Problem:
Suppose a building worth 2000 per month to its owner. A cloth
merchant is ready to pay a monthly rent of 2500, whereas a bank
offers to pay 3000 per month. Find the core allocation of this
game.
Solution:
First we have to formulate the characteristic function.
Suppose players are A, B and C, where A is owner, C is cloth
merchant and B is the bank. vA = 2000, vB = 0, vC = 0,
vAC = 2500, vAB = 3000, vBC = 0 and vABC = 3000.
xA ≥ 2000, xB ≥ 0, xC ≥ 0.
xA + xC ≥ 2500, xA + xB ≥ 3000, xB + xC ≥ 0
xA + xC + xB ≥ 3000.
Substituting xB + xC ≥ 0 in xA + xC + xB = 3000, since core
allocation is an imputation we have, xA ≤ 3000 And we have
xA ≥ 2000. So we get that
2000 ≤ xA ≤ 3000.
Again we have xB ≤ 500 , by substituting xA + xC ≥ 2500 in
xA + xC + xB = 3000. This implies that 0 ≤ xB ≤ 500.
We have xC ≤ 0 , by substituting xA + xB ≥ 3000 in
xA + xC + xB = 3000. This implies that 0 = xC .
Thus, core allocations are
2500 ≤ xA ≤ 3000 , 0 ≤ xB ≤ 500, and xC = 0 .
Problem:
Suppose there are four players {A1 , A2 , A3 , A4 }, suppose there are
two disjoint set L and R of these players. R = {A1 , A2 , } and
L = {A3 , A4 , }. Players of set R has right shoe and players in set L
has left shoe. A pair of shoe contains two shoes - left and right. A
pair of shoe worth 1 and if there is only left or only right, it has no
value. Consider set S = {A1 , A3 , A4 }, so one right shoe.
S ∩ R = {A1 } and L ∩ S = {A3 , A4 }, two left shoes. If there is
cooperation between the players in S, then it will have one pair of
shoe. The minimum number of left or right shoes determine the
number of pairs. So the worth of coalition is min{|S ∩ R|, |S ∩ L|},
when |S| ≥ 2.
We get the following characteristic function.
(
0, if |S| ∈ {0, 1},
v (s) =
min{|S ∩ R|, |S ∩ L|}, if |S| ≥ 2.
v (N) = min{|R|, |L|}.
We need to find core allocation.
coalitions v()
∅ 0
{1} v ({1}) = 0
{2} v ({2}) = 0
{3} v ({3}) = 0
{4} v ({4}) = 0
{1, 2} v ({1, 2}) = 0
{1, 3} v ({1, 3}) = 1
{1, 4} v ({2, 3}) = 1
{2, 3} v ({2, 3}) = 1
{2, 4} v ({2, 4}) = 1
{4, 3} v ({4, 3}) = 0
{1, 2, 3} v ({1, 2, 3}) = 1
{1, 2, 4} v ({1, 2, 4}) = 1
coalitions v()
{1, 3, 4} v ({1, 4, 3}) = 1
{2, 3, 4} v ({4, 2, 3}) = 1
{1, 2, 3, 4} v ({1, 4, 2, 3}) = 2
xi ≥ 0, i = 1, 2, 3, 4.
x1 + x3 ≥ 1, x1 + x4 ≥ 1, x2 + x3 ≥ 1, x2 + x4 ≥ 1,
x1 + x2 ≥ 0, x4 + x3 ≥ 0.
x1 + x2 + x3 ≥ 1, x1 + x2 + x4 ≥ 1, x1 + x4 + x3 ≥ 1,
x4 + x2 + x3 ≥ 1.
x1 + x2 + x3 + x4 ≥ 2
We have x1 + x3 ≤ 1, by substituting x2 + x4 ≥ 1 in
x1 + x2 + x3 + x4 = 2. And we have x1 + x3 ≥ 1, so x1 + x3 = 1
Similarly we have x2 + x4 ≤ 1 and we have x2 + x4 ≥ 1, so
x2 + x4 = 1.
We also have x1 + x4 = 1 and x2 + x3 = 1.
We also get x1 ≤ 1 by substituting x4 + x2 + x3 ≥ 1 in
x1 + x2 + x3 + x4 = 2. Similarly we have x2 ≤ 1, x3 ≤ 1, x4 ≤ 1.
From x1 + x3 = 1, x2 + x4 = 1, x1 + x4 = 1 x2 + x3 = 1 and
x1 + x2 + x3 + x4 = 2. We have x1 = x2 and x3 = x4 .
The core allocations are
0 ≤ xi ≤ 1, i = 1, 2, 3, 4, x1 = x2 , x3 = x4 , and
x1 + x2 + x3 + x4 = 2.
