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Solution Manual for Excursions in Modern Mathematics, 9th Edition

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Solution Manual for Excursions in Modern Mathematics, 9th Edition - Page 1 preview imageSOLUTIONSMANUALDALEBUSKE,PH.D.St. Cloud State UniversityEXCURSIONS INMODERNMATHEMATICSNINTHEDITIONPeter TannenbaumCalifornia State University–Fresno
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Solution Manual for Excursions in Modern Mathematics, 9th Edition - Page 2 preview image
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Solution Manual for Excursions in Modern Mathematics, 9th Edition - Page 3 preview imageiiiTable of ContentsSolutionsChapter 11Chapter 221Chapter 344Chapter 470Chapter 591Chapter 6104Chapter 7120Chapter 8130Chapter 9148Chapter 10160Chapter 11173Chapter 12191Chapter 13211Chapter 14224Chapter 15233Chapter 16247Chapter 17261
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Solution Manual for Excursions in Modern Mathematics, 9th Edition - Page 4 preview imageChapter 1WALKING1.1.Ballots and Preference Schedules1.Number of voters5353231st choiceAACDDB2nd choiceBDECCE3rd choiceCBDBBA4th choiceDCAEAC5thchoiceEEBAEDThis schedule was constructed by noting, for example, that there were five ballots listing candidateCas thefirst preference, candidateEas the second preference, candidateDas the third preference, candidateAas thefourth preference, and candidateBas the last preference.2.Number of voters45621st choiceABCA2nd choiceDCAC3rd choiceBDDD4thchoiceCABB3.(a)5 + 5 + 3 + 3 + 3 + 2 = 21(b)11. There are 21 votes all together. A majority is more than half of the votes, or at least 11.(c)Chavez. Argand has 3 last-place votes, Brandt has 5 last-place votes, Chavez has no last-place votes,Dietz has 3 last-place votes, and Epstein has 5 + 3 + 2 = 10 last-place votes.4.(a)202 + 160 + 153 + 145 + 125 + 110 + 108 + 102 + 55 = 1160(b)581; There are 1160 votes all together. A majority is more than half of the votes, or at least 581.(c)Alicia. She has no last-place votes. Note that Brandy has 125 + 110 + 55 = 290 last-place votes, Cleohas 202 + 145 + 102 = 449 last-place votes, and Dionne has 160 + 153 + 108 = 421 last-place votes.5.Number of voters3736241351st choiceBABEC2nd choiceEBABE3rd choiceADDCA4th choiceCCEAD5thchoiceDECDBHere Brownstein was listed first by 37 voters. Those same 37 voters listed Easton as their second choice,Alvarez as their third choice, Clarkson as their fourth choice, and Dax as their last choice.6.Number of voters14108741st choiceBBADE2nd choiceADBCB3rd choiceEAEBA4th choiceCEDEC5thchoiceDCCAD
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Solution Manual for Excursions in Modern Mathematics, 9th Edition - Page 5 preview image2Chapter 1:The Mathematics of Elections7.Number of voters1410874A23153B11232C55524D42415E34341Here 14 voters had the same preference ballot listingBas their first choice,Aas their second choice,Eastheir third choice,Das their fourth choice, andCas their fifth and last choice.8.Number of voters373624135A12524B31241C24315D53152E454339.Number of voters2554807651st choiceLCM2nd choiceMML3rd choiceCLC(0.17)(1500) = 255; (0.32)(500) = 480; The remaining voters (51% of 1500 or 1500-255-480=765) preferMthe most,Cthe least, so thatLis their second choice.10.Number of voters4509002256751st choiceABCC2nd choiceCCBA3rd choiceBAAB100% - 20% - 40% = 40% of the voters number 225 + 675 = 900. So, ifNrepresents the total number ofvoters, then(0.40)900N. This means there areN= 2250 total voters.20% of 2250 is 450 (these votershave preference ballotsA,C,B). 40% of 2250 is 900 (these voters have preference ballotsB,C,A).1.2.Plurality Method11.(a)C. Ahas 15 first-place votes.Bhas 11 + 8 + 1 = 20 first-place votes.Chas 27 first-place votes.Dhas 9first-place votes.Chas the most first-place votes with 27 and wins the election.(b)C,B,A,D. Candidates are ranked according to the number of first-place votes they received (27, 20, 15,and 9 forC, B, A, andDrespectively).12.(a)D. Ahas 21 first-place votes.Bhas 18 first-place votes.Chas 10 + 1 = 11 first-place votes.Dhas 29first-place votes.Dhas the most first-place votes with 29 and wins the election.(b)D,A,B,C.13.(a)C. Ahas 5 first-place votes.Bhas 4 + 2 = 6 first-place votes.Chas 6 + 2 + 2 + 2 = 12 first-place votes.Dhas no first-place votes.Chas the most first-place votes with 12 and wins the election.(b)C,B,A,D. Candidates are ranked according to the number of first-place votes they received (12, 6, 5,and 0 forC, B, A, andDrespectively).14.(a)B. Ahas 6 + 3 = 9 first-place votes.Bhas 6 + 5 + 3 = 14 first-place votes.Chas no first-place votes.Dhas 4 first-place votes.Bhas the most first-place votes with 14 and wins the election.
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Solution Manual for Excursions in Modern Mathematics, 9th Edition - Page 6 preview imageISM:Excursions in Modern Mathematics, 9E3(b)B,A,D,C. Candidates are ranked according to the number of first-place votes they received (14, 9, 4and 0 forB, A, D, andCrespectively).15.(a)D. Ahas 11% of the first-place votes.Bhas 14% of the first-place votes.Chas 24% of the first-placevotes.Dhas 23% + 19% + 9% = 51% of the first-place votes.Ehas no first-place votes.Dhas thelargest percentage of first-place votes with 51% and wins the election.(b)D,C,B,A,E. Candidates are ranked according to the percentage of first-place votes they received (51%,24%, 14%, 11% and 0% forD, C, B, A, andErespectively).16.(a)C. Ahas 12% of the first-place votes.Bhas 15% of the first-place votes.Chas 25% + 10% + 9% + 8%= 52% of the first-place votes.Dhas no first-place votes.Ehas 21% of the first-place votes.Chas thelargest percentage of first-place votes with 52% and wins the election.(b)C,E,B,A,D. Candidates are ranked according to the percentage of first-place votes they received (52%,21%, 15%, 12%, and 0% forC, E, B, AandDrespectively).17.(a)A. Ahas 5 + 3 = 8 first-place votes.Bhas 3 first-place votes.Chas 5 first-place votes.Dhas 3 + 2 = 5first-place votes.Ehas no first-place votes.Ahas the most first-place votes with 8 and wins the election.(b)A,C,D,B, E.Candidates are ranked according to the number of first-place votes they received (8, 5, 5,3, and 0 forA, C, D, B, andErespectively). Since both candidatesCandDhave 5 first-place votes, thetie in ranking is broken by looking at last-place votes. SinceChas no last-place votes andDhas 3 last-place votes, candidateCis ranked above candidateD.18.(a)A. Ahas 153 + 102 + 55 = 310 first-place votes.Bhas 202 + 108 = 310 first-place votes.Chas 160 +110 = 270 first-place votes.Dhas 145 + 125 = 270 first-place votes. BothAandBhave the most first-place votes with 310 so the tie is broken using last-place votes.Ahas no last-place votes.Bhas 125 +110 + 55 = 290 last-place votes.SoAwins the election.(b)A,B, D, C.Candidates are ranked according to the number of first-place votes they received (310, 310,270 and 270 forA, B, D, andCrespectively). In part (a), we saw that the tie betweenAandBwasbroken in favor ofA. Since both candidatesCandDhave 270 first-place votes, the tie in ranking isbroken by looking at last-place votes. SinceChas 202 + 145 + 102 = 449 last-place votes andDhas160 + 153 + 108 = 421 last-place votes, candidateDis ranked above candidateC.19.(a)A. Ahas 5 + 3 = 8 first-place votes.Bhas 3 first-place votes.Chas 5 first-place votes.Dhas 3 + 2 = 5first-place votes.Ehas no first-place votes.Ahas the most first-place votes with 8 and wins the election.(Note: This is exactly the same as in Exercise 17(a).)(b)A,C,D,B, E.Candidates are ranked according to the number of first-place votes they received (8, 5, 5,3, and 0 forA, C, D, B, andErespectively). Since both candidatesCandDhave 5 first-place votes, thetie in ranking is broken by a head-to-head comparison between the two. But candidateCis ranked higherthanDon 5 + 5 + 3 = 13 of the 21 ballots (a majority). Therefore, candidateCis ranked above candidateD.20.(a)B. Ahas 153 + 102 + 55 = 310 first-place votes.Bhas 202 + 108 = 310 first-place votes.Chas 160 +110 = 270 first-place votes.Dhas 145 + 125 = 270 first-place votes. BothAandBhave the most first-place votes with 310 so the tie is broken by head-to-head comparison. But candidateAis ranked higherthanBon 153 + 125 + 110 + 102 + 55 = 545 of the 1160 ballots (less than a majority).SoBwins thetiebreaker and the election.
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Solution Manual for Excursions in Modern Mathematics, 9th Edition - Page 7 preview image4Chapter 1:The Mathematics of Elections(b)B, A, D, C.Candidates are ranked according to the number of first-place votes they received (310, 310,270 and 270 forB,A,D, andCrespectively). In part (a), we saw that the tie betweenAandBwasbroken in favor ofB. Since both candidatesCandDhave 270 first-place votes, the tie in ranking isbroken by head-to-head comparison. CandidateCis ranked higher thanDon 160 + 153 + 110 + 108 =531 of the 1160 ballots (less than a majority). So in the final ranking, candidateDis ranked abovecandidateC.1.3.Borda Count21.(a)Ahas4153(981)211127163uuuupoints.Bhas4(1181)3152(279)10197uuuupoints.Chas42730281(151191)160uuuupoints.Dhas493(2711)2(151)18190uuuupoints.The winner isB.(b)B,D,A,C.Candidates are ranked according to the number of Borda points they received.22.(a)Ahas4213182(2910)11217uuuupoints.Bhas4183(101)221129176uuuupoints.Chas4(101)3(2921)21810230uuuupoints.Dhas42930211(211810)167uuuupoints.The winner isC.(b)C, A, B, D.Candidates are ranked according to the number of Borda points they received.23.(a)Ahas45322(62)1(422)50uuuupoints.Bhas4(42)3(22)221(65)51uuuupoints.Chas4(6222)302(542)1070uuuupoints.Dhas403(6542)221(22)59uuuupoints.The winner isC.(b)C, D, B, A.Candidates are ranked according to the number of Borda points they received.24.(a)Ahas4(63)3(43)261574uuuupoints.Bhas4(653)33201(64)75uuuupoints.Chas403(665)2(433)1071uuuupoints.Dhas44302(65)1(633)50uuuupoints.The winner is B.(b)B, A, C, D.Candidates are ranked according to the number of Borda points they received.25.Here we can use a total of 100 voters for simplicity.Ahas5114(242319)3(149)2010388uuuuupoints.Bhas514403(2411)2231(199)249uuuuupoints.Chas5244(14119)32321910363uuuuupoints.Dhas5(23199)40302141(2411)318uuuuupoints.Ehas50403192(24119)1(2314)182uuuuupoints.The ranking (according to Borda points) isA, C, D, B, E.
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Solution Manual for Excursions in Modern Mathematics, 9th Edition - Page 8 preview imageISM:Excursions in Modern Mathematics, 9E526.We use a total of 100 voters for simplicity.Ahas51240392(252110)1(158)222uuuuupoints.Bhas515493(2112)281(2510)261uuuuupoints.Chas5(251098)40302151(2112)323uuuuupoints.Dhas504(21151210)3(258)2019340uuuuupoints.Ehas5214(258)3(1510)2(129)10354uuuuupoints.The ranking (according to Borda points) isE, D, C, B, A.27.Cooper had349228013161023uuupoints.Gordon had337243212751250uuupoints.Mariota had37882741222534uuupoints.The ranking (according to Borda points) is Mariota (2534), Gordon (1250), and Cooper (1023).28.Hernandez had713417302010159uuuuupoints.Kluberhad717411322010169uuuuupoints.Lester had7040332151746uuuuupoints.Salehad7042319251378uuuuupoints.Scherzerhad704034261832uuuuupoints.As in Example 1.12, this uses a modified Borda count. In this case, first-place votes count 7 points ratherthan the usual 5.The ranking (according to modified Borda points) is Kluber (169), Hernandez (159), Sale(78), Lester (46), and Sherzer (32).29.Each ballot has4131211110uuuupoints that are awarded to candidates according to the Bordacount. With 110 voters, there are a total of110101100uBorda points. SoDhas 1100 – 320 – 290 – 180 =310 Borda points. The ranking is thusA(320),D(310),B(290), andC(180).30.Each ballot has714131211117uuuuupoints that are awarded to candidates according to theBorda count. With 50 voters, there are a total of5017850uBorda points. SoEhas 850 – 152 – 133 – 191– 175 = 199 Borda points. The ranking is thusE(199),C(191),D(175),A(152), andB(133).1.4.Plurality-with-Elimination31.(a)Ais the winner. Round 1:CandidateABCDNumber of first-place votes1520279Dis eliminated.Round 2: The 9 first-place votes originally going toDnow go toA.CandidateABCDNumber of first-place votes242027Bis eliminated.Round 3: There are 8 + 1 = 9 first-place votes originally going toBthat now go toA. There are also 11first-place votes going toBthat would now go toD. But, since D is already eliminated, these 11 first-place votes go toA.CandidateABCDNumber of first-place votes4427CandidateAnow has a majority of the first-place votes and is declared the winner.(b)A complete ranking of the candidates can be found by noting in part (a) when each candidate waseliminated. SinceDwas eliminated first, it is ranked last. SinceBwas eliminated next, it is ranked nextto last. The final ranking is henceA, C, B, D.
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Solution Manual for Excursions in Modern Mathematics, 9th Edition - Page 9 preview image6Chapter 1:The Mathematics of Elections32.(a)Bis the winner. Round 1:CandidateABCDNumber of first-place votes21181129Cis eliminated.Round 2: The 11 first-place votes originally going toCwould next go toB.CandidateABCDNumber of first-place votes212929Round 3: The 21 first-place votes going toAwould next go toC. ButChas been eliminated. So these21 first-place votes go toB.CandidateABCDNumber of first-place votes5029CandidateBnow has a majority of the first-place votes and is declared the winner.(b)A complete ranking of the candidates can be found by noting in part (a) when each candidate waseliminated. SinceCwas eliminated first, it is ranked last. SinceAwas eliminated next, it is ranked nextto last. The final ranking is henceB, D, A, C.33.(a)Cis the winner. Round 1:CandidateABCDNumber of first-place votes56120CandidateChas a majority of the first-place votes and is declared the winner.(b)To determine a ranking, we ignore the fact thatCwins and at the end of round 1,Dis the first candidateeliminated.Round 2: No first-place votes are changed.CandidateABCDNumber of first-place votes5612Ais eliminated.Round 3: There are 5 first-place votes originally going toAthat now go toC.CandidateABCDNumber of first-place votes617The final ranking isC, B, A, D.34.(a)Bis the winner. Round 1:CandidateABCDNumber of first-place votes91404CandidateBhas a majority of the first-place votes and is declared the winner.(b)To determine a ranking, we ignore the fact thatBwins and at the end of round 1,Cis the first candidateeliminated.Round 2: No first-place votes are changed.CandidateABCDNumber of first-place votes9144Dis eliminated.Round 3: There are 4 first-place votes originally going toDthat now go toA.CandidateABCDNumber of first-place votes1314The final ranking isB, A, D, C.
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Solution Manual for Excursions in Modern Mathematics, 9th Edition - Page 10 preview imageISM:Excursions in Modern Mathematics, 9E735.Round 1:CandidateABCDENumber of first-place votes88760Eis eliminated.Round 2: No first-place votes change.CandidateABCDENumber of first-place votes8876Dis eliminated.Round 3: There are 4 first-place votes forDthat move to candidateCand there are 2 first-place votes forDthat move to candidateB.CandidateABCDENumber of first-place votes81011Ais eliminated.Round 4: There are 5 first-place votes forAthat move to candidateBand there are 3 first-place votes forAthat move to candidateC(sinceDandEhave both been eliminated).CandidateABCDENumber of first-place votes1514Bnow has a majority of the first-place votes and is declared the winner. The final ranking isB, C, A, D, E.36.Round 1:CandidateABCDENumber of first-place votes8411125Bis eliminated.Round 2: There are 4 first-place votes forBthat move to candidateE.CandidateABCDENumber of first-place votes811129Ais eliminated.Round 3: There are 5 first-place votes forAthat move to candidateE(sinceBis eliminated),2 first-placevotes forAthat move to candidateD(sinceBis eliminated), and 1 first-place vote forAthat moves toC.CandidateABCDENumber of first-place votes121414Cis eliminated.Round 4: There are 6 first-place votes forCthat move to candidateE(sinceAis eliminated), 5 first-placevotes forCthat move to candidateD(since bothAandBare eliminated), and there is 1 first-place vote forCthat moves to candidateE(Cearned this vote earlier when candidateAwas eliminated).CandidateABCDENumber of first-place votes1921Enow has a majority of the first-place votes and is declared the winner. The final ranking isE, D, C, A, B.37.(a)Dis the winner. Round 1:CandidateABCDEPercentage of first-place votes111424510CandidateDhas a majority of the first-place votes and is declared the winner.(b)To determine a ranking, we ignore the fact thatDwins and at the end of round 1,Eis the first candidateeliminated.Round 2: No first-place votes are changed.
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Solution Manual for Excursions in Modern Mathematics, 9th Edition - Page 11 preview image8Chapter 1:The Mathematics of ElectionsCandidateABCDEPercentage of first-place votes11142451Ais eliminated.Round 3: The 11% of the first-place votes that went toAnow go toC.CandidateABCDEPercentage of first-place votes143551Bis eliminated.Round 4: The 14% of the first-place votes that went toBnow go toC.CandidateABCDEPercentage of first-place votes4951A complete ranking of the candidates can be found by noting when each candidate was eliminated. Thefinal ranking is henceD, C, B, A, E.38.(a)Cis the winner. Round 1:CandidateABCDEPercentage of first-place votes121552021CandidateChas a majority of the first-place votes and is declared the winner.(b)To determine a ranking, we ignore the fact thatCwins and at the end of round 1,Dis the first candidateeliminated.Round 2: No first-place votes are changed.CandidateABCDEPercentage of first-place votes12155221Ais eliminated.Round 3: The 12% of the first-place votes that went toAnow go toB(sinceDhas already beeneliminated).CandidateABCDEPercentage of first-place votes275221Eis eliminated.Round 4: The 14% of the first-place votes that went toEnow go toB(sinceDhas already beeneliminated).CandidateABCDEPercentage of first-place votes4852A complete ranking of the candidates can be found by noting when each candidate was eliminated. Thefinal ranking is henceC, B, E, A, D.39.Round 1:CandidateABCDENumber of first-place votes88760CandidatesE, D,andCare all eliminated.Round 2: There are 4 first-place votes forDthat go toB(sinceChas been eliminated). There are 2 first-place votes forDthat go toB. There are 7 first-place votes forCthat go toA(since bothDandEareeliminated).CandidateABCDENumber of first-place votes1514Anow has a majority of the first-place votes and is declared the winner.
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Solution Manual for Excursions in Modern Mathematics, 9th Edition - Page 12 preview imageISM:Excursions in Modern Mathematics, 9E940.Round 1:CandidateABCDENumber of first-place votes8411125CandidatesB, E,andAare all eliminated.Round 2: There are 4 first-place votes forBthat go toC(sinceEhas been eliminated). There are 5 first-placevotes forEthat go toD(sinceAhas been eliminated). There are 5 first-place votes forAthat go toC(sincebothBandEare eliminated), there are 2 first-place votes forAthat go toD(sinceBis eliminated), and 1first-place vote forAthat goes toC.CandidateABCDENumber of first-place votes2119Cnow has a majority of the first-place votes and is declared the winner.1.5.Pairwise Comparisons41.(a)CandidateDis the winner.AversusB: 15 + 9 = 24 votes to 27 + 11 + 8 + 1 = 47 votes (Bwins).Bgets 1 point.AversusC: 15 + 11 + 9 + 8 + 1 = 44 votes to 27 votes (Awins).Agets 1 point.AversusD: 15 + 8 + 1 = 24 votes to 27 + 11 + 9 = 47 votes (Dwins).Dgets 1 point.BversusC: 15 + 11 + 9 + 8 + 1 = 44 votes to 27 votes (Bwins).Bgets 1 point.BversusD: 15 + 11 + 8 + 1 = 35 votes to 27 + 9 = 36 votes (Dwins).Dgets 1 point.CversusD: 27 + 8 = 35 votes to 15 + 11 + 9 + 1 = 36 votes. (Dwins).Dgets 1 point.The final tally is 1 point forA, 2 points forB, 0 points forC, and 3 points forD.(b)A complete ranking for the candidates is found by tallying points. In this case, the final ranking isD(3points),B(2 points),A(1 point), andC(0 points).42.(a)CandidateCis the winner.AversusB: 29 + 21 = 50 votes to 18 + 10 + 1 = 29 votes (Awins).Agets 1 point.AversusC: 21 + 18 = 39 votes to 29 + 10 + 1 = 40 votes (Cwins).Cgets 1 point.AversusD: 21 + 18 + 10 = 49 votes to 29 + 1 = 30 votes (Awins).Agets 1 point.BversusC: 18 votes to 29 + 21 + 10 + 1 = 61 votes (Cwins).Cgets 1 point.BversusD: 21 + 18 + 10 + 1 = 50 votes to 29 votes (Bwins).Bgets 1 point.CversusD: 21 + 18 + 10 + 1 = 50 votes to 29 votes. (Cwins).Cgets 1 point.The final tally is 2 points forA, 1 point forB, 3 points forC, and 0 points forD.(b)A complete ranking for the candidates is found by tallying points. In this case, the final ranking isC(3points),A(2 points),B(1 point), andD(0 points).43.(a)CandidateCis the winner.AversusB: 6 + 5 = 11 votes to 4 + 2 + 2 + 2 + 2 = 12 votes (Bwins).Bgets 1 point.AversusC: 5 + 2 = 7 votes to 6 + 4 + 2 + 2 + 2 = 16 votes (Cwins).Cgets 1 point.AversusD: 5 + 2 + 2 = 9 votes to 6 + 4 + 2 + 2 = 14 votes (Dwins).Dgets 1 point.BversusC: 4 + 2 = 6 votes to 6 + 5 + 2 + 2 + 2 = 17 votes (Cwins).Cgets 1 point.BversusD: 4 + 2 + 2 + 2 = 10 votes to 6 + 5 + 2 = 13 votes (Dwins).Dgets 1 point.CversusD: 6 + 2 + 2 + 2 + 2 = 10 votes to 5 + 4 = 9 votes. (Cwins).Cgets 1 point.The final tally is 0 points forA, 1 point forB, 3 points forC, and 2 points forD.(b)A complete ranking for the candidates is found by tallying points. In this case, the final ranking isC(3points),D(2 points),B(1 point), andA(0 points).
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Solution Manual for Excursions in Modern Mathematics, 9th Edition - Page 13 preview image10Chapter 1:The Mathematics of Elections44.(a)CandidateBis the winner.AversusB: 6 + 4 + 3 = 13 votes to 6 + 5 + 3 = 14 votes (Bwins).Bgets 1 point.AversusC: 6 + 4 + 3 + 3 = 16 votes to 6 + 5 = 11 votes (Awins).Agets 1 point.AversusD: 6 + 6 + 3 + 3 = 18 votes to 5 + 4 = 9 votes (Awins).Agets 1 point.BversusC: 6 + 5 + 3 = 14 votes to 6 + 4 + 3 = 13 votes (Bwins).Bgets 1 point.BversusD: 6 + 5 + 3 = 14 votes to 6 + 4 + 3 = 13 votes (Bwins).Bgets 1 point.CversusD: 6 + 6 + 5 + 3 + 3 = 23 votes to 4 votes. (Cwins).Cgets 1 point.The final tally is 2 points forA, 3 points forB, 1 point forC, and 0 points forD.(b)The final ranking isB(3 points),A(2 points),C(1 point), andD(0 points).45.CandidateDis the winner.AversusB: 24% + 23% + 19% + 11% + 9% = 86% of the votes to 14% of the votes (Awins).AversusC: 23% + 19% + 11% = 53% of the votes to 24% + 14% + 9% = 47% of the votes (Awins).AversusD: 24% + 14% + 11% = 49% of the votes to 23% + 19% + 9% = 51% of the votes (Dwins).AversusE: 24% + 23% + 19% + 14% + 11% + 9% = 100% of the votes to 0% of the votes (Awins).BversusC: 14% of the votes to 24% + 23% + 19% + 11% + 9% = 86% of the votes (Cwins).BversusD: 24% + 14% + 11% = 49% of the votes to 23% + 19% + 9% = 51% of the votes (Dwins).BversusE: 24% + 23% + 14% + 11% = 72% of the votes to 19% + 9% = 28% of the votes (Bwins).CversusD: 24% + 14% + 11% = 49% of the votes to 23% + 19% + 9% = 51% of the votes (Dwins).CversusE: 24% + 23% + 14% + 11% + 9% = 81% of the votes to 19% of the votes (Cwins).DversusE: 23% + 19% + 14% + 9% = 65% of the votes to 24% + 11% = 35% of the votes (Dwins).The final tally is 3 points forA, 1 point forB, 2 points forC, 4 points forD, and 0 points forE.46.CandidateCis the winner.AversusB: 25% + 12% + 10% = 47% of the votes to 21% + 15% + 9% + 8% = 53% of the votes (Bwins).AversusC: 21% + 12% = 33% of the votes to 25% + 15% + 10% + 9% + 8% = 67% of the votes (Cwins).AversusD: 12% + 9% = 21% of the votes to 25% + 21% + 15% + 10% + 8% = 79% of the votes (Dwins).AversusE: 12% + 9% = 21% of the votes to 25% + 21% + 15% + 10% + 8% = 79% of the votes (Ewins).BversusC: 21% + 15% + 12% = 48% of the votes to 25% + 10% + 9% + 8% = 52% of the votes (Cwins).BversusD: 15% + 9% = 24% of the votes to 25% + 21% + 12% + 10% + 8% = 76% of the votes (Dwins).BversusE: 15% + 12% + 9% = 36% of the votes to 25% + 21% + 10% + 8% = 64% of the votes (Ewins).CversusD: 25% + 10% + 9% + 8% = 52% of the votes to 21% + 15% + 12% = 48% of the votes (Cwins).CversusE: 25% + 10% + 9% + 8% = 52% of the votes to 21% + 15% + 12% = 48% of the votes (Cwins).DversusE: 15% + 12% + 10% = 37% of the votes to 25% + 21% + 9% + 8% = 63% of the votes (Ewins).The final tally is 0 points forA, 1 point forB, 4 points forC, 2 points forD, and 3 points forE.47.AversusB: 7 + 5 + 3 = 15 votes to 8 + 4 + 2 = 14 votes (Awins).Agets 1 point.AversusC: 8 + 5 + 3 = 16 votes to 7 + 4 + 2 = 13 votes (Awins).Agets 1 point.AversusD: 8 + 5 + 3 = 16 votes to 7 + 4 + 2 = 13 votes (Awins).Agets 1 point.AversusE: 5 + 3 + 2 = 10 votes to 8 + 7 + 4 = 19 votes (Ewins).Egets 1 point.BversusC: 8 + 5 + 2 = 15 votes to 7 + 4 + 3 = 14 votes (Bwins).Bgets 1 point.BversusD: 8 + 5 = 13 votes to 7 + 4 + 3 + 2 = 16 votes (Dwins).Dgets 1 point.BversusE: 8 + 5 + 4 + 2 = 19 votes to 7 + 3 = 10 votes (Bwins).Bgets 1 point.CversusD: 8 + 7 + 5 = 20 votes to 4 + 3 + 2 = 9 votes (Cwins).Cgets 1 point.CversusE: 7 + 5 + 4 + 2 = 18 votes to 8 + 3 = 11 votes (Cwins).Cgets 1 point.DversusE: 5 + 4 + 3 + 2 = 14 votes to 8 + 7 = 15 votes (Ewins).Egets 1 point.The final tally is 3 points forA, 2 points forB, 2 points forC, 1 point forD, and 2 points forE.NowB,C, andEeach have 2 points. In head-to-head comparisons,BbeatsCandBbeatsEso thatBisranked higher thanCandE. Also,CbeatsEsoCis ranked higher thanEas well. The final ranking is thusA, B, C, E, D.
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Solution Manual for Excursions in Modern Mathematics, 9th Edition - Page 14 preview imageISM:Excursions in Modern Mathematics, 9E1148.AversusB: 6 + 5 + 5 + 5 + 5 + 2 + 1 = 29 votes to 11 votes (Awins).Agets 1 point.AversusC: 7 + 5 + 5 + 2 + 1 = 20 votes to 20 votes (tie).AandCeach get ½ point.AversusD: 6 + 5 + 5 + 5 + 2 + 1 = 24 votes to 16 votes (Awins).Agets 1 point.AversusE: 7 + 6 + 5 + 5 + 5 + 2 + 1 = 31 votes to 9 votes (Awins).Agets 1 point.BversusC: 7 + 5 + 5 + 4 + 2 = 23 votes to 17 votes (Bwins).Bgets 1 point.BversusD: 6 + 5 + 5 + 4 + 2 + 1 = 23 votes to 17 votes (Bwins).Bgets 1 point.BversusE: 7 + 5 + 5 + 4 + 2 = 23 votes to 17 votes (Bwins).Bgets 1 point.CversusD: 6 + 5 + 5 + 4 + 1 = 21 votes to 19 votes (Cwins).Cgets 1 point.CversusE: 7 + 6 + 5 + 5 + 1 = 24 votes to 16 votes (Cwins).Cgets 1 point.DversusE: 7 + 5 + 5 + 2 = 19 votes to 21 votes (Ewins).Egets 1 point.The final tally is 3.5 points forA, 3 points forB, 2.5 points forC, 0 points forD, and 1 point forE.The final ranking is thusA, B, C, E, D.49.(a)With five candidates, there are a total of 4 + 3 + 2 + 1 = 10 pairwise comparisons.Each candidate is partof 4 of these (one against each other candidate). So, to find the number of points each candidate earns,we simply subtract the losses from 4. The 10 points are distributed as follows:Ewins11 2points,Dwins12 2points,Cgets 3 points,Bgets 2 points, andAgets the remaining11101232122point.SoAloses 3 pairwise comparisons.(b)CandidateC, with 3 points, is the winner. [The complete ranking isC, D, B, E, A.]50.(a)Since there are a total of(65) / 215upairwise comparisons,Fmust have won 15 – 1 – 2 – 2 -13 2-12 2= 4 of them (Aearned 1 point,BandCeach earned 2 points,Dearned13 2, andEearned12 2points). This meansFlost1 pairwise comparison.(b)CandidateF, with 4 points, is the winner.1.6.Fairness Criteria51.First, we determine the winner using the Borda count.Ahas4630201(23)29uuuupoints.Bhas4236231032uuuupoints.Chas4332261030uuuupoints.Dhas4033221619uuuupoints.So candidateBis the winner using Borda count. However, candidateAhas 6 of the 11 votes (a majority) andbeats all three other candidates (B,C, andD) in head-to-head comparisons. That is, candidateAis aCondorcet candidate. Since this candidate did not win using the Borda count, this is a violation of theCondorcet criterion.52.First, we determine the winner is candidateBusing the plurality-with-elimination method (see Exercise32(a)). In Exercise 42, however, we saw that candidateCwas a Condorcet candidate (beating each othercandidate in head-to-head comparisons and earning 3 points in the process). Since candidateCdid not winusing plurality-with-elimination, this is a violation of the Condorcet criterion.
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Solution Manual for Excursions in Modern Mathematics, 9th Edition - Page 15 preview image12Chapter 1:The Mathematics of Elections53.The winner of this election is candidateRusing the plurality method. NowFis clearly a nonwinningcandidate. RemovingFas a candidate leaves the following preference table.Number of voters494831st choiceRHH2nd choiceHSS3rd choiceOOO4th choiceSRRIn a recount, candidateHwould be the winner using the plurality method. This is a violation of the IIAcriterion.54.First, we determine the winner using the Borda count.Ahas41430201(10841)79uuuupoints.Bhas443(1410)2(81)10106uuuupoints.Chas4(101)382(144)10104uuuupoints.Dhas483(41)21011481uuuupoints.So candidateB(Boris) is the winner using Borda count. NowD(Dave) is clearly a nonwinning (irrelevant)candidate. RemovingDas a candidate leaves the following preference table.Number of voters14108411st choiceACCBC2nd choiceBBBCB3rd choiceCAAAAWe now recount using the Borda count.Ahas314201(10841)65uuupoints.Bhas342(141081)1078uuupoints.Chas3(1081)2411479uuupoints.In a recount, candidateC(Carmen!) would be the winner using Borda count, a violation of the IIA criterion.55.First, we determine the winner using plurality-with-elimination. Round 1:CandidateABCDENumber of first-place votes83550CandidateEis eliminated. Round 2: No votes are shifted.CandidateABCDENumber of first-place votes8355CandidateBis eliminated. Round 3: The 3 first-place votes forBnow go toA(sinceEhas been eliminated).CandidateABCDENumber of first-place votes1155CandidateAnow has a majority (11 of the 21 votes) and is declared the winner. NowCis clearly anonwinning (irrelevant) candidate. RemovingCas a candidate leaves the following preference table.Number of voters5533321st choiceAEADBD2nd choiceBDDBEB3rd choiceDBBEAA4th choiceEAEADEWe now recount using the plurality-with-elimination.
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Solution Manual for Excursions in Modern Mathematics, 9th Edition - Page 16 preview imageISM:Excursions in Modern Mathematics, 9E13Round 1:CandidateABDENumber of first-place votes8355CandidateBis eliminated.Round 2: The 3 first-place votes that went toBnow shift toE.CandidateABDENumber of first-place votes858CandidateDis eliminated.Round 3:Dhad 5 first-place votes. Of these, 3 go toEand 2 go toA(sinceBwas eliminated).CandidateABDENumber of first-place votes1011CandidateEnow has a majority (11 of the 21 votes) and is declared the winner. Remember that candidateAwas the winner before candidateCwas removed. This is a violation of the IIA criterion.56.IfXhas a majority of the first-place votes, thenXwill win every pairwise comparison (it is ranked above allother candidates on more than half the ballots) and is, therefore, the winner under the method of pairwisecomparisons.57.IfXis the Condorcet candidate, then by definitionXwins every pairwise comparison and is, therefore, thewinner under the method of pairwise comparisons.58.When a voter moves a candidate up in his or her ballot the number of first place votes for that candidate eitherincreases or stays the same. It follows that ifXhad a plurality of the first place votes and a voter changes hisor her ballot to rankXhigher, thenXstill has a plurality.59.When a voter moves a candidate up in his or her ballot, that candidate’s Borda points increase. It follows thatifXhad the most Borda points and a voter changes his or her ballot to rankXhigher, thenXstill has the mostBorda points.60.When a voter moves a candidate up in his or her ballot it can’t hurt the candidate in a pairwise comparison—the candidate wins the same pairwise comparisons as before and possibly a few more. It follows that ifXwonthe most pairwise comparisons and a voter changes his or her ballot to rankXhigher, thenXstill wins themost pairwise comparisons.JOGGING61.Suppose the two candidates areAandBand thatAgetsafirst-place votes andBgetsbfirst-place votes andsuppose thata>b. ThenAhas a majority of the votes and the preference schedule isNumber of votersab1st choiceAB2nd choiceBAIt is clear that candidateAwins the election under the plurality method, the plurality-with-eliminationmethod, and the method of pairwise comparisons. Under the Borda count method,Agets 2a+bpoints whileBgets 2b+apoints. Sincea>b, 2a+b> 2b+aand so againAwins the election.62.In this variation of the Borda count each candidate gets 1 less point per ballot. It follows that ifNis thenumber of voters, each candidate getsNfewer points than he or she would under the standard Borda countmethod. Since each candidate’s total points gets decreased by the same numberN, the ranking of thecandidates remains the same.63.The number of points under this variation is complementary to the number of points under the standard Bordacount method: a first place is worth 1 point instead ofN, a second place is worth 2 points instead ofN– 1,…,a last place is worthNpoints instead of 1. It follows that having the fewest points here is equivalent to havingthe most points under the standard Borda count method, having the second fewest is equivalent to having thesecond most, and so on.
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