Select to build groups by (analytic) sum - sql

Question

Please help me to build a sql select to assign (software development) tasks to a software release. Actually this is a fictive example to solve my real business specific problem.
I have a relation Tasks:
ID Effort_In_Days
3 3
1 2
6 2
2 1
4 1
5 1
I want to distribute the Tasks to releases which are at most 2 days long (tasks longer than 2 shall still be put into one release). In my real problem I have much more "days" available to distribute "tasks" to. Expected output:
Release Task_ID
1 3
2 1
3 6
4 2
4 4
5 5
I think I need to use analytic functions, something with sum(effort_in_days) over and so on, to get the result. But I'm I haven't used analytic functions much and didn't find an example that's close enough to my specific problem. I need to build groups (releases) if a sum (>= 2) is reached.

Answer 1

I would do something like:
with data as (
select 3 ID, 3 Effort_In_Days from dual union all
select 1 ID, 2 Effort_In_Days from dual union all
select 6 ID, 2 Effort_In_Days from dual union all
select 2 ID, 1 Effort_In_Days from dual union all
select 4 ID, 1 Effort_In_Days from dual union all
select 5 ID, 1 Effort_In_Days from dual
)
select id, effort_in_days, tmp, ceil(tmp/2) release
from (
select id, effort_in_days, sum(least(effort_in_days, 2)) over (order by effort_in_days desc rows unbounded preceding) tmp
from data
);
Which results in:
ID EFFORT_IN_DAYS TMP RELEASE
---------- -------------- ---------- ----------
3 3 2 1
1 2 4 2
6 2 6 3
2 1 7 4
4 1 8 4
5 1 9 5
Basically, I am using least() to convert everything over 2 down to 2. Then I am putting all rows in descending order by that value and starting to assign releases. Since they are in descending order with a max value of 2, I know I need to assign a new release every time when I get to a multiple of 2.
Note that if you had fractional values, you could end up with releases that do not have a full 2 days assigned (as opposed to having over 2 days assigned), which may or may not meet your needs.
Also note that I am only showing all columns in my output to make it easier to see what the code is actually doing.

Answer 2

This is an example of a bin-packing problem (see here). There is not an optimal solution in SQL, that I am aware of, except in some boundary cases. For instance, if all the tasks have the same length or if all the tasks are >= 2, then there is an easy-to-find optimal solution.
A greedy algorithm works pretty well. This is to put a given record in the first bin where it fits, probably going through the list in descending size order.
If your problem is really as you state it, then the greedy algorithm will work to produce an optimal solution. That is, if the maximum value is 2 and the efforts are integers. There might even be a way to calculate the solution in SQL in this case.
Otherwise, you will need pl/sql code to achieve an approximate solution.

Answer 3

SQL Fiddle
Oracle 11g R2 Schema Setup:
CREATE TABLE data AS
select 3 ID, 3 Effort_In_Days from dual union all
select 1 ID, 2 Effort_In_Days from dual union all
select 6 ID, 2 Effort_In_Days from dual union all
select 2 ID, 1 Effort_In_Days from dual union all
select 4 ID, 1 Effort_In_Days from dual union all
select 5 ID, 1 Effort_In_Days from dual union all
select 9 ID, 2 Effort_In_Days from dual union all
select 7 ID, 1 Effort_In_Days from dual union all
select 8 ID, 1 Effort_In_Days from dual;
Query 1:
Give the rows an index so that they can be kept in order easily;
Assign groups to the rows where the Effort_In_Days is 1 so that all adjacent rows with Effort_In_Days of 1 are in the same group and rows separated by higher values for Effort_In_Days are in different groups;
Assign a cost of 1 to each row where the Effort_In_Days is higher than 1 or where Effort_In_Days is 1 and the row has an odd row number within the group; then
Finally, the release is the sum of all the costs for the row and all preceding rows.
Like this:
WITH indexes AS (
SELECT ID,
Effort_In_Days,
ROWNUM AS idx
FROM Data
),
groups AS (
SELECT ID,
Effort_In_Days,
idx,
CASE Effort_In_Days
WHEN 1
THEN idx - ROW_NUMBER() OVER ( PARTITION BY Effort_In_Days ORDER BY idx )
END AS grp
FROM indexes
ORDER BY idx
),
costs AS (
SELECT ID,
Effort_In_Days,
idx,
CASE Effort_In_Days
WHEN 1
THEN MOD( ROW_NUMBER() OVER ( PARTITION BY grp ORDER BY idx ), 2 )
ELSE 1
END AS cost
FROM groups
ORDER BY idx
)
SELECT ID,
Effort_In_Days,
SUM( cost ) OVER ( ORDER BY idx ROWS BETWEEN UNBOUNDED PRECEDING AND CURRENT ROW ) AS Release
FROM costs
ORDER BY idx
Results:
| ID | EFFORT_IN_DAYS | RELEASE |
|----|----------------|---------|
| 3 | 3 | 1 |
| 1 | 2 | 2 |
| 6 | 2 | 3 |
| 2 | 1 | 4 |
| 4 | 1 | 4 |
| 5 | 1 | 5 |
| 9 | 2 | 6 |
| 7 | 1 | 7 |
| 8 | 1 | 7 |