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The Hamilton and Jefferson Method for Apportionment

The Hamilton and Jefferson Method for Apportionment. Ideal Ratio. Example 1 A 989 B 855 C 694 D 462. If there are 30 seats to hand out then the ideal ratio would be found by. Quotas. Each class size divided by the ideal ratio. Example 1 A 989 B 855 C 694 D 462. A = 989 ÷ 100 = 9.89

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The Hamilton and Jefferson Method for Apportionment

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  1. The Hamilton and Jefferson Method for Apportionment

  2. Ideal Ratio • Example 1 • A 989 • B 855 • C 694 • D 462 If there are 30 seats to hand out then the ideal ratio would be found by

  3. Quotas • Each class size divided by the ideal ratio. • Example 1 • A 989 • B 855 • C 694 • D 462 • A = 989 ÷ 100 = 9.89 • B = 855 ÷ 100 = 8.55 • C = 694 ÷ 100 = 6.94 • D = 462 ÷ 100 = 4.62

  4. Hamilton Initial Distribution Truncate the quotas to find the starting place. Truncate – is to round down or to drop the decimal part

  5. Hamilton Final Distribution The remaining seats go to the class with the largest decimal part of the quota. With each class getting at most one additional seat

  6. Jefferson Meathod • Instead of giving the remaining seats to the class with the largest decimal part. Find how many people each representative will represent. • Do this by finding the Jefferson adjusted ratio.

  7. Jefferson Initial Distribution Divide each class size by one more than the trunc

  8. Jefferson Final Distribution Give the class with an adjusted ratio that is the closest to the ideal ratio gets the first additional seat. Calculate another adjusted rate for that class and use this new A.R. to help determine who gets any additional seats. A class may get more than one additional seat. 98.9 95 99.14

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