Chapter 15: Apportionment Part 5: Webster’s Method

To understand the differences, we’ll need some notation: 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 <q> 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, <6.8> = 7 and <3.4446> = 3.

Remember s = p/h where p = total population and h = house size. 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 pi = population of state i.

Webster’s Method Once we find each state’s quota, we give that state an initial apportionment equal to <q>. That is, we’ll round the quota q to the integer <q>. 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.

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 name population A 1646 B 6936 C 154 D 2091 E 685 F 988 total 12,500 In this case, the standard divisor is 12500/250 = 50.

Example 1: Webster’s Method State population q = p/s ni A 1646 1646/50 = 32.92 33 B 6936 6936/50 = 138.72 139 C 154 154/50 = 3.08 3 D 2091 2091/50 = 41.82 42 E 685 685/50 = 13.7 14 F 988 988/50 = 19.76 20 total 12,500 250 251 Calculate each state’s quota, q. Notice that rounding the normal way produces a total apportionment that is actually more than the available seats. Therefore, we must find a modified divisor that will yield the desired total when rounding the normal way…

Example 1: Webster’s Method State population q = p/s ni A 1646 1646/50 = 32.92 33 B 6936 6936/50 = 138.72 139 C 154 154/50 = 3.08 3 D 2091 2091/50 = 41.82 42 E 685 685/50 = 13.7 14 F 988 988/50 = 19.76 20 total 12,500 250 251 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.

Example 1: Webster’s Method State population q = p/s ni Modified quota Rounding A 1646 1646/50 = 32.92 33 1646/50.1 = 32.85 B 6936 6936/50 = 138.72 139 6936/50.1 = 138.44 138 C 154 154/50 = 3.08 3 154/50.1 = 3.07 D 2091 2091/50 = 41.82 42 2091/50.1 = 41.74 E 685 685/50 = 13.7 14 685/50.1 = 13.67 F 988 988/50 = 19.76 20 988/50.1 = 19.72 total 12,500 251 250

Example 1: Webster’s Method Final answer State population q = p/s ni Modified quota Rounding A 1646 1646/50 = 32.92 33 1646/50.1 = 32.85 B 6936 6936/50 = 138.72 139 6936/50.1 = 138.44 138 C 154 154/50 = 3.08 3 154/50.1 = 3.07 D 2091 2091/50 = 41.82 42 2091/50.1 = 41.74 E 685 685/50 = 13.7 14 685/50.1 = 13.67 F 988 988/50 = 19.76 20 988/50.1 = 19.72 total 12,500 251 250

Example 2: Webster’s Method 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? State population A 453 B 367 C 697 total 1517 The standard divisor is s = p/h = 1517/75 = 20.2267 q = pi/s that is, (state pop.)/(std. divisor) What is the quota for each state ?

Example 2: Webster’s Method 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. State population q A 453 22.39618 B 367 18.14436 C 697 34.45946 total 1517 75

Example 2: Webster’s Method 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. State population q ni = <q> A 453 22.39618 22 B 367 18.14436 18 C 697 34.45946 34 total 1517 75 74 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.

Example 2: Webster’s Method 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. State population q ni = <q> A 453 22.39618 22 B 367 18.14436 18 C 697 34.45946 34 total 1517 75 74 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.

Example 2: Webster’s Method Final answer … State population q ni = <q> Modified quotas Rounding A 453 22.39618 22 453/20.15 = 22.48 B 367 18.14436 18 367/20.15 = 18.21 C 697 34.45946 34 697/20.15 = 34.59 35 total 1517 75 74 The last column is the final result. The answer is A gets 22 seats, B gets 18 and C gets 35 seats.