September 3, 2012
In the previous article in this series, a question was posed regarding how many SQL statements would be required to generate the following report, which displays in Microsoft Excel:
A significant number of items regarding the report require explanation:
- Employee IDs and employee names appear in column A (the left-most column). There could be many employees, but only those employees reporting labor transactions for machining operations in production batches (work order lots) that completed during one of the weeks should be listed.
- Part IDs, part descriptions, and operations (Sub n OP nn in the above example report) are listed in column B (the second column from the left). The number of operations listed for an employee could range from one to possibly 50 (or more, depending on the date range of the report), but that list will only include machining type operations (not welding, blasting, painting, laser cutting, etc.).
- Total hours are reported in column C. The second output row for an employee shows the sum of the numbers directly below for the employee. The numbers on the 3rd, 6th, 9th, etc. rows for the employee in column C are the sum of the hours the employee worked in the operation for all manufacturing batches that completed in the report time period.
- The first row for each employee, in columns E though H in the above report (could very well extend beyond column H, depending on the number of weeks selected), are the Monday through Sunday dates for the week.
- The second row for each employee, in columns E through H in the above report show the weighted average efficiency for the employee in that week. The weighting is based on the percentage of hours the employee spent in the operations, for those operations in batches that completed during the week. This number of hours is calculated, but not printed in the report.
- The numbers on the 4th, 7th, 10th, etc. rows for each employee, in columns E through H, shows the average number of hours the operation required, for those batches that completed during the week. If an employee did not work in the operation one week, but did another, then a blank should appear in the week that the employee did not work in the operation. All employees that worked in the operation for the week will have the same average printed. For example, Cary Grant and James Stewart both worked on WIDGET101 operation Sub 0 OP 30. The average hours for the first week is 10.50, and that value appears for both employees. During the second week, only James Stewart worked on that operation, and the average hours for that week decreased to 7.56.
- The numbers of on the 3rd, 6th, 9th, etc. rows for each employee, shows the ratio of the engineering standard hours divided by the average hours for the operation in that week. The same values will appear for all employees reporting time to the operation in the week.
- The percentage in column I is the simple average of the weighted averages to the left. There may be causes where a blank will appear for one of the weighted averages for an employee, so it is not possible to simply sum the weighted averages and divide by the number of weeks.
This appeared to be a simple report when it was just a sketch on a sheet of paper!
How many SQL statement executions will be necessary to generate the above report? Before answering that question, let’s take another look at how the data is organized and related between the tables:
OK, so the above diagram might be a little confusing.
The engineering standard hours are found by querying the OPERATION table with the WORKORDER_TYPE=’W’ and the WORKORDER_BASE_ID equal to the part ID that is of interest (the WORK_ORDER table also contains engineering standard information, accessed by searching for TYPE=’M’ and the BASE_ID equal to the part ID tht is of interest). In theory, there could be multiple engineering standards for the same part ID and operation, which is why we might need to join to the PART table to make certain that the current engineering standard is retrieved, but we will ignore that potential issue for now. It would be a good idea to also check the WORK_ORDER table to obtain a distinct list of part IDs that had a close date in the time range that is of interest (there is no sense in forcing the database RDBMS to retrieve the engineering standards for parts that have not been produced in the last 17 years), so we will need to specify that the TYPE column in the WORK_ORDER table is equal to W and that the CLOSE_DATE is in the date range of the report. Many operations are NOT machining operations, and we can only determine the type of operation by determining the type machine that is selected for use, so we must also query the SHOP_RESOURCE table to determine if the operation is processed at a machining center.
The actual production hours are found by querying the LABOR_TICKET table with the the WORKORDER_TYPE=’W’ (to eliminate spurious indirect labor) and joined to the SHOP_RESOURCE table to make certain that the operation is a machining type operation (as explained above). To determine the employees’ name, the LABOR_TICKET table is joined to the EMPLOYEE table. To determine the part ID that is produced in the labor ticket transaction, the LABOR_TICKET table must be joined to the WORK_ORDER table, with the TYPE column in the WORK_ORDER table set to ‘W’ and the SUB_ID column in the WORK_ORDER table set to ’0′ so that the main part ID for the work order is returned rather than a subordinate part ID. To retrieve the description of the part ID that is produced, the WORK_ORDER table must be joined to the PART table.
That sketch of a report on a sheet of paper sure looked simple, but the degree of difficulty is all relative (or relational).
How many query executions will be required to produce the report that will appear in Microsoft Excel? Should we retrieve the entire database over the network and process the data client-side? I don’t program in the Java programming language, but I am guessing that the majority of Java programmers would not attempt to retrieve the entire database. Would the number of SQL statement executions depend on the number of employees? Would the number of SQL statement executions depend on the number of different part operations whose batches closed in a particular week? Would the number of SQL statement executions depend on the number weeks included in the report?
In the words of the TV show “Name that Tune“, I can name that tune in 1 note. Because there are many to many joins between the various table, it is not possible to construct a single simple SQL statement that retrieves the correct result. However, it is possible to construct a single complex query using inline views or subquery factoring (WITH blocks), along with analytic functions to retrieve the correct, necessary information to the client for presentation in Microsoft Excel. The correct answer to the above question must:
- Minimize the amount of data flowing across the network. SQL*Net compression may help, but of course that can only possibly help if more than one row is retrieved at a time. Ideally, the array fetch size should be set to a large value (possibly 100 up to the maximum of 5,000) to make the best use of SQL*Net compression.
- Minimize as best as possible the performance impact on the database server, the database server is a shared resource without unlimited capacity. Setting a reasonably large array fetch size may help reduce the number of memory accesses (specifically consistent gets) on the database server. Make certain that available indexes are usable by the query – be careful about using functions (such as TRUNC) on date type columns in the WHERE clause.
- When a query accesses many tables, and also when there are multiple inline views (or factored subqueries), the Oracle optimizer may not always find the optimal join order. The _OPTIMIZER_MAX_PERMUTATIONS hidden parameter defaults to a value of 2,000, which of course restricts the number of possible join permutations attempted per query block in a query, so with many tables in a query block (more than 6, for instance), the limit for the maximum number of permutations may easily be hit. Statistics must be reasonably accurate for cardinality estimates to be close to accurate, otherwise the query optimizer may select not only an inappropriate join order but also an inappropriate join type. Histograms may help (or harm) the query optimizer, indexes may help (or harm), establishing default cardinality and cost for custom PL/SQL functions may help, and adding hints to help control the cardinality estimates or join orders may also help.
- Lastly, and most importantly, the queries must provide correct information.
I struggled with this problem a bit. The report is intended to measure efficiency, so it would be somewhat laughable if the report performed inefficiently. How could I minimize execution time for this particular report? If Oracle Database is licensed by CPU core (or CPU socket in the case of the Standard Edition), does it make sense to perform all of the computations on the database server, or does it make sense to off-load some of that processing to the client’s dual, quad, or eight core CPU? If the report were rendered by a shared resource, such as a web server, application server, or Citrix server, would I want that same CPU offloading as I would have planned with a dedicated client computer? The report shows the information in a (modified) pivot table format, should I use the PIVOT function that was introduced with Oracle Database 11.1; or maybe I should use client-side control break logic that was taught in one of my university-level structured programming classes years ago?
That sketch of a report on a sheet of paper sure looked simple, now I need a control break (must be simple one key solution for that problem – check your keyboard :-) ).
It is possible to group the manufacturing lots by the Monday of the work week of the completion date (CLOSE_DATE) with the following function call:
Looks fairly simple. If I had a date range of four weeks in mind, I might be inclined to do something like this in the WHERE clause of the query:
CLOSE_DATE BETWEEN (NEXT_DAY(TO_DATE('20-AUG-2012','DD-MON-YYYY'),'MONDAY')-28) AND (NEXT_DAY(TO_DATE('20-AUG-2012','DD-MON-YYYY'),'MONDAY')-1)
However, I did not do that. Why not? One reason is that the Oracle query optimizer would likely have a hard time determining the approximate number of rows that would be returned from the table with the CLOSE_DATE column. The NEXT_DAY(TO_DATE()) combination hides the actual date range of interest, so the optimizer falls back to using default cardinality percentages for that predicate. Secondly, what happens when the user of the program is interested in something other than four weeks – the solution may be cumbersome to maintain. Thirdly, the above did not make use of bind variables, so every time the date range is changed, the query optimizer will perform a hard parse of the query. A better approach calculates the date range on the client-side and submits the WHERE clause like this:
CLOSE_DATE BETWEEN TO_DATE('30-JUL-2012','DD-MON-YYYY') AND TO_DATE('26-AUG-2012','DD-MON-YYYY')
Or, better yet using bind variables:
CLOSE_DATE BETWEEN :START_DATE AND :END_DATE
So, how many query executions will this report require? Just one query, executed a single time. The particular programming environment that I used requires ? to appear in bind variable positions, so when you see a ? just read it as either :START_DATE or :END_DATE.
SELECT L.EMPLOYEE_ID, E.LAST_NAME||', '||E.FIRST_NAME EMPLOYEE_NAME, L.PART_ID, L.WORKORDER_SUB_ID, L.OPERATION_SEQ_NO, L.DESCRIPTION, L.CLOSE_WEEK, L.PART_EFF_WEEK, L.AVG_HRS_PC, L.EMP_ACCUM_EFF_WEEK, SUM(L.HOURS_WORKED) OVER (PARTITION BY L.EMPLOYEE_ID) EMP_HOURS_WORKED, SUM(L.HOURS_WORKED) OVER (PARTITION BY L.EMPLOYEE_ID,L.PART_ID, L.WORKORDER_SUB_ID,L.OPERATION_SEQ_NO) EMP_PART_HOURS_WORKED, SUM(L.EMP_ACCUM_EFF_WEEK) OVER (PARTITION BY L.EMPLOYEE_ID,L.CLOSE_WEEK) EMP_EFF_WEEK FROM (SELECT O.PART_ID, O.DESCRIPTION, O.WORKORDER_SUB_ID, O.OPERATION_SEQ_NO, O.CLOSE_WEEK, O.RECEIVED_QTY, O.ACT_RUN_HRS, L.EMPLOYEE_ID, L.HOURS_WORKED, ROUND(O.ACT_RUN_HRS/O.RECEIVED_QTY,2) AVG_HRS_PC, ENG.ENG_HRS, ROUND(DECODE((O.ACT_RUN_HRS/O.RECEIVED_QTY),0,1,ENG.ENG_HRS/(O.ACT_RUN_HRS/O.RECEIVED_QTY)),4) PART_EFF_WEEK, SUM(L.HOURS_WORKED) OVER (PARTITION BY L.EMPLOYEE_ID,L.CLOSE_WEEK) EMP_HRS_WEEK, ROUND(L.HOURS_WORKED/(SUM(L.HOURS_WORKED) OVER (PARTITION BY L.EMPLOYEE_ID,L.CLOSE_WEEK)),4) EMP_PERCENT_WEEK, ROUND((L.HOURS_WORKED/(SUM(L.HOURS_WORKED) OVER (PARTITION BY L.EMPLOYEE_ID,L.CLOSE_WEEK))) * DECODE((O.ACT_RUN_HRS/O.RECEIVED_QTY),0,1,ENG.ENG_HRS/(O.ACT_RUN_HRS/O.RECEIVED_QTY)),4) EMP_ACCUM_EFF_WEEK FROM (SELECT PART_ID, DESCRIPTION, WORKORDER_SUB_ID, OPERATION_SEQ_NO, NEXT_DAY(TRUNC(CLOSE_DATE),'MONDAY')-7 CLOSE_WEEK, SUM(RECEIVED_QTY) RECEIVED_QTY, SUM(ACT_RUN_HRS) ACT_RUN_HRS FROM (SELECT WO.TYPE, WO.BASE_ID, WO.LOT_ID, WO.SPLIT_ID, WO.RECEIVED_QTY, WO.CLOSE_DATE, SUM(O.ACT_SETUP_HRS+O.ACT_RUN_HRS) ACT_RUN_HRS, WO.PART_ID, O.WORKORDER_SUB_ID, O.SEQUENCE_NO OPERATION_SEQ_NO, P.DESCRIPTION FROM WORK_ORDER WO, PART P, OPERATION O, SHOP_RESOURCE SR WHERE WO.TYPE='W' AND WO.SUB_ID='0' AND WO.CLOSE_DATE BETWEEN ? AND ? AND WO.RECEIVED_QTY>0 AND WO.TYPE=O.WORKORDER_TYPE AND WO.BASE_ID=O.WORKORDER_BASE_ID AND WO.LOT_ID=O.WORKORDER_LOT_ID AND WO.SPLIT_ID=O.WORKORDER_SPLIT_ID AND WO.PART_ID=P.ID AND O.RESOURCE_ID=SR.ID AND SR.BUILDING_ID='Machine' AND SR.LOGICAL_GROUP IN ('10 TURNING CENTERS','20 VERTICAL M.C.','30 630MM PALLET HMC','40 800MM-1000MM PALLET HMC','50 1000MM PALLET HMC','55 LARGE 2-TABLE HMC/HBM','60 SINGLE TABLE HBM','60 SINGLE TABLE HDM','70 LARGE FLOOR MACHINES') GROUP BY WO.TYPE, WO.BASE_ID, WO.LOT_ID, WO.SPLIT_ID, WO.CLOSE_DATE, WO.RECEIVED_QTY, WO.PART_ID, O.WORKORDER_SUB_ID, O.SEQUENCE_NO, P.DESCRIPTION) GROUP BY PART_ID, DESCRIPTION, WORKORDER_SUB_ID, OPERATION_SEQ_NO, NEXT_DAY(TRUNC(CLOSE_DATE),'MONDAY')-7) O, (SELECT /*+ LEADING(WO LT) */ WO.PART_ID, LT.EMPLOYEE_ID, LT.WORKORDER_SUB_ID, LT.OPERATION_SEQ_NO, NEXT_DAY(TRUNC(CLOSE_DATE),'MONDAY')-7 CLOSE_WEEK, SUM(LT.HOURS_WORKED) HOURS_WORKED FROM (SELECT WO.TYPE, WO.BASE_ID, WO.LOT_ID, WO.SPLIT_ID, WO.CLOSE_DATE, WO.PART_ID FROM WORK_ORDER WO, OPERATION O, SHOP_RESOURCE SR WHERE WO.TYPE='W' AND WO.SUB_ID='0' AND WO.CLOSE_DATE BETWEEN ? AND ? AND WO.RECEIVED_QTY>0 AND WO.TYPE=O.WORKORDER_TYPE AND WO.BASE_ID=O.WORKORDER_BASE_ID AND WO.LOT_ID=O.WORKORDER_LOT_ID AND WO.SPLIT_ID=O.WORKORDER_SPLIT_ID AND O.RESOURCE_ID=SR.ID AND SR.BUILDING_ID='Machine' AND SR.LOGICAL_GROUP IN ('10 TURNING CENTERS','20 VERTICAL M.C.','30 630MM PALLET HMC','40 800MM-1000MM PALLET HMC','50 1000MM PALLET HMC','55 LARGE 2-TABLE HMC/HBM','60 SINGLE TABLE HBM','60 SINGLE TABLE HDM','70 LARGE FLOOR MACHINES') GROUP BY WO.TYPE, WO.BASE_ID, WO.LOT_ID, WO.SPLIT_ID, WO.CLOSE_DATE, WO.PART_ID) WO, LABOR_TICKET LT, SHOP_RESOURCE SR WHERE WO.TYPE=LT.WORKORDER_TYPE AND WO.BASE_ID=LT.WORKORDER_BASE_ID AND WO.LOT_ID=LT.WORKORDER_LOT_ID AND WO.SPLIT_ID=LT.WORKORDER_SPLIT_ID AND LT.TYPE IN ('R','S') AND LT.HOURS_WORKED<>0 AND SR.LOGICAL_GROUP IN ('10 TURNING CENTERS','20 VERTICAL M.C.','30 630MM PALLET HMC','40 800MM-1000MM PALLET HMC','50 1000MM PALLET HMC','55 LARGE 2-TABLE HMC/HBM','60 SINGLE TABLE HBM','60 SINGLE TABLE HDM','70 LARGE FLOOR MACHINES') AND LT.RESOURCE_ID=SR.ID GROUP BY WO.PART_ID, LT.WORKORDER_SUB_ID, LT.OPERATION_SEQ_NO, LT.EMPLOYEE_ID, NEXT_DAY(TRUNC(CLOSE_DATE),'MONDAY')-7) L, (SELECT /*+ LEADING(WO) */ WO2.TYPE, WO2.BASE_ID, O.WORKORDER_SUB_ID, O.SEQUENCE_NO OPERATION_SEQ_NO, SUM(O.SETUP_HRS+O.RUN_HRS) ENG_HRS, WO.PART_ID FROM (SELECT WO.PART_ID FROM WORK_ORDER WO WHERE WO.TYPE='W' AND WO.SUB_ID='0' AND WO.CLOSE_DATE BETWEEN ? AND ? AND WO.RECEIVED_QTY>0 GROUP BY WO.PART_ID) WO, WORK_ORDER WO2, OPERATION O, SHOP_RESOURCE SR WHERE WO2.TYPE='M' AND WO.PART_ID=WO2.BASE_ID AND WO2.LOT_ID='0' AND WO2.SPLIT_ID='0' AND WO2.SUB_ID='0' AND WO2.TYPE=O.WORKORDER_TYPE AND WO2.BASE_ID=O.WORKORDER_BASE_ID AND WO2.LOT_ID=O.WORKORDER_LOT_ID AND WO2.SPLIT_ID=O.WORKORDER_SPLIT_ID AND O.RESOURCE_ID=SR.ID AND SR.BUILDING_ID='Machine' AND SR.LOGICAL_GROUP IN ('10 TURNING CENTERS','20 VERTICAL M.C.','30 630MM PALLET HMC','40 800MM-1000MM PALLET HMC','50 1000MM PALLET HMC','55 LARGE 2-TABLE HMC/HBM','60 SINGLE TABLE HBM','60 SINGLE TABLE HDM','70 LARGE FLOOR MACHINES') GROUP BY WO2.TYPE, WO2.BASE_ID, WO.PART_ID, O.WORKORDER_SUB_ID, O.SEQUENCE_NO) ENG WHERE O.PART_ID=L.PART_ID AND O.WORKORDER_SUB_ID=L.WORKORDER_SUB_ID AND O.OPERATION_SEQ_NO=L.OPERATION_SEQ_NO AND O.CLOSE_WEEK=L.CLOSE_WEEK AND O.PART_ID=ENG.PART_ID AND O.WORKORDER_SUB_ID=ENG.WORKORDER_SUB_ID AND O.OPERATION_SEQ_NO=ENG.OPERATION_SEQ_NO) L, EMPLOYEE E WHERE L.EMPLOYEE_ID=E.ID ORDER BY L.EMPLOYEE_ID, L.PART_ID, TO_NUMBER(L.WORKORDER_SUB_ID), L.OPERATION_SEQ_NO, L.CLOSE_WEEK;
I suppose that I could have performed more of the processing on the database server by sliding the above query into an inline view and used the PIVOT function, and/or additional analytic functions to calculate the employee’s weighted efficiency average per week and the simple averages of the wieghted averages – doing so would have eliminated some potentially complex client-side logic that needed to be programmed a single time, but counter point is that every time the report executed, it would require a more of the database server’s resources than were absolutely required.
That sketch of a report on a sheet of paper sure looked simple, but it turned into a three part blog article series. Interesting, now the person would like to analyze the data by part ID, listing the employees working on the operations for the part. Do I smell another three part blog article series? Sorry, no – I just changed the ORDER BY clause so that the PART_ID column was listed first, and made a small change to the client-side control break logic. Now the person wants to analyze 104 weeks worth of data rather than just four weeks, and only a single part ID or employee. Too late, those changes were anticipated, and included in the original specification in the client-side programming, implemented with a simple change to the query and pre-planning for a variable number of weeks in the report.