This example uses real data on commercial burglaries over a 12 month period to illustrate how you can calculate the level of repeat victimisation. The total number of commercial burglaries is 459.
You can use the following steps for any crime/incident data that you may have, for example, domestic burglary, racial crime or vehicle crime. The same approach can be used for crime, incidents or a combination of the two.
Step 1. How many crimes do you have recorded?
In this example, there are 459 commercial burglaries.
Step 2. How many victims do you have?
If each of the 459 burglaries were suffered by a different business, we would expect to have 459 victims. However, we know that crime is not spread evenly. You can examine your records to see how often the same business names and addresses crop up. You may have to look at several fields to check whether it is the same victim.
In this case, records show that there are 250 victims altogether.
Step 3. How many repeat crimes do you have?
If you have 459 burglaries and 250 victims then there are 209 repeat burglaries (subtract 250 from 459)
Step 4. So how much of your burglary problem is down to repeat victimisation?
Two hundred and nine of the commercial burglaries are repeats, therefore 46% (divide 209 by 459 multiplied by 100) of the burglary problem is accounted for by repeat victimisation. This tells you the scale of the problem, which can inform decisions about committing resources. The figure is also useful when monitoring performance.
While knowing about the overall figure for repeats is useful, it does not provide detailed enough information to help target resources most effectively. To do this you need to know exactly where the greatest risks of repeat victimisation are.
Step 1. Group victims by number of victimisations
When working out how many victims you have overall (point 2 above) you have listed your victims noting the number of times they had been victimised.
For example:
Shoe Stop 2 crimes Fruity Bites 6 crimes All-Nite Take Away 2 crimes Using this information you can group victims by the number of victimisations. In this example this can be set out as follows:
|
No. of victimisations (a) |
No. of businesses (victims) (b) |
|
1 |
153 |
|
2 |
50 |
|
3 |
20 |
|
4 |
10 |
|
5 |
8 |
|
6 |
5 |
|
9 |
4 |
Total |
250 |
Columns (a) and (b) together show that of the 250 victimised businesses, 153 were burgled once, and therefore 97 more than once (with 50 burgled twice, 20 three time and so on).
Remember……
Only count victims once. In the table a business or victim can only appear once: for example, a business victimised 5 times will not be included in levels 1-4
Do not hide crime by amalgamating it. It is tempting to group the higher levels of victimisation together, for example to put 5+ victimisations rather than 5,6 and 9 as in the table, because there are fewer victims. However, this might hide some of the worst levels of victimisation – in this case that 4 businesses were each burgled 9 times.
Step 2. How do the repeat crimes spread out?
You know that there were 209 repeat burglaries. But, how do these spread out across the 97 repeat victims. It could be:
Lots of businesses each experiencing a few crime, or
A few businesses experiencing lots of crimes
Knowing the spread gives a better idea of the likely impact you can have on repeat victimisation and the scale of the resources required.
There are two steps to this:
Step 2.1 Multiply column (a) by column (b)
Building on the above boxes gives the information below:
No. of victimisations
(a)
No. of businesses (victims)
(b)
No. of burglaries (crimes)
(a x b) = (c)
1
153
153
2
50
100
3
20
60
4
10
40
5
8
40
6
5
30
9
4
36
250
459
Column (c) shows the number of burglaries for each level of victimisation. This includes first and subsequent crimes.
Step 2.2 Subtract the first crimes to find the number of repeat crimes
As a business is only represented once in the table the numbers in column (b) represent the number of first time crimes. For example, 20 businesses have been burgled three times, and together account for 60 burglaries. Discounting the first time burglaries (20) leaves 40 repeat burglaries. Again building on the existing columns produces the information below:
No. of victimisations
(a)
No. of businesses (victims)
(b)
No. of burglaries (crimes)
(a x b = (c)
No. of repeats
(c – b) = (d)
1
153
153
0
2
50
100
50
3
20
60
40
4
10
40
30
5
8
40
32
6
5
30
25
9
4
36
32
250
459
209
The data in this table can help managers to make informed decisions about the optimal use of resources. Detailed analysis of this kind reveals the potential crime reductions achievable by intervening at each of the victimisation levels.
If we could effectively protect all businesses after the first burglary, then we would reduce overall burglaries by 46%. Protection after the second burglary would lead to a reduction of 24%, after the third burglary of 14% and so on.
The table also shows that even if we had only prevented the four most heavily victimised businesses from being burgled after the first offence, we could have reduced overall burglaries by 7%.
Because we know that risk escalates with each victimisation, this information is valuable in directing both your prevention and detection efforts. The heavily victimised businesses, for example, are obvious candidates for measures with detection opportunities, such as alarms.