# The University of Texas

CS 327e

Fall 2001

### Lab 1 Key

1a. For each table, identify, where possible,

a. The primary key

Table CHARTER: CHAR_TRIP

Table AIRCRAFT: AC_NUM

Table MODEL: MOD_CODE

Table PILOT: EMP_NUM

Table EMPLOYEE: EMP_NUM

Table CUSTOMER: CUS_CODE

b. A super key

Table CHARTER: CHAR_TRIP + CHAR_DATE

Table AIRCRAFT: AC_NUM + MOD-CODE

Table MODEL: MOD_CODE + MOD_NAME

Table PILOT: EMP_NUM + PIL_LICENSE

Table EMPLOYEE: EMP_NUM + EMP_DOB

Table CUSTOMER: CUS_CODE + CUS_LNAME

c. A candidate key

Table CHARTER: No practical candidate key is available. For example,

### CHAR_DATE + CHAR_DESTINATION + AC_NUMBER + CHAR_PILOT + CHAR_COPILOT

will still not necessarily yield unique matches, because it is possible to fly an aircraft to the same destination twice on one date with the same pilot and copilot. You could, of course, present the argument that the combination of all the attributes would yield a unique outcome.

Table AIRCRAFT: See the previous discussion

Table MODEL: See the previous discussion

Table PILOT: See the previous discussion

Table EMPLOYEE: See the previous discussion. But Perhaps the combination of

### EMP_LNAME + EMP_FNAME + EMP_INITIAL + EMP_DOB

will yield an acceptable candidate key.

Table CUSTOMER: See the previous discussion

d. The foreign key(s)

Table CHARTER: CHAR_PILOT (references PILOT)

CHAR_COPILOT (references PILOT)

AC_NUMBER (references AIRCRAFT)

CUS_CODE (references CUSTOMER)

Table AIRCRAFT: MOD_CODE

Table MODEL: NONE

Table PILOT: EMP_NUM (references EMPLOYEE)

Table EMPLOYEE: NONE

Table CUSTOMER: NONE

e. A secondary key.

Table CHARTER: CHAR_DATE + AC_NUMBER + CHAR_DESTINATION

Table AIRCRAFT: MOD_CODE

Table MODEL: MOD_MANUFACTURER + MOD_NAME

Table PILOT: PIL_LICENSE + PIL_MED_DATE

Table EMPLOYEE: EMP_LNAME + EMP_FNAME + EMP_DOB

Table CUSTOMER: CUS_LNAME + CUS_FNAME + CUS_PHONE

1b. Create the Entity Relationship diagram. Hint: Take a look at the table contents. You will discover that an AIRCRAFT can be used to fly many CHARTER trips, but that each CHARTER trip is flown by one AIRCRAFT. Similarly, you will discover that a MODEL references many AIRCRAFT, but each AIRCRAFT references a single MODEL, and so on.

Teacher's Note: Database design properly precedes the creation of table structures and contents. However, we have found that our students generally find it much easier to understand design concepts and requirements after they have examined some data. By observing actual attribute values, the abstract notion of relationships becomes "real" and, therefore, more easily understood. So, throughout this chapter and in the chapter's question and problem sets, we have shown the table structures and contents first and then the E-R diagram is constructed to reflect the available information. Actually, this process is too often typical of real world projects in which "documentation after the fact" and "reverse engineering" are an unfortunate reality. Do stress that proper design always begins at the conceptual level and that table structures and contents are done at the implementation stage... after the design is judged to be correct. We will follow the appropriate procedures when we examine design issues in greater detail in Chapter 4.

After carefully exploring the CH2_AVIA database's contents, note the following relationships:

¨ a CUSTOMER can request many CHARTER trips, but each CHARTER trip is requested by one CUSTOMER.

¨ Each CHARTER trip requires one AIRCRAFT, but an AIRCRAFT may be used to fly many different CHARTER trips.

¨ Each AIRCRAFT is of a certain MODEL and, because the charter company may have many aircraft of a given model, each MODEL may reference many AIRCRAFT.

¨ a PILOT may pilot many CHARTER trips, but each CHARTER trip is piloted by one PILOT.

¨ a PILOT may serve as a copilot on a CHARTER trip.

¨ each PILOT is an EMPLOYEE (but not all EMPLOYEEs are PILOTs!).

The E-R diagram reflects this list of relationships:

Teacher's Note: The orientation of the E-R diagram's components is immaterial. Show the students that the E-R diagram is just as valid if the entities are re-arranged. The key is the proper identification of all the relationships. However, it is considered to be good form to avoid crossing relationship lines.

The above diagram is there for reference only (not required for lab1).

1c. Produce the output shown in Figure P3.5 for aircraft 2778V. Note that this output includes data from the CHARTER, EMPLOYEE, and MODEL tables.

Hint: Two of the joins pass through other tables. For example, the “connection” between CHARTER and MODEL requires the existence of AIRCRAFT, because the CHARTER table does not contain a foreign key to MODEL. However, CHARTER does contain a foreign key to AIRCRAFT, which contains a foreign key to MODEL.

Figure P3.5 Selected Output from the CHARTER, EMPLOYEE,

and MODEL Tables

SELECT CHAR_DATE,CHARTER.AC_NUMBER,MOD_NAME,EMP_LNAME

FROM CHARTER,AIRCRAFT,MODEL,PILOT,EMPLOYEE

WHERE CHARTER.AC_NUMBER = AIRCRAFT.AC_NUMBER AND

AIRCRAFT.MOD_CODE = MODEL.MOD_CODE AND

PILOT.EMP_NUM = EMPLOYEE.EMP_NUM AND

CHARTER.CHAR_PILOT = PILOT.EMP_NUM AND

CHARTER.AC_NUMBER = '2778V';

1d. Create the SQL query that will produce a list of customers who have an unpaid balance. The required output is shown in Figure P3.11. Note that the balances are listed in descending order.

FIGURE P3.11 A List of Customers with Unpaid Balances

SELECT CUS_LNAME, CUS_FNAME, CUS_INITIAL, CUS_BALANCE

FROM CUSTOMER

WHERE CUS_BALANCE <> 0

ORDER BY CUS_BALANCE DESC;

1e. Using the CHARTER table as the source, group the aircraft data, and then use the SQL functions to produce the output shown in Figure 3.13. (We used utility software to modify the headers, so your headers may have a different “look.”)

FIGURE P3.13 The Aircraft Data Summary Statement

SELECT CHARTER.AC_NUMBER, COUNT(CHARTER.AC_NUMBER),

SUM(CHAR_DISTANCE), AVG(CHAR_DISTANCE),

SUM(CHAR_HOURS_FLOWN), AVG(CHAR_HOURS_FLOWN)

FROM AIRCRAFT, CHARTER

### WHERE AIRCRAFT.AC_NUMBER = CHARTER.AC_NUMBER

GROUP BY CHARTER.AC_NUMBER;

Lab1

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