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Arbitration procedures with multiple arbitrators
European Journal of Operational Research 217 (2012) 198–203
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European Journal of Operational Research journal homepage: www.elsevier.com/locate/ejor
Decision Support
Arbitration procedures with multiple arbitrators Vladimir Mazalov a,⇑, Julia Tokareva b a b
Institute of Applied Mathematical Research, Karelian Research Center of Russian Academy of Sciences, Pushkinskaya St. 11, Petrozavodsk 185610, Russia Zabaikalsky State Humanitarian Pedagogical University named after N. Tchernishevsky, Babushkina St., 129, Chita 672007, Russia
a r t i c l e
i n f o
Article history: Received 22 February 2011 Accepted 6 September 2011 Available online 16 September 2011 Keywords: Group decision and negotiation Final-offer arbitration Multiple arbitrators Equilibrium
a b s t r a c t We consider two final-offer arbitration procedures in the case where there is more than one arbitrator. Two players, labeled 1 and 2 and interpreted here as Labor and Management, respectively, are in dispute about an increase in the wage rate. They submit final offers to a Referee. There are N arbitrators. Each of the arbitrators has her own assessment and selects the offer which is closest to her assessment. After that each arbitrator informs the Referee about her decision. The Referee counts the votes and declares the player obtaining the most votes to be the winner. Under the second arbitration scheme, the Referee takes into account only the assessments which lie between the players’ offers. The game is modeled as a zerosum game. The Nash equilibrium in this arbitration game is derived. Ó 2011 Elsevier B.V. All rights reserved.
1. Introduction Two players, labeled 1 and 2 and interpreted here as Labor and Management, respectively, are in dispute about an increase in the wage rate. Player 1 prefers a high rate, player 2 a low rate. Players 1 and 2 simultaneously submit the offers x 2 R and y 2 R, respectively. If x 6 y, then the agreement will be (x + y)/2. Otherwise, the parties call in an arbitrator, labeled A. Assume that the assessment of the arbitrator can be described by a random variable a. There are two standard arbitration procedures which can be used to solve the conflict. Under the conventional arbitration procedure (CA), the final decision is a. Under the second procedure called final-offer arbitration (FOA) which was proposed by Stevens (1966), after observing the offers x and y, the arbitrator simply chooses the offer which is closer to a. The FOA and CA procedures and their modifications were investigated in papers by Farber (1980), Chatterjee (1981), Brams and Merill (1983), Brams (1990), Samuelson (1991), Kilgour (1994), Zeng (2003), and Mazalov et al. (2006). Brams and Merill (1986) suggest a combination of these procedures. Under their scheme, the arbitrator is allowed to choose a settlement which does not lie between the two offers; otherwise, if the arbitrator’s assessment is somewhere between the two FOA offers, then she must choose one or the other of these offers. It is supposed that the players are informed about the arbitrator’s assessment a. Gerchak et al. (2004) consider arbitration with partial information. In this case, the assessment of the arbitrator is unknown to the players and they only know the final decision.
Based on the history of k arbitrations and assuming that a1, . . . , ak are random variables with a distribution which depends on an unknown parameter h, they estimate the arbitrator’s hidden judgements. Hurley (2003) examined final-offer arbitration with a panel of arbitrators. The assessments of arbitrators are modeled by random variables with identical distribution functions. Given the offers x and y, the Referee picks the offer that is closest to the average or median, as required, of the arbitrators’ assessments. It is shown that the procedure based on the mean gives a smaller difference between offers than does the procedure based on the median, and the difference between the equilibrium offers becomes smaller as the number of arbitrators increases. We apply the approach of multiple arbitrators to final-offer arbitration with non-identical arbitrators and its modification. We first consider final-offer arbitration with three arbitrators and then consider the general case. In Section 3 we introduce a new arbitration model with multiple arbitrators. Under this procedure, the Referee takes into account the only assessments of the arbitrators which lie between the players’ offers. In the case of both players obtaining an equal number of votes the Referee’s decision is (x + y)/2. For each procedure, the payoff function H(x, y) and the equilibrium conditions are derived. The optimal offers x⁄ and y⁄ are computed for cases where the arbitrators’ assessments are either all uniformly or all normally distributed. Finally, we analyse the equilibrium when the parameters of these distributions are assumed to be random variables. 2. Arbitration with multiple arbitrators
⇑ Corresponding author. E-mail addresses: (J. Tokareva).
[email protected] (V. Mazalov),
[email protected]
0377-2217/$ - see front matter Ó 2011 Elsevier B.V. All rights reserved. doi:10.1016/j.ejor.2011.09.014
Imagine that Players 1 and 2 submit their offers x and y to a Referee. We assume that there are an odd number, N = 2n 1, of
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arbitrators. Each arbitrator has her own assessment. After observing the offers, each arbitrator votes on which offer should be selected. Suppose that each arbitrator selects the offer which is closest to her assessment and informs the Referee about her decision. The Referee counts the votes and declares the player obtaining the most votes (at least n votes) to be the winner. Let the assessments of the arbitrators be independent random variables a1, . . . , a2n1 with CDFs F1(a), . . . , F2n1(a). Both players have the same beliefs about the arbitrators’ assessments and these beliefs are common knowledge. For y < x the FOA procedure states that the Referee prefers the offer y if at least n arbitrators have assessments 6(x + y)/2. Otherwise, the Referee chooses the offer x. 2.1. Model with three arbitrators
where
nn oo x þ yo n xþy [ ai ; aj 6 < ak ; i – j – k a1 ; a2 ; a3 6 2 2 x þ y x þ y x þ y ¼ F1 F2 F3 2 2 2 x þ y x þ y x þ y F2 1 F3 þ F1 2 2 2 x þ y x þ y x þ y 1 F2 F3 þ F1 2 2 2 x þ y x þ y x þ y þ 1 F1 F2 F3 : 2 2 2
ð3Þ
Denote the expression in square brackets in (3) as G(m). Thus, we can rewrite these equations as
(
1 x y ¼ GðmÞ ;
x þ y ¼ 2m; or
(
1 x ¼ m þ 2GðmÞ ; 1 y ¼ m 2GðmÞ ;
Example 1. Suppose that the assessments of the arbitrators are uniformly distributed in the intervals [1, 1],pffiffi[0, 1] and [1, 2], respectively. The solution of Eq. (2) is m ¼ 321. Consequently, pffiffi 5 33 GðmÞ ¼ 6 , and the equilibrium strategies are
8 pffiffiffi 4 > < x ¼ 11 2 3 1 0:896; pffiffiffi > 1 : y ¼ 11 3 3 7 0:164:
It is easy to see that the second-order optimality condition is satisfied. Notice that if m is the median of the distribution F1, then (2) can be presented as
or
F 2 ðmÞ þ F 3 ðmÞ ¼ 1: Hence, if the equality F2(m) + F3(m) = 1 is satisfied at the point m, then the median m is the solution of (2). For instance, if F1, F2, F3 are normal distributions with equal standard deviations and mean values k, 0, k where k is any real number, then the solution of (2) is m = 0.
ð1Þ
where for convenience we write Fi = Fi((x + y)/2), i = 1, 2, 3. The first order condition for the equilibrium in this game is @ H(x, y)/@x = 0, @H(x, y)/@y = 0. This yields
@H ¼ 1 ðF 1 F 2 þ F 1 F 3 þ F 2 F 3 2F 1 F 2 F 3 Þ @x y x 0 F 1 ðF 2 þ F 3 Þ þ F 02 ðF 1 þ F 3 Þ þ F 03 ðF 1 þ F 2 Þ þ 2 2ðF 01 F 2 F 3 þ F 1 F 02 F 3 þ F 1 F 2 F 03 Þ ¼ 0
Example 2. Let F1, F2,F3 normal distributions with variance 1 and mean values 2, 0, 1, respectively. Eq. (2) gives m = 0.141. It follows that G(m) = 0.483. Hence the optimal strategies are x⁄ = 0.894, y⁄ = 1.176. Eqs. (2) and (3) can be simplified if the medians of all the distributions F1, F2, F3 are equal. Denote this common median by m. In this case, m satisfies Eq. (2). Eq. (3) yields
xy¼
and
2 ; F 01 ðmÞ þ F 02 ðmÞ þ F 03 ðmÞ
or
@H ¼ ðF 1 F 2 þ F 1 F 3 þ F 2 F 3 2F 1 F 2 F 3 Þ @y y x 0 þ F 1 ðF 2 þ F 3 Þ þ F 02 ðF 1 þ F 3 Þ þ F 03 ðF 1 þ F 2 Þ 2 2ðF 01 F 2 F 3 þ F 1 F 02 F 3 þ F 1 F 2 F 03 Þ ¼ 0:
xy 1 ¼ ; 2 f1 ðmÞ þ f2 ðmÞ þ f3 ðmÞ where f1, f2, f3 are the densities of F1, F2, F3, respectively. Finally, we obtain the equilibrium in this case
Subtracting the second of these equations from the first gives (x + y)/2 = m, where m satisfies the equation
1 : 2 ð2Þ
F 1 ðmÞF 2 ðmÞ þ F 1 ðmÞF 3 ðmÞ þ F 2 ðmÞF 3 ðmÞ 2F 1 ðmÞF 2 ðmÞF 3 ðmÞ ¼
Moreover,
þF 03 ðmÞðF 1 ðmÞ þ F 2 ðmÞÞ 2ðF 01 ðmÞF 2 ðmÞF 3 ðmÞ 1 þF 1 ðmÞF 02 ðmÞF 3 ðmÞ þ F 1 ðmÞF 2 ðmÞF 03 ðmÞÞ :
1 1 1 F 2 ðmÞ þ F 3 ðmÞ þ F 2 ðmÞF 3 ðmÞ F 2 ðmÞF 3 ðmÞ ¼ ; 2 2 2
Simplifying, we obtain
Hðx; yÞ ¼ x þ ðy xÞðF 1 F 2 þ F 1 F 3 þ F 2 F 3 2F 1 F 2 F 3 Þ;
x y ¼ F 01 ðmÞðF 2 ðmÞ þ F 3 ðmÞÞ þ F 02 ðmÞðF 1 ðmÞ þ F 3 ðmÞÞ
where m satisfies Eq. (2).
Let us derive the solution in the case of three arbitrators. In this case, the Referee chooses y, if a1, a2, a3 6 (x + y)/2, or if a1, a2 6 (x + y)/2, or if a1, a3 6 (x + y)/2, or if a2, a3 6 (x + y)/2. Otherwise, the Referee chooses the offer x. So, the payoff function in this case is nn oo x þ yo n xþy < ak ; i – j – k Hðx; yÞ ¼ yP a1 ; a2 ; a3 6 [ ai ; aj 6 2 2 nn oo x þ yo n xþy þ xP a1 ; a2 ; a3 P [ ai ; aj P > ak ; i – j – k ; 2 2
P
y ¼ m
1 ; f1 ðmÞ þ f2 ðmÞ þ f3 ðmÞ
x ¼ m þ
1 : f1 ðmÞ þ f2 ðmÞ þ f3 ðmÞ ð4Þ
For instance, if a1, a2, a3 are normal random variables with the same mean value m and different standard deviations r1, r2, r3, we obtain
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pffiffiffiffiffiffiffi 2p y ¼m 1 ; 1 1 þ r1 r2 þ r3
pffiffiffiffiffiffiffi 2p x ¼mþ 1 : 1 1 þ r1 r2 þ r3
y ¼ m D;
where
"
If a1, a2, a3 are uniformly distributed in [0, 1], then
y ¼
1 1 1 ¼ ; 2 3 6
x ¼
x ¼ m þ D;
D¼
1 1 5 þ ¼ : 2 3 6
Notice that this equilibrium differs from the solution y⁄ = 0, x⁄ = 1 in the game with one arbitrator (Gibbons, 1992).
2n n
#1
2n3 2X n1 1 fi ðmÞ : 2 i¼1
From this theorem, it follows that in the case of normal distributions F1, . . . , F2n1 with the same mean value m and different standard deviations ri, i = 1, . . . , 2n 1, the optimal offers are y⁄ = m D, x⁄ = m + D with
In the arbitration game with 2n 1 > 3 arbitrators the payoff function is
pffiffiffiffiffiffiffi 2p 1
D¼ : 2n 12n 1 8 1 2 r1 þ þ rN n
Hðx; yÞ ¼ x þ ðy xÞ 0
When the arbitrators’ assessments are i.i.d., we obtain the following result.
2.2. Model with 2n 1 arbitrators
1
[
B PB @
ai1 ; . . . ; aik 6
i1 ––ik k¼n;nþ1;...;2n1
C xþy < aikþ1 ; . . . ; ai2n1 C A; 2
Corollary 2. Let F1 = F2 = = F2n1 = F. Then the equilibrium in the arbitration game is
y ¼ m D;
or
X
Hðx; yÞ ¼ x þ ðy xÞ
F i1 ; . . . ; F ik F ikþ1 ; . . . ; F i2n1 ;
ð5Þ
i1 ––ik k¼n;nþ1;...;2n1
x þ y x þ y @H ¼1F þ ðy xÞF 0 ¼ 0; @x 2 2
F i1 ; . . . ; F ik F ikþ1 ; . . . ; F i2n1 :
ð6Þ
F i1 ; . . . ; F ik F ikþ1 ; . . . ; F i2n1 ¼
i1 ––ik k¼n;nþ1;...;2n1
1 2
2n 1 1 1 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ pffiffiffiffiffiffiffi : 2 pn n 1 2p2n
2n
Hence,
20
i1 ––ik k¼n;nþ1;...;2n1
C7 7 F i1 ; . . . ; F ik F ikþ1 ; . . . ; F i2n1 C A5 :
ð8Þ
We can determine the optimal values x and y from Eqs. 7 and 8. Let us suppose that the arbitrators’ assessments have a common median m. It is easy to see that m satisfies Eq. (7). This immediately yields
xþy ¼ m: 2
as n tends to 1. We see that the difference between equilibrium offers becomes smaller as the number of arbitrators increases. 3. Modified arbitration with multiple arbitrators
10 31 X
pffiffiffi 2pr 1 pffiffiffi ! 0 8 n
ð7Þ
and the difference x y is equal to
B 16 6B 2 4@
pffiffiffiffiffiffiffi 1 2pr
D¼ : 2n 12n 8n 2 n
D
From these equations, we immediately obtain that (x + y)/2 = m satisfies the equation
xy¼
For normal distributions with a common standard deviation r, (10) yields
4
i1 ––ik k¼n;nþ1;...;2n1
X
#1
2n3 1 nf ðmÞ : 2 n
2n
This coincides with the result of paper (Hurley, 2003).
where we denote
x þ y ¼ F 2
"
Using Stirling’s formula, we obtain the following estimate
x þ y x þ y @H ¼F þ ðy xÞF 0 ¼ 0; @y 2 2
F¼
x ¼ m þ D;
where
D¼
where F ¼ 1 F and Fi = Fi((x + y)/2), i = 1, . . . , 2n 1. In the sum (5) we have one term containing no F i ; 2n 1 terms including only one
2n 1 F i , and terms including a product of the form F j1 . . . F jk . k Differentiating (5), we obtain the following system of equations
X
ð10Þ
ð9Þ
The proof of the following theorem is given in Appendix A. Theorem 1. Let the medians of all the distribution functions F1, . . . , F2n1 be equal to m. In this case, the equilibrium in the arbitration game is
In this section we consider a new arbitration procedure with multiple arbitrators and a Referee who takes into account only the assessments of the arbitrators which belong to interval between the player’s offers, i.e. [y, x], when x P y. Assume that the assessments of the arbitrators are i.i.d. random variables a1, . . . , a2n1 with CDF F(a) and density f(a). Assume that y < x, and k arbitrators prefer the offer y, i.e. y 6 ai 6 (x + y)/2, i = i1, . . . , ik and m arbitrators prefer the offer x, i.e. (x + y)/2 6 ai 6 x, i = i1, . . . , im. Here, k + m 6 2n 1. After counting the votes, the Referee selects the offer y if k > m, and the offer x if k < m. Otherwise, the Referee’s decision is (x + y)/2. First, we find the solution in the arbitration game with one arbitrator. If the arbitrators’ assessment belongs to the interval [y, (x + y)/2], then the final decision is y. If the assessment belongs ((x + y)/2, x] then the decision is x. Otherwise, the decision is (x + y)/ 2. Thus, the payoff function in this game is
V. Mazalov, J. Tokareva / European Journal of Operational Research 217 (2012) 198–203
x þ y x þ y x þ y Hðx; yÞ ¼ y F FðyÞ þ x FðxÞ F þ ðFðyÞ 2 2 2 þ 1 FðxÞÞ:
xþy ð1 p1 p2 Þ 2 x þ y x y 3 þ v u3 þ 3½v 2 ð1 v Þ u2 ð1 uÞ ¼¼ 2 2 þ3ðv uÞð1 u v Þ2 :
Hðx; yÞ ¼ yp1 þ xp2 þ
We can derive the equilibrium from the first-order conditions
x þ y FðxÞ þ FðyÞ x þ y 1 @H x y ¼ f ðxÞ f þ F þ ¼ 0; @x 2 2 2 2 2 x þ y FðxÞ þ FðyÞ x þ y 1 @H x y ¼ f ðyÞ f þF þ ¼ 0: @y 2 2 2 2 2
201
The first-order condition for equilibrium is @Hðx;yÞ ¼ 0; @Hðx;yÞ ¼ 0. @x @y For a distribution which is symmetric with respect to the point 0, the conditions x = y and u(x, y) = v(x, y) must be satisfied. Hence, the equilibrium value of x satisfies the equation
1 þ xð6v 2 6v þ 3Þðf ðxÞ f ð0ÞÞ ¼ 0; 2 where
Example 3. Let a have the standard normal distribution N(0, 1). The symmetry of the problem yields x = y and we can rewrite the equation for the equilibrium in the form x2
xð1 e 2 Þ ¼
rffiffiffiffi
p 2
v ðxÞ ¼ FðxÞ Fð0Þ;
f ðxÞ ¼ F 0 ðxÞ:
Example 4. Let F(x) be the standard normal distribution N(0,1). Then equilibrium condition gives x⁄ 1.352. Hence, the equilibrium is
:
From this, we obtain
y 1:352;
y 1:667;
We see that the difference between the equilibrium offers is smaller than in the model with one arbitrator.
x 1:667:
It is not difficult to check that this is the equilibrium of the game. Notice, that the equilibrium under the original FOA procedure (Gibbons, 1992) is
rffiffiffiffi
y ¼
p 2
1:253;
x ¼
rffiffiffiffi
p 2
1:253:
We see that this new arbitration procedure leads to the players proposing values which differ more than under the FOA procedure. However, as the number of arbitrators increases, the difference between the equilibrium offers falls.
x 1:352:
3.2. Arbitration with 2n 1 arbitrators Now imagine that there are 2n 1 arbitrators. In Appendix B it is shown that the final decision of the Referee is y with probability
p1 ¼
n n1 X X 2n 1 2n 1 i un1þi ð1 uÞni þ u n1þi i i¼1 i¼1
i1 X 2n 1 i v j ð1 u v Þ2n1ij j j¼0
3.1. Model with three arbitrators
and x with probability
Let the arbitration panel consist of three arbitrators. Assume that the assessments of the arbitrators are i.i.d random variables a1, a2, a3 with continuous CDF F(a). xþy Let and xþyy < x. Define the intervals L and R by L ¼ y; 2 R ¼ 2 ; x , and set
p2 ¼
u¼F
x þ y 2
FðyÞ;
v ¼ FðxÞ F
x þ y : 2
n n1 X X 2n 1 2n 1 n1þi v ð1 v Þni þ vi n1þi i i¼1 i¼1
i1 X 2n 1 i j u ð1 u v Þ2n1ij : j j¼0
In all other cases the decision is (x + y)/2. Hence, the payoff in this game is
"
n 2n 1 xþy xy X þ ðv n1þi ð1 v Þni 2 2 n 1 þ i i¼1
i1 n1 X 2n 1 X 2n 1 i un1þi ð1 uÞni Þ þ i j i¼1 j¼0 1 u v Þ2n1ij ðv i uj v j ui Þ : ð11Þ
Hðx; yÞ ¼ Thus, the final decision of the Referee is y in the following three cases: Firstly, when the assessments of all the arbitrators lie in L. The probability of this is u3. Secondly, when the assessments of any two arbitrators are in L, but the third assessment is not. The probability of this is 3u2(1 u). Thirdly, when the assessment of one of them is in L, but the assessments of the other arbitrators are not in L or R. The probability of this is 3u(1 u v)2. So, the probability of the final decision being y is
p1 ¼ u3 þ 3u2 ð1 uÞ þ 3uð1 u v Þ2 : Analogously, the decision of the Referee is x with probability
p2 ¼ v 3 þ 3v 2 ð1 v Þ þ 3v ð1 u v Þ2 : In all other cases the decision is (x + y)/2. The probability of this is 1 p1 p2. Hence, the payoff function is
Denote the expression in square brackets in (11) by S(v, u). We can see that S(v, u) is an anti-symmetric function, i.e. S(v, u) = S(u, v). Hence, for v = u
Sðu; uÞ ¼ 0: The first-order equilibrium conditions for (x, y) is
@H 1 1 x y @S 1 x þ y @S 1 x þ y f ðxÞ f ¼ þ Sðv ; uÞ þ þ f ¼0 @x 2 2 2 @v 2 2 @u 2 2 and
@H 1 1 x y @S 1 x þ y @S 1 x þ y ¼ Sðv ; uÞ f þ f ðyÞ f ¼ 0: @y 2 2 2 @v 2 2 @u 2 2
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V. Mazalov, J. Tokareva / European Journal of Operational Research 217 (2012) 198–203
Table 1 Equilibrium strategies for various numbers of arbitrators. n x⁄
1 1.667
2 1.352
3 1.203
4 1.111
5 1.047
Now consider the case where the CDF F(x) is symmetric with respect to zero. Then, by symmetry, at equilibrium the equalities x = y and u = v hold. Note that S(v, u) = S(u, v). This yields @Sðv ;uÞ vÞ ¼ @Sðu; . Thus, for u = v and x = y we obtain @v @u
1 @Sðv ; v Þ þ xðf ðxÞ f ð0ÞÞ ¼ 0: 2 @v
ð12Þ
From (11), we obtain
n 2n 1 @Sðv ; v Þ X ¼ v nþi2 ð1 2vÞn1i ððn 1 þ iÞð1 v Þ ðn iÞv Þ @v n1þi i¼1
i1 n1 X 2n 1 X 2n 1 i ð13Þ þ ð1 2v Þ2n1ij ði jÞv iþj1 : i j i¼1 j¼0 Eqs. (12) and (13) can be used to numerically calculate the equilibrium strategies in this game. In Table 1, we present the results of numerical calculations for various values of n. We see that the difference between equilibrium offers decreases when the number of arbitrators increases. 4. Conclusion In this paper we have considered two arbitration procedures for settling labor disputes involving multiple arbitrators. Each arbitrator follows FOA rules and the Referee takes the final decision on the basis of majority voting. We introduce a new arbitration procedure in which the Referee does not take into account assessments of the arbitrators which do not lie between the offers. The equilibrium strategies of the players under both procedures are derived in analytical form. We estimate the expected difference between the equilibrium offers and show that this difference becomes smaller as the number of arbitrators increases. This can be used to determine the optimal number of arbitrators. This work was supported by the Division of Mathematical Sciences of the Russian Academy of Sciences and the Analytical program ‘‘Development of Scientific Potential of the Higher School’’ (project 1.8.10). Appendix A
10
0
X i1 ––ik k¼n;nþ1;...;2n1
2n1 X
Fi
Y
7 0.962
8 0.932
9 0.906
10 0.884
!0 Fj
;
j–i
i¼1
contains 2n 1 terms with f1, of which 2n 2 are positive and one is negative. The derivative 2n1 X
FiFj
i;j¼1;i–j
Y
!0 Fk
;
k–i;j
2n 1 2n 1 terms with f1, of which ð2n 2Þ are 2 2 positive and 2n 2 are negative. Arguing in the same way, for any k 6 n 1 in the derivative contains
X
F i1 ; . . . ; F ik
i1 ––ik
!0
Y
Fj
;
j–i1 ;...;ik
2n 1 there are terms with f1. Among them k
2n 1 2n 2 2n 2 terms are positive and terms k k1 k1 are negative. So the coefficient of f1 (and consequently of each fi) is equal to
s ¼ 1 þ ð2n 2 1Þ þ
2n 2 2 ; n2
2n 1 2
2ð2n 2Þ þ þ
2n 1 n1
Simplifying this yields
s¼
2n : n
Now we can present (A1) in the form
s
2n1 2n1 2n1 2n1 X X 2n 1 1 fi ðmÞ ¼ fi ðmÞ; 2 2 n i¼1 i¼1
and (8) yields
Proof of Theorem 1. Let us simplify Eq. (8). Consider the value of the derivative
B B @
6 1.000
C F i1 ; . . . ; F ik ð1 F ikþ1 Þ; . . . ; ð1 F i2n1 ÞC A;
#1
2n2 2n1 X 1 fi ðmÞ : 2 n i¼1
2n
ðA2Þ
(A2) and (9) establish the theorem.
ðA1Þ
at the point m, where m is the median of the distribution Fi. So, Fi(m) = 1/2, i = 1, . . . , 2n 1. Hence, the absolute value of each term in this sum is of the following form
2n2 x þ y0 1 2n1 1 fi ðmÞ ¼ fi ðmÞ; 2 2 2
xy¼
"
i ¼ 1; . . . ; 2n 1:
Such a term is positive when we differentiate Fi and negative when we differentiate F i . From the form of (A1), it is sufficient to calculate the coefficients of the terms which include f1(m). The derivative (F1. . .F2n1)0 contains only one term with f1 ¼ F 01 . The derivative
Appendix B Consider all the possible results of voting such that the final decision of the Referee is y. Firstly, suppose that all 2n 1 assessments of the arbitrators lie in the interval L. The probability of this is equal to u2n1. Secondly, suppose that 2n 2 arbitrators’ assessments belong to L and one of them does not. The probability of this is
2n 1 u2n2 ð1 uÞ. 2n 2 Thirdly, suppose 2n 3 arbitrators’ assessments are in L and the other two are not in L. The probability of this is
2n 1 u2n3 ð1 uÞ2 . 2n 3
V. Mazalov, J. Tokareva / European Journal of Operational Research 217 (2012) 198–203
and so on, until the n-th such case where n arbitrators’ assessments are in L and the other n 1 are not in L. The probability of
2n 1 n this is u ð1 uÞn1 . n Moreover, the final decision is also y in the following cases. If n 1 arbitrators’ assessments are in L (the probability of this
2n 1 n1 is u Þ, and among the other n, exactly j assessments ben1 long to R and the remaining n j assessments are neither in L nor in R, where j = 0, . . . , n 2. The probability of this is
ð1 u v Þn þ
n
þ
1
v ð1 u v Þn1 þ
n
n2
v n2 ð1 u v Þ2 :
Or, if n 2 arbitrators’ assessments are in L (the probability of this
2n 1 n2 is u Þ, and among the other n + 1, exactly j assessments n2 belong to R and the remaining n + 1 j assessments are neither in L nor in R, where j = 0, . . . , n 3. The probability of this is
ð1 u v Þnþ1 þ
nþ1
v ð1 u v Þn þ 1
nþ1 v n3 ð1 u v Þ4 : þ n3
and so on. Finally the decision is y, if one of the assessments belongs to L
2n 1 (with probability u), and the other 2n 2 assessments 1 are neither in L nor in R. The probability of this is (1 u v)2n2. Consequently, the probability of the final decision being y is equal to
p1 ¼
n X 2n 1 un1þi ð1 uÞni n1þi i¼1
i1 n1 X 2n 1 i X 2n 1 i þ v j ð1 u v Þ2n1ij : u i j i¼1 j¼0
203
Arguing in the same way, we can find the probability that the decision is x
p2 ¼
n X 2n 1 n1þi
n1 X 2n 1
v n1þi ð1 v Þni
i¼1
þ
i¼1
i
vi
i1 X 2n 1 i j u ð1 u v Þ2n1ij : j j¼0
References Brams, S.J., 1990. Negotiation Games: Applying Game Theory to Bargaining and Arbitration p. 280. Routledge, New York. Brams, S.J., Merill, S., 1983. Equilibrium strategies for final-offer arbitration: There is no median convergence. Management Science 29 (8), 927–941. Brams, S.J., Merill, S., 1986. Binding versus final-offer arbitration: A combination is best. Management Science 32 (10), 1346–1355. Chatterjee, K., 1981. Comparison of arbitration procedures: Models with complete and incomplete information. IEEE Transactions on Systems, Man, and Cybernetics 11 (2), 101–109. Farber, H., 1980. An analysis of final-offer arbitration. Journal of Conflict Resolution 24, 683–705. Gerchak, Y., Greenstein, E., Weissman, I., 2004. Estimating arbitrator’s hidden judgement in final offer arbitration. Group Decision and Negotiation 13 (3), 291–298. Gibbons, R., 1992. A Primer in Game Theory p. 273. Prentice Hall, N.Y.. Hurley, W.J., 2003. Effects of multiple arbitrators on final-offer arbitration settlements. European Journal of Operational Research 145, 660–664. Kilgour, D.M., 1994. Game-theoretic properties of final-offer arbitration. Group Decision and Negotiation 3, 285–301. Mazalov, V.V., Mentcher, A.E., Tokareva, J.S., 2006. Equilibrium in a discrete arbitration procedure. Scientiae Mathematicae Japonicae 63 (3), 283–288. Samuelson, W.F., 1991. Final-offer arbitration under incomplete information. Management Science 37 (10), 1234–1247. Stevens, C.M., 1966. Is compulsory arbitration compatible with bargaining. Industrial Relations 5, 38–52. Zeng, D.-Z., 2003. An amendment to final-offer arbitration. Mathematical Social Sciences 46 (1), 9–19.
Contents lists available at SciVerse ScienceDirect
European Journal of Operational Research journal homepage: www.elsevier.com/locate/ejor
Decision Support
Arbitration procedures with multiple arbitrators Vladimir Mazalov a,⇑, Julia Tokareva b a b
Institute of Applied Mathematical Research, Karelian Research Center of Russian Academy of Sciences, Pushkinskaya St. 11, Petrozavodsk 185610, Russia Zabaikalsky State Humanitarian Pedagogical University named after N. Tchernishevsky, Babushkina St., 129, Chita 672007, Russia
a r t i c l e
i n f o
Article history: Received 22 February 2011 Accepted 6 September 2011 Available online 16 September 2011 Keywords: Group decision and negotiation Final-offer arbitration Multiple arbitrators Equilibrium
a b s t r a c t We consider two final-offer arbitration procedures in the case where there is more than one arbitrator. Two players, labeled 1 and 2 and interpreted here as Labor and Management, respectively, are in dispute about an increase in the wage rate. They submit final offers to a Referee. There are N arbitrators. Each of the arbitrators has her own assessment and selects the offer which is closest to her assessment. After that each arbitrator informs the Referee about her decision. The Referee counts the votes and declares the player obtaining the most votes to be the winner. Under the second arbitration scheme, the Referee takes into account only the assessments which lie between the players’ offers. The game is modeled as a zerosum game. The Nash equilibrium in this arbitration game is derived. Ó 2011 Elsevier B.V. All rights reserved.
1. Introduction Two players, labeled 1 and 2 and interpreted here as Labor and Management, respectively, are in dispute about an increase in the wage rate. Player 1 prefers a high rate, player 2 a low rate. Players 1 and 2 simultaneously submit the offers x 2 R and y 2 R, respectively. If x 6 y, then the agreement will be (x + y)/2. Otherwise, the parties call in an arbitrator, labeled A. Assume that the assessment of the arbitrator can be described by a random variable a. There are two standard arbitration procedures which can be used to solve the conflict. Under the conventional arbitration procedure (CA), the final decision is a. Under the second procedure called final-offer arbitration (FOA) which was proposed by Stevens (1966), after observing the offers x and y, the arbitrator simply chooses the offer which is closer to a. The FOA and CA procedures and their modifications were investigated in papers by Farber (1980), Chatterjee (1981), Brams and Merill (1983), Brams (1990), Samuelson (1991), Kilgour (1994), Zeng (2003), and Mazalov et al. (2006). Brams and Merill (1986) suggest a combination of these procedures. Under their scheme, the arbitrator is allowed to choose a settlement which does not lie between the two offers; otherwise, if the arbitrator’s assessment is somewhere between the two FOA offers, then she must choose one or the other of these offers. It is supposed that the players are informed about the arbitrator’s assessment a. Gerchak et al. (2004) consider arbitration with partial information. In this case, the assessment of the arbitrator is unknown to the players and they only know the final decision.
Based on the history of k arbitrations and assuming that a1, . . . , ak are random variables with a distribution which depends on an unknown parameter h, they estimate the arbitrator’s hidden judgements. Hurley (2003) examined final-offer arbitration with a panel of arbitrators. The assessments of arbitrators are modeled by random variables with identical distribution functions. Given the offers x and y, the Referee picks the offer that is closest to the average or median, as required, of the arbitrators’ assessments. It is shown that the procedure based on the mean gives a smaller difference between offers than does the procedure based on the median, and the difference between the equilibrium offers becomes smaller as the number of arbitrators increases. We apply the approach of multiple arbitrators to final-offer arbitration with non-identical arbitrators and its modification. We first consider final-offer arbitration with three arbitrators and then consider the general case. In Section 3 we introduce a new arbitration model with multiple arbitrators. Under this procedure, the Referee takes into account the only assessments of the arbitrators which lie between the players’ offers. In the case of both players obtaining an equal number of votes the Referee’s decision is (x + y)/2. For each procedure, the payoff function H(x, y) and the equilibrium conditions are derived. The optimal offers x⁄ and y⁄ are computed for cases where the arbitrators’ assessments are either all uniformly or all normally distributed. Finally, we analyse the equilibrium when the parameters of these distributions are assumed to be random variables. 2. Arbitration with multiple arbitrators
⇑ Corresponding author. E-mail addresses: (J. Tokareva).
[email protected] (V. Mazalov),
[email protected]
0377-2217/$ - see front matter Ó 2011 Elsevier B.V. All rights reserved. doi:10.1016/j.ejor.2011.09.014
Imagine that Players 1 and 2 submit their offers x and y to a Referee. We assume that there are an odd number, N = 2n 1, of
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V. Mazalov, J. Tokareva / European Journal of Operational Research 217 (2012) 198–203
arbitrators. Each arbitrator has her own assessment. After observing the offers, each arbitrator votes on which offer should be selected. Suppose that each arbitrator selects the offer which is closest to her assessment and informs the Referee about her decision. The Referee counts the votes and declares the player obtaining the most votes (at least n votes) to be the winner. Let the assessments of the arbitrators be independent random variables a1, . . . , a2n1 with CDFs F1(a), . . . , F2n1(a). Both players have the same beliefs about the arbitrators’ assessments and these beliefs are common knowledge. For y < x the FOA procedure states that the Referee prefers the offer y if at least n arbitrators have assessments 6(x + y)/2. Otherwise, the Referee chooses the offer x. 2.1. Model with three arbitrators
where
nn oo x þ yo n xþy [ ai ; aj 6 < ak ; i – j – k a1 ; a2 ; a3 6 2 2 x þ y x þ y x þ y ¼ F1 F2 F3 2 2 2 x þ y x þ y x þ y F2 1 F3 þ F1 2 2 2 x þ y x þ y x þ y 1 F2 F3 þ F1 2 2 2 x þ y x þ y x þ y þ 1 F1 F2 F3 : 2 2 2
ð3Þ
Denote the expression in square brackets in (3) as G(m). Thus, we can rewrite these equations as
(
1 x y ¼ GðmÞ ;
x þ y ¼ 2m; or
(
1 x ¼ m þ 2GðmÞ ; 1 y ¼ m 2GðmÞ ;
Example 1. Suppose that the assessments of the arbitrators are uniformly distributed in the intervals [1, 1],pffiffi[0, 1] and [1, 2], respectively. The solution of Eq. (2) is m ¼ 321. Consequently, pffiffi 5 33 GðmÞ ¼ 6 , and the equilibrium strategies are
8 pffiffiffi 4 > < x ¼ 11 2 3 1 0:896; pffiffiffi > 1 : y ¼ 11 3 3 7 0:164:
It is easy to see that the second-order optimality condition is satisfied. Notice that if m is the median of the distribution F1, then (2) can be presented as
or
F 2 ðmÞ þ F 3 ðmÞ ¼ 1: Hence, if the equality F2(m) + F3(m) = 1 is satisfied at the point m, then the median m is the solution of (2). For instance, if F1, F2, F3 are normal distributions with equal standard deviations and mean values k, 0, k where k is any real number, then the solution of (2) is m = 0.
ð1Þ
where for convenience we write Fi = Fi((x + y)/2), i = 1, 2, 3. The first order condition for the equilibrium in this game is @ H(x, y)/@x = 0, @H(x, y)/@y = 0. This yields
@H ¼ 1 ðF 1 F 2 þ F 1 F 3 þ F 2 F 3 2F 1 F 2 F 3 Þ @x y x 0 F 1 ðF 2 þ F 3 Þ þ F 02 ðF 1 þ F 3 Þ þ F 03 ðF 1 þ F 2 Þ þ 2 2ðF 01 F 2 F 3 þ F 1 F 02 F 3 þ F 1 F 2 F 03 Þ ¼ 0
Example 2. Let F1, F2,F3 normal distributions with variance 1 and mean values 2, 0, 1, respectively. Eq. (2) gives m = 0.141. It follows that G(m) = 0.483. Hence the optimal strategies are x⁄ = 0.894, y⁄ = 1.176. Eqs. (2) and (3) can be simplified if the medians of all the distributions F1, F2, F3 are equal. Denote this common median by m. In this case, m satisfies Eq. (2). Eq. (3) yields
xy¼
and
2 ; F 01 ðmÞ þ F 02 ðmÞ þ F 03 ðmÞ
or
@H ¼ ðF 1 F 2 þ F 1 F 3 þ F 2 F 3 2F 1 F 2 F 3 Þ @y y x 0 þ F 1 ðF 2 þ F 3 Þ þ F 02 ðF 1 þ F 3 Þ þ F 03 ðF 1 þ F 2 Þ 2 2ðF 01 F 2 F 3 þ F 1 F 02 F 3 þ F 1 F 2 F 03 Þ ¼ 0:
xy 1 ¼ ; 2 f1 ðmÞ þ f2 ðmÞ þ f3 ðmÞ where f1, f2, f3 are the densities of F1, F2, F3, respectively. Finally, we obtain the equilibrium in this case
Subtracting the second of these equations from the first gives (x + y)/2 = m, where m satisfies the equation
1 : 2 ð2Þ
F 1 ðmÞF 2 ðmÞ þ F 1 ðmÞF 3 ðmÞ þ F 2 ðmÞF 3 ðmÞ 2F 1 ðmÞF 2 ðmÞF 3 ðmÞ ¼
Moreover,
þF 03 ðmÞðF 1 ðmÞ þ F 2 ðmÞÞ 2ðF 01 ðmÞF 2 ðmÞF 3 ðmÞ 1 þF 1 ðmÞF 02 ðmÞF 3 ðmÞ þ F 1 ðmÞF 2 ðmÞF 03 ðmÞÞ :
1 1 1 F 2 ðmÞ þ F 3 ðmÞ þ F 2 ðmÞF 3 ðmÞ F 2 ðmÞF 3 ðmÞ ¼ ; 2 2 2
Simplifying, we obtain
Hðx; yÞ ¼ x þ ðy xÞðF 1 F 2 þ F 1 F 3 þ F 2 F 3 2F 1 F 2 F 3 Þ;
x y ¼ F 01 ðmÞðF 2 ðmÞ þ F 3 ðmÞÞ þ F 02 ðmÞðF 1 ðmÞ þ F 3 ðmÞÞ
where m satisfies Eq. (2).
Let us derive the solution in the case of three arbitrators. In this case, the Referee chooses y, if a1, a2, a3 6 (x + y)/2, or if a1, a2 6 (x + y)/2, or if a1, a3 6 (x + y)/2, or if a2, a3 6 (x + y)/2. Otherwise, the Referee chooses the offer x. So, the payoff function in this case is nn oo x þ yo n xþy < ak ; i – j – k Hðx; yÞ ¼ yP a1 ; a2 ; a3 6 [ ai ; aj 6 2 2 nn oo x þ yo n xþy þ xP a1 ; a2 ; a3 P [ ai ; aj P > ak ; i – j – k ; 2 2
P
y ¼ m
1 ; f1 ðmÞ þ f2 ðmÞ þ f3 ðmÞ
x ¼ m þ
1 : f1 ðmÞ þ f2 ðmÞ þ f3 ðmÞ ð4Þ
For instance, if a1, a2, a3 are normal random variables with the same mean value m and different standard deviations r1, r2, r3, we obtain
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V. Mazalov, J. Tokareva / European Journal of Operational Research 217 (2012) 198–203
pffiffiffiffiffiffiffi 2p y ¼m 1 ; 1 1 þ r1 r2 þ r3
pffiffiffiffiffiffiffi 2p x ¼mþ 1 : 1 1 þ r1 r2 þ r3
y ¼ m D;
where
"
If a1, a2, a3 are uniformly distributed in [0, 1], then
y ¼
1 1 1 ¼ ; 2 3 6
x ¼
x ¼ m þ D;
D¼
1 1 5 þ ¼ : 2 3 6
Notice that this equilibrium differs from the solution y⁄ = 0, x⁄ = 1 in the game with one arbitrator (Gibbons, 1992).
2n n
#1
2n3 2X n1 1 fi ðmÞ : 2 i¼1
From this theorem, it follows that in the case of normal distributions F1, . . . , F2n1 with the same mean value m and different standard deviations ri, i = 1, . . . , 2n 1, the optimal offers are y⁄ = m D, x⁄ = m + D with
In the arbitration game with 2n 1 > 3 arbitrators the payoff function is
pffiffiffiffiffiffiffi 2p 1
D¼ : 2n 12n 1 8 1 2 r1 þ þ rN n
Hðx; yÞ ¼ x þ ðy xÞ 0
When the arbitrators’ assessments are i.i.d., we obtain the following result.
2.2. Model with 2n 1 arbitrators
1
[
B PB @
ai1 ; . . . ; aik 6
i1 ––ik k¼n;nþ1;...;2n1
C xþy < aikþ1 ; . . . ; ai2n1 C A; 2
Corollary 2. Let F1 = F2 = = F2n1 = F. Then the equilibrium in the arbitration game is
y ¼ m D;
or
X
Hðx; yÞ ¼ x þ ðy xÞ
F i1 ; . . . ; F ik F ikþ1 ; . . . ; F i2n1 ;
ð5Þ
i1 ––ik k¼n;nþ1;...;2n1
x þ y x þ y @H ¼1F þ ðy xÞF 0 ¼ 0; @x 2 2
F i1 ; . . . ; F ik F ikþ1 ; . . . ; F i2n1 :
ð6Þ
F i1 ; . . . ; F ik F ikþ1 ; . . . ; F i2n1 ¼
i1 ––ik k¼n;nþ1;...;2n1
1 2
2n 1 1 1 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ pffiffiffiffiffiffiffi : 2 pn n 1 2p2n
2n
Hence,
20
i1 ––ik k¼n;nþ1;...;2n1
C7 7 F i1 ; . . . ; F ik F ikþ1 ; . . . ; F i2n1 C A5 :
ð8Þ
We can determine the optimal values x and y from Eqs. 7 and 8. Let us suppose that the arbitrators’ assessments have a common median m. It is easy to see that m satisfies Eq. (7). This immediately yields
xþy ¼ m: 2
as n tends to 1. We see that the difference between equilibrium offers becomes smaller as the number of arbitrators increases. 3. Modified arbitration with multiple arbitrators
10 31 X
pffiffiffi 2pr 1 pffiffiffi ! 0 8 n
ð7Þ
and the difference x y is equal to
B 16 6B 2 4@
pffiffiffiffiffiffiffi 1 2pr
D¼ : 2n 12n 8n 2 n
D
From these equations, we immediately obtain that (x + y)/2 = m satisfies the equation
xy¼
For normal distributions with a common standard deviation r, (10) yields
4
i1 ––ik k¼n;nþ1;...;2n1
X
#1
2n3 1 nf ðmÞ : 2 n
2n
This coincides with the result of paper (Hurley, 2003).
where we denote
x þ y ¼ F 2
"
Using Stirling’s formula, we obtain the following estimate
x þ y x þ y @H ¼F þ ðy xÞF 0 ¼ 0; @y 2 2
F¼
x ¼ m þ D;
where
D¼
where F ¼ 1 F and Fi = Fi((x + y)/2), i = 1, . . . , 2n 1. In the sum (5) we have one term containing no F i ; 2n 1 terms including only one
2n 1 F i , and terms including a product of the form F j1 . . . F jk . k Differentiating (5), we obtain the following system of equations
X
ð10Þ
ð9Þ
The proof of the following theorem is given in Appendix A. Theorem 1. Let the medians of all the distribution functions F1, . . . , F2n1 be equal to m. In this case, the equilibrium in the arbitration game is
In this section we consider a new arbitration procedure with multiple arbitrators and a Referee who takes into account only the assessments of the arbitrators which belong to interval between the player’s offers, i.e. [y, x], when x P y. Assume that the assessments of the arbitrators are i.i.d. random variables a1, . . . , a2n1 with CDF F(a) and density f(a). Assume that y < x, and k arbitrators prefer the offer y, i.e. y 6 ai 6 (x + y)/2, i = i1, . . . , ik and m arbitrators prefer the offer x, i.e. (x + y)/2 6 ai 6 x, i = i1, . . . , im. Here, k + m 6 2n 1. After counting the votes, the Referee selects the offer y if k > m, and the offer x if k < m. Otherwise, the Referee’s decision is (x + y)/2. First, we find the solution in the arbitration game with one arbitrator. If the arbitrators’ assessment belongs to the interval [y, (x + y)/2], then the final decision is y. If the assessment belongs ((x + y)/2, x] then the decision is x. Otherwise, the decision is (x + y)/ 2. Thus, the payoff function in this game is
V. Mazalov, J. Tokareva / European Journal of Operational Research 217 (2012) 198–203
x þ y x þ y x þ y Hðx; yÞ ¼ y F FðyÞ þ x FðxÞ F þ ðFðyÞ 2 2 2 þ 1 FðxÞÞ:
xþy ð1 p1 p2 Þ 2 x þ y x y 3 þ v u3 þ 3½v 2 ð1 v Þ u2 ð1 uÞ ¼¼ 2 2 þ3ðv uÞð1 u v Þ2 :
Hðx; yÞ ¼ yp1 þ xp2 þ
We can derive the equilibrium from the first-order conditions
x þ y FðxÞ þ FðyÞ x þ y 1 @H x y ¼ f ðxÞ f þ F þ ¼ 0; @x 2 2 2 2 2 x þ y FðxÞ þ FðyÞ x þ y 1 @H x y ¼ f ðyÞ f þF þ ¼ 0: @y 2 2 2 2 2
201
The first-order condition for equilibrium is @Hðx;yÞ ¼ 0; @Hðx;yÞ ¼ 0. @x @y For a distribution which is symmetric with respect to the point 0, the conditions x = y and u(x, y) = v(x, y) must be satisfied. Hence, the equilibrium value of x satisfies the equation
1 þ xð6v 2 6v þ 3Þðf ðxÞ f ð0ÞÞ ¼ 0; 2 where
Example 3. Let a have the standard normal distribution N(0, 1). The symmetry of the problem yields x = y and we can rewrite the equation for the equilibrium in the form x2
xð1 e 2 Þ ¼
rffiffiffiffi
p 2
v ðxÞ ¼ FðxÞ Fð0Þ;
f ðxÞ ¼ F 0 ðxÞ:
Example 4. Let F(x) be the standard normal distribution N(0,1). Then equilibrium condition gives x⁄ 1.352. Hence, the equilibrium is
:
From this, we obtain
y 1:352;
y 1:667;
We see that the difference between the equilibrium offers is smaller than in the model with one arbitrator.
x 1:667:
It is not difficult to check that this is the equilibrium of the game. Notice, that the equilibrium under the original FOA procedure (Gibbons, 1992) is
rffiffiffiffi
y ¼
p 2
1:253;
x ¼
rffiffiffiffi
p 2
1:253:
We see that this new arbitration procedure leads to the players proposing values which differ more than under the FOA procedure. However, as the number of arbitrators increases, the difference between the equilibrium offers falls.
x 1:352:
3.2. Arbitration with 2n 1 arbitrators Now imagine that there are 2n 1 arbitrators. In Appendix B it is shown that the final decision of the Referee is y with probability
p1 ¼
n n1 X X 2n 1 2n 1 i un1þi ð1 uÞni þ u n1þi i i¼1 i¼1
i1 X 2n 1 i v j ð1 u v Þ2n1ij j j¼0
3.1. Model with three arbitrators
and x with probability
Let the arbitration panel consist of three arbitrators. Assume that the assessments of the arbitrators are i.i.d random variables a1, a2, a3 with continuous CDF F(a). xþy Let and xþyy < x. Define the intervals L and R by L ¼ y; 2 R ¼ 2 ; x , and set
p2 ¼
u¼F
x þ y 2
FðyÞ;
v ¼ FðxÞ F
x þ y : 2
n n1 X X 2n 1 2n 1 n1þi v ð1 v Þni þ vi n1þi i i¼1 i¼1
i1 X 2n 1 i j u ð1 u v Þ2n1ij : j j¼0
In all other cases the decision is (x + y)/2. Hence, the payoff in this game is
"
n 2n 1 xþy xy X þ ðv n1þi ð1 v Þni 2 2 n 1 þ i i¼1
i1 n1 X 2n 1 X 2n 1 i un1þi ð1 uÞni Þ þ i j i¼1 j¼0 1 u v Þ2n1ij ðv i uj v j ui Þ : ð11Þ
Hðx; yÞ ¼ Thus, the final decision of the Referee is y in the following three cases: Firstly, when the assessments of all the arbitrators lie in L. The probability of this is u3. Secondly, when the assessments of any two arbitrators are in L, but the third assessment is not. The probability of this is 3u2(1 u). Thirdly, when the assessment of one of them is in L, but the assessments of the other arbitrators are not in L or R. The probability of this is 3u(1 u v)2. So, the probability of the final decision being y is
p1 ¼ u3 þ 3u2 ð1 uÞ þ 3uð1 u v Þ2 : Analogously, the decision of the Referee is x with probability
p2 ¼ v 3 þ 3v 2 ð1 v Þ þ 3v ð1 u v Þ2 : In all other cases the decision is (x + y)/2. The probability of this is 1 p1 p2. Hence, the payoff function is
Denote the expression in square brackets in (11) by S(v, u). We can see that S(v, u) is an anti-symmetric function, i.e. S(v, u) = S(u, v). Hence, for v = u
Sðu; uÞ ¼ 0: The first-order equilibrium conditions for (x, y) is
@H 1 1 x y @S 1 x þ y @S 1 x þ y f ðxÞ f ¼ þ Sðv ; uÞ þ þ f ¼0 @x 2 2 2 @v 2 2 @u 2 2 and
@H 1 1 x y @S 1 x þ y @S 1 x þ y ¼ Sðv ; uÞ f þ f ðyÞ f ¼ 0: @y 2 2 2 @v 2 2 @u 2 2
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V. Mazalov, J. Tokareva / European Journal of Operational Research 217 (2012) 198–203
Table 1 Equilibrium strategies for various numbers of arbitrators. n x⁄
1 1.667
2 1.352
3 1.203
4 1.111
5 1.047
Now consider the case where the CDF F(x) is symmetric with respect to zero. Then, by symmetry, at equilibrium the equalities x = y and u = v hold. Note that S(v, u) = S(u, v). This yields @Sðv ;uÞ vÞ ¼ @Sðu; . Thus, for u = v and x = y we obtain @v @u
1 @Sðv ; v Þ þ xðf ðxÞ f ð0ÞÞ ¼ 0: 2 @v
ð12Þ
From (11), we obtain
n 2n 1 @Sðv ; v Þ X ¼ v nþi2 ð1 2vÞn1i ððn 1 þ iÞð1 v Þ ðn iÞv Þ @v n1þi i¼1
i1 n1 X 2n 1 X 2n 1 i ð13Þ þ ð1 2v Þ2n1ij ði jÞv iþj1 : i j i¼1 j¼0 Eqs. (12) and (13) can be used to numerically calculate the equilibrium strategies in this game. In Table 1, we present the results of numerical calculations for various values of n. We see that the difference between equilibrium offers decreases when the number of arbitrators increases. 4. Conclusion In this paper we have considered two arbitration procedures for settling labor disputes involving multiple arbitrators. Each arbitrator follows FOA rules and the Referee takes the final decision on the basis of majority voting. We introduce a new arbitration procedure in which the Referee does not take into account assessments of the arbitrators which do not lie between the offers. The equilibrium strategies of the players under both procedures are derived in analytical form. We estimate the expected difference between the equilibrium offers and show that this difference becomes smaller as the number of arbitrators increases. This can be used to determine the optimal number of arbitrators. This work was supported by the Division of Mathematical Sciences of the Russian Academy of Sciences and the Analytical program ‘‘Development of Scientific Potential of the Higher School’’ (project 1.8.10). Appendix A
10
0
X i1 ––ik k¼n;nþ1;...;2n1
2n1 X
Fi
Y
7 0.962
8 0.932
9 0.906
10 0.884
!0 Fj
;
j–i
i¼1
contains 2n 1 terms with f1, of which 2n 2 are positive and one is negative. The derivative 2n1 X
FiFj
i;j¼1;i–j
Y
!0 Fk
;
k–i;j
2n 1 2n 1 terms with f1, of which ð2n 2Þ are 2 2 positive and 2n 2 are negative. Arguing in the same way, for any k 6 n 1 in the derivative contains
X
F i1 ; . . . ; F ik
i1 ––ik
!0
Y
Fj
;
j–i1 ;...;ik
2n 1 there are terms with f1. Among them k
2n 1 2n 2 2n 2 terms are positive and terms k k1 k1 are negative. So the coefficient of f1 (and consequently of each fi) is equal to
s ¼ 1 þ ð2n 2 1Þ þ
2n 2 2 ; n2
2n 1 2
2ð2n 2Þ þ þ
2n 1 n1
Simplifying this yields
s¼
2n : n
Now we can present (A1) in the form
s
2n1 2n1 2n1 2n1 X X 2n 1 1 fi ðmÞ ¼ fi ðmÞ; 2 2 n i¼1 i¼1
and (8) yields
Proof of Theorem 1. Let us simplify Eq. (8). Consider the value of the derivative
B B @
6 1.000
C F i1 ; . . . ; F ik ð1 F ikþ1 Þ; . . . ; ð1 F i2n1 ÞC A;
#1
2n2 2n1 X 1 fi ðmÞ : 2 n i¼1
2n
ðA2Þ
(A2) and (9) establish the theorem.
ðA1Þ
at the point m, where m is the median of the distribution Fi. So, Fi(m) = 1/2, i = 1, . . . , 2n 1. Hence, the absolute value of each term in this sum is of the following form
2n2 x þ y0 1 2n1 1 fi ðmÞ ¼ fi ðmÞ; 2 2 2
xy¼
"
i ¼ 1; . . . ; 2n 1:
Such a term is positive when we differentiate Fi and negative when we differentiate F i . From the form of (A1), it is sufficient to calculate the coefficients of the terms which include f1(m). The derivative (F1. . .F2n1)0 contains only one term with f1 ¼ F 01 . The derivative
Appendix B Consider all the possible results of voting such that the final decision of the Referee is y. Firstly, suppose that all 2n 1 assessments of the arbitrators lie in the interval L. The probability of this is equal to u2n1. Secondly, suppose that 2n 2 arbitrators’ assessments belong to L and one of them does not. The probability of this is
2n 1 u2n2 ð1 uÞ. 2n 2 Thirdly, suppose 2n 3 arbitrators’ assessments are in L and the other two are not in L. The probability of this is
2n 1 u2n3 ð1 uÞ2 . 2n 3
V. Mazalov, J. Tokareva / European Journal of Operational Research 217 (2012) 198–203
and so on, until the n-th such case where n arbitrators’ assessments are in L and the other n 1 are not in L. The probability of
2n 1 n this is u ð1 uÞn1 . n Moreover, the final decision is also y in the following cases. If n 1 arbitrators’ assessments are in L (the probability of this
2n 1 n1 is u Þ, and among the other n, exactly j assessments ben1 long to R and the remaining n j assessments are neither in L nor in R, where j = 0, . . . , n 2. The probability of this is
ð1 u v Þn þ
n
þ
1
v ð1 u v Þn1 þ
n
n2
v n2 ð1 u v Þ2 :
Or, if n 2 arbitrators’ assessments are in L (the probability of this
2n 1 n2 is u Þ, and among the other n + 1, exactly j assessments n2 belong to R and the remaining n + 1 j assessments are neither in L nor in R, where j = 0, . . . , n 3. The probability of this is
ð1 u v Þnþ1 þ
nþ1
v ð1 u v Þn þ 1
nþ1 v n3 ð1 u v Þ4 : þ n3
and so on. Finally the decision is y, if one of the assessments belongs to L
2n 1 (with probability u), and the other 2n 2 assessments 1 are neither in L nor in R. The probability of this is (1 u v)2n2. Consequently, the probability of the final decision being y is equal to
p1 ¼
n X 2n 1 un1þi ð1 uÞni n1þi i¼1
i1 n1 X 2n 1 i X 2n 1 i þ v j ð1 u v Þ2n1ij : u i j i¼1 j¼0
203
Arguing in the same way, we can find the probability that the decision is x
p2 ¼
n X 2n 1 n1þi
n1 X 2n 1
v n1þi ð1 v Þni
i¼1
þ
i¼1
i
vi
i1 X 2n 1 i j u ð1 u v Þ2n1ij : j j¼0
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