Calculating Implied Results for New Boundaries

This page first posted 29 October 2006

This article describes how we calculate the new-boundary implied results for the 2005 General Election. We set up the problem; describe the datasets involved; and give the calculation algorithm. The same algorithm has been used to calculated the implied results for the 2010 General Election under the 2013 Constituency Boundary review.

1. A problem of allocation

We have to calculate the implied results of the 2005 election under the new boundaries. We assume that the same people would have voted and that they would have voted the same way. This may not be strictly true, especially if tactical voting is important, but it is hard to work with any other assumption. So the total number of votes for each party must be the same. Indeed, the total number of votes for each party cast within one old seat must be the same.

The problem is one of allocation. We must allocate an old seat's votes amongst the fragments of new seats which make it up. And all the totals must match. There are many possible ways of doing this, and there is no definitive right answer. So we must make reasonable guesses. Here's one possible method:

Simple Homogeneous Method This method works by assuming that each old seat is politically homogeneous (evenly spread). Suppose an old seat is split up into 3 fragments of new seats. The fragments, by electorate, make up 65%, 20% and 15% of the old seat respectively. Then we assume that the number of votes for each party in the first fragment is just 65% of the votes cast for each party in the whole old seat. And similarly for the other fragments.

This has the advantage that it is a valid allocation scheme. The sum of the implied votes in each fragment will equal the old seat totals. It is also simple to calculate as the only data required is the electorate in each fragment. But it is quite approximate. Many seats are not homogeneous and contain areas which differ from the rest of the seat. These are often on the borders of the seat and are quite likely to be transferred to a new neighbouring seat. So the Homogeneous Method will not do.

Naive Local Elections Method Another approach is to add up the local election results for the wards in each fragment. This allows for the non-homogeneity of seats, but has a serious problem. The sum of votes over the fragments will now equal the sum of the local election results in the old seat. But that will be (often quite) different from the actual 2005 election result in the old seat. So this naive method is not a valid allocation scheme.

The challenge is to find a method of allocation which has regard to the local election results, but is smart enough to match back to old seat totals. But before we get that far, we have to be clear about the data that is available to us.

2. Data, data, data

For this project data is no laughing matter. The quantities of data required are significant and their acquisition is not straightforward. This section desribes the main data elements and their sources. Where possible the data are available elsewhere on the site for inspection. We have done our best to maintain their integrity, but please let us know of any errors.

2.1 Election Result 2005 - the old seat results

This dataset has already been acquired. It was compiled and cross-checked from various sources, notably UK national newspapers, the BBC election website, district returning officers, and other websites. The principal dataset was entered twice from two different sources and cross-checked for errors. Users of this website have also been very helpful in pointing out any discrepancies.

For each old seat, we store the following information

This dataset is available as a flatfile.

2.2 New seat definitions

The Boundary Commissions for England and for Wales have published their final recommendations for the new constituencies. These definitions are in terms of local authority district and borough wards. It is a rule that every new seat is composed of a collection of undivided local wards. The wards may not all be in the same district, but they are all intact. The Boundary Commissions publish the ward composition of each new seat on their website, along with their electorates as at 2000.

We also need to know which, for each ward, which old seat it lies in. The Boundary Commission for Wales website shows the old seat against each ward in its data spreadsheets. The Boundary Commission for England does not have that information on its website, but it makes the spreadsheets available on request. (Their spreadsheets are ENGLAND.XLS, which shows the old seats and new wards, and ENGLAND_NEW_CONSTITUENCIES.XLS which shows the new seats and new wards.) Those spreadsheets also contain the electorates at 2000 and often at 2005. Where the 2005 electorate was available, it was preferred.

Combining the data gives us, for each new ward, its electorate (usually at 2005), its old seat, and its new seat. A complication appears, because the wards themselves have been redistricted since the last boundary review in 1995. This means that old seats are not made up of entire new wards. Where a new ward is split between two (or three or four) old seats, the parts are named "Wardname [1]", "Wardname [2]", and so on. The Boundary Commission usually gives the electorate separately for each ward sub-division, which is helpful.

Generally, the Boundary Commissions' data are accurate, but it has not usually been possible to cross-check it. Some ward names may be misspelt, but we have given priority to the Boundary Commission over the local authorities in the event of discrepancy.

This ward data are available in the Ward Breakdown links in the Region Table, for example North.

2.3 Local election results

There are 8,900 local district wards in England and Wales, spread over 376 local authorities. These are district councils, borough councils, London boroughs, metropolitan borough councils, or unitary councils. We do not consider county councils, because their wards are larger and do not match constituency boundaries. Although each council (usually) publishes its own election results, the formats are widely varying and there is no centralised or standardised dataset.

Some old seats are unchanged, so their allocation is trivial and the local election results for those wards are not needed. This reduces the amount of data required to 294 councils and 5,900 wards.

There were three main sources for the local electoral data:

The full list of 325 councils used, the election year, and the data source is available here.

Some standards were set for how to handle results. Votes were divided into four parties: Conservative, Labour, Lib Dem, and Other. (Plus Nationalist in Wales.) Only official party candidates were recognised for Con, Lab and Lib. Candidates with no party affiliation were classed as Other. In seats where candidates were elected unopposed, the convention was used of a negative number indicating the number of seats won (-1 for one seat, -2 for two seats, and so on).

In multi-member wards, there may be two or three candidates from each party. We took the sum of all those candidates' votes, but we recognise there is a valid alternative to use the most popular candidate's votes only. The question of which method to use is balanced. If each party puts up the same number of candidates in the ward, there will not be much difference between the methods. But often one (dominant) party will put up more candidates than the other parties. There are hard cases on both sides, and it is not possible to be definitive on the superiority of either method.

An example of a single ward's data looks like:

DistrictWardElect
2005
Old SeatNew Seat Con
Votes
Lab
Votes
Lib
Votes
Oth
Votes
West WiltshireMelksham North4,294 DevizesChippenham06131038122

The data is from West Wiltshire District Council which had its election in 2003. They published the results on their website. The ward of Melksham North has an electorate of 4,294 (in 2005 according to the Boundary Commission). The ward was in the old seat of Devizes and is now in the new seat of Chippenham. At the local election, there were no Conservative candidates and the Lib Dems beat Labour convincingly. There were also 122 votes cast for other parties (Green Party candidate). The data does not show how many councillors were elected (it was two), or which parties they represent (both Lib Dems).

Whilst viewing the data, we notice that many areas of the country have a strong tradition of independent councillors. Often independent councillors are elected unopposed. We have classed these councillors as Other, and they will be a further challenge to the allocation scheme below.

The results dataset comprises 7,599 wards in 325 councils. The ward results data are also available in the Ward Breakdown links in the Region Table, for example North.

3. Calculation Method

Consider a single old seat which is made up of a number, N, of wards. (If any wards are sub-divided we assume homogeneity within a ward.) There are n parties. Let Then we define the allocated electorate EA(k), to be the deemed electorate in each ward k so that the total electorate adds up to E0 as

EA(k) = E(k) × E0 / ΣjE(j).

We also define the normalised votes CN(i,k) for each ward by assuming that general election turnout is constant across all the wards in the old seat

CN(i,k) = TU × EA(k) × C(i,k) / Σj C(i,j),

where both sums are taken over j = 1,...,N. These normalised votes have the useful property that adding them up over all the parties and all the wards gives the same voter turnout as the old seat in 2005. Note that unopposed seats are also handled well by this formula. Even though C(i,k) is negative, it will imply a 100% support for the winning party in this ward.

But we still have to adjust the normalised votes CN(i,k) so that the sum for each party i over all the wards equals the general election vote P(i). This we do using a version of the Transition Model which we use to apply opinion poll changes to seats. The idea is to apply a linear transform to the wards to get the allocated votes CA(i,k),

CA(i,k) = ΣjTM(j,i) × CN(j,k),

where TM(j,i) is the transition matrix which says how likely a voter from party j is to change to party i. It is constructed as follows.

With this definition of TM, the resulting allocated local votes CA(i,k) have three good properties: they are all non-negative; the sum over the wards for each party matches the general election vote for that party; and the turnout is equal across wards. Also in the trivial case where there was only one ward making up the entire old seat, the allocated votes CA(i,1) would equal the actual votes P(i).

4. Example calculation

Here is a worked example on the imaginary old seat of Easton in the district of Easthampton. The general election result was:

Easton
MP Lee Eastman  (LAB)
Electorate64,900 Turnout57.01%
2005 Actual Votes2005 Share
LAB14,59639.45%
CON10,73529.02%
LIB9,79726.48%
OTH1,8695.05%
LAB Majority3,86110.44%

The old seat is made up of just three wards, whose results are:

DistrictWardElect
2005
Old SeatNew Seat Con
Votes
Lab
Votes
Lib
Votes
Oth
Votes
EasthamptonAbbey18,000EastonNewSeat1000-1
EasthamptonCastle23,000EastonNewSeat23,6202,2301,9500
Easthampton Central 25,000Easton NewSeat3 2,9404,2001,2600

Note that the independent candidate in Abbey ward was elected unopposed. We now follow the calculation steps described above.

Electorate allocation: We allocate the old seat electorate of 64,900 onto the three wards in proportion to their electorates. This gives allocated ward electorates of Abbey 17,700, Castle 22,617 and Central 24,583.

Normalised votes: Following the vote normalisation formula we get

DistrictWardElect
2005
Old SeatNew Seat Con
Votes
Lab
Votes
Lib
Votes
Oth
Votes
EasthamptonAbbey17,700EastonNewSeat100010,090
EasthamptonCastle22,617EastonNewSeat25,9843,6863,2230
Easthampton Central 24,583Easton NewSeat3 4,9057,0072,1020

We can now calculate the local vote totals W(j) to be Con 10,889, Lab 10,693, Lib 5,325, and Oth 10,090. Compared with the general election, Con and Oth are too high, and Lab and Lib are too low.

Transition Matrix: The Transition Matrix is

ConLabLibOth
Con98.58%0.66%0.75%0
Lab0100%00
Lib00100%0
Oth037.97%43.51%18.52%

When we apply it to the normalised votes, we get the allocated vote table

DistrictWardElect
2005
Old SeatNew Seat Con
Votes
Lab
Votes
Lib
Votes
Oth
Votes
EasthamptonAbbey17,700EastonNewSeat103,8314,3901,869
EasthamptonCastle22,617EastonNewSeat25,8993,7253,2680
Easthampton Central 24,583Easton NewSeat3 4,8367,0392,1390
Total10,73514,5969,7971,869

The vote totals now match the general election result exactly.

This is a slightly forced example showing an extreme situation. The Conservative vote is very strongly concentrated in Castle and Central wards, which is why they get no votes in Abbey ward. The Lib Dems score poorly in Castle and Central wards, so we deduce that the independent is Liberal-leaning and there are many Lib Dem voters in Abbey.

5. Summary

We have shown a method of allocating general election results into the constituent wards of each old seat. This method preserves the totals of votes cast for each party in each old seat. We can then calculate the implied result for each new seat simply by adding up the allocated votes over the wards in the new seat.