Recall the notation from the previous webpage:

n = the number of voters

k = the number of candidates which will be elected (which is the number of voting districts if voting districts are used)

v = the number of votes which each voter casts (for at-large voting, v=1 for voting districts).

x = the number of voters in your party (hence n-x is the number of candidates in the oppositon party, we are assuming two parties or factions).

w = the number of candidates which your party elects (hence k-w is the number of candidates which the oppositon elects, we are assuming two parties or factions).

A faction could get 2 candidates elected if it had 2002 voters, 3 if it had 3003 voters, etc. If there are only two factions, dividing the number of voters in each by 1001 [n/(k+1) rounded up] and then then rounding the resultant quotients down will usually account for all 9 [k] candidates to be elected. (The rounding down reduces k+1 to k candidates that are elected.) The other cases may entail ties which we shall not consider.

There are two ways to approach this problem: 1) how many candiates can a party get elected, if their number of voters is known, and 2) how many voters do they need to get a specified number of candidates elected. We shall first consider the first case where the number of voters in each party is known. This can addressed by constructing a table.

- x vs. min(n-x, (v × (n-x))/k
- min(x, (v × x)/2) vs. min(n-x, (v × (n-x))/(k-1)
- min(x, (v × x)/3) vs. min(n-x, (v × (n-x))/(k-2)
- . . .
- min(x, (v × x)/(k-2)) vs. min(n-x, (v × (n-x))/3
- min(x, (v × x)/(k-1)) vs. min(n-x, (v × (n-x))/2
- min(x, (v × x)/k) vs. n-x

An example of spreading votes among more than 6 candidates which is used below follows. It is for spreading 6 × 5500 votes among 7 candidates. In this case there cannot be just two blocs of majority voters, but overlapping candidate sets. For example, if Alfred, Bob, Celeste, Daphne, Eunice, Frank and George are to be elected; 642 voters vote for ABCDEF, 643 voters vote for BCDEFG, 643 voters vote for ACDEFG, 643 voters vote for ABDEFG, 643 voters vote for ABCEFG, 643 voters vote for ABCDFG, and 643 voters vote for ABCDEG.

As an example, let n=10000, x = 2500, v = 6, k = 9

looking at the first juxtaposition (vs.) above, you can only give 2500 votes to a candidate, but the opposition can spread its 6 × 7500 = 45000 votes among 9 candidates giving 5000 votes to each. Hence the opposition cangive 9 candidates 5000 votes each, but you can only give 2500 votes to a candidate, and the opposition can elect 9 candidates.

Next consider n=10000, x = 4500, v = 6, k = 9

The second juxtaposition above gives 4500 > 4125 = 6 × 5500 / 8, which means you can get 2 candidates elected if they spread their votes among 8 candidates.

however, the third juxtaposition above gives 4500 < 4714.28 = 6 × 5500 / 7, which means that the opposition could give more votes to 7 candidates than you could give to three, so they can get 7 candidates elected.

Note that these comparisons are base on a total of k+1 candidates. The faction that can give more votes to their block of candidates can elect at least that many candidates. You will always reach a point where the inequality sign reverses as you go down the table (unless one party can elect all k candidates), and the numbers of candidates each party can elect will sum to k (for one party the number will be before the inequality reversed, and for the other it will be after).

The majority party can always elect at least v candidates, since no voter can cast more than one vote for a candidate. Further, if the majority party has M voters, it can always elect at least min(k, v × M/ (n-M)) because it can spread its votes among that many candidates and give them more votes than (n-M). However, if v < k/2 both the minority and majority parties may need to spread their votes among more than v candidates, and further analysis is required.

In the above examples v > k/2 (6 > 9/2), and the above bound provideds the results. With M = 7500, 6 × 7500/2500 = 18, min(9, 18) = 9. With M = 5500, 6 × 5500/4500 = 7.33, and the majority can elect 7 candidates.

[I am leaving this paragraph in because I feel it completes the analysis, yet believe some students would be best to omit it since it contains technical details which distract from the main idea.] Further analysis is needed if the voters cast only 2 votes for 9 candidates to be elected. Then if there were 7500 majority faction voters and 2500 minority faction voters 2 × 7500 ÷ 2500 = 6, suggesting that the majority faction could get only 5 or 6 (depending on how one deals with ties) candidates elected. But if the minority faction elects three candidates, then it can only provide (2 × 2500)/3 = 1667 votes per candidate (e.g., if Xavier, Yolanda, and Zelda are to be elected; 1333 voters choose XY, 1334 voters choose XZ, and 1334 voters choose YZ). Thus to elect 6 candidates the majority faction only needs to exceed 1667 votes; 2 × 7500 ÷ 1667 = 8.99. These calculations show that if the minority faction restricted their votes to two candidates, only 5 or 6 majority candidates could exceed that number, but because 9 candidates are elected, they could elect 7 candidates because they only need to exceed 0 votes for the remaining positions. If the minority party spread their votes among three candidates, then the majority party could have 8 candidates with more votes. So the majority party could get 7 candidates elected. More concretely, the minority faction could only give two candidates more than 1667 votes, the majority faction could give eight candidates more than 1667 votes, so the majority faction could get at least 7 candidates elected (2 + 8 = 10 > 9, so one faction will get one fewer candidate elected). The minority faction could give two candidates 2500 votes, the majority faction could give only 6 candidates 2500 votes, so the minority faction could get at least two candidates elected.

The thrust of these calculations is that in order for a minority faction to gain representation, they must be a large minority (near 50%) and/or have few votes per voter.

The second question is how many voters a party needs to get a specified number of candidates elected. If v=1, the answer is 1+n/(k+1), rounding up is often necessary. As indicated above, twice as many voters are are needed to elect two candidates, thrice as many to elect three candidates ... .

if v=k, then the majority takes all, one needs a majority to get one candidate elected, in which case on can get all k candidates elected.

The case 1 < v < k is less trivial. [I believe the next three paragraphs are a nice solution to this question and provide a complementary perspective on the table above, but I shall only expect you to know the cases w=1, w=k, and w=majority. The cases w=1 and w=k have been treated above. The case w=majority can be solved by invoking symmetry: one faction must win a majority, and it will obviously be whichever faction has a majority. This is will not work if ties can ocur (e.g., k is even), but will suffice for any problems I give you on a test.] In order for a party to get w candidates elected, its number of voters x must satisfy min(x, xv/w) > min((n-x), (n-x)v/(k-w+1). The quotients in the minimum functions entail spreading their votes equally among the winning candidates, the minimum function (i.e., x and (n-x)) entails that voters can only cast one vote per candidate. cf. the table above.

For example, with n=10,000 voters and k=9 candidates will be elected and v=5 votes per voter, how many voters are needed to elect w=3 candidates? We note first that v/w = 5/3 > 1, so min(x, xv/w) = x. Similarly, v/(k-w+1) = 5/(9-3+1) < 1, so min ((n-x), (n-x)v/(k-w+1)) = (n-x)(5/7). Hence one needs to solve x > (n-x)(5/7) which provides x = 4167.

With n=10,000 voters, k=9 candidates elected, and v=5 votes per voter, how many voters are needed to elect w=6 candidates? v/w = 5/6 < 1, so min(x, xv/w) = xv/w = (5/6)x. Similarly, v/(k-w+1) = 5/4 > 1, so min((n-x), v/(k-w+1)) = (n-x). Hence one needs to solve (5/6)x > (n-x) which provides x = 5455.

Exercise: If there are 25,000 voters and 11 candidates will be elected, how large must a minority be to be assured of getting at least one representative elected if each voter has 1 vote? 6 votes? 11 votes?

to be assured of getting at least 6 representatives elected if each voter has 1 vote? 6 votes? 11 votes?

to be assured of getting at least 11 representatives elected if each voter has 1 vote? 6 votes? 11 votes?

If there are 15,500 majority faction voters and 9500 minority faction voters and 11 candidates are elected, how many candidates can the minority faction get elected if each voter casts 1 vote? 6 votes? 11 votes?

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