The Oil Cartel and Misallocation of Production

Abstract

The oil cartel’s activity causes distortions of the oil industry structure resulting from the spatial misallocation of production. In the long term, production shifting from the low-cost cartel to the high-cost competitive fringe brings about a misallocation of global oil resources. In this chapter we consider a two-region model of resource extraction with investment in reserves development, in which the global allocation of oil resources is determined endogenously. We show that if the cartel’s advantage in production costs over the competitive fringe is considerable, the distorting effects of the cartel are substantial, and the efficiency losses on the industry level may be significant. The cartel captures a part of consumer benefits at the cost of significant losses from resource misallocation and deadweight losses. These losses, however, can be reduced due to technological changes improving the competitiveness of the fringe.

This chapter is based on the unfinished paper by Barry Ickes and Georgy Trofimov.

Appendices

Appendices

1.1 A.1 Condition (10.20)

Investment in creation of the initial reserve stock is

$$ {i}_a={z}_a{s}_a={z}_a{x}_a/{\delta}_a{\xi}_a={z}_a{x}_a{\left(2{\delta}_ar{z}_a\right)}^{-1/2}={x}_a{\left(2r{z}_a/{\delta}_a\right)}^{1/2}/2r={x}_a{\xi}_a/2r, $$

since ξ_a = x_a/δ_as_a and due to (10.11). From Eq. (10.18), the firm’s net present value (10.19) is

$$ {V}_a={r}^{-1}{\pi}_a-{i}_a={x}_a^2/2r{\delta}_a{s}_a-{x}_a{\xi}_a/2r={x}_a\left({\xi}_a-{\xi}_a\right)/2r=0. $$

1.2 A.2 Equations (10.27), (10.28)

From Eqs. (10.22), (10.24), the first-order condition for the cartel is:

$$ {P}_t+{x}_{At}\frac{\partial {P}_t}{\partial {x}_{At}}-\frac{x_{At}}{\delta_a{s}_{At}}-{z}_a=0. $$

We use the fact that ∂P_t/∂x_At =  − 1/(μ + δ_bs_Bt) and rearrange the left-hand side:

$$ {\displaystyle \begin{array}{l}{P}_t+{x}_{At}\frac{\partial {P}_t}{\partial {x}_{At}}-\frac{x_{At}}{\delta_a{s}_{At}}-{z}_a={P}_t-\frac{x_{At}}{\mu +{\delta}_b{s}_{Bt}}-\frac{x_{At}}{\delta_a{s}_{At}}-{z}_a\\ {}\kern5em ={P}_t-\frac{\left(\mu +{\delta}_a{s}_{At}+{\delta}_b{s}_{Bt}\right){x}_{At}}{\left(\mu +{\delta}_b{s}_{Bt}\right){\delta}_a{s}_{At}}-{z}_a={P}_t-\frac{x_{At}}{\left({\sigma}_{Bt}+{\sigma}_{Dt}\right){\delta}_a{s}_{At}}-{z}_a\\ {}\kern5em ={P}_t-\frac{\xi_{At}}{\sigma_{Bt}+{\sigma}_{Dt}}-{z}_a.\end{array}} $$

This yields (10.27).

Consider Eq. (10.28). From Eqs. (10.8), (10.27), the supply functions are:

$$ {x}_{At}={\delta}_a{s}_{At}\left(1-{\sigma}_{At}\right)\left({P}_t-{z}_a\right),{x}_{Bt}={\delta}_b{s}_{Bt}\left({P}_t-{z}_b\right). $$

Inserting these functions into the market-clearing condition (10.3) implies the equation for P_t:

$$ {\sigma}_{At}{E}_t^C\left(1-{\sigma}_{At}\right)\left({P}_t-{z}_a\right)+{\sigma}_{Bt}{E}_t^C\left({P}_t-{z}_b\right)=\mu \left(\alpha -{P}_t\right), $$

because ${\delta}_j{s}_{Jt}={\sigma}_{Jt}{E}_t^C$ for J = A, B, D = αμ. This equation can be rewritten as

$$ {\sigma}_{At}\left(1-{\sigma}_{At}\right)\left({P}_t-{z}_a\right)+{\sigma}_{Bt}\left({P}_t-{z}_b\right)={\sigma}_{Dt}\left(\alpha -{P}_t\right). $$

and the market-clearing cartel price is equal to

$$ {\displaystyle \begin{array}{l}{P}_t=\frac{\sigma_{At}\left(1-{\sigma}_{At}\right){z}_a+{\sigma}_{Bt}{z}_b+{\sigma}_{Dt}\alpha }{\sigma_{At}\left(1-{\sigma}_{At}\right)+{\sigma}_{Bt}+{\sigma}_{Dt}}=\frac{\sigma_{At}\left(1-{\sigma}_{At}\right){z}_a+{\sigma}_{Bt}{z}_b+{\sigma}_{Dt}\alpha }{1-{\sigma}_{At}^2}\\ {}\kern5em ={\theta}_{At}{z}_a+{\theta}_{Bt}{z}_b+{\theta}_{Dt}\alpha .\end{array}} $$

1.3 A.3 Proposition 10.2

The cartel price Eqs. (10.30) and (10.31) are compatible if σ_A satisfies:

$$ {\xi}_b+{z}_b=\frac{\xi_a}{1-{\sigma}_A}+{z}_a $$

or 1 − σ_A = ξ_a/(ξ_b + z_b − z_a), implying (10.33):

$$ {\sigma}_A=1-\frac{\xi_a}{\xi_b+{z}_b-{z}_a}=\frac{\Delta P}{\xi_a+\Delta P}, $$

since ΔP = Δξ + Δz, where Δξ = ξ_b − ξ_a, Δz = z_b − z_a.

Equation (10.31) is compatible with (10.32) represented as

$$ P=\frac{\sigma_A\left(1-{\sigma}_A\right){z}_a+{\sigma}_B{z}_b+{\sigma}_D\alpha }{1-{\sigma}_A^2}, $$

if the following equation is fulfilled:

$$ \frac{\xi_a}{1-{\sigma}_A}+{z}_a=\frac{\sigma_A\left(1-{\sigma}_A\right){z}_a+{\sigma}_B{z}_b+{\sigma}_D\alpha }{1-{\sigma}_A^2}. $$

Rearrange both sides:

$$ \left(1+{\sigma}_A\right){\xi}_a+\left(1-{\sigma}_A^2\right){z}_a={\sigma}_A\left(1-{\sigma}_A\right){z}_a+{\sigma}_B{z}_b+{\sigma}_D\alpha $$

$$ \left(1+{\sigma}_A\right){\xi}_a+\left(1-{\sigma}_A\right){z}_a={\sigma}_B{z}_b+{\sigma}_D\alpha $$

or, since σ_D = 1 − σ_A − σ_B,

$$ \left(1+{\sigma}_A\right){\xi}_a-\left(1-{\sigma}_A\right)\left(\alpha -{z}_a\right)={\sigma}_B\left({z}_b-\alpha \right). $$

From Eq. (10.31), ξ_a = (1 − σ_A)(P − z_a), hence

$$ {\sigma}_B=\left(1-{\sigma}_A\right)\frac{\alpha -{z}_a-\left(1+{\sigma}_A\right)\left(P-{z}_a\right)}{\alpha -{z}_b}. $$

We have:

$$ \left(1+{\sigma}_A\right)\left(P-{z}_a\right)=\frac{\xi_a+2\Delta P}{\xi_a+\Delta P}\left(P-{z}_a\right)={\xi}_a+2\Delta P=-{z}_a-p+2P $$

due to (10.33) and since p = ξ_a + z_a.

Consequently

$$ {\sigma}_B=\left(1-{\sigma}_A\right)\frac{\alpha +p-2P}{\alpha -{z}_b}=\left(1-{\sigma}_A\right)\frac{2\left(\overline{P}-P\right)}{\alpha -{z}_b}. $$

1.4 A.4 The Pure Monopoly Case

Equating the right-hand sides of Eqs. (10.31′) and (10.32′) yields the following equation on σ_A:

$$ \frac{\xi_a}{1-{\sigma}_A}+{z}_a=\frac{\sigma_A}{1+{\sigma}_A}{z}_a+\frac{\alpha }{1+{\sigma}_A}. $$

Rearrange the terms:

$$ \frac{\xi_a}{1-{\sigma}_A}=\frac{\alpha -{z}_a}{1+{\sigma}_A} $$

$$ \left(1+{\sigma}_A\right){\xi}_a=\left(1-{\sigma}_A\right)\left(\alpha -{z}_a\right) $$

and obtain:

$$ {\sigma}_A=\frac{\alpha -{z}_a-{\xi}_a}{\alpha -{z}_a+{\xi}_a}=\frac{\alpha -p}{\alpha -p+2{\xi}_a}. $$

1.5 A.5 Proposition 10.3

We have it from condition (10.29) that ξ_J = x_J/δ_js_J = ξ_j implying that x_J = δ_js_Jξ_j for J = A, B and j = a, b. This yields x_J = (σ_J/σ_D)μξ_j because σ_J = δ_js_J/E^C and σ_D = μ/E^C.

1.6 A.6 Equations (10.41), (10.42)

From Eqs. (10.31), (10.40):

$$ {\displaystyle \begin{array}{l}{\pi}_A={Px}_A-{c}_a\left({x}_A,{s}_A\right)-{i}_A=\left(\frac{\xi_a}{1-{\sigma}_A}+{z}_a\right){x}_A-\frac{x_A^2}{2{\delta}_a{s}_A}-{z}_a{x}_A=\frac{\xi_a{x}_A}{1-{\sigma}_A}-\frac{x_A^2}{2{\delta}_a{s}_A}\\ {}\kern4em =\frac{\xi_a{x}_A}{1-{\sigma}_A}-\frac{\xi_a{x}_A}{2}=\left(\frac{1}{1-{\sigma}_A}-\frac{1}{2}\right){\xi}_a{x}_A=\frac{1+{\sigma}_A}{1-{\sigma}_A}\cdot \frac{\xi_a{x}_A}{2}\kern0.5em .\end{array}} $$

The cartel value in period 0 is V_A = r⁻¹π_A − z_as_A. Since x_A = δ_as_Aξ_a and from Eq. (10.29) we have:

$$ {z}_a{s}_A={z}_a\frac{x_A}{\delta_a{\xi}_a}=\frac{z_a{x}_A}{{\left(2{\delta}_ar{z}_a\right)}^{1/2}}. $$

(10.50)

Hence

$$ {\displaystyle \begin{array}{l}{V}_A={r}^{-1}\frac{1+{\sigma}_A}{1-{\sigma}_A}\cdot \frac{\xi_a{x}_A}{2}-\frac{z_a{x}_A}{{\left(2{\delta}_a{rz}_a\right)}^{1/2}}=\left(\frac{1+{\sigma}_A}{1-{\sigma}_A}\cdot \frac{\xi_a}{2}-\frac{z_ar}{{\left(2{\delta}_a{rz}_a\right)}^{1/2}}\right){x}_A/r\\ {}\kern5em =\left(\frac{1+{\sigma}_A}{1-{\sigma}_A}\cdot \frac{\xi_a}{2}-\frac{{\left(2{rz}_a/{\delta}_a\right)}^{1/2}}{2}\right){x}_A/r=\left(\frac{1+{\sigma}_A}{1-{\sigma}_A}\cdot \frac{\xi_a}{2}-\frac{\xi_a}{2}\right){x}_A/r\\ {}\kern5em =\left(\frac{1+{\sigma}_A}{1-{\sigma}_A}-1\right){x}_A{\xi}_a/2r=\frac{2{\sigma}_A{\xi}_a}{1-{\sigma}_A}{x}_A/2r=\frac{\sigma_A{\xi}_a}{1-{\sigma}_A}{x}_A/r={x}_A\Delta P/r,\end{array}} $$

because, from Eq. (10.33),

$$ \Delta P=\frac{\sigma_A{\xi}_a}{1-{\sigma}_A}. $$

Equations (10.50), (10.29) imply that

$$ {i}_{0A}={z}_a{s}_A={z}_a{x}_A{\left(2{\delta}_ar{z}_a\right)}^{-1/2}={x}_A{\left(2r{z}_a/{\delta}_a\right)}^{1/2}{(2r)}^{-1}={r}^{-1}{x}_A{\xi}_a/2 $$

$$ ={r}^{-1}{x}_A\left(p-{z}_a\right)/2. $$

This means that the initial investment by the cartel equals the present value of the cartel’s net competitive profit shown by triangle z_apG in Fig. 10.5.

1.7 A.7 Equation (10.44)

The loss of consumer surplus due to cartel pricing is

$$ {\displaystyle \begin{array}{l}\varDelta Q={Q}^C-Q=\mu \left({\left(\alpha -P\right)}^2-{\left(\alpha -p\right)}^2\right)/2r=\mu \left(p-P\right)\left(\left(\alpha -P\right)+\left(\alpha -p\right)\right)/2r\\ {}\kern1.6em =-\varDelta P\left(D(P)+D(p)\right)/2r.\end{array}} $$

(10.51)

From Eqs. (10.42), (10.51) and (10.3):

$$ \varDelta {W}^C=\varDelta Q+\varDelta {V}_A=-\left(D(P)+D(p)\right)\varDelta P/2r+{x}_A\varDelta P/r=-\left(D(P)+D(p)-2{x}_A\right)\varDelta P/2r=-\left({x}_B+{x}_a-{x}_A\right)\varDelta P/2r=-\left({x}_B-\varDelta {x}_A\right)\varDelta P/2r. $$

From Eq. (10.38), −Δx_A = x_B − ΔD and, consequently, ΔW^C =  − (2x_B − ΔD)ΔP/2r =  − (x_B + (μ/2)ΔP)ΔP/r =  − (x_BΔP + (μ/2)(ΔP)²)/r, since −ΔD = μΔP.

1.8 A.8 Equations (10.47)–(10.49)

From Proposition 10.2, the demand share is

$$ {\sigma}_D=\left(1-{\sigma}_A\right)\left(1-\phi \right)=\left(1-{\sigma}_A\right)\frac{P+\Delta P-{z}_b}{\alpha -{z}_b}=\left(1-{\sigma}_A\right)\frac{\xi_b+\Delta P}{\alpha -{z}_b}. $$

(10.52)

Equations (10.42), (10.39), (10.33), (10.52) imply that

$$ \kern1em \varDelta {V}_A={x}_A\varDelta P/r=\mu \frac{\sigma_A}{r{\sigma}_D}{\xi}_a\varDelta P=\mu \frac{\sigma_A\left(\alpha -{z}_b\right)}{r\left(1-{\sigma}_A\right)\left({\xi}_b+\varDelta P\right)}{\xi}_a\varDelta P=\mu \frac{\Delta P\left(\alpha -{z}_b\right)}{r{\xi}_a\left({\xi}_b+\varDelta P\right)}{\xi}_a\varDelta P=\frac{\mu \left(\alpha -{z}_b\right){\left(\Delta P\right)}^2}{r\left(2{\xi}_b+{z}_b-{\xi}_a-{z}_a\right)}=\frac{\mu \left(\alpha -{z}_b\right){\left(\Delta P\right)}^2}{2r\left({\xi}_b+{z}_b-0.5{\xi}_a-0.5\left({z}_b+{z}_a\right)\right)}=\frac{\mu \left(\alpha -{z}_b\right){\left(\Delta P\right)}^2}{2r\left(P-\tilde{z}-0.5{\xi}_a\right)}. $$

From Eq. (10.21), W = μ(α − p)²/2r, hence

$$ \frac{\Delta {V}_A}{W}=\frac{\left(\alpha -{z}_b\right){\left(P-p\right)}^2}{\left(P-\overset{\sim }{z}-0.5{\xi}_a\right){\left(\alpha -p\right)}^2}. $$

From Eqs. (10.45), (10.34), (10.39), (10.52) we have it that

$$ -\Delta {W}_1={x}_B\Delta P/r=\mu \frac{\sigma_B}{r{\sigma}_D}{\xi}_b\Delta P=\mu \frac{2\left(\overline{P}-P\right)}{r\left({\xi}_b+\Delta P\right)}{\xi}_b\Delta P=\frac{\mu \left(\overline{P}-P\right){\xi}_b\Delta P}{r\left(P-\overset{\sim }{z}-0.5{\xi}_a\right)}. $$

Consequently

$$ \frac{\Delta {W}_1}{W}=-\frac{2{\xi}_b\left(\overline{P}-P\right)\left(P-p\right)}{\left(P-\overset{\sim }{z}-0.5{\xi}_a\right){\left(\alpha -p\right)}^2}. $$

From Eqs. (10.46), (10.21), ΔW₂ =  − μ(ΔP)²/2r and

$$ \frac{\Delta {W}_2}{W}=-\frac{{\left(P-p\right)}^2}{{\left(\alpha -p\right)}^2}. $$