Apportionment: the pool grew by one and a claimant lost its only unit

Three claimants hold 53, 42 and 5 percent of a pool of 10 whole units: 5, 4 and 1 by largest remainder. Grow the pool to 11 and the smallest claimant's only unit goes to the others, though its claim never moved. Round at the half against one divisor and the unit stays.

Live result

Growing the pool from 10 to 11 under largest remainder takes the only unit from C, whose claim never moved.

at 10 unitsA 5 · B 4 · C 1
at 11 unitsA 6 · B 5 · C 0
Loses at 11C 1 to 0

Rounding at the half takes no unit from any claimant across pools 1 to 30.

Givens

Positive claims set each share of the total.

Claimant 1
Claimant name 1A
Claim A
Claimant 2
Claimant name 2B
Claim B
Claimant 3
Claimant name 3C
Claim C

At least 3 claimants are required.

Pool of whole units: Pool of whole units; A whole count from 1 to 500.

Allocation rule: Allocation rule

Units held as the pool grows

Units heldTyped poolUnit lost

Each step shows whole units held; a gap marks a tie, and the outline identifies a claimant who loses a unit.

One more unit, two rules

Claimant	Largest remainder at 10	Largest remainder at 11	At the half at 10	At the half at 11
A	5	6	5	6
B	4	5	4	4
C	1	0	1	1

A tie leaves the allocation unresolved; name order decides nothing.

What the two rules do

A quota gives each claimant its exact share of the pool, but whole units cannot follow every fraction.

Largest remainder gives each claimant its quota’s whole part, then distributes leftovers to the largest fractions. A larger pool reshuffles those fractions: two claimants can overtake a third while only one new unit exists.

Rounding at the half divides every claim by one common divisor and rounds each quotient. Lowering the divisor grows the total until it matches the pool. Every quotient moves forward, so no claimant loses a unit.

Two more failures of the same rule

New claimant: adding a claimant with a matching share of a larger pool can move a unit between existing claimants.

Population: a faster growing claim can lose a unit to a slower growing one because leftover rankings compare fractions.

The boundary where this stops

The page compares declared rules on your claims without choosing an allocation or ranking claimants.

Here is the maths

The formulas

Each quota multiplies the claim’s share by the pool:

q_i=\frac{c_i\,P}{\sum_j c_j}

where $q_i$ is the quota of one claimant, $c_i$ its claim, $P$ the pool of whole units and $j$ runs over every claimant.

Whole parts go first, then leftovers go to the largest fractions:

r_i=q_i-\lfloor q_i\rfloor

a_i=\lfloor q_i\rfloor+e_i,\qquad e_i\in\{0,1\},\qquad \sum_i e_i=P-\sum_i\lfloor q_i\rfloor

\begin{gathered}a_i=\lfloor q_i\rfloor+e_i, \\ e_i\in\{0,1\}, \\ \sum_i e_i=P-\sum_i\lfloor q_i\rfloor\end{gathered}

where $r_i$ is the fraction, $a_i$ the units held and $e_i$ one for a winning fraction, zero otherwise. A tie at the cut leaves the allocation unresolved.

One divisor makes the rounded quotients sum to the pool:

\sum_i\left\lfloor\frac{c_i}{d}+\frac{1}{2}\right\rfloor=P

where $d$ is the common divisor, found by bisection. The working interval shares one allocation; its midpoint is reported. A binding half tie remains unresolved.

A worked example

The smallest claim loses its only unit by largest remainder and keeps it by rounding at the half: claims 53, 42 and 5, pools 10 then 11.

At a pool of 10 the quotas are:

q_A=\frac{53\times 10}{100}=5.3,\qquad q_B=\frac{42\times 10}{100}=4.2,\qquad q_C=\frac{5\times 10}{100}=0.5

\begin{gathered}q_A=\frac{53\times 10}{100}=5.3, \\ q_B=\frac{42\times 10}{100}=4.2, \\ q_C=\frac{5\times 10}{100}=0.5\end{gathered}

Whole parts total 9, leaving one unit:

\lfloor 5.3\rfloor+\lfloor 4.2\rfloor+\lfloor 0.5\rfloor=5+4+0=9,\qquad 10-9=1

\begin{gathered}\lfloor 5.3\rfloor+\lfloor 4.2\rfloor+\lfloor 0.5\rfloor=5+4+0=9, \\ 10-9=1\end{gathered}

The third claimant has the largest fraction:

r_C=0.5>r_A=0.3>r_B=0.2,\qquad (a_A,a_B,a_C)=(5,4,1)

\begin{gathered}r_C=0.5>r_A=0.3>r_B=0.2, \\ (a_A,a_B,a_C)=(5,4,1)\end{gathered}

At a pool of 11 the quotas are:

q_A=\frac{53\times 11}{100}=5.83,\qquad q_B=\frac{42\times 11}{100}=4.62,\qquad q_C=\frac{5\times 11}{100}=0.55

\begin{gathered}q_A=\frac{53\times 11}{100}=5.83, \\ q_B=\frac{42\times 11}{100}=4.62, \\ q_C=\frac{5\times 11}{100}=0.55\end{gathered}

Whole parts still total 9, leaving two units:

\lfloor 5.83\rfloor+\lfloor 4.62\rfloor+\lfloor 0.55\rfloor=9,\qquad 11-9=2

\begin{gathered}\lfloor 5.83\rfloor+\lfloor 4.62\rfloor+\lfloor 0.55\rfloor=9, \\ 11-9=2\end{gathered}

The first two fractions overtake the third:

r_A=0.83>r_B=0.62>r_C=0.55,\qquad (a_A,a_B,a_C)=(6,5,0)

\begin{gathered}r_A=0.83>r_B=0.62>r_C=0.55, \\ (a_A,a_B,a_C)=(6,5,0)\end{gathered}

At 10, rounding works with a divisor from about 9.64 to 10:

d=9.8:\qquad \left\lfloor\frac{53}{9.8}+\frac{1}{2}\right\rfloor=5,\qquad \left\lfloor\frac{42}{9.8}+\frac{1}{2}\right\rfloor=4,\qquad \left\lfloor\frac{5}{9.8}+\frac{1}{2}\right\rfloor=1

\begin{gathered}d=9.8: \\ \left\lfloor\frac{53}{9.8}+\frac{1}{2}\right\rfloor=5, \\ \left\lfloor\frac{42}{9.8}+\frac{1}{2}\right\rfloor=4, \\ \left\lfloor\frac{5}{9.8}+\frac{1}{2}\right\rfloor=1\end{gathered}

At 11, a divisor from about 9.33 to 9.64 preserves the third claimant’s unit:

d=9.5:\qquad \left\lfloor\frac{53}{9.5}+\frac{1}{2}\right\rfloor=6,\qquad \left\lfloor\frac{42}{9.5}+\frac{1}{2}\right\rfloor=4,\qquad \left\lfloor\frac{5}{9.5}+\frac{1}{2}\right\rfloor=1

\begin{gathered}d=9.5: \\ \left\lfloor\frac{53}{9.5}+\frac{1}{2}\right\rfloor=6, \\ \left\lfloor\frac{42}{9.5}+\frac{1}{2}\right\rfloor=4, \\ \left\lfloor\frac{5}{9.5}+\frac{1}{2}\right\rfloor=1\end{gathered}

Show the full maths

Find a divisor and its whole interval

Adding the rounded quotients gives a total that can only fall as the divisor rises:

T(d)=\sum_i\left\lfloor\frac{c_i}{d}+\frac12\right\rfloor

where

T(d)

is the total units at that divisor.

Start with a lower divisor giving at least the pool and an upper divisor giving no units:

\ell_0=\frac{\sum_i c_i}{P+N}\qquad h_0=2\max_i c_i+1\qquad T(\ell_0)\ge P\qquad T(h_0)=0

\begin{gathered}\ell_0=\frac{\sum_i c_i}{P+N} \\ h_0=2\max_i c_i+1 \\ T(\ell_0)\ge P \\ T(h_0)=0\end{gathered}

where

\ell

is the lower search bound and

N

the number of claimants and

h

the upper search bound.

Bisect the current search bracket:

m_t=\frac{\ell_t+h_t}{2}

where

m_t

is the trial divisor and

t

the search step.

Keep the half that straddles the last divisor with enough units:

(\ell_{t+1},h_{t+1})=\begin{cases}(m_t,h_t)&T(m_t)\ge P\\(\ell_t,m_t)&T(m_t)<P\end{cases}

The model repeats this step 200 times. If the total jumps past the required pool at a shared half threshold, it leaves the tie unresolved.

For a resolved allocation, a quotient must stay inside its rounding band:

a_i-\frac12\le\frac{c_i}{d}<a_i+\frac12

Intersect those bands to find every positive divisor giving the same allocation:

L=\max_i\frac{c_i}{a_i+\frac12}\qquad H=\min_{i:a_i>0}\frac{c_i}{a_i-\frac12}\qquad L<d\le H

\begin{gathered}L=\max_i\frac{c_i}{a_i+\frac12} \\ H=\min_{i:a_i>0}\frac{c_i}{a_i-\frac12} \\ L<d\le H\end{gathered}

where

L

is the lower endpoint and

H

the upper endpoint. A claimant with no units imposes no upper bound. The lower endpoint is excluded because reaching it rounds a quotient up; the upper endpoint is included under rounding at the half.

Inside this interval no quotient crosses a rounding threshold, so every claimant keeps the same whole count. The page reports the interior midpoint:

d=\frac{L+H}{2}

Comparisons use nine decimal places; a collapsed interval remains unresolved.

For the worked claims at a pool of 10, the exact interval and its midpoint are:

\frac{106}{11}<d\le10\qquad d=\frac{108}{11}\qquad (a_A,a_B,a_C)=(5,4,1)

\begin{gathered}\frac{106}{11}<d\le10 \\ d=\frac{108}{11} \\ (a_A,a_B,a_C)=(5,4,1)\end{gathered}

At 11, the next interval gives the larger pool while preserving the smallest claimant’s unit:

\frac{28}{3}<d\le\frac{106}{11}\qquad d=\frac{313}{33}\qquad (a_A,a_B,a_C)=(6,4,1)

\begin{gathered}\frac{28}{3}<d\le\frac{106}{11} \\ d=\frac{313}{33} \\ (a_A,a_B,a_C)=(6,4,1)\end{gathered}

The summary’s example divisors lie inside these intervals.

What this does not model

No simulation of the other two failures. No claim weights or forecasts. Comparisons use nine decimal places; ties remain unresolved.

What the sweep checks

Both rules are checked from 1 through the typed pool plus 20; the finding names the first loss at or after that pool. Divisor rules avoid all three failures. Rounding at the half uniquely avoids average bias toward large or small claimants. No rule combines quota compliance with consistency as claims change.

Sources

Michel Balinski and H. Peyton Young, Fair Representation, explains pool growth, population changes and divisor immunity, unbiased rounding, quota limits and proportional division.

Scale: Whole units, linear
Basis: Both rules computed at every whole pool size from 1 through the typed pool plus 20, from the typed claims alone
Assumptions: Claims are held fixed while the pool grows, and the typed order of the rows decides nothing
Model version: 1.0.0

The whole record prints. Your browser can save it as a PDF instead of printing it.

These figures compare your claims under two rules; they neither establish anyone’s entitlement nor recommend an allocation or a rule.

Apportionment record

Apportionment: the pool grew by one and a claimant lost its only unit

Accepted inputs

Allocation rule: Largest remainder
Pool: 10
Claimant 1: A, 53
Claimant 2: B, 42
Claimant 3: C, 5

Results

Units at 10: A 5, B 4, C 1
Units at 11: A 6, B 5, C 0
First loss at or after the typed pool: C, 10 to 11

Record fields

Calculation version: 1.0.0
Boundary: The page compares declared rules on your claims without choosing an allocation or ranking claimants.
Sources: Michel Balinski and H. Peyton Young, Fair Representation, explains pool growth, population changes and divisor immunity, unbiased rounding, quota limits and proportional division.

Givens

Units held as the pool grows

One more unit, two rules

What the two rules do

Two more failures of the same rule

Here is the maths

The formulas

A worked example

Find a divisor and its whole interval

What this does not model

What the sweep checks

Sources

Related

Accepted inputs

Results

Record fields