<p>For notational convenience, I'll write <span class="math-container">$D$</span> instead of <span class="math-container">$D(p)$</span>.</p>
<p>We have the problem:
<span class="math-container">$$
\max_{s_2, p} (p - c_1)(1 - s_2) D \text{ s.t. } (p - c_2) s_2 D \ge \overline{\Pi}_2.
$$</span>
The first order conditions give (where <span class="math-container">$\lambda$</span> is the Lagrange multiplier):
<span class="math-container">$$
\begin{align*}
&amp;(1 - s_2) D + (p - c_1)(1 - s_2)D' - \lambda s_2 D - \lambda (p - c_2)s_2 D' = 0,\\
&amp;-(p - c_1) D - \lambda (p - c_2) D = 0,\\
&amp;(p - c_2) s_2 D = \overline{\Pi}_2.
\end{align*}
$$</span>
The second condition tells us that <span class="math-container">$\lambda = - \dfrac{p - c_1}{p - c_2}$</span>. Substitute into the first condition.</p>
<p><span class="math-container">$$
(1 - s_2)D + (p - c_1)(1 - s_2) D' + \frac{(p - c_1)}{(p - c_2)} s_2 D + (p - c_1)s_2 D' = 0.
$$</span>
Gathering terms:
<span class="math-container">$$
\frac{D}{p - c_2}\left((1 - s_2)(p - c_2) + s_2(p - c_1)\right) + D'(p - c_1) = 0.
$$</span></p>
<p>The third first order condition gives:
<span class="math-container">$$
s_2 = \frac{\overline{\Pi}_2}{(p - c_2)D},
$$</span>
So:
<span class="math-container">$$
(1 - s_2)(p - c_2) + s_2(p - c_1) = (p - c_2) - \frac{\overline{\Pi}_2}{D} + \frac{\overline{\Pi}_2 (p - c_1)}{D(p - c_2)}
$$</span>
Using this, we get the condition:
<span class="math-container">$$
\begin{align*}
0 &amp;= 
\frac{D}{p - c_2}\left((p - c_2) - \frac{\overline{\Pi}_2}{D} + \frac{\overline{\Pi}_2 (p - c_1)}{D(p - c_2)}\right) + D'(p - c_1),\\
&amp;= D - \frac{\overline{\Pi}_2}{p - c_2} + \frac{\overline{\Pi}_2(p - c_1)}{(p - c_2)^2} + D'(p - c_1),\\
&amp;= D + D'(p - c_1) + \overline{\Pi}_2 \frac{c_2 - c_1}{(p - c_2)^2}.
\end{align*}
$$</span></p>

CC BY-SA 4.0
2024-01-22T12:35:59.370Z
Thank you! I did not see that there was a simplification by factoring. Nice job!
Matteo Bulgarelli
CC BY-SA 4.0
2024-01-22T09:32:46.647Z
2024-01-22T09:32:46.647Z
tdm
<p>I am trying to solve exercise 6.1 of the famous book from Jean Tirole, The Theory of Industrial of Organization. I am stuck on a part he deems &quot;straightforward&quot;:</p>
<p>&quot;Suppose there are two firms, with unit costs <span class="math-container">$c_1 &lt; c_2$</span>. Let <span class="math-container">$p^m(c)$</span> denote the monopoly price for unit cost <span class="math-container">$c$</span>; it maximizes <span class="math-container">$(p-c)D(p)$</span>. If the two firms could get together and sign a contract, they would let firm 1 produce everything and charge <span class="math-container">$p^m(c_1)$</span>. This would maximize industry profits. The &quot;pie&quot; could then be divided between the two firms through an arbitrary lump-sum transfer from firm 1 to firm 2. But suppose that it is illegal for firms to overtly agree and use side payments. We can determine the set of industry allocations that are efficient for the two firms, <em>constrained</em> by the fact that side payments are prohibited [...]. To this purpose, we fix a profit target <span class="math-container">$\bar\Pi_2$</span> for firm 2 in the interval <span class="math-container">$[0, \Pi^m(c_2)]$</span> and look for profit-sharing agreements (without transfers) in which both firms charge the same price <span class="math-container">$p$</span>. They choose market shares <span class="math-container">$s_1$</span> and <span class="math-container">$s_2$</span> such that <span class="math-container">$s_1 + s_2 = 1$</span>. The interpretation of these market shares is that firm <span class="math-container">$i$</span> produces exactly <span class="math-container">$q_i = s_iD(p)$</span>; if <span class="math-container">$s_i&lt;1/2$</span>, the consumers who go to firm <span class="math-container">$i$</span> and are rationed buy from firm <span class="math-container">$j$</span>. Given the profit target <span class="math-container">$\bar\Pi_2$</span> for firm 2, the efficient allocation is a choice of price <span class="math-container">$p$</span> and market shares <span class="math-container">$s_1$</span> and <span class="math-container">$s_2$</span> so as to solve:</p>
<p><span class="math-container">\begin{align}
\max_{p, s_1, s_2} &amp; (p-c_1)s_1 D(p) \\
s.t.               &amp; (p-c_2)s_2 D(p) \geq \bar\Pi_2 \\
                   &amp; s_1 + s_2 = 1
\end{align}</span></p>
<p>Suppose that the profit function (p-c)D(p) is concave for all <span class="math-container">$p$</span> and all <span class="math-container">$c$</span>.</p>
<p>(i) After substituting <span class="math-container">$s_1$</span>, obtain the first-order conditions Show that</p>
<p><span class="math-container">$p^m(c_1) \leq p \leq p^m(c_2)$</span></p>
<p>Show that the objective function is concave (I have done this, idea is to realize from the FOCs that <span class="math-container">$D(p)+(p-c_1)D(p) \leq 0 $</span> and that <span class="math-container">$D(p)+ (p-c_2)D(p) \geq 0$</span>, and then you prove the result by contradiction by showing that it can't neither be that <span class="math-container">$p^* &lt; p^m(c_1)$</span> nor that <span class="math-container">$p &gt; p^m(c_2)$</span>, then for the second point, it's just tedious algebra).</p>
<p>(ii) Show that</p>
<p><span class="math-container">$$(p-c_1)D'(p) + D(p) + \frac{(c_2 - c_1)\bar\Pi_2}{(p-c_2)^2} = 0 $$</span></p>
<p>This is the one I can't solve... Any help?</p>
<p>I reached this point:</p>
<p><span class="math-container">$$(p-c_1)D'(p) + D(p) + \frac{(p-c_1)\bar\Pi_2}{(p-c_2)^2}\frac{(p-c_2)D'(p) + D(p)}{D(p)-\bar\Pi_2/((p-c_2)D(p))} = 0 $$</span></p>

<industrial-organisation>
Exercise 6.1 Theory of IO (Jean Tirole's book)
Matteo Bulgarelli
# Exercise 6.1 Theory of IO (Jean Tirole's book)

(Asked by: username_0 on 2024-01-11)

I am trying to solve exercise 6.1 of the famous book from Jean Tirole, The Theory of Industrial of Organization. I am stuck on a part he deems "straightforward":

"Suppose there are two firms, with unit costs $c\_1 < c\_2$. Let $p^m(c)$ denote the monopoly price for unit cost $c$; it maximizes $(p-c)D(p)$. If the two firms could get together and sign a contract, they would let firm 1 produce everything and charge $p^m(c\_1)$. This would maximize industry profits. The "pie" could then be divided between the two firms through an arbitrary lump-sum transfer from firm 1 to firm 2. But suppose that it is illegal for firms to overtly agree and use side payments. We can determine the set of industry allocations that are efficient for the two firms, *constrained* by the fact that side payments are prohibited [...]. To this purpose, we fix a profit target $\bar\Pi\_2$ for firm 2 in the interval $[0, \Pi^m(c\_2)]$ and look for profit-sharing agreements (without transfers) in which both firms charge the same price $p$. They choose market shares $s\_1$ and $s\_2$ such that $s\_1 + s\_2 = 1$. The interpretation of these market shares is that firm $i$ produces exactly $q\_i = s\_iD(p)$; if $s\_i<1/2$, the consumers who go to firm $i$ and are rationed buy from firm $j$. Given the profit target $\bar\Pi\_2$ for firm 2, the efficient allocation is a choice of price $p$ and market shares $s\_1$ and $s\_2$ so as to solve:

\begin{align}
\max\_{p, s\_1, s\_2} & (p-c\_1)s\_1 D(p) \\
s.t. & (p-c\_2)s\_2 D(p) \geq \bar\Pi\_2 \\
& s\_1 + s\_2 = 1
\end{align}

Suppose that the profit function (p-c)D(p) is concave for all $p$ and all $c$.

(i) After substituting $s\_1$, obtain the first-order conditions Show that

$p^m(c\_1) \leq p \leq p^m(c\_2)$

Show that the objective function is concave (I have done this, idea is to realize from the FOCs that $D(p)+(p-c\_1)D(p) \leq 0 $ and that $D(p)+ (p-c\_2)D(p) \geq 0$, and then you prove the result by contradiction by showing that it can't neither be that $p^\* < p^m(c\_1)$ nor that $p > p^m(c\_2)$, then for the second point, it's just tedious algebra).

(ii) Show that

$$(p-c\_1)D'(p) + D(p) + \frac{(c\_2 - c\_1)\bar\Pi\_2}{(p-c\_2)^2} = 0 $$

This is the one I can't solve... Any help?

I reached this point:

$$(p-c\_1)D'(p) + D(p) + \frac{(p-c\_1)\bar\Pi\_2}{(p-c\_2)^2}\frac{(p-c\_2)D'(p) + D(p)}{D(p)-\bar\Pi\_2/((p-c\_2)D(p))} = 0 $$

**Tags:** industrial-organisation


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## 1 answers

### Answer by username_1 on 2024-01-22. Score: 2 | Accepted answer

For notational convenience, I'll write $D$ instead of $D(p)$.

We have the problem:
$$
\max\_{s\_2, p} (p - c\_1)(1 - s\_2) D \text{ s.t. } (p - c\_2) s\_2 D \ge \overline{\Pi}\_2.
$$
The first order conditions give (where $\lambda$ is the Lagrange multiplier):
$$
\begin{align\*}
&(1 - s\_2) D + (p - c\_1)(1 - s\_2)D' - \lambda s\_2 D - \lambda (p - c\_2)s\_2 D' = 0,\\
&-(p - c\_1) D - \lambda (p - c\_2) D = 0,\\
&(p - c\_2) s\_2 D = \overline{\Pi}\_2.
\end{align\*}
$$
The second condition tells us that $\lambda = - \dfrac{p - c\_1}{p - c\_2}$. Substitute into the first condition.

$$
(1 - s\_2)D + (p - c\_1)(1 - s\_2) D' + \frac{(p - c\_1)}{(p - c\_2)} s\_2 D + (p - c\_1)s\_2 D' = 0.
$$
Gathering terms:
$$
\frac{D}{p - c\_2}\left((1 - s\_2)(p - c\_2) + s\_2(p - c\_1)\right) + D'(p - c\_1) = 0.
$$

The third first order condition gives:
$$
s\_2 = \frac{\overline{\Pi}\_2}{(p - c\_2)D},
$$
So:
$$
(1 - s\_2)(p - c\_2) + s\_2(p - c\_1) = (p - c\_2) - \frac{\overline{\Pi}\_2}{D} + \frac{\overline{\Pi}\_2 (p - c\_1)}{D(p - c\_2)}
$$
Using this, we get the condition:
$$
\begin{align\*}
0 &=
\frac{D}{p - c\_2}\left((p - c\_2) - \frac{\overline{\Pi}\_2}{D} + \frac{\overline{\Pi}\_2 (p - c\_1)}{D(p - c\_2)}\right) + D'(p - c\_1),\\
&= D - \frac{\overline{\Pi}\_2}{p - c\_2} + \frac{\overline{\Pi}\_2(p - c\_1)}{(p - c\_2)^2} + D'(p - c\_1),\\
&= D + D'(p - c\_1) + \overline{\Pi}\_2 \frac{c\_2 - c\_1}{(p - c\_2)^2}.
\end{align\*}
$$

> **Comments:**
> **username_0**: Thank you! I did not see that there was a simplification by factoring. Nice job!  

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