Exercise #4
Jefferson's Method
(State pop natquota initial )
(Tanglewood 7871 9.91637 9 )
(Park Hill 7122 8.97273 8 )
(Cityview 5894 7.42562 7 )
(Berkeley 5124 6.45553 6 )
(Ridglea 5003 6.30309 6 )
(Westcliff 2323 2.92666 2 )
(Total 33337 (793.738) 38 )
(State threshhold final )
(Tanglewood 787.100 10 )
(Park Hill 791.333 9 )
(Cityview 736.750 8 )
(Berkeley 732 6 )
(Ridglea 714.714 6 )
(Westcliff 774.333 3 )
(Total (736.750) 42 )
Webster's Method
(State pop natquota initial )
(Tanglewood 7871 9.91637 10 )
(Park Hill 7122 8.97273 9 )
(Cityview 5894 7.42562 7 )
(Berkeley 5124 6.45553 6 )
(Ridglea 5003 6.30309 6 )
(Westcliff 2323 2.92666 3 )
(Total 33337 (793.738) 41 )
(State threshhold final )
(Tanglewood 749.619 10 )
(Park Hill 749.684 9 )
(Cityview 785.867 7 )
(Berkeley 788.308 7 )
(Ridglea 769.692 6 )
(Westcliff 663.714 3 )
(Total (788.308) 42 )
Exercise #6
Jefferson's Method
(State pop natquota initial )
(Kent 122709 3.52236 3 )
(New Castle 474838 13.6302 13 )
(Sussex 134034 3.84744 3 )
(Total 731581 (34837.190) 19 )
(State threshhold final )
(Kent 30677.2 3 )
(New Castle 33917 14 )
(Sussex 33508.5 4 )
(Total (33508.500) 21 )
Webster's Method
(State pop natquota initial )
(Kent 122709 3.52236 4 )
(New Castle 474838 13.6302 14 )
(Sussex 134034 3.84744 4 )
(Total 731581 (34837.190) 22 )
(State threshhold final )
(Kent 35059.7 3 )
(New Castle 35173.2 14 )
(Sussex 38295.4 4 )
(Total (35059.715) 21 )
Hamilton's Method
(State pop natquota initial )
(Kent 122709 3.52236 3 )
(New Castle 474838 13.6302 13 )
(Sussex 134034 3.84744 3 )
(Total 731581 (34837.190) 19 )
(State frac final )
(Kent 0.522356 3 )
(New Castle 0.630204 14 )
(Sussex 0.847440 4 )
(Total 21 )
Lowndes's Method
(State pop natquota initial )
(Kent 122709 3.52236 3 )
(New Castle 474838 13.6302 13 )
(Sussex 134034 3.84744 3 )
(Total 731581 (34837.190) 19 )
(State relfrac final )
(Kent 0.174119 4 )
(New Castle 4.847720E-2 13 )
(Sussex 0.282480 4 )
(Total 21 )
Exercise #11
Jefferson's Method
(State pop natquota initial )
(ANC 12237655 252.659 252 )
(NP 3983690 82.2473 82 )
(IFP 2058294 42.4956 42 )
(FF 424555 8.76537 8 )
(DP 338426 6.98715 6 )
(PAC 243478 5.02685 5 )
(ACDP 88104 1.81900 1 )
(Total 19374202 (48435.505) 396 )
(State threshhold final )
(ANC 48370.2 254 )
(NP 47996.3 83 )
(IFP 47867.3 42 )
(FF 47172.8 8 )
(DP 48346.6 7 )
(PAC 40579.7 5 )
(ACDP 44052 1 )
(Total (47996.265) 400 )
Webster's Method
(State pop natquota initial )
(ANC 12237655 252.659 253 )
(NP 3983690 82.2473 82 )
(IFP 2058294 42.4956 42 )
(FF 424555 8.76537 9 )
(DP 338426 6.98715 7 )
(PAC 243478 5.02685 5 )
(ACDP 88104 1.81900 2 )
(Total 19374202 (48435.505) 400 )
(State threshhold final )
(ANC 48274.8 253 )
(NP 48287.2 82 )
(IFP 48430.4 42 )
(FF 44690.0 9 )
(DP 45123.5 7 )
(PAC 44268.7 5 )
(ACDP 35241.6 2 )
(Total (48435.505) 400 )
Hamilton's Method
(State pop natquota initial )
(ANC 12237655 252.659 252 )
(NP 3983690 82.2473 82 )
(IFP 2058294 42.4956 42 )
(FF 424555 8.76537 8 )
(DP 338426 6.98715 6 )
(PAC 243478 5.02685 5 )
(ACDP 88104 1.81900 1 )
(Total 19374202 (48435.505) 396 )
(State frac final )
(ANC 0.658768 253 )
(NP 0.247310 82 )
(IFP 0.495562 42 )
(FF 0.765367 9 )
(DP 0.987147 7 )
(PAC 2.684962E-2 5 )
(ACDP 0.818996 2 )
(Total 400 )
Lowndes's Method
(State pop natquota initial )
(ANC 12237655 252.659 252 )
(NP 3983690 82.2473 82 )
(IFP 2058294 42.4956 42 )
(FF 424555 8.76537 8 )
(DP 338426 6.98715 6 )
(PAC 243478 5.02685 5 )
(ACDP 88104 1.81900 1 )
(Total 19374202 (48435.505) 396 )
(State relfrac final )
(ANC 2.614157E-3 252 )
(NP 3.015977E-3 82 )
(IFP 1.179909E-2 43 )
(FF 9.567093E-2 9 )
(DP 0.164525 7 )
(PAC 5.369924E-3 5 )
(ACDP 0.818996 2 )
(Total 400 )
"
e: Yes, using Jefferson's method the ANC ended up with 2 additional votes.
"
"
f: Yes, Lowndes' method agreed with the actual allocation.
"
"
20. Explain why a divisor that is too small to give the correct number of seats
using Jeffereson's method is also too small to use with Webster's method.
Using the same divisor, Webster's method will allocate at least as many seats
as Jefferson's method. Thus, a divisor that is too small for Jefferson's
method, allocating too many seats, will also be too small for Webster's.
"
Exercise #22
"
22. Find an example with two states for which Jefferson's method and Webster's
method give different apportionments.
"
Jefferson's Method
(State pop natquota initial )
(A 849 8.49000 8 )
(B 151 1.51000 1 )
(Total 1000 (100.000) 9 )
(State threshhold final )
(A 94.3333 9 )
(B 75.5000 1 )
(Total (94.333) 10 )
Webster's Method
(State pop natquota initial )
(A 849 8.49000 8 )
(B 151 1.51000 2 )
(Total 1000 (100.000) 10 )
(State threshhold final )
(A 99.8824 8 )
(B 60.4000 2 )
(Total (100.000) 10 )
)