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Apportionment: Huntington-Hill Method - YouTube
Channel: Mathispower4u
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- WELCOME TO A LESSON
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ON THE HUNTINGTON-HILL METHOD
OF APPORTIONMENT.
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IN 1920 NO NEW APPORTIONMENT
WAS DONE
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BECAUSE CONGRESS COULDN'T AGREE
ON A METHOD TO BE USED.
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THEY APPOINTED A COMMITTEE
OF MATHEMATICIANS
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TO INVESTIGATE
AN APPORTIONMENT METHOD.
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THEY RECOMMENDED
THE HUNTINGTON-HILL METHOD.
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HOWEVER, THEY CONTINUED TO USE
WEBSTER'S METHOD IN 1931.
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AFTER A SECOND REPORT
RECOMMENDING HUNTINGTON-HILL
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IT WAS ADOPTED IN 1941
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AND IT'S THE CURRENT METHOD OF
APPORTIONMENT USED IN CONGRESS.
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THE HUNTINGTON-HILL METHOD
IS SIMILAR TO WEBSTER'S METHOD,
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BUT ATTEMPTS TO MINIMIZE
THE PERCENT DIFFERENCES
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OF HOW MANY PEOPLE EACH
REPRESENTATIVE WILL REPRESENT.
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TO APPLY THE HUNTINGTON-HILL
METHOD,
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STEP ONE,
WE DETERMINE HOW MANY PEOPLE
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EACH REPRESENTATIVE
SHOULD REPRESENT.
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WE DO THIS BY DIVIDING THE TOTAL
POPULATION OF ALL THE STATES
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BY THE TOTAL NUMBER
OF REPRESENTATIVES.
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THIS ANSWER IS CALLED
THE STANDARD DEVISOR
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OR JUST THE DEVISOR.
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STEP TWO, WE DIVIDE EACH STATE'S
POPULATION BY THE DEVISOR
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TO DETERMINE HOW MANY
REPRESENTATIVES IT SHOULD HAVE.
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WE RECORD THIS ANSWER
TO SEVERAL DECIMAL PLACES.
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THIS ANSWER IS CALLED THE QUOTA.
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STEP THREE, WE CUT OFF
THE DECIMAL PART OF THE QUOTA
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TO OBTAIN WHAT'S CALLED
THE LOWER QUOTA,
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AND WE'LL CALL THIS N.
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NEXT WE COMPUTE THE SQUARE ROOT
OF N x THE QUANTITY N + 1,
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WHICH IS THE GEOMETRIC MEAN
OF THE LOWER QUOTA
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AND ONE VALUE HIGHER.
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STEP FOUR, IF THE QUOTA IS
LARGER THAN THE GEOMETRIC MEAN
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WE ROUND UP THE QUOTA.
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IF THE QUOTA IS SMALLER THAN
THE GEOMETRIC MEAN
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WE ROUND DOWN THE QUOTA.
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THEN WE ADD THE RESULTING
WHOLE NUMBERS
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TO GET THE INITIAL ALLOCATION.
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STEP FIVE,
IF THE TOTAL FROM STEP FOUR
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IS LESS THAN THE TOTAL NUMBER
OF REPRESENTATIVES
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WE REDUCE THE DEVISOR
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AND RECALCULATE THE QUOTA
AND ALLOCATION.
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IF THE TOTAL FROM STEP FOUR
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IS LARGER THAN THE TOTAL NUMBER
OF REPRESENTATIVES
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THEN WE INCREASE THE DEVISOR
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AND RECALCULATE THE QUOTA
AND ALLOCATION.
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WE CONTINUE DOING THIS
UNTIL THE TOTAL IN STEP FOUR
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IS EQUAL TO THE TOTAL NUMBER
OF REPRESENTATIVES.
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THE DEVISOR WE END UP WITH IS
CALLED THE MODIFIED DEVISOR
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OR ADJUSTED DEVISOR.
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LET'S TAKE A LOOK AT AN EXAMPLE.
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A COLLEGE OFFERS TUTORING
IN MATH, ENGLISH, CHEMISTRY,
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AND BIOLOGY.
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THE NUMBER OF STUDENTS ENROLLED
IN EACH SUBJECT IS LISTED BELOW.
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THE COLLEGE CAN ONLY AFFORD
TO HIRE 22 TUTORS.
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USING HUNTINGTON-HILL'S METHOD
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DETERMINE THE APPORTIONMENT
OF THE TUTORS.
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SO FOR THE FIRST STEP WE WANT
TO FIND THE STANDARD DEVISOR.
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SO WE FIND THE SUM OF THE ENTIRE
ENROLLMENT, WHICH IS 780,
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AND WE DIVIDE BY THE NUMBER
OF TUTORS, WHICH IS 22.
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SO THE STANDARD DEVISOR
IS APPROXIMATELY 35.4545.
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AND NOW TO FIND THE QUOTA
FOR EACH DISCIPLINE
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WE TAKE EACH ENROLLMENT AND
DIVIDE BY THE STANDARD DEVISOR.
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NOTICE HOW IT'S ALREADY BEEN
DONE,
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BUT LET'S GO AHEAD
AND CHECK AT LEAST ONE OF THEM.
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FOR MATH WE WOULD TAKE 380
AND DIVIDE BY 35.4545
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WHICH GIVES US THE QUOTA
OF APPROXIMATELY 10.718.
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WE WOULD DO THE SAME FOR
ENGLISH, CHEMISTRY, AND BIOLOGY
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TO GET THE REMAINING QUOTAS.
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AND NOW TO FIND THE LOWER QUOTA
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WE REMOVE THE DECIMAL PART
OF THE QUOTA.
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SO WE HAVE 10, 6, 2, AND 1.
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FROM HERE WE'LL FIND
THE GEOMETRIC MEAN
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WHERE THE LOWER QUOTA
IS ACTUALLY N.
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WE WANT TO FIND THE SQUARE ROOT
OF N x THE QUANTITY AND + 1.
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SO TO FIND THE GEOMETRIC MEAN
FOR MATH, THE FIRST ONE,
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WE WOULD HAVE THE SQUARE ROOT
OF 10 x 10 + 1 OR x 11.
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SO THE GEOMETRIC MEAN
IS APPROXIMATELY 10.488.
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WHEN THE LOWER QUOTA IS 6
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THE GEOMETRIC MEAN WOULD BE
THE SQUARE ROOT OF 6 x 7
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OR APPROXIMATELY 6.481.
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WHEN THE LOWER QUOTA IS 2
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THE GEOMETRIC MEAN WOULD BE
THE SQUARE ROOT OF 2 x 3,
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WHICH IS APPROXIMATELY 2.449.
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AND FINALLY,
WHEN THE LOWER QUOTA IS 1
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THE GEOMETRIC MEAN WOULD BE
THE SQUARE ROOT OF 1 x 2,
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WHICH IS APPROXIMATELY 1.414.
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SO THIS IS HOW WE FIND
THE GEOMETRIC MEANS.
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AND NOW FOR THE NEXT STEP
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WE WANT TO COMPARE THE QUOTA
IN THE GEOMETRIC MEAN.
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IF THE QUOTA IS LARGER
WE WOULD ROUND THE QUOTA UP.
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IF THE GEOMETRIC MEAN IS LARGER
WE WOULD ROUND THE QUOTA DOWN.
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SO FOR MATH, NOTICE
HOW THE QUOTA IS LARGER,
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FOR ENGLISH
THE QUOTA'S ALSO LARGER,
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FOR CHEMISTRY THE QUOTA
IS STILL LARGER.
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AND FINALLY, FOR BIOLOGY
THE QUOTA IS STILL LARGER,
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WHICH MEANS IN EACH CASE
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WE'LL ROUND THE QUOTA UP
FOR OUR INITIAL ALLOCATION.
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SO WE'LL ROUND THE MATH QUOTA
TO 11,
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WE'LL ROUND THE ENGLISH QUOTA
TO 7, CHEMISTRY QUOTA TO 3,
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AND THE BIOLOGY QUOTA TO 2.
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NOW WE'LL FIND THIS SUM
AND COMPARE TO 22.
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NOTICE HOW THE SUM
IS ACTUALLY 23,
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IT'S TOO HIGH BECAUSE
WE ONLY HAVE 22 TUTORS.
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SO NOW WE'LL HAVE TO--
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SO NOW WE'LL HAVE TO MODIFY
THE DEVISOR,
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OR IN THIS CASE
INCREASE THE DEVISOR
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AND THEN REPEAT THE PROCESS.
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SO LET'S INCREASE THE DEVISOR
TO 37.
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NOTICE HOW THE QUOTAS
HAVE CHANGED
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BECAUSE NOW WE HAVE TO DIVIDE
THE ENROLLMENT BY 37.
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FOR EXAMPLE, FOR MATH WE WOULD
HAVE 380 DIVIDED BY 37,
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WHICH GIVES US APPROXIMATELY
10.270.
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FOR ENGLISH WE'D HAVE 240
DIVIDED BY 37,
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WHICH GIVES US APPROXIMATELY
6.486 AND SO-ON.
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SO, AGAIN,
TO FIND THE LOWER QUOTA
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WE WOULD REMOVE
THE DECIMAL PART,
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WHICH WOULD BE 10, 6, 2, AND 1.
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NOTICE HOW THIS LOWER QUOTA
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IS THE SAME AS THE PREVIOUS
LOWER QUOTA,
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WHICH MEANS OUR GEOMETRIC MEAN
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WOULD ACTUALLY BE THE SAME
AS BEFORE.
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FOR MATH IT WOULD BE
THE SQUARE ROOT OF 10 x 11,
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FOR ENGLISH IT WOULD BE
THE SQUARE ROOT OF 6 x 7,
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FOR CHEMISTRY THE SQUARE ROOT
OF 2 x 3,
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AND FOR BIOLOGY
THE SQUARE ROOT OF 1 x 2,
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WHICH AS WE SAW BEFORE, GIVES US
THESE DECIMAL VALUES HERE.
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AND, AGAIN, FROM HERE WE'RE
GOING TO COMPARE THE QUOTA
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AND THE GEOMETRIC MEAN.
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IF THE QUOTA IS LARGER
WE ROUND UP.
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IF THE GEOMETRIC MEAN IS LARGER
WE ROUND DOWN.
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NOTICE THIS TIME FOR MATH
THE GEOMETRIC MEAN IS GREATER.
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FOR ENGLISH THE QUOTA
IS JUST SLIGHTLY LARGER.
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FOR CHEMISTRY
THE QUOTA'S LARGER.
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AND FOR BIOLOGY
THE QUOTA'S ALSO LARGER.
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WHICH MEANS, WE'LL NOW ROUND
THE MATH QUOTA DOWN TO TEN,
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ROUND THE ENGLISH, CHEMISTRY,
AND BIOLOGY QUOTAS UP.
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SO WE'D HAVE 7, 3, AND 2.
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NOTICE HOW NOW WHEN WE FIND
THE SUM OF THE ALLOCATIONS
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IT IS EXACTLY 22, THE NUMBER
OF TUTORS THAT WE HAVE.
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AND, THEREFORE, WE'RE DONE.
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THIS IS OUR FINAL ALLOCATION
USING THE HUNTINGTON-HILL METHOD
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OF APPORTIONMENT.
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I HOPE YOU FOUND THIS HELPFUL.
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