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The largest remainder method (also known as Hare–Niemeyer method, Hamilton method or as Vinton's method^{[1]}) is one way of allocating seats proportionally for representative assemblies with party list voting systems. It contrasts with various highest averages methods (also known as divisor methods).
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Apportionment: Hamilton's Method

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 WELCOME TO A LESSON ON HAMILTON'S METHOD OF APPORTIONMENT. ALEXANDER HAMILTON PROPOSED THE METHOD THAT NOW BEARS HIS NAME. HIS METHOD WAS APPROVED BY CONGRESS IN 1791, BUT WAS VETOED BY PRESIDENT WASHINGTON. IT WAS LATER ADOPTED IN 1852, AND USED THROUGH 1911. SINCE HE WAS INTERESTED IN THE QUESTION OF CONGRESSIONAL REPRESENTATION, WE'LL USE THE LANGUAGE OF STATES AND REPRESENTATIVES. HAMILTON'S METHOD PROVIDES A PROCEDURE TO DETERMINE HOW MANY REPRESENTATIVES EACH STATE SHOULD RECEIVE. SO THE FIRST STEP IN HAMILTON'S METHOD IS TO 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 DIVISOR, OR DIVISOR. STEP TWO, WE DIVIDE EACH STATE'S POPULATION BY THE DIVISOR TO DETERMINE HOW MANY REPRESENTATIVES IT SHOULD HAVE. WE WOULD RECORD THIS ANSWER TO SEVERAL DECIMAL PLACES, AND THIS ANSWER IS CALLED THE QUOTA. SINCE WE CAN ONLY ALLOCATE WHOLE REPRESENTATIVES, HAMILTON RESOLVES THE WHOLE NUMBER PROBLEM AS FOLLOWS: STEP THREE, WE CUT OFF THE DECIMAL PARTS OF ALL THE QUOTAS, AND THESE ARE CALLED THE LOWER QUOTAS. THEN WE ADD THE LOWER QUOTAS. THIS SUM WILL ALWAYS BE LESS THAN OR EQUAL TO THE TOTAL NUMBER OF REPRESENTATIVES. STEP FOUR, ASSUMING THAT THE TOTAL FROM STEP THREE WAS LESS THAN THE TOTAL NUMBER OF REPRESENTATIVES, ASSIGN THE REMAINING REPRESENTATIVES ONE EACH TO THE STATES WHOSE DECIMAL PART OF THE QUOTA WERE LARGEST UNTIL THE DESIRED TOTAL IS REACHED. WE DO WANT TO MAKE SURE THAT EACH STATE ENDS UP WITH AT LEAST ONE REPRESENTATIVE. LET'S TAKE A LOOK AT OUR FIRST EXAMPLE. AGAIN, THE FIRST STEP, WE WANT TO FIND THE DIVISOR, OR STANDARD DIVISOR. SO WE TAKE THE SUM OF THE POPULATION FROM ALL THE STATES, WHICH IS 189,000, AND DIVIDE BY THE TOTAL OF SEATS IN CONGRESS, WHICH IS 30. NOTICE HOW THIS GIVES US A DIVISOR OF 6,300. AND NOW, TO FIND THE QUOTAS WE TAKE EACH STATE POPULATION AND DIVIDE BY 6,300. SO FOR STATE "A" THE QUOTA WOULD BE 27,500 DIVIDED BY 6,300. IF WE ROUND TO FOUR DECIMAL PLACES, THE QUOTA FOR STATE "A" WOULD BE APPROXIMATELY 4.3651. FOR STATE B WE WOULD HAVE 38,300 DIVIDED BY 6,300, SO THE QUOTA FOR STATE B WOULD BE APPROXIMATELY 6.0794, AND SO ON. TO SAVE SOME TIME WE WON'T SHOW ALL THIS DIVISION. SO HERE ARE OUR QUOTAS TO FOUR DECIMAL PLACES, AND NOW FOR THE INITIAL APPORTIONMENT OR INITIAL ASSIGNMENT OF THE SEATS IN CONGRESS, WE REMOVE THE DECIMAL PARTS OF THE QUOTA. WHICH MEANS STATE "A" WOULD RECEIVE 4, STATE B WOULD RECEIVE 6, STATE C WOULD RECEIVE 7, AND STATE D WOULD RECEIVE 12. WE NOTICE HOW THE SUM HERE IS 29, AND WE HAVE A TOTAL OF 30 SEATS. WE NOW ASSIGN THE REMAINING SEAT TO THE STATE WHOSE QUOTA HAS THE LARGEST DECIMAL PART. NOTICE STATE "A" HAS THE LARGEST DECIMAL PART AT .3651, AND THEREFORE STATE "A" RECEIVES ONE MORE SEAT. WHICH MEANS FOR THE FINAL APPORTIONMENT STATE "A" RECEIVES 5 SEATS, STATE B RECEIVES 6, STATE C RECEIVES 7, AND STATE D RECEIVES 12. NOTICE HERE THE TOTAL IS 30. WE'VE USED ALL THE SEATS IN CONGRESS. AND HERE'S THE RESULT IN A NICE TABLE. LET'S TAKE A LOOK AT A SECOND EXAMPLE. HERE A TEACHER WISHES TO DISTRIBUTE 10 IDENTICAL PIECES OF CANDY AMONG 4 STUDENTS, BASED UPON HOW MANY PAGES OF A BOOK THEY READ LAST MONTH, USING HAMILTON'S METHOD. THE TABLE BELOW LISTS THE TOTAL NUMBER OF PAGES READ BY EACH STUDENT. SO IN THIS PARTICULAR QUESTION WE'RE ASKED TO FIND THE DIVISOR, THE QUOTA FOR ANTONIO, AND THE INITIAL APPORTIONMENT FOR ANTONIO. BUT WE'LL ACTUALLY GO AHEAD AND GO THROUGH THIS ENTIRE PROCESS FOR THIS PROBLEM. SO THE FIRST STEP IS TO FIND THE DIVISOR, SO WE'LL SET THIS UP AS A TABLE AS WE SEE HERE. NOTICE HOW WE ALREADY FOUND THE SUM OF THE TOTAL NUMBER OF PAGES, WHICH IS 1,120, AND THERE ARE 10 PIECES OF CANDY TO APPORTION. SO OUR DIVISOR IS 1,120 DIVIDED BY 10, OR 112. SO TO FIND THE QUOTA WE'LL TAKE THE NUMBER OF PAGES EACH STUDENT READ AND DIVIDE BY 112. SO THE QUOTE FOR ALAN WOULD BE 580 DIVIDED BY 112, WHICH WOULD BE APPROXIMATELY 5.1786. FOR ANTONIO THE QUOTA WOULD BE 230 DIVIDED BY 112, GIVING A QUOTA OF APPROXIMATELY 2.0536, AND SO ON. SO AGAIN, HERE ARE OUR QUOTAS. WE FOUND FROM THE PREVIOUS SLIDE THE DIVISOR IS 112, AND THE QUOTA FOR ANTONIO WE NOW KNOW IS 2.0536. AND NOW FOR THE INITIAL APPORTIONMENT WE REMOVE THE DECIMALS FROM THE QUOTA. SO ALAN WOULD RECEIVE 5 PIECES, ANTONIO WOULD RECEIVE 2, ALEX WOULD RECEIVE 1, AND LUCAS WOULD ALSO RECEIVE 1. SO THE INITIAL APPORTIONMENT FOR ANTONIO, WHICH IS WHAT OUR QUESTION ASKED FOR, IS 2. BUT NOTICE HOW THERE ARE 10 PIECES OF CANDY, AND THIS ONLY ADDS TO 9. SO GOING BACK TO THE ORIGINAL QUOTAS, SINCE ALEX HAS THE LARGEST DECIMAL PART IN HIS QUOTA, HE WOULD RECEIVE THE EXTRA PIECE OF CANDY. AND THEREFORE THE FINAL APPORTIONMENT WOULD BE 5, 2, 2, AND 1. NOTICE HOW HERE THE TOTAL IS 10 PIECES, SO WE'VE USED ALL THE CANDY. AND AGAIN, HERE IS THE RESULT IN A NICE TABLE. HAMILTON'S METHOD DOES SATISFY WHAT IS CALLED THE QUOTA RULE. THE QUOTA RULE SAYS THAT THE FINAL NUMBER OF REPRESENTATIVES A STATE GETS SHOULD BE WITHIN ONE OF THAT STATE'S QUOTA. SINCE WE'RE DEALING WITH WHOLE NUMBERS FOR OUR FINAL ANSWERS, THAT MEANS THAT EACH STATE SHOULD EITHER GO UP TO THE NEXT WHOLE NUMBER ABOVE ITS QUOTA OR GO DOWN TO THE NEXT WHOLE NUMBER BELOW ITS QUOTA. AND THEN FINALLY, THERE IS SOME CONTROVERSY WHEN USING HAMILTON'S METHOD. AFTER SEEING HAMILTON'S METHOD, MANY PEOPLE FIND THAT IT MAKES SENSE, AND IT'S NOT THAT DIFFICULT TO USE. SO WHY WOULD ANYONE WANT TO USE ANOTHER METHOD? WELL, THE PROBLEM IS THAT HAMILTON'S METHOD IS SUBJECT TO SEVERAL WHAT WE CALL PARADOXES. THREE OF THEM HAPPENED ON SEPARATE OCCASIONS WHEN HAMILTON'S METHOD WAS USED TO APPORTION THE UNITED STATES HOUSE OF REPRESENTATIVES. AND THOSE THREE PARADOXES ARE NUMBER ONE, THE ALABAMA PARADOX, TWO, THE NEW STATES PARADOX, AND THREE, THE POPULATION PARADOX. WE'LL TALK ABOUT EACH OF THESE IN FUTURE LESSONS. I HOPE YOU FOUND THIS HELPFUL.
Method
The largest remainder method requires the numbers of votes for each party to be divided by a quota representing the number of votes required for a seat (i.e. usually the total number of votes cast divided by the number of seats, or some similar formula). The result for each party will usually consist of an integer part plus a fractional remainder. Each party is first allocated a number of seats equal to their integer. This will generally leave some remainder seats unallocated: the parties are then ranked on the basis of the fractional remainders, and the parties with the largest remainders are each allocated one additional seat until all the seats have been allocated. This gives the method its name.
Quotas
There are several possibilities for the quota. The most common are: the Hare quota and the Droop quota. The use of a particular quota with the largest remainders method is often abbreviated as "LR[quota name]", such as "LRDroop".^{[2]}
The Hare (or simple) quota is defined as follows
It is used for legislative elections in Russia (with a 5% exclusion threshold since 2016), Ukraine (5% threshold), Bulgaria (4% threshold), Lithuania (5% threshold for party and 7% threshold for coalition), Tunisia,^{[3]} Taiwan (5% threshold), Namibia and Hong Kong. The Hamilton method of apportionment is actually a largestremainder method which uses the Hare Quota. It is named after Alexander Hamilton, who invented the largestremainder method in 1792.^{[4]} It was first adopted to apportion the U.S. House of Representatives every ten years between 1852 and 1900.
The Droop quota is the integer part of
and is applied in elections in South Africa. The HagenbachBischoff quota is virtually identical, being
either used as a fraction or rounded up.
The Hare quota tends to be slightly more generous to less popular parties and the Droop quota to more popular parties. This means that Hare can arguably be considered more proportional than Droop quota.^{[5]}^{[6]}^{[7]}^{[8]}^{[9]} However, an example shows that the Hare quota can fail to guarantee that a party with a majority of votes will earn at least half of the seats (though even the Droop quota can very rarely do so).
The Imperiali quota
is rarely used since it suffers from the defect that it might result in more seats being allocated than there are available (this can also occur with the HagenbachBischoff quota but it is very unlikely, and it is impossible with the Hare and Droop quotas). This will certainly happen if there are only two parties. In such a case, it is usual to increase the quota until the number of candidates elected is equal to the number of seats available, in effect changing the voting system to the Jefferson apportionment formula.
Examples
These examples take an election to allocate 10 seats where there are 100,000 votes.
Hare quota
Party  Yellows  Whites  Reds  Greens  Blues  Pinks  Total 

Votes  47,000  16,000  15,800  12,000  6,100  3,100  100,000 
Seats  10  
Hare Quota  10,000  
Votes/Quota  4.70  1.60  1.58  1.20  0.61  0.31  
Automatic seats  4  1  1  1  0  0  7 
Remainder  0.70  0.60  0.58  0.20  0.61  0.31  
Highestremainder seats  1  1  0  0  1  0  3 
Total seats  5  2  1  1  1  0  10 
Droop quota
Party  Yellows  Whites  Reds  Greens  Blues  Pinks  Total 

Votes  47,000  16,000  15,800  12,000  6,100  3,100  100,000 
Seats  10+1=11  
Droop quota  9,091  
Votes/quota  5.170  1.760  1.738  1.320  0.671  0.341  
Automatic seats  5  1  1  1  0  0  8 
Remainder  0.170  0.760  0.738  0.320  0.671  0.341  
Highestremainder seats  0  1  1  0  0  0  2 
Total seats  5  2  2  1  0  0  10 
Pros and cons
It is relatively easy for a voter to understand how the largest remainder method allocates seats. The Hare quota gives an advantage to smaller parties while the Droop quota favours larger parties.^{[10]} However, whether a list gets an extra seat or not may well depend on how the remaining votes are distributed among other parties: it is quite possible for a party to make a slight percentage gain yet lose a seat if the votes for other parties also change. A related feature is that increasing the number of seats may cause a party to lose a seat (the socalled Alabama paradox). The highest averages methods avoid this latter paradox; but since no apportionment method is entirely free from paradox,^{[11]} they introduce others like quota violation (see Quota rule).^{[12]}
Technical evaluation and paradoxes
The largest remainder method satisfies the quota rule (each party's seats amount to its ideal share of seats, either rounded up or rounded down) and was designed to satisfy that criterion. However, that comes at the cost of paradoxical behaviour. The Alabama paradox is exhibited when an increase in seats apportioned leads to a decrease in the number of seats allocated to a certain party. In the example below, when the number of seats to be allocated is increased from 25 to 26 (with the number of votes held constant), parties D and E counterintuitively end up with fewer seats.
With 25 seats, the results are:
Party  A  B  C  D  E  F  Total 

Votes  1500  1500  900  500  500  200  5100 
Seats  25  
Hare quota  204  
Quotas received  7.35  7.35  4.41  2.45  2.45  0.98  
Automatic seats  7  7  4  2  2  0  22 
Remainder  0.35  0.35  0.41  0.45  0.45  0.98  
Surplus seats  0  0  0  1  1  1  3 
Total seats  7  7  4  3  3  1  25 
With 26 seats, the results are:
Party  A  B  C  D  E  F  Total 

Votes  1500  1500  900  500  500  200  5100 
Seats  26  
Hare quota  196  
Quotas received  7.65  7.65  4.59  2.55  2.55  1.02  
Automatic seats  7  7  4  2  2  1  23 
Remainder  0.65  0.65  0.59  0.55  0.55  0.02  
Surplus seats  1  1  1  0  0  0  3 
Total seats  8  8  5  2  2  1  26 
References
 ^ Tannenbaum, Peter (2010). Excursions in Modern Mathematics. New York: Prentice Hall. p. 128. ISBN 9780321568038.
 ^ Gallagher, Michael; Mitchell, Paul (20050915). The Politics of Electoral Systems. OUP Oxford. ISBN 9780191531514.
 ^ "2". Proposed Basic Law on Elections and Referendums  Tunisia (Nonofficial translation to English). International IDEA. 26 January 2014. p. 25. Retrieved 9 August 2015.
 ^ Eerik Lagerspetz (26 November 2015). Social Choice and Democratic Values. Studies in Choice and Welfare. Springer. ISBN 9783319232614. Retrieved 20170817.
 ^ "Archived copy" (PDF). Archived from the original (PDF) on 20150924. Retrieved 20110508.
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: CS1 maint: archived copy as title (link)  ^ "Archived copy" (PDF). Archived from the original (PDF) on 20070926. Retrieved 20070926.
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: CS1 maint: archived copy as title (link)  ^ "Notes on the Political Consequences of Electoral Laws by Lijphart, Arend, American Political Science Review Vol. 84, No 2 1990". Archived from the original on 20060516. Retrieved 20060516.
 ^ "Lipjhart on PR formulas".
 ^ See for example the 2012 election in Hong Kong Island where the DAB ran as two lists and gained twice as many seats as the singlelist Civic despite receiving fewer votes in total: New York Times report
 ^ Balinski, Michel; H. Peyton Young (1982). Fair Representation: Meeting the Ideal of One Man, One Vote. Yale Univ Pr. ISBN 0300027249.
 ^ Messner; et al. "RangeVoting: Apportionment and rounding schemes". Retrieved 20140202.