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The Hamilton and Jefferson Method for Apportionment

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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

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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

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Hamilton Initial Distribution Class sizeQuotaTrunc A9899.899 B8558.558 C6946.946 D4624.624 27 Truncate the quotas to find the starting place.

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Hamilton Final Distribution Class sizeQuotaTruncHamilton A9899.89910 B8558.5588 C6946.9467 D4624.6245 2730 The remaining seats go to the class with the largest decimal part of the quota. With each class getting at most one additional seat

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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.

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Jefferson Initial Distribution Class sizeQuotaTruncJ.A.R. A9899.899989÷1098.9 B8558.558855÷995 C6946.946694÷799.14 D4624.624462÷592.4 27 Divide each class size by one more than the trunc

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Jefferson Final Distribution Class sizeQuotaTruncJ.A.R. Jefferson A9899.89998.989.9110 B8558.558959 C6946.94699.1486.757 D4624.62492.44 2730 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.

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4.1 Apportionment Problems

4.1 Apportionment Problems

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