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Populations 56, 35, 26, 15, 6 Total population p = 138 House size h = 200 Standard divisor s = p/h = .69. Jefferson Method Example. Total: 197. ← must fill 3 seats. 10. 9 p 5. 8. p 5 bumped by 1 when s reaches 0.667. .667. ← s →. .690. What happens to p 5 as we lower s?. 23.
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Populations 56, 35, 26, 15, 6 Total population p = 138 House size h = 200 Standard divisor s = p/h = .69 Jefferson Method Example Total: 197 ← must fill 3 seats
10 9 p5 8 p5 bumped by 1 when s reaches 0.667 .667 ← s → .690 What happens to p5 as we lower s?
23 10 22 p4 9 p5 21 8 p4 bumped up at these two s values .652 .682 .667 What happens to p4 as we lower s? ← s → .690
39 23 10 9 p5 38 p3 22 p4 21 8 37 .667 .684 .652 .682 .667 What happens to p3 as we lower s? ← s → .690
10 52 23 39 9 p5 22 p4 51 p2 38 p3 8 21 37 50 .673 .667 .684 .652 .682 .667 What happens to p2 as we lower s? .686 ← s → .690
.675 10 39 52 23 83 51 p2 82 p1 38 p3 22 p4 9 p5 8 81 21 37 50 .673 .667 .684 .652 .682 .667 What happens to p1 as we lower s? .683 .686 ← s → .690
.675 10 39 52 23 83 51 p2 82 p1 38 p3 22 p4 9 p5 8 81 21 37 50 .673 .667 .684 .652 .682 .667 Since we need 3 more seats, the first three that are bumped up get the seats .683 .686 ← s → .690
Final Result • p2 bumped to 51 • p1 bumped to 82 • p3 bumped to 38
.675 10 39 52 23 83 51 p2 82 p1 38 p3 22 p4 9 p5 8 81 21 37 50 .673 .667 .684 .652 .682 .667 Need a quick way to determine these divisor values .683 .686 ← s → .690
.675 10 39 52 23 83 51 p2 82 p1 38 p3 22 p4 9 p5 8 81 21 37 50 .673 .667 .684 .652 .682 .667 p1, with population 56, gets a bump to 82 when s = 56/82 .683 .686 ← s → .690
.675 10 39 52 23 83 51 p2 82 p1 38 p3 22 p4 9 p5 8 81 21 37 50 .673 .667 .684 .652 .682 .667 p1, with population 56, gets a bump to 82 when s = 56/82 56/82 .686 ← s → .690
56/83 10 83 52 23 39 56/82 9 p5 22 p4 51 p2 82 p1 38 p3 81 8 50 37 21 .673 .667 .684 .652 .682 .667 p1, with population 56, gets a bump to 83 when s = 56/83 .686 ← s → .690
56/83 10 83 52 23 39 56/82 9 p5 22 p4 51 p2 82 p1 38 p3 81 8 50 37 21 .673 .667 .684 .652 .682 .667 p2, with population 35, gets a bump to 51 when s = 35/51 35/51 ← s → .690
56/83 10 83 52 23 39 56/82 9 p5 22 p4 51 p2 82 p1 38 p3 81 8 50 37 21 35/52 35/51 26/39 26/38 15/23 15/22 and so on… 6/9 ← s → .690
pi pi di = di = ni + 1 ni + 2 For a population pi with house seats ni, the divisor needed to get to ni+1 seats is To get to ni+2 seats: And so on …
In some cases, some populations get 2 seats before other can get 1. Consider the following example with four states, where p = 1012, h=100 and s = 10.12 Total: 97
71 21 70 p1 4 p4 20 p2 7 p3 69 19 6 3 9.97 10.11 9.57 10.05 9.43 9.25 ← s → 10.12 Again, need 3 seats, so the first three bumps get the seats p1 bumped twice before p3, p4 bumped once
71 21 70 p1 4 p4 20 p2 7 p3 69 19 6 3 9.97 10.11 9.57 10.05 9.43 9.25 ← s → 10.12 Again, need 3 seats, so the first three bumps get the seats In fact…
73 72 9.69 9.83 9.97 10.11 21 71 20 p2 70 p1 7 p3 4 p4 19 6 3 69 9.57 10.05 9.43 9.25 ← s → 10.12 p1 would be bumped many times before p3, p4
Final Result • p1 bumped twice to 71 • p2 bumped to 20
Another example with four states, where p = 1000, h=100 and s = 10 Total: 97
96 p1 95 2 p2 2 p3 2 p4 94 1 1 1 … 98 97 9.68 9.78 9.89 9.99 9 8.5 8 ← s → 10 In this case p1 is the only one that is bumped
Final Result • p1 bumped three times to 97