Problem #2
"Hamilton's Method"
(State pop natquota initial )
(WPL 43.9000 6.58500 6 )
(IES 40.9000 6.13500 6 )
(Interstate 15.2000 2.28000 2 )
(Total 100.000 (6.667) 14 )
(State frac final )
(WPL 0.585000 7 )
(IES 0.135000 6 )
(Interstate 0.280000 2 )
(Total 15 )
"Lowndes's Method"
(State pop natquota initial )
(WPL 43.9000 6.58500 6 )
(IES 40.9000 6.13500 6 )
(Interstate 15.2000 2.28000 2 )
(Total 100.000 (6.667) 14 )
(State relfrac final )
(WPL 9.750000E-2 6 )
(IES 2.250000E-2 6 )
(Interstate 0.140000 3 )
(Total 15 )
Problem #5
"Hamilton's Method"
(State pop natquota initial )
(1 7478 9.30076 9 )
(2 9003 11.1975 11 )
(3 5397 6.71252 6 )
(4 8825 10.9761 10 )
(5 3562 4.43024 4 )
(6 5936 7.38290 7 )
(Total 40201 (804.020) 47 )
(State frac final )
(1 0.300764 9 )
(2 0.197483 11 )
(3 0.712520 7 )
(4 0.976095 11 )
(5 0.430238 5 )
(6 0.382901 7 )
(Total 50 )
"Lowndes's Method"
(State pop natquota initial )
(1 7478 9.30076 9 )
(2 9003 11.1975 11 )
(3 5397 6.71252 6 )
(4 8825 10.9761 10 )
(5 3562 4.43024 4 )
(6 5936 7.38290 7 )
(Total 40201 (804.020) 47 )
(State relfrac final )
(1 3.341818E-2 9 )
(2 1.795297E-2 11 )
(3 0.118753 7 )
(4 9.760951E-2 11 )
(5 0.107560 5 )
(6 5.470013E-2 7 )
(Total 50 )
Problem #11
We did this in class!
See your notes. Kevin did it on the board in class, too.
Problem #12
Part a.
"Hamilton's Method"
(State pop natquota initial )
(first 1 0.750000 0 )
(second 3 2.25000 2 )
(Total 4 (1.333) 2 )
(State frac final )
(first 0.750000 1 )
(second 0.250000 2 )
(Total 3 )
"Lowndes's Method"
(State pop natquota initial )
(first 1 0.750000 0 )
(second 3 2.25000 2 )
(Total 4 (1.333) 2 )
(State relfrac final )
(first undefined 1 )
(second 0.125000 2 )
(Total 3 )
Part b.
"Hamilton's Method"
(State pop natquota initial )
(first 1 0.272727 0 )
(second 10 2.72727 2 )
(Total 11 (3.667) 2 )
(State frac final )
(first 0.272727 0 )
(second 0.727273 3 )
(Total 3 )
"Lowndes's Method"
(State pop natquota initial )
(first 1 0.272727 0 )
(second 10 2.72727 2 )
(Total 11 (3.667) 2 )
(State relfrac final )
(first undefined 1 )
(second 0.363636 2 )
(Total 3 )
Problem #14
Since state A is smaller than state B, the initial allocation of alpha for A
and beta for B satisfy
alpha <= beta
Suppose Lowndes' method rounds A's quota down, and B's quota up; this means
that the relative fractional parts satisfy
frac A frac B
------ <= ------
alpha beta
We need to show that Hamilton's method will not simultaneously round A's quota
up, and B's quota down, or
frac A <= frac B
From
frac A frac B
------ <= ------
alpha beta
we can multiply both sides by alpha and conclude that
frac B
frac A <= ------ * alpha
beta
but
alpha
----- <= 1
beta
(since beta >= alpha), so
frac B
frac A <= ------ * alpha <= frac B
beta