Huntington–Hill method

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. 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 $\scriptstyle D={\sqrt {n(n+1)}}$, 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.

The method is how the United States House of Representatives assigns the number of representative seats to each state – the purpose for which it was devised. It is credited to Edward Vermilye Huntington and Joseph Adna Hill.

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

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

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:

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

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:

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

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).

It can also be skipped if the Huntington-Hill system is used in the nationwide stage of a national remnant system, because the only qualified parties are those which obtained seats at the subnational stage.

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

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

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 Sainte-Laguë 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 a single-stage PR election under the Huntington–Hill system, however, the first stage would be to calculate which parties are eligible for an initial seat.

This could be done by excluding any parties which polled less than the Hare quota, and giving every party which polled at least the Hare quota one seat. The Hare quota is calculated by dividing the number of votes cast (230,000) by the number of seats (8), which in this case gives a qualification value of 28,750 votes.

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

Each eligible party is assigned one seat. With all the initial seats assigned, the remaining five seats are distributed by a priority number calculated as follows. 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 5 highest entries, marked with asterisks, range from 70,711 down to 28,868. 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	Initial seats	Seats won (*)	Total Seats	Proportionate seats
Party A	70,711*	40,825*	28,868*	22,361	18,257	15,430	13,363	11,785	1	3	4	3.5
Party B	56,569*	32,660*	23,094	17,889	14,606	12,344	10,690	9,428	1	2	3	2.8
Party C	21,213	12,247	8,660	6,708	5,477	4,629	4,009	3,536	1	0	1	1.0
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)	+/–
width=2px bgcolor="Template:Likud/meta/color" |	Likud			985,408	30	30	31	30.50	0
	Zionist Union			786,313	24	24	25	24.50	0
width=2px bgcolor="Template:Joint List/meta/color"|	Joint List			446,583	13	13	14	13.49	0
bgcolor="Template:Yesh Atid/meta/color" |	Yesh Atid			371,602	11	11	12	11.49	0
bgcolor="Template:Kulanu/meta/color" |	Kulanu			315,360	10	9	10	9.49	–1
bgcolor="Template:The Jewish Home/meta/color" |	The Jewish Home			283,910	8	9	10	9.49	+1
bgcolor="Template:Shas/meta/color" |	Shas			241,613	7	7	8	7.48	0
bgcolor="Template:Yisrael Beiteinu/meta/color" |	Yisrael Beiteinu			214,906	6	6	7	6.48	0
bgcolor="Template:United Torah Judaism/meta/color" |	United Torah Judaism			210,143	6	6	7	6.48	0
bgcolor="Template:Meretz/meta/color" |	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.