ন্যাশনাল টেলিকমিউনিকেশন মনিটরিং সেন্টার, স্বরাষ্ট্র মন্ত্রনালয়
Post: Assistant Maintenance Engineer (AME)
Subject: ICT, Exam Date: 06/06/2022, Exam Taker: BPSC
Given Data:
– Bandwidth per voice channel = 4 kHz
– Number of voice channels = 10
– Guard band between adjacent channels = 500 Hz
Each voice channel requires 4 kHz, so for 10 channels:
Total bandwidth for voice channels = 10 × 4 = 40 kHz
There are 9 guard bands (since guard bands exist between adjacent channels):
Total guard band = 9 × 500 = 4500 Hz = 4.5 kHz
Total Bandwidth = Bandwidth for channels + Total guard band
= 40 kHz + 4.5 kHz = 44.5 kHz
- Step 1: SYN (Synchronization)
- Client → Server: The client initiates the connection by sending a SYN packet to the server.
- The SYN packet contains a randomly generated sequence number.
- Step 2: SYN-ACK (Synchronization and Acknowledgment)
- Server → Client: The server responds with a SYN-ACK packet.
- The packet contains:
- ACK flag to acknowledge the client’s SYN.
- SYN flag with the server’s own sequence number.
- The server also advertises its window size and maximum segment size.
- Step 3: ACK (Acknowledgment)
- Client → Server: The client sends an ACK packet confirming receipt of the SYN-ACK.
- After this step, the TCP connection is established and data transmission begins.
- Step 1: SYN (Synchronization)
- Client → Server: Client server-এর কাছে একটি SYN packet পাঠিয়ে connection শুরু করে।
- এই SYN packet-এ একটি randomly generated sequence number থাকে।
- Step 2: SYN-ACK (Synchronization and Acknowledgment)
- Server → Client: Server, client-এর SYN পাওয়ার পর একটি SYN-ACK packet পাঠায়।
- এই packet-এ থাকে:
- ACK flag, যা client-এর SYN acknowledge করে।
- SYN flag, যা server-এর sequence number নির্দেশ করে।
- Server তার window size এবং maximum segment size-ও পাঠায়।
- Step 3: ACK (Acknowledgment)
- Client → Server: Client, server-এর SYN-ACK পাওয়ার পর একটি ACK packet পাঠায়।
- এই ধাপের পর TCP connection সম্পূর্ণভাবে establish হয় এবং data transmission শুরু হয়।
(i) Number of subnets that will be created
(ii) Number of hosts per subnet
(iii)Broadcast address in the network of the current IP address.(6 marks)
Given Data:
– IP Address: 172.16.68.230
– Original Subnet Mask: 255.255.0.0 (/16)
– New Subnet Mask: 255.255.240.0 (/20)
1. Number of Subnets: The original subnet mask is /16, and the new mask is /20. The difference of 4 bits (20 – 16) allows us to create 2⁴ = 16 subnets.
2. Number of Hosts per Subnet: With a /20 mask, there are 12 bits for host addresses. The number of usable hosts per subnet is 2¹² – 2 = 4094. The subtraction of 2 accounts for the network and broadcast addresses.
3. Broadcast Address: The subnet containing 172.16.68.230 starts at 172.16.64.0 (as /20 subnets increase in steps of 16). The last address in this range is 172.16.79.255, which is the broadcast address.
Final Results:
| Parameter | Value |
|---|---|
| Number of Subnets | 16 |
| Number of Hosts per Subnet | 4094 |
| Broadcast Address | 172.16.79.255 |
Stepwise Transformation of Infix to Postfix Expression
Given Infix Expression: ((5+6) * 2) / (9 – 3)
| Step | Symbol Read | Stack | Postfix Expression |
|---|---|---|---|
| 1 | ( | ( | |
| 2 | ( | (( | |
| 3 | 5 | (( | 5 |
| 4 | + | ((+ | 5 |
| 5 | 6 | ((+ | 5 6 |
| 6 | ) | ( | 5 6 + |
| 7 | * | (* | 5 6 + |
| 8 | 2 | (* | 5 6 + 2 |
| 9 | ) | 5 6 + 2 * | |
| 10 | / | / | 5 6 + 2 * |
| 11 | ( | / ( | 5 6 + 2 * |
| 12 | 9 | / ( | 5 6 + 2 * 9 |
| 13 | – | / (- | 5 6 + 2 * 9 |
| 14 | 3 | / (- | 5 6 + 2 * 9 3 |
| 15 | ) | / | 5 6 + 2 * 9 3 – |
| 16 | End | 5 6 + 2 * 9 3 – / |
Final Postfix Expression: 5 6 + 2 * 9 3 – /
| Step | Sorted List |
|---|---|
| Convert to all same digits | [321, 150, 235, 065, 573, 789, 928, 542] |
| Sorting by 1’s place | [150, 321, 542, 573, 235, 065, 928, 789] |
| Sorting by 10’s place | [321, 928, 235, 542, 150, 065, 573, 789] |
| Sorting by 100’s place | [065, 150, 235, 321, 542, 573, 789, 928] |
Final Sorted List: [65, 150, 235, 321, 542, 573, 789, 928]
