Conditional counting of late payments monthly, including reset to zero - sql

Question

I have a table with the following structure:
EOM ID Principal Pay_plan cum_Payments
2019-12-31 AY4525 25000.000000 796.000000 936.000000
2020-01-31 AY4525 25000.000000 1592.000000 936.000000
2020-02-29 AY4525 25000.000000 2388.000000 936.000000
2020-03-31 AY4525 25000.000000 3184.000000 3184.00000
2020-04-30 AY4525 25000.000000 3980.000000 3980.00000
2020-05-31 AY4525 25000.000000 4776.000000 3980.00000
2020-06-30 AY4525 25000.000000 5572.000000 3980.00000
2020-04-30 KD4525 35000.000000 500.000000 500.000000
2020-05-31 KD4525 35000.000000 1000.000000 1000.00000
2020-06-30 KD4525 35000.000000 1500.000000 1000.00000
2020-07-31 KD4525 35000.000000 2000.000000 2500.00000
So I have a cumulative payment plan and cumulative payments for unique client IDs per month. Now I want to add a column which starts counting months that the client is late with payments, hence when pay_plan > cum_payments:
EOM ID Principal Pay_plan cum_Payments months_Late
2019-12-31 AY4525 25000.000000 796.000000 936.000000 0
2020-01-31 AY4525 25000.000000 1592.000000 936.000000 1
2020-02-29 AY4525 25000.000000 2388.000000 936.000000 2
2020-03-31 AY4525 25000.000000 3184.000000 3184.00000 0
2020-04-30 AY4525 25000.000000 3980.000000 3980.00000 0
2020-05-31 AY4525 25000.000000 4776.000000 3980.00000 1
2020-06-30 AY4525 25000.000000 5572.000000 3980.00000 2
2020-04-30 KD4525 35000.000000 500.000000 500.000000 0
2020-05-31 KD4525 35000.000000 1000.000000 1000.00000 0
2020-06-30 KD4525 35000.000000 1500.000000 1000.00000 1
2020-07-31 KD4525 35000.000000 2000.000000 2500.00000 0
The counter must be reset when pay_plan = cum_payments again. I have tried numerous ways of doing this with OVER(), but not found a solid solution. Anyone got an idea how to solve this?

Answer 1

This is a gaps-and-islands problem. The islands are when the cumulative amount is less than the planned amount. So, you can use a cumulative sum to define the islands and then row_number():
select t.*,
(case when cum_payments >= pay_plan then 0
else row_number() over (partition by id, grp order by eom) - 1
end) as months_late
from (select t.*,
sum(case when cum_payments >= pay_plan then 1 else 0 end) over (partition by id order by eom) as grp
from t
) t;
Here is a db<>fiddle.
You can handle the situation where the first payment is late by using:
select t.*,
(case when cum_payments >= pay_plan then 0
else row_number() over (partition by id, grp order by eom) - 1 +
(case when min(eom) over (partition by id) = min(eom) over (partition by id, grp) and
first_value(cum_payments) over (partition by id, grp order by eom) < first_value(pay_plan) over (partition by id, grp order by eom)
then 1 else 0
end)
end) as months_late
from (select t.*,
SUM(case when cum_payments >= pay_plan then 1 else 0 end) over (partition by id order by eom) as grp
from t
) t
I actually left this logic out of the above, because it seems inelegant. There might be a better solution, but it does not readily occur to me. Here is the revised db<>fiddle.

Answer 2

If I understand your logic correct, Months_late value for your last row should be 0 and if it is correct, you can use this below logic to achieve your requirement-
Demo Here
Let cum_Payments will always increase or same per day for a specific ID
SELECT *,
(
SELECT COUNT(*)
FROM your_table B
WHERE B.cum_Payments = A.cum_Payments
AND B.EOM < A.EOM
AND B.ID = A.ID
) Months_late
FROM your_table A
ORDER BY ID,EOM
This following query will return the exact result as you are looking for. It is true that the query is bit heavy for a table with huge data but it is acceptable if you have one time use of this query. In case or more use, you can think about creating a view for improving query execution performance.
Demo2 Here
SELECT *,
(
SELECT count(1) FROM your_table b
WHERE b.id = a.id AND b.eom <= a.eom
AND b.eom >
(
ISNULL
(
(
SELECT MAX(eom) FROM your_table c
WHERE c.id = a.id AND (c.pay_plan - c.cum_payments) = 0
AND c.eom <= a.eom
)
,
(
SELECT MIN(eom) FROM your_table d
WHERE d.id = a.id
)
)
)
)
FROM your_table a
ORDER BY id, eom