## Thinking Mathematically (6th Edition)

Published by Pearson

# Chapter 13 - Voting and Apportionment - 13.4 Flaws of Apportionment Methods - Exercise Set 13.4 - Page 885: 2

#### Answer

(a) Using Hamilton's method, each school is apportioned the following number of computers: School A is apportioned 20 computers. School B is apportioned 18 computers. School C is apportioned 16 computers. School D is apportioned 3 computers. (b) The Alabama paradox occurs. Initially, with 57 computers, School D was allocated 3 computers. After the number of computers increased to 58, School D was allocated only 2 computers.

#### Work Step by Step

(a) We can find the standard divisor. $standard~divisor = \frac{total ~enrollment}{number~of~ computers}$ $standard~divisor = \frac{14,250}{57}$ $standard~divisor = 250$ We can find each school's standard quota. The standard quota of each school is the school's enrollment divided by the standard divisor. School A: $standard ~quota = \frac{enrollment}{standard~divisor}$ $standard~quota = \frac{5040}{250}$ $standard~quota = 20.16$ School B: $standard ~quota = \frac{enrollment}{standard~divisor}$ $standard~quota = \frac{4560}{250}$ $standard~quota = 18.24$ School C: $standard ~quota = \frac{enrollment}{standard~divisor}$ $standard~quota = \frac{4040}{250}$ $standard~quota = 16.16$ School D: $standard ~quota = \frac{enrollment}{standard~divisor}$ $standard~quota = \frac{610}{250}$ $standard~quota = 2.44$ Hamilton's method is an apportionment method that involves rounding each standard quota down to the nearest whole number. Surplus computers are given, one at a time, to the schools with the largest decimal parts in their standard quotas until there are no more surplus computers. Initially, each school is apportioned its lower quota. School A is apportioned 20 computers. School B is apportioned 18 computers. School C is apportioned 16 computers. School D is apportioned 2 computers. We can find the total number of computers which have been apportioned. total = 20 + 18 + 16 + 2 = 56 computers Since there is a total of 57 computers, there is one surplus computer. One more computer is given to School D because it has the largest decimal part (0.44) in its standard quota. Using Hamilton's method, each school is apportioned the following number of computers: School A is apportioned 20 computers. School B is apportioned 18 computers. School C is apportioned 16 computers. School D is apportioned 2 + 1 = 3 computers. (b) We can find the standard divisor. $standard~divisor = \frac{total ~enrollment}{number~of~ computers}$ $standard~divisor = \frac{14,250}{58}$ $standard~divisor = 245.69$ We can find each school's standard quota. The standard quota of each school is the school's enrollment divided by the standard divisor. School A: $standard ~quota = \frac{enrollment}{standard~divisor}$ $standard~quota = \frac{5040}{245.69}$ $standard~quota = 20.51$ School B: $standard ~quota = \frac{enrollment}{standard~divisor}$ $standard~quota = \frac{4560}{245.69}$ $standard~quota = 18.56$ School C: $standard ~quota = \frac{enrollment}{standard~divisor}$ $standard~quota = \frac{4040}{245.69}$ $standard~quota = 16.44$ School D: $standard ~quota = \frac{enrollment}{standard~divisor}$ $standard~quota = \frac{610}{245.69}$ $standard~quota = 2.48$ Hamilton's method is an apportionment method that involves rounding each standard quota down to the nearest whole number. Surplus computers are given, one at a time, to the schools with the largest decimal parts in their standard quotas until there are no more surplus computers. Initially, each school is apportioned its lower quota. School A is apportioned 20 computers. School B is apportioned 18 computers. School C is apportioned 16 computers. School D is apportioned 2 computers. We can find the total number of computers which have been apportioned. total = 20 + 18 + 16 + 2 = 56 computers Since there is a total of 58 computers, there are two surplus computers. The first computer is given to School B because it has the largest decimal part (0.56) in its standard quota. The second computer is given to School A because it has the second largest decimal part (0.51) in its standard quota. Using Hamilton's method, each school is apportioned the following number of computers: School A is apportioned 20 + 1 = 21 computers. School B is apportioned 18 + 1 = 19 computers. School C is apportioned 16 computers. School D is apportioned 2 computers. We can see that the Alabama paradox occurs. Initially, with 57 computers, School D was allocated 3 computers. After the number of computers increased to 58, School D was allocated only 2 computers. Therefore, the Alabama paradox occurs.

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.