4.5 Huntington-Hill Method

In 1920, no new apportionment was done, because Congress couldn’t agree on the method to be used. They appointed a committee of mathematicians to investigate, and they recommended the Huntington-Hill Method. They continued to use Webster’s method in 1931, but after a second report recommending Huntington-Hill, it was adopted in 1941 and is the current method of apportionment used in Congress.

The Huntington-Hill Method is similar to Webster’s method, but attempts to minimize the percent differences of how many people each representative will represent.

Note

Huntington-Hill Method

Determine how many people each representative should represent. Do this by dividing the total population of all the states by the total number of representatives. This answer is called the standard divisor.
Divide each state’s population by the divisor to determine how many representatives it should have. Record this answer to several decimal places. This answer is called the quota.
Cut off the decimal part of the quota to obtain the lower quota, which we’ll call $n$ . Compute $\sqrt{n (n + 1)}$ , which is the geometric mean of the lower quota and one value higher.
If the quota is larger than the geometric mean, round up the quota; if the quota is smaller than the geometric mean, round down the quota. Add up the resulting whole numbers to get the initial allocation.
If the total from Step 4 was less than the total number of representatives, reduce the divisor and recalculate the quota and allocation. If the total from step 4 was larger than the total number of representatives, increase the divisor and recalculate the quota and allocation. Continue doing this until the total in Step 4 is equal to the total number of representatives. The divisor we end up using is called the modified divisor or adjusted divisor.

Example 1

Again, Delaware, with an initial divisor of 21,900.82927:

County	Population	Quota	Lower Quota	Geom Mean	Initial
Kent	162,310	7.4111	7	7.48	7
New Castle	538,479	24.5872	24	24.49	25
Sussex	197,145	9.0017	9	9.49	9
Total	$897, 934$				$41$

Show solution

This gives the required total, so we’re done.

Example 2

Again, Rhode Island, with an initial divisor of 14,034.22667:

County	Population	Quota	Lower Quota	Geom Mean	Initial
Bristol	49,875	3.5538	3	3.46	4
Kent	166,158	11.8395	11	11.49	12
Newport	82,888	5.9061	5	5.48	6
Providence	626,667	44.6528	44	44.50	45
Washington	126,979	9.0478	9	9.49	9
Total	$1, 052, 567$				$76$

Show solution

This is too many, so we need to increase the divisor. Let’s try 14,100:

County	Population	Quota	Lower Quota	Geom Mean	Initial
Bristol	49,875	3.5372	3	3.46	4
Kent	166,158	11.7843	11	11.49	12
Newport	82,888	5.8786	5	5.48	6
Providence	626,667	44.4445	44	44.50	44
Washington	126,979	9.0056	9	9.49	9
Total	$1, 052, 567$				$75$

This works, so we’re done.

In both these cases, the apportionment produced by the Huntington-Hill method was the same as those from Webster’s method.

Example 3

Consider a small country with 5 states, two of which are much larger than the others. We need to apportion 70 representatives. We will apportion using both Webster’s method and the Huntington-Hill method.

# Math in Society 4.5 -- One population table, every apportionment method
# Hamilton and Lowndes share out leftovers; Jefferson, Adams, Webster and
# Huntington-Hill hunt for a modified divisor. Here the divisor methods are run in
# their equivalent "hand out one seat at a time to the highest priority" form.
from math import sqrt

pops = {"A": 300500, "B": 200000, "C": 50000, "D": 38000, "E": 21500}
SEATS = 70          # EDIT: the house size, or edit the populations above

def leftover_method(p, s, lowndes):
    divisor = sum(p.values()) / s
    quota = {k: v / divisor for k, v in p.items()}
    got = {k: int(q) for k, q in quota.items()}
    rank = (lambda k: (-(quota[k] - got[k]) / max(got[k], 1), k)) if lowndes \
        else (lambda k: (-(quota[k] - got[k]), k))
    for k in sorted(p, key=rank)[:s - sum(got.values())]:
        got[k] += 1
    return got

def divisor_method(p, s, priority):
    got = {k: 0 for k in p}
    for _ in range(s):
        got[max(p, key=lambda k: (priority(p[k], got[k]), p[k]))] += 1
    return got

BIG = float("inf")
methods = {
    "Hamilton":  lambda p, s: leftover_method(p, s, False),
    "Lowndes":   lambda p, s: leftover_method(p, s, True),
    "Jefferson": lambda p, s: divisor_method(p, s, lambda w, n: w / (n + 1)),
    "Adams":     lambda p, s: divisor_method(p, s, lambda w, n: BIG if n == 0 else w / n),
    "Webster":   lambda p, s: divisor_method(p, s, lambda w, n: w / (n + 0.5)),
    "Hunt-Hill": lambda p, s: divisor_method(p, s,
                    lambda w, n: BIG if n == 0 else w / sqrt(n * (n + 1))),
}

std = sum(pops.values()) / SEATS
names = list(pops)
print(f"{SEATS} seats, standard divisor {sum(pops.values()):,} / {SEATS} = {std:,.3f}\n")
print(f"{'':<11}" + "".join(f"{k:>9}" for k in names) + f"{'total':>9}")
print(f"{'population':<11}" + "".join(f"{pops[k]:>9,}" for k in names))
print(f"{'quota':<11}" + "".join(f"{pops[k]/std:>9.3f}" for k in names))
print("-" * (11 + 9 * (len(names) + 1)))
results = {}
for m, fn in methods.items():
    results[m] = fn(pops, SEATS)
    flags = [k for k in names if not (pops[k]/std - 1 < results[m][k] < pops[k]/std + 1)]
    note = "   quota rule VIOLATED: " + ", ".join(flags) if flags else ""
    print(f"{m:<11}" + "".join(f"{results[m][k]:>9}" for k in names)
          + f"{sum(results[m].values()):>9}{note}")

print("\nSame people, same seats, different answers:")
for k in names:
    answers = {results[m][k] for m in methods}
    if len(answers) > 1:
        print(f"   State {k}: quota {pops[k]/std:>7.3f}"
              f"  ->  anywhere from {min(answers)} to {max(answers)} seats")
print("\nAdams rounds everything up and so leans small; Jefferson cuts everything off")
print("and so leans large; Webster and Huntington-Hill sit in between, and here they")
print("disagree by one seat -- Webster gives it to the largest state, Huntington-Hill")
print("to the smallest.")
print("Try it: set SEATS = 71 and re-run, then 69. The disagreements move around.")

State	Population
A	300,500
B	200,000
C	50,000
D	38,000
E	21,500

Show solution

1. The total population is 610,000. Dividing this by the 70 representatives gives the divisor: 8714.286.

2. Dividing each state’s population by the divisor gives the quotas.

State	Population	Quota
A	300,500	34.48361
B	200,000	22.95082
C	50,000	5.737705
D	38,000	4.360656
E	21,500	2.467213

Webster’s Method

3. Using Webster’s method, we round each quota to the nearest whole number

State	Population	Quota	Initial
A	300,500	34.48361	34
B	200,000	22.95082	23
C	50,000	5.737705	6
D	38,000	4.360656	4
E	21,500	2.467213	2

4. Adding these up, they only total 69 representatives, so we adjust the divisor down. Adjusting the divisor down to 8700 gives an updated allocation totaling 70 representatives

State	Population	Quota	Initial
A	300,500	34.54023	35
B	200,000	22.98851	23
C	50,000	5.747126	6
D	38,000	4.367816	4
E	21,500	2.471264	2

Huntington-Hill Method

3. Using the Huntington-Hill method, we round down to find the lower quota, then calculate the geometric mean based on each lower quota. If the quota is less than the geometric mean, we round down; if the quota is more than the geometric mean, we round up.

State	Population	Quota	Lower Quota	Geometric Mean	Initial
A	300,500	34.48361	34	34.49638	34
B	200,000	22.95082	22	22.49444	23
C	50,000	5.737705	5	5.477226	6
D	38,000	4.360656	4	4.472136	4
E	21,500	2.467213	2	2.44949	3

These allocations add up to 70, so we’re done.

Notice that this allocation is different than that produced by Webster’s method. In this case, state E got the extra seat instead of state A.

Adapted from Math in Society by David Lippman, hosted on LibreTexts (math.libretexts.org) and licensed under CC BY-SA 3.0. Changes were made. License: CC-BY-SA-3.0.