Solution Manual for Excursions in Modern Mathematics, 9th Edition

Excursions in Modern Mathematics 9th Edition Tannenbaum Solutions Manual Full Download: http://testbanklive.com/download/excursions-in-modern-mathematics-9th-edition-tannenbaum-solutions-manual/ Chapter 2 WALKING 2.1. Weighted Voting 1. (a) A generic weighted voting system with N = 5 players is described using notation [q : w1 , w2 , w3 , w4 , w5 ] where q represents the value of the quota and wi represents the weight of player Pi . In this case, the players are the partners, w1 15 , w2 12 , w3 w4 10 , and w5 3 . Since the total number of votes is 15 + 12 + 10 + 10 + 3 = 50 and the quota is determined by a simple majority (more than 50% of the total number of votes), q = 26. That is, the partnership can be described by [26 :15,12,10,10,3] . 1 §2· (b) Since ¨ ¸ 50 33 , we choose the quota q as the smallest integer larger than this value which is 34. 3 ©3¹ The partnership can thus be described by [34 :15,12,10,10,3] . 2. (a) There are 100 votes (each one representing 1% ownership). A simple majority requires more than half of that or q = 51. That is, the partnership can be described by [51: 30, 25, 20,16,9] . 2 §2· (b) Since ¨ ¸100 66 , we choose the quota q as the smallest integer larger than this value which is 67. 3 ©3¹ The partnership can thus be described by [67 : 30, 25, 20,16,9] . 3. (a) The quota must be more than half of the total number of votes. This system has 6 + 4 + 3 + 3 + 2 + 2 = 20 total votes. Since 50% of 20 is 10, the smallest possible quota would be 11. Note: q = 10 is not sufficient. If 10 votes are cast in favor and 10 cast against a motion, that motion should not pass. (b) The largest value q can take is 20, the total number of votes. (c) 3 u 20 15, so the value of the quota q would be 15. 4 (d) The value of the quota q would be strictly larger than 15. That is, 16. 4. (a) The quota must be more than half of the total votes. This system has 10 + 6 + 5 + 4 + 2 = 27 total votes. 1 u 27 13.5, so the smallest value q can take is 14. 2 (b) The largest value q can take is 27, the total number of votes. (c) 2 u 27 18 3 (d) 19 5. (a) P1 , the player with the most votes, does not have veto power since the other players combined have 3 + 3 + 2 = 8 votes and can successfully pass a motion without him. The other players can’t have veto power either then as they have fewer votes and hence less (or equal) power. So no players have veto power in this system. Copyright © 2018 Pearson Education, Inc. Full download all chapters instantly please go to Solutions Manual, Test Bank site: testbanklive.com 22 Chapter 2: The Mathematics of Power (b) P1 does have veto power here since the other players combined have only 3 + 3 + 2 = 8 votes and cannot successfully pass a motion without him. P2 , on the other hand, does not have veto power since the other players combined have the 4 + 3 + 2 = 9 votes necessary to meet the quota. Since P3 and P4 have the same or fewer votes than P2 , it follows that only P1 has veto power. (c) P4 , the player with the fewest votes, does not have veto power here since the other players combined have 4 + 3 + 3 = 10 votes and can pass a motion without him. P3 , on the other hand, does have veto power since the other players combined have only 4 + 3 + 2 = 9 votes. Since P1 and P2 have the same or more votes than P3 , it follows that P1 , P2 , and P3 all have veto power. (d) P4 , the player with the fewest votes, has veto power since the other players combined have only 4 + 3 + 3 = 10 votes and cannot pass a motion without him. Since the other players all have the same or more votes than P4 , it follows that all players have veto power in this system. 6. (a) P1 does have veto power here since the other players combined have only 4 + 2 + 1 = 7 votes and cannot pass a motion without him. P2 , on the other hand, does not have veto power since the other players combined have the 8 + 2 + 1 = 11 votes necessary to meet the quota. Since P3 and P4 have the same or fewer votes than P2 , it follows that only P1 has veto power. (b) Based on (a), we know that P1 still has veto power since only the quota has changed (increased). It would now be even more difficult for the other players to pass a motion without P1 . P2 also has veto power here since the other players combined have 8 + 2 + 1 = 11 votes and cannot pass a motion without him. P3 , on the other hand, does not have veto power since the other players combined have 8 + 4 + 1 = 13 votes. Since P4 has fewer votes than P3 , it follows that only P1 and P2 have veto power. (c) Based on (a) and (b), we know that P1 and P2 still have veto power since only the quota has changed (increased). P3 now also has veto power here since the other players combined have only 8 + 4 + 1 = 13 votes and cannot pass a motion without him. P4 , on the other hand, does not have veto power since the other players combined have 8 + 4 + 2 = 14 votes. (d) P4 , the player with the fewest votes, has veto power since the other players combined have only 8 + 4 + 2 = 14 votes and cannot pass a motion without him. Since the other players all have the same or more votes than P4 , it follows that all four players have veto power in this system. 7. (a) In order for all three players to have veto power, the player having the fewest votes (the weakest) must have veto power. In order for that to happen, the quota q must be strictly larger than 7 + 5 = 12. The smallest value of q for which this is true is q = 13. (b) In order for P2 to have veto power, the quota q must be strictly larger than 7 + 3 = 10. The smallest value of q for which this is true is q = 11. [Note: When q = 11, we note that P3 does not have veto power since the other two players have 7 + 5 = 12 votes.] 8. (a) In order for all five players to have veto power, the player having the fewest votes (the weakest) must have veto power. In order for that to happen, the quota q must be strictly larger than 10 + 8 + 6 + 4 = 28. The smallest value of q for which this is true is q = 29. (b) In order for P3 to have veto power, the quota q must be strictly larger than 10 + 8 + 4 + 2 = 24. The smallest value of q for which this is true is q = 25. [Note: When q = 25, we note that P4 does not have veto power since the other four players have 10 + 8 + 6 + 2 = 26 votes.] Copyright © 2018 Pearson Education, Inc. ISM: Excursions in Modern Mathematics, 9E 23 9. To determine the number of votes each player has, write the weighted voting system as [49: 4x, 2x, x, x]. (a) If the quota is defined as a simple majority of the votes, then x is the largest integer satisfying 4x 2x x x which means that 8 x 98 or x 12.25 . So, x = 12 and the system can be 49 ! 2 described as [49: 48, 24, 12, 12]. (b) If the quota is defined as more than two-thirds of the votes, then x is the largest integer satisfying 2 4x 2x x x 49 ! which means that 16 x 147 or x 9.1875 . So, x = 9 and the system can be 3 described as [49: 36, 18, 9, 9]. (c) If the quota is defined as more than three-fourths of the votes, then x is the largest integer satisfying 3 4x 2x x x which means that 24 x 196 or x 8.167 . So, x = 8 and the system can be 49 ! 4 described as [49: 32, 16, 8, 8]. 10. (a) [121: 96, 48, 48, 24, 12, 12]; if the quota is defined as a simple majority of the votes, then x is the largest 8x 4 x 4 x 2 x x x . integer satisfying 121 ! 2 (b) [121: 72, 36, 36, 18, 9, 9]; if the quota is defined as more than two-thirds of the votes, then x is the largest 2 8x 4 x 4 x 2 x x x . integer satisfying 121 ! 3 (c) [121: 64, 32, 32, 16, 8, 8]; if the quota is defined as more than three-fourths of the votes, then x is the 3 8x 4 x 4 x 2 x x x . largest integer satisfying 121 ! 4 2.2. Banzhaf Power 11. (a) w1 w3 7 + 3 = 10 (b) Since (7 + 5 + 3) / 2 = 7.5, the smallest allowed value of the quota q in this system is 8. For {P1 , P3 } to be a winning coalition, the quota q could be at most 10. So, the values of the quota for which {P1 , P3 } is winning are 8, 9, and 10. (c) Since 7 + 5 + 3 = 15, the largest allowed value of the quota q in this system is 15. For {P1 , P3 } to be a losing coalition, the quota q must be strictly greater than its weight, 10. So, the values of the quota for which {P1 , P3 } is losing are 11, 12, 13, 14, and 15. 12. (a) 8 + 6 + 4 = 18 (b) Since (10 + 8 + 6 + 4 + 2) / 2 = 15, the smallest allowed value of the quota q in this system is 16. For {P2 , P3 , P4 } to be a winning coalition, the quota q could be at most 18. So, the values of the quota for which {P2 , P3 , P4 } is winning are 16, 17, and 18. (c) Since 10 + 8 + 6 + 4 + 2 = 30, the largest allowed value of the quota q in this system is 30. For {P2 , P3 , P4 } to be a losing coalition, the quota q must be strictly greater than its weight, 18. So, the values of the quota for which {P2 , P3 , P4 } is losing are integer values from 19 to 30. Copyright © 2018 Pearson Education, Inc. 24 Chapter 2: The Mathematics of Power 13. P1 is critical (underlined) three times; P2 is critical three times; P3 is critical once ; P4 is critical once. The total number of times the players are critical is 8 (number of underlines). The Banzhaf power distribution is E1 3 / 8; E 2 3 / 8; E 3 1/ 8; E 4 1/ 8 . 14. P1 is critical (underlined) seven times; P2 is critical five times; P3 is critical three times; P4 is critical three times; P5 is critical one time. The total number of times the players are critical (all underlines) is 19. The Banzhaf power distribution is E1 7 /19; E2 3 /19; E4 5 /19; E3 3 /19; E5 1/19. 15. (a) P1 is critical since the other players only have 5 + 2 = 7 votes. P2 is also critical since the other players only have 6 + 2 = 8 votes. However, P4 is not critical since the other two players have 6 + 5 = 11 (more than q = 10). (b) The winning coalitions are those whose weights are 10 or more. These are: {P1 , P2 }, {P1 , P3 }, {P1 , P2 , P3 }, {P1 , P2 , P4 }, {P1 , P3 , P4 }, {P2 , P3 , P4 }, {P1 , P2 , P3 , P4 }. (c) We underline the critical players in each winning coalition: {P1 , P2 }, {P1 , P3 }, {P1 , P2 , P3 }, {P1 , P2 , P4 }, {P1 , P3 , P4 }, {P2 , P3 , P4 }, {P1 , P2 , P3 , P4 }. Then, it follows that E1 5 ; E2 12 E3 3 ; E4 12 1 . 12 16. (a) All the players are critical in this coalition since the total weight of the coalition (3 + 1 + 1 = 5) is exactly the same as the quota. If any one player were to leave the coalition, the remaining players would not have enough votes to meet the quota. (b) The winning coalitions are those whose weights are 5 or more. These are: {P1 , P2 }, {P1 , P2 , P3 }, {P1 , P2 , P4 }, {P1 , P3 , P4 }, {P1 , P2 , P3 , P4 }. (c) We underline the critical players in each winning coalition: {P1 , P2 }, {P1 , P2 , P3 }, {P1 , P2 , P4 }, {P1 , P3 , P4 }, {P1 , P2 , P3 , P4 }. Then, it follows that E1 5 ; E2 10 3 ; E3 10 1 ; E4 10 1 . 10 17. (a) The winning coalitions (with critical players underlined) are: {P1 , P2 } , {P1 , P3 } , and {P1 , P2 , P3 } . P1 is critical three times; P2 is critical one time; P3 is critical one time. The total number of times the players are critical is 5. The Banzhaf power distribution is E1 3 / 5; E2 1/ 5; E3 1/ 5. (b) The winning coalitions (with critical players underlined) are: {P1 , P2 } , {P1 , P3 } , and {P1 , P2 , P3 } . P1 is critical three times; P2 is critical one time; P3 is critical one time. The total number of times the players are critical is 5. The Banzhaf power distribution is E1 3 / 5; E2 1/ 5; E3 1/ 5. The distributions in (a) and (b) are the same. 18. (a) The winning coalitions (with critical players underlined) are: {P1 , P2 } and {P1 , P2 , P3 } . P1 is critical twice; P2 is critical twice; P3 is never critical. The total number of times the players are critical is 4. The Banzhaf power distribution is E1 2 / 4; E 2 2 / 4; E 3 0 / 4. Copyright © 2018 Pearson Education, Inc. ISM: Excursions in Modern Mathematics, 9E 25 (b) The winning coalitions (with critical players underlined) are: {P1 , P2 } and {P1 , P2 , P3 } . P1 is critical twice; P2 is critical twice; P3 is never critical. The total number of times the players are critical is 4. The Banzhaf power distribution is E1 2 / 4; E 2 2 / 4; E 3 0 / 4. The distributions in (a) and (b) are the same. 19. (a) The winning coalitions (with critical players underlined) are: {P1 , P2 , P3 } , {P1 , P2 , P4 } , {P1 , P2 , P5 } , {P1 , P3 , P4 } , {P1 , P2 , P3 , P4 } , {P1 , P2 , P3 , P5 } , {P1 , P2 , P4 , P5 } , {P1 , P3 , P4 , P5 } , {P2 , P3 , P4 , P5 } , {P1 , P2 , P3 , P4 , P5 } . The Banzhaf power distribution is E1 8 / 24; E 2 E 5 2 / 24. 6 / 24; E 3 4 / 24; E 4 4 / 24; (b) The situation is like (a) except that {P1 , P2 , P5 } , {P1 , P3 , P4 } and {P2 , P3 , P4 , P5 } are now losing coalitions. In addition, P2 is now critical in {P1 , P2 , P3 , P4 } , P3 is now critical in {P1 , P2 , P3 , P5 } , P4 is now critical in {P1 , P2 , P4 , P5 } , P5 is now critical in {P1 , P3 , P4 , P5 } , and P1 is now critical in the grand coalition {P1 , P2 , P3 , P4 , P5 } . The winning coalitions (with critical players underlined) are now: {P1 , P2 , P3 } , {P1 , P2 , P4 } , {P1 , P2 , P3 , P4 } , {P1 , P2 , P3 , P5 } , {P1 , P2 , P4 , P5 } , {P1 , P3 , P4 , P5 } , {P1 , P2 , P3 , P4 , P5 } . The Banzhaf power distribution is E1 7 /19; E 2 5 /19; E 3 3 /19; E 4 3 /19; E 5 1/19. (c) This situation is like (b) with the following exceptions: {P1 , P2 , P4 } and {P1 , P3 , P4 , P5 } are now losing coalitions; P3 is critical in {P1 , P2 , P3 , P4 } ; P5 is critical in {P1 , P2 , P4 , P5 } ; and P2 is critical in {P1 , P2 , P3 , P4 , P5 } . The winning coalitions (with critical players underlined) are now: {P1 , P2 , P3 } , {P1 , P2 , P3 , P4 } , {P1 , P2 , P3 , P5 } , {P1 , P2 , P4 , P5 } , {P1 , P2 , P3 , P4 , P5 } . The Banzhaf power distribution is E1 5 /15; E 2 5 /15; E 3 3 /15; E 4 1/15; E 5 1/15. (d) Since the quota equals the total number of votes in the system, the only winning coalition is the grand coalition and every player is critical in that coalition. The Banzhaf power distribution is easy to calculate in the case since all players share power equally. It is E1 1/ 5; E 2 1/ 5; E 3 1/ 5; E 4 1/ 5; E 5 1/ 5. 20. (a) P1 is a dictator and the other players are dummies. Thus P1 is the only critical player in each winning coalition. The Banzhaf power distribution is E1 1; E 2 E 3 E 4 0. (b) The winning coalitions (with critical players underlined) are: {P1 , P2 } , {P1 , P3 } , {P1 , P4 } , {P1 , P2 , P3 } , {P1 , P2 , P4 } , {P1 , P3 , P4 } , {P1 , P2 , P3 , P4 } . The Banzhaf power distribution is E1 E2 1 10 10%; E 3 1 10 10%; E 4 1 10 7 10 70%; 10%. (c) This situation is like (b) with the following exceptions: {P1 , P4 } is now a losing coalition; P1 and P2 are both critical in {P1 , P2 , P4 }; P1 and P3 are both critical in {P1 , P3 , P4 } . The winning coalitions (with critical players underlined) are now: {P1 , P2 } , {P1 , P3 } , {P1 , P2 , P3 } , {P1 , P2 , P4 } , {P1 , P3 , P4 } , {P1 , P2 , P3 , P4 } . The Banzhaf power distribution is E1 E4 6 10 60%; E 2 0. Copyright © 2018 Pearson Education, Inc. 2 10 20%; E 3 2 10 20%; 26 Chapter 2: The Mathematics of Power (d) The winning coalitions (with critical players underlined) are now: {P1 , P2 , P3 } , {P1 , P2 , P3 , P4 } . The 2 ; E2 6 Banzhaf power distribution is E1 2 ; E3 6 2 ; E4 6 0. 21. (a) A player is critical in a coalition if that coalition without the player is not on the list of winning coalitions. So, in this case, the critical players are underlined below. {P1 , P2 } , {P1 , P3 } , {P1 , P2 , P3 } (b) P1 is critical three times; P2 is critical one time; P3 is critical one time. The total number of times all players are critical is 3 + 1 + 1 = 5. The Banzhaf power distribution is E1 3 / 5; E2 1/ 5; E3 1/ 5. 22. (a) A player is critical in a coalition if that coalition without the player is not on the list of winning coalitions. {P1 , P2 }, {P1 , P2 , P3 }, {P1 , P2 , P4 }, {P1 , P2 , P3 , P4 } (b) E1 4 / 8 , E2 4 / 8 , E3 E4 0 23. (a) The winning coalitions (with critical players underlined) are {P1 , P2 } , {P1 , P3 } , {P2 , P3 } , {P1 , P2 , P3 } , {P1 , P2 , P4 } , {P1 , P2 , P5 } , {P1 , P2 , P6 } , {P1 , P3 , P4 } , {P1 , P3 , P5 } , {P1 , P3 , P6 } , {P2 , P3 , P4 } , {P2 , P3 , P5 } , {P2 , P3 , P6 } . (b) These winning coalitions (with critical players underlined) are {P1 , P2 , P4 } , {P1 , P3 , P4 } , {P2 , P3 , P4 } , {P1 , P2 , P3 , P4 } , {P1 , P2 , P4 , P5 } , {P1 , P2 , P4 , P6 } , {P1 , P3 , P4 , P5 } , {P1 , P3 , P4 , P6 } , {P2 , P3 , P4 , P5 } , {P2 , P3 , P4 , P6 } , {P1 , P2 , P3 , P4 , P5 } , {P1 , P2 , P3 , P4 , P6 } , {P1 , P2 , P4 , P5 , P6 } , {P1 , P3 , P4 , P5 , P6 } , {P2 , P3 , P4 , P5 , P6 } , {P1 , P2 , P3 , P4 , P5 , P6 } . (c) P4 is never a critical player since every time it is part of a winning coalition, that coalition is a winning coalition without P4 as well. So, E 4 0. (d) A similar argument to that used in part (c) shows that P5 and P6 are also dummies. One could also argue that any player with fewer votes than P4 , a dummy, will also be a dummy. So, P4 , P5 , and P6 will never be critical — they all have zero power. The only winning coalitions with only two players are {P1 , P2 }, {P1 , P3 }, and {P2 , P3 }; and both players are critical in each of those coalitions. All other winning coalitions consist of one of these coalitions plus additional players, and the only critical players will be the ones from the two-player coalition. So P1 , P2 , and P3 will be critical in every winning coalition they are in, and they will all be in the same number of winning coalitions, so they all have the same power. Thus, the Banzhaf power distribution is E1 E2 1/ 3; E3 1/ 3; E 4 0; E 5 0; E 6 1/ 3; 0. 24. (a) Three-player winning coalitions (with critical players underlined): {P1 , P2 , P3 }, {P1 , P2 , P4 }, {P1 , P2 , P5 }, {P1 , P3 , P4 }, {P2 , P3 , P4 }. (b) Four-player winning coalitions (with critical players underlined): {P1 , P2 , P3 , P4 }, {P1 , P2 , P3 , P5 }, {P1 , P2 , P3 , P6 }, {P1 , P2 , P4 , P5 }, {P1 , P2 , P4 , P6 }, {P1 , P2 , P5 , P6 }, {P1 , P3 , P4 , P5 }, {P1 , P3 , P4 , P6 }, {P1 , P3 , P5 , P6 }, {P2 , P3 , P4 , P5 }, {P2 , P3 , P4 , P6 }. Copyright © 2018 Pearson Education, Inc. ISM: Excursions in Modern Mathematics, 9E 27 (c) Five-player winning coalitions (with critical players underlined): {P1 , P2 , P3 , P4 , P5 }, {P1 , P2 , P3 , P4 , P6 }, {P1 , P2 , P3 , P5 , P6 }, {P1 , P2 , P4 , P5 , P6 }, {P1 , P3 , P4 , P5 , P6 }, {P2 , P3 , P4 , P5 , P6 }. 15 13 11 9 3 ; E2 ; E3 ; E4 ; E5 ; E6 52 52 52 52 52 winning coalition. However, it has no critical players. (d) E1 1 . Note: The grand coalition is also a 52 25. (a) { A, B} , { A, C} , {B, C} , { A, B, C} , { A, B, D} , { A, C , D} , {B, C , D} , { A, B, C , D} (b) A, B, and C have Banzhaf power index of 4/12 each; D is a dummy. (D is never a critical player, and the other three clearly have equal power.) 26. (a) { A, B} , { A, C} , { A, D} , { A, B, C} , { A, B, D} , { A, C , D} , {B, C , D} , { A, B, C , D} (b) A has Banzhaf power index of 6/12; B, C, and D have Banzhaf power index of 2/12 each. 2.3. Shapley-Shubik Power 27. P1 is pivotal (underlined) ten times; P2 is pivotal (again, underlined) ten times; P3 is pivotal twice ; P4 is pivotal twice. The total number of times the players are pivotal is 4! = 24 (number of underlines). The Shapley-Shubik power distribution is V 1 10 / 24; V 2 10 / 24; V 3 2 / 24; V 4 2 / 24 . 28. P1 is pivotal (underlined) ten times; P2 is pivotal six times; P3 is pivotal six times ; P4 is pivotal twice. The total number of times the players are pivotal is 4! = 24. The Shapley-Shubik power distribution is V 1 10 / 24; V 2 6 / 24; V 3 6 / 24; V 4 2 / 24 . 29. (a) There are 3! = 6 sequential coalitions of the three players. Each pivotal player is underlined. P1 , P2 , P3 !, P1 , P3 , P2 !, P2 , P1 , P3 !, P2 , P3 , P1 !, P3 , P1 , P2 !, P3 , P2 , P1 ! (b) P1 is pivotal four times; P2 is pivotal one time; P3 is pivotal one time. The Shapley-Shubik power distribution is V 1 4 / 6; V 2 1/ 6; V 3 1/ 6. 30. (a) There are 3! = 6 sequential coalitions of the three players. Each pivotal player is underlined. P1 , P2 , P3 !, P1 , P3 , P2 !, P2 , P1 , P3 !, P2 , P3 , P1 !, P3 , P1 , P2 !, P3 , P2 , P1 ! (b) V 1 2 / 6; V 2 2 / 6; V 3 2 / 6. 31. (a) Since P1 is a dictator, V 1 1, V 2 0, V 3 0, and V 4 0. (b) There are 4! = 24 sequential coalitions of the four players. Each pivotal player is underlined. P1 , P2 , P3 , P4 !, P1 , P2 , P4 , P3 !, P1 , P3 , P2 , P4 !, P1 , P3 , P4 , P2 !, P1 , P4 , P2 , P3 !, P1 , P4 , P3 , P2 !, P2 , P1 , P3 , P4 !, P2 , P1 , P4 , P3 !, P2 , P3 , P1 , P4 !, P2 , P3 , P4 , P1 !, P2 , P4 , P1 , P3 !, P2 , P4 , P3 , P1 !, P3 , P1 , P2 , P4 !, P3 , P1 , P4 , P2 !, P3 , P2 , P1 , P4 !, P3 , P2 , P4 , P1 !, P3 , P4 , P1 , P2 !, P3 , P4 , P2 , P1 !, P4 , P1 , P2 , P3 !, P4 , P1 , P3 , P2 !, P4 , P2 , P1 , P3 !, P4 , P2 , P3 , P1 !, P4 , P3 , P1 , P2 !, P4 , P3 , P2 , P1 ! P1 is pivotal 16 times; P2 is pivotal 4 times; P3 is pivotal 4 times; P4 is pivotal 0 times. The ShapleyShubik power distribution is V 1 16 / 24; V 2 4 / 24; V 3 4 / 24; V 4 0. Copyright © 2018 Pearson Education, Inc. 28 Chapter 2: The Mathematics of Power (c) The only way a motion will pass is if P1 and P2 both support it. In fact, the second of these players that appears in a sequential coalition will be the pivotal player in that coalition. It follows that V 1 12 / 24 1/ 2, V 2 12 / 24 1/ 2, V 3 0, and V 4 0 . (d) Because the quota is so high, the only way a motion will pass is if P1 , P2 , and P3 all support it. In fact, the third of these players that appears in a sequential coalition will always be the pivotal player in that coalition. It follows that V 1 8 / 24 1/ 3, V 2 8 / 24 1/ 3, V 3 8 / 24 1/ 3, and V 4 0 . 32. (a) Since P1 is a dictator, V 1 1, V2 0 , V3 0 , V4 0. (b) P1 , P2 , P3 , P4 !, P1 , P2 , P4 , P3 !, P1 , P3 , P2 , P4 !, P1 , P3 , P4 , P2 !, P1 , P4 , P2 , P3 !, P1 , P4 , P3 , P2 !, P2 , P1 , P3 , P4 !, P2 , P1 , P4 , P3 !, P2 , P3 , P1 , P4 !, P2 , P3 , P4 , P1 !, P2 , P4 , P1 , P3 !, P2 , P4 , P3 , P1 !, P3 , P1 , P2 , P4 !, P3 , P1 , P4 , P2 !, P3 , P2 , P1 , P4 !, P3 , P2 , P4 , P1 !, P3 , P4 , P1 , P2 !, P3 , P4 , P2 , P1 !, P4 , P1 , P2 , P3 !, P4 , P1 , P3 , P2 !, P4 , P2 , P1 , P3 !, P4 , P2 , P3 , P1 !, P4 , P3 , P1 , P2 !, P4 , P3 , P2 , P1 ! P1 is pivotal 18 times; P2 is pivotal 2 times; P3 is pivotal 2 times; P4 is pivotal 2 times. The Shapley-Shubik power distribution is V 1 9 /12 , V 2 1/12 , V 3 1/12 , V 4 1/12 . (c) V 1 1/ 2 , V 2 1/ 2 , V 3 0 , V 4 0 (The second of P1 or P2 to vote is always the pivotal player in any of the 24 sequential coalitions. This happens 12 times in each case.) (d) V 1 1 / 3 , V 2 1/ 3 , V 3 1/ 3 , V 4 0 (The third of P1 , P2 , or P3 to vote is always the pivotal player in any of the 24 sequential coalitions. This happens 8 times in each case.) 33. (a) There are 4! = 24 sequential coalitions of the four players. Each pivotal player is underlined. P1 , P2 , P3 , P4 !, P1 , P2 , P4 , P3 !, P1 , P3 , P2 , P4 !, P1 , P3 , P4 , P2 !, P1 , P4 , P2 , P3 !, P1 , P4 , P3 , P2 !, P2 , P1 , P3 , P4 !, P2 , P1 , P4 , P3 !, P2 , P3 , P1 , P4 !, P2 , P3 , P4 , P1 !, P2 , P4 , P1 , P3 !, P2 , P4 , P3 , P1 !, P3 , P1 , P2 , P4 !, P3 , P1 , P4 , P2 !, P3 , P2 , P1 , P4 !, P3 , P2 , P4 , P1 !, P3 , P4 , P1 , P2 !, P3 , P4 , P2 , P1 !, P4 , P1 , P2 , P3 !, P4 , P1 , P3 , P2 !, P4 , P2 , P1 , P3 !, P4 , P2 , P3 , P1 !, P4 , P3 , P1 , P2 !, P4 , P3 , P2 , P1 ! P1 is pivotal in 10 coalitions; P2 is pivotal in six coalitions; P3 is pivotal in six coalitions; P4 is pivotal in two coalitions. The Shapley-Shubik power distribution is V 1 10 / 24; V 2 6 / 24; V 3 6 / 24; V4 2 / 24. (b) This is the same situation as in (a) – there is essentially no difference between 51 and 59 because the players’ votes are all multiples of 10. The Shapley-Shubik power distribution is thus still V 1 10 / 24; V 2 6 / 24; V 3 6 / 24; V 4 2 / 24. (c) This is also the same situation as in (a) – any time a group of players has 51 votes, they must have 60 votes. The Shapley-Shubik power distribution is thus still V 1 10 / 24; V 2 6 / 24; V 3 6 / 24; V4 2 / 24. Copyright © 2018 Pearson Education, Inc. ISM: Excursions in Modern Mathematics, 9E 29 34. (a) V 1 18 / 24; V 2 2 / 24; V 3 2 / 24; V 4 2 / 24. The sequential coalitions (with pivotal player in each coalition underlined) are: P1 , P2 , P3 , P4 !, P1 , P2 , P4 , P3 !, P1 , P3 , P2 , P4 !, P1 , P3 , P4 , P2 !, P1 , P4 , P2 , P3 !, P1 , P4 , P3 , P2 !, P2 , P1 , P3 , P4 !, P2 , P1 , P4 , P3 !, P2 , P3 , P1 , P4 !, P2 , P3 , P4 , P1 !, P2 , P4 , P1 , P3 !, P2 , P4 , P3 , P1 !, P3 , P1 , P2 , P4 !, P3 , P1 , P4 , P2 !, P3 , P2 , P1 , P4 !, P3 , P2 , P4 , P1 !, P3 , P4 , P1 , P2 !, P3 , P4 , P2 , P1 !, P4 , P1 , P2 , P3 !, P4 , P1 , P3 , P2 !, P4 , P2 , P1 , P3 !, P4 , P2 , P3 , P1 !, P4 , P3 , P1 , P2 !, P4 , P3 , P2 , P1 ! (b) V 1 18 / 24; V 2 2 / 24; V 3 2 / 24; V 4 This is the same situation as in (a). 2 / 24. (c) This is also the same situation as in (a) – any time a group of players has 41 votes, they must have 50 votes. The Shapley-Shubik power distribution is thus still V 1 12 / 24; V 2 4 / 24; V 3 4 / 24; V4 4 / 24. 35. (a) There are 3! = 6 sequential coalitions of the three players. P1 , P2 , P3 !, P1 , P3 , P2 !, P2 , P1 , P3 !, P2 , P3 , P1 !, P3 , P1 , P2 !, P3 , P2 , P1 ! (The second player listed will always be pivotal unless the first two players listed are P2 and P3 since that is not a winning coalition. In that case, P1 is pivotal.) (b) P1 is pivotal four times; P2 is pivotal one time; P3 is pivotal one time. The Shapley-Shubik power distribution is V 1 4 / 6; V 2 1/ 6; V 3 1/ 6. 36. (a) P1 , P2 , P3 ! , P1 , P3 , P2 ! , P2 , P1 , P3 ! , P2 , P3 , P1 ! , P3 , P1 , P2 ! , P3 , P2 , P1 ! (The third player listed will always be pivotal unless the first two players listed are P1 and P2 since that is a winning coalition. In that case, the second of these players is pivotal.) (b) V 1 1/ 2 , V 2 1/ 2 , V 3 0 37. We proceed to identify pivotal players by moving down each column. We start with the first column. Since P2 is pivotal in the sequential coalition P1 , P2 , P3 , P4 ! , we know that ^ P1 , P2 ` form a winning coalition. This tells us that P2 is also pivotal in the second sequential coalition P1 , P2 , P4 , P3 ! listed in the first column. A similar argument tells us that P3 is pivotal in the fourth sequential coalition P1 , P3 , P4 , P2 ! listed in the first column. In the second column, P1 is clearly pivotal in the sequential coalition P2 , P1 , P3 , P4 ! since we know that ^P1 , P2 ` form a winning coalition. Similarly, P1 is pivotal in the sequential coalition P2 , P1 , P4 , P3 ! . In the third column, P1 is pivotal in the sequential coalition P3 , P1 , P2 , P4 ! since the third sequential coalition in the first column ( P1 , P3 , P2 , P4 ! ) identified ^ P1 , P3 ` as a winning coalition. Similarly, P1 is pivotal in the sequential coalition P3 , P1 , P4 , P2 ! . Since P1 was pivotal in the sequential coalition P2 , P3 , P1 , P4 ! back in the second column, it must also be the case that P1 is pivotal in the sequential coalition P3 , P2 , P1 , P4 ! . Similarly, P4 is pivotal in the sequential coalition P3 , P2 , P4 , P1 ! since P4 was Copyright © 2018 Pearson Education, Inc. 30 Chapter 2: The Mathematics of Power also pivotal in the sequential coalition P2 , P3 , P4 , P1 ! back in the second column. In the fourth column, the third player listed will always be pivotal since the first two players when listed first in the opposite order (in an earlier column) never contain a pivotal player and the fourth player was never pivotal in the first three columns. The final listing of pivotal players is found below. P1 , P2 , P3 , P4 !, P2 , P1 , P3 , P4 !, P3 , P1 , P2 , P4 !, P4 , P1 , P2 , P3 !, P1 , P2 , P4 , P3 !, P2 , P1 , P4 , P3 !, P3 , P1 , P4 , P2 !, P4 , P1 , P3 , P2 !, P1 , P3 , P2 , P4 !, P2 , P3 , P1 , P4 !, P3 , P2 , P1 , P4 !, P4 , P2 , P1 , P3 !, P1 , P3 , P4 , P2 !, P2 , P3 , P4 , P1 !, P3 , P2 , P4 , P1 !, P4 , P2 , P3 , P1 !, P1 , P4 , P2 , P3 !, P2 , P4 , P1 , P3 !, P3 , P4 , P1 , P2 !, P4 , P3 , P1 , P2 !, P1 , P4 , P3 , P2 !, P2 , P4 , P3 , P1 !, P3 , P4 , P2 , P1 !, P4 , P3 , P2 , P1 ! The Shapley-Shubik distribution is then V 1 10 / 24; V 2 6 / 24; V 3 6 / 24; V 4 2 / 24. 38. We identify pivotal players by moving down each column. Since P2 is pivotal in the sequential coalition P1 , P2 , P3 , P4 ! , we know that ^ P1 , P2 ` form a winning coalition. This tells us that P2 is also pivotal in the second sequential coalition P1 , P2 , P4 , P3 ! . In the second column, P1 is clearly pivotal in the sequential coalition P2 , P1 , P3 , P4 ! since we know that ^P1 , P2 ` form a winning coalition. Similarly, P1 is pivotal in the sequential coalition P2 , P1 , P4 , P3 ! . Since P1 is pivotal in P2 , P3 , P4 , P1 ! , we know that P1 must be pivotal in P2 , P4 , P3 , P1 ! too. In the third column, P2 is pivotal in the sequential coalition P3 , P1 , P2 , P4 ! according to the third sequential coalition in the first column ( P1 , P3 , P2 , P4 ! ). Similarly, P4 is pivotal in the sequential coalition P3 , P1 , P4 , P2 ! according to the fourth sequential coalition in the first column ( P1 , P3 , P4 , P2 ! ). P1 is pivotal in the sequential coalition P3 , P2 , P1 , P4 ! according to the third sequential coalition in the second column ( P2 , P3 , P1 , P4 ! ). P1 is pivotal in the sequential coalitions P3 , P2 , P4 , P1 ! and P3 , P4 , P2 , P1 ! according to the fourth sequential coalition in the second column ( P2 , P3 , P4 , P1 ! ). In the fourth column, P2 is pivotal in the sequential coalition P4 , P1 , P2 , P3 ! according to the fifth sequential coalition in the first column ( P1 , P4 , P2 , P3 ! ). Also, P3 is pivotal in the sequential coalition P4 , P1 , P3 , P2 ! according to the sixth sequential coalition in the first column ( P1 , P4 , P3 , P2 ! ). We have P1 pivotal in the sequential coalition P4 , P2 , P1 , P3 ! according to the fifth sequential coalition in the second column ( P2 , P4 , P1 , P3 ! ). We also have P1 pivotal in the sequential coalitions P4 , P2 , P3 , P1 ! and P4 , P3 , P2 , P1 ! according to the fourth sequential coalition in the second column ( P2 , P3 , P4 , P1 ! ). Copyright © 2018 Pearson Education, Inc. ISM: Excursions in Modern Mathematics, 9E 31 Lastly, we see that P1 pivotal in the sequential coalition P4 , P3 , P1 , P2 ! according to the fifth sequential coalition in the third column ( P3 , P4 , P1 , P2 ! ). The final listing of pivotal players is found below. P1 , P2 , P3 , P4 !, P2 , P1 , P3 , P4 !, P3 , P1 , P2 , P4 !, P4 , P1 , P2 , P3 !, P1 , P2 , P4 , P3 !, P2 , P1 , P4 , P3 !, P3 , P1 , P4 , P2 !, P4 , P1 , P3 , P2 !, P1 , P3 , P2 , P4 !, P2 , P3 , P1 , P4 !, P3 , P2 , P1 , P4 !, P4 , P2 , P1 , P3 !, P1 , P3 , P4 , P2 !, P2 , P3 , P4 , P1 !, P3 , P2 , P4 , P1 !, P4 , P2 , P3 , P1 !, P1 , P4 , P2 , P3 !, P2 , P4 , P1 , P3 !, P3 , P4 , P1 , P2 !, P4 , P3 , P1 , P2 !, P1 , P4 , P3 , P2 !, P2 , P4 , P3 , P1 !, P3 , P4 , P2 , P1 !, P4 , P3 , P2 , P1 ! The Shapley-Shubik distribution is then V 1 14 / 24; V 2 6 / 24; V 3 2 / 24; V 4 2 / 24. 2.4. Subsets and Permutations 39. (a) A set with N elements has 2 N subsets. So, A has 210 1024 subsets. (b) There is only one subset, namely the empty set { }, having less than one element. So, using part (a), the number of subsets of A having one or more elements is 1024 – 1 = 1023. (c) Each element of A can be used to form a subset containing exactly one element. So, there are exactly 10 such subsets. (d) The number of subsets of A having two or more elements is the number of subsets of A not having either 0 or 1 element. From parts (a), (b) and (c), we calculate this number as 1024 – 1 – 10 = 1013. 40. (a) A set with N elements has 2 N subsets. So, A has 212 4096 subsets. (b) There is only one subset, namely the empty set { }, having less than one element. So, using part (a), the number of subsets of A having one or more elements is 4096 – 1 = 4095. (c) Each element of A can be used to form a subset containing exactly one element. So, there are exactly 12 such subsets. (d) The number of subsets of A having two or more elements is the number of subsets of A not having either 0 or 1 element. From parts (a), (b) and (c), we calculate this number as 4096 – 1 – 12= 4083. 41. (a) A weighted voting system with N players has 2 N 1 coalitions. So, a system with 10 players has 210 1 1023 coalitions. (b) The number of coalitions with two or more players is all of the coalitions minus the number of those having at most one player. But we know that there are exactly 10 coalitions consisting of exactly one player. So, the number of coalitions with two or more players is 1023 – 10 = 1013. See also Exercise 39(d). 42. (a) A weighted voting system with N players has 2 N 1 coalitions. So, a system with 12 players has 212 1 4095 coalitions. (b) 4095 – 12 = 4083 43. (a) 26 1 63 coalitions (b) There are 25 1 31 coalitions of the remaining five players P2 , P3 , P4 , P5 , and P6 . These are exactly those coalitions that do not include P1 . Copyright © 2018 Pearson Education, Inc. 32 Chapter 2: The Mathematics of Power (c) As in (b), there are 25 1 31 coalitions of the remaining five players P1 , P2 , P4 , P5 , and P6 . These are exactly those coalitions that do not include P3 . (d) There are 24 1 15 coalitions of the remaining four players P2 , P4 , P5 , P6 . These are exactly those coalitions that do not include either P1 or P3 . (e) 16 coalitions include P1 and P3 (all 24 1 15 coalitions of the remaining four players P2 , P4 , P5 , P6 together with the empty coalition could be combined with these two players to form such a coalition). 44. (a) 25 1 31 coalitions (b) There are 24 1 15 coalitions of the remaining four players P2 , P3 , P4 , and P5 . These are exactly those coalitions that do not include P1 . (c) As in (b), there are 24 1 15 coalitions of the remaining four players P1 , P2 , P3 , and P4 . These are exactly those coalitions that do not include P5 . (d) There are 23 1 7 coalitions of the remaining three players P2 , P3 , P4 . These are exactly those coalitions that do not include P1 or P5 . (e) 8 coalitions include P1 and P5 (all 23 1 7 coalitions of the remaining three players P2 , P3 , P4 together with the empty coalition could be combined with these two players to form such a coalition). 45. (a) 13! = 6,227,020,800 (b) 18! 6,402,373,705,728,000 | 6.402374 u 1015 (c) 25! 15,511,210,043,330,985,984,000,000 | 1.551121u 1025 (d) There are 25! sequential coalitions of 25 players. 1 second 1 hour 1 day 1 year u 25! sequential coaltions u u u 1, 000, 000, 000, 000 sequential coalitions 3600 seconds 24 hours 365 days | 491,857 years 46. (a) 12! = 479,001,600 (b) 15! 1,307,674,368,000 | 1.307674 u 1012 (c) 20! 2,432,902,008,176,640,000 | 2.432902 u 1018 (d) There are 20! sequential coalitions of 20 players. 1 second 1 hour 1 day 1 year u u u 20! sequential coalitions u 1, 000, 000, 000 sequential coalitions 3600 seconds 24 hours 365 days | 77 years 47. (a) (b) 13! 3! 13! 3!10! 6, 227, 020,800 6 6, 227, 020,800 6 u 3, 628,800 1, 037,836,800 286 Copyright © 2018 Pearson Education, Inc. ISM: Excursions in Modern Mathematics, 9E (c) 13! 4!9! 6, 227, 020,800 24 u 362,880 715 (d) 13! 5!8! 6, 227, 020,800 120 u 40,320 1287 48. (a) 12! 2! 479, 001, 600 2 239,500,800 (b) 12! 2!10! 479, 001, 600 2 u 3, 628,800 66 (c) 12! 3!9! 479, 001, 600 6 u 362,880 220 (d) 12! 4!8! 479, 001, 600 24 u 40,320 495 49. (a) 10! 10 u 9 u 8 u u 3 u 2 u 1 10 u 9! 10! So, 9! 10 3, 628,800 10 362,880 . (b) 11! 11u 10 u 9 u u 3 u 2 u 1 11u 10! 11! 11u 10! So, 11 . 10! 10! (c) 11! 11u 10 u 9 u u 3 u 2 u 1 11u 10 u 9! 11u 10 u 9! 11u 10 110 . 9! So, 11! 9! (d) 9! 6! 9 u 8 u 7 u 6! 6! (e) 101! 99! 9u8u 7 504 101u 100 u 99! 101u 100 10,100 99! 50. (a) 20! 20 u 19! so 19! 20! 20 2, 432,902, 008,176, 640, 000 20 121,645,100,408,832,000 (b) 20 (c) 201! 199! 201u 200 u 199! 199! 201u 200 40, 200 (d) 990 11!/ 8! (11u 10 u 9 u 8!) / 8! 11u 10 u 9 990 51. (a) 7! 5040 sequential coalitions Copyright © 2018 Pearson Education, Inc. 33 34 Chapter 2: The Mathematics of Power (b) There are 6! 720 sequential coalitions of the remaining six players P1 , P2 , P3 , P4 , P5 and P6 . These correspond exactly to those seven-player coalitions that have P7 as the first player. (c) As in (b), there are 6! 720 sequential coalitions of the remaining six players P1 , P2 , P3 , P4 , P5 and P6 . These correspond exactly to those seven-player coalitions that have P7 as the last player. (d) 5040 – 720 = 4320 sequential coalitions do not have P1 listed as the first player. 52. (a) 6! = 720 (b) There are 5! 120 sequential coalitions of the remaining five players P1 , P2 , P3 , P5 and P6 . These correspond exactly to those six-player coalitions that have P4 as the last player. (c) 5! = 120; see (b) (d) 720 – 120 = 600 53. (a) First, note that P1 is pivotal in all sequential coalitions except when it is the first player. By 51(d), P1 is pivotal in 4320 of the 5040 sequential coalitions. (b) Based on the results of (a), V 1 4320 / 5040 6/7. (c) Since players P2 , P3 , P4 , P5 , P6 and P7 share the remaining power equally, V 1 V2 V3 V4 V5 V6 V7 4320 / 5040 6 / 7 , . [Note also that 1/42 is 1/7 divided into 6 equal parts.] 120 / 5040 1/ 42 V5 V 6 1/ 30 (the other players share power equally) 54. (a) 600 (see Exercise 52(d)) (b) 600/720 = 5/6 (c) V 1 5/ 6 , V2 V3 V4 JOGGING 55. (a) Suppose that a winning coalition that contains P is not a winning coalition without P. Then P would be a critical player in that coalition, contradicting the fact that P is a dummy. (b) P is a dummy œ P is never critical œ the numerator of its Banzhaf power index is 0 œ it’s Banzhaf power index is 0. (c) Suppose P is not a dummy. Then, P is critical in some winning coalition. Let S denote the other players in that winning coalition. The sequential coalition with the players in S first (in any order), followed by P and then followed by the remaining players has P as its pivotal player. Thus, P’s Shapley-Shubik power index is not 0. Conversely, if P’s Shapley-Shubik power index is not 0, then P is pivotal in some sequential coalition. A coalition consisting of P together with the players preceding P in that sequential coalition is a winning coalition and P is a critical player in it. Thus, P is not a dummy. 56. (a) P5 is a dummy. It takes (at least) three of the first four players to pass a motion. P5 ‘ s vote doesn’t make any difference. (b) E1 V1 1/ 4; E2 V2 1/ 4; E3 V3 1/ 4; E4 V4 1/ 4; E5 V5 0 (c) q = 21, q = 31, q = 41. (d) Since we assume that the weights are listed in non-increasing order, P5 is a dummy if w = 1, 2, 3. Copyright © 2018 Pearson Education, Inc. ISM: Excursions in Modern Mathematics, 9E 35 57. (a) The quota must be at least half of the total number of votes and not more than the total number of votes. 7 d q d 13 . (b) For q = 7 or q = 8, P1 is a dictator because {P1} is a winning coalition. (c) For q = 9, only P1 has veto power since P2 and P3 together have just 5 votes. (d) For 10 d q d 12, both P1 and P2 have veto power since no motion can pass without both of their votes. For q = 13, all three players have veto power. (e) For q = 7 or q = 8, both P2 and P3 are dummies because P1 is a dictator. For 10 d q d 12, P3 is a dummy since all winning coalitions contain {P1 , P2 } which is itself a winning coalition. 58. (a) 5 d w d 9 ; Since we assume that the weights are listed in non-increasing order, we have 5 d w . In w 5 2 1 9 d w 5 2 1 , i.e., w + 8 0 (c = 0 would make no sense), then the sum of the weights of all the players in the coalition (including Pk ) is at least cq but the sum of the weights of all the players in the coalition except Pk is less than cq. Therefore Pk is critical in the same coalition in the weighted voting system [cq : cw1 , cw2 , , cwN ]. Since the critical players are the same in both weighted voting systems, the Banzhaf power distributions will be the same. Copyright © 2018 Pearson Education, Inc. 38 Chapter 2: The Mathematics of Power 68. (a) Both have Shapley-Shubik power distribution V 1 1/ 2; V 2 1/ 6; V 3 1/ 6; V 4 1/ 6. (b) In the weighted voting system [q : w1 , w2 , , wN ], Pk is pivotal in the sequential coalition P1 , P2 , , Pk , , PN means w1 w2 wk t q but w1 w2 wk 1 q . In the weighted voting system [cq : cw1 , cw2 , , cwN ] , Pk is pivotal in the sequential coalition P1 , P2 , , Pk , , PN means cw1 cw2 cwk t cq but cw1 cw2 cwk 1 cq . These two statements are equivalent since cw1 cw2 cwk c w1 w2 wk t cq if and only if w1 w2 wk t q , and cw1 cw2 cwk 1 c w1 w2 wk 1 cq if and only if w1 w2 wk 1 q . This same reasoning applies to any sequential coalition and so the pivotal players are exactly the same. 69. (a) A player is critical in a coalition if that coalition without the player is not on the list of winning coalitions. The critical players in the winning coalitions are underlined: {P1 , P2 , P3 }, {P1 , P2 , P4 }, {P1 , P3 , P4 }, {P1 , P2 , P3 , P4 } . So the Banzhaf power distribution is E1 4 /10 , E2 E3 E4 2 /10 . (b) A player P is pivotal in a sequential coalition if the players that appear before P in that sequential coalition do not, as a group, appear on the list of winning coalitions, but would appear on the list of winning coalitions if P were included. For example, P3 is pivotal in P1 , P2 , P3 , P4 ! since {P1 , P2 } is not on the list of winning coalitions but {P1 , P2 , P3 } is on that list. The pivotal players in each sequential coalition is underlined: P1 , P2 , P3 , P4 !, P1 , P2 , P4 , P3 !, P1 , P3 , P2 , P4 !, P1 , P3 , P4 , P2 !, P1 , P4 , P2 , P3 !, P1 , P4 , P3 , P2 !, P2 , P1 , P3 , P4 !, P2 , P1 , P4 , P3 !, P2 , P3 , P1 , P4 !, P2 , P3 , P4 , P1 !, P2 , P4 , P1 , P3 !, P2 , P4 , P3 , P1 !, P3 , P1 , P2 , P4 !, P3 , P1 , P4 , P2 !, P3 , P2 , P1 , P4 !, P3 , P2 , P4 , P1 !, P3 , P4 , P1 , P2 !, P3 , P4 , P2 , P1 !, P4 , P1 , P2 , P3 !, P4 , P1 , P3 , P2 !, P4 , P2 , P1 , P3 !, P4 , P2 , P3 , P1 !, P4 , P3 , P1 , P2 !, P4 , P3 , P2 , P1 ! So, the Shapley-Shubik power indices are V 1 12 / 24 1/ 2 , V 2 V3 V4 4 / 24 1/ 6 . 70. (a) [5: 2,1,1,1,1,1,1] (b) The mayor has a Shapley-Shubik power index of 2/7. (Of the 7! sequential coalitions, the mayor is pivotal anytime he/she is in the 4th slot or in the 5th slot, and there are 6! of each kind.) Each of the 1 § 2· 5 . council members has a Shapley-Shubik power index of ¨1 ¸ 6 © 7 ¹ 42 RUNNING 71. Write the weighted voting system as [q: 8x, 4x, 2x, x]. The total number of votes is 15x. If x is even, then so 15 x 15 x is 15x. Since the quota is a simple majority, q 1 . But, 1 7.5 x 1 d 8 x since x t 2 . So, P1 2 2 15 x 1 . (having 8x votes) is a dictator. If x is odd, then so is 15x. Since the quota is a simple majority, q 2 15 x 1 1 But, 7.5 x d 8 x since x t 1 . So, again, P1 (having 8x votes) is a dictator. 2 2 Copyright © 2018 Pearson Education, Inc. ISM: Excursions in Modern Mathematics, 9E 39 72. Since P2 is a pivotal player in the sequential coalition P1 , P2 , P3 ! , it is clear that {P1 , P2 } and {P1 , P2 , P3 } are winning coalitions, but {P1} is a losing coalition. Similarly, we construct the following chart. Pivotal Player and Sequential Coalition Winning Coalitions Losing Coalitions P1 , P2 , P3 ! {P1 , P2 } , {P1 , P2 , P3 } {P1} P1 , P3 , P2 ! {P1 , P3 } , {P1 , P2 , P3 } {P1} P2 , P1 , P3 ! {P1 , P2 } , {P1 , P2 , P3 } {P2 } P2 , P3 , P1 ! {P1 , P2 , P3 } {P2 } , {P3 } , {P2 , P3 } P3 , P1 , P2 ! {P1 , P3 } , {P1 , P2 , P3 } {P3 } P3 , P2 , P1 ! {P1 , P2 , P3 } {P2 } , {P3 } , {P2 , P3 } Based on this, the winning coalitions with critical players underlined are {P1 , P2 } , {P1 , P3 } , and {P1 , P2 , P3 } . It follows that the Banzhaf power distribution is E1 3 / 5; E 2 1/ 5; E 3 1/ 5 . 73. (a) [4: 2, 1, 1, 1] or [9: 5, 2, 2, 2] are among the possible answers. (b) The sequential coalitions (with pivotal players underlined) are: H , A1 , A2 , A3 !, H , A1 , A3 , A2 !, H , A2 , A1 , A3 !, H , A2 , A3 , A1 !, H , A3 , A1 , A2 !, H , A3 , A2 , A1 !, A1 , H , A2 , A3 !, A1 , H , A3 , A2 !, A1 , A2 , H , A3 !, A1 , A2 , A3 , H !, A1 , A3 , H , A2 !, A1 , A3 , A2 , H !, A2 , H , A1 , A3 !, A2 , H , A3 , A1 !, A2 , A1 , H , A3 !, A2 , A1 , A3 , H !, A2 , A3 , H , A1 !, A2 , A3 , A1 , H !, A3 , H , A1 , A2 !, A3 , H , A2 , A1 !, A3 , A1 , H , A2 !, A3 , A1 , A2 , H !, A3 , A2 , H , A1 !, A3 , A2 , A1 , H ! . H is pivotal in 12 coalitions; A1 is pivotal in four coalitions; A2 is pivotal in four coalitions; A3 is pivotal in four coalitions. The Shapley-Shubik power distribution is V H VA 1 VA 2 VA 3 12 / 24 1/ 2; 4 / 24 1/ 6 . 74. There are N! ways that P1 (the senior partner) can be the first player in a sequential coalition (this is the number of sequential coalitions consisting of the other N players). When this happens, the senior partner is not pivotal (the second player listed in the sequential coalition is instead). There are ( N 1)! N ! = N !( N 1 1) = N ˜ N ! ways that P1 is not the first player in a sequential coalition. In each of these cases, P1 is the pivotal player. It follows that the Shapley-Shubik power index of P1 (the senior partner) is N ˜N! = ( N 1)! N ˜N! N = . Since the other N players divide the remaining power equally, the junior partners each ( N 1) ˜ N ! N 1 N N 1 N 1 1 N N N 1 1 1 have a Shapley-Shubik power index of . N N N N 1 75. (a) The losing coalitions are {P1}, {P2 }, and {P3 }. The complements of these coalitions are {P2 , P3 }, {P1 , P3 }, and {P1 , P2 } respectively, all of which are winning coalitions. Copyright © 2018 Pearson Education, Inc. 40 Chapter 2: The Mathematics of Power (b) The losing coalitions are {P1}, {P2 }, {P3 }, {P4 }, {P2 , P3 }, {P2 , P4 }, and {P3 , P4 }. The complements of these coalitions are {P2 , P3 , P4 }, {P1 , P3 , P4 }, {P1 , P2 , P4 }, {P1 , P2 , P3 }, {P1 , P4 }, {P1 , P3 }, and {P1 , P2 } respectively, all of which are winning coalitions. (c) If P is a dictator, the losing coalitions are all the coalitions without P; the winning coalitions are all the coalitions that include P. The complement of any coalition without P (losing) is a coalition with P (winning). (d) Take the grand coalition out of the picture for a moment. Of the remaining 2 N 2 coalitions, half are losing coalitions and half are winning coalitions, since each losing coalition pairs up with a winning coalition (its complement). Half of 2 N 2 is 2 N 1 1. In addition, we have the grand coalition (always a winning coalition). Thus, the total number of winning coalitions is 2 N 1. 76. The mayor has power index 5/13 and each of the four other council members has a power index of 2/13. The ^ ` ^M , P , P ` , ^M , P , P ` , ^M , P , P ` , ^M , P , P ` , ^M , P , P ` , ^M , P , P , P ` , ^M , P , P , P ` , ^M , P , P , P ` , ^M , P , P , P ` , ^ P , P , P , P ` , winning coalitions (with critical players underlined) are: M , P2 , P3 , 3 5 4 2 5 3 4 2 3 5 2 4 2 5 4 2 3 4 5 5 3 2 3 4 4 5 ^M , P2 , P3 , P4 , P5 ` . 77. (a) According to Table 2-11, the Banzhaf index of California in the Electoral College is E California The relative voting weight of California is wCalifornia for California is then S California (b) S H 1 S LB 0.114 . 55 | 0.1022 . The relative Banzhaf voting power 538 0.114 | 1.115 . 0.1022 1/ 3 115 1/ 3 115 , SH2 , S OB 31/115 93 31/115 93 0 0 0 , S GC 0. 2 /115 2 /115 1/ 3 28 /115 115 , S NH 84 0 21/115 0, 78. (a) The possible coalitions are all coalitions with A but not P in them or with P but not A in them. If we call B and C the players with 4 and 3 votes respectively, the possible coalitions are: ^ A` , ^ A, B` , ^ A, C` , ^ A, B, C` and ^P` , ^P, B` , ^P, C` , ^P, B, C` . The winning coalitions (with critical players underlined) are: ^ A, B` , ^ A, C` , ^ A, B, C` , ^ P, B, C` . The Banzhaf power distribution in this case is E A 3 / 8; E B 2 / 8; EC 2 / 8; E D 1/ 8 . (b) The possible coalitions under these circumstances are all subsets of the players that contain either A or P but not both. There are 2 N 2 subsets that contain neither A nor P (all possible subsets of the N – 2 other players). If we throw A into each of these subsets, we get all the coalitions that have A but not P in them—a total of 2 N 2 coalitions. If we throw P into each of the 2 N 2 subsets we get all the coalitions that have P but not A in them. Adding these two lists gives 2 ˜ 2 N 2 2 N 1 coalitions. (c) Consider the same weighted voting system discussed in part (a) but without restrictions. The Banzhaf 5 3 3 1 . When A becomes the antagonist power distribution in this case is E A ; EB ; EC ; ED 12 12 12 12 of P, A’s Banzhaf power index goes down (to 3/8) and P’s Banzhaf power index goes up (to 1/8). If we reverse the roles of A and P (A becomes the “player” and P his “antagonist”, the Banzhaf power index calculations remain the same, and in this case it is the antagonist’s (P) power index that goes up and the player’s (A) power that goes down. Copyright © 2018 Pearson Education, Inc. ISM: Excursions in Modern Mathematics, 9E 41 (d) Once P realizes that A will always vote against him, he can vote exactly the opposite way of his true opinion. This puts A’s votes behind P’s true opinion. If A has more votes than P, this strategy essentially increases P’s power at the expense of A. 79. (a) [4: 3,1,1,1]; P1 is pivotal in every sequential coalition except the six sequential coalitions in which P1 is the first member. So, the Shapley-Shubik power index of P1 is (24 – 6)/24 = 18/24 = 3/4. P2 , P3 , and P4 share power equally and have Shapley-Shubik power index of 1/12. (b) In a weighted voting system with 4 players there are 24 sequential coalitions — each player is the first member in exactly 6 sequential coalitions. The only way the first member of a coalition can be pivotal is if she is a dictator. Consequently, if a player is not a dictator, she can be pivotal in at most 24 – 6 = 18 sequential coalitions and so that player’s Shapley-Shubik power index can be at most 18/24 = 3/4. (c) In a weighted voting system with N players there are N! sequential coalitions — each player is the first member in exactly (N–1)! sequential coalitions. The only way the first member of a coalition can be pivotal is if she is a dictator. Consequently, if a player is not a dictator, she can be pivotal in at most N! – (N–1)! = (N–1)!(N–1) sequential coalitions and so that player’s Shapley-Shubik power index can be at most (N–1)!(N–1)/N! = (N–1)/N. (d) [N: N–1, 1, 1, 1,…, 1]; Here N is the number of players as well as the quota. P1 is pivotal in every sequential coalition except for the (N–1)! sequential coalitions in which P1 is the first member. So the Shapley-Shubik power index of P1 is [N! – (N–1)!]/N! = (N–1)/N. 80. (a) A player with veto power is critical in every winning coalition. Therefore that player must have Banzhaf power index of at least as much as any other player. Since the total Banzhaf power indexes of all N players is 1, they cannot all be less than 1/N. (b) A player with veto power is pivotal in every sequential coalition in which that player is the last player. There are (N – 1)! such sequential coalitions. Consequently, the player must have Shapley-Shubik power index of at least (N – 1)!/N! = 1/N. Copyright © 2018 Pearson Education, Inc. 42 Chapter 2: The Mathematics of Power APPLET BYTES 81. Power in the first Electoral College is summarized in the table below. A quota of q = 67 is used. State Virginia Number of Electors 21 Banzhaf Power 17.2034% ShapleyShubik Power 17.3049% Massachusetts 16 12.2842% 12.4892% Pennsylvania 15 11.4728% 11.6567% New York 12 8.9970% 9.0568% North Carolina 12 8.9970% 9.0568% Connecticut 9 6.6458% 6.5945% Maryland 8 5.9173% 5.8680% South Carolina 8 5.9173% 5.8680% New Jersey 7 5.1474% 5.0971% New Hampshire 6 4.2761% 4.2236% Georgia 4 2.9160% 2.8355% Kentucky 4 2.9160% 2.8355% Rhode Island 4 2.9160% 2.8355% Vermont 3 2.1968% 2.1390% Delaware 3 2.1968% 2.1390% 82. Slightly increasing the quota from 58 produces more desirable results (in terms of reducing the gap). One quota to be considered is 65 (Note: This is slightly larger than 31 + 31 + 2 = 64). q = 65 produces the following gaps which are within the 4% threshold. Another interesting quota, though not one meeting the 4% threshold for gaps, is q = 67. Player Weight 31 Weight as a % 26.96% Banzhaf Power 26% Hempstead 1 Gap 0.96% Hempstead 2 31 26.96% 26% 0.96% Oyster Bay 28 24.35% 22% 2.35% N. Hempstead 21 18.26% 22% 3.74% Long Beach 2 1.74% 2% 0.26% Glen Cove 2 1.74% 2% 0.26% Quota 65 Copyright © 2018 Pearson Education, Inc. Excursions in Modern Mathematics 9th Edition Tannenbaum Solutions Manual Full Download: http://testbanklive.com/download/excursions-in-modern-mathematics-9th-edition-tannenbaum-solutions-manual/ ISM: Excursions in Modern Mathematics, 9E 43 83. (a) The following table describes Banzhaf power vs. the weight of each player (as a percentage) in the weighted voting system [65: 30, 28, 22, 15, 7, 6]. Player Weight 30 Weight as a % 27.7778% Banzhaf Power 28.8462% Hempstead 1 Gap 1.0684% Hempstead 2 28 25.9259% 25.0000% 0.9259% Oyster Bay 22 20.3704% 21.1538% 0.7834% N. Hempstead 15 13.8889% 17.3077% 3.4188% Long Beach 7 6.4815% 5.7692% 0.7123% Glen Cove 6 5.5556% 1.9231% 3.6325% Quota 65 (b) The largest gap for q = 65 is 3.6325%. The average value of the gaps for q = 65 is 1.7569%. (c) Experimentation with various quotas produce the results given in the table below. The best largest and average gaps occur for any value of the quota between 60 and 63. Quotas below q = 59 or above q = 65 produce even larger gaps. Weight Weight as a % Banzhaf Power (q=65) Gap Banzhaf Power (q=64) Gap Banzhaf Power (q=60-63) Gap Banzhaf Power (q=59) Gap 30 27.78% 28.85% 1.07% 30.77% 2.99% 27.78% 0.00% 29.63% 1.85% 28 25.93% 25.00% 0.93% 26.92% 1.00% 27.78% 1.85% 25.93% 0.00% 22 20.37% 21.15% 0.78% 19.23% 1.14% 20.37% 0.00% 22.22% 1.85% 15 13.89% 17.31% 3.42% 15.38% 1.49% 12.96% 0.93% 11.11% 2.78% 7 6.48% 5.77% 0.71% 3.85% 2.64% 5.56% 0.92% 7.41% 0.93% 6 5.56% 1.92% 3.64% 3.85% 1.71% 5.56% 0.00% 3.70% 1.86% MAX 3.64% 2.99% 1.85% 2.78% AVG 1.76% 1.83% 0.62% 1.54% Copyright © 2018 Pearson Education, Inc. Full download all chapters instantly please go to Solutions Manual, Test Bank site: testbanklive.com