June 23, 2024

Database Systems and Application Question and Answers

Database Systems
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Database Systems and Application Question and Answers


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1.


2.i). A. In following table first column is showing transactions which are in wait mode as the resource required by it are held by the transactions in second column.
Transaction no 1 Tansaction no 2 and 5 Transaction no 2
Transaction no 3 and 4 Transaction no 3
Transaction no 5 and 6 Transaction no 4
Transaction no 5 Transaction no 5
Transaction no 2 Transaction no 6 Transaction no 7 Transaction so 7 Transaction no 5 Transaction no 8 Transaction no 7 and 5 Let us assume each transaction takes t sec to complete. How long does transaction 8 have to wait for completion? Justify our answer with proper diagram and reason.

ANSWER:-

To determine how long Transaction 8 (Txn 8) has to wait for completion, we can construct a wait-for graph based on the given information. This graph will help us visualize the dependencies between transactions.

Here’s the wait-for graph:

Txn 1 —–> Txn 2, Txn 5
Txn 2 —–> Txn 3, Txn 4
Txn 3 —–> Txn 5, Txn 6
Txn 4 —–> Txn 5
Txn 5 —–> Txn 2
Txn 6 —–> Txn 7
Txn 7 —–> Txn 5
Txn 8 —–> Txn 7, Txn 5

In this graph, each arrow represents a “waits-for” relationship, indicating that a transaction on the left side of the arrow is waiting for the transaction(s) on the right side to release the resources it needs.

Now, let’s analyze the dependencies:

  1. Txn 8 is directly waiting for Txn 7 and Txn 5.
  2. Txn 7 is waiting for Txn 5.
  3. Txn 5 is waiting for Txn 2.

To calculate how long Txn 8 has to wait for completion, we need to follow the chain of dependencies:

Txn 8 –> Txn 7 –> Txn 5 –> Txn 2

Txn 2 is the last transaction in the chain, and it needs to complete before Txn 5 can release its resources, allowing Txn 7 and then Txn 8 to proceed.

Given that each transaction takes t seconds to complete, Txn 2 must complete first. Therefore, Txn 8 will have to wait for the completion of Txn 2.

So, the time Txn 8 has to wait for completion is t seconds, and this wait is justified by the dependency chain in the wait-for graph.


3.Consider a disk with a sector size of 512 bytes, 2000 tracks per surface, 50 sectors per track, five double-sided platters, and average seek time of 10 msec.

1. What is the capacity of a track in bytes? What is the capacity of each surface? What is the capacity of the disk?

2. How many cylinders does the disk have?

3. Give examples of valid block sizes. Is 256 bytes a valid block size? 2048? 51200?

4. If the disk platters rotate at 5400 rpm (revolutions per minute), what is the maximum rotational delay?

5. If one track of data can be transferred per revolution, what is the transfer rate?

ANSWER:-

1. Capacity of a track (in bytes) = sector size * sectors per track = 512 bytes * 50 sectors = 25,600 bytes.
Since there are five double-sided platters, each platter has two surfaces. Therefore, the capacity of each surface is the same as the capacity of a track, which is 25,600 bytes.
Total capacity of the disk (all surfaces) = capacity per surface * number of surfaces = 25,600 bytes * 2 * 5 = 256,000 bytes.

2. A cylinder is a set of tracks that are at the same position on each platter. In this case, there are 2000 tracks per surface, and you have 5 double-sided platters.

So, the total number of cylinders on the disk = 2000 (tracks per surface) * 5 (platters) = 10,000 cylinders.
3. A valid block size should be a multiple of the sector size (512 bytes). So, 256 bytes is not valid since it’s not a multiple of 512. Both 2048 bytes and 51200 bytes are valid block sizes because they are multiples of 512.

4.The maximum rotational delay is the time it takes for the desired sector to rotate under the disk head. It can be calculated as half of the time for one complete rotation (since on average, you will have to wait for half of the rotation).
Rotational delay = (1 / (rotational speed in revolutions per minute)) / 2 = (1 / 5400) / 2 ≈ 0.0000926 minutes, which is approximately 0.005556 seconds (or about 5.56 milliseconds).

5. The transfer rate is the amount of data transferred per unit of time. Since one track is transferred per revolution, and there are 2000 tracks per surface, each with a capacity of 25,600 bytes (as calculated in question 1), the transfer rate is:
Transfer rate = (track capacity) * (tracks per surface) * (rotational speed in revolutions per minute) = 25,600 bytes * 2000 * 5400 rpm = 2,764,800,000 bytes per minute (or about 2.64 GB per minute).
These answers should help you understand the characteristics of the disk system you described.


4.(A). i)Write an EXTERNAL XML Document Type Declaration (DTD) to define a

root-element as wase_wims_dbsa_2021 and

element-declarations having bits_id, name, department, address, phone. 

ii) Following query will List employee number, employee name, their department_number, department Name, project_number and project_location of all employees working as managers.

元 e.empus, e anamee.depine, d.diame, p.pno, p.location (Op.location=’billare’ (EMPLOYEE) deption.depe 617-2

DEPARTMENT à deptno-p.deptnum PROJECT)

Draw a Query tree for the above Relational Algebraic expression.

ANSWER:-

i) Here’s an external XML Document Type Declaration (DTD) to define the specified elements:

xml
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<!DOCTYPE wase_wims_dbsa_2021 [
<!ELEMENT wase_wims_dbsa_2021 (bits_id, name, department, address, phone)>
<!ELEMENT bits_id (#PCDATA)>
<!ELEMENT name (#PCDATA)>
<!ELEMENT department (#PCDATA)>
<!ELEMENT address (#PCDATA)>
<!ELEMENT phone (#PCDATA)>
]>

4.(B).B). Consider the data storage design where 5 disks are maintained in 3 different sites. Due to some technical issues Disk 2 and Disk 3 losses data. Given below RAID configuration, recover the data in disks 2 and 3.
Disk 4 is the redundant parity storage for Disks 1 and 2.
Disk 5 is the redundant parity storage for Disks 2 and 3.

ANSWER:- 

To recover the data in Disk 2 and Disk 3 using the provided RAID configuration, you can use the XOR operation between the data disks and their respective parity disks. Here’s how you can recover the data:

Let’s assume:

Data in Disk 1: 10101101 Data in Disk 4 (parity for Disks 1 and 2): 00110011 Data in Disk 5 (parity for Disks 2 and 3): 01101010

To recover Disk 2, you can perform the XOR operation between Disk 1, Disk 4, and Disk 5:

Disk 2 = Data in Disk 1 XOR Data in Disk 4 XOR Data in Disk 5

Disk 2 = 10101101 XOR 00110011 XOR 01101010

Now, calculate the XOR:

Disk 2 = 11110100

So, the recovered data in Disk 2 is 11110100.

To recover Disk 3, you can perform the XOR operation between Disk 2 (which we just recovered), Data in Disk 5, and Data in Disk 4:

Disk 3 = Disk 2 XOR Data in Disk 5 XOR Data in Disk 4

Disk 3 = 11110100 XOR 01101010 XOR 00110011

Now, calculate the XOR:

Disk 3 = 10101101

So, the recovered data in Disk 3 is 10101101.

Now, you have successfully recovered the data for Disk 2 and Disk 3:

Data in Disk 2: 11110100 Data in Disk 3: 10101101


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R


5.(A). Normalize the following table. Show all work and clearly indicate the primary and foreign keys.
R(elevator_no,building_no,building name,capacity, staff_no, first_name, last_name, date_examined)
Functional Dependencies:

1. elevator_no-building_no.capacity

2. building_no→ building_name

3. staff no→ first_name.last_name

4. elevator no.staff_no→ date examined

Normalize table R up to 3NF.

ANSWER:- To normalize the table R up to 3NF (Third Normal Form), we’ll follow these steps:

Step 1: Identify the Candidate Keys

A candidate key is a minimal set of attributes that uniquely identifies each tuple (row) in the table.
From the given functional dependencies, we can identify the following candidate keys:

{elevator_no, building_no}
{staff_no}
{elevator_no, staff_no}
Step 2: Eliminate Partial Dependencies (1NF)

Ensure that the table is in First Normal Form (1NF) by removing any partial dependencies.
Our table R already appears to be in 1NF, as there are no attributes with multiple values.

Step 3: Eliminate Transitive Dependencies (2NF)

Ensure that the table is in Second Normal Form (2NF) by removing any transitive dependencies.
2NF decomposition:
R1(elevator_no, building_no, capacity)
R2(building_no, building_name)
R3(staff_no, first_name, last_name)
R4(elevator_no, staff_no, date_examined)

Step 4: Eliminate Dependencies on Partial Candidate Keys (3NF)

Ensure that the table is in Third Normal Form (3NF) by removing any dependencies on partial candidate keys.
3NF decomposition:
R1(elevator_no, building_no, capacity)
R2(building_no, building_name)
R3(staff_no, first_name, last_name)
R4(elevator_no, staff_no, date_examined)

Now, let’s identify the primary and foreign keys for each normalized table:

R1:
Primary Key: {elevator_no, building_no}
Foreign Key(s): None

R2:
Primary Key: {building_no}
Foreign Key(s): None

R3:
Primary Key: {staff_no}
Foreign Key(s): None

R4:
Primary Key: {elevator_no, staff_no}
Foreign Key(s): None

These tables are now in 3NF, and the primary and foreign keys have been indicated for each of them.

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R

5.(B).Consider the following relations for a database that keeps track of business trips of Salespersons in a sales office:
SALESPERSON (SSN, Name, start year. Dept_no)
TRIP (SSN, From_city, To_city. Departure Date. Return_Date, Trip_ID)
EXPENSE (Trip ID. Account#, Amount)
Specify the following queries in relational algebra and SQL:
(i) Give the details (all attributes of TRIP) for trips that exceeded $2000 in expenses.

(ii) Print the SSN of salesman who took trips to Honolulu

ANSWER:- Give the details (all attributes of TRIP) for trips that exceeded $2000 in expenses.

Relational Algebra:

sql
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π * (σ Amount > 2000 (EXPENSE ⨝ TRIP))
SQL:

sql
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SELECT *
FROM TRIP
WHERE Trip_ID IN (SELECT Trip_ID FROM EXPENSE WHERE Amount > 2000);

(ii) Print the SSN of salesmen who took trips to Honolulu.

Relational Algebra:

sql
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π SSN (σ To_city = ‘Honolulu’ (TRIP))
SQL:

sql
Copy code
SELECT SSN
FROM TRIP
WHERE To_city = ‘Honolulu’;
These queries should give you the desired results in both relational algebra and SQL.


6.(A). Consider the three transactions T1, T2, and T3, and the schedules S1 and S2 given below. Draw the serializability (precedence) graphs for S1 and S2, and state whether each schedule is serializable or not. If a schedule is serializable, write down the equivalent serial schedule(s).

Tl:rl (P); r1 (R); wl (P);

T2: 12 (R): 12 (Q); w2 (R); w2 (R) ;

T3: 13 (P): 13 (Q); w3 (Q);

$1:rl (P): 12 rl (R); r3 (P); 13 (Q); wl (P); w3 (Q); Qw2 w2 (Q);

S2: r1 (P); 12 13 (P); r1 (R); r2 (Q); r3 (Q); wl (P); w2 (R); w3 (Q); w2 (Q);

a).Draw the serializability (precedence) graphs for S1 and S2.

b).State whether each schedule S1, S2 is serializable or not.

c).If a schedule is serializable, write down the equivalent serial schedule(s).

ANSWER:-

a) Let’s first draw the serializability (precedence) graphs for schedules S1 and S2.

For Schedule S1:

T1 –(rl(P))–> T2
|                             |
(rl(R)) (w2(R))
|                          |
(rl(P)) (w2(R))
|                          |
(rl(P)) (w2(Q))
|                          |
(wl(P)) |
|          |
(w3(Q)) |
| |
| |
| |
| |
| |
For Schedule S2:

T1 –(rl(P))–> T2 –(rl(R))–> T1
|                                                        |
(rl(R)) (r2(Q))
|                       |
(rl(P)) (r3(Q))
|                       |
(rl(P)) (wl(P))
|                        |
(wl(P)) (w3(Q))
|                        |
(w2(R)) (w2(Q))
|                        |
(w3(Q)) |
| |
| |
b) Now, let’s determine whether each schedule S1 and S2 is serializable:
Schedule S1 is not serializable because it contains a cycle in the precedence graph.
Schedule S2 is serializable because there are no cycles in the precedence graph.

c) Since Schedule S1 is not serializable, there are no equivalent serial schedules to provide. For Schedule S2, since it is serializable, the equivalent serial schedules would be all the possible orders in which the transactions could be executed without violating the precedence graph. These serial schedules could include different orderings of T1, T2, and T3 as long as the precedence relationships are maintained.


7.(A). XYZ Enterprises is a thriving e-commerce company that offers a wide range of products through its online platform. As the company continues to expand its product offerings, it recognizes the need to efficiently manage and present product information to its customers. To address this, XYZ Enterprises has decided to adopt an XML data model for its product catalog.
a) What are the challenges involved with XYZ enterprises?
b) What can be the possible solutions for these challenges?
c) What are the benefits achieved?

ANSWER:-

a) Challenges for XYZ Enterprises:

  1. Data Integration: Integrating diverse product data from multiple sources into a unified XML format can be complex.
  2. Data Accuracy: Ensuring the accuracy of product information, such as pricing, availability, and descriptions, is crucial.
  3. Scalability: As the company expands its product range, managing and updating XML data at scale can become challenging.
  4. Data Security: Protecting sensitive product data from unauthorized access or data breaches is a concern.
  5. User Experience: Presenting XML-based product information in a user-friendly and responsive manner is essential for customer satisfaction.

b) Possible Solutions:

  1. Data Integration Tools: Use data integration tools or platforms to streamline the process of importing and converting data into XML.
  2. Data Validation: Implement data validation checks to ensure the accuracy and consistency of product information.
  3. Content Management System (CMS): Invest in a CMS that supports XML, making it easier to manage and update product data.
  4. Encryption and Access Control: Implement robust security measures, including encryption and access control, to safeguard XML product data.
  5. Responsive Web Design: Develop a responsive website or app to provide an optimal user experience across devices.

c) Benefits Achieved:

  1. Efficient Data Management: XML allows structured data storage, making it easier to organize and manage product information.
  2. Data Consistency: With XML, maintaining consistent and standardized product data becomes more straightforward.
  3. Scalability: XML can handle a growing product catalog, accommodating new products seamlessly.
  4. Security: Implementing security measures protects sensitive product data from unauthorized access.
  5. Improved User Experience: XML-based product information can be presented in a user-friendly and customizable way, enhancing the customer experience.

7.(B). Construct a B+ tree for (11, 14, 17, 20, 27, 31, 41, 35, 29, 30, 38, 52) with n=4. Then perform the following operations in the order they are given:
a) Delete 38
b) Delete 31
c) Insert 25
d) Insert 23
e) Delete 27

ANSWER:-

To construct a B+ tree with a maximum degree of n=4 for the given data (11, 14, 17, 20, 27, 31, 41, 35, 29, 30, 38, 52), we will start by creating the initial tree. Afterward, we will perform the requested operations.

Initial B+ Tree:
27
/  |  \
14    20 31   41
/  \     |    /  \    | \
11 17 29 30 35 38 52
a) Delete 38: Delete 38 from the tree. Since it’s in a leaf node, we simply remove it. No further adjustments are needed. Updated Tree:
27
/ | \
14 20 31 41
/ \ |     /    \ | \
11 17 29 30 35 52
b) Delete 31: Delete 31 from the tree. This causes the leaf node to be underutilized, so we redistribute keys. Updated Tree:
27
/ | \
14 20 35 41
/ \   |     |   | \
11 17 29 30 52
c) Insert 25: Insert 25 into the tree. Since the leaf node is full, we need to split it and propagate the middle key upwards. Updated Tree:
27
/ | \
20 27 35 41
/ \   |     |   | \
14 17 25 29 30 52
/ \
11 20
d) Insert 23: Insert 23 into the tree. Since the leaf node is full, we need to split it and propagate the middle key upwards. Updated Tree:
27
/ | \
25 27 35 41
/ \  |    |   | \
20 23 29 30 52
/ \  |    |
14 17 20
/ \
11 17
e) Delete 27: Delete 27 from the tree. This causes the leaf node to be underutilized, so we redistribute keys. Updated Tree:
29
/ | \
25 30 35 41
/ \   |   | \
20 23 52
/ \
14 17
/ \
11 17