Computer Science Engineering (CSE) Exam > Computer Science Engineering (CSE) Questions > Directions: A 3000 km long trunk is used to t...
Start Learning for Free
Directions: A 3000 km long trunk is used to transmit frames using a Go-Back-N protocol. The propagation speed is 6 μ sec/km and the trunk data rate is 1.5 Mbps. We ignore the time taken to receive the bits in the acknowledgement. Frame size is 64 Bytes.
What is the maximum number of bits of the sequence number?
Correct answer is '7'. Can you explain this answer?
| FREE This question is part of | Download PDF Attempt this Test |
Most Upvoted Answer
Directions: A 3000 km long trunk is used to transmit frames using a Go...
× 10^5 km/s and the frame size is 1000 bits. The transmission rate is 1 Mbps.
1. Calculate the total time required for a frame to travel from the sender to the receiver.
To calculate the total time required for a frame to travel from the sender to the receiver, we need to consider both the transmission time and the propagation time.
Transmission Time:
The transmission time can be calculated using the formula: Transmission Time = Frame Size / Transmission Rate.
In this case, the frame size is 1000 bits and the transmission rate is 1 Mbps (1 million bits per second).
So, the transmission time = 1000 bits / 1 Mbps = 0.001 seconds.
Propagation Time:
The propagation time can be calculated using the formula: Propagation Time = Distance / Propagation Speed.
In this case, the distance is 3000 km and the propagation speed is 6 × 10^5 km/s (600,000 km/s).
So, the propagation time = 3000 km / (6 × 10^5 km/s) = 5 seconds.
Total Time:
The total time required for a frame to travel from the sender to the receiver is the sum of the transmission time and the propagation time.
Total Time = Transmission Time + Propagation Time = 0.001 seconds + 5 seconds = 5.001 seconds.
Therefore, it takes approximately 5.001 seconds for a frame to travel from the sender to the receiver.
2. Calculate the maximum number of frames that can be in transit at any given time.
The maximum number of frames that can be in transit at any given time is determined by the window size of the Go-Back-N protocol. The window size represents the maximum number of unacknowledged frames that can be sent by the sender before waiting for acknowledgments.
The window size can be calculated using the formula: Window Size = (Propagation Time / Transmission Time) + 1.
In this case, the propagation time is 5 seconds and the transmission time is 0.001 seconds (calculated in question 1).
So, the window size = (5 seconds / 0.001 seconds) + 1 = 5001.
Therefore, the maximum number of frames that can be in transit at any given time is 5001.
1. Calculate the total time required for a frame to travel from the sender to the receiver.
To calculate the total time required for a frame to travel from the sender to the receiver, we need to consider both the transmission time and the propagation time.
Transmission Time:
The transmission time can be calculated using the formula: Transmission Time = Frame Size / Transmission Rate.
In this case, the frame size is 1000 bits and the transmission rate is 1 Mbps (1 million bits per second).
So, the transmission time = 1000 bits / 1 Mbps = 0.001 seconds.
Propagation Time:
The propagation time can be calculated using the formula: Propagation Time = Distance / Propagation Speed.
In this case, the distance is 3000 km and the propagation speed is 6 × 10^5 km/s (600,000 km/s).
So, the propagation time = 3000 km / (6 × 10^5 km/s) = 5 seconds.
Total Time:
The total time required for a frame to travel from the sender to the receiver is the sum of the transmission time and the propagation time.
Total Time = Transmission Time + Propagation Time = 0.001 seconds + 5 seconds = 5.001 seconds.
Therefore, it takes approximately 5.001 seconds for a frame to travel from the sender to the receiver.
2. Calculate the maximum number of frames that can be in transit at any given time.
The maximum number of frames that can be in transit at any given time is determined by the window size of the Go-Back-N protocol. The window size represents the maximum number of unacknowledged frames that can be sent by the sender before waiting for acknowledgments.
The window size can be calculated using the formula: Window Size = (Propagation Time / Transmission Time) + 1.
In this case, the propagation time is 5 seconds and the transmission time is 0.001 seconds (calculated in question 1).
So, the window size = (5 seconds / 0.001 seconds) + 1 = 5001.
Therefore, the maximum number of frames that can be in transit at any given time is 5001.
Free Test
FREE
| Start Free Test |
Community Answer
Directions: A 3000 km long trunk is used to transmit frames using a Go...
Propagation time, TP = 6 μ sec/km × 3000 km = 18 ms
Speed = 1.5 Mbps
Transmission time for a frame, Tf = 64 byte × 8/1.5 Mbps
= 0.3 ms
Total transmission time for a frame and its acknowledgement,
TF = 2 × TP + Tf
TF = 2 × 18 ms + 0.3 ms = 36.3 ms
Size of window = 36.3/0.3 = 121
Hence, 27 = 128 > 121, so the maximum numbers of bits for the sequence is 7.
Speed = 1.5 Mbps
Transmission time for a frame, Tf = 64 byte × 8/1.5 Mbps
= 0.3 ms
Total transmission time for a frame and its acknowledgement,
TF = 2 × TP + Tf
TF = 2 × 18 ms + 0.3 ms = 36.3 ms
Size of window = 36.3/0.3 = 121
Hence, 27 = 128 > 121, so the maximum numbers of bits for the sequence is 7.
Attention Computer Science Engineering (CSE) Students!
To make sure you are not studying endlessly, EduRev has designed Computer Science Engineering (CSE) study material, with Structured Courses, Videos, & Test Series. Plus get personalized analysis, doubt solving and improvement plans to achieve a great score in Computer Science Engineering (CSE).
|
Explore Courses for Computer Science Engineering (CSE) exam
|
|
Similar Computer Science Engineering (CSE) Doubts
Top Courses for Computer Science Engineering (CSE)View all
Directions: A 3000 km long trunk is used to transmit frames using a Go-Back-N protocol. The propagation speed is 6 μ sec/kmand the trunk data rate is 1.5 Mbps. We ignore the time taken to receive the bits in the acknowledgement. Frame size is 64 Bytes.What is the maximum number of bits of the sequence number?Correct answer is '7'. Can you explain this answer?
Question Description
Directions: A 3000 km long trunk is used to transmit frames using a Go-Back-N protocol. The propagation speed is 6 μ sec/kmand the trunk data rate is 1.5 Mbps. We ignore the time taken to receive the bits in the acknowledgement. Frame size is 64 Bytes.What is the maximum number of bits of the sequence number?Correct answer is '7'. Can you explain this answer? for Computer Science Engineering (CSE) 2024 is part of Computer Science Engineering (CSE) preparation. The Question and answers have been prepared according to the Computer Science Engineering (CSE) exam syllabus. Information about Directions: A 3000 km long trunk is used to transmit frames using a Go-Back-N protocol. The propagation speed is 6 μ sec/kmand the trunk data rate is 1.5 Mbps. We ignore the time taken to receive the bits in the acknowledgement. Frame size is 64 Bytes.What is the maximum number of bits of the sequence number?Correct answer is '7'. Can you explain this answer? covers all topics & solutions for Computer Science Engineering (CSE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Directions: A 3000 km long trunk is used to transmit frames using a Go-Back-N protocol. The propagation speed is 6 μ sec/kmand the trunk data rate is 1.5 Mbps. We ignore the time taken to receive the bits in the acknowledgement. Frame size is 64 Bytes.What is the maximum number of bits of the sequence number?Correct answer is '7'. Can you explain this answer?.
Directions: A 3000 km long trunk is used to transmit frames using a Go-Back-N protocol. The propagation speed is 6 μ sec/kmand the trunk data rate is 1.5 Mbps. We ignore the time taken to receive the bits in the acknowledgement. Frame size is 64 Bytes.What is the maximum number of bits of the sequence number?Correct answer is '7'. Can you explain this answer? for Computer Science Engineering (CSE) 2024 is part of Computer Science Engineering (CSE) preparation. The Question and answers have been prepared according to the Computer Science Engineering (CSE) exam syllabus. Information about Directions: A 3000 km long trunk is used to transmit frames using a Go-Back-N protocol. The propagation speed is 6 μ sec/kmand the trunk data rate is 1.5 Mbps. We ignore the time taken to receive the bits in the acknowledgement. Frame size is 64 Bytes.What is the maximum number of bits of the sequence number?Correct answer is '7'. Can you explain this answer? covers all topics & solutions for Computer Science Engineering (CSE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Directions: A 3000 km long trunk is used to transmit frames using a Go-Back-N protocol. The propagation speed is 6 μ sec/kmand the trunk data rate is 1.5 Mbps. We ignore the time taken to receive the bits in the acknowledgement. Frame size is 64 Bytes.What is the maximum number of bits of the sequence number?Correct answer is '7'. Can you explain this answer?.
Solutions for Directions: A 3000 km long trunk is used to transmit frames using a Go-Back-N protocol. The propagation speed is 6 μ sec/kmand the trunk data rate is 1.5 Mbps. We ignore the time taken to receive the bits in the acknowledgement. Frame size is 64 Bytes.What is the maximum number of bits of the sequence number?Correct answer is '7'. Can you explain this answer? in English & in Hindi are available as part of our courses for Computer Science Engineering (CSE).
Download more important topics, notes, lectures and mock test series for Computer Science Engineering (CSE) Exam by signing up for free.
Here you can find the meaning of Directions: A 3000 km long trunk is used to transmit frames using a Go-Back-N protocol. The propagation speed is 6 μ sec/kmand the trunk data rate is 1.5 Mbps. We ignore the time taken to receive the bits in the acknowledgement. Frame size is 64 Bytes.What is the maximum number of bits of the sequence number?Correct answer is '7'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of
Directions: A 3000 km long trunk is used to transmit frames using a Go-Back-N protocol. The propagation speed is 6 μ sec/kmand the trunk data rate is 1.5 Mbps. We ignore the time taken to receive the bits in the acknowledgement. Frame size is 64 Bytes.What is the maximum number of bits of the sequence number?Correct answer is '7'. Can you explain this answer?, a detailed solution for Directions: A 3000 km long trunk is used to transmit frames using a Go-Back-N protocol. The propagation speed is 6 μ sec/kmand the trunk data rate is 1.5 Mbps. We ignore the time taken to receive the bits in the acknowledgement. Frame size is 64 Bytes.What is the maximum number of bits of the sequence number?Correct answer is '7'. Can you explain this answer? has been provided alongside types of Directions: A 3000 km long trunk is used to transmit frames using a Go-Back-N protocol. The propagation speed is 6 μ sec/kmand the trunk data rate is 1.5 Mbps. We ignore the time taken to receive the bits in the acknowledgement. Frame size is 64 Bytes.What is the maximum number of bits of the sequence number?Correct answer is '7'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice Directions: A 3000 km long trunk is used to transmit frames using a Go-Back-N protocol. The propagation speed is 6 μ sec/kmand the trunk data rate is 1.5 Mbps. We ignore the time taken to receive the bits in the acknowledgement. Frame size is 64 Bytes.What is the maximum number of bits of the sequence number?Correct answer is '7'. Can you explain this answer? tests, examples and also practice Computer Science Engineering (CSE) tests.
|
|
Explore Courses for Computer Science Engineering (CSE) exam
|
|
Suggested Free Tests
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
|
© EduRev
|
Education Revolution
|
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup