1. Chapter 15: Apportionment Part 5: Webster’s Method

2. Webster’s Method Webster’s method is similar to Jefferson’s method. The first step is the same in both methods. To understand the differences, we’ll need some notation: Let represent the integer obtained by rounding the real number q in “the usual way.” That is, if the decimal part of q is greater or equal to.5, then we round q up to the next greatest integer – otherwise, we round it down. For example, = 7 and = 3.

3. Webster’s Method We begin Webster’s method by calculating the standard divisor, s, just as with Jefferson’s method. Remember s = p/h where p = total population and h = house size. Then we calculate each state’s quota: where p i = population of state i.

4. Webster’s Method 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. 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.

5. Example 1: Webster’s Method Let’s determine the apportionment by the Webster Method for the fictional country introduced in a previous example: Suppose a country has 6 states with populations as given in the table. Also, suppose there are 250 seats in the house of representatives for this country. State namepopulation A1646 B6936 C154 D2091 E685 F988 total12,500 In this case, the standard divisor is 12500/250 = 50.

6. Example 1: Webster’s Method Statepopulationq = p/snini A16461646/50 = 32.92 33 B69366936/50 = 138.72 139 C154154/50 = 3.08 3 D20912091/50 = 41.82 42 E685685/50 = 13.7 14 F988988/50 = 19.76 20 total12,500250251 1.Calculate each state’s quota, q. 2.Notice that rounding the normal way produces a total apportionment that is actually more than the available seats. 3.Therefore, we must find a modified divisor that will yield the desired total when rounding the normal way…

7. Example 1: Webster’s Method Statepopulationq = p/snini A16461646/50 = 32.92 33 B69366936/50 = 138.72 139 C154154/50 = 3.08 3 D20912091/50 = 41.82 42 E685685/50 = 13.7 14 F988988/50 = 19.76 20 total12,500250251 After some experimentation… We discover that a modified divisor of d=50.1 will work. Notice that to make the total a little lower, we needed to find a modified divisor a little larger than the standard divisor.

8. Example 1: Webster’s Method Statepopulationq = p/snini Modified quota Rounding A16461646/50 = 32.92 331646/50.1 = 32.85 33 B69366936/50 = 138.72 1396936/50.1 = 138.44 138 C154154/50 = 3.08 3154/50.1 = 3.07 3 D20912091/50 = 41.82 422091/50.1 = 41.74 42 E685685/50 = 13.7 14685/50.1 = 13.67 14 F988988/50 = 19.76 20988/50.1 = 19.72 20 total12,500251250

9. Example 1: Webster’s Method Statepopulationq = p/snini Modified quota Rounding A16461646/50 = 32.92 331646/50.1 = 32.85 33 B69366936/50 = 138.72 1396936/50.1 = 138.44 138 C154154/50 = 3.08 3154/50.1 = 3.07 3 D20912091/50 = 41.82 422091/50.1 = 41.74 42 E685685/50 = 13.7 14685/50.1 = 13.67 14 F988988/50 = 19.76 20988/50.1 = 19.72 20 total12,500251250 Final answer

10. Example 2: Webster’s Method Statepopulation A 453 B 367 C 697 total 1517 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. Suppose that this country has a house of representatives with 75 seats. What is the standard divisor? The standard divisor is s = p/h = 1517/75 = 20.2267 What is the quota for each state ? q = p i /s that is, (state pop.)/(std. divisor)

11. Example 2: Webster’s Method Statepopulationq A 453 22.39618 B 367 18.14436 C 697 34.45946 total 1517 75 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. Notice the total sum of all values of q is the house size, but of course, we expect each state to get an integer number of seats. We now will use Webster’s Method to find the final apportionment.

12. Example 2: Webster’s Method Statepopulationqn i = A 453 22.3961822 B 367 18.1443618 C 697 34.4594634 total 1517 7574 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. The initial apportionments are found by rounding the quota for each state in the usual way. However, as a result, we have one seat still to be assigned. Note that in the previous example we had to take a seat away. In this example, we will add a seat.

13. Example 2: Webster’s Method Statepopulationqn i = A 453 22.3961822 B 367 18.1443618 C 697 34.4594634 total 1517 7574 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. Because we need to add a seat, we will search for a slightly smaller modified divisor. We were using a standard divisor of 1517/75 = 20.2267. After some experimentation, we discover that using a modified divisor of 20.15 will produce the desired total.

14. Example 2: Webster’s Method Statepopulationqn i = Modified quotas Rounding A 45322.3961822453/20.15 = 22.48 22 B 36718.1443618367/20.15 = 18.21 18 C 69734.4594634697/20.15 = 34.59 35 total 1517 757475 Final answer … The last column is the final result. The answer is A gets 22 seats, B gets 18 and C gets 35 seats.