1. Independent Practice Problem There are 15 scholarships to be apportioned among 231 English majors, 502 History majors, and 355 Psychology majors. How would they be apportioned if you used the Jefferson Method? What would change if you used these methods if the number of scholarships was moved to 16?

2. Further Practice The student council has 20 members at Central High for the sophomore, junior, and senior classes that have 459, 244 and 197 students respectively. Using the Hamilton method, how many seats would each class be entitled to? What happens if the number of student council members is moved to 21 to avoid tie votes?

3. Continued At the beginning of the second semester, the enrollment has changed to 460, 274 and 196 students in each class and a re-apportionment of seats is called for by the senior class. Using the Hamilton method, re-calculate the apportionment of seats….. Do you see a problem?

4. Warm-Up Problem The Banana Republic has states Apure (3,310,000); Barinas (2,670,000); Carabobo (1,330,000); and Dolores (690,000) and 160 seats in the legislature, with populations in parentheses. Use the Jefferson Method to apportion the seats.

5. Practice Problem A mathematics department has 30 teaching assistants to be divided among three courses, according to their respective enrollments. Suppose the enrollments are as follows. Apportion the teaching assistants among the three courses using the Hamilton Method. Then Re-Apportion the TAs using 31 teaching assistants. Course College Algebra StatisticsLiberal Arts Math Total Enrollment9785003221800

6. Paradoxes Associated with Apportionment Methods Violating the Quota Rule Paradoxes The Alabama Paradox: An increase in the total number of seats to be apportioned causes a state to lose a seat. The Population Paradox: An increase in a state’s population causes it to lose a seat. The New States Paradox: Adding a new state with its fair share of seats affects the number of seats apportioned to other states.

7. Webster Method Step 1.Find a number D such that when each state's modified quota (state's population / D) is rounded the conventional way (to the nearest integer), the total is the exact number of seats to be apportioned. Step 2.Apportion to each state its modified quota rounded to the nearest integer (relative to the arithmetic mean).

8. Webster Example A college has 6 dorm complexes on campus: A, B, C, D, E and F. Their respective populations are 1646; 6936; 154; 2091; 685; and 988. College uses a 25-member student advisory board to create guidelines and rules for the dorm complexes. If they choose to use the Webster Method of Apportionment to assign the dorm complexes their seats on the board, how many seats would each complex be assigned?

9. Another Webster Example A county is composed of four districts: A, B, C and D. Their populations are 210; 1082; 311; and 284 respectively. The county commission has 18 seats. Use the Webster Method to find the final apportionment for each District.

10. Warm-Up Problem For another example, let’s consider a country with 3 states, named A, B and C. Suppose the populations of each state are as given below A--453 B--367 C--697 Suppose that this country has a house of representatives with 75 seats. What is the standard divisor? Find the Apportionment for each state using the Webster Method.

11. continued What is the quota for each state ? Is this the correct number of seats? If it is we are finished. If not, what do you think we should do…..besides give up?

12. continued If the sum of the Quotas is NOT equal to the correct number of seats to be apportioned, then, by TRIAL and ERROR, find a number (a modified divisor) to use in the place of the standard divisor so that when the modified quota for each state is rounded based on the arithmetic mean, the sum of all the rounded quotas is the exact number of seats available.

13. Webster Summarized Once we find each state’s quota, we give that state an initial apportionment equal to. That is, we’ll round the quota q to the integer using traditional rounding techniques. At this point, if the total apportionment we’ve assigned equals the house size then we are done. However, we may have already assigned more than the available seats or there may be extra seats available that have not yet been assigned. In either case (too many seats assigned or not enough) we will determine a modified divisor that yield the required total apportionment when rounding the normal way.