Funding natural monopoly infrastructure expansion: auctions versus regulated uniform access prices

Abstract

This paper characterizes equilibrium behavior in an auction where firms bid a unit price in order to fund and gain access to a new piece of infrastructure. Firms need access to this new infrastructure to sell their output in a competitive market. The new facility is built only if enough revenue is generated through the sale of access to the firms. We show by means of an example that this auction mechanism can dominate in terms of efficiency the standard mechanism where a uniform access price is determined ex-ante by a regulator.

Appendix

1.1 Proof of Proposition 1

Proof

The objective of the regulator is to set the optimal access fees $\alpha _{i}^{*}$ and $\alpha _{j}^{*}$ so that the facility is built whenever it is efficient to do so and to recover exactly the construction cost K. Therefore, the regulator maximizes the probability of completion, that is,

$$\begin{aligned}\rho =F(\hat{p}-\alpha _i)F(\hat{p}-\alpha _j), \end{aligned}$$

subject to the following three conditions,

$$\begin{aligned}&\left( \hat{p}-\hat{c}-\alpha _{i}^{*}\right) \bar{q}_{i}\ge 0,\\&\left( \hat{p}-\hat{c}-\alpha _{j}^{*}\right) \bar{q}_{j}\ge 0,\\&\alpha _{i}^{*}\bar{q}_{i}+\alpha _{j}^{*}\bar{q} _{j} =K. \end{aligned}$$

Substituting from the last condition into the objective function for each $\alpha$ and then differentiating gives the following first-order conditions.

$$\begin{aligned}&-f(\hat{p}-\alpha _i) F\left( \hat{p}-\frac{K-\alpha _i\bar{q}_i}{\bar{q}_j}\right) \nonumber \\&\quad + \frac{\bar{q}_i}{\bar{q}_j} f\left( \hat{p}-\frac{K-\alpha _i\bar{q}_i}{\bar{q}_j}\right) F(\hat{p}-\alpha _i)=0 \end{aligned}$$

(27)

$$\begin{aligned}&-f(\hat{p}-\alpha _j) F\left( \hat{p}-\frac{K-\alpha _j\bar{q}_j}{\bar{q}_i}\right) \nonumber \\&\quad + \frac{\bar{q}_j}{\bar{q}_i} f\left( \hat{p}-\frac{K-\alpha _j\bar{q}_j}{\bar{q}_i}\right) F(\hat{p}-\alpha _j)=0 \end{aligned}$$

(28)

Rearranging the above two terms would give Eqs. (3) and (4). $\hfill\square$

1.2 Proof of Corollary 1

Proof

When capacities are equal it is easy to check that both first order conditions becomes identical. That is the optimal access charges are the same. Then substituting equal access charges to the last condition, $\alpha _{i}^{*}\bar{q}_{i}+\alpha _{j}^{*}\bar{q}_{j} =K$, gives $\alpha _i = \alpha _j = \frac{K}{\bar{q}_{i}+\bar{q}_{j}}$. $\hfill\square$

1.3 Proof of Corollary 2

Proof

Substituting uniform distribution in the first-order conditions gives the above access charges. $\hfill\square$

1.4 Proof of Proposition 2

Proof

If both firm follow the same strategy in a symmetric equilibrium we have,

$$\begin{aligned}&F\big ( \alpha ^{-1} \big (\frac{K-\alpha (c) \bar{q}_i}{\bar{q}_j}\big )\big )+ \frac{\bar{q}_i}{\bar{q}_j}(\hat{p} -c - \alpha (c) ) f\big ( \alpha ^{-1} \big (\frac{K-\alpha (c) \bar{q}_i}{\bar{q}_j}\big )\big )(\alpha ^{-1})'\nonumber \\&\quad \times \big (\frac{K-\alpha (c) \bar{q}_i}{\bar{q}_j}\big )=0 \end{aligned}$$

(29)

Equation (29) is not an ordinary differential equation and cannot be solved with standard computation methods. Thus, we proceed by transforming the differential equation given by Eq. (29) into two ordinary differential equations.

$$\begin{aligned} \psi (c)= \alpha ^{-1} \big (\frac{K-\alpha (c) \bar{q}_i}{\bar{q}_j}\big ). \end{aligned}$$

(30)

Then we can rewrite Eq. (29) as follows,

$$\begin{aligned} F\big ( \psi (c)\big )+ \frac{\bar{q}_i}{\bar{q}_j}(\hat{p} -c - \alpha (c) ) f\big ( \psi (c)\big )(\alpha ^{-1})'\big (\alpha (\psi (c))\big )=0. \end{aligned}$$

(31)

Because $(\alpha ^{-1})\big (\alpha (\psi (c))\big )=\psi (c)$, it is straightforward to show,

$$\begin{aligned}&(\alpha ^{-1})'\big (\alpha (\psi (c))\big )\alpha '(\psi (c))=1. \end{aligned}$$

Thus the first-order condition becomes,

$$\begin{aligned} -\frac{ \frac{\bar{q}_i}{\bar{q}_j}(\hat{p} -c - \alpha (c) ) f\big ( \psi (c)\big )}{F\big ( \psi (c)\big )}=\alpha '(\psi (c)). \end{aligned}$$

(32)

When $\bar{q}_i=\bar{q}_j$ we have, $\psi (c)= \alpha ^{-1} \big (\frac{K}{\bar{q}_i}-\alpha (c) \big )$. Define $A=\frac{K}{\bar{q}_i}=\frac{K}{\bar{q}_j}$, then we have,

$$\begin{aligned} A =\alpha (\psi (c)) + \alpha (c). \end{aligned}$$

(33)

Using the same equation we can also write, $A=\alpha (\psi (\psi (c))) + \alpha (\psi (c))$. Thus we have, $\alpha (\psi (\psi (c))) = \alpha (c)$ or $\psi (\psi (c)) = c$. Applying this to Eq. (32) results in

$$\begin{aligned} \alpha '(c)=\frac{ (-\hat{p} +\psi (c) + A - \alpha (c) ) f( c)}{F(c)} \end{aligned}$$

(34)

By differentiating Eq. (33) we get another equation as follows,

$$\begin{aligned} \psi '(c) =\frac{- \alpha '(c)}{\alpha '(\psi (c))}= \frac{ (-\hat{p} +\psi (c) + A - \alpha (c) ) f( c) F(\psi (c))}{(\hat{p} -c - \alpha (c) ) f\big ( \psi (c)\big )F(c)}. \end{aligned}$$

(35)

$\hfill\square$

1.5 Proof of Proposition 3

Proof

If $\hat{p}-c_i\ge \frac{K}{2\bar{q}}$, then neither of the firms submit a bid less than $\frac{K}{2\bar{q}}$, because according to the payment rule this guarantees the payoff of zero. If one firm follows $\alpha ^*$, then the other firm’s best strategy is to also follow $\alpha ^*$, because bidding more does not change the probability of completion while reduces the profit and bidding less guarantees zero. Thus, $\alpha ^*$ is an equilibrium.

Now if a firm submits a bid equal to $\frac{K}{2\bar{q}}$ then it must be the case with $\hat{p}-c_i \ge \frac{K}{2\bar{q}}$, otherwise it is better off by not bidding. $\hfill\square$