Workstation & Microcomputer Facilities

Team 5Asmita Karandikar, Zach Levine,

Michelle Lew, Brian Loo, Mike Lovejoy, Yue Tu

Friday, December 7th, 2007

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Overview

1. Organization summary2. Simplified EER diagram3. Relational Design (Schema)4. Normalization5. Queries in SQL and Access6. Query snapshot7. Conclusion

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Organization Summary

• UC Berkeley Microcomputer Facilities• 12 locations, 7 open to public• Approximately 300 computers total• Approximately 100 full- and part-

time staff

Dwinelle MF Evans MF Moffitt MF

UC Berkeley Microcomputer FacilitiesDatabase

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Relation Design (Schema)Microcomputer Facilities Database

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Relation Design (Continued)

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Relation Design (Continued)

19. Teaches (FID2b, C_S_Y_ID9) 20. Uses (UID2, Accessory_ID4, Location_ID4) 21. Enrolls_In (SID2a, C_S_Y_ID9) 22. Placed_At (Location_ID5,C_S_Y_ID9) 23. Unit(Unit_Number) 24. Maintains_Comp (WMF_Staff_ID2d, Computer_ID8, Location_ID8) 25. Reserves (Timeslot_ID6, C_S_Y_ID9) 26. Includes (Timeslot_ID6, Shift_ID7, Location_ID7) 27. Logs_Into (Computer_ID5, Location_ID5, UID2, Date, Time) 28. Logs_Out_Of (Computer_ID5, Location_ID5, UID2, Date, Time) 29. Works_T(WMF_Staff_ID2d, Timeslot_ID6, Location_ID5)

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Access Relationship View

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Access Interface Example 1 (Edit/Add Computer)

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Access Interface Example II (Request and Assign Repair)

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Functional Dependencies Normalization #1: The following relation, used to provide information about each class, is in 2NF: Course (C_S_ Y_ID, Course_ID, Semester, Year, Department, Course_Number, Course_ Name, Location_ID5, WMF_Staff_ID_Super2di2, Temp_Course_ID) This is in 2NF because all attributes can be derived from the primary key, C_S_Y_ID, but some of them can also be derived from other attributes that are not the primary key.

Therefore, to normalize this relationship as 3NF, the relationship could be changed into the following relationships: R1(C_S_ Y_ID, Course_ID, Semester, Year) R2 (C_S_Y_ID, Location_ID5, WMF_Staff_ID_Super2di2, Temp_ Course_ID) R3(Course_ID, Department, Course_ Number, Course_ Name)

However, R1 is not in BCNF form because the Course_ID, Semester, and Year could be used to determine the C_S_Y_ID. So in order for this to be in BCNF form, R1 should be changed into the following three relations: R11(C_S_ Y_ID, Course_ID) R12(C_S_Y_ID, Semester) R13(C_S_Y_ID, Year)

Functional Dependencies (cont’d)

However, this relation is not in BCNF because the Service_ Tag and the Express_Service_Code are also unique to each. To make BCNF, modify as follows: R1(Location_ID5, Computer_ID, Computer_Type, Model, Operating_System, Installation_Date) R2(Location_ID5, Computer_ID, Service_Tag) R3(Location_ID5, Computer_ID, Express_Service_Code)

Functional Dependencies (cont’d)

Normalization #2: The following relation, which provides information about each computer, is in 3NF: Computer (Location_ID5, Computer_ ID, Computer_Type, Model, Operating_System, Service_Tag, Express_Service_Code, Installation_Date)

Functional Dependencies (cont’d)

Normalization #3: The below relation, which gives information about each waitlist, is in BCNF: Waitlist (Waitlist_ ID, Location_ID5, Operating_System)

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Query- Budget Analysis Budget: How much money is spent on employees in the previous month compared to the previous 3 months and the same month the previous year? For these same five months, what is the ratio of total employee cost per student user? For the last month, what would be the additional cost of increasing wages by $0.50, and of extending lab hours by 1, 2, 3, or 4 hours? What is the three-month moving average of total money spent on employees each month over the previous year? - This combines a client-requested query with a ratio that can analyze the productivity of the

lab. - The additional hours are only part-time employees. - The additional hours simply extend the part-time employees who worked the last shift. - Full-time employees simply receive Salary/2000 per hour worked. - Full-time employees work only one shift per day. - This query meets the needs of our client very closely. Budget decisions come down to

simple number comparisons. When evaluating facility costs, labor represents an enormous percentage (along with electricity and maintenance). Forecasting overall changes in the budget by increasing (or hypothetically decreasing) wages helps the management make decisions quickly. Additionally, the parameter, ($ of wages)/ (student served), gives a general perspective of the dollar value of providing computing services to the student.

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Query – Budget Analysis (cont’d)

SQL> SELECT (COUNT(i.TimeSlot_ID)*p.Wage + COUNT(w.Shift_ID)*(f.Salary/ 240)) as Total_Cost, Total_Cost/ (COUNT(l.Date)), (COUNT(i.TimeSlot_ID))*(0.25) as Wage_Increase_of_25, (COUNT(i.TimeSlot_ID))*(0.5) as Wage_Increase_of_50, MONTH(t.Date) as Month, YEAR(t.Date) as Year FROM Part_Time p, Includes i, Works w, TimeSlot t, Staff s, Logs_Into l, Full_Time f

WHERE i.Shift_ID = w.Shift_ID AND (w.WMF_Staff_ID = p.WMF_Staff_ID OR w.WMF_Staff_ID = f.WMF_Staff_ID) AND MONTH(t.Date) = (SELECT MONTH(t.Date) FROM Timeslot t WHERE ((MONTH(t.Date) = MONTH(Date()) – 1 OR MONTH(t.Date) = MONTH(Date()) – 2 OR MONTH(t.Date) = MONTH(Date()) – 3

OR MONTH(t.Date) = MONTH(Date()) – 4) AND YEAR(t.Date) = YEAR(Date())) OR YEAR(t.Date) =

(SELECT (YEAR(t.Date) FROM Timeslot t WHERE YEAR(t.Date) = YEAR(Date()) – 1 AND MONTH(t.Date) = MONTH(Date()) – 1))

AND MONTH(l.Date) =

Query – Budget Analysis (cont’d)

(SELECT MONTH(l.Date) FROM Logs_Into l WHERE (MONTH(l.Date) = MONTH(Date()) – 1 OR MONTH(l.Date) = MONTH(Date()) – 2 OR MONTH(l.Date) = MONTH(Date()) – 3

OR MONTH(l.Date) = MONTH(Date()) – 4) AND YEAR(l.Date) = YEAR(Date()) OR YEAR(l.Date) =

(SELECT YEAR(l.Date) FROM Logs_Into l WHERE YEAR(l.Date) = YEAR(Date()) – 1 AND MONTH(l.Date) = MONTH(Date()) – 1))

ORDER BY Month DESC, Year DESC;

Query – Budget Analysis (cont’d)

SQL> SELECT COUNT(i.TimeSlot_ID)*p.Wage as Additional_Cost1, 2*Additional_Cost1 as Additional_Cost2, 3*Additional_Cost1 as Additional_Cost3, 4*Additional_Cost1 as Additional_Cost4

FROM Includes i, Part_Time p, TimeSlot t WHERE NOT EXISTS HOUR(t.Date) + 1 AND t.TimeSlot_ID = i.TimeSlot_ID AND i.Location_ID = w.Location_ID AND w.WMF_Staff_ID = p.WMF_Staff_ID;

Query – Budget Analysis (cont’d)

5 separate queries: SELECT (Count(t.TimeSlot_ID)*p.Wage+Count(w.TimeSlot_ID)*(f.Salary/ 2000)) AS Total_Cost, (Count(t.TimeSlot_ID)*(p.Wage+0.5)+Count(w.TimeSlot_ID)*(f.Salary/ 2000)) AS Total_Cost1, count(l.date) AS mycount0 FROM Part_Time AS p, Works_T AS w, TimeSlot AS t, Staff AS s, Logs_Into AS l, Full_Time AS f WHERE (((w.WMF_Staff_ID)=p.WMF_Staff_ID Or (w.WMF_Staff_ID)=f.WMF_Staff_ID) And ((w.Timeslot_ID)=t.Timeslot_ID) And ((Month(t.Date))=Month(Date()))) GROUP BY w.Timeslot_ID, p.Wage, f.Salary;

One query to bring them all together: SELECT Sum([Query 2a].Total_Cost) AS Total_Employee_Cost, Sum([Query 2b].Total_Cost_1Mo) AS TEC1Mo, Sum([Query 2c].Total_Cost_2Mo) AS TEC_2Mo, Sum([Query 2d].Total_Cost_3Mo) AS TEC3Mo, Sum([Query 2e].Total_Cost_1Yr) AS TEC1Yr, Max([Query 2a].mycount0) AS Count0, Max([Query 2b].mycount1) AS Count1, Max([Query 2c].mycount2) AS Count2, Max([Query 2d].mycount3) AS Count3, Max([Query 2e].mycount1yr) AS Count_1Yr, Sum([Query 2a].Total_Cost1) AS Total_Employee_Cost_50, Sum([Query 2b].Total_Cost_1Mo1) AS TEC1Mo50, Sum([Query 2c].Total_Cost_2Mo1) AS TEC_2Mo50, Sum([Query 2d].Total_Cost_3Mo1) AS TEC3Mo50, Sum([Query 2e].Total_Cost_1Yr1) AS TEC1Y50 FROM [Query 2a], [Query 2b], [Query 2c], [Query 2d], [Query 2e];

Query – Budget Analysis (Prt Sc)

Query – Budget Analysis Report

Query – Maintenance Mgmt

Maintenance: Rank the computers by number of total maintenance issues by location, then by Operating System, then by Model. What is the average (and variance of) number of repairs per computer for each location and operating system? What is the mean time to failure for each computer, sorted by location, then Operating System (Mac or PC)? - This query will ignore any computers that have not experienced any maintenance

issues. - We will enter the resulting data into Excel, which will determine the distribution of

the TTF. - Date includes a time, and a computer cannot have simultaneous repair requests.

Query – Maintenance MgmtQuery 4 1. Rank the computers by number of total maintenance issues. SELECT c.Computer_ID as Computer_ID, COUNT(r.Date) as Repairs, c.Location_ID as

Location, c.Operating_System as Operating_System, c.Model as Model FROM Computer c, Requsts_Repairs_From r WHERE c.Computer_ID = r.Computer_ID AND c.Location_ID = r.Location_ID GROUP BY c.Operating_System, c.Location_ID, c.Computer_ID, c.Model;

Query – Maintenance Mgmt (cont’d)

2a. What is the average number of repairs per computer for each location? SELECT c.Location_ID as Location, COUNT(r.Date)/ COUNT(c.Computer_ID) as

Repair_Rate_L FROM Requsts_Repairs_From r, Computer c WHERE c.Computer_ID = r.Computer_ID AND c.Location_ID = r.Location_ID GROUP BY Location; 2b. What is the average number of repairs per computer for each operating system? SELECT c.Operating_System as Operating_System,

COUNT(r.Date)/COUNT(c.Computer_ID) as Repair_Rate_OS FROM Requsts_Repairs_From r, Computer c WHERE c.Computer_ID = r.Computer_ID AND c.Location_ID = r.Location_ID GROUP BY Operating_System;

Query – Maintenance Mgmt (cont’d)

3. What is the mean time to failure for each computer, sorted by location, then Operating System (Mac or PC)?

SELECT c.Computer_ID AS Computer_ID, c.Location_ID AS Location, (MAX(r.Date)-

c.Installation_Date)/ COUNT(r.Date) AS Mean_TTF FROM Requsts_Repairs_From r, Computer c WHERE c.Computer_ID = r.Computer_ID AND c.Location_ID = r.Location_ID GROUP BY Operating_System, c.Computer_ID, c.Location_ID, c.Installation_Date;

Query – Maintenance Mgmt(Prt Sc)

Query – Maintenance Mgmt (Prt Sc)

Query – Maintenance Mgmt (Prt Sc)

Query – Maintenance Mgmt (Prt Sc)

Query – Maintenance Mgmt Report

Query – Human Resources 1. Human Resources: What is the average length of employment grouped by location by

position for all staff position, with variance? For each position, what is the average total length of employment with the MCF, with variance? For each position, what is the average and variance of the number of previous positions held by employees, and what is the most common previous position? For each position, what is the average student level (part-time staff)? SQL> Select AVG(s.termination_date-s.hiring_date), VARIANCE(s.termination_date- s.hiring_date), From Staff s, Group By WMF_Staff_ID SQL> Select AVG(SUM(Termination_Date-Hiring_Date)) , VARIANCE(SUM(Termination_Date-Hiring_Date)), AVG(Count(Distinct WMF_Staff_ID), VARIANCE(Count(Distinct WMF_Staff_ID) FROM Staff s, Maintenance m, Supervisor sup, Scheduler sch, Other oth, Part_Time pt WHERE s.SSN=m.SSN AND s.SSN=sup.SSN AND s.SSN=sch.SSN AND s.SSN=oth.SSN AND s.SSN=pt.SSN SQL> Select AVG(pt.rank) FROM Part_Time pt GROUP BY WMF_Staff_ID

Query – Scheduling

1. Give a ranked list of part-time employees who have worked a timeslot the most number of times over the last month, for each timeslot. SELECT pt.WMF_Staff_ID AS WMF_Staff_ID, HOUR(t.Hour) AS Timeslot_Start_Hour, COUNT(t.Hour) AS Number_of_Times_Worked FROM Part_Time AS pt, Timeslot AS t, Works_T AS w WHERE MONTH(t.Date) = MONTH(Date()) AND w.Timeslot_ID = t.Timeslot_ID AND w.WMF_Staff_ID = pt.WMF_Staff_ID GROUP BY pt.WMF_Staff_ID, t.Hour, t.Timeslot_ID;

Query – Scheduling

2. Who has never worked a timeslot that had a waitlist? SELECT pt.WMF_Staff_ID AS WMF_Staff_ID FROM Part_Time AS pt WHERE NOT EXISTS (SELECT * FROM Works_T w, Enters e, Timeslot t WHERE pt.WMF_Staff_ID = w.WMF_Staff_ID AND Hour(e.Time) = t.Hour AND w.Timeslot_ID = t.Timeslot_ID AND t.Date = e.Date);

Query – Scheduling 3. Who has worked a beginning or ending shift (first or last of the open period) during the month? -Timeslot_IDs are consecutive SELECT pt.WMF_Staff_ID AS WMF_Staff_ID, w.Location_ID AS Location FROM Part_Time AS pt, Works_T AS w, Timeslot AS t WHERE pt.WMF_Staff_ID = w.WMF_Staff_ID AND w.Timeslot_ID = t.Timeslot_ID AND (EXISTS SELECT * FROM Works_T AS w WHERE t.Timeslot_ID-1 = w.Timeslot_ID AND w.Location_ID = Location AND NOT EXISTS w.WMF_Staff_ID) OR (EXISTS SELECT * FROM Works_T AS w WHERE t.Timeslot_ID+1 = w.Timeslot_ID AND w.Location_ID = Location AND NOT EXISTS w.WMF_Staff_ID);

Query – Service Levels1. Service Levels: What is the approximate current wait time for each location? What is the average wait time and variance for a given day? What is the utilization rate over a one-week period? Rank each parameter by location. - Helping to create a wait time board, this query helps distribute students among the available resources. - Service level, and specifically wait time, is arguably the most important parameter to which the management can gauge each facility's effectiveness.

Query – Service Levels SQL> IF COUNT(w.waitlist_ id)>2 FROM Waitlist w, Enters e, Logs_ Into li, Computer c THEN Select AVG(li.time-e.time)*COUNT(Waiting) WHERE e.UID=li.UID AND Year(e.date)=Year(current_ timestamp) AND Year(e.month)=Year(current_ timestamp) AND GROUP BY Location, Operating_System DESC ELSE Print ‘Wait time is ‘Count(w.waitlist_ id)*4’

SQL> SELECT AVG(li.time-e.time), VARIANCE(li.time-e.time) FROM Enters e, Logs_ Into li WHERE e.UID=li.UID GROUP BY Date, Location, Operating_System SQL> Select (COUNT(li.computer_ id)/COUNT(c.computer_ id)), From Logs_ Into li, Computer c, Timeslot t

Where li.time BETWEEN MIN(t.time) AND MAX(t.time) AND t.date=”3/17/06” GROUP BY Location_ ID IN DESC, Operating_System

SQL> Select SUM(lo.time-li.time)/1080

From Logs_ Out_Of lo, Logs_ Into li, Computer c Where lo.date=li.date AND lo.UID = li.UID AND lo.Computer_ ID = li.Computer_ ID GROUP by

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Questions?