Section 14.7:
Problem 1. Let (x1, x2, x3, x4) be a point in the core. Then x1, x2,
x3,
and x4 must satisfy:
(1) x1 + x2 + x375
(2) x1 + x2 + x475
(3) x1 + x3 + x475
(4) x2 + x3 + x475
(5) x1 + x2 + x3 + x4100
(6) x3 + x460
xi0.
Since (x1, x2, x3, x4) must be an imputation we also require that
(7) x1 + x2 +x3 + x4 = 100.
Adding (1)-(4) yields 3(x1 + x2 + x3 + x4)300
or x1 + x2 + x3 + x4100.
By (7) we now know that (1)-(4) must all hold with equality.
Thus, any point in the core must satisfy x1 = x2 = x3 = x4 =25.
This, however, violates (6), so the core is empty.
Problem 2. Inequality (6) in solution to Problem 1 is now
x3 + x450.
Since (25, 25, 25, 25) satisfies this inequality the core is the
point (25, 25, 25, 25).
Problem 3a. v({ }) = 0, v({1}) = v({2}) = v({3}) = -2,
v({1,2}) = v({2,3}) = v({1,3}) = 2, v({1,2,3}) = 3.
Problem 3b. An imputation (x1, x2, x3) must satisfy
x1 -2, x2-2, x3-2 and
(1) x1 + x2 + x3 = 3.
For an imputation (x1, x2, x3) to be in the core it must satisfy
(2) x1 + x22
(3) x1 + x32
(4) x2 + x32
(5) x1 + x2 + x33.
Adding (2)-(4) implies that 2(x1 + x2 + x3)6 or x1 + x2 + x33.
By (1), however (2)-(4) must all hold with equality. Solving
(2)-(4) shows that the core consists of the point (1, 1, 1).
Problem 3c. Shapley value to player 1 is
(2/6)(v({1} - v({ })) + 1/6(v({1,2}) - v({2})
+(1/6)(v({1,3}) - v({3}) + (2/6)(v({1,2,3}) - v({2,3})
= (1/3)(-2) + (1/6)4 + (1/6)(4) + (1/3)(1) = 1.
By symmetry the Shapley value to Player 2 = Shapley value to
player 3 = 1.
Problem 7. Cost of first 2,000 feet = $40,000 is divided equally
among 2000 landings. Thus each Type 1 plane should pay 40,000/2000
= $20. The cost of the next 1000 feet ($20,000) is divided among
700 + 500 + 200 = 1400 landings. Thus each Type 2 plane should pay
$20 + (20,000/1400) = $34.29. The cost of the next 1000 feet
($20,000) is equally divided among 500 + 200 = 700 landings. Thus
each Type 3 plane should pay $34.29 + (20,000/700) = $62.86. The
cost of the last 1000 feet ($20,000) is divided among 200
landings. Thus each Type 4 plane should pay $62.86 + (20,000/200)
= $162.86.
Problem 8a. If (x1, x2, x3) is in the core, then
(1) x10.2
(2) x20
(3) x30
(4) x2 + x31.8
(5) x1 + x31.6
(6) x1 + x21.5
(7) x1 + x2 + x3 = 2.
Adding (4), (5), and (6) yields 2(x1 + x2 + x3)4.9 or
x1 + x2 + x32.45. This contradicts (7), so the core is empty.
Problem 8b.
Order of Arrival
Player 1
Player 2 Player 3
Adds
Adds
Adds
----------------------------123
0.2
1.3
0.5
----------------------------132
0.2
0.4
1.4
----------------------------213
1.5
0.0
0.5
----------------------------231
0.2
0.0
1.8
----------------------------312
1.6
0.4
0.0
----------------------------321
0.2
1.8
0.0
----------------------------Sum
3.9
3.9
4.2
----------------------------On the average, Player 1 adds 3.9/6 = .65. On the average, Player
2 adds 3.9/6 = .65, On the average, Player 3 adds 4.2/6 = .70.
Problem 8c. (1.05, .45, .55) dominates (1, 1/2, 1/2) through the
coalition of players 1 and 3 because players 1 and 3 are entitled
to at least 1.6 units if they band together.
Problem 11. We find the characteristic function to be
v({1}) = 75, v({2}) =85, v({3}) = 62, v({1,2}) = 200, v({1,3}) =
177, v({2,3}) = 187 . v({1,2,3}) = 302. (x1, x2, x3) will be in the
core if it satisfies
x1._75,
x285,
x362,
x1 + x2200,
x1 + x3177,
x2 + x3187,
x1 + x2 + x3 = 302.
Thus, if c1 + c2 + c3 = 2, then (100 + c1, 100 + c2, 100 + c3) will
be in the core. This shows that the core consists of an infinite
number of imputations.
We find the Shapley value form the following table.
Order of Arrival
Payoff to 1 Payoff to 2 Payoff to 3
123
75
125
102
132
75
125
102
213
115
85
102
231
115
85
102
312
115
125
62
321
115
125
62
Average
610/6
670/6
532/6
Thus, Doctor 1 receives $101,67; Doctor 2 receives $111,67; Doctor
3 receives $88,67. Each doctor receives a value equal to his
revenues-costs-(total overhead/3), which is what we would expect.
CHAPTER 14 REVIEW PROBLEMS:
Problem 6a. v({ }) = 0, v({1}) = v({49}) = v({50}) = 0,
v({1,50}) = v({49,50}) = 1, v({1,49}) = 0, v({1,49,50)} = 1.
Problem 6b. Let x1 = reward to 1% shareholder x49 = reward to 49%
shareholder, and x50 = reward to 50% shareholder. Then x1, x49, and
x50 must satisfy x10, x490, and x500 and
(1) x1 + x501
(2) x49 + x501
(3) x1 + x490
(4) x1 + x49 + x50 = 1.
Adding (1)-(3) yields 2(x1 + x49 + x50)2. Then (4) implies that
(1)-(3) all hold with equality. Thus x1 = x49 = 0 and x50 = 1 is the
unique point in the core. Thus core emphasizes the importance of
the most important (50% shareholder) player.
Problem 6c.
Order of Arrival
1, 49, 50
1, 50, 49
49, 1, 50
49, 50, 1
50, 1, 49
50, 49, 1
Sum
1% Adds
49% Adds
50% Adds
---------------------------------0
0
1
---------------------------------0
0
1
---------------------------------0
0
1
---------------------------------0
0
1
---------------------------------1
0
0
---------------------------------0
1
0
---------------------------------1
1
4
Thus, Shapley value to 1% shareholder = 1/6
Shapley value to 49% shareholder = 1/6
Shapley value to 50% shareholder = 4/6.
Problem 6d. x1 = 1/2, x49 = 0, x50 = 1/2 dominates x1 = x49 = x50 =
1/3 through players 1% and 50%.