A general model of price competition with soft capacity constraints

Abstract

We propose a general model of oligopoly with firms relying on a two factor production function. In a first stage, firms choose a certain fixed factor level. In the second stage, firms compete on price and adjust the variable factor to satisfy all the demand. When the factors are substitutable, the capacity constraint is “soft,” implying a convex cost function in the second stage. We show that there exists a continuum of subgame perfect equilibria in pure strategies, whatever the returns to scale. Among them, a payoff dominant one can always be selected. The equilibrium price may increase with the number of firms.

Appendices

Appendix A: Proof of Lemmas 1 to 5

1.1 Proof of Lemma 1

Implicit differentiation of $\hat{v}$ yields Eqs. (2) to (6).

The quasi-concavity of f means that:

$$\begin{aligned} -\hat{v}_{zz}\hat{v}_y^2+2\hat{v}_{zy}\hat{v}_z\hat{v}_y-\hat{v}_{yy}\hat{v}_z^2=f_{zz}f_v<0. \end{aligned}$$

(14)

This proves that v is quasi-convex.

Moreover, it is easy to check that:

$$\begin{aligned} \left| \begin{array}{cc} \hat{v}_{yy} &{}\hat{v}_{yz}\\ \hat{v}_{zy} &{}\hat{v}_{zz} \end{array} \right| =\frac{1}{f_v^4}\left| \begin{array}{cc} f_{zz} &{}f_{zv}\\ f_{vz} &{}f_{vv} \end{array} \right| . \end{aligned}$$

If f is concave, then this determinant is necessarily positive. The second-order pure derivatives of $\hat{v}$ are also positive [cf. (4) and (5)], proving part (2) of Lemma.

1.2 Proof of Lemma 2

The first step is to expand $\Omega$ and $\Omega _p$

$$\begin{aligned} \Omega (p,z,m)=\left( \frac{m-1}{m}\right) D(p)p-w_2\left( \hat{v}(D(p),z)-\hat{v}\left( \frac{D(p)}{m},z\right) \right) . \end{aligned}$$

Differentiating gives:

$$\begin{aligned} \begin{aligned} \Omega _p(p,z,m)&=\left( \frac{m-1}{m}\right) D(p)+D'(p)\left[ p-w_2\hat{v}_y(D(p),z)\right. \\&\quad \left. -\frac{1}{m}\left( p-w_2\hat{v}_y\left( \frac{D(p)}{m},z\right) \right) \right] . \end{aligned} \end{aligned}$$

(15)

We are now going to prove existence.

For a given z and m, $\Omega (0,z,m)=-w_2 \left( \hat{v}(Q_\mathrm{max},z)-\hat{v}\left( \frac{Q_\mathrm{max}}{m},z\right) \right) <0$ (because $\hat{v}_y>0$ and $Q_\mathrm{max}>Q_\mathrm{max}/m$). We also have $\Omega (p_\mathrm{max},z,m)=0$ with $\Omega _{p^-}(p_\mathrm{max},z,m)=D'^{-}(p_\mathrm{max}) \frac{m-1}{m}p_\mathrm{max}<0$ (with $D'^{-}$ being the left derivative of the demand function). $\Omega$ is continuous in p over the interval $[0,p_\mathrm{max}]$, initially negative and finally converging to zero from above. This implies that there is necessarily a $\bar{p}(z,m)\in (0,p_\mathrm{max})$ that solves $\Omega (p,z,m)=0$

We now prove the uniqueness of $\bar{p}(z,m)$ in $(0,p_\mathrm{max})$. Over this interval, we have $D(p)>D(p)/m>0$. Moreover, the strict convexity of $\hat{v}$ implies that:

$$\begin{aligned} \hat{v}_y\left( \frac{D(p)}{m},z\right)<\frac{\hat{v}(D(p),z)-\hat{v}\left( \frac{D(p)}{m},z\right) }{D(p)-\frac{D(p)}{m}} <\hat{v}_y(D(p),z). \end{aligned}$$

From the definition of $\bar{p}$, $\frac{m-1}{m}D(\bar{p})\bar{p}=w_2\left( \hat{v}(D(\bar{p}),z)-\hat{v}\left( \frac{D(\bar{p})}{m},z\right) \right)$ and thus

$$\begin{aligned} w_2\hat{v}_y\left( \frac{D(\bar{p})}{m},z\right)<\bar{p}<w_2 \hat{v}_y(D(\bar{p}),z). \end{aligned}$$

(16)

Finally, considering Eq. (15), it is now obvious that $\Omega _p(\bar{p},z, m)>0$. This means that, in the interval $(0,p_\mathrm{max})$, $\Omega$ can only intercept the x-axis from below. And since $\Omega$ is a continuous functions, this can only happen once.

1.3 Proof of Lemma 3

It is easy to check that ${\hat{\pi }}(0,z,m)<-w_1z$, ${\hat{\pi }}(p_\mathrm{max},z,m)=-w_1z$ and $\hat{\pi }_{p^{-}}(p_\mathrm{max},z,m)<0$. Then, the strict concavity of $\hat{\pi }$ in p implies that $\hat{p}$ exists and is unique. Implicit differentiation of $\hat{\pi }$ for a given z yields

$$\begin{aligned} \left. \frac{\mathrm {d}\hat{p}}{\mathrm {d}m}\right| _{dz=0}=\frac{1}{m} \frac{D(\hat{p})\left( \hat{p}- w_2\hat{v}_y\left( \frac{D(\hat{p})}{m},z\right) \right) }{D(\hat{p})+D'(\hat{p})(\hat{p}- w_2\hat{v}_y(\frac{D(\hat{p})}{m},z))}<0. \end{aligned}$$

(17)

For $p<p_\mathrm{max}$, we have $\frac{D(p)}{m}>D(p_\mathrm{max})=0$. The strict convexity of $\hat{v}$ then implies that $\hat{v}\left( \frac{D(p)}{m},z\right) -0<\left( \frac{D(p)}{m}-0\right) \hat{v}_y\left( \frac{D(p)}{m},z\right)$. By definition, $\hat{p}$ is such that $\hat{p}\frac{D(\hat{p})}{m}=w_2\hat{v}\left( \frac{D(\hat{p})}{m},z\right)$ and then $\hat{p}<w_2\hat{v}_y\left( \frac{D(\hat{p})}{m},z\right)$, which gives the sign of the implicit derivative and proves that $\hat{p}$ decreases with m. Thus, for $m\ge 2$, $\hat{p}(z,m)<\hat{p}(z,1)$ and $\hat{\pi }(\hat{p}(z,m),z,1)<-w_1z$. It follows that for $m \ge 2$, $\Omega (\hat{p}(z,m),z,m)<0$, implying $\hat{p}(z,m) <\bar{p}(z,m)$.

1.4 Proof of Lemma 4

It is easy to verify that ${{\hat{\pi }}}_p(0,z,m)>0$ and ${{\hat{\pi }}}_{p^{-}}(p_\mathrm{max},z,m)<0$. ${{\hat{\pi }}}_p$ is continuous, ensuring that the program has an interior maximum. The strict concavity of ${{\hat{\pi }}}$ with p ensures that the maximum is unique. Because ${{\hat{\pi }}}_p({\hat{p}},z,m)>0$, $p^*(z,m)>{\hat{p}}(z,m)$

1.5 Proof of Lemma 5

At $\hat{p}$, $\frac{D(\hat{p})}{m}\hat{p}-w_2\hat{v} (\frac{D(\hat{p})}{m},z)=0$.

The derivative of the above expression with respect to z is:

$$\begin{aligned} \left. \frac{\mathrm {d}\hat{p}}{\mathrm {d}z}\right| _{dm=0}=\frac{w_2 \hat{v}_z (\frac{D(\hat{p})}{m},z)}{\frac{D'(\hat{p})}{m}\hat{p}+\frac{D(\hat{p})}{m}-w_2\frac{D'(\hat{p})}{m}\hat{v}_y (\frac{D(\hat{p})}{m},z)}<0. \end{aligned}$$

From Eq. (17), we have

$$\begin{aligned} \left. \frac{\mathrm {d}\hat{p}}{\mathrm {d}m}\right| _{dz=0}<0. \end{aligned}$$

At $\bar{p}$, we have $\Omega (\bar{p},z,m)=0$. The derivatives of the above equality with respect to z and m are:

$$\begin{aligned} \left. \frac{\mathrm {d}\bar{p}}{\mathrm {d}z}\right| _{dm=0}=-\frac{\Omega _z(\bar{p},z,m)}{\Omega _p(\bar{p},z,m)}=\frac{w_2\left( \hat{v}_z(D(\bar{p}),z)-\hat{v}_z(\frac{D(\bar{p})}{m},z)\right) }{\Omega _p(\bar{p},z,m)}, \end{aligned}$$

which is $<0$ since $\Omega _p(\bar{p},z,m)>0$ and $\hat{v}_z<0, \hat{v}_{yz}<0$.

$$\begin{aligned} \left. \frac{\mathrm {d}\bar{p}}{\mathrm {d}m}\right| _{dz=0}=-\frac{\Omega _m(\bar{p},z,m)}{\Omega _p(\bar{p},z,m)}=-\frac{\frac{D(\bar{p})}{m^2}(\bar{p}-w_2\hat{v}_y(\frac{D(\bar{p})}{m},z))}{\Omega _p(\bar{p},z,m)}, \end{aligned}$$

which is $<0$ since $\Omega _p(\bar{p},z,m)>0$ and from Eq. (16), $\bar{p}>w_2\hat{v}_y(\frac{D(\bar{p})}{m},z)$.

Finally, we obtain,

$$\begin{aligned} \left. \frac{\mathrm {d}{p^*}}{\mathrm {d}z}\right| _{dm=0}=-\frac{\hat{\pi }_{pz}(p^{*},z,m)}{\hat{\pi }_{pp}(p^{*},z,m)}=w_2\frac{D'(p^*)}{m}\frac{\hat{v}_{yz}(\frac{D(p^*)}{m},z)}{\hat{\pi }_{pp}(p^{*},z,m)}<0 \end{aligned}$$

and

$$\begin{aligned} \left. \frac{\mathrm {d}{p^*}}{\mathrm {d}m}\right| _{dz=0}=-\frac{\hat{\pi }_{pm}(p^{*},z,m)}{\hat{\pi }_{pp}(p^{*},z,m)}=-w_2\frac{D'(p^*)}{m^3}\frac{D(p^*)\hat{v}_{yy}(\frac{D(p^*)}{m},z)}{\hat{\pi }_{pp}(p^{*},z,m)}<0. \end{aligned}$$

Appendix B: Proof of Proposition 1

1.1 Methodological statement

• 1. Price competition is cursed by discontinuity of the payoff function in the vicinity of the equilibrium. In the literature, the use of mixed strategies solves partially the problem. Dasgupta and Maskin (1986) provide some results that, relying on weaker forms of continuity, prove the existence and uniqueness of mixed strategies equilibria for some classes of discontinuous games that can be applied to price competition with capacity constraint. In this paper, we address the problem of the existence of pure strategies equilibria in line with Dastidar (1995). We want to point out the methodological switch he proposes, restarting the reasoning from scratch, relying only on the definition of Nash equilibrium, a definition that requires no assumption of any kind for continuity of the payoff function.⁹ Within this approach, the only requirement is to build a clever partition of the outcome space, that allows to test for a finite number of classes of unilateral profitable deviations from any possible outcome. To solve the equilibrium in the second stage, we adopt exactly the same line of reasoning, and the purpose of Lemmas 2 to 5 is to provide this “clever” partition of the outcomes space.

• 2. However, conversely to Dastidar (1995), as most models of capacity constraint, our model is a sequential two-stage game, with firms playing simultaneously at each stage, i.e., a game of imperfect information. Moreover, conversely to the approach in mixed strategies, we have a multiplicity of the equilibria in the second stage instead of uniqueness, a property that makes impossible to reduce the whole game to a simple strategic one. Our innovation to resolve this problem is thus to extend the Dastidarian methodological move to the case of subgame perfect Nash equilibrium.

• 3. We know that, in general, in a sequential game of complete but imperfect information, when we have multiplicity of equilibria in some subgame, subgame perfection may failed to achieve sensible equilibrium prediction. It is usual in this case to introduce more sophisticated equilibrium concept and refinement relying on some kink of belief or expectation rules, etc. However, in our special setting, despite the multiplicity of equilibria, some subgame perfect equilibria exists and we will base our equilibrium prediction on this underlying concept.

1.2 1. Core of the proof of Proposition 1

An outcome $(\vec {z}^{\,e},\vec {p}^{\,e})$ with $\vec {z}^{\,e}=(z_1^{\,e},\ldots ,z_n^{\,e})$ and $\vec {p}^{\,e}=(p_1^{\,e},\ldots ,p_n^{\,e})$ is a subgame perfect Nash equilibrium if and only if:

1. $(\vec {\mathbf {z}}^{\,e},\vec {p}^{\,e})$ is an Nash equilibrium in price in the second stage (the bold notation indicates that the value is given at the stage of reasoning):

$$\begin{aligned} \forall i \in \{1,\ldots ,n\}, \forall p_i \in [0,p_\mathrm{max}], \pi _i(p^{\,e}_i,\vec {\mathbf {p}}^{\,e}_{-i},\vec {\mathbf {z}}^{\,e})\ge \pi _i(p_i, \vec {\mathbf {p}}^{\,e}_{-i},\vec {\mathbf {z}}^{\,e}). \end{aligned}$$

2. $(\vec {z}^{\,e},\vec {p}^{\,e})$ is an Nash equilibrium of the whole game:

$$\begin{aligned}&\forall i \in \{1,\ldots ,n\}, \forall p_i \in [0,p_\mathrm{max}], \forall z_i \in {{\mathbb {R}}}^{+}, \pi _i((z^{\,e}_i,p^{\,e}_i),\vec {\mathbf {z}} ^{\,e}_{-i},\vec {\mathbf {p}}^{\,e}_{-i})\\&\quad \ge \pi _i((z_i,p_i),\vec {\mathbf {z}} ^{\,e}_{-i},\vec {\mathbf {p}}^{\,e}_{-i}). \end{aligned}$$

We are interested in outcomes in which all n firms operate in the market, meaning that we consider the possibility for an outcome in which all firms are setting the same price $p^N$ to be SPNE.

Step 1: $p^N$ has to be a Nash equilibrium in the second stage. The firms have no incentive to unilaterally deviate in the second stage as long as $p^N \in {\bigcap }_{i} \, [\hat{p}(z_i,n), \bar{p}(z_i,n)] \ne \emptyset$. In other words, the necessity for the outcome to be a Nash equilibrium in price in the second stage prevents the possibility for the firms to play too different levels of fixed factor in the first stage. More precisely, for all i, $z_i$ has to belong in the interval $[{\bar{z}} (p^N,n),{\hat{z}} (p^N,n)]$ with ${\bar{z}}(p^N,n)$ such that ${\bar{p}}(z,n)=p^N$ and ${\hat{z}}(p^N,n)$ such that ${\hat{p}}(z,n)=p^N$ (that are both unique, thanks to Lemma 5).

Among all the outcomes that are Nash equilibria in the second stage (Step 1), with all firms operating the market, we are now going to eliminate those who are subject to profitable unilateral deviation in stage 1. Because the strategy of player i in stage 1 is of dimension 2, firms can deviate in either dimension or both and we will have to consider all the possible deviations.

Step 2: By definition of program ${{\mathcal {P}}}_1\left( p,n\right)$:

$$\begin{aligned}&\forall i \in \{1,\ldots ,n\}, \forall z_i,z_{-i} \in [{\bar{z}} (p^N,n),{\hat{z}} (p^N,n)],\\&\quad \pi _i((z^{*}(p^N,n),p^{N}),\vec {\mathbf {z}} _{-i},\vec {\mathbf {p}}^{N}_{-i})\ge \pi _i((z_i,p^N),\vec {\mathbf {z}} _{-i},\vec {\mathbf {p}}^{N}_{-i}). \end{aligned}$$

It means that for firm i, the strategy $(z^N,p^{N})$ (with $z^N=z^{*}(p^N,n)$) played in the first stage dominates all other strategy that lead to $p^N$ being a Nash equilibria in the second stage with all firm operating the market. All the firms having the same technology, we conclude that a necessary condition for an outcome of the game with all n firms quoting $p^N$ to be a SPNE is that all firms choose the same level of fixed factor $(z^{*}(p^N,n)$ solution of ${{\mathcal {P}}}_1\left( p,n\right)$. That corresponds to the first condition (i.e., optimality condition) of Proposition 1.

Step 3: Now we are going to test whether an outcome in which all firms play $(z^{*}(p^N,n),p^{N})$ is subject to profitable unilateral deviation $(z^d,p^d)$ by firm i, with $p^d \ne p^N$, under condition that $p^d$ can be sustained as a Nash equilibrium in the second stage (not necessarily with all firms operating the market).

Step 3.1 Let us start by all the unilateral deviation such that $z^d \in {{\mathbb {R}}}^{+}$ and $p^d\in (p^N,p_\mathrm{max}]$. In this case, the firm sells nothing and the profit is $-w_1 z^d$. The “best” of these unilateral deviation is $z^d=0$ that earns zero profit. This deviation is not profitable as long as ${{\hat{\pi }}}(p^N,z^N,n)\ge 0$. That imposes the second condition (i.e., profitability) of Proposition 1

Step 3.2 Let us start by all the unilateral deviation such that $z^d \in {{\mathbb {R}}}^{+}$ and $p^d\in [0,{\hat{p}} (z^N,n))$. In this case, the firm will operate the market alone in the second stage. The profit corresponding to these unilaterally deviating strategies is upper bounded by profit corresponding to $(\underset{z}{{\text {argmax}}},\quad {{\hat{\pi }}}({\hat{p}}(z^N,n),z,1),1),{\hat{p}}(z^N,n))$. That corresponds to the third condition (i.e., nonexistence of limit pricing strategy) of Proposition 1

All those conditions are necessary for having a SNPE and must be satisfied simultaneously.

1.3 2. Core of the proof of Proposition 2

We start with an outcome that verifies all the three conditions of Proposition 1 and we consider remaining possible deviations.

Step 3.3 Finally, let us consider the case with $p^d\in [{\hat{p}} (z^N,n),p^N)$ and $z^d \in {{\mathbb {R}}}+$. The case $z^d<{\bar{z}}(p^d,n)$ or $z^d>{\hat{z}}(p^N,n)$ can be excluded because they are incompatible with the possibility for $p^d$ to be sustain as a Nash equilibrium in the second stage. For a given $p^d$, all the remaining unilateral deviations of that kind are dominated by $(z^{*}(p^d,n),p^d)$. Thus, as long as $\forall p \in [{\hat{p}} (z^N,n),p^N), \quad \pi (p^N,z^{*}(p^N,n),n)\overset{\text {def}}{=}\Pi (p^N,n)\ge \pi (p,z^{*}(p,n),n)\overset{\text {def}}{=}\Pi (p,n)$ these unilateral deviations are unprofitable.

Step 4: When the supplementary condition of Proposition 2 is satisfied, Steps 2, 3.1, 3.2 and 3.3 define necessary conditions for $(z^N,p^N)$ to be a SPNE with all firms operating the market. But taken together, they exhausts all the possible case implying that they are also sufficient.

Remark

The condition described in case 3.3 is operative only for outcomes associated with the non-decreasing part (in p) of the function $\Pi$. For outcomes associated with the decreasing part (in p) of $\Pi$, subgame perfection is not powerful enough to obtain an equilibrium, even if those outcomes can be appropriately ranked relying on function $\Pi$. And of course, they will be dominated by the outcome that is associated with the price p that maximizes $\Pi$ for a given n, which is a SPNE. That will be the core argument of Proposition 3.