# Jefferson Method of Apportionment-Help Please!

• Sep 7th 2009, 06:05 PM
TNB
Hello. I'm on my last math homework problem for the night, and I would like to finish it so I can get a good grade on it tomorrow at school. Now to to the problem at hand...

I am being asked by my textbook to apportion 100 congressional seats among 3 states (ever so originally named states a, b, and c..) by the Jefferson method of apportionment. It seems to have stumped me.
How would I go about doing this?
Oh, and by the way, here's my data:

Population
for each State:
----------------
A: 647 Ideal Ratio: 10 Quota: 64.7, 24.7, 10.6
B: 247
C: 106

(Bow)-Arigato.
• Sep 8th 2009, 04:22 AM
aidan
Could you define:
Quote:

the Jefferson method of apportionment
It seems to have me stumped since I'm not taking the class.

2nd question:
What is the: Ideal Ratio: 10 Quota: 64.7, 24.7, 10.6 values for states B & C?

.
• Sep 12th 2009, 06:20 AM
HallsofIvy
I googled on "Jefferson Apportionment Method" and got a description:
"Recall that the Jefferson apportionment method works as follows. Begin by giving each state one representative. Then, give the next representative to the state with the smallest value of (ai+1)/pi, where ai is the number of seats that state i has gotten so far, and pi is the population of state i. Keep going until all the representatives are apportioned."

Giving one representative to each state means that ai= 1 for all three states. "(ai+1)/pi" is 2/647 for state A, 2/247 for state B, and 2/106 for state C. Obviously state A with the largest population has the smallest fraction so give one representative to state A.

Now A has 2 reps, B has 1 and C has 1 so we look at 3/647, 2/247, 2/106. 3/647 is still smallest so we give another representative to A. Now A has 3 reps, B 1 and C 1 so we look at 4/647, 2/247, 2/106. 4/647 is still smallest so we give another representative to A. Now A has 4 reps, B 1 and C 1 so we look at 5/647, 2/247, 2/106. 5/647 is still smallest so we give another representative to A. Now A has 5 reps, B 1 and C 1 so we look at 6/647, 2/247, 2/106. Now 2/247 is smallest so we give B another representative. Now A has 5 reps, B 2 and C 1 so we look at 6/647, 3/247, 2/106. 6/647 is the smallest of those so we give another representative to A. Now A has 6 reps, B 2 and C 1 so we look at 6/647, 3/247, and 2/106. 6/647 is the smallest of those so we give A another representative. Now A has 7 reps, B 2 and C 1.

Oh, Blast!! I was thinking that the problem was to apportion 10 representatives and it was fairly easy. But I don't want to have to do this 90 more times!

Instead note that the entire population is 1000 (nice!) so if we could apportion fractional seats, A would get $\frac{647}{1000}(100)= 64.7$ seats, B would get $\frac{247}{1000}(10)= 24.7$ seats, and C would get $\frac{106}{1000}(10)= 10.6$ seats. That is the "quota" referred to.

Okay, since we cannot have fractional seats, lets start by giving A 64 seats, B 24 seats and C 10 seats. Then we have given out 98 seats and need to allocate the remaining two. That's not too bad. (ai+ 1)/pi is 65/647= .1004 approx for A, 25/247= .101 for B, and 11/102= 0.108 for C. The smallest of those is 65/647 so A gets another seat. Now A has 65 seats, B has 24 seats and C has 10 seats. Now we look at 66/647= 0.102, 24/247= .101, and 11/102= 0.108. Now 24/237 is smallest so we give the remaining seat to B.

A gets 65 seats, B gets 25 seats and C gets 10 seats.

By the way, as to the "originality" of naming the state "a", "b", and "c", there is a fairly famous British math book that has problems about the town "Ayling", "Beeling", and "Ceiling". Apparently "Ayling", at least, is an actual town in England. Now, that's clever!
• Sep 12th 2009, 10:37 AM
TNB
kk
Thanks, but I kinda figured this out...