operates can best be understood with some numerical examples. The examples below are for a hypothetical four-member constituency containing four equal-sized districts. The constituency is represented by four elected members, although elected members are understood to represent the whole constituency, not just the individual districts. The two examples are virtually identical, except that in the second example, the number of votes secured by the Blue Party is slightly higher in District 4 relative to District 3. Seat allocation by party is determined using the Hare quota, largest remainder method. Example 1 Four seats must be filled. The Red Party received 1670 votes, Blues 910, Yellows 890 and Greens 530. The Red Party thus wins two seats, while the Blue and Yellow Parties win one seat each. Party candidates John, James, Hughes and Joshua, shown in bold, are elected. Note that District 3 elects two candidates, while District 4 elects none, but in principle this is not a problem, because all winners represent their party for the whole constituency, not the individual district. Example 2 Four seats must be filled. The Red Party received 1650 votes, Blues 930, Yellows 890 and Greens 530. As before, the Red Party wins two seats while the Blue and Yellow Parties win one seat each. Party candidates John, James, Hughes and Mary are elected. In this case, the front-runner, Anne, loses her race in favour of the runner-up, but again this is in principle not a problem, because Mary represents the Blue Party for the constituency as a whole and was the strongest of the Blue Party candidates. See also List of democracy and elections-related topics References Party-list proportional representation