Rupali Bank PLC
Post: Assistant Network Engineer
Exam Date: 15-01-2022, Exam Taker: CTI
(i) What is the capacity of disk pack?
(ii) If the format overhead is 32 bytes per sector, what is the formatted disk space?
(iii)If the disk is rotating at 3600 RPM, what is the data transfer rate?
What is the capacity of the disk pack?
Answer:
Capacity of disk pack = Total number of surfaces × Number of tracks per surface × Number of sectors per track × Number of bytes per sector
= 16 × 128 × 256 × 512 bytes = 256 MB
If the format overhead is 32 bytes per sector, what is the formatted disk space?
Answer:
Total number of sectors = Total number of surfaces × Number of tracks per surface × Number of sectors per track
= 16 × 128 × 256 sectors = 219 sectors
Number of bits required to address the sector = 19 bits
Formatting overhead = Total number of sectors × Overhead per sector
= 219 × 32 bytes = 219 × 25 bytes = 224 bytes = 16 MB
Formatted disk space = Total disk space – Formatting overhead
= 256 MB – 16 MB = 240 MB
If the disk is rotating at 3600 RPM, what is the data transfer rate?
Answer:
Number of rotations in one second = (3600 / 60) rotations/sec = 60 rotations/sec
Data transfer rate = Number of heads × Capacity of one track × Number of rotations in one second
= 16 × (256 × 512 bytes) × 60
= 24 × 28 × 29 × 60 bytes/sec
= 60 × 221 bytes/sec = 120 Mbps
2. Job arrival time and execution time of Operating system tasks table is given, find out
| Job | Arrival Time | Execution Time |
|---|---|---|
| A | 0 | 10 |
| B | 3 | 7 |
| C | 5 | 3 |
Average waiting time
Transmission rate = (1.5 Mbps) / 16 = (1.5 x 106) / 16 Kbps
Now, the time to send data is:
Time = file size / transmission rate + delay
Substituting the values:
Time = (2 × 500,000 × 16) / (1.5 × 106) + 300 / 1000 = 10.7 sec
In order: 20, 30,35, 40,45, 50, 55, 65, 75
Preorder: 50, 40, 30, 20, 35, 45, 65, 55, 70
50
/ \
40 65
/ \ / \
30 45 55 75
/ \
20 35
Procedure sumNodes(Node* root)
int sumNodes(Node* root)
{
if (root == NULL)
return 0;
return root->data + sumNodes(root->left) + sumNodes(root->right);
}Sample Input: 192.168.0.0
Sample Output: Class C
#include <iostream>
#include <string>
#include <sstream>
using namespace std;
// Function to determine the class of an IPv4 address
string getIPAddressClass(const string& ipAddress) {
stringstream ss(ipAddress);
string token;
// Get the first octet
getline(ss, token, '.');
// Convert first octet to integer
int firstOctet = stoi(token);
// Determine class
if (firstOctet >= 1 && firstOctet <= 127) {
return "Class A";
}
else if (firstOctet >= 128 && firstOctet <= 191) {
return "Class B";
}
else if (firstOctet >= 192 && firstOctet <= 223) {
return "Class C";
}
else if (firstOctet >= 224 && firstOctet <= 239) {
return "Class D";
}
else if (firstOctet >= 240 && firstOctet <= 255) {
return "Class E";
}
else {
return "Invalid IP address";
}
}
int main() {
string ipAddress;
cin >> ipAddress;
string ipAddressClass = getIPAddressClass(ipAddress);
cout << "IP Address: " << ipAddress
<< " belongs to " << ipAddressClass << endl;
return 0;
}
First, calculate the number of required bits for each subnet:
- For 4000 hosts: Needs at least 12 bits for host portion (212 = 4096).
- For 2000 hosts: Needs at least 11 bits for host portion (211 = 2048).
- For 6000 hosts: Needs at least 13 bits for host portion (213 = 8192).
- For 8000 hosts: Needs at least 13 bits for host portion (213 = 8192).
Subnet Calculations:
Department C (6000 hosts)
Subnet Mask: 172.16.0.0/19 (because 32 – 13 = 19 bits for network portion).
Subnet Range: 172.16.0.0 – 172.16.31.255
First IP: 172.16.0.1
Last IP: 172.16.31.254
Department D (8000 hosts)
Subnet Mask: 172.16.32.0/19 (same as above).
Subnet Range: 172.16.32.0 – 172.16.63.255
First IP: 172.16.32.1
Last IP: 172.16.63.254
Department A (4000 hosts)
Subnet Mask: 172.16.64.0/20 (because 32 – 12 = 20 bits for network portion).
Subnet Range: 172.16.64.0 – 172.16.79.255
First IP: 172.16.64.1
Last IP: 172.16.79.254
Department B (2000 hosts)
Subnet Mask: 172.16.80.0/21 (because 32 – 11 = 21 bits for network portion).
Subnet Range: 172.16.80.0 – 172.16.95.255
First IP: 172.16.80.1
Last IP: 172.16.95.254
(i) Find out the logical expression in Sum of product for 2bit binary input.
i) For 2 bit binary, we can find 00, 01, 10, 11 and the equivalent decimal value is
0,1,2,3.
Now, For Binary= 00 (Decimal= 0)= abcdef is high
For Binary= 01 (Decimal=1)= bc is high
For Binary= 10 (Decimal=2) = abged is high
For Binary=11 (Decimal=3)= abgcd is high
(i) Draw ER diagram
(ii) Convert the ER diagram to relations(Table)
ER Diagram → Relational Tables Conversion
Below is the relational schema (tables) derived from the ER diagram.
1. Coach Table
Coach(Coach_ID, Coach_name)
* Primary Key: Coach_ID
2. Team Table
Team(Team_ID, Team_name, Coach_ID)
* Primary Key: Team_ID
* Foreign Key: Coach_ID → Coach(Coach_ID)
3. Player Table
Player(Player_ID, Player_name, Player_position, Team_ID)
* Primary Key: Player_ID
* Foreign Key: Team_ID → Team(Team_ID)
4. Stadium Table
Stadium(Stadium_ID, Stadium_name, Stadium_location)
* Primary Key: Stadium_ID
5. Game Table
Game(Game_no, Game_name, Game_time)
* Primary Key: Game_no
6. Match Table
Match(Match_ID, Match_date, Match_result, Stadium_ID, Game_no)
* Primary Key: Match_ID
* Foreign Key:
* Stadium_ID → Stadium(Stadium_ID)
* Game_no → Game(Game_no)
(a) Increased High Costs
(b) Draining Battery of devices
(c) Increased infrastructure development cost
8. Below tables are given, Employee (employee_id, name, salary, department) Leave
(employee_id, date, reason, no_leaves) Holiday (Date, description)
SELECT employee_id, COUNT(employee_id)
FROM Leave
GROUP BY employee_id;
SELECT *
FROM employee
WHERE employee_id IN (
SELECT employee_id
FROM leave
GROUP BY employee_id
HAVING COUNT(employee_id) > 5
);
