To satisfy the disjoint constraint in the the specialization {STAFF, FUCULTY, STUDENT
ASSISTANT} of EMPLOYEE entity type. Just NOT NULL constraint CHECK constraint Before insert or update Trigger Foreign key with a NOT NULL constraint .
To satisfy the total completeness constraint in the specialization {STAFF, FUCULTY, STUDENT
ASSISTANT} of EMPLOYEE entity type NOT NULL constraint In the application layer (required field) Before insert or update Trigger Both a and b.
After mapping the above EER into relational DB schema, the STUDENT_ASSISTANT relation
will include: Two foreign keys referencing the EMPLOYEE and PERSON relations. One foreign key referencing the EMPLOYEE relation. One foreign key referencing the STUDENT relation. Two foreign keys referencing the EMPLOYEE and STUDENT relations.
Assuming the structural constraint (4, 35) that is defined on the participation of
DEPARTMENT entity type in the relationship WORKS FOR with the EMPLOYEE entity type,
answer questions 1 and 3.
It means that Each department must have at least 4 employees and at most 35 employees Each department must have more than 4 and less than 35 employees. Each department must have more than 4 employees. Each department must have less than 35 employees.
Respecting insertion, updating and deletion The MIN part of constraint (5,20) can be satisfied
in: Just Database layer Just Application layer Both of database and application layers. Non of the above.
The MAX part of constraint (4, 35) can be satisfied in: Database layer using stored procedures Database layer using triggers Database layer using index Database layer using stored functions.
To satisfy the participation (1,1) of EMPLOYEE entity type in relationship WORKS FOR: Just NOT NULL constraint CHECK constraint ) Foreign key with a Unique constraint Foreign key with a NOT NULL constraint .
To simplify a complex query about detailed employees data combined and compared with
Projects totals we can use: Views of aggregated data Stored Functions Stored procedures Both a and b .
To simplify a complex query about detailed employees data combined and compared with
Projects and departments totals we can use: Views of aggregated data Stored Functions Stored procedures Both a and b .
In the shown relation scheme emp_dept: There is insertion anomaly only. There are insertion and modification anomalies. There are insertion, modification, and deletion anomalies. There are no update anomalies.
Joining relations on attributes that are not foreign key, primary key combination: Always produces spurious tuples give valid information. Always produces spurious tuples give invalid information. Might produce spurious tuples give valid information. Might produce spurious tuples give invalid information.
A functional dependency X → Y between two sets of attributes X and Y that are subsets of R
specifies a constraint on the possible tuples that can form a relation state r of R. The
constraint is that: for any two tuples t1 and t2 in r that have t1[X] = t2[X], they must also have t1[Y] = t2[Y] and vice versa. have t1[X] = t2[X], they must also have t1[Y] ≠t2[Y] . have t1[X] = t2[X], they must also have t1[Y] = t2[Y] . have t1[Y] = t2[Y], they must also have t1[X] = t2[X].
A functional dependency X → Y between two sets of attributes X and Y that are subsets of R
specifies a constraint on the possible tuples that can form a relation state r of R. The
constraint is that: for any two tuples t1 and t2 in r that have t1[X] = t2[X], they must also have t1[Y] = t2[Y] and vice versa. have t1[X] = t2[X], they must also have t1[Y] ≠t2[Y] . have t1[X] = t2[X], they must also have t1[Y] = t2[Y] and may vice versa. have t1[Y] = t2[Y], they must also have t1[X] = t2[X].
if whenever two tuples t1 and t2 with t1[X] = t2[X], they also have t1[Y] = t2[Y] and vice versa
then: X is a key attribute and Y is a non-key attribute. X is a key attribute and Y is a partial key attribute. Both of X and Y are key attributes. None of the above.
First normal form (1NF): Avoids update anomalies. Avoids partial functional dependencies. Avoids transitive functional dependencies. Avoid multivalued and composite attributes.
Second normal form (2NF): Avoids update anomalies. Avoids partial functional dependencies. Avoids transitive functional dependencies All of the above.
Third normal form (3NF): It is not necessarily in 2NF. It must be in 1NF and 2NF Avoids transitive functional dependencies. BOTH b and c.
Third normal form (3NF): Avoids composite attributes Avoids partial functional dependencies. Avoids transitive functional dependencies. All of the above.
Which one of the following is true about primary index : It is dense index It is defined on anon-key attribute It has number of rows equal the number of data file blocks. None of the above.
Which one of the following is true about primary index : It is dense index It is defined on anon-key attribute It is a space index. None of the above.
Which one of the following is true about primary index : It is dense index It is defined on a key attribute that the data file is physically ordered on it It can be defined on anon-key attribute None of the above.
Which one of the following is true about clustering index : It is dense index It is defined on anon-key attribute that the data file is physically ordered on it It has number of rows equal the number of data file blocks None of the above.
Which one of the following is true about clustering index : It is a dense index It is defined on a key attribute that the data file is physically ordered on it It has number of its rows equal the number of distinct index field values All of the above.
Which one of the following is true about clustering index : It is a dense index It is defined on a key attribute that the data file is physically ordered on it It is a space index None of the above.
Every time attribute A appears, it is matched with the same value of attribute B, but not
the same value of attribute C. Therefore, it is true that: A → B. A → C A → (B,C) (B,C) → A.
The different classes of relations created by the technique for preventing modification
anomalies are called: normal forms. referential integrity constraints. functional dependencies. None of the above is correct.
Which one of the following is true about secondary index : It is a dense index if it defined on a non-key attribute It can be defined on a key or non-key attribute It is a space index if it defined on a key attribute None of the above.
For questions 27 to 29, consider the relation R = {A, B, C, D, E, F, G, H, I, J} and the set of
functional dependencies F = { {A}→{D, E}, {B}→{F}, {F}→{G,H}, {D}→{I, J}, {A, B}→{C} }.
and the decomposition D = {R1, R2, R3} for the relation schema R in which
R1 = {A, B, C, D, E}, R2 = {B, F, G, H}, R3 = {D, I, J}
Decomposition D, with respect to F, has the dependency preservation and lossless join properties has the dependency preservation but not lossless join properties. has the lossless join but not dependency preservation properties. does not have lossless join nor dependency preservation properties.
For questions 27 to 29, consider the relation R = {A, B, C, D, E, F, G, H, I, J} and the set of
functional dependencies F = { {A}→{D, E}, {B}→{F}, {F}→{G,H}, {D}→{I, J}, {A, B}→{C} }.
and the decomposition D = {R1, R2, R3} for the relation schema R in which
R1 = {A, B, C, D, E}, R2 = {B, F, G, H}, R3 = {D, I, J}
R1 in decomposition D is in 1NF 2NF 3NF BCNF.
For questions 27 to 29, consider the relation R = {A, B, C, D, E, F, G, H, I, J} and the set of
functional dependencies F = { {A}→{D, E}, {B}→{F}, {F}→{G,H}, {D}→{I, J}, {A, B}→{C} }.
and the decomposition D = {R1, R2, R3} for the relation schema R in which
R1 = {A, B, C, D, E}, R2 = {B, F, G, H}, R3 = {D, I, J}
R2 in decomposition D is in 1NF 2NF 3NF BCNF.
Files with variable length records might be produced by Storing records with optional fields. Storing records with variable-length fields Storing records of different types. All the above.
. A functional dependency between two or more non-key attributes is called Transitive dependency Partial transitive dependency Functional dependency Partial functional dependency.
What is ACID properties of Transactions? Atomicity, Consistency, Isolation, Database Atomicity, Consistency, Isolation, Durability Atomicity, Consistency, Inconsistent, Durability Automatically, Concurrency, Isolation, Durability.
. A transaction completes its execution is said to be Saved Loaded Rolled Committed.
For questions 29 to 36, consider a file for employees has 20,000 of unspanned fixed length
records of size 500 bytes stored on a disk with block size 1024 bytes. The field emp_no is the
primary key. An index is constructed on another key field emp_civil_id with 10 bytes length with
a block pointer 5 bytes
How many blocks are needed to store the file? 9,766 blocks with blocking factor 2.048. 6,667 blocks with blocking factor 3 10,000 blocks with blocking factor 2. 1,000 blocks with blocking factor 20.
For questions 29 to 36, consider a file for employees has 20,000 of unspanned fixed length
records of size 500 bytes stored on a disk with block size 1024 bytes. The field emp_no is the
primary key. An index is constructed on another key field emp_civil_id with 10 bytes length with
a block pointer 5 bytes.
If the file is ordered on emp_id, it can be searched for a specific emp_id using Binary search with average 14 block reads. Binary search with average 13 block reads. Sequential search with average 5,000 block reads. Sequential search with average 500 block reads.
If the file is hashed on emp_no as a hash key and the first records whose hash leading binary
digits: 0010, 0110, 0011 and 1010 are loaded into the file in the given order. Suppose the
bucket size is one block, using the extendible hashing: There are 3 buckets with local depths 2, 2, 2 and global depth 2. There are 2 buckets with local depths 2, 2 and global depth 2. There are 3 buckets with local depths 2, 2, 1 and global depth 2. There are 3 buckets with local depths 2, 1, 1 and global depth 2.
The difference between an index and the directory of extendible/dynamic hashing is: Nothing An index is much larger than the directory. Searching in a directory is based on the hash value of the field while searching in an
index is based on the field value itself. b) and c).
for the employee file described above, suppose that the index emp_civil_id is a B+ tree
index of order p=3; after adding 4 records with emp_civil_id 5, 25, 4, and 29 with that
order, the final tree will have: 1 internal node with value 5 and 2 leaf nodes. 1 internal node with value 25 and 2 leaf nodes. 1 internal node with value 5 and 3 leaf nodes. 1 internal node with value 25 and 3 leaf nodes.
for the employee file described above, suppose that the index emp_civil_id is a B tree
index of order p=3; after adding 4 records with emp_civil_id 5, 25, 4, and 29 with that
order, the final tree will have: 1 internal node with value 5 and 2 leaf nodes (4,5) , (25,29 ). 1 internal node with value 25 and 2 leaf nodes (4,5) , (25,29 ). 1 internal node with value 5 and 2 leaf nodes (4, ) , (25, 29 ). 1 internal node with value 25 and 2 leaf nodes (4, ) , (25,29 ).
Suppose the employees file is ordered on the primary key, if the index on emp_civil_id is
a single-level one, it is: A secondary dense index using block anchors A primary dense index using block anchors. A secondary dense index using record pointers. A secondary non-dens index using record pointers.
Consider the following interleaved transactions and choose the problem that can
happen: Dirty read problem. Wrong summary problem. Lost update problem Update anomalies problem.
Consider the following interleaved transactions and choose the problem that can
happen Dirty read problem Wrong summary problem. Lost update problem Update anomalies problem.
For some relations, changing the data can have undesirable consequences called: referential integrity constraints modification anomalies normal forms transitive dependencies.
Consider the following interleaved transactions and choose the problem that can happen
if T1 finaly makes a rollback: Dirty read problem Wrong summary problem. Lost update problem Update anomalies problem.
If every non-key attribute is functionally dependent on the primary key, the relation will
be in First Normal Form Second Normal Form Third Normal Form Fourth Formal Form.
In DBMS FD stands for_______ Facilitate data Functional data Facilitate dependency Functional dependency.
A relation that has no partial dependencies is in which normal form First Second Third BCNF.
When the values in one or more attributes being used as a foreign key must exist in
another set of one or more attributes in another table, we have created a(n:( transitive dependency. insertion anomaly. referential integrity constraint. normal form.
A functional dependency is a relationship between or among: tables rows relations attributes.
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