数理解析研究所講究録
第 1604 巻 2008 年 42-49

42

Bargaining Model on the Plane
Vladimir V. Mazalov
Institute of Applied Mathematical Research
Karelian Research Center of Russian Academy of Sciences
Pushkinskaya str. 11, Petrozavodsk 185910, Russia
e-mail: vmazalov@krc.karelia.ru
Julia S. Tokareva
Zabaikalsky State Humanitarian Pedagogical University
named after N.Tchernishevsky
Babushkin str. 129, Chita 672007, Russia
e-mail: jtokareva2@mail.ru
Abstact

Non-coalition non-zero-sum game related with arbitration procedure is considered. Twoplayers(firlns) propose
solutions on the plane. The arbtrator has his own solution which is inodelling by random variables in the circle on the plane. As a solution of theconflict we use here thefinal offer arbitration procedure (FOA). The objective is the construction ofNash equilibrium. For
two cases of thc random solution of the arbitrator we found the equilibriumin pure stratcgics.
$\backslash SOI\iota\iota e$

Key words: final offer arbitration, equilibrium, optimal strategies.

AMS Subject Classification: $91A05,91A80$

1Introduction

Imagine the market where two different firms (players I and II) propose
their expected
costs for the building construction. The administration have choose one proposal.
The
first firm can build the building with costfor the timc
and second firm-withcostand time .
the solution of the arbitrator is random variable distributed in theunit circle
with probablility density in polar coordinates. The objective of the players isto receive the order for the building.
$x_{1}$$x_{2}$$y_{1}$$y_{2}$$Let_{l}$$z$$f(r, \theta)$43
2

Solution of the gameLet the pointon the plane corresponds to the offer of the player I andcorresponds to the offer II and is the solution of the arbitrator. The arbitrator prefers
a firm which is closer to the point .
Let us divide the circle for two parts and . The first part consists of the pointsof the circle the distance from which to the pointis smaler than to the point. The second part is opposite case.
$(x_{1\backslash }y_{1})$$(x_{2}, y_{2})$

$A$

$A$

$s_{1}$$s_{2}$$(x_{1}, y_{1})$$(x_{2_{j}}y_{2})$Fig. 1Then the payoffs of the firms I and II are$H_{1}(r_{1}, \theta_{1}, r_{2}\eta\theta_{2})=h_{1}(r_{1}, \theta_{1})\cdot\mu(S_{1})$,$H_{2}(r_{1}, \theta_{\rceil}, r_{2}, \theta_{2})=h_{2}(r_{2}, \theta_{2})\cdot\mu(S_{2})=h_{2}(r_{2}, \theta_{2})\cdot(1-\mu(S_{1}))$

,

(1)

wherc
$\mu(S_{i})=\int_{s}f(r_{7}\theta)drd\theta$$r_{j}$$\theta_{i}$is the probablity measure of$S_{i}$,

is radius-vector of the point,is the angle between $Ox$ and radius-vector ,is payoff of the i-th firm if her proposal is closer to the arbitrator
solution.
Let us find Nash equilibrium in this game. We find it from the conditions$(x_{j}, \uparrow\int i)$$r_{i}$$h_{i}(r_{i}, \theta_{i})$,$\frac{\partial H_{1}}{\partial r_{1}}=\frac{\partial H_{2}}{\partial r_{2}}=0$

where$\frac{\partial H_{i}}{\partial r_{t}}=\frac{\partial h_{i}(r_{i},\theta_{i})}{\partial r_{i}}\cdot\mu(S_{i})+\frac{\partial\mu(S_{i})}{\partial r_{i}}\cdot h_{i}(r_{i}, \theta_{i})$Consider two different cases of distribution of the arbitrator.

.

(2)

44
3

Distribution of the \dagger \dagger center\dagger \dagger type

Let the distribution density of random variables

in polar coordinates has form:$z$$f(r, \theta)=\frac{3(1-r)}{\pi}$(3).

Consider the payoffs when the first firm maximises the first coordinate and the second
firm-second coordinate.
Then the payoffs of the players are$H_{1}=x_{1}(1-\mu(S_{2}))=r_{1}(1-\mu(S_{2}))$ ,$H_{2}=y_{2}\mu(S_{2})=r_{2}\mu(S_{2})$

.

(4)

Fig.2

Calculate the measure of$S_{2}$. Denote$R= \frac{r^{\frac{)}{\underline\prime y}}-r^{\frac{‘)}{1}}}{2\sqrt{r_{2}^{2}+r_{1}^{2}}}$.

We obtain$\mu(S_{2})=2_{arcsi}/_{nR}^{/2}\pi d\theta_{\frac{1^{1}R}{si_{11}\theta}}r\frac{3(1-r)}{\pi}dr=\frac{6}{\pi}(\frac{\pi}{12}-\frac{1}{6}$arcsin $R+ \frac{R^{3}}{12}\ln|1+\sqrt{1-R^{2}}|-$

45$- \frac{R^{3}}{12}\ln|1-\sqrt{1-R^{2}}|-\frac{R}{3}\sqrt{1-R^{2}}$(5 )Consequently, the payoffs of players have the form:$H_{1}=r_{1}( \frac{1}{2}+\frac{1}{\pi}$arcsin$R- \frac{R^{3}}{2\pi}$

In $|1+\sqrt{1-R^{2}}|+$$+ \frac{R^{3},}{2\pi}\ln|1-\sqrt{1-R^{2}}|+\frac{2R}{\pi}\sqrt{1-R^{2}}$$H_{2}=r_{2}( \frac{1}{2}-\frac{1}{\pi}$,arcsin $R+ \frac{R^{3}}{2\pi}\ln|1+\sqrt{1-R^{2}}|-$$- \frac{R^{3}}{2\pi}\ln|1-\sqrt{1-R^{2}}|-\frac{2R}{\pi}\sqrt{1-R^{2}}$Differentiating the first function in$\frac{\partial H_{1}}{\partial r_{1}}=\frac{1}{2}+\frac{1}{\pi}$arcsin$R- \frac{R^{3}}{2\pi}$$+ \frac{2R}{\pi}\sqrt{1-R^{2}}+\frac{3r_{1}}{2\pi}$$\frac{\partial H_{2}}{\partial r_{2}}=\frac{1}{2}-\frac{1}{\pi}$arcsin$r_{1}$and second function in$r_{2}$we obtain

In $|1+ \sqrt{1-R^{2}}|+\frac{R^{3}}{2\pi}$ in $|1-\sqrt{1-R^{2}}|+$$(2\sqrt{1-R^{2}}R_{r_{1}}’+R^{2}\cdot R_{r_{1}}’$$R+ \frac{R^{3}}{2\pi}$(6). in $|1-\sqrt{1-R^{2}}|)$ ,In $|1+ \sqrt{1-R^{2}}|-\frac{R^{3}}{2\pi}$ In $|1-\sqrt{1-R^{2}}|-$$- \frac{2R}{\pi}\sqrt{1-R^{2}}-\frac{3_{\mathcal{T}_{2}’}}{2\pi}(2\sqrt{1-R^{2}}R_{r_{2}}’+R^{2}\cdot R_{r_{2}}’\cdot\ln|1-\sqrt{1-R^{2}}|)$

,

(7)

where
$R_{r_{1}}’=- \frac{r_{1}^{3}+3r_{1}r_{2}^{2}}{2\sqrt{(r_{1}^{2}+r_{2}^{2})^{3}}}$$R_{r_{2}}’= \frac{r_{2}^{3}+3r_{2}r_{1}^{2}}{2\sqrt{(r_{1}^{2}+r_{2}^{2})^{3}}}$,.

Symmetry of the problem yields $r_{1}=r_{2}$ hence,$R=0,$$R_{r_{1}}’.=- \frac{1}{\sqrt{2}}=-R_{r_{2}}’$.

It follows$\frac{\partial H_{1}}{\partial r_{1}}r_{2}=r_{1}=\frac{1}{2}-\frac{3r_{1}}{\sqrt{2}\pi}$,$\frac{\partial H_{2}}{\partial r_{2}}r_{1}=r_{2}=\frac{1}{2}-\frac{3r_{2}}{\sqrt{2}\pi}$

.

(8)

46
Condition$\frac{\partial H_{i}}{\partial r_{i}}=0$yields$r_{i}= \frac{\sqrt{2}\pi}{6}$.

Consequently,$\theta_{1}=0,$$\theta_{2}=\frac{\pi}{2},$$x_{1}= \frac{\sqrt{2}\pi}{6},$$x_{2}=0,$Find the second derivative of the function$H_{1}$$y_{1}=0$$y_{2}= \frac{\sqrt{2}\pi}{6}$

,.

(9)

:

$H_{l_{r_{I}r_{1}}}’’= \frac{3}{\pi}(2\sqrt{1-R^{2}}\cdot R_{\tau_{1}}’.+R^{2}\cdot R_{71}’\cdot\ln|1-\sqrt{1-R^{2}}|)+$$+ \frac{3r_{1}}{2\pi}(2R\cdot(R’)^{2}\cdot\ln|1-\sqrt{1-R^{2}}|+R^{2}\cdot R_{r_{1}r_{1}}’’\cdot\ln|1-\sqrt{1-R^{2}}|-$$-R \cdot(R’)^{2}\frac{1-\sqrt{1-R^{2}}}{\sqrt{1-R^{2}}}$(10)

For $r_{1}=r_{2}$ we obtain$H_{1_{r_{1}r_{1}}}’’= \frac{-6}{\pi\sqrt{2}}<0$

Consequently,
the function$r_{1}= \frac{\sqrt{2}\pi}{6}$is the local maximum of.

$H_{1}$. Analogously, $H”,<0$ and forachieves the maximum.
The expected payoff is equal to$r_{2}= \frac{\sqrt{2}\pi}{6}$$H_{2}$$H= \frac{\sqrt{2}\pi}{12}$4.

Distribution of the center and seaside“ type$1\uparrow$

Let now the distribution density has form
$f(r,$ $\theta)=a+\frac{3}{2}(\pi^{-1}-a)r$ ,(11)

is some parameter. Consider the case when the first firm is interested tomaximise $(y-x)$ and the second- $(x-y)$ . Then the payofTs of the players arewhere$a$47$H_{1}=(y_{1}-x_{1})(1-\mu(S_{2}))$ ,$H_{2}=(x_{2}-y2)\mu(S_{2})$ .

(12)

Fig.3
The identities$y_{1^{-\prime}}x_{1}=r_{1} \sin\frac{3\pi}{4}-r_{1}\cos\frac{3\pi}{4}=r_{1}\frac{\sqrt{2}}{2}-r_{1}\frac{\sqrt{2}}{2}=\sqrt{2}r_{1}$$x_{2}-y_{2}=r_{2} \cos(-\frac{\pi}{4})+r_{2}\sin(-\frac{\pi}{4})=\sqrt{2}r_{2}$

,,

(13)

give the payoffs of the form
$H_{2}=2/0^{\arccos^{z_{\overline{2}}\perp}1}/_{2co6}rA^{\wedge}r_{\theta}\lrcorner dr\cdot rr-r_{d\theta}$$= \sqrt{2}r_{2}[\frac{1}{\pi}$arccos. $(a+ \frac{3}{2}(\pi^{-1}-a)r)\cdot\sqrt{2}r_{2}=$$\frac{r_{2}-r_{1}}{2}-\frac{1}{32}(\frac{1}{\pi}-a)(r_{2}-r_{1})^{3}\ln\frac{2+\sqrt{4-(\gamma_{2}-r}\overline{1)^{2}}}{2-\sqrt{4-(r_{2}-r_{1}}\overline{)^{2}}}+$$+ \frac{1}{8}(\frac{1}{\pi}-a)(r_{2}-r_{1})\sqrt{4-(r_{2}-r_{1})^{2}}]$,

48$H_{1}= \sqrt{2}r_{1}[1-\frac{1}{\pi}$arccos$\frac{r_{2}-r_{1}}{2}+\frac{1}{32}(\frac{1}{\pi}-a)(7_{2^{-}}r_{1})^{3}\ln|\frac{2+\sqrt{4-(r_{2}-r_{1})^{2}}}{2-\sqrt{4-(r_{2}-r_{1})^{2}}}|-$$- \frac{1}{8}(\frac{1}{\pi}-a)(r_{2}-r_{1})\sqrt{4-(r_{2}-r_{1})^{2}}]$

Calculate the partial derivative$\frac{\partial}{\partial}r_{i}H\lrcorner$.

(14)

and let it is equal to zero:$\frac{\partial H_{1}}{\partial r_{1}}=\sqrt{2}[1-\frac{1}{\pi}$arccos$\frac{r_{2}-r_{1}}{2}+$$+ \frac{1}{32}(\frac{1}{\pi}-a)(r_{2}-r_{1})^{3}\ln\frac{2+\sqrt{4-(r_{2}-r_{1})^{2}}}{2-\sqrt{4-(r_{2}-r_{1})\sim)}}$$- \frac{1}{8}(\frac{1}{\pi}-a)(r_{2}-r_{1})\sqrt{4-(7_{2}-\gamma_{1})^{2}}]+\sqrt{2}\prime r_{1}[\frac{-1}{\pi\sqrt{4-(r_{2}-r_{1})^{2}}}+$$+ \frac{1}{8}(\frac{1}{\pi}-a)\sqrt{4-(r_{2}-r_{1})^{2}}-\frac{3}{32}(\frac{1}{\pi}-a)(r_{2}-r_{1})^{2}\ln\frac{2+\sqrt{4-(r_{2}-r_{1})^{2}}}{2-\sqrt{4-(r_{2}-r_{1})^{2}}}$$\frac{\partial H_{2}}{\partial r_{2}}=\sqrt{2}[\frac{1}{\pi}$arccos$\frac{r_{2}-r_{1}}{2}+\frac{1}{\backslash 32}(\frac{1}{\pi}-a)(r_{2}-r_{1})^{3}\ln\frac{2+\sqrt{4-(r_{2}-r_{1})^{2}}}{2-\sqrt{4-(r_{2}-r_{1})^{2}}}-$$- \frac{1}{8}(\frac{1}{\pi}-a)(r_{2}-r_{1})\sqrt{4-(72^{-r_{1})^{2}}}]-\sqrt{2}r_{1}[\frac{-1}{\pi\sqrt{4-(r_{2}-r_{1})^{2}}}+$$+ \frac{1}{8}(\frac{1}{\pi}-a)\sqrt{4-(r_{2}-r_{1})^{2}}-\frac{3}{32}(\frac{1}{\pi}-a)(r_{2}-r_{1})^{2}\ln\frac{2+\sqrt{4-(r_{2}-r_{1})^{2}}}{2-\sqrt{4-(r_{2}-r_{1})^{2}}}$

It yields$r_{i}= \frac{2\pi}{1+\pi a}$.

From here we obtain$\theta_{1}=\frac{3\pi}{4},$$x_{1}=- \frac{\sqrt{2}\pi}{1+\pi a},$$\theta_{2}=-\frac{\pi}{4})x_{2}=\frac{\sqrt{2}\pi}{1+\pi a},$$y_{1}= \frac{\sqrt{2}\pi}{1+\pi a}$,$y_{2}=- \frac{\sqrt{2}\pi}{1+\pi a}$

.

(16)

49
The expected payoff of the firm is equal to$H= \frac{\sqrt{2}\pi}{1+\pi 0}$

The condition $r\geq 0$ gives the restriction forvalues of for some .
$r$$a$.

as$a>- \frac{1}{\pi}$. In the table you find the$a$

Table

References[1] Farber H. An analysis of final-offer arbitration, Journal of conflict resolution,
24 (1980), 683-705.
[2] Mazalov V.,Sakaguchi M. Location Game on the Plane, International Gaine
theory Review, 5 (2003), 13-25.