MA 111-001 & 002

Homework #2 (Chapter 1: #14, 16, 20, 22, 24, 26, 28, 30, 34, 36, 40, 52, 56) Solutions

**14.** **a.** In order for B to win the election outright, we have to give B enough votes to beat C, since C is the leader (25 votes), then split the remaining 5 votes between B and C, making sure B always stays at least one ahead. So B will get 3 more for a total of 28 votes.

** b. **D will need to collect 9 votes to pass C, and then we can split the remaining votes between C and D, making sure D always has at least one more. In this way, D will receive 11 more votes, for a total of 20 votes.

**16.** **a. **If there are 4 candidates, and the votes were split evenly, each would get 256, and there would be 1 vote left over. So to have a plurality, you need at least 257 votes.

** b.** We begin by supposing each candidate receives the same number of votes, splitting 1025 into 8 piles of 128, with one vote left over, which would go to one candidate to get a plurality. So there would have to be 8 candidates in the election.

** c.** There must be 5 candidates in the election.

**20.** **a. **Candidate A has 74 points, B has 51, C has 69 and D has 77, so D wins.

** b.** With 27 voters, C has a majority of the first place votes, but does not win.

** c.** C is the Condorcet winner, but does not win the election with the Borda method.

**22.** A wins with 28.4% of the points.

**24.** **a. **If a candidate received all first place votes, he would get 20(5)=100 points.

** b.** If a candidate received all last place votes, he would get 20(1)=20 points.

** c.** Each ballot has 5+4+3+2+1=15 points.

** d.** Twenty ballots, each having 15 points, gives 300 total points.

** e.** Summing all of the points earned by the first four candidates gives 251 points. Out of the 300 total points, this leaves 49 for candidate E.

**26.** With 5 candidates, each ballot has 15 points available. Forty of these ballots means we have 600 total points. Summing the points for candidates A thru D gives 453. So E wins with the 147 remaining points.

**28.** With 1240 total voters, 621 are needed for a majority. Initially no one has a majority, and with only 270 first place votes, C has the least number and gets eliminated. This leaves A with 420 , B with 490 and D with 330 (after redistributing those votes that originally belonged to C). Since there is still no majority candidate, we eliminate D, who has the fewest first place votes, and recount. This gives B a majority with 645 votes.

**30.** **a. **With 27 voters, 14 are needed for a majority. So candidate C wins.

** b.** We can determine the winner without eliminating anyone because C has a majority of the first place votes in the beginning.

** c.** Because we take the candidate with the majority of first place votes (if one exists) as the winner in PWE, it satisfies the majority criterion.

**34.** **a. **With 17 voters, 9 votes are needed for a majority. So after two eliminations, C wins.

** b. **After two eliminations, B wins this time.

** c.** This is a violation of the Monotonicity criterion.

**36.** With 3 points, C wins.

**40.** **a. **Each candidate is compared to 5 other candidates. So if A loses 3, this means he wins 2, so he gets 2 points. Similarly, B and C each have 4 points, and D and E each have 3.5 points.

** b.** Candidates B and C tie in this election.

**52.** Using (N+1)N/2, with N equal to 3220, we find this sum equal to 5,185, 810.

**56.** **a. **We want to sum all of the numbers from 20 to 1, so we use the formula from 52, and get 210 matches.

** b.** With 6 matches per hour and 12 hours per day, we can fit 72 matches into each day. Two days will not be enough, and so we will need the hall for 3 days, which will give us time for 216 total matches.

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