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.

Problem2. 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).

Problem3a.v({ }) = 0, v({1}) = v({2}) = v({3}) = -2,

v({1,2}) = v({2,3}) = v({1,3}) = 2, v({1,2,3}) = 3.

Problem3b.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).

Problem3c. 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.

Problem7. 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.

Problem8a.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

12375125102

13275125102

21311585102

23111585102

312115125 62

321115125 62

Average 610/6 670/6532/6

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% Adds 49% Adds 50% Adds

1, 49, 50 0 0 1

1, 50, 49 0 0 1

49, 1, 50 0 0 1

49, 50, 1 0 0 1

50, 1, 49 1 0 0

50, 49, 1 0 1 0

Sum 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%.