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Apportionment :. A Mathematical Dilemma for Congress and the Electoral College (abbreviated version). 4. Standard Divisor Before we introduce some of the ways that question has been answered, we need to define a useful concept called the standard divisor.
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Apportionment: A Mathematical Dilemma for Congress and the Electoral College (abbreviated version)
4. Standard Divisor Before we introduce some of the ways that question has been answered, we need to define a useful concept called the standard divisor. • recall that we got the IDEAL QUOTA by calculating… Standard Divisor
5. Hypothetical Example (again) In this case, the standard divisor = (26,000 ÷ 26) = 1,000 Divide state pop by the standard divisor to get the ideal quota. An assembly has 26 seats StatePopulationIdeal Quota Mass Bay9,061 9,061 ÷ 1,000 = 9.061 Connecticut 7,179 7,179 ÷ 1,000 = 7.179 Rhode Island 5,259 5,259 ÷ 1,000 = 5.259 New Hampshire 3,319 3,319 ÷ 1,000 = 3.319 Maine 1,1821,182 ÷ 1,000 =1.182 TOTAL 26,000 26.
Hamilton’s Method Give each state the whole number in its ideal quota (making sure that each state has 1 seat), then assign remaining seats to those states with the largest fraction. 26 seats So we find the state with the largest fraction. StatePopulationIdeal Quotawhole number Mass Bay9,061 9.061 9 Connecticut 7,179 7.179 7 Rhode Island 5,259 5.259 5 New Hampshire 3,319 3.319 4 Maine 1,1821.1821 TOTAL 26,000 26. 26 And give this state 1 seat.
2. Jefferson’s Method Find a divisor so that the whole numbers contained in the quotients of the states sum to the required total. Then assign each state its whole number. 1. Start with the standard divisor as the divisor. Recall:ideal quota = state pop÷standard divisor StatePopulationIdeal Quota Mass Bay9,061 9.061 =9,061 ÷ 1,000 Connecticut 7,179 7.179 =7,179 ÷ 1,000 Rhode Island 5,259 5.259 =5,259 ÷ 1,000 New Hampshire 3,319 3.319 =3,319 ÷ 1,000 Maine 1,1821.182=1,182 ÷ 1,000 TOTAL 26,000 26.
2. Jefferson’s Method 1. Start with the standard divisor 2. Add the whole numbers in the ideal quota (watch them). 3. Adjust the divisor until the whole numbers add to desired # seats (=26). 4. Assign the whole numbers to each state. StatePopulationQuotaWhole Number Mass Bay9,061 10.068 10 Connecticut 7,179 7.977 7 Rhode Island 5,259 5.843 5 New Hampshire 3,319 3.688 3 Maine 1,1821.313 1 TOTAL 26,000 26
Hill-Huntington Method • Find the ideal quota. • Calculate the lower quota (LQ) and upper quota (UQ). This is the whole number of the ideal quota and the whole number plus 1. • Calculate the geometric mean of the lower and upper quotas (= SQRT(LQ×UP)). • If the ideal quota is less than the geometric mean, assign the state its lower quota. If the ideal quota is greater than or equal to the geometric mean, assign the state it’s upper quota. • Add assigned quota. If it doesn’t add to the desired number of seats, alter the standard divisor.
3. Hill-Huntington Method • Find the ideal quota. • Calculate the lower quota (LQ) and upper quota (UQ). • Calculate the geometric mean = sqrt(LQ × UQ). • If the ideal quota is less than the geometric mean, assign LQ. If the ideal quota is greater than the geometric mean, assign UQ. • If quota, doesn’t add to required seats, adjust common divisor. Divisor = 960 … changes all the figures in blue. Adds to 26! StatePopulationQuotaLQUQG.MeanQuota Mass Bay9,061 9.439 9 10 9.487 9 Connecticut 7,179 7.478 7 8 7.483 7 Rhode Island 5,259 5.478 5 6 5.477 6 New Hampshire 3,319 3.457 3 4 3.464 3 Maine 1,1821.231 1 2 1.414 1 TOTAL 26,000 26.000 26