1. Douglas Properties & Management
Mishaal Alruwaili
Spencer Lauber
Linh Vu

2. Project Description Housing company that operates in Spokane, WA
South Hill, North Hill, Logan, Riverside
Large number of properties that include residential spaces including:
Standalone houses
Apartment complexes
Townhouses

3. Project Description
Douglas Properties & Management has lease agreements for each room in each house/apartment
Keeps track of all the individual tenants in the units

4. Project Objectives
Due to the complexity of Douglas’s information, we have designed a database model that increases the effectiveness and efficiency of managing data
Increase efficiency in viewing the properties and maintaining all units in each property
Easily track renters’ information including location, payments
Differentiate Douglas form other competitors in the housing industry

5. System Development Life Cycle
Analysis
Design: Three major parts
Tenant
Property
Employee
Implementation
Tenant, Payment, Contract
Property, Room
Employee (Administrative, Landscaping)
(Look-up tables: Neighborhood, Neighborhood Properties)
Maintenance
Continuously add more data to the database immediately after any changes

6. ERD Model TENANT & PROPERTIES

7. EMPLOYEE & LOOK-UP TABLES

8. Third Normal Form

9. SQL Commands (Payment table)
DROP TABLE payment CASCADE CONSTRAINTS;
CREATE TABLE payment
(tenant_id NUMBER(5),
contract_id NUMBER(5),
date_paid DATE,
on_time CHAR(1),
amount_paid NUMBER(6,2),
CONSTRAINT payment_cpk PRIMARY KEY (Tenant_ID, Contract_ID, Date_Paid),
CONSTRAINT payment_On_Time_ck CHECK (On_Time IN ('Y','N')),
CONSTRAINT payment_contract_id_fk FOREIGN KEY (contract_id)
REFERENCES contract (contract_id),
CONSTRAINT payment_tenant_id_fk FOREIGN KEY (tenant_id) REFERENCES tenant (tenant_id));

10. SQL Commands (Payment table)
INSERT INTO payment (tenant_id, contract_id, date_paid, on_time, amount_paid) VALUES ('11001', '22001', '01-JUN-2013', 'Y', '400');

11. SQL Query 1
Listing of all our tenants addresses to send them a notice of inspection
SELECT t.fname, t.lname, k.room_id, p.address, p.city, p.state, p.zip
FROM tenant t, room r, property p, contract k
WHERE k.room_id = r.room_id
AND k.tenant_id = t.tenant_id
AND r.property_id = p.property_id
ORDER BY lname;

12. SQL Query 1 Result FNAME LNAME ROOM_ID ADDRESS CITY ST ZIP KELLY
ALDERMAN
10086
560 UPLAND WY
SPOKANE WA
99209
SERAFINA
BUI
10078
110 WELLESY ST
MUI
CHAN
10095
550 PORTOLA AVE
99258
JIAYU
CHEN
10057
992 GRAND AVE
99203
GREG
CHU
10003
229 BALDWIN AVE
99207
JAY
CHUI
10067
4500 DIVISION ST
KENDRA
COCKERHAM
10080
TIM DO
10055
201 LAMONTE ST
FRANCIS GO
10060
QUYEN HA
10001
CRYSTAL
HARTLEY
10056
KELSEY
HOOD
10022
1020 SINTO AVE
LERA
HUAN
10045
428 AUGUSTA AVE
ANAS
HUSSAIN
10087

13. SQL Query 2 Number of rooms in each building and the type of building.
SELECT p.address, p.building_type, COUNT(r.room_id) "NUMBER_OF_ROOMS“
FROM property p, room r
WHERE r.property_id = p.property_id
GROUP BY p.building_type, p.address;

14. SQL Query 2 Result ADDRESS BUILDING_TYPE NUMBER_OF_ROOMS 992 GRAND AVE
APARTMENTS 8
4500 DIVISION ST 9
560 UPLAND WY
STAND-ALONE 4
550 PORTOLA AVE 2
428 AUGUSTA AVE 11
229 BALDWIN AVE
201 LAMONTE ST 3
110 WELLESY ST 12
4545 SUPERIOR DR 10
780 FULTON ST
450 MOANA WY
808 NORA AVE 5
1020 SINTO AVE 15
455 37TH AVE

15. SQL Query 3 Number of rooms still available in each building.
SELECT p.address, p.building_type, COUNT(r.room_id) "AVAILABLE_ROOMS“
FROM property p, room r
WHERE r.property_id = p.property_id
AND r.room_id NOT IN
(SELECT k.room_id
FROM contract k)
GROUP BY p.building_type, p.address;

16. SQL Query 3 Result ADDRESS BUILDING_TYPE AVAILABLE_ROOMS
4500 DIVISION ST
APARTMENTS 5
992 GRAND AVE 6
560 UPLAND WY
STAND-ALONE 2
428 AUGUSTA AVE 10
110 WELLESY ST 8
229 BALDWIN AVE
4545 SUPERIOR DR 9
450 MOANA WY 4
780 FULTON ST 1
1020 SINTO AVE 12
808 NORA AVE
408 INDIANA DR

17. SQL Query 4 Which rooms are available in 110 Wellesey?
SELECT r.room_id, r.square_footage
FROM room r, property p
WHERE r.property_id = p.property_id
AND p.address = '110 WELLESY ST‘
AND r.room_id NOT IN
(SELECT k.room_id FROM contract k);

18. SQL Query 4 Result ROOM_ID SQUARE_FOOTAGE 10082 160 10083 170 10076
10074
150
10075
140
10084
120
10077
10079

19. SQL Query 5 Renters who has a half year contract
SELECT t.lname || ', ' || t.fname "NAME", t.tenant_id AS "TENANT_NUMBER“
FROM tenant t, contract k
WHERE t.tenant_id = k.tenant_id
AND (k.date_end - k.date_begin) = 213
ORDER BY t.lname;

20. SQL Query 5 Result NAME TENANT_NUMBER CHAN, MUI 11025 CHEN, JIAYU
11011
CHU, GREG
11017
DO, TIM
11014
GO, FRANCIS
11022
HA, QUYEN
11003
HARTLEY, CRYSTAL
11012
HUSSAIN, ANAS
11007
JASSEM, FAHAD
11004
KIM, ROBIN
11016
PETERSON, JASON
11015
SIPP, TINA
11021
STOREBO, JAKE
11008
TIUR, GRACE
11026

21. SQL Query 6 Average hourly wage of landscaping employees.
SELECT TO_CHAR(AVG(hourly_wage), '$999.99') FROM landscaping_employee;

22. SQL Query 6 Result
AVERAGE HOURLY WAGE
$13.12

23. SQL Query 7
Which Landscaping employee makes the highest hourly wage among all other employees, and in which neighborhood is he/she assigned to?
SELECT DISTINCT e.employee_id, e.fname, e.lname, TO_CHAR(lse.hourly_wage, '$999.99'), lse.neighborhood_id
FROM employee e, landscaping_employee lse WHERE e.employee_id = lse.lsemployee_id
AND lse.hourly_wage =
(SELECT MAX(hourly_wage)
FROM landscaping_employee);

24. SQL Query 7 Result EMPLOYEE_ID FNAME LNAME HOURLY WAGE NEIGHBORHOOD_ID17
JENETTE
JEFFERIES
$17.80 4

25. SQL Query 8
Name, monthly rent and contract id of tenant who lives in room 11007
SELECT r.room_id, t.fname, t.lname, t.tenant_id, TO_CHAR(c.monthly_rent_due, '$999.99'), c.contract_id
FROM room r, contract c, tenant t
WHERE t.tenant_id = c.tenant_id AND c.room_id = r.room_id
AND t.tenant_id = 11007;

26. SQL Query 8 Result ROOM_ID FNAME LNAME TENANT_ID RENT_DUE CONTRACT_ID
10087
ANAS
HUSSAIN
11007
$400.00
22007

27. SQL Query 9
List all tenants who have made a late payment, includes: id, names, number of late payments they have made
SELECT DISTINCT t.tenant_id, t.fname, t.lname, COUNT(on_time) AS "# OF LATE PAYMENTS" FROM tenant t, payment p
WHERE t.tenant_id = p.tenant_id
AND p.on_time = 'N‘
GROUP BY t.tenant_id, t.fname, t.lname;

28. SQL Query 9 Result TENANT_ID FNAME LNAME # OF LATE PAYMENTS 11013
JANEAL
JAMES 1
11027
CAMERON
LEA 2
11005
TOMOMI
SAKATA
11006
CELINA OH
11008
JAKE
STOREBO
11011
JIAYU
CHEN
11028
KENDRA
COCKERHAM
11004
FAHAD
JASSEM
11007
ANAS
HUSSAIN
11010
JAY
CHUI
11022
FRANCIS GO
11009
KELLY
ALDERMAN
11021
TINA
SIPP
11020
KIERRA
IRVIN

29. Managerial and organizational impacts
Making Douglas Properties & Management operations run much more efficiently
Improve the ease and relevance of data retrieval
Efficiently analyze historical data to find out important information
Manage the company more efficiently

30. Conclusion
Have a chance to experience a real life example of interacting with a decent sized database system
Teamwork
Minimize conflicts in data entry stage
Time saving
Minimize database anomalies
Encounter several challenges because of transitive dependencies

31. Questions ?