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# Huntington–Hill method

The HuntingtonHill method (sometimes method of equal proportions) is a highest averages method for assigning seats in a legislature to political parties or states.[1] Since 1941, this method has been used to apportion the 435 seats in the United States House of Representatives following the completion of each decennial census.[2][3]

The method minimizes the relative difference in the number of constituents represented by each legislator. In other words, the method selects the algorithm such that no transfer of a seat from one state to another can reduce the percent error in representation for both states.[1]

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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.

## Apportionment method

In this method, as a first step, each of the 50 states is given its one guaranteed seat in the House of Representatives, leaving 385 seats to assign. The remaining seats are allocated one at a time, to the state with the highest average district population, to bring its district population down. However, it is not clear if we should calculate the average before or after allocating an additional seat, and the two procedures give different results. Huntington-Hill uses a continuity correction as a compromise, given by taking the geometric mean of both divisors, i.e.:[4]

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

where P is the population of the state, and n is the number of seats it currently holds before the possible allocation of the next seat.

Consider the reapportionment following the 2010 U.S. census: after every state is given one seat:

1. The largest value of A1 corresponds to the largest state, California, which is allocated seat 51.
2. The 52nd seat goes to Texas, the 2nd largest state, because its A1 priority value is larger than the An of any other state.
3. he 53rd seat goes back to California because its A2 priority value is larger than the An of any other state.
4. The 54th seat goes to New York because its A1 priority value is larger than the An of any other state at this point.

This process continues until all remaining seats are assigned. Each time a state is assigned a seat, n is incremented by 1, causing its priority value to be reduced.

## Division by zero

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.[5] In the U.S. House of Representatives, this is ensured by guaranteeing each state at least one seat;[5] in party-list representation, small parties would likely be eliminated using some electoral threshold, or the first divisor can be modified.

## Examples

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 2  • 1 ≈ 1.41, then by approximately 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 another seat.

Denominator 1·2
1.41
2·3
2.45
3·4
3.46
4·5
4.47
5·6
5.48
6·7
6.48
7·8
7.48
8·9
8.49
Initial
seats
Seats
won (*)
Total
Seats
Ideal
seats
Party A 70,711* 40,825* 28,868* 22,361 18,257 15,430 13,363 11,785 1 3 4 3.8
Party B 56,569* 32,660* 23,094 17,889 14,606 12,344 10,690 9,428 1 2 3 3.0
Party C 21,213 12,247 8,660 6,708 5,477 4,629 4,009 3,536 1 0 1 1.1

### Knesset example

The Knesset (Israel's unicameral legislature), are elected by party-list representation with apportionment by the D'Hondt method.[a] 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:

(hypothetical) (actual)
Last Priority[b] Next Priority[c] Seats Seats
Likud 985,408 33408 32313 30 30 0
Zionist Union 786,313 33468 32101 24 24 0
Joint List 446,583 35755 33103 13 13 0
Yesh Atid 371,602 35431 32344 11 11 0
Kulanu 315,360 37166 33242 9 10 –1
The Jewish Home 283,910 33459 29927 9 8 +1
Shas 241,613 37282 32287 7 7 0
Yisrael Beiteinu 214,906 39236 33161 6 6 0
United Torah Judaism 210,143 38367 32426 6 6 0
Meretz 165,529 37013 30221 5 5 0
Source: CEC

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

## Notes

1. ^ a b The method used for the 20th Knesset was actually a modified D'Hondt, called the Bader-Ofer method. This modification allows for spare vote agreements between parties.[6]
2. ^ This is each party's last priority number which resulted in a seat being gained by the party. Likud gained the last seat (the 120th seat allocated). Each priority number in this column is greater than any priority number in the Next Priority column.
3. ^ This is each party's next priority number which would result in a seat being gained by the party. Kulanu would have gained the next seat (if there were 121 seats in the Knesset). Each priority number in this column is less than any priority number in the Last Priority column.

## References

1. ^ a b "Congressional Apportionment". NationalAtlas.gov. Archived from the original on 2009-02-28. Retrieved 2009-02-14.
2. ^
3. ^ "Computing Apportionment". United States Census Bureau. Retrieved 2021-04-26.
4. ^ Pukelsheim, Friedrich (2017), Pukelsheim, Friedrich (ed.), "Divisor Methods of Apportionment: Divide and Round", Proportional Representation: Apportionment Methods and Their Applications, Cham: Springer International Publishing, pp. 71–93, doi:10.1007/978-3-319-64707-4_4, ISBN 978-3-319-64707-4, retrieved 2021-09-01
5. ^ a b Pukelsheim, Friedrich (2017), Pukelsheim, Friedrich (ed.), "Divisor Methods of Apportionment: Divide and Round", Proportional Representation: Apportionment Methods and Their Applications, Cham: Springer International Publishing, pp. 71–93, doi:10.1007/978-3-319-64707-4_4, ISBN 978-3-319-64707-4, retrieved 2021-09-01
6. ^ "With Bader-Ofer method, not every ballot counts". The Jerusalem Post. Retrieved 2021-05-04.