To determine There are five parts to be determined in this question and these are: (a) To find apportionment using Hamilton’s method. (b) To find apportionment using Jefferson’s method. (c) To find apportionment using Webster’s method. (d) To find apportionment using Hill-Huntington’s method. (e) Compare results of various methods Explanation Given: Total 100 delegates are to be apportioned among three states A, B, C and D with populations 1234, 3498, 2267 and 5558 respectively. Formula Used: Standard divisor = total population number of delegates Standard quota = state population standard divisor Calculations: Number of delegates to be apportioned = 100 Total population of the four states = 1234 + 3498 + 2267 + 5558 = 12557 Calculate standard divisor Standard divisor = 12557 100 = 125.57 Calculate Standard quota for each state Standard quota = state population 125.57 (These calculations are shown in the third row of the table below) Give each state the number of delegates found by rounding its standard quota down. This number is the state’s minimum quota. (These calculations are shown in the fourth row of the table below) Find the sum total of the delegates so obtained. You will find that two delegate are left to be apportioned. State A B C D Total Population 1234 3498 2267 5558 12557 Standard quota 1234 125.57 = 9.83 3498 125.57 = 27.86 2267 125.57 = 18.05 1234 125.57 = 44.26 100 Minimum quota 9 27 18 44 98 Part (a): Hamilton’s Method of apportionment Find each state’s fractional remainder – fraction that remains in the standard quota after subtracting the minimum quota. Give each extra delegate to the state with highest fractional remainder. (These calculations are shown in the fifth row of the table below) The final apportionment is given in the sixth row of the table below: State A B C D Total Population 1234 3498 2267 5558 12557 Standard quota 1234 125.57 = 9.83 3498 125.57 = 27.86 2267 125.57 = 18.05 1234 125.57 = 44.26 100 Minimum quota 9 27 18 44 98 Fractional remainder 0.83 (2 nd largest) 0.86 (largest) 0.05 0.26 1 Final Apportionment 10 28 18 44 100 Thus, the apportionment to the four states A, B, C and D are 10, 28, 18 and 44 respectively. Part (b): Jefferson’s Method of apportionment Since, there are two extra seats, so according to this method you have to choose a modified divisor that is lower than the standard divisor. The state population is divided by modified divisor to find modified quota. Round these modified quotas to find new set of minimum quotas. The last row in the table below shows the new minimum quotas with a modified divisor of 123.5. Observe that this new set of minimum quota uses all 100 seats. Hence, the last row is the final apportionment. State A B C D Total Population 1234 3498 2267 5558 12557 Standard quota 1234 125.57 = 9.83 3498 125.57 = 27.86 2267 125.57 = 18.05 1234 125.57 = 44.26 100 Minimum quota 9 27 18 44 98 Modified quota 1234 123.5 = 9.99 3498 123.5 = 28.32 2267 123.5 = 18.36 5558 123.5 = 45 101.68 Minimum quota 9 28 18 45 100 Thus, the apportionment to the four states A, B, C and D are 9, 28, 18 and 45 respectively...

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ISBN: 9781269927116

Author: Bennett

Publisher: PEARSON

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Chapter 12.C, Problem 42E

To determine

There are five parts to be determined in this question and these are:

(a) To find apportionment using Hamilton’s method.

(b) To find apportionment using Jefferson’s method.

(c) To find apportionment using Webster’s method.

(d) To find apportionment using Hill-Huntington’s method.

(e) Compare results of various methods

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Chapter 12.C, Problem 41E

Chapter 12.C, Problem 43E

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