The process of grouping a large number of numeric values into a small number of categories is sometimes referred to as binning. In RFM analysis, the bins are the ranked categories. You can use the Binning tab to modify the method used to assign recency, frequency, and monetary values to those bins.
Nested. In nested binning, a simple rank is assigned to recency values. Within each recency rank, customers are then assigned a frequency rank, and within each frequency rank, customer are assigned a monetary rank. This tends to provide a more even distribution of combined RFM scores, but it has the disadvantage of making frequency and monetary rank scores more difficult to interpret. For example, a frequency rank of 5 for a customer with a recency rank of 5 may not mean the same thing as a frequency rank of 5 for a customer with a recency rank of 4, since the frequency rank is dependent on the recency rank.
Independent. Simple ranks are assigned to recency, frequency, and monetary values. The three ranks are assigned independently. The interpretation of each of the three RFM components is therefore unambiguous; a frequency score of 5 for one customer means the same as a frequency score of 5 for another customer, regardless of their recency scores. For smaller samples, this has the disadvantage of resulting in a less even distribution of combined RFM scores.
Number of Bins
The number of categories (bins) to use for each component to create RFM scores. The total number of possible combined RFM scores is the product of the three values. For example, 5 recency bins, 4 frequency bins, and 3 monetary bins would create a total of 60 possible combined RFM scores, ranging from 111 to 543.
- The default is 5 for each component, which will create 125 possible combined RFM scores, ranging from 111 to 555.
- The maximum number of bins allowed for each score component is nine.
A "tie" is simply two or more equal recency, frequency, or monetary values. Ideally, you want to have approximately the same number of customers in each bin, but a large number of tied values can affect the bin distribution. There are two alternatives for handling ties:
- Assign ties to the same bin. This method always assigns tied values to the same bin, regardless of how this affects the bin distribution. This provides a consistent binning method: If two customers have the same recency value, then they will always be assigned the same recency score. In an extreme example, however, you might have 1,000 customers, with 500 of them making their most recent purchase on the same date. In a 5-bin ranking, 50% of the customers would therefore receive a recency score of 5, instead of the ideal value of 20%.
Note that with the nested binning method "consistency" is somewhat more complicated for frequency and monetary scores, since frequency scores are assigned within recency score bins, and monetary scores are assigned within frequency score bins. So two customers with the same frequency value may not have the same frequency score if they don't also have the same recency score, regardless of how tied values are handled.
- Randomly assign ties. This ensures an even bin distribution
by assigning a very small random variance factor to ties prior to
ranking; so for the purpose of assigning values to the ranked bins,
there are no tied values. This process has no effect on the original
values. It is only used to disambiguate ties. While this produces
an even bin distribution (approximately the same number of customers
in each bin), it can result in completely different score results
for customers who appear to have similar or identical recency, frequency,
and/or monetary values -- particularly if the total number of customers
is relatively small and/or the number of ties is relatively high.
Table 1. Assign Ties to Same Bin vs. Randomly Assign Ties ID Most Recent Purchase (Recency) Assign Ties to Same Bin Randomly Assign Ties 1 10/29/2006 5 5 2 10/28/2006 4 4 3 10/28/2006 4 4 4 10/28/2006 4 5 5 10/28/2006 4 3 6 9/21/2006 3 3 7 9/21/2006 3 2 8 8/13/2006 2 2 9 8/13/2006 2 1 10 6/20/2006 1 1
- In this example, assigning ties to the same bin results in an uneven bin distribution: 5 (10%), 4 (40%), 3 (20%), 2 (20%), 1 (10%).
- Randomly assigning ties results in 20% in each bin, but to achieve
this result the four cases with a date value of 10/28/2006 are assigned
to 3 different bins, and the 2 cases with a date value of 8/13/2006
are also assigned to different bins.
Note that the manner in which ties are assigned to different bins is entirely random (within the constraints of the end result being an equal number of cases in each bin). If you computed a second set of scores using the same method, the ranking for any particular case with a tied value could change. For example, the recency rankings of 5 and 3 for cases 4 and 5 respectively might be switched the second time.