Calculate using previous column and rows

Question

Hello TNO members,
I have a complicated problem I need to solve, how ever I am missing knowledge about calculating using previous rows and columns in current select.
Test data
with t as (
  select 1 as box, 1 as box_group, 10 as max_qty from dual union all
  select 2, 1, 15 from dual union all
  select 3, 1, 40 from dual union all
  select 4, 1, 45 from dual union all
  select 5, 2, 15 from dual union all
  select 6, 2, 20 from dual union all
  select 7, 2, 20 from dual union all
  select 8, 3, 20 from dual)Expected Output result
box box_group max_qty assigned_from_60
1   1         10      10  
2   1         15      15
3   1         40      17
4   1         45      18
5   2         15      0
6   2         20      0
7   2         20      0
8   3         20      0The problem:
In total 60 items are shared among the boxes in the same group, ordered by the lowest max_qty.
10 items can be assign to each box in group 1. (Used items: 40)
The remaining 20 items will be assigned to the other boxes in group 1.
5 more items can be assign to each box in group 1 (Used items: 15)
The remaining 15 items will be assigned to the remaining boxes in group 1.
2 more items can be assign to each box in group 1 (used items: 4)
One item remains. When items cannot be shared equally among the remaining boxes, ordered by the highest max_quantity, assign +1 till no item remains.
My solution in steps:
1. Calculate max_qty difference. How can I calculate the difference between the max_qty from box 1 and 2? Tricky is not to calculate the difference between different groups.
This means output result should be something like
box box_group max_qty qty_dif
1   1         10      10  
2   1         15      5
3   1         40      25
4   1         45      5
5   2         15      15
6   2         20      5
7   2         20      0
8   3         20      202. Remaining boxes in the same group. I want to know how many boxes are in the same group. Especially the remaining boxes when the current max_quantity is filled.
Using the following code does not result in the correct output, what is wrong or missing here?
count(*) over(partition by box_group order by max_qty asc range between current row and unbounded following) This means output result should be something like
box box_group max_qty qty_dif rem_boxes
1   1         10      10      4  
2   1         15      5       3
3   1         40      25      2
4   1         45      5       1
5   2         15      15      3
6   2         20      5       2
7   2         20      0       1
8   3         20      20      13. Calculate costs. This one is faily easy rem_boxes * qty_dif (*per row*)
This means output result should be something like
box box_group max_qty qty_dif rem_boxes cost
1   1         10      10      4         40
2   1         15      5       3         15
3   1         40      25      2         50
4   1         45      5       1         5
5   2         15      15      3         45
6   2         20      5       2         10
7   2         20      0       1         0
8   3         20      20      1         204. Calculate rem_items. 60 - (rem_boxes * qty_dif of box 1) - (rem_boxes * qty_dif of box 2) - (rem_boxes * qty_dif
of box n). How can I calculate using results of previous rows? (*all, not per group*)
This means output result should be something like
box box_group max_qty qty_dif rem_boxes cost rem_items
1   1         10      10      4         40   20
2   1         15      5       3         15   5
3   1         40      25      2         50   -45
4   1         45      5       1         5    -50
5   2         15      15      3         45   -95
6   2         20      5       2         10   -105
7   2         20      0       1         0    -105
8   3         20      20      1         20   -1255. Assign full quantity. For each row check if rem_items > 0 then 1 else 0
This means output result should be something like
box box_group max_qty qty_dif rem_boxes cost rem_items assign
1   1         10      10      4         40   20        1
2   1         15      5       3         15   5         1
3   1         40      25      2         50   -45       0
4   1         45      5       1         5    -50       0
5   2         15      15      3         45   -95       0
6   2         20      5       2         10   -105      0
7   2         20      0       1         0    -105      0
8   3         20      20      1         20   -125      06. Calculate assign quantity attemp 1. Calculate assign quantity of remaining boxes per group
When assign = 1 then max_qty else pervious a_qty (*within same group*)
This means output result should be something like
box box_group max_qty qty_dif rem_boxes cost rem_items assign a_qty
1   1         10      10      4         40   20        1       10
2   1         15      5       3         15   5         1       15
3   1         40      25      2         50   -45       0       15
4   1         45      5       1         5    -50       0       15
5   2         15      15      3         45   -95       0       0
6   2         20      5       2         10   -105      0       0
7   2         20      0       1         0    -105      0       0
8   3         20      20      1         20   -125      0       0How to solve the rest, I do not know yet. Any other suggestion to solve this problem, is welcome.
Since I'm not really a professional this is what I tried till now
with z as (
  select 1 as box, 1 as box_group, 10 as max_qty from dual union all
  select 2, 1, 15 from dual union all
  select 3, 1, 40 from dual union all
  select 4, 1, 45 from dual union all
  select 5, 2, 15 from dual union all
  select 6, 2, 20 from dual union all
  select 7, 2, 20 from dual union all
  select 8, 3, 20 from dual)
select u.*,
       case
         when u.assign = 2 then u.max_qty
         when u.assign = 1 then 0
         when u.assign = 0 then 0
       end as assigned_qty
from
    select v.*,
           case
             when 60 - sum(v.max_qty) over (order by v.box_group, v.max_qty, v.box) >= 0
               and v.rem_items_before >= 0 then 2
             when 60 - sum(v.max_qty) over (order by v.box_group, v.max_qty, v.box) < 0
               and v.rem_items_before > 0 then 1 else 0
           end as assign
    from
        select w.*,
               w.rem_items_after + w.max_qty as rem_items_before
        from
            select x.*,
                   60 - x.qty_assigned as rem_items_after
            from  
                select y.*,
                       y.max_qty * y.rem_boxes as total_cost,
                       sum(y.max_qty) over (order by y.box_group, y.max_qty, y.box) as qty_assigned
                from 
                    select z.*,
                           count(*) over (partition by z.box_group order by z.max_qty, z.box asc range between current row and unbounded following) as rem_boxes
                    from z
                  ) y
              ) x
          ) w
      ) v
  ) uKind regards,
Metro
Edited by: 858378 on 30-mei-2011 4:39
Edited by: 858378 on 30-mei-2011 5:05

Answer 1

Hi, Metro,
858378 wrote:
Hello, thanks for your help so far.
The course I am in, teaches you how to use the basic pl sql language such are selecting from tables.Is it about PL/SQL or SQL?
I have learned things about
- SELECT
- FROM
- WHERE
- GROUP BY
- ORDER BY
- SUB SELECTION IN SELECT
- SUB SELECTION IN FROM
- SUM, COUNT, MIN, MAX
- CASES
- INNER, OUTER, LEFT, FULL, CROSS JOINSAll of these are parts of the SQL language, not PL/SQL.
We are now at partitioning.Are you specifically at partitioning, or iare you at a point where the book talks about analytic functions, which sometimes, but not always, have a PARTTION BY clause?
It's written in Dutch, I tried to translate it, so it might not be well written English.
Excercise 192
Distribution center 'One by one' wants to automate the distribution process.
Items are distributed to the boxes one by one. Sorry, I can't figure out what Exercise 192 is, based on just that.
It is similar to the next excercise, but this above was a lot more straight to the point.
*Excercise 193*
Distribution center 'All in one box' wants to automate the distribution process.
One of the major changes in this process is to distribute items equally to all boxes in the same group.
This means starting from the lowest quantity, assign the lowest quantity to all boxes in the same group if possible.
If this is not possible distribute the amount of items divided by the number of boxes in the same group. If the amount of items per box is lower than 1 and not 0.
Divide the remaining items per box ordered by the box with the highest quantity, till there are no items left.
When it is possible to distribute the lowest quantity to all boxes, move up to the next box in the same group.
When all boxes in the same group are filled to their maximum quantity, move up to the next group and repeat this process. ...
So Exercise 193 is what you asked yesterday, right?
A) Order the following information by box_group and max_qty as described.
B) Calculate the distribution results for 60, 120, 170 items.
When I have 60 items, the following output result should be
box box_group max_qty assigned_from_60
1   1         10      10  
2   1         15      15
3   1         40      17
4   1         45      18
5   2         15      0
6   2         20      0
7   2         20      0
8   3         20      0Box 1 and 2 can be filled completely as you can fill atleast 10 to box 1,2,3,4 and an additional 5 to box 2, 3 and 4.
The last item goes to the box 4.
when I have 120 items
box box_group max_qty assigned_from_120
1   1         10      10  
2   1         15      15
3   1         40      40
4   1         45      45
5   2         15      3
6   2         20      3
7   2         20      4
8   3         20      0
Based on what you posted, it seems like the following should be equally acceptable:
box box_group max_qty assigned_from_120
1   1         10      10  
2   1         15      15
3   1         40      40
4   1         45      45
5   2         15      3
6   2         20      4
7   2         20      3
8   3         20      0That is, the last 10 items have to be distributed among the 3 boxes in box_group=2 as equally as possible. So one box will get 4 items and the others will get 3. The extra item will go to the box with the highest max_qty, but in this case, there is a tie: box 6 has just as much of a claim to having the highest max_qty as box 7. The line marked "***** Add if needed *****" in the query blow guarantees that, in case of a tie like this, the box with the higher box value will be considered "larger" than another box with the same max_qty.
when I have 170 items
box box_group max_qty assigned_from_170
1   1         10      10  
2   1         15      15
3   1         40      40
4   1         45      45
5   2         15      15
6   2         20      20
7   2         20      20
8   3         20      5
I accidentally posted the wrong query yesterday. This is what I should have posted:
WITH     cntr     AS
     SELECT     LEVEL     AS n
     FROM     (  SELECT  MAX (max_qty)     AS max_max_qty
             FROM    z
--             WHERE   box_group     = :target_box_group          -- *****  Removed  *****
     CONNECT BY  LEVEL  <= max_max_qty
,     got_r_num     AS
     SELECT     z.box
     ,     c.n
     ,     ROW_NUMBER () OVER ( ORDER BY  box_group          -- *****  Added  *****
                               ,            c.n
                         ,         z.max_qty     DESC
                         ,            box          DESC     -- ***** Add if needed  *****
                       )     AS r_num
     FROM     cntr     c
     JOIN     z          ON c.n     <= z.max_qty
--     WHERE     z.box_group     = :target_box_group               -- *****  Removed  *****
,     got_assigned     AS
     SELECT       box
     ,       COUNT (*)     AS assigned
     FROM       got_r_num
     WHERE       r_num     <= :total_items
     GROUP BY  box
SELECT     z.*,     NVL (a.assigned, 0)     AS assigned
FROM          z
LEFT OUTER JOIN     got_assigned     a  ON     z.box     = a.box
ORDER BY     z.box_group
,          z.max_qty
;Yesterday, I described how you need to remove 2 lines and add 1, but the code I posted was the unchanged query. The query above is what I should have posted then. Look for comments beginning "*****" above for the changes. I apologize for my mistake.
This query gets the results you posted for all 3 values of :total_items that you posted. If it doesn't work for some other value, or some other data, post the new values and the correct results you want from them, and point out where the query above is wrong.
C) Sum the maximum quantities per group
D) Get the amount of boxes in each group
E) Create a column for remaining boxes per group
F) For each distinct quantity, count the amount of boxes with this quantity
G) Calculate how many items are required to fill all boxes in the same group
H) Create a plan how to solve this distribution problem described in the introduction?
I) Solve this problem using your plan.Are these the steps that the book suggests using?
I don't understand that approach. It might be a good way to solve the problem without using a computer. It might be a good way to solve the problem using a procedural language, such as PL/SQL. It might be one way of solving the problem in SQL, but I think it will be more complicated and less efficient than what I psoted.
The approach above is iterative; that is, you repeat certain steps, with different values. For example, you distribute a certain number of items to all boxes in a box_group. The you remove the smallest box(es) from the group, and repeat, distributing the remaining items among the remianing boxes. That's not hard to do in a language like PL/SQL, where you have loops and variables. In SQL, the closest thing to that is the MODEL clause. I'm sure you could write a MODEL solution to this problem, but, if your book hasn't mentioned MODEL yet, then that's certainly not what it's expecting you to do.
Even using the approach in steps A) trhough G) above, I don't see how a PARTITION BY would help.