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

2. 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.

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.

4. 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.

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 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.

6. Example 1: Webster’s Method
State
population
q = p/s ni A
1646
1646/50 = 32.92 33 B
6936
6936/50 =
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…

7. Example 1: Webster’s Method
State
population
q = p/s ni A
1646
1646/50 = 32.92 33 B
6936
6936/50 =
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.

8. 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 =
139
6936/50.1 =
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

9. 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 =
139
6936/50.1 =
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

10. 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 =
q = pi/s
that is, (state pop.)/(std. divisor)
What is the quota for each state ?

11. 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 B
367 C
697
total
1517 75

12. 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 B
367 18 C
697 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.

13. 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 B
367 18 C
697 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 =
After some experimentation, we discover that using a modified divisor of will produce the desired total.

14. Example 2: Webster’s Method
Final answer …
State
population q
ni = <q>
Modified quotas
Rounding A
453 22
453/20.15 = 22.48 B
367 18
367/20.15 = 18.21 C
697 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.