The impact of transportation asymmetry on the choice of a spatial price policy

Abstract

This paper studies the strategic choice of a spatial price policy between mill pricing and uniform-delivered pricing (UDP) under transportation asymmetry. Mill pricing in the model emerges as an equilibrium price policy, but it contains different types of strategic interaction, depending on the level of discrepancy between two firms’ transportation rates. If the transportation rate discrepancy is not too great, then mill pricing is a dominant strategy. When the discrepancy is large enough, mill pricing is viewed as a M-matching strategy, whereby the low-transportation-cost firm prefers matching the rival’s strategy, but the high-transportation-cost firm does not do so. There is no “Prisoner’s Dilemma” like the argument that Thisse and Vives (Am Econ Rev 78(1):122–137, 1988, AER) proposes, and there is no robustness for firms to choose a UDP policy like Kats and Thisse (in: Ohta and Thisse (eds) Does economic space matter? St Martin’s, New York, 1993) do. Our study matches the current trend of technology advancement in transportation.

Appendices

Appendix 1: Firm 2’s equilibrium profits under U–U competition

Recall that $t_2=1$ and firm 2 is located at the right endpoint of the line market. Consider firm 2’s undercutting strategy, whereby $p_2<p_1$. If so, firm 2 sells its product further to location ${\hat{y}}_1$ where its delivered price equals delivered cost, that is, $p_2 = c + 1\cdot (1-{\hat{y}}_1) \Rightarrow {\hat{y}}_1 = 1+c-p_2$. Under this situation for any value of $p_2<p_1$, firm 2’s profit $\pi _2^+$ from undercutting 1’s price is

$$\begin{aligned} \pi _2^+ = \int _{{\hat{y}}_1}^1 (p_2 - c-(1-x)){\text {d}}x = \frac{( p_2 - c )^2}{2}. \end{aligned}$$

(14)

If firm 2 uses a concession strategy with $p_2\ge p_1$, then firm 2 serves a place until it reaches the location ${\hat{y}}_2$ of the marginal consumer who is indifferent towards buying from firm 2 or firm 1. One thus has $p_2 = c + t_1\cdot {\hat{y}}_2 \Rightarrow {\hat{y}}_2 = (p_2-c)/t_1$. Firm 2’s profit under a conceding strategy is then

$$\begin{aligned} \pi _2^- = \int ^1_{{\hat{y}}_2} (p_2 - c -1\cdot (1-x)){\text {d}}x =\frac{(c+t_1-p_2)[(2t_1+1)p_2-t_1-c-2t_1c]}{2t_1^2}. \end{aligned}$$

(15)

Firm 2’s mixed strategy is expressed in terms of the cumulative distribution function $G(p_2)$, which is firm 2’s probability of using a conceding strategy:

$$\begin{aligned} G(p_2) = \int ^{p_2}_0\,{\text {d}}G(p_1) = \text{ Prob } ( 0 < p_1 \le p_2 ). \end{aligned}$$

The expected payoff to firm 2 when it offers price $p_2$ is

$$\begin{aligned} E[\pi _2] = \pi _2^{-} \cdot G(p_2) + \pi _2^{+} \cdot ( 1 - G(p_2)). \end{aligned}$$

Since $G(p_2)$ represents the mixed strategy for firm 2, in equilibrium it must satisfy the probability distribution of $p_2 \in (0, \infty )$ in its support for $E[\pi _2]=V_2$, where $V_2$ is the value of firm 2’s profits. This implies

$$\begin{aligned} V_2=\pi _2^{-} \cdot G(p_2) + \pi _2^{+} \cdot ( 1 - G(p_2)). \end{aligned}$$

(16)

Solving this expression for $G(p_2)$, we obtain

$$\begin{aligned} G(p_2) = \frac{V_2 - \pi _2^{+}}{ \pi _2^{-} - \pi _2^{+}}. \end{aligned}$$

Let $p_a > 0$ be the highest price that firm 2 will ever offer in a mixed-strategy equilibrium. Under this situation, $G(p_a) = 1$, representing that firm 2 absolutely adopts a conceding strategy. Thus, from (16) we obtain

$$\begin{aligned} V_2 = \pi _2^{-}(p_a) = \frac{(c+t_1-p_a)[(2t_1+1)p_a-t_1-c-2t_1c]}{2t_1^2}. \end{aligned}$$

(17)

One can maximize $V_2$ by choosing $p_a$. Solving for the first-order condition $\partial V_2/\partial p_a =0$ and then verifying the second-order condition provide firm 2’s best highest price commanded as below:

$$\begin{aligned} p_a^* =\frac{t_1^2+2t_1c+(t_1+c)}{2t_1+1}. \end{aligned}$$

(18)

Plugging the solution $p_a^*$ into the value of the game in (17) yields:

$$\begin{aligned} V_2^* = \frac{t_1^2}{2(2t_1+1)}. \end{aligned}$$

Let us denote $p_b$ as the lowest price that firm 2 will ever offer. Under this situation, $G(p_b)=0$, and firm 2 absolutely uses an undercutting strategy. If so, then $V_2=\pi _2^{+}= ( p_b - c )^2/2$. In equilibrium, $V_2=V_2^*$. This implies

$$\begin{aligned} \frac{( p_b - c )^2}{2}= \frac{t_1^2}{2(2t_1+1)}. \end{aligned}$$

The lowest optimal price commanded by firm 2 is thus

$$\begin{aligned} p_b^* = c + t_1\sqrt{\frac{1}{2t_1+1}}. \end{aligned}$$

(19)

From (18) and (19), one obtains the range of $p_2$ to sustain a mixed-strategy equilibrium as

$$\begin{aligned} \frac{t_1^2+2t_1c+(t_1+c)}{2t_1+1} \le p_2 \le c + t_1\sqrt{\frac{1}{2t_1+1}}. \end{aligned}$$

(20)

If $t_1=1$, i.e., under the situation of transportation symmetry, then firm 2’s range of $p_2$ in (20) is the same as that of firm 1’s range $p_1$ in (9). Given (20) we obtain firm 2’s mixed strategies in equilibrium as below:

$$\begin{aligned} G(p_2) = \frac{V_2^* - \pi _2^{+}}{ \pi _2^{-} - \pi _2^{+}} = \frac{\frac{t_1^2}{2(2t_1+1)} - \frac{( p_2 - c )^2}{2} }{\frac{(c+t_1-p_2)[(2t_1+1)p_2-t_1-c-2t_1c]}{2t_1^2} - \frac{( p_2 - c )^2}{2}}. \end{aligned}$$

Appendix 2: Both firms’ equilibrium profits under U–M competition

Recall that $t_2=1$. Let firm 1 commit to adopt UDP, while firm 2 uses mill pricing. Under this situation an indifferent consumer is located at location ${\hat{x}}=1+p_2-p_1$. Both firms’ profits are, respectively,

$$\begin{aligned} \pi _1= & {} \int _0^{{\hat{x}}} ( p_1 - c -t_1x )\, {\text {d}}x,\\ \pi _2= & {} ( p_2 - c )(1- {\hat{x}}) = ( p_1 - c ) ( p_1 - p_2). \end{aligned}$$

Both firm’s first-order conditions are, respectively, $\partial \pi _1/\partial p_1 =(1+t_1)p_2-(2+t_1)+t_1+c+1 = 0$ and $\partial \pi _2/\partial p_2 =p_1-2p_2+c = 0$. Simultaneously solving the two equations provides equilibrium prices $p_1^*$ and $p_2^*$:

$$\begin{aligned} p_1^* = \frac{2t_1+2+t_1c+3c}{t_1+3},\; p_2^* = \frac{1+t_1+t_1c+3c}{t_1+3},\; {\hat{x}}^* = \frac{2}{t_1+3}. \end{aligned}$$

The equilibrium profits are thus, respectively,

$$\begin{aligned} \pi _1(U,M)=\frac{2(t_1+2)}{(t_1+3)^2},\;\pi _2(U,M) =\frac{(t_1+1)^2}{(t_1+3)^2}. \end{aligned}$$