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 (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 the congressional districts.^[1] When envisioned as a proportional electoral system, this 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 this party list (the upper quota).

Although no legislature uses this method of apportionment to assign seats to parties after an election, 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.^[2]

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}}({\frac {{\mbox{total}}\;{\mbox{votes}}}{\mbox{100}}})

Ever 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}}({\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

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. However, most legislatures using party-list proportional representation distribute seats in several large multi-member districts; a notable exception is the Israeli Knesset, all of which 120 seats are assigned nationwide at-large. In this Example, the results of the Israeli legislative election, 2015 would be used as a demonstration of how the Huntington–Hill system would be used to allocate seats between parties after an election.

Unlike in the U.S. House of Representatives, where every state is guaranteed one seat, the first stage would be to calculate which parties are eligible for seats in the legislature. This would be done by dividing the number of votes cast (4,254,738) by 100 and multiplying the results by 3.25 (the exclusion threshold of the Knesset), which gives us a qualification value of 138,279, and excluding any party which polled less than 138,279 votes. This leaves us only with the ten most voted for parties: The Likud (985,408 votes), the Zionist Union (786,313 votes), the Joint List (446,583 votes), Yesh Atid (371,602 votes), Kulanu (315,360 votes), The Jewish Home (283,910 votes), Shas (241,613 votes), Yisrael Beiteinu (214,906 votes), United Torah Judaism (210,143 votes), and Meretz (165,529 votes).

The next stage is to determine the initial number of seats assigned to the Parties. As the Knesset employs a 3.25% exclusion threshold, and 3.25 times 1.2 (120 ÷ 100) equals 3.9, which is not a whole number, each party starts off 4 seats and with a divisor 4.47 (the square root of the product of 4, the number seats currently assigned, and 5, the number of seats that would next be assigned). Each party's number of votes is divided by its D, to produce a priority value for each party. Since all of the Ds are equal for this first seat, the party with the largest vote share would win (Likud).

Now, the Likud's D would be changed to 5.48 (the square root of 5 × 6), and we would repeat this process. In this case the Likud would again win the next seat because its vote share (985,408) divided by its current D (5.48) is greater than the Zionist Union's similarly calculated priority value (786,313 ÷ 4.47) for priority values of 179910 and 175825, respectively).

This process is repeated until all 120 Knesset seats have been assigned to the 10 parties. If the number of Knesset seats were equal in size to the number of votes case for those ten parties, this method would guarantee that the appointments would equal the vote shares of each party.

The Huntington–Hill system is only used to apportion House seats according to state population in the U.S. House of Representatives. Had the Knesset used the Huntington–Hill Method, rather than the Bader-Ofer method currently used, to apportion seats following the 2015 elections, the 120 seats in the 20th Knesset would have been apportioned as follows:

Party				Votes	Seats (Actual results, Bader-Ofer)	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
		of which Labor			19	19	20	19.49	0
		of which Hatnuah			5	5	6	5.48	0
			of which Green Movement		1	1	2	1.41	0
	Joint List			446,583	13	13	14	13.49	0
		of which Hadash			5	5	6	5.48	0
		of which Balad			3	3	4	3.46	0
		of which Ra'am			3	3	4	3.46	0
		of which Ta'al			2	2	3	2.45	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

↑ "Congressional Apportionment". NationalAtlas.gov. Retrieved 2009-02-14.
↑ "The History of Apportionment in America". American Mathematical Society. Retrieved 2009-02-15.

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