Strategic bypass deterrence

Abstract

In liberalized network industries, competitors can either compete for service using the existing infrastructure (access) or deploy their own capacity (bypass). We revisit this make-or-buy problem making two contributions to the literature. First we analyze both the profit maximizing behavior of an incumbent and the welfare maximizing behavior when the entrant chooses between access and bypass. Second, we extend the baseline model studied in the literature by allowing for fixed costs of network installation. By analogy to the literature on strategic entry deterrence, we distinguish three régimes of blockaded bypass, deterred bypass and accommodated bypass depending on the entrant’s unit cost. We show that the make-or-buy decision of the entrant is not necessarily technologically efficient: when bypass is chosen, it is always the cheapest option but access may be chosen when it is not cost effective.

A The linear model

In this Appendix, we use the linear demand functions defined in Eq. (1) to derive the explicit functional forms for the equilibrium prices, quantities and profits and to check our comparative static results. Lemmas are proven using these explicit formulae. The linear model allows us to derive the thresholds for the access charge $\omega ^e$ and $\omega ^l$ and for the cost $c^D$, $c^B$, $c^W$ and $c^I$ as they are the solution to second degree equations. These expressions are used for our numerical simulations reproduced in Tables 1 and 2.

Equilibrium prices and quantities under access Profits under access are defined as:

$$\begin{aligned} \pi _1^a(p_1, p_2, w)= & {} (p_1-c_1)(1-p_1+\delta p_2)+(w-c_1) (1-p_2+\delta p_1)-f_1, \\ \pi _2^a(p_1, p_2, w)= & {} (p_2-w) (1-p_2+\delta p_1). \end{aligned}$$

From the profit functions, we can derive the unique equilibrium prices and the corresponding quantities for any given w:

$$\begin{aligned} {\tilde{p}}_1^a= & {} \frac{2+\delta }{4-\delta ^2}+\frac{2(1-\delta )}{4-\delta ^2}c_1+\frac{3\delta }{4-\delta ^2} w,\\ {\tilde{p}}_2^a= & {} \frac{2+\delta }{4-\delta ^2}+\frac{\delta (1-\delta )}{4-\delta ^2}c_1+\frac{(2+\delta ^2)}{4-\delta ^2} w,\\ {\tilde{q}}_1^a= & {} \frac{2+\delta }{4-\delta ^2}-\frac{(1-\delta )(2-\delta ^2)}{4-\delta ^2}c_1-\frac{\delta (1-\delta ^2)}{4-\delta ^2} w, \\ {\tilde{q}}_2^a= & {} \frac{2+\delta }{4-\delta ^2}+\frac{\delta (1-\delta )}{4-\delta ^2}c_1-\frac{2 (1-\delta ^2))}{4-\delta ^2} w. \\ \end{aligned}$$

It is straightforward to check that equilibrium prices are increasing in w and the corresponding equilibrium quantities are decreasing. The equilibrium profits are given by:

$$\begin{aligned} {\tilde{\pi }}_1^a= & {} \frac{[2-c_1(2+2 \delta - \delta ^2)+\delta + 3 w \delta ][2 +\delta -\delta w (1-\delta ^2) -c_1(1-\delta )(2-\delta ^2)]}{(4-\delta ^2)^2} \\&+ \frac{[(w-c_1)(4-\delta ^2)][2 +\delta +c_1 \delta (1-\delta )- 2 w(1-\delta ^2)]}{(4-\delta ^2)^2}, \\ {\tilde{\pi }}_2^a= & {} \frac{(2+\delta +\delta (1-\delta ) c_1 -2(1-\delta ^2)w)^2}{(4-\delta ^2)^2}. \end{aligned}$$

The profit functions are quadratic in w, ${\tilde{\pi }}_1^a$ is concave in w and $\frac{\partial {\tilde{\pi }}_2^a}{ \partial w}<0$. The profit maximizing access charge $w^*$ is defined as:

$$\begin{aligned} w^* =\frac{8 + \delta ^3}{2(1-\delta )(8+\delta ^2)} +\frac{(1-\delta )(8+ 2 \delta ^2 -\delta ^3)}{2(1-\delta )(8+\delta ^2)}c_1>c_1. \end{aligned}$$

Equilibrium prices and quantities under bypass Profits under bypass are defined as:

$$\begin{aligned} \pi _1^b(p_1, p_2)= & {} (p_1-c_1)(1-p_1+\delta p_2)-f_1, \\ \pi _2^b(p_1, p_2)= & {} (p_2-c_2) (1-p_2+\delta p_1)-f_2. \end{aligned}$$

Solving for the linear demand model, equilibrium prices, quantities and profits are given by:

$$\begin{aligned} {\tilde{p}}_1^b= & {} \frac{2+\delta }{4-\delta ^2}+\frac{2}{4-\delta ^2}c_1+\frac{\delta }{4-\delta ^2}c_2,\\ {\tilde{p}}_2^b= & {} \frac{2+\delta }{4-\delta ^2}+\frac{\delta }{4-\delta ^2}c_1+\frac{2}{4-\delta ^2}c_2,\\ {\tilde{q}}_1^b= & {} \frac{2+\delta }{4-\delta ^2}-\frac{(2 - \delta ^2)}{4-\delta ^2}c_1+\frac{\delta }{4-\delta ^2}c_2, \\ {\tilde{q}}_2^b= & {} \frac{2+\delta }{4-\delta ^2}+\frac{\delta }{4-\delta ^2}c_1-\frac{(2 - \delta ^2)}{4-\delta ^2}c_2. \\ {\tilde{\pi }}_1^b= & {} \frac{(2+\delta -(2-\delta ^2)c_1+\delta c_2)^2}{(4-\delta ^2)^2}-f_1, \\ {\tilde{\pi }}_2^b= & {} \frac{(2+\delta -(2-\delta ^2)c_2+\delta c_1)^2}{(4-\delta ^2)^2}-f_2. \end{aligned}$$

And the standard comparative static results apply: $\frac{\partial {\tilde{p}}_i^b}{\partial c_i}>\frac{\partial {\tilde{p}}_i^b}{\partial c_j}>0$, $\frac{\partial {\tilde{\pi }}_i^b}{\partial c_i}<0$ and $\frac{\partial {\tilde{\pi }}_i^b}{\partial c_j}>0$.

Proof of Lemma 1

The proof of Lemma 1 can be easily done by replacing w and $c_2$ by x in ${\tilde{p}}_i^a$ and ${\tilde{p}}_i^b$. Then, we have that:

$$\begin{aligned} {\tilde{p}}_1^a-{\tilde{p}}_1^b=\frac{2\delta }{4-\delta ^2}(c_1-x),\,\, {\tilde{p}}_2^a-{\tilde{p}}_2^b=\frac{\delta ^2}{4-\delta ^2}(c_1-x). \end{aligned}$$

And the lemma is proven. $\square$

Limit and equivalent access charges Solving Eq. (10), we find the limit access charge $\omega ^l$:

$$\begin{aligned} \omega ^l= \frac{2+{\delta }+ c_1 {\delta }(1-{\delta }) - \sqrt{(2 + {\delta }- (2-{\delta }^2)c_2+ {\delta }c_1)^2 - f_2(4- {\delta }^2)^2}}{2(1-{\delta }^2)}. \end{aligned}$$

And $\omega ^l$ is increasing in both $c_2$ and $f_2$. When the entrant has no fixed cost ($f_2=0$), then $\omega ^l= \frac{c_2(2-\delta ^2) - c_1 \delta ^2}{2(1-\delta ^2)}$ and it is easy to check that $\omega ^l> c_2$ if $c_2> c_1$ and $\omega ^l< c_2$ if $c_2< c_1$. The equivalent access charge is the solution to Eq. (11) but it is not reproduced here as the expression is complicated and has no value-added.

Proof of Lemma 2

To check that Lemma 2 is satisfied, note that:

$$\begin{aligned} \frac{\partial {\tilde{\pi }}_2^a(x)}{\partial x} - \frac{\partial {\tilde{\pi }}_2^b(x)}{\partial x}= & {} {\tilde{q}}_2^b - {\tilde{q}}_2^a + \delta \left[ \frac{\partial {\tilde{p}}_1^a}{\partial x} - \frac{\partial {\tilde{p}}_1^b}{\partial x}\right] \\= & {} \frac{\delta }{4-\delta ^2} \left[ (2-\delta )c_1 - \delta x + 2 \delta \right] \\> & {} 0. \end{aligned}$$

$\square$

Proof of Lemma 3

We check that Lemma 3 is satisfied by noting that $\frac{\partial {\tilde{\pi }}_1^{a}(w) }{\partial w} \frac{\partial w^l(x)}{\partial x}- \frac{\partial {\tilde{\pi }}_1^{b}(x)}{\partial x}$ is a decreasing function of x. Hence, $\frac{\partial {\tilde{\pi }}_1^{a}(w) }{\partial w} \frac{\partial w^l(x)}{\partial x}- \frac{\partial {\tilde{\pi }}_1^{b}(x)}{\partial x} > 0$ if and only if the function is positive when $x = c_1$ and ${\tilde{w}}_2(x)= c_1$. Computations show that, at this point,

$$\begin{aligned} \frac{\partial {\tilde{\pi }}_1^{a}(w) }{\partial w} \frac{\partial w^l(x)}{\partial x}- \frac{\partial {\tilde{\pi }}_1^{b}(x)}{\partial x}= & {} -\,4 \delta (1-\delta ^2)(2+\delta ) + (2-\delta ^2)(2+\delta )(2\delta +\delta ^3+1) \\&+ \,c_1 \left[ 1-\frac{2(1-\delta ^2)(1+\delta ^2)}{(2-\delta ^2)} \right. \\&\left. +\, (2-\delta ^2)(1-\delta )(2+3\delta -4 \delta ^2-4\delta ^3+\delta ^4)\right] \end{aligned}$$

$\square$

which is a linear function of $c_1$ and is positive for all $0< \delta < 1$ both at $c_1=0$ and at $c_1=1$, showing that the lemma is satisfied.

Welfare and regulation

It is easy to check that the welfare under access is concave in the access charge:

$$\begin{aligned} \frac{\partial W^a}{\partial w} = \frac{1-\delta ^2}{(4 - \delta ^2)^2} \left[ -(2+\delta )^2 - w(4+5 \delta ^2) +(8+2 \delta ^2 - \delta ^3) c_1\right] . \end{aligned}$$

We can thus identify an access charge ${\hat{w}}$ that maximizes the welfare function $W^a$:

$$\begin{aligned} \hat{w} = \frac{(8+2\delta ^2 - \delta ^3) c_1-(2+\delta )^2}{(4+5 \delta ^2)}<c_1. \end{aligned}$$

(17)

However, the incumbent’s profit $\pi _1^a({\hat{w}})$ is negative and the access charge ${\hat{w}}$ does not satisfy the constraint ${\hat{w}}\ge \omega _0$ even for $f_1=0$:

$$\begin{aligned} {\tilde{\pi }}_1^a(\omega _0)=-\frac{(1-c_1(1-\delta ))^2(12+\delta (28+\delta (20+\delta (17+4\delta ))))}{(4+5 \delta ^2)^2}<0. \end{aligned}$$

The welfare under bypass is decreasing in the entrant’s cost $c_2$:

$$\begin{aligned} \frac{\partial W^b}{\partial c_2} = \frac{1}{(4 - \delta ^2)^2} \left[ -(2+\delta )(2-\delta -\delta ^2) - 2 c_1 \delta (2-\delta ^2) + c_2 (\delta ^2 + (2-\delta )^2\right] <0. \end{aligned}$$