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Huntington–Hill method

From Wikipedia, the free encyclopedia

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 , 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]

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.[3]

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  • Apportionment: Huntington-Hill Method
  • Huntington-Hill method 1
  • Apportionment - Huntington-Hill Method
  • Huntington-Hill method 2
  • Comparing Huntington-Hill to Webster's method

Transcription

- WELCOME TO A LESSON ON THE HUNTINGTON-HILL METHOD OF APPORTIONMENT. IN 1920 NO NEW APPORTIONMENT WAS DONE BECAUSE CONGRESS COULDN'T AGREE ON A METHOD TO BE USED. THEY APPOINTED A COMMITTEE OF MATHEMATICIANS TO INVESTIGATE AN APPORTIONMENT METHOD. THEY RECOMMENDED THE HUNTINGTON-HILL METHOD. HOWEVER, THEY CONTINUED TO USE WEBSTER'S METHOD IN 1931. AFTER A SECOND REPORT RECOMMENDING HUNTINGTON-HILL IT WAS ADOPTED IN 1941 AND IT'S THE CURRENT METHOD OF APPORTIONMENT USED IN CONGRESS. THE HUNTINGTON-HILL METHOD IS SIMILAR TO WEBSTER'S METHOD, BUT ATTEMPTS TO MINIMIZE THE PERCENT DIFFERENCES OF HOW MANY PEOPLE EACH REPRESENTATIVE WILL REPRESENT. TO APPLY THE HUNTINGTON-HILL METHOD, STEP ONE, WE DETERMINE HOW MANY PEOPLE EACH REPRESENTATIVE SHOULD REPRESENT. WE DO THIS BY DIVIDING THE TOTAL POPULATION OF ALL THE STATES BY THE TOTAL NUMBER OF REPRESENTATIVES. THIS ANSWER IS CALLED THE STANDARD DEVISOR OR JUST THE DEVISOR. STEP TWO, WE DIVIDE EACH STATE'S POPULATION BY THE DEVISOR TO DETERMINE HOW MANY REPRESENTATIVES IT SHOULD HAVE. WE RECORD THIS ANSWER TO SEVERAL DECIMAL PLACES. THIS ANSWER IS CALLED THE QUOTA. STEP THREE, WE CUT OFF THE DECIMAL PART OF THE QUOTA TO OBTAIN WHAT'S CALLED THE LOWER QUOTA, AND WE'LL CALL THIS N. NEXT WE COMPUTE THE SQUARE ROOT OF N x THE QUANTITY N + 1, WHICH IS THE GEOMETRIC MEAN OF THE LOWER QUOTA AND ONE VALUE HIGHER. STEP FOUR, IF THE QUOTA IS LARGER THAN THE GEOMETRIC MEAN WE ROUND UP THE QUOTA. IF THE QUOTA IS SMALLER THAN THE GEOMETRIC MEAN WE ROUND DOWN THE QUOTA. THEN WE ADD THE RESULTING WHOLE NUMBERS TO GET THE INITIAL ALLOCATION. STEP FIVE, IF THE TOTAL FROM STEP FOUR IS LESS THAN THE TOTAL NUMBER OF REPRESENTATIVES WE REDUCE THE DEVISOR AND RECALCULATE THE QUOTA AND ALLOCATION. IF THE TOTAL FROM STEP FOUR IS LARGER THAN THE TOTAL NUMBER OF REPRESENTATIVES THEN WE INCREASE THE DEVISOR AND RECALCULATE THE QUOTA AND ALLOCATION. WE CONTINUE DOING THIS UNTIL THE TOTAL IN STEP FOUR IS EQUAL TO THE TOTAL NUMBER OF REPRESENTATIVES. THE DEVISOR WE END UP WITH IS CALLED THE MODIFIED DEVISOR OR ADJUSTED DEVISOR. LET'S TAKE A LOOK AT AN EXAMPLE. A COLLEGE OFFERS TUTORING IN MATH, ENGLISH, CHEMISTRY, AND BIOLOGY. THE NUMBER OF STUDENTS ENROLLED IN EACH SUBJECT IS LISTED BELOW. THE COLLEGE CAN ONLY AFFORD TO HIRE 22 TUTORS. USING HUNTINGTON-HILL'S METHOD DETERMINE THE APPORTIONMENT OF THE TUTORS. SO FOR THE FIRST STEP WE WANT TO FIND THE STANDARD DEVISOR. SO WE FIND THE SUM OF THE ENTIRE ENROLLMENT, WHICH IS 780, AND WE DIVIDE BY THE NUMBER OF TUTORS, WHICH IS 22. SO THE STANDARD DEVISOR IS APPROXIMATELY 35.4545. AND NOW TO FIND THE QUOTA FOR EACH DISCIPLINE WE TAKE EACH ENROLLMENT AND DIVIDE BY THE STANDARD DEVISOR. NOTICE HOW IT'S ALREADY BEEN DONE, BUT LET'S GO AHEAD AND CHECK AT LEAST ONE OF THEM. FOR MATH WE WOULD TAKE 380 AND DIVIDE BY 35.4545 WHICH GIVES US THE QUOTA OF APPROXIMATELY 10.718. WE WOULD DO THE SAME FOR ENGLISH, CHEMISTRY, AND BIOLOGY TO GET THE REMAINING QUOTAS. AND NOW TO FIND THE LOWER QUOTA WE REMOVE THE DECIMAL PART OF THE QUOTA. SO WE HAVE 10, 6, 2, AND 1. FROM HERE WE'LL FIND THE GEOMETRIC MEAN WHERE THE LOWER QUOTA IS ACTUALLY N. WE WANT TO FIND THE SQUARE ROOT OF N x THE QUANTITY AND + 1. SO TO FIND THE GEOMETRIC MEAN FOR MATH, THE FIRST ONE, WE WOULD HAVE THE SQUARE ROOT OF 10 x 10 + 1 OR x 11. SO THE GEOMETRIC MEAN IS APPROXIMATELY 10.488. WHEN THE LOWER QUOTA IS 6 THE GEOMETRIC MEAN WOULD BE THE SQUARE ROOT OF 6 x 7 OR APPROXIMATELY 6.481. WHEN THE LOWER QUOTA IS 2 THE GEOMETRIC MEAN WOULD BE THE SQUARE ROOT OF 2 x 3, WHICH IS APPROXIMATELY 2.449. AND FINALLY, WHEN THE LOWER QUOTA IS 1 THE GEOMETRIC MEAN WOULD BE THE SQUARE ROOT OF 1 x 2, WHICH IS APPROXIMATELY 1.414. SO THIS IS HOW WE FIND THE GEOMETRIC MEANS. AND NOW FOR THE NEXT STEP WE WANT TO COMPARE THE QUOTA IN THE GEOMETRIC MEAN. IF THE QUOTA IS LARGER WE WOULD ROUND THE QUOTA UP. IF THE GEOMETRIC MEAN IS LARGER WE WOULD ROUND THE QUOTA DOWN. SO FOR MATH, NOTICE HOW THE QUOTA IS LARGER, FOR ENGLISH THE QUOTA'S ALSO LARGER, FOR CHEMISTRY THE QUOTA IS STILL LARGER. AND FINALLY, FOR BIOLOGY THE QUOTA IS STILL LARGER, WHICH MEANS IN EACH CASE WE'LL ROUND THE QUOTA UP FOR OUR INITIAL ALLOCATION. SO WE'LL ROUND THE MATH QUOTA TO 11, WE'LL ROUND THE ENGLISH QUOTA TO 7, CHEMISTRY QUOTA TO 3, AND THE BIOLOGY QUOTA TO 2. NOW WE'LL FIND THIS SUM AND COMPARE TO 22. NOTICE HOW THE SUM IS ACTUALLY 23, IT'S TOO HIGH BECAUSE WE ONLY HAVE 22 TUTORS. SO NOW WE'LL HAVE TO-- SO NOW WE'LL HAVE TO MODIFY THE DEVISOR, OR IN THIS CASE INCREASE THE DEVISOR AND THEN REPEAT THE PROCESS. SO LET'S INCREASE THE DEVISOR TO 37. NOTICE HOW THE QUOTAS HAVE CHANGED BECAUSE NOW WE HAVE TO DIVIDE THE ENROLLMENT BY 37. FOR EXAMPLE, FOR MATH WE WOULD HAVE 380 DIVIDED BY 37, WHICH GIVES US APPROXIMATELY 10.270. FOR ENGLISH WE'D HAVE 240 DIVIDED BY 37, WHICH GIVES US APPROXIMATELY 6.486 AND SO-ON. SO, AGAIN, TO FIND THE LOWER QUOTA WE WOULD REMOVE THE DECIMAL PART, WHICH WOULD BE 10, 6, 2, AND 1. NOTICE HOW THIS LOWER QUOTA IS THE SAME AS THE PREVIOUS LOWER QUOTA, WHICH MEANS OUR GEOMETRIC MEAN WOULD ACTUALLY BE THE SAME AS BEFORE. FOR MATH IT WOULD BE THE SQUARE ROOT OF 10 x 11, FOR ENGLISH IT WOULD BE THE SQUARE ROOT OF 6 x 7, FOR CHEMISTRY THE SQUARE ROOT OF 2 x 3, AND FOR BIOLOGY THE SQUARE ROOT OF 1 x 2, WHICH AS WE SAW BEFORE, GIVES US THESE DECIMAL VALUES HERE. AND, AGAIN, FROM HERE WE'RE GOING TO COMPARE THE QUOTA AND THE GEOMETRIC MEAN. IF THE QUOTA IS LARGER WE ROUND UP. IF THE GEOMETRIC MEAN IS LARGER WE ROUND DOWN. NOTICE THIS TIME FOR MATH THE GEOMETRIC MEAN IS GREATER. FOR ENGLISH THE QUOTA IS JUST SLIGHTLY LARGER. FOR CHEMISTRY THE QUOTA'S LARGER. AND FOR BIOLOGY THE QUOTA'S ALSO LARGER. WHICH MEANS, WE'LL NOW ROUND THE MATH QUOTA DOWN TO TEN, ROUND THE ENGLISH, CHEMISTRY, AND BIOLOGY QUOTAS UP. SO WE'D HAVE 7, 3, AND 2. NOTICE HOW NOW WHEN WE FIND THE SUM OF THE ALLOCATIONS IT IS EXACTLY 22, THE NUMBER OF TUTORS THAT WE HAVE. AND, THEREFORE, WE'RE DONE. THIS IS OUR FINAL ALLOCATION USING THE HUNTINGTON-HILL METHOD OF APPORTIONMENT. I HOPE YOU FOUND THIS HELPFUL.  

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

,

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:

.

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:

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

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)[4]

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) +/–
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. Archived from the original on 2009-02-28. Retrieved 2009-02-14.
  2. ^ Draft House of Lords Reform Bill: report session 2010-12, Vol. 2. Google Books. 23 April 2012. ISBN 9780108475801. Retrieved 6 November 2017.
  3. ^ "The History of Apportionment in America". American Mathematical Society. Retrieved 2009-02-15.
  4. ^ Note the slight favouring of the largest party over the smallest (if we subtract Proportionate seats from Total Seats Party A - the largest party - gets the highest total (0.5) while Party B only gets 0.2, Party C gets 0 and Party D gets -0.7)
This page was last edited on 11 November 2020, at 19:54
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