \right\} . \end{aligned}$$

We have with this sequence of stopping times

$$\begin{aligned} \mathbb {E}\left[ e^{-\rho \left( T \wedge \tau _n\right) } w\left( X^x_{T \wedge \tau _n}\right) \right]= & {} w(x) + \mathbb {E}\left[ \int _0^{T \wedge \tau _n} e^{- \rho u} \left[ \mathcal{L}^{q_u}w\left( X^x_u\right) - \rho w\left( X^x_u\right) \right] du\right] \\\le & {} w(x) - \mathbb {E}\left[ \int _0^{T \wedge \tau _n} e^{- \rho u} f\left( X^x_u, q_u\right) du \right] . \end{aligned}$$

By using dominated convergence theorem and send n to infinity

$$\begin{aligned} \mathbb {E}\left[ e^{-\rho T} w\left( X^x_{T }\right) \right]\le & {} w(x) - \mathbb {E}\left[ \int _0^{T } e^{- \rho u} f\left( X^x_u, q_u\right) du \right] . \end{aligned}$$

(6.7)

By sending T to infinity and using the dominated convergence theorem we have for any strategy $q \in \mathcal{A}(x)$

$$\begin{aligned} w(x)\ge & {} \mathbb {E}\left[ \int _0^{\infty } e^{- \rho u} f\left( X^x_u, q_u\right) du \right] , \end{aligned}$$

which implies $w(x) \ge v(x)$ for any $x \in \mathbb {R}_+ \times [0, Q^D] \times [0, Q^D]$.

(ii) By repeating the above arguments and observing that the control ${\hat{q}}$ achieves equality (6.7), we have

$$\begin{aligned} \mathbb {E}\left[ e^{-\rho T} w\left( {\hat{X}}^{x}_T\right) \right]= & {} w(x) - \mathbb {E}\left[ \int _0^T e^{-\rho u} f\left( {\hat{X}}^{x}_u, {\hat{q}}\left( {\hat{X}}^{x}_u\right) \right) du \right] . \end{aligned}$$

By sending T to infinity and from (4.8) we get

$$\begin{aligned} w(x)\le & {} \mathbb {E}\left[ \int _0^\infty e^{-\rho u} f\left( {\hat{X}}^{x}_u, {\hat{q}}\left( {\hat{X}}^{x}_u\right) \right) du \right] , \end{aligned}$$

where the left side term is by definition $J(x,{\hat{q}})$, thus $w(x) =J(x,{\hat{q}})$. $\square$