<p>Note: The story behind the problem is still confusing, but I will merely focus on your utility functions.</p>

<blockquote>
  <p><strong>Edit</strong></p>
</blockquote>

<p>In the problem, it seems that both are minimizing costs: $A$ is doing so via $c$, and $B$ via $\alpha$.  I assume you want a functional form that yields a closed-form solution (and not corner solutions).  In that case, you may want $U$'s to be quasi-concave (i.e. to have a unique maximum).  Namely, you want
\begin{eqnarray*}
\frac{\partial U(A)}{\partial c} &gt;0,&amp;&amp;  \frac{\partial^2 U(A)}{\partial c^2} &lt;0, 
\\
\frac{\partial U(B)}{\partial \alpha} &gt;0,&amp;&amp;  \frac{\partial^2 U(B)}{\partial \alpha^2} &lt;0. 
\end{eqnarray*}</p>

<p>It seems that $f(1-\alpha) = - \ln(1-\alpha)$ satisfies those inequalities.</p>

CC BY-SA 4.0
2018-06-25T13:18:05.643Z
Exactly You are right until "the problem boils down to finding response functions of each player." and I missed a point here that instead of $log(1+P)$ it is $log(1+\alpha P)$,because it is using only an $\alpha$ share resource of P. Now the best response is not independent and similarly the best response of company A is also not independent of $\alpha$.
user7341333
CC BY-SA 4.0
2018-06-25T15:32:08.863Z
@user7341333  Please, make corresponding changes in your OP
Green.H
CC BY-SA 4.0
2018-06-27T05:39:29.687Z
using $-ln(1-\alpha)$ gives me the following equation :                                                            $\frac{p}{ln(2)(1+\alpha p)}+p*c*ln(1-\alpha)-p*c\frac{\alpha}{1-\alpha}=0  $                              How can I solve it for $\alpha$, even if I am able to solve it, for the value of $c$using backward induction  in utility of Company B, if I replace the value of $\alpha$ it will become too complex to solve.
user7341333
CC BY-SA 4.0
2018-06-27T09:06:23.837Z
@user7341333 You cant get an explicit form solution, but you can show that a solution exists
Green.H
CC BY-SA 4.0
2018-06-27T09:33:43.833Z
also, your derivative is wrong: the first term should be $\frac{p}{1+\alpha p}$
Green.H
CC BY-SA 4.0
2018-06-24T10:10:55.983Z
2018-06-25T15:40:36.623Z
2018-06-25T15:40:36.623Z
Green.H
<p>Company A needs to provide service $p$ to company B. However if the company B will reserve the $(1-\alpha)$ fraction of resource for the serving customers of company A, company A will provide a discount proportional to ($1-\alpha$) to company B. </p>

<p>The utility function of company A is composed of benefit obtained from company B given as </p>

<p>$$U(A)=\alpha*p*c*f(1-\alpha)$$</p>

<p>c here is the unit cost of resource $p$ and this function is maximized in c. </p>

<p>The utility of company B is based on benefit-cost function and given as : </p>

<p>$$U(B)=log(1+\alpha p)-\alpha *p * c * f(1-\alpha)$$</p>

<p>Which is maximized in $\alpha$</p>

<p>I need a function$f(1-\alpha)$ which reduces the price $p*c$ .</p>

<p>I have considered the following functions </p>

<p>$$exp(-(1-\alpha))$$
 But I could not find the value of $\alpha$ that would maximize the utility of company B</p>

<p>$$\frac{\alpha p}{(1+(1-\alpha)^2)}$$
But this function also does not works.</p>

<p>Any ideas for function $f(1-\alpha)$</p>

CC BY-SA 4.0
2018-06-21T12:53:27.423Z
Have you tried $f(1-\alpha)=-(1-\alpha)^2$?
Weierstraß Ramirez
CC BY-SA 4.0
2018-06-21T14:32:35.673Z
yes this gives a negative value of $p$
user7341333
CC BY-SA 4.0
2018-06-23T17:23:39.423Z
I don't understand the second sentence
ahorn
<game-theory><discount>
Discount function
user7341333
# Discount function

(Asked by: username_0 on 2018-06-21)

Company A needs to provide service $p$ to company B. However if the company B will reserve the $(1-\alpha)$ fraction of resource for the serving customers of company A, company A will provide a discount proportional to ($1-\alpha$) to company B.

The utility function of company A is composed of benefit obtained from company B given as

$$U(A)=\alpha\*p\*c\*f(1-\alpha)$$

c here is the unit cost of resource $p$ and this function is maximized in c.

The utility of company B is based on benefit-cost function and given as :

$$U(B)=log(1+\alpha p)-\alpha \*p \* c \* f(1-\alpha)$$

Which is maximized in $\alpha$

I need a function$f(1-\alpha)$ which reduces the price $p\*c$ .

I have considered the following functions

$$exp(-(1-\alpha))$$
But I could not find the value of $\alpha$ that would maximize the utility of company B

$$\frac{\alpha p}{(1+(1-\alpha)^2)}$$
But this function also does not works.

Any ideas for function $f(1-\alpha)$

**Tags:** game-theory, discount
> **Comments:**
> **username_1**: Have you tried $f(1-\alpha)=-(1-\alpha)^2$?  
> **username_0**: yes this gives a negative value of $p$  
> **username_2**: I don't understand the second sentence  

---

## 1 answers

### Answer by username_3 on 2018-06-24. Score: 1

Note: The story behind the problem is still confusing, but I will merely focus on your utility functions.

> **Edit**

In the problem, it seems that both are minimizing costs: $A$ is doing so via $c$, and $B$ via $\alpha$. I assume you want a functional form that yields a closed-form solution (and not corner solutions). In that case, you may want $U$'s to be quasi-concave (i.e. to have a unique maximum). Namely, you want
\begin{eqnarray\*}
\frac{\partial U(A)}{\partial c} >0,&& \frac{\partial^2 U(A)}{\partial c^2} <0,
\\
\frac{\partial U(B)}{\partial \alpha} >0,&& \frac{\partial^2 U(B)}{\partial \alpha^2} <0.
\end{eqnarray\*}

It seems that $f(1-\alpha) = - \ln(1-\alpha)$ satisfies those inequalities.

> **Comments:**
> **username_0**: Exactly You are right until "the problem boils down to finding response functions of each player." and I missed a point here that instead of $log(1+P)$ it is $log(1+\alpha P)$,because it is using only an $\alpha$ share resource of P. Now the best response is not independent and similarly the best response of company A is also not independent of $\alpha$.  
> **username_3**: @user7341333  Please, make corresponding changes in your OP  
> **username_0**: using $-ln(1-\alpha)$ gives me the following equation :                                                            $\frac{p}{ln(2)(1+\alpha p)}+p*c*ln(1-\alpha)-p*c\frac{\alpha}{1-\alpha}=0  $                              How can I solve it for $\alpha$, even if I am able to solve it, for the value of $c$using backward induction  in utility of Company B, if I replace the value of $\alpha$ it will become too complex to solve.  
> **username_3**: @user7341333 You cant get an explicit form solution, but you can show that a solution exists  
> **username_3**: also, your derivative is wrong: the first term should be $\frac{p}{1+\alpha p}$  

---

