<p>Here is the maximization problem solved by the company :</p>

<p>\begin{eqnarray*} \max_{x_h, p_h, x_l, p_l} &amp; p_h + p_l - x_h - x_l \\ \text{s.t} &amp;  \theta_hv(x_h) - p_h \geq 0 \tag{IR$_h$} \\ &amp;  \theta_lv(x_l) - p_l \geq 0 \tag{IR$_l$} \\ &amp;  \theta_hv(x_h) - p_h \geq \theta_hv(x_l) - p_l \tag{IC$_h$} \\ &amp;  \theta_lv(x_l) - p_l \geq \theta_lv(x_h) - p_h \tag{IC$_l$}\end{eqnarray*}</p>

<p>Now we solve the above maximization problem for $v(x) = \ln x$. Given that $\theta_l &lt; \theta_h$, it is easy to see that only IC$_h$ and IR$_l$ will be binding in optimum, and we can rewrite the optimization problem as :</p>

<p>\begin{eqnarray*} \max_{x_h, p_h, x_l, p_l} &amp; p_h + p_l - x_h - x_l \\ \text{s.t} &amp;   \theta_l\ln x_l - p_l = 0 \tag{IR$_l$} \\ &amp;  \theta_h\ln x_h - p_h = \theta_h\ln x_l - p_l \tag{IC$_h$} \end{eqnarray*}</p>

<p>Substituting the constraints in the objective, we can rewrite the objective as follows :</p>

<p>\begin{eqnarray*} \max_{x_h, x_l} &amp; \ \ \theta_h\ln x_h + (2\theta_l-\theta_h)\ln x_l - x_h - x_l  \end{eqnarray*}</p>

<p>Further assuming $\theta_h + 1 &lt; 2\theta_l$, and solving the problem yields the following:</p>

<p>\begin{eqnarray*} x_h^* &amp; = &amp; \theta_h \\  x_l^* &amp; = &amp; 2\theta_l-\theta_h \end{eqnarray*}
Consequently,
\begin{eqnarray*} p_l^* &amp; = &amp; \theta_l \ln(2\theta_l-\theta_h) \\  p_h^* &amp; = &amp; \theta_h\ln \theta_h + (\theta_l-\theta_h)\ln (2\theta_l-\theta_h) \end{eqnarray*}</p>

CC BY-SA 4.0
2018-05-15T03:38:25.863Z
2018-05-15T03:38:25.863Z
Amit
<p>I have a problem with finding a way how to solve this example: We have the company, which has two types of customers indexed as $H$ and $L$. We assume: $θ_H &gt; θ_L$. Quantity$(Q)$ of $H$ indexed customers is in population. Payout of type of the customer$(i)$, which buys products($x$) for a price($p$) is $U_i=θ_iv(x)−p$. Company determines quantity and price for that quantity for every type of a customer ($x_i,p_i$). Costs of the company are $C(x_h,x_l)=x_h+x_l$. Company does not know type of the customer and maximizes its profit.</p>

<blockquote>
  <p>Find maximization problem which company solves (with restriction). Solve this example with $v(x)=\ln x$.</p>
</blockquote>

CC BY-SA 4.0
2018-05-14T18:53:58.187Z
The function $v$ does not occur in your description of the model.
Michael Greinecker
CC BY-SA 4.0
2018-05-14T21:24:42.890Z
The function $v$ is in $U_i$ payout. I  editted it.
Dominik Chovanec
CC BY-SA 4.0
2018-05-14T21:30:47.727Z
You could start writing the incentive compatibility constraints that guarantee that a type $i$ customer chooses the contract $(x_i,p_i)$.There should also be some individual rationality constraints. Is the outside option $0$?
Michael Greinecker
CC BY-SA 4.0
2018-05-14T21:37:09.220Z
I think that outside option should be $0$. But I don't know the meaning of the θ, is it some willingness to pay?
Dominik Chovanec
CC BY-SA 4.0
2018-05-14T21:39:50.550Z
Pretty much. The $\theta_i$ guarantee that for every $x$, the marginal utility when measured in terms of the numeraire good (m) differs by the same proportion.
Michael Greinecker
CC BY-SA 4.0
2018-05-14T21:47:16.877Z
And what about information, that H indexed customers are in quantity($Q$) in population, could it be some signal for the company?
Dominik Chovanec
<game-theory><extensive-games>
Game theory - Write a maximization problem (with restriction)
Dominik Chovanec
# Game theory - Write a maximization problem (with restriction)

(Asked by: username_0 on 2018-05-14)

I have a problem with finding a way how to solve this example: We have the company, which has two types of customers indexed as $H$ and $L$. We assume: $θ\_H > θ\_L$. Quantity$(Q)$ of $H$ indexed customers is in population. Payout of type of the customer$(i)$, which buys products($x$) for a price($p$) is $U\_i=θ\_iv(x)−p$. Company determines quantity and price for that quantity for every type of a customer ($x\_i,p\_i$). Costs of the company are $C(x\_h,x\_l)=x\_h+x\_l$. Company does not know type of the customer and maximizes its profit.

> Find maximization problem which company solves (with restriction). Solve this example with $v(x)=\ln x$.

**Tags:** game-theory, extensive-games

> **Comments:**
> **username_1**: The function $v$ does not occur in your description of the model.  
> **username_0**: The function $v$ is in $U_i$ payout. I  editted it.  
> **username_1**: You could start writing the incentive compatibility constraints that guarantee that a type $i$ customer chooses the contract $(x_i,p_i)$.There should also be some individual rationality constraints. Is the outside option $0$?  
> **username_0**: I think that outside option should be $0$. But I don't know the meaning of the θ, is it some willingness to pay?  
> **username_1**: Pretty much. The $\theta_i$ guarantee that for every $x$, the marginal utility when measured in terms of the numeraire good (m) differs by the same proportion.  
> **username_0**: And what about information, that H indexed customers are in quantity($Q$) in population, could it be some signal for the company?  

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## 1 answers

### Answer by username_2 on 2018-05-15. Score: 4 | Accepted answer

Here is the maximization problem solved by the company :

\begin{eqnarray\*} \max\_{x\_h, p\_h, x\_l, p\_l} & p\_h + p\_l - x\_h - x\_l \\ \text{s.t} & \theta\_hv(x\_h) - p\_h \geq 0 \tag{IR$\_h$} \\ & \theta\_lv(x\_l) - p\_l \geq 0 \tag{IR$\_l$} \\ & \theta\_hv(x\_h) - p\_h \geq \theta\_hv(x\_l) - p\_l \tag{IC$\_h$} \\ & \theta\_lv(x\_l) - p\_l \geq \theta\_lv(x\_h) - p\_h \tag{IC$\_l$}\end{eqnarray\*}

Now we solve the above maximization problem for $v(x) = \ln x$. Given that $\theta\_l < \theta\_h$, it is easy to see that only IC$\_h$ and IR$\_l$ will be binding in optimum, and we can rewrite the optimization problem as :

\begin{eqnarray\*} \max\_{x\_h, p\_h, x\_l, p\_l} & p\_h + p\_l - x\_h - x\_l \\ \text{s.t} & \theta\_l\ln x\_l - p\_l = 0 \tag{IR$\_l$} \\ & \theta\_h\ln x\_h - p\_h = \theta\_h\ln x\_l - p\_l \tag{IC$\_h$} \end{eqnarray\*}

Substituting the constraints in the objective, we can rewrite the objective as follows :

\begin{eqnarray\*} \max\_{x\_h, x\_l} & \ \ \theta\_h\ln x\_h + (2\theta\_l-\theta\_h)\ln x\_l - x\_h - x\_l \end{eqnarray\*}

Further assuming $\theta\_h + 1 < 2\theta\_l$, and solving the problem yields the following:

\begin{eqnarray\*} x\_h^\* & = & \theta\_h \\ x\_l^\* & = & 2\theta\_l-\theta\_h \end{eqnarray\*}
Consequently,
\begin{eqnarray\*} p\_l^\* & = & \theta\_l \ln(2\theta\_l-\theta\_h) \\ p\_h^\* & = & \theta\_h\ln \theta\_h + (\theta\_l-\theta\_h)\ln (2\theta\_l-\theta\_h) \end{eqnarray\*}


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