Duopoly models with a joint capacity constraint

Abstract

This paper extends two recent models of duopoly with a joint capacity constraint (Nie and Chen in Econ Model 29:1715–1721, 2012; Nie in J Econ 112(3):283–294, 2014) by relaxing their crucial assumption that the game has an equilibrium which is unique and interior. Instead, this paper tackles a more basic issue on when their assumption applies. Our investigation is undertaken by using the best-response curve analysis. Both standard duopoly and mixed duopoly are considered, and both simultaneous moves and sequential moves are investigated. We find that their assumption applies only when the constraint is sufficiently weak, otherwise the equilibrium is either non-unique or corner.

Appendix

Proof of Lemma 1

Without any input constraint, firm i’s best responses in quantity is given by (6). Therefore, part (i) of the lemma holds if the joint capacity constraint is never binding. To see this, we check that $\frac{a-c_i -q_j }{2}+q_j =\frac{a-c_i +q_j }{2}<R$ if $q_j <a-c_i$ and $R\ge a-c_i$.

For part (ii) of the lemma, we note that the first line of (7) is obvious. What remains to be shown is that the joint capacity constraint is binding if $2R-\left( {a-c_i } \right)<q_j <R$. Indeed, $\frac{a-c_i -q_j }{2}+q_j =\frac{a-c_i +q_j }{2}>R$ if $2R-\left( {a-c_i } \right) <q_j$. It follows that firm i’s unconstrained optimal choice of $\frac{a-c_i -q_j }{2}$ is not feasible when $q_j >2R-\left( {a-c_i } \right)$, but is feasible when $q_j \le 2R-\left( {a-c_i } \right)$.

Part (iii) follows easily from part (ii) by noting that when $R\le \left( {a-c_i } \right) /2$, $2R-\left( {a-c_i } \right) \le 0$, implying that the third line of (7) is vacuous. $\square$

Proof of Proposition 1

Results in this proposition are easily obtained by combining the firms’ best-response curves presented in Lemma 1.

If $R\ge a-c_2$ then, by part (i) of Lemma 1, both firms’ best-response curves are the same as in the standard model without an input constraint. Therefore, the equilibrium is the same, which is given by (9). If $\frac{2a-c_1 -c_2 }{3}\le R<a-c_2$, at least firm 2’s best-response curve will be affected by the joint capacity constraint. However, the outputs in (9) are still on both firms’ best response curves and moreover the two curves intersect only at this point. This proves part (i) of the proposition.

If $R<\frac{2a-c_1 -c_2 }{3}$, the two firms’ best-response curves always overlap. From Lemma 1, the part of the line $q_1 +q_2 =R$ that lies in between the standard (unrestricted) best-response curves for the two firms is now always part of both firms’ best-response curves. In fact, the points in this line segment are the only overlap points of the two best-response curves. They represent all of the Cournot–Nash equilibrium points. Depending on the level of R, we have the remaining three parts of the proposition. In part (ii), R is less restrictive so that the above line segment lies entirely inside the first quadrant. The ranges for $q_1$ and $q_2$ in (10) are found by finding the intersection points between the line $q_1 +q_2 =R$ and the two firms’ unrestricted best-response lines. In part (iii), R is more restrictive so that one end of the line segment is on the vertical axis. This case is given by (11). Finally in part (iv), R is most restrictive. In this case, the line segment reaches both axes. This case is given by (12). $\square$

Proof of Proposition 2

We first prove that ${\bar{Q}}$ defined by (14) is greater than the total quantity in the unrestricted Stackelberg equilibrium, i.e., ${\bar{Q}} >q_1^S +q_2^S$, where $q_1^S$ and $q_2^S$ are given in (13).

Substituting (13) into (14), we have

$$\begin{aligned} \left( {a-c_1 -{\bar{Q}} } \right) {\bar{Q}} =\left( {a-c_1 -\frac{3a-2c_1 -c_2 }{4}} \right) \frac{a-2c_1 +c_2 }{2}=\frac{\left( {a-2c_1 +c_2 } \right) ^{2}}{8} \end{aligned}$$

Denote $z=\frac{{\bar{Q}} }{a-c_1 }$ and $\varepsilon =\frac{c_1 -c_2 }{a-c_1 }$. Note that assumptions about the parameters in the main text imply that $0\le \varepsilon <1$. The above equation can be rewritten as

$$\begin{aligned} \left( {1-z} \right) z=\frac{\left( {1-\varepsilon } \right) ^{2}}{8} \end{aligned}$$

Solving this equation for z gives $z=\frac{2+\sqrt{4-2\left( {1-\varepsilon } \right) ^{2}}}{4}$.¹⁵ It follows that

$$\begin{aligned} {\bar{Q}} =\frac{2+\sqrt{4-2\left( {1-\varepsilon } \right) ^{2}}}{4}\left( {a-c_1 } \right) . \end{aligned}$$

Applying this and (13), we have

$$\begin{aligned} {\bar{Q}} -\left( {q_1^S +q_2^S } \right)= & {} \frac{2+\sqrt{4-2\left( {1-\varepsilon } \right) ^{2}}}{4}\left( {a-c_1 } \right) -\frac{3+\varepsilon }{4}\left( {a-c_1 } \right) \\= & {} \frac{\sqrt{4-2\left( {1-\varepsilon } \right) ^{2}}-1-\in }{4}\left( {a-c_1 } \right) . \end{aligned}$$

Since $4-2\left( {1-\varepsilon } \right) ^{2}-\left( {1+\varepsilon } \right) ^{2}=\left( {1+3\varepsilon } \right) \left( {1-\varepsilon } \right) >0$, the last term above is positive, implying that ${\bar{Q}} >q_1^S +q_2^S$.

Now consider part (i) of Proposition 2. If $R\ge {\bar{Q}}$ then the unrestricted Stackelberg equilibrium point $\left( {q_1^S ,q_2^S } \right)$ is below the joint capacity constraint (4) since ${\bar{Q}} >q_1^S +q_2^S$. It follows that the unrestricted Stackelberg equilibrium is still feasible under the joint capacity constraint (4). Moreover, the leader firm 1’s iso-profit curve passing the unrestricted Stackelberg equilibrium point lies entirely below the joint capacity constraint. This is because (a) every point on the decreasing portion of firm 1’s inverted U-shaped iso-profit curve (given by the equation $\left( {a-c_1 -q_1 -q_2 } \right) q_1 =\left( {a-c_1 -{\bar{Q}} } \right) {\bar{Q}} )$ has a slope between 0 and $-1$, implying that firm 1’s iso-profit curve is always flatter than the joint capacity constraint (which has a constant slope of $-1$); and (b) the intersection point of firm 1’s iso-profit curve with the $q_1$-axis $\left( {{\bar{Q}} ,0}\right)$ lies on or below the joint capacity constraint. Hence, in the case of $R\ge {\bar{Q}}$, firm 1 has no incentive to raise its output (to ${\bar{Q}}$) to drive firm 2 out of the market.

The facts established above together imply that in the case of $R\ge {\bar{Q}}$ the Stackelberg equilibrium under the joint capacity constraint (4) is the same as the unrestricted Stackelberg equilibrium.¹⁶

Consider next part (ii) of Proposition 2. If $R<{\bar{Q}}$ then firm 1’s iso-profit curve passing the unrestricted Stackelberg equilibrium point always crosses the joint capacity constraint from below. The same argument about comparing slopes presented above easily establishes that firm 1’s best output choice is equal to R, resulting in an output of zero for the follower firm 2. $\square$

Proof of Proposition 3

If $R>a-c_1$ then firm 1’s best-response curve is the same as in the standard model without a joint constraint, while firm 2’s best-response curve is at most affected on the segment for $q_1 >R$ which does not affect its intersection point with firm 1’s best-response curve. Hence, the equilibrium is the same as in the unrestricted equilibrium, which is given by (18). This proves part (i) of the proposition.

If $\left( {a-c_2 } \right) /2<R\le a-c_1$, the two firms’ best-response curves always overlap. From (17), the while line segment of $q_1 +q_2 =R$ between the two axes is part of firm 1’s best-response curve. The line segment of $q_1 +q_2 =R$ between the $q_1$ axis and the intersection point between $q_1 +q_2 =R$ and firm 2’s unrestricted best-response curve is part of firm 2’s best-response curve. These overlapped points represent all of the Nash equilibrium points, as given by (19). This proves part (ii) of the proposition.

Finally, for $R\le \left( {a-c_2 } \right) /2$, the whole line segment of $q_1 +q_2 =R$ between the two axes represents the intersection points of both firms’ response curves, resulting in multiple Nash equilibria as given by (20). This proves part (iii) of the proposition. $\square$