# Proportional approval voting

Proportional approval voting (PAV) is an electoral system which is an extension of approval voting to multiple-winner elections. It applies proportional representation principles with a ballot which is no more complicated than ballots for plurality voting. It allows each voter to vote for as many or as few candidates as they choose. The system was invented by Thorvald N. Thiele.[1] [2] [3] It was rediscovered by Forest Simmons in 2001,[4] who coined the name "proportional approval voting".

## Description

PAV works by looking at how "satisfied" each voter is with each potential result or outcome of the election. The calculated satisfaction with any particular result for an individual voter is a function of how many of the elected candidates the individual originally voted for.[5] Under PAV, to calculate the satisfaction of an individual, only the elected candidates that the individual voted for are counted - the unsuccessful candidates that they voted for, as well as the elected candidates that they did not vote for are not taken into account. Assuming that an individual voted for n candidates that were successful, their satisfaction would be calculated using the formula[4]

${\displaystyle 1+{\frac {1}{2}}+{\frac {1}{3}}+\cdots +{\frac {1}{n}}}$

Summing the satisfaction across all voters with any potential result gives the population's total satisfaction with that result. The total satisfaction is calculated for every possible set of candidates, and the set of candidates with the highest total satisfaction is deemed to be the winning set.

In an election with only one winner, PAV operates in exactly the same way as normal approval voting. If on the other hand, each voter voted exclusively for all of the candidates within a single party, PAV would function in the same way as the D'Hondt method of party-list proportional representation.

Counting the votes in PAV is NP-hard, making it a very computationally demanding voting method as the number of candidates and seats increase.[6] If there were c candidates and s seats, then there would be

${\displaystyle {\frac {c!}{s!(c-s)!}}}$

combinations of candidates to compare with each election,[7] for example if there were 24 candidates for 4 seats, there would be 10,626 combinations to calculate total satisfaction for. An election requiring this many calculations would necessitate vote counting by computer.

## Example

2 seats to be filled, four candidates: Andrea (A), Brad (B), Carter (C), and Delilah (D), and 30 voters. The ballots are:

• 5: AB
• 17: AC
• 8: D

There are 6 possible results: AB, AC, AD, BC, BD, and CD.

AB AC AD BC BD CD
Voters approving at least 1 successful candidate (satisfaction of 1 for 1st approved candidate) 22 22 30 22 13 25
Voters approving at least 2 successful candidates (satisfaction of 1/2 for 2nd approved candidate) 5 17 0 0 0 0
Total satisfaction 24.5 30.5 30 22 13 25

Andrea and Carter are elected.