final presentation “home architect & contractor database”

Presentation Outline1.0 Project Summary

Client DescriptionClient Needs

2.0 Progress SummarySimplified EER DiagramRelational Schema and View in MS Access

3.0 Final Analysis and ImplementationQueriesNormalization

1.1 Client Description

More Than a Carpenter, Inc. Residential design/build contractor

specializing in architectural design, remodeling, and historical restoration.

They offer a broad range of services ranging from small handiwork to new home design and construction.

They hire subcontractors to complete each stage of their designs.

1.0 Project Summary

1.2 Client NeedsCurrent Database

List of MS Outlook contacts to track client info

Bookkeeping software to manage individual projects

Missing FunctionalityNo way of analyzing data related to

customers and contractsNo way to relate advertising efforts to salesNo way to analyze the company’s success in

terms of presentations and sales calls

1.0 Project Summary

1.3 ApproachOur goal for the Home Architect and

Contractor database was to create a flexible system that would track data about the following:AdvertisingClientsEmployeesSales calls, visits, and presentationsSub-contractorsPotential and current contractsContract stages and company resources used

on them

1.0 Project Summary

2.0 Progress Summary

2.1 DP Review I – Simplified EER Diagram

WORD_OF_MOUTH

d,t

OTHER REALTOR

ADVERTISEMENT

SIGN

WEB

d,t

VEHICLE

VEH_INST

instance_of

PAMPHLET

LAWNSIGN_TYPEBILLBOARD_TYPE

d,t

BILL_INST LAWN_INST

instance_ofinstance_of

CONTRACT

NEW_CONSTRUCTION

HANDWORK

RENOVATION

d,t

comes_from

PROPOSAL

requires

PERMIT

works_on

SUB-CONTRACTOR EQUIPMENT

needs

MATERIAL

includes STAGE

needs

CLIENT LEAD

d,t

CUSTOMER

signs

referred_by

created for

PROPERTY

Associated_with

Is_for yields

EMPLOYEE

DESIGNER

OFFICE_STAFF o,p

designs_for

Liason_to

for a makesVISIT

for a createsBUDGET

givesPRESENTATIONfor a

Supervises

2.2 DP Review II – Relational Schema

2.0 Progress Summary

1. ADVERTISEMENT (AID, Date_published)

1a. WORD_OF_MOUTH (WMID, AID1, Fname, Lname)

1aa. REALTOR (Realtor_ID, WMID1a, Company)

1ab. OTHER_WM (OWMID, AID1a, Description)

1b. WEB (Web_ID, AID1, Website, Web_Address, Monthly_Cost)

1c. PAMPHLET (Pamphlet_ID, AID1, Type_of_pamphlet, Color, Location, Cost, Count)

1d. VEHICLE (Vehicle_ID, AID1, VIID1d1, Manufacturer, Model, Year, Color)

1d1. VEH_INST (VIID)

1e. SIGN (Sign_ID, AID1, Location, Size)

1ea. LAWNSIGN_TYPE (LSTID, Sign_ID1e, LSIID1ea1, Lawn_sign_type, Cost)

1ea1. LAWN_INST (LSIID)

1eb. BILLBOARD_TYPE (Billboard_ID, Sign_ID1e, BIID1eb1, Billboard_type, Monthly_Cost)

1eb1. BILL_INST (BIID, AID1eb)

2. COMES_FROM(AID1, Proposal_ID12)

3. CONTRACT (CID, Proposal_ID12, Est_time, Type, Description, Date_signed)

3a. HANDWORK (CID3, Room)

3b. RENOVATION (CID3, Room)

3c. NEW_CONSTRUCTION (CID3, description)

4. PERMIT (Permit_ID, City, Ordinance, Date_applied, Date_received)

5. REQUIRES (Permit_ID4, CID3)

6. CUSTOMER (CusID, RcusID, Fname, Lname, MI, Street_address, City, State, Zipcode)

6a. LEAD (LeadID, CusID5)

6b. CLIENT (Client_ID, CusID5, CID3, Num_contracts)

7. REFERED_BY(CusID6, RcusID6)

8. PHONE_NUM (CusID6, Phone_number)

9. EMAIL (CusID6, Email_address)

10. EMPLOYEE(ESSN, SupSSN, Lname, Fname, MI, Salary, Email, Phone, Start_Date)

10a. OFFICE STAFF(OSID, ESSN10, Branch, Office_No) 10b. DESIGNER (Designer_ID, ESSN10, Skill, Software) 11. SUPERVISES(ESSN10, SupSSN10) 12. PROPOSAL(Proposal_ID, PID20, Name, Description, PDay,

PMonth, PYear, PID20, CusID6) 13. VISIT(Visit_ID, Date, Time_Spent) 14. BUDGET(Budget_ID, Proposal_ID12, Total_Cost) 15. PRESENTATION(Presentation_ID, Date, Time Spent,

ClientResponse, Who_Presented_To) 16. STAGE(Stage_ID, CID3, Level, Forecasted_Start_Date,

Actual_Start_Date, Finish_Date, Est_cost, Actual_cost, Customer_rating, Sub_ID17)

17. SUB_CONTRACTOR(Sub_ID, Name_of_company, Representative, Email, Phone)

18. EQUIPMENT(EID, Name, Quantity, Type, Date_Acquired) 19. MATERIAL(MID, Name, Amount, Cost, Date_Acquired) 20. PROPERTY(PID, Name, Street_address, City, Country,

Zipcode, Type, Built_date) 21. PROP_VISIT(Proposal_ID12, Visit_ID13) 22. EMP_VISIT(ESSN10, Visit_ID13) 23. EMP_BUDGET(ESSN10, Budget_ID14) 24. EMP_PRES(ESSN10, Presentation_ID15) PROP_PRES(Proposal_ID12, Presentation_ID15) 26. YIELDS(Proposal_ID12, CID3) 27. DESIGNS_FOR(Proposal_ID12, ESSN10b) 28. EQUIPMENTS_STAGE(Stage_ID16,EID18) 29. MATERIALS_STAGE(Stage_ID16, MID19) 30. PROPERTY_CONTRACT(PID20,CID3) 31. LIAISON_TO(Proposal_ID12, ESSN10a) 32. PROP_ADV (Proposal_ID12, AID1)

2.3 DP Review III – MS Access Relationship View

3.0 Final Analysis and Implementation

Query 1: Customer Referrals

“Home Architect & Contractor” Database

Description: Which customers are most successful at referring new clients?

Output: Referral statistics for each customer.

Benefit: By finding our most valuable clients in terms of customer referrals, we can reward those clients who provide us with the most business. This will encourage more referrals and strengthen customer relationships.

3.0 Final Analysis & implementation

Query 1 Code and Implementation

SELECT r.RcusID, Sum(b.Total_Cost) AS TotRev, Avg(b.Total_Cost) AS AvgRev, Var(b.Total_Cost) AS VarRev, Count(b.Total_Cost) AS NumRefContracts FROM Referred_By AS r, Customer AS cu, Proposal AS p, Budget AS b, Contract AS co WHERE cu.CusID = r.CusID AND cu.CusID = p.CusID AND p.Proposal_ID = b.Proposal_ID

AND co.Proposal_ID = p.Proposal_ID GROUP BY r.RcusID ORDER BY Sum(b.Total_Cost) DESC;

“Home Architect & Contractor” Database3.0 Final Analysis & implementation

cc

Query 2: Employee EffectivenessDescription: For each employee, produce a rating that takes into account the number of presentations delivered and the number of visits made, with respect to the number of contracts struck as a result. Output: Employee’s SSN, Rating

Reason: This query is one way to evaluate the performance of our employees. Through this query we are able to place a rating on all our employees. This will ensure that it is fair and accurately reflects the employee’s performance. We will use this query to assess whether an individual employee will receive a raise.


Query 2: Code and Implementation

SELECT ep.ESSN, Count(ep.Presentation_ID) AS countPresentationContract

FROM EMP_PRES ep, Proposal pr, Contract c, Prop_Pres pp

WHERE ep.Presentation_ID= pp.Presentation_ID

AND pr.Proposal_ID=pp.Proposal_ID

AND pp.Proposal_ID=c.Proposal_ID

GROUP BY ep.ESSNORDER BY ep.ESSN; SELECT ev.ESSN, count(ev.Visit_ID) AS

countVisitContractFROM EMP_VISIT ev, Proposal pr,

Contract c, Prop_Visit pvWHERE ev.Visit_ID = pv.Visit_ID AND pr.Proposal_ID =

pv.Proposal_ID AND pv.Proposal_ID =

c.Proposal_IDGROUP BY ev.ESSNORDER BY ev.ESSN;

SELECT ep.ESSN, count(ep.Presentation_ID) AS countPresentationFROM EMP_PRES epWHERE EXISTS (SELECT * FROM Query2a)GROUP BY ep.ESSNORDER BY ep.ESSN; SELECT ev.ESSN, Count(ev.Visit_ID) AS countVisitFROM EMP_VISIT evWHERE EXISTS (SELECT * FROM Query2b)GROUP BY ev.ESSNORDER BY ev.ESSN;


Query 2: Code, Putting it all togetherSELECT e.ESSN,

0.5*{(Sum(q2a.countPresentationContract)) /

(Sum(q2c.countPresentation))}

+0.5*{(Sum(q2b.CountVisitContract))

/Sum((q2d.CountVisit))} AS EmpRatingFROM Employee e, Query2a q2a,

Query2b q2b, Query2c q2c, Query2d q2dWHERE e.ESSN=q2a.ESSN AND q2a.ESSN=q2b.ESSN AND q2a.ESSN=q2c.ESSN AND q2b.ESSN=q2d.ESSNGROUP BY e.ESSNORDER BY

0.5*{(Sum(q2a.countPresentationContract)) /

(Sum(q2c.countPresentation))}

+0.5*{(Sum(q2b.CountVisitContract)) / Sum((q2d.CountVisit))}

AS EmpRating DESC;


cc

Query 3: Advertising ValueDescription: For each advertising category,

calculate its return on investment (ROI).

Output: (Profit – Cost)/Cost Benefit: This query allows our client to see

which forms of advertising are providing revenue and which are resulting in a loss.


Query 3: Code

SELECT pa.Pamphlet_ID ( sum(b.Total_cost) – pa.Quantity*pa.Cost )/( pa.Quantity*pa.Cost) AS PamphletRating

FROM Pamphlet pa, Budget b, Advertisement a, Proposal pr, Contract c, Prop_Adv pa

WHERE pa.AID = a.AID AND a.AID = pa.AID AND pa.Proposal_ID = pr.Proposal_ID AND pr.Proposal_ID = b.Proposal_ID AND pr.Proposal_ID = c.Proposal_IDGROUP BY pa.Pamphlet_IDORDER BY ( sum(b.Total_cost) – pa.Quantity*pa.Cost

)/( pa.Quantity*pa.Cost) DESC;


cc

Query 4: Proposal ForecastDescription: How many proposals can we expect

to receive this month based on how many we’ve received in the last several months.

Output: A table showing how many proposals were received in each of the last twelve months. A cubic-spline forecast is made using this data.

Benefit: This information will help our client budget for the next month, develop a short-term business strategy, and assess the feasibility of future business expansion.

“Home Architect & Contractor” Database 3.0 Final Analysis & implementation

Query 4: Cubic InterpolationUsing the SQL output, we can use this

result and interpolate the demand for the next year.

We use a piecewise continuous function, passing through each of the values of the table. The form of the equation is as follows,

We also need to put boundaries in the 1st and 2nd order derivative,

dxxcxxbxxaxS iii )()()()( 23

)(')(' 1 ii xSxS )('')('' 1 ii xSxS

Query 4: CodeForecasting for the following month from all previous data: SELECT Month, Year, Count(Proposal_ID) FROM Proposal GROUP BY Month, Year ORDER BY Year ASC, Month ASC;

Forecasting for an upcoming month given data for that month from previous years:

SELECT Month, Year, Count (Proposal_ID) FROM Proposal WHERE Month = ‘insertmonthhere’ GROUP BY Year ORDER BY Year ASC;

Constraints can be added to the “SELECT” and/or “WHERE” statements if we are interested in looking at a narrower period of time to forecast demand, or in examining timespans other than months.

cc

Query 5: Sub-contractor RatingDescription: Which subcontractors complete jobs

most cost-effectively?

Output: A rating based upon the actual cost of completing a project relative to the cost stated in the contract, for each project in which a subcontractor takes part.

Benefit: Serves as one criterion with which the company can determine which subcontractors to pursue more involved business relationships with and to recommend to customers. Consequently, provides an incentive to subcontractors to make more accurate estimates.


Query 5: Code and ImplementationSELECT STAGE.Stage_ID, STAGE.Est_Cost, STAGE.Actual_Cost, 1-

Abs([Est_Cost][Actual_Cost])/[Actual_Cost] AS Stage_RatingFROM STAGEGROUP BY STAGE.Stage_ID, STAGE.Est_Cost, STAGE.Actual_Cost, 1-

Abs([Est_Cost]-[Actual_Cost])/[Actual_Cost]HAVING (((2*[Actual_Cost])>=[Est_Cost]))UNION SELECT STAGE.Stage_ID, STAGE.Actual_Cost, STAGE.Est_Cost, 0 AS

StageRatingFROM STAGEGROUP BY STAGE.Stage_ID, STAGE.Actual_Cost, STAGE.Est_CostHAVING (((2*[Actual_Cost])<=[Est_Cost]));

cc

Normalization AnalysisFirst Normal Form

6. CUSTOMER (CusID, Proposal_ID12, RcusID, Fname, Lname, MI, Street_address, City, State, Zipcode)

8. PHONE_NUM (CusID6, Phone_number)9. EMAIL (CusID6, Email_address) Customer is in first normal form because all

the variables are atomic (indivisible) and single valued. The relation, PHONE_NUM, allows customers to have multiple phone numbers.


cc

Normalization AnalysisSecond Normal Form 16. STAGE(Stage_ID, CID3, Level,

Forecasted_Start_Date, Actual_Start_Date, Finish_Date, Est_cost, Actual_cost, Customer_rating, Sub_ID17)

STAGE is second normal form because there are

no partial dependencies. Stage_ID and CID3

make up a composite key so this case is non-trivial. All other attributes depend on the composite key entirely and not a part of it.


cc

Normalization AnalysisThird Normal Form 10. EMPLOYEE(ESSN, SupSSN, Lname,

Fname, MI, Salary, Email, Phone, Start_Date)

EMPLOYEE is in third normal form because every nonprime attribute is dependent on every CK in R and there are no transitive dependencies.


Questions?

“Home Architect & Contractor” Database

final presentation “home architect & contractor database”

Documents

client client

sign sign

type billboard

project summary1

type lstid

client descriptionmore

progress summary1

project summary2