Alberta Ingenuity & CMASTE
Lesson 4: Customer Attrition Analysis (Teachers’ Resource)
Purpose: Analysis of customer attrition (how long customers stay with a company and
why they leave) is very important to subscription-based businesses such as cable TV,
insurance, credit cards, magazine subscription, and more recently, cell phone programs
and residential utilities. For a large company millions of database records of customers
must be analyzed. This is where the Machine Learning strategy of data mining takes
over. Data mining uses computer-based queries to find patterns and relevant information
in a large data set.
Problem: Computer programs use a set of algorithms to analyze the customer data for
hazard probability (the chance that subscriber who has survived a certain amount of time
is going to stop, cancel, or expire before the next unit of time) and also for survival
probability, the chance that a random customer will still be with the company after a
specific amount of time. We want to see if there is a mathematical pattern in this analysis
that we can use to make inferences and projections about customer attrition.
Hypothesis: The data mining technique will provide rapid feedback about customer
behavior and a way to quantify customer loyalty.
Design: The assumption is that time is discrete, in units of days, months, etc., whereas
traditional statistical analysis treats time as continuous. The hazards in a typical
subscription business are shown in Figure 1 as: 1) customers who don’t start
2) customers who start but never pay, which occurs at about 60 days and 3) customers
who stop when the promotion ends, which occurs at about 90 days.
(Figure 1)
We can define hazard probability as P 
number who succumbed to the risk
.
population at risk during that time
See Appendix for the data mining program used to calculate these probabilities.
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The gradual decline in hazards over time implies that the longer a customer stays with a
company, the less likely they are to leave.
Procedure:
1) From Figure 1, estimate these hazard probabilities as percentages:
a) probability of a customer not starting the subscription ________
Answer: P ~ 4%
b) probability of a customer being lost due to non-payment _________
Answer: P ~ 10%
c) probability of a customer being lost at the end of the promotion _________
Answer: P ~ 6%
This leads us to the concept of survivor probability or retention rate, which is a more
holistic picture of the likelihood that a random customer will stay with a company to a
certain point in time. Rather than using a probability formula here, we will use graph and
data analysis to help quantify survivor probability.
Figure 2 shows three examples of survival curves for credit card users for a particular
company: the top curve is for customers who start as card holders, paying customers who
are charged automatically every month, while the bottom curve is for customers who did
not have a credit card previously and who are billed monthly and pay by cheque. The
middle curve is the “average” of the other two curves.
(Figure 2)
We notice that the curves have the basic shape of an exponential decay curve and we can
study them from a Math 30 P perspective with equation type y  a(b) x or from a Math 31
point of view with equation type y  c  e x .
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2) a) Using the tables of data read from the graph of the upper survivor (retention)
curve, generate a regression equation of the form y  a(b) x . ________________
t
(months)
Survivor
Prob.
(%)
0
1
2
3
4
5
6
7
8
9
10
11
12
97
92
88
77
68
62
57
54
50
48
45
43
42
Answer: y  95.34(0.9276) x
b) Use your equation from part a) to calculate:
i) the retention probability after 18 months
ii) the number of months or
days until the retention
probability was 40%
(graphically)
P  95.34(0.9276)18
Answer:
 24.65%
Answer:
Y1  95.34(0.9276) x
Y2  40
Window
x[0, 20, 2] y[0, 100, 10]
Intersecti on occurs when
x  11.557 months
c) (* for Math 31 students) use an algebraic process to rewrite your regression
equation in the form y  c  e x
Answer:
y  95.34(0.9276) x
y  95.34(e z )
 (0.9276) x  e z
ln( 0.9276) x  ln( e z )
x  ln( 0.9276)  z  ln( e)
 z  x  ln( 0.9276)
z  0.075 x
y  95.34(e 0.075x )
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dy
for your equation in part c)
dx
and specifically the value of the derivative at t = 5 months. Explain the
meaning of that number with respect to the data.
Answer:
y   95.34e 0.075x  0.075
d) (* for Math 31 students) Find the derivative
 7.1505e 0.075x
y (5)  7.1505e 0.075(5)
 4.91, which means that at 5 months we are losing about 4.91% of
customers per month.
e) Using the tables of data read from the graph of the lower survivor curve,
generate a regression equation of the form y  a(b) x . ___________________
t
(months)
Survivor
Prob.
(%)
0
1
2
3
4
5
6
7
8
9
10
11
12
97
74
59
30
21
19
17
15
14
13
12
11
10
Answer: y  65.97(0.8338) x
f) Use your equation from part e) to calculate:
i) the retention probability after 18 months
ii) the number of months or
days until the retention
probability was 50%
(graphically)
18
P  65.97(0.8338)
Answer:
Answer:
P  2.5%
Y1  65.97(0.8338) x
Y2  50
Window
x[0, 12, 1] y[0, 100, 10]
Intersecti on occurs when
x  1.525 months
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g) (* for Math 31 students) use an algebraic process to rewrite your
regression equation in the form y  c  e x
y  65.97(0.8338) x
y  65.97(e) z
 0.8338 x  e z
Answer:
ln( 0.8338) x  ln( e) z
x  ln( 0.8338)  z  ln( e)
 z  x  ln( 0.8338)
z  0.18 x
 y  65.97(e 0.18 x )
dy
for your equation in
dx
part g) and specifically the value of the derivative at t = 5 months.
Explain the meaning of that number with respect to the data.
h) (* for Math 31 students) Find the derivative
Answer:
y   65.97e .18 x  .18
 11.8746e .18 x
y (5)  11.8746e .18(5)
 4.83 , which means that at 5 months we are losing about 4.83% of
customers per month.
2) Did you notice the steep drop off in the lower graph at about 60 days; what
factor(s) do you think might account for that?
Answer: This drop off would correspond to non-payment of credit card bills for those
customers new to credit cards and having to pay by cheque.
3) Why do you think the upper graph does not have the same drop-off at that time?
Answer: These customers had a credit card previously, so they are used to monthly
payments. As well, their payment is done by direct debit, which is more convenient.
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We can quantify the difference between the two groups in another way, using a common
industry measure called customer half-life or median customer lifetime. This is the
time when exactly half or 50% of the original customers would still be subscribers.
(Math 30 P students are familiar with this term from work with radioactive decay
problems.) Figure 3 shows this for each of our two groups.
(Figure 3)
4) From Figure 3 identify the customer half-life in days, for each group.
Answer: The half-life for the “credit card group” appears to be about 245 days, and
the half-life for the “non-credit card group” appears to be about 65 days.
Evaluation: The data mining techniques provide very useful information that can be
analyzed mathematically, leading to inferences about subscription customers and their
retention rates that would aid in decision-making for the company.
Synthesis: Data mining is very versatile and can be adapted for a wide variety of
different industries.
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Appendix: Data Mining program to calculate hazard probabilities:
Calculating Hazards in a Database
Let's take a closer look at how survival data mining works with a database-in this case,
running with Oracle. Assume that a database contains one row for each customer with the
following information:



Start_date
Stop_date (NULL is not stopped)
Other interesting variables such as stop reason, channel, and so on.
How is this data used to calculate hazards? Oracle extensions make the full calculation
possible. The first thing is to calculate the time with company (tenure) and the stop flag:
SELECT ((case when stop_date is NULL then < today >
else stop_date end) - start_date) as tenure,
(case when stop_date is NULL then 0
else 1 end) as is_stopped
FROM customers
The next step is to aggregate these fields by time with the company or tenure. This gives
the number of customers with exactly each amount of time and the number that stopped
at a certain time (some customers with that tenure will still be active):
SELECT ((case when stop_date is NULL then < today >
else stop_date end) - start_date) as tenure,
count(*) as pop_at_t,
sum(case when stop_date is NULL then 0
else 1 end) as num_stopped
FROM customers
GROUP BY ((case when stop_date is NULL then < today >
else stop_date end) - start_date)
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At this point, you could continue the calculation in a spreadsheet. However, the analytic
functions make it possible to calculate the total population at risk, and thus the hazard.
The total population is the sum of pop_at_t for all times greater than or equal to t. The
hazard is num_stopped divided by this total. The following query does this calculation
SELECT tenure,
sum(pop_at_t) over
(order by tenure desc range unbounded preceding),
num_stopped /
(sum(pop_at_t) over
(order by tenure desc range unbounded preceding))
FROM < subquery >
GROUP BY tenure
ORDER BY tenure
*I felt that it was worthwhile for teachers and/or students to see the type of program used
in a machine learning setting. It is not expected that the program be used or even
understood by teachers and/or students.
Sources
1) http://www.intelligententerprise.com, Data Mining and Hazard Survival, Linoff,
Gordon S., 2004
2) http://www.cs.ualberta.ca/~greiner/C-466/SLIDES/syllabus.html
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