Math Help - Jefferson Method of Apportionment-Help Please!

1. Jefferson Method of Apportionment-Help Please!

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

-Arigato.

2. Could you define:
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?

.

3. 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!

4. kk

Thanks, but I kinda figured this out...