What happens when Webster's method can't narrow down to a working divisor?

Question

I've been poking around and without trying, managed to bump into some scenarios where Webster's method gives inconclusive results. The fact I managed it more than once without trying implies to me that this is a highly probable situation, though maybe only with small or rounded numbers.

For example, with 20 seats available, and states with populations of: 10000, 5000, 9000, 1000, 3000, 2000, 8000 (total: 38,000)

State	Population	Standard Q	Modified Q	Rounded Q (2,000.00000000)	Final (2,000.00000000)		Rounded Q (2,000.00000001)	Final (2,000.00000001)
A	10000	5.263157895	5	5	5		5	5
B	5000	2.631578947	2.5	3	3		2	2
C	9000	4.736842105	4.5	5	5		4	4
D	1000	0.526315789	0.5	1	1		0	1
E	3000	1.578947368	1.5	2	2		1	1
F	2000	1.052631579	1	1	1		1	1
G	8000	4.210526316	4	4	4		4	4
total	38000	20	19	17	21 (too high)		17	18 (too low)

I can't tell what it is about these numbers but the tiniest fraction of a change to the divisor will always swing between the result being too high or too low, with the a perfect result being impossible to hit.

I'm wondering primarily if there's a lesser-known official next-step for dealing with this situation as written in Webster's original algorithm, or a common follow-up or modified step for dealing with this situation that someone else has come up with?

Otherwise, is there a way of telling in advance if the numbers will work or not?

Answer 1

There is no standard solution to this. Websters method assumes that if you change the modified divisor by sufficiently small amounts, you can achieve any desired total.

However in the example you give, with a modified divisor of 2000, and several constituencies with a population that are exactly odd multiples of 1000, you get several constituents that switch from rounding up to rounding down exactly at 2000, which causes the total to switch from 21 (at MD=2000) to 17 (at MD=2001)

This is unlikely to occur in "real life" as it is unlikely that multiple constituents will all share a large common factor. With a more "natural" distribution of populations, I was able to achieve apportionment using Webster's method:

State	Population	Standard Q	Modified Q(1985)	Rounded Q	Final
A	10016	5.258847002	5.045843829	5	5
B	4966	2.607371627	2.501763224	3	3
C	8910	4.678147643	4.488664987	4	4
D	959	0.503517799	0.4831234257	0	1
E	3078	1.616087367	1.550629723	2	2
F	2065	1.084217158	1.040302267	1	1
G	8098	4.251811404	4.079596977	4	4
total	38092	20	19.18992443	19	20

(You'll note that I had to aim for "19" to allow me to give a single seat to D in this example)

One alternate solution is to use a different rounding scheme. Balanced rounding always rounds towards odd values, so 1.5 rounds down to 1 but 2.5 rounds up to 3. This isn't guaranteed to work but it does do so in your example (note E has been rounded down to 1):

State	Population	Standard Q	Modified Q(2000)	Rounded Q	Final
A	10000	5.263157895	5	5	5
B	5000	2.631578947	2.5	3	3
C	9000	4.736842105	4.5	5	5
D	1000	0.5263157895	0.5	1	1
E	3000	1.578947368	1.5	1	1
F	2000	1.052631579	1	1	1
G	8000	4.210526316	4	4	4
total	38000	20	19	20	20