Transcript The Hamilton and Jefferson Method for Apportionment Ideal Ratio Ideal R atio  T otal P opulation N um ber of Seats • • • • • Example 1 A 989 B 855 C.

The Hamilton and
Jefferson Method for
Apportionment
Ideal Ratio
Ideal R atio 
T otal P opulation
N um ber of Seats
•
•
•
•
•
Example 1
A 989
B 855
C 694
D 462
If there are 30 seats to hand out then
the ideal ratio would be found by
IR 
989  855  694  462
30
 100
Quotas
• Each class size divided by the ideal ratio.
•
•
•
•
•
Example 1
A 989 • A = 989 ÷ 100 = 9.89
B 855 • B = 855 ÷ 100 = 8.55
C 694 • C = 694 ÷ 100 = 6.94
D 462 • D = 462 ÷ 100 = 4.62
Hamilton Initial Distribution
Truncate the quotas to find the starting place.
A
B
C
D
Class size
989
855
694
462
Quota
9.89
8.55
6.94
4.62
Trunc
9
8
6
4
27
Hamilton Final Distribution
The remaining seats go to the class with the largest
decimal part of the quota. With each class getting at most
one additional seat
A
B
C
D
Class size
989
855
694
462
Quota
9.89
8.55
6.94
4.62
Trunc
9
8
6
4
Hamilton
10
8
7
5
27
30
Jefferson Meathod
• Instead of giving the remaining seats to the
class with the largest decimal part. Find
how many people each representative will
represent.
• Do this by finding the Jefferson adjusted
ratio.
Jefferson Initial Distribution
Divide each class size by one more than the trunc
A
B
C
D
Class size
989
855
694
462
Quota Trunc
9.89
9
8.55
8
6.94
6
4.62
4
27
J.A.R.
989÷10
855÷9
694÷7
462÷5
J.A.R.
98.9
95
99.14
92.4
Jefferson Final Distribution
Give the class with an adjusted ratio that is the closest to the ideal
ratio gets the first additional seat. Calculate another adjusted rate for
that class and use this new A.R. to help determine who gets any
additional seats. A class may get more than one additional seat.
A
B
C
D
Class size
989
855
694
462
Quota Trunc J.A.R. J.A.R. Jefferson
9.89
9
98.9 89.91
10
8.55
8
95
9
6.94
6
99.14 86.75
7
4.62
4
92.4
4
27
30