Largest remainder method
||An automated process has detected links on this page on the local or global blacklist.|
|This article needs additional citations for verification. (November 2011)|
|Part of the Politics series|
The largest remainder method (also known as Hare-Niemeyer method or as Vinton's method1) is one way of allocating seats proportionally for representative assemblies with party list voting systems. It contrasts with the highest averages method.
The largest remainder method requires the numbers of votes for each party to be divided by a quota representing the number of votes required for a seat (i.e. usually the total number of votes cast divided by the number of seats, or some similar formula). The result for each party will usually consist of an integer part plus a fractional remainder. Each party is first allocated a number of seats equal to their integer. This will generally leave some seats unallocated: the parties are then ranked on the basis of the fractional remainders, and the parties with the largest remainders are each allocated one additional seat until all the seats have been allocated. This gives the method its name.
The Hare (or simple) Quota is defined as follows
The Hamilton method of apportionment is actually a largest-remainder method which uses the Hare Quota. It is named after Alexander Hamilton, who invented the largest-remainder method in 1792. It is used for legislative elections in Russia (with a 7% exclusion threshold since 2007), Ukraine (3% threshold), Namibia and Hong Kong. It was historically applied for congressional apportionment in the United States during the 19th century.
The Droop quota is the integer part of
and is applied in elections in South Africa. The Hagenbach-Bischoff quota is virtually identical, being
either used as a fraction or rounded up.
The Hare quota tends to be slightly more generous to less popular parties and the Droop quota to more popular parties, and can arguably be considered more proportional than Droop quota2 although it is more likely to give fewer than half the seats to a list with more than half the vote.
The Imperiali quota
is rarely used since it suffers from the defect that it might result in more seats being allocated than there are available (this can also occur with the Hagenbach-Bischoff quota but it is very unlikely, and it is impossible with the Hare and Droop quotas). This will certainly happen if there are only two parties. In such a case, it is usual to increase the quota until the number of candidates elected is equal to the number of seats available, in effect changing the voting system to the Jefferson apportionment formula (see D'Hondt method).
These examples take an election to allocate 10 seats where there are 100,000 votes.
|Highest Remainder Seats||1||1||0||0||1||0||3|
|Highest Remainder Seats||0||1||1||0||0||0||2|
- It is relatively easy for a voter to understand how the largest remainder method allocates seats. The Hare quota gives an advantage to smaller parties while the Droop quota favours larger parties.3 However, whether a list gets an extra seat or not may well depend on how the remaining votes are distributed among other parties: it is quite possible for a party to make a slight percentage gain yet lose a seat if the votes for other parties also change. A related feature is that increasing the number of seats may cause a party to lose a seat (the so-called Alabama paradox). The highest averages methods avoid this latter paradox but since no apportionment method is entirely free from paradox,4 they introduce others like quota violation.5
The largest remainder method is the only apportionment that satisfies the quota rule; in fact, it is designed to satisfy this criterion. However, it comes at the cost of paradoxical behaviour. The Alabama paradox is exhibited when an increase in seats apportioned leads to a decrease in the number of seats allocated to a certain party. Suppose 25 seats are to be apportioned between 6 parties with votes cast in the proportions 1500:1500:900:500:500:200. The two parties with 500 votes get three seats each. Now allocate 26 seats, and it will be found that the these parties get only two seats apiece.
With 25 seats, we get:
With 26 seats, we have:
- Tannenbaum, Peter (2010). Excursions in Modern Mathematics. New York: Prentice Hall. p. 128. ISBN 978-0-321-56803-8.
- See the following references:     .
- See for example the 2012 election in Hong Kong Island where the DAB ran as two lists and gained twice as many seats as the single-list Civic despite receiving fewer votes in total: New York Times report
- Balinski, Michel; H. Peyton Young (1982). Fair Representation: Meeting the Ideal of One Man, One Vote. Yale Univ Pr. ISBN 0-300-02724-9.
- Messner et al. "RangeVoting: Apportionment and rounding schemes". Retrieved 2014-02-02.