Proceedings of The National Conference
On Undergraduate Research (NCUR) 2005
Virginia Military Institute and
Washington and Lee University
Lexington, Virginia
April 21 - 23, 2005
Graphical Approach and Property Tests in Fair Division Problems with
Indivisible Objects and Monetary Compensation
Aeron Huang
Department of Mathematics
Goshen College
1700 South Main Street
Goshen, Indiana 46526, USA
Faculty Advisor: Dr. David Housman
Abstract
This paper introduces a graphical approach toward the study of several properties in a fair division problem with
indivisible objects and monetary compensation. The properties include efficiency, envy-freeness, individual rational
and individual stand-alone. With observations into the dynamics of the set of allocations affected by the change of
entries in an object valuation matrix, three property tests are discovered.
Key words: fair division, allocation, object valuation matrix, efficient, envy-free, individual stand-alone,
individual rational,
1. Concepts and notations
Fair division problems involve dividing an object or a bundle of objects among a group of players fairly. The objects
to be discussed in this paper are indivisible; in other words, any subset of players cannot share one single object or
obtain part of the object as a result of the fair allocation. In addition, money compensation is allowed. An example
for this would be an inheritance problem: three siblings trying to fairly allocate an estate consisting of four objects.
Assume that the total number of player is m, and the total number of objects available is n.
A fair allocation is based on bidding results. Before using a particular method to acquire fairness, each player has to
bid a value for each object. Name u pi player i’s bid on object p. In compensation, players pay a certain amount of
money for the object they win. This amount of money goes to a money pot that is to be divided among players at the
end of the allocation process. An allocation ( S , M ) , as the result, consists of 1) a partition of the set of
objects S1 , S 2 ...S m each of which is distributed to the subscripted player; 2) real numbers
m
which is the net amount of receipt for each player in monetary term. In addition,
M
i 1
i
M 1 , M 2 ...M m each of
 0 . Note that in this
paper, the second subscript is always the player to evaluate the objects or player who wins the objects represented in
the first subscript
For the allocation ( S , M ) the valuation of player i is,
v(i ) 
u
pS i
pi
 Mi
In fact, in player j’s perspective,
u
pN
pj
is a different number, since player i and player j could bid differently. We
define
Vij 
player j’s evaluation on the objects player i wins.
u
pSi
pj
Vii is thus i’s evaluation on the object he wins.
Also,
v(ij )  Vij  M i
would be the total value player i wins in the perspective of j, including money terms.
We can summarize a fair division problem with an m m square matrix. Rows of this matrix are objects or bundles
of objects, each of which is won by one certain player. Columns of the matrix are players. The object valuation
matrix looks like:
V11 V12
V V
 21 22
V  .

 .
Vm1 .

.
.





V( m 1) m 
Vmm 
V1m
.
.
.
.
. Vm ( m 1)
Along the diagonal of this matrix are the winning bids of each player. All other off-diagonal entries are some
player’s evaluation on the object (bundle of objects) won by the highest bidder in the row of that object (bundle of
objects). This form is helpful in analyzing envy-free solutions as well as individual rational and stand-alone
solutions, with the aid of the graphic approach, all of which to be discussed later in this paper.
An allocation is efficient if no other allocation is able to make every player at least as well of and some player
strictly better off. It can be shown that in order for an allocation to be efficient, the highest bidder of an object
should be awarded that object. In other words, for player i and j, it must satisfy that Vii  Vij .
An allocation is individual rational if each player values his share of the allocation at least one-mth of his value of
the total estate:
v(ii ) 
1 n
  u pi . An allocation is individual stand-alone if each player values his own share of
m p 1
the allocation no more than his value of the total estate: v(ii ) 
n
u
p 1
rational and individual stand-alone must satisfy
pi
. An allocation that satisfies both individual
n
1 n
  u pi  v(ii )   u pi for each player i.
m p 1
p 1
An allocation is envy-free if no player would prefer another’s bundle to his or her own: That is,
v(ij )  v( jj)
 j . In a fair division problem with two players A and B, two inequalities are required to ensure
envy-freeness: A not-envy B ( v ( BA )  v ( AA) ), and B not-envy A ( v( AB)  v( BB ) ). For an n-player game,
for all players i
n(n  1) inequalities are required to ensure envy-freeness; this can be shown by multiplying the number of columns
(n) and the entries in each row except for entry along the diagonal (n-1) from the object valuation matrix.
2. An example
Suppose in a fair division problem players are A, B and C, and objects are a, b, c and d. Here’s a diagram of their
biddings:
a
b
c
d
A
B
C
28
5
15
10
11
3
26
3
17
14
23
6
Table 1: 3 player bidding scheme
Each entry in the table is a bid. For example,
u aA  28 and u cB  3 . A possible allocation is to give player A object
a and c; give player B object b and d. Meanwhile, we ask A to pay 54 and B to pay 34 to the money pot. Next, when
dividing the pot among A, B and C, A gets 8, B gets 50 and C gets 30. Therefore, the net monetary gains are:
M A  54  8  46 , M B  34  50  16 and M C  30 . Both B and C are gaining money from this
allocation, while A has to pay. Also note that
thus v( A) 
u
p  a ,c
M
i  A , B ,C
pA
i
 46  16  30  0 . Total gain of player A is
 M A  28  26  46  8 . Also in this example, VBA  ubA  u dA  10  14  24 , items
that B wins are worth 24 in total in A’s perspective;
V AA  u aA  u cA  28  26  54 , A values 54 on his own
winning objects; v( BA)  VBA M B 24  16  40 , A values B’s total gain as 40, including money term. The
Object Valuation Matrix would look like:
54 8 32
V  24 34 9 
 0 0 0 
This is an efficient allocation, since the diagonal entries of the matrix are their row max, which means the object is
rewarded to the highest bidder. For example, VAA  VAB , and VBB  VBC . If we instead give object a to player B,
then the new Object Valuation matrix would look like;
26 3 17 
V  52 39 24
 0 0 0 
As we can see,
VBB  VBA , an inefficient allocation just occurred.
For A to be individual rational, A must be getting at least one-third of what he thinks the entire estate worth:
1
v( A)  (54  24  0)  26 ; any final value of allocation less than 26 will not be individual irrational for player
3
A. For player A to be individual stand alone, VA  (54  24  0)  78 , thus, any final value of allocation that’s
greater than 78 wouldn’t satisfy individual stand-alone property for player A. To satisfy both properties, it must satisfy
that 26  v( A)  78 .
In this above allocation: v( BA)  40  8  v( AA) , thus A will envy B. However, if we give 50 to A and 8 to B
M A  54  50  4 ,
and M B  34  8  26 ,as a result, v( AA)  54  4  50  2  24  26  v( BA) , A would not envy B
from the money pot, and this change the money allocation to:
anymore.
3. The graphical approach
It would be interesting to synthesize the properties efficiency, individual rational and stand-alone, and envy-free of a
fair division problem with a single device. We can achieve this with a graphical tool. This graphical approach is
constrained to a three-player game and it is served as a tool to lead to some observation of nature between allocation
result and the bidding of the players. From there on, higher dimensional result is conjectured and proved.
Construct a coordinate system in three-dimensional space. Choose the axis to be v( A), v( B), v(C ) . Thus, every point
in this coordinate system is the valuation of possible allocation. All allocations that sum up to a constant, that is,
v( A)  v( B)  v(C )  u would be a plane in the 3-D coordinate system. In addition, suppose that only positive
values are considered. As we can see, the intersection of the plane and the coordinate system compose a triangle
(Figure 1).
v(C)
C
B
v(B)
A
v(A)
Figure 1
At three verticies of the triangle, are the maximun value each player can get out of the allocation. For example at point
A, player A gets the most he can from the allocation and player B and C get nothing. An efficient allocation lies on the
plane out-most in the positive direction.
We need to do a transformation from the three-dimensional space to the two-dimensional surface of the triangle.
Suppose any point ( x, y ) in the plane can be affinely transformed from one point (v( A), v( B ), v(C )) in the 3-D
plane. Inversely, any point (v( A), v( B ), v(C )) in the 3-D system can be affinely transformed from one ( x, y ) on
the 2-D plane:
v( A)  a b 
g 
v( B)    c d    x    h 

 
  y  
v(C )  e f     i 
Suppose points in 3-D space are v( A)  (v,0,0) , v ( B )  (0, v,0) , v (C )  (0,0, v ) , if the maximum value each
player can get is
v . Suppose points in the 2-D plane are: A(0,0) , B(
2
3
v,0) , C (
v
3
, v) . We are regulating this
triangle to be an equilateral triangle.
Perform the affine transformation, we can solve for the transformation matrix:

3

2
v( A)  
v( B)    3

  2
v(C ) 

 0
1
 
2
v 
1   x  
   0

2   y  
0


1 
To plot the graph, we just need to substitute v( A), v( B), v(C ) with the function of x and y.
A graphing routine was made with Maple to illustrate areas of intersection of the several inequalities which are
generated from the properties of envy-freeness, individual rational and individual stand-alone. (A copy of the routine
is listed in appendix 2)
3. Results from envy-free property:
Three sets of inequalities are needed to condition envy-freeness in a three player game:
Set 1, in figure 3(a):
Line 1: Player A not envy player B:
v( BA)  v( AA)  v( A)  M B  VBA  v( A)  (v( B)  VBB )  VBA
Line 2: Player B not envy player A: v( AB)  v( BB )  v( B)  M A  VAB  v( B)  (v( A)  VAA )  VAB
Set 2, in figure 3(b):
Line 3: Player A not envy player C:
v(CA)  v( AA)  v( A)  M C  VCA  v( A)  (v(C )  VCC )  VCA
Line 4: Player C not envy player A:
v( AC)  v(CC)  v(C )  M A  V AC  v(C )  (v( A)  V AA )  V AC
Set 3, in figure 3(c):
Line 5: Player B not envy player C:
v(CB)  v( BB )  v( B)  M C  VCB  v( B)  (v(C )  VCC )  VCB
Line 6: Player C not envy player B:
v( BC )  v(CC)  v(C )  M B  VBB  v(C )  (v( B)  VBB )  VBC
Applying the affine transformation, we obtain the following graphs. Note that lines in these graphs are the equability
forms of the inequalities. The area between a line and its correspondent point is the group of allocations that satisfies
that particular envy-free inequality.
2
C
1
C
C
5
3
4
B
A
6
B
A
B
A
(a)
(b)
(c)
Figure 2
In fact, the areas of intersection of these six inequalities, as one can tell from the form of these inequalities, are only
dependent upon entries of the object valuation matrix:
V AA V AB V AC 
V

 BA VBB VBC 
VCA VCB VCC 
of this three-player game. Changes of entries from the matrix can affect the area of intersection dramatically. For
example, if we start with the bidding scheme shown in figure 3(a), as VCB increase (figure 3(b) and 3(c)), B is more
likely to envy C, leaving the shaded envy-free area smaller and smaller.
(a)
60 30 30
Oa  30 60 30
30 30 60
(b)
60 30 30
Ob  30 60 30
30 40 60
(c)
60 30 30
Oc  30 60 30
30 59 60
Figure 3
Considering the complexity of taking into account all six inequalities in a three-player game, or, n(n-1) inequalities
in an n-player game, it might be helpful to know when some of the inequalities are irrelevant to the solution of set of
envy-freeness (see figure 4). In fact, there is a way to check, at least in a three-player game:
Figure 4
Theorem 1 (Property Test 1) : To determine whether an allocation is envy-free, we need not check whether player
i not envy player j, if and only if the object valuation matrix satisfies: Vki  V ji  Vkk  V jk .
Proof:
i) To prove that if Vki  V ji  Vkk  V jk , then i not envy player j is irrelevant to the solution set of envyfreeness, assume v(i ), v( j ), v(k ) are the valuations of an allocation satisfying all envy-free constraints except for i
not envy j. It follows that:
v(ii )  v(ki)
 Vki  M k
 V ji  Vkk  V jk  M k
because
Vki  V ji  Vkk  V jk
because
v(kk)  v( jk )
 V ji  v(kk)  v( jk )  M j
 V ji  M j
 v( ji)
Therefore, any allocation that satisfies Vki  V ji  Vkk  V jk and all envy-free allocations but i not envy j, should
also satisfy i not envy j.
ii) To prove by contrapositive that if i not envy j is irrelevant, then Vki  V ji  Vkk  V jk , assume
Vki  V ji  Vkk  V jk . Also, choose the following allocation point ( v(i ), v( j ), v(k ) ), which is obtained through
calculating the intersection of the equation, i not envy k and the equation k not envy j.
1
v(i )  (V jk  2Vki  u  V jj  2Vkk )
3
1
v( j )  (2V jk  Vki  u  2V jj  Vkk )
3
1
v(k )  (V jk  Vki  u  V jj  Vkk )
3
This allocation satisfies all five envy-free inequalities except for i not envy
j since:
Note that v(i )  v( j )  v(k )  u , verify that this is an allocation.
By substitution: v(i )  v(k )  Vki
 Vkk , v(k )  v( j )  V jk  V jj
since V jj  V jk and Vkk  Vkj
v( j )  v(k )  V jk  V jj  Vkj  Vkk
then  V jk  V jj  Vkj  Vkk  0
v( j )  v(i )  V jk  Vki  V jj  Vkk  Vij  Vii since V jj  V jk , Vkk  Vki , and Vii  Vij
then,  V jk  Vki  V jj  Vkk  Vij  Vii  0
Vkk  Vki , Vii  Vik
then,  Vki  Vkk  Vik  Vii  0
v(k )  v(i)  Vki  Vkk  Vik  Vii
since
Also since,
v(ii )  v( ji)  v(ii )  v( jj)  V jj  V ji  v(i )  v( j )  V ji  V jj
Therefore,
v(ii )  v(ki),
v(kk)  v( jk)
v( jj)  v(kj),
v( jj)  v(ij ),
v(kk)  v(ik ),
Substitute v (i ) and v ( j ) ,
v(ii )  v( ji)  v(i )  v( j )  V ji  V jj  V jk  Vki  V jj  Vkk  V ji  V jj  V jk  Vki  Vkk  V ji
According to the assumption Vki  V ji  Vkk  V jk  Vki  V ji  Vkk  V jk  0 , and so v (ii )  v ( ji)  0 .
Thus, v(ii )  v( ji)
Therefore, allocation v(i ), v( j ), v(k ) violates the envy-free relation: i not envy j. Therefore, if i not envy j is
irrelevant then
Vk i  V ji  Vk k  V jk .
4. Results from individual rational and individual stand-alone property:
Again, graphical approach is used to study the property of individual rational and individual stand-alone. The result of
individual rational solutions is in the shape of a triangle similar to the original triangle that contains all the allocation
points (example in Figure 5(a)). The result of individual stand-alone is in the shape of an upside-down triangle
(example in Figure 5(b))
(a)
(b)
If we start with three-player fair division problem with the object valuation matrix:
Figure 5
V AA V AB V AC 
V

 BA VBB VBC 
VCA VCB VCC 
An individual rational solution would satisfy:
An individual stand-alone must satisfy:
1
v( A)  (V AA  VBA  VCA )
3
1
v( B)  (V AB  VBB  VCB )
3
1
v(C )  (V AC  VBC  VCC )
3
v( A)  V AA  VBA  VCA
v( B)  V AB  VBB  VCB
v(C )  V AC  VBC  VCC
The set of individual stand-alone solutions and the set of individual rational solutions always intersect, and they can
never completely overlap each other. There are cases, though, when one is the subset of the other.
Theorem 2. For three-player game, the set of individual stand-alone solutions is a subset of the set of rational
solutions, if and only if Vik  V jk  3V ji  3Vki  3Vij  3Vkj  2Vkk for the permutations of ( i, j , k )=(1,2,3),
(3,1,2) and (2,3,1)
Proof:
i) Suppose Vik  V jk  3V ji  3Vki  3Vij  3Vkj  2Vkk , and valuation v of allocation is individual stand-alone.
We will prove that v(i ), v( j ), v(k ) is individual rational.
Since v is individual stand-alone,
v(i )  Vii  V ji  Vki
1)
v( j )  Vij  V jj  Vkj
2)
1) + 2)  v(i )  v( j )  Vii  V ji  Vki  Vij  V jj  Vkj
3)
since
v(k )  Vkk  M k
4)
3) + 4)  v(i )  v( j )  v(k )  Vii  V jj  Vkk  V ji  Vki  Vij  Vkj  M k
since
v(i )  v( j )  v(k )  Vii  V jj  Vkk
then,
0  V ji  Vki  Vij  Vkj  M k
and
0  3V ji  3Vki  3Vij  3Vkj  3M k  3V ji  3Vki  3Vij  3Vkj  3M k  
where
 is a non-negative real number.
we assumed that,
Vik  V jk  3V ji  3Vki  3Vij  3Vkj  2Vkk
which is
5)
Vik  V jk  Vkk  3Vkk  (3V ji  3Vki  3Vij  3Vkj )
combining this with 5), it yields that
Vik  V jk  Vkk  3Vkk  (3M k   )  3Vkk  3M k    3v(k )  
thus
Vik  V jk  Vkk  3v(k )
and
1
v(k )  (Vik  V jk  Vkk )
3
since this holds for arbitrary k, v is individual rational
ii) Suppose Vik  V jk  3V ji  3Vki  3Vij  3Vkj  2Vkk , we will exhibit an allocation that is individual standalone but not individual rational.
v(i )  Vii  V ji  Vki
1)
v( j )  Vij  V jj  Vkj
2)
v(k )  u  v(i )  v( j )
Clearly, v is an allocation and the individual stand-alone constraint for I and j are satisfied,. That the standalone constraint for player k holds will follow from the following argument showing that the individual
rational constraint for player k is violated.
1) + 2)  v(i )  v( j )  Vii  V ji  Vki  Vij  V jj  Vkj
3)
since
v(k )  Vkk  M k
4)
3) + 4)  v(i )  v( j )  v(k )  Vii  V jj  Vkk  V ji  Vki  Vij  Vkj  M k
since
v(i )  v( j )  v(k )  Vii  V jj  Vkk
then,
0  V ji  Vki  Vij  Vkj  M k
and
0  3V ji  3Vki  3Vij  3Vkj  3M k  3V ji  3Vki  3Vij  3Vkj  3M k
we assumed that,
Vik  V jk  3V ji  3Vki  3Vij  3Vkj  2Vkk
which is
Vik  V jk  Vkk  3Vkk  (3V ji  3Vki  3Vij  3Vkj )
combining this with 5), it yields that
Vik  V jk  Vkk  3Vkk  (3M k )  3Vkk  3M k  3v(k )
thus
Vik  V jk  Vkk  3v(k )
and
1
v(k )  (Vik  V jk  Vkk )
3
thus this allocation is not individual rational for player k.
5)
Theorem 3. For player i, j and k, the set of individual rational solutions is a subset of the set of individual standalone solutions, if and only if 3Vik  3V jk  Vki  V ji  Vij  Vki  2Vii  2V jj
Proof:
i) Suppose 3Vik  3V jk  Vki  V ji  Vij  Vki  2Vii  2V jj , and the valuations v for the allocations is
individual rational, we will prove that v(i ), v( j ), v(k ) is individual stand-alone.
1
v(i)  (Vii  V ji  Vki )  3v(i)  Vii  V ji  Vki
3
 3(Vii  M i )  Vii  V ji  Vki
 2Vii  3M i  V ji  Vki
1)
1
v( j )  (Vij  V jj  Vkj )  3v( j )  Vij  V jj  Vkj
3
 3(V jj  M j )  Vij  V jj  Vkj
 2V jj  3M j  Vij  Vkj
1)  2)
2)
 2Vii  3M i  2V jj  3M j  V ji  Vki  Vij  Vkj
 2Vii  2V jj  (V ji  Vki  Vij  Vkj )  3M i  3M j
 2Vii  2V jj  (V ji  Vki  Vij  Vkj )  3M i  3M j  
where
3)
 is a non-negative real number
We assumed that:
3Vik  3V jk  Vki  V ji  Vij  Vki  2Vii  2V jj
 3Vik  3V jk  2Vii  2V jj  (Vki  V ji  Vij  Vki )
 3Vik  3V jk  3M i  3M j  
Since equation 3)
 3Vik  3V jk  3( M i  M j )    0
 3Vik  3V jk  3( M i  M j )  3M k  3M k  3(v(k )  Vkk )
 3Vik  3V jk  3v(k )  3Vkk
 v(k )  Vik  V jk  Vkk
Since M i  M j  M k  0
For any arbitrary k, v is individual stand-alone
ii) ) Suppose 3Vik  3V jk  Vki  V ji  Vij  Vki  2Vii  2V jj , we will exhibit an allocation that is individual
rational but not individual stand-alone. Clearly, v is an allocation and the individual rational constraints are satisfied.
That the rational constraint for player k holds will follow from the following argument showing that individual
stand-alone constraint for player k is violated.
1
v(i)  (Vii  V ji  Vki )  3v(i)  Vii  V ji  Vki
3
 3(Vii  M i )  Vii  V ji  Vki
 2Vii  3M i  V ji  Vki
1)
1
v( j )  (Vij  V jj  Vkj )  3v( j )  Vij  V jj  Vkj
3
 3(V jj  M j )  Vij  V jj  Vkj
 2V jj  3M j  Vij  Vkj
1)  2)
2)
 2Vii  3M i  2V jj  3M j  V ji  Vki  Vij  Vkj
 2Vii  2V jj  (V ji  Vki  Vij  Vkj )  3M i  3M j
3)
We assumed that:
3Vik  3V jk  Vki  V ji  Vij  Vki  2Vii  2V jj
 3Vik  3V jk  2Vii  2V jj  (Vki  V ji  Vij  Vki )
 3Vik  3V jk  3M i  3M j
Since equation 3)
 3Vik  3V jk  3( M i  M j )  0
 3Vik  3V jk  3( M i  M j )  3M k  3M k  3(v(k )  Vkk )
 3Vik  3V jk  3v(k )  3Vkk
 v(k )  Vik  V jk  Vkk
Since M i  M j  M k  0
thus this allocation is not individual stand-alone for player k.
5. Conclusion
Using the graphs of a three-player fair division problem helps understanding the dynamics of each property
discussed as well as capturing the interactions among each property. The object valuation matrix is a way of
simplifying a fair division problem into a compact form aiding the formularization of the properties discussed.
Theorem 1 includes a property test on the envy-free property. It helps avoiding unnecessary envy-free constraints in
order to simplify the fair-division problem. Theorem 2 and theorem 3 include property tests on individual standalone and individual rational property. These two tests help avoid unnecessary individual stand-alone or individual
rational constraints by finding out whether one is a subset of the other. Further work in this topic includes extending
the property tests to an n-player game based on the 3-player game result; and exploring other properties in fairdivision problems that are not listed in this paper.
6. Acknowledgement
This work was completed during the 2004 Goshen College Maple Scholars Program. Funding for this work was
obtained through the Mathematical Association of America's Strengthening Underrepresented Minority Mathematics
Achievement Program, funded by the National Security Agency and the National Science Foundation.
7. Appendix -- Maple graphing routine
> with(plots):
> A:=matrix([[50,5,5],[5,50,5],[5,5,50]]):
U:=(A[1,1]+A[2,2]+A[3,3]):
“A” matrix can be adjusted according to needs.
1. Envy-Free Efficiency Graph
The labeling of ineq is not correspondent to the lableing of graph, thus relableing is needed.
> ineq1:=subs(V[A]=-sqrt(3)/2*x-0.5*y+U,V[C]=y,V[B]=sqrt(3)/2*x-1/2*y,V[A]>=V[B]-A[2,2]+A[2,1]);
ineq2:=subs(V[A]=-sqrt(3)/2*x-0.5*y+U,V[C]=y,V[B]=sqrt(3)/2*x-1/2*y,V[A]>=V[C]-A[3,3]+A[3,1]):
ineq3:=subs(V[A]=-sqrt(3)/2*x-0.5*y+U,V[C]=y,V[B]=sqrt(3)/2*x-1/2*y,V[B]>=V[A]-A[1,1]+A[1,2]):
ineq4:=subs(V[A]=-sqrt(3)/2*x-0.5*y+U,V[C]=y,V[B]=sqrt(3)/2*x-1/2*y,V[B]>=V[C]-A[3,3]+A[3,2]):
ineq5:=subs(V[A]=-sqrt(3)/2*x-0.5*y+U,V[C]=y,V[B]=sqrt(3)/2*x-1/2*y,V[C]>=V[A]-A[1,1]+A[1,3]):
ineq6:=subs(V[A]=-sqrt(3)/2*x-0.5*y+U,V[C]=y,V[B]=sqrt(3)/2*x-1/2*y,V[C]>=V[B]-A[2,2]+A[2,3]):
> U;
inequal({
ineq1,ineq2,ineq3,ineq4,ineq5,ineq6,
y>=0,
y<=sqrt(3)*x,
y<=-sqrt(3)*x+2*U},
x=0..2*U/sqrt(3),y=0..U,
scaling=CONSTRAINED,
optionsclosed=(color=red, thickness=1),
optionsexcluded=(color=white)):
2. Individual Rational Graph
Individual rational says that each player's gain must be at least one third of the entire estate value in his or her view.
> ineq7:=subs(V[A]=-sqrt(3)/2*x-0.5*y+U,V[C]=y,V[B]=sqrt(3)/2*x1/2*y,V[A]>=1/3*(A[1,1]+A[2,1]+A[3,1]));
ineq8:=subs(V[A]=-sqrt(3)/2*x-0.5*y+U,V[C]=y,V[B]=sqrt(3)/2*x1/2*y,V[B]>=1/3*(A[1,2]+A[2,2]+A[3,2]));
ineq9:=subs(V[A]=-sqrt(3)/2*x-0.5*y+U,V[C]=y,V[B]=sqrt(3)/2*x1/2*y,V[C]>=1/3*(A[1,3]+A[2,3]+A[3,3]));
> inequal({
ineq7,ineq8,ineq9,
y>=0,
y<=sqrt(3)*x,
y<=-sqrt(3)*x+2*U},
x=0..2*U/sqrt(3),y=0..U,
scaling=CONSTRAINED,
optionsclosed=(color=red, thickness=1),
optionsexcluded=(color=white)):
3. Individual Stand-Alone Graph
individule stand alone says that each player does not get more than what he thinks the entire estate worth
> ineq10:=subs(V[A]=-sqrt(3)/2*x-0.5*y+U,V[C]=y,V[B]=sqrt(3)/2*x1/2*y,V[A]<=(A[1,1]+A[2,1]+A[3,1]));
ineq11:=subs(V[A]=-sqrt(3)/2*x-0.5*y+U,V[C]=y,V[B]=sqrt(3)/2*x1/2*y,V[B]<=(A[1,2]+A[2,2]+A[3,2]));
ineq12:=subs(V[A]=-sqrt(3)/2*x-0.5*y+U,V[C]=y,V[B]=sqrt(3)/2*x1/2*y,V[C]<=(A[1,3]+A[2,3]+A[3,3]));
> inequal({
ineq10,ineq11,ineq12,
y>=0,
y<=sqrt(3)*x,
y<=-sqrt(3)*x+2*U},
x=0..2*U/sqrt(3),y=0..U,
scaling=CONSTRAINED,
optionsclosed=(color=red, thickness=1),
optionsexcluded=(color=white)):