EXAMPLE 15 Sharing Diamonds by the Webster Method
Tim and Louise have convinced Winnie to go with the Webster method because the Jefferson method is biased in her favor. The quotas are 78.26 diamonds for Winnie, 17.39 for Louise, and 4.35 for Tim; these are rounded to get the tentative shares of 78, 17, and 4 diamonds, respectively, which add up to 99. The standard divisor was 0.23, and because we need to apportion one more seat, we will have to try a smaller divisor. Our trials are shown in Table 14.15.
595
| Divisors | |||||||
|---|---|---|---|---|---|---|---|
| Participant | Tickets | 0.227 | 0.228 | 0.229 | |||
| Winnie | 18 | 79.2952 | 79 | 78.9474 | 79 | 78.6026 | 79 |
| Louise | 4 | 17.6211 | 18 | 17.5439 | 18 | 17.4672 | 17 |
| Tim | 1 | 4.4053 | 4 | 4.3860 | 4 | 4.3668 | 4 |
| Total | 23 | 101 | 101 | 100 | |||
The first divisor that we tried was 0.227. The apportionment quotients are in the left column under that divisor in the table, and to their right are the rounded values. Because the total number of “seats” with that divisor turned out to be 101 diamonds, we tried the divisor 0.228, with the same result. The divisor 0.229 produced our goal, 100 diamonds. We never tried any divisor greater than 0.23 because we knew from the start that 0.23 is too large. The result was Winnie’s preferred apportionment: 79 diamonds for her, 17 for Louise, and 4 for Tim.