# Huntington–Hill method

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The Huntington–Hill method of apportionment assigns seats by finding a modified divisor D such that each constituency's priority quotient (its population divided by D), using the geometric mean of the lower and upper quota for the divisor, yields the correct number of seats that minimizes the percentage differences in the size of subconstituencies.[1] When envisioned as a proportional electoral system, it is effectively a highest averages method of party-list proportional representation in which the divisors are given by ${\displaystyle \scriptstyle D={\sqrt {n(n+1)}}}$, n being the number of seats a state or party is currently allocated in the apportionment process (the lower quota) and n+1 is the number of seats the state or party would have if it is assigned to the party list (the upper quota).

Although no legislature uses this method of apportionment to assign seats to parties after an election, it was considered for House of Lords elections under the ill-fated House of Lords Reform Bill,[2] and the United States House of Representatives uses it to assign the number of representative seats to each state for which it was devised.

The method is credited to Edward Vermilye Huntington and Joseph Adna Hill.[3]

## Allocation

In a legislative election under the Huntington–Hill method, after the votes have been tallied, the qualification value would be calculated. This step is necessary because in an election, unlike in a legislative apportionment, not all parties are always guaranteed at least one seat. If the legislature concerned has no exclusion threshold, the qualification value would be the Hare quota, or

${\displaystyle {\frac {{\mbox{total}}\;{\mbox{votes}}}{{\mbox{total}}\;{\mbox{seats}}}}}$,

where

• total votes is the total valid poll; that is, the number of valid (unspoilt) votes cast in an election.
• total seats is the total number of seats to be filled in the election.

In legislatures which use an exclusion threshold, the qualification value would be:

${\displaystyle {\mbox{exclusion}}\;{\mbox{threshold}}\;{\mbox{[percentage]}}({\frac {{\mbox{total}}\;{\mbox{votes}}}{\mbox{100}}})}$.

Every party polling votes equal to or greater than the qualification value would be given an initial number of seats, again varying if whether or not there is a threshold:

In legislatures which do not use an exclusion threshold, the initial number would be 1, but in legislatures which do, the initial number of seats would be:

${\displaystyle {\mbox{exclusion}}\;{\mbox{threshold}}\;{\mbox{[percentage]}}({\frac {{\mbox{total}}\;{\mbox{seats}}}{\mbox{100}}})}$

with all fractional remainders being rounded up.

In legislatures elected under a mixed-member proportional system, the initial number of seats would be further modified by adding the number of single-member district seats won by the party before any allocation.

Determining the qualification value is not necessary when distributing seats in a legislature among states pursuant to census results, where all states are guaranteed a fixed number of seats, either one (as in the US) or a greater number, which may be uniform (as in Brazil) or vary between states (as in Canada).

After all qualified parties or states received their initial seats, successive quotients are calculated, as in other Highest Averages methods, for each qualified party or state, and seats would be repeatedly allocated to the party or state having the highest quotient until there are no more seats to allocate. The formula of quotients calculated under the Huntington-Hill method is

${\displaystyle A_{n}={\frac {V}{\sqrt {s(s+1)}}}}$

where:

• V is the population of the state or the total number of votes that party received, and
• s is the number of seats that the state or party has been allocated so far.

### Example

Even though the Huntington–Hill system was designed to distribute seats in a legislature among states pursuant to census results, it can also be used, when putting parties in the place of states and votes in place of population, for the mathematically equivalent task of distributing seats among parties pursuant to an election results in a party-list proportional representation system. A party-list PR system requires large multi-member districts to function effectively.

In this example, 230,000 voters decide the disposition of 8 seats among 4 parties. Unlike the d'Hondt and Saintë-Lague systems, which allow the allocation of seats by calculating successive quotients right away, the Huntington–Hill system requires each party or state have at least one seat to avoid a division by zero error. In the U.S. House of Representatives, this is ensured by guaranteeing each state at least one seat; in PR election under the Huntington–Hill system, however, the first stage would be to calculate which parties are eligible for an initial seat.

This would be done by dividing the number of votes cast (230,000) by 8, which gives a qualification value of 28,750, excluding any party which polled less than 28,750 votes, and giving every party which polled at least 28,750 votes one seat.

Denominator Votes Is the party eligible or disqualified?
Party A 100,000 Eligible
Party B 80,000 Eligible
Party C 30,000 Eligible
Threshold 28,750
Party D 20,000 Disqualified

With all the initial seats assigned, each eligible party's (Parties A, B, and C) total votes is divided by 1.41 (the square root of the product of 1, the number of seats currently assigned, and 2, the number of seats that would next be assigned), then by 2.45, 3.46, 4.47, 5.48, 6.48, 7.48, and 8.49. The 8 highest entries, marked with asterisks, range from 70,711 down to 21,213. For each, the corresponding party gets a seat.

For comparison, the "Proportionate seats" column shows the exact fractional numbers of seats due, calculated in proportion to the number of votes received. (For example, 100,000/230,000 × 8 = 3.48) The slight favouring of the largest party over the smallest is apparent.

Denominator 1.41 2.45 3.46 4.47 5.48 6.48 7.48 8.49 Seats
won (*)
Proportionate
seats
Party A 70,711* 40,825* 28,868* 22,361* 18,257 15,430 13,363 11,785 4 3.4
Party B 56,569* 32,660* 23,094* 17,889 14,606 12,344 10,690 9,428 3 2.8
Party C 21,213* 12,247 8,660 6,708 5,477 4,629 4,009 3,536 1 1.1
Party D Disqualified 0 0.7

If the number of seats was equal in size to the number of votes case for the qualified parties, this method would guarantee that the appointments would equal the vote shares of each party.

In this example, the results of the apportionment is identical to one under the d'Hondt system. However, as the District magnitude increases, differences emerge: all 120 members of the Knesset, Israel's unicameral legislature, are elected under the d'Hondt method. Had the Huntington–Hill method, rather than the d'Hondt method, been used to apportion seats following the elections to the 20th Knesset, held in 2015, the 120 seats in the 20th Knesset would have been apportioned as follows:

Party Votes Seats (actual results, d'Hondt) Seats (hypothetical results, Huntington–Hill) Upper quota (Huntington–Hill) Geometric mean (Huntington–Hill) +/–
Likud 985,408 30 30 31 30.50 0
Zionist Union 786,313 24 24 25 24.50 0
Joint List 446,583 13 13 14 13.49 0
Yesh Atid 371,602 11 11 12 11.49 0
Kulanu 315,360 10 9 10 9.49 –1
The Jewish Home 283,910 8 9 10 9.49 +1
Shas 241,613 7 7 8 7.48 0
Yisrael Beiteinu 214,906 6 6 7 6.48 0
United Torah Judaism 210,143 6 6 7 6.48 0
Meretz 165,529 5 5 6 5.48 0
Source: CEC

Compared with the actual apportionment, Kulanu would have lost one seat, while The Jewish Home would have gained one seat.

## References

1. ^ "Congressional Apportionment". NationalAtlas.gov. Retrieved 2009-02-14.
2. ^ "Draft House of Lords Reform Bill: report session 2010-12, Vol. 2". Google Books. Retrieved 6 November 2017.
3. ^ "The History of Apportionment in America". American Mathematical Society. Retrieved 2009-02-15.