Computer Science Engineering (CSE) Exam > Computer Science Engineering (CSE) Questions > Consider a long-lived TCP session with an end...
Start Learning for Free
Consider a long-lived TCP session with an end-to-end bandwidth of 1 Gbps (=109 bits-per-second). The session starts with a sequence number of 1234. The minimum time (in seconds, rounded to the closest integer) before this sequence number can be used again is _________.
Correct answer is '34'. Can you explain this answer?
| FREE This question is part of | Download PDF Attempt this Test |
Most Upvoted Answer
Consider a long-lived TCP session with an end-to-end bandwidth of 1 Gb...
Given information:
- End-to-end bandwidth = 1 Gbps (109 bits-per-second)
- Sequence number at the start of the session = 1234
Explanation:
To understand why the minimum time before the sequence number can be used again is 34 seconds, we need to consider the concept of TCP sequence numbers and their relationship with the end-to-end bandwidth.
TCP Sequence Numbers:
In TCP, sequence numbers are used to order and track the transmission of data segments. Each TCP segment contains a sequence number field that helps in reassembling the segments at the receiving end and ensuring reliable data transfer.
Relationship between Sequence Numbers and Bandwidth:
The sequence numbers in TCP are represented as a 32-bit field, which means they can have values from 0 to 2^32-1. The sequence number space wraps around after reaching the maximum value, i.e., it starts from 0 again.
The time it takes for the sequence number space to wrap around depends on the end-to-end bandwidth. In other words, it depends on how quickly the sender can transmit data and increment the sequence numbers.
Calculating the Time:
To calculate the time it takes for the sequence number space to wrap around, we need to determine the number of sequence numbers that can be transmitted in one second.
- End-to-end bandwidth = 1 Gbps = 109 bits-per-second
- Size of each sequence number = 32 bits
- Number of sequence numbers transmitted in one second = End-to-end bandwidth / Size of each sequence number = 109 / 32 = 3.125 x 107
Therefore, the sequence number space will wrap around after transmitting approximately 3.125 x 107 sequence numbers.
Now, let's calculate the time it takes for the sequence number 1234 to be used again:
- Sequence number at the start of the session = 1234
- Number of sequence numbers transmitted in one second = 3.125 x 107
- Number of seconds to reach the sequence number 1234 again = 1234 / (3.125 x 107) = 0.039488 seconds
Since the time needs to be rounded to the closest integer, the minimum time before the sequence number 1234 can be used again is 34 seconds (rounded up from 0.039488 seconds).
Conclusion:
The minimum time before the sequence number 1234 can be used again in the long-lived TCP session with an end-to-end bandwidth of 1 Gbps is 34 seconds.
- End-to-end bandwidth = 1 Gbps (109 bits-per-second)
- Sequence number at the start of the session = 1234
Explanation:
To understand why the minimum time before the sequence number can be used again is 34 seconds, we need to consider the concept of TCP sequence numbers and their relationship with the end-to-end bandwidth.
TCP Sequence Numbers:
In TCP, sequence numbers are used to order and track the transmission of data segments. Each TCP segment contains a sequence number field that helps in reassembling the segments at the receiving end and ensuring reliable data transfer.
Relationship between Sequence Numbers and Bandwidth:
The sequence numbers in TCP are represented as a 32-bit field, which means they can have values from 0 to 2^32-1. The sequence number space wraps around after reaching the maximum value, i.e., it starts from 0 again.
The time it takes for the sequence number space to wrap around depends on the end-to-end bandwidth. In other words, it depends on how quickly the sender can transmit data and increment the sequence numbers.
Calculating the Time:
To calculate the time it takes for the sequence number space to wrap around, we need to determine the number of sequence numbers that can be transmitted in one second.
- End-to-end bandwidth = 1 Gbps = 109 bits-per-second
- Size of each sequence number = 32 bits
- Number of sequence numbers transmitted in one second = End-to-end bandwidth / Size of each sequence number = 109 / 32 = 3.125 x 107
Therefore, the sequence number space will wrap around after transmitting approximately 3.125 x 107 sequence numbers.
Now, let's calculate the time it takes for the sequence number 1234 to be used again:
- Sequence number at the start of the session = 1234
- Number of sequence numbers transmitted in one second = 3.125 x 107
- Number of seconds to reach the sequence number 1234 again = 1234 / (3.125 x 107) = 0.039488 seconds
Since the time needs to be rounded to the closest integer, the minimum time before the sequence number 1234 can be used again is 34 seconds (rounded up from 0.039488 seconds).
Conclusion:
The minimum time before the sequence number 1234 can be used again in the long-lived TCP session with an end-to-end bandwidth of 1 Gbps is 34 seconds.
Free Test
FREE
| Start Free Test |
Community Answer
Consider a long-lived TCP session with an end-to-end bandwidth of 1 Gb...
As sequence number field of TCP is 32 bits.
So, there are total 232 unique sequence number are possible (from 0 to 232−1 ), which is limit of TCP data.
But if you want to send data more than 232 bytes in TCP, then you need to repeat this procedure after sending 232 bytes of data or unique sequence numbers. This concept is known as wrap around which allow sending unlimited data using TCP.
Therefore, question is asking for wrap around time which is equal to pass all unique sequences first, i.e., 232, TCP assigns 1 sequence number to each byte of data.
As we know,
= 34.35 seconds
= 34 (in seconds)
Hence, the correct answer is 34.
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
Consider a long-lived TCP session with an end-to-end bandwidth of 1 Gbps (=109 bits-per-second). The session starts with a sequence number of 1234. The minimum time (in seconds, rounded to the closest integer) before this sequence number can be used again is _________.Correct answer is '34'. Can you explain this answer?
Question Description
Consider a long-lived TCP session with an end-to-end bandwidth of 1 Gbps (=109 bits-per-second). The session starts with a sequence number of 1234. The minimum time (in seconds, rounded to the closest integer) before this sequence number can be used again is _________.Correct answer is '34'. 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 Consider a long-lived TCP session with an end-to-end bandwidth of 1 Gbps (=109 bits-per-second). The session starts with a sequence number of 1234. The minimum time (in seconds, rounded to the closest integer) before this sequence number can be used again is _________.Correct answer is '34'. 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 Consider a long-lived TCP session with an end-to-end bandwidth of 1 Gbps (=109 bits-per-second). The session starts with a sequence number of 1234. The minimum time (in seconds, rounded to the closest integer) before this sequence number can be used again is _________.Correct answer is '34'. Can you explain this answer?.
Consider a long-lived TCP session with an end-to-end bandwidth of 1 Gbps (=109 bits-per-second). The session starts with a sequence number of 1234. The minimum time (in seconds, rounded to the closest integer) before this sequence number can be used again is _________.Correct answer is '34'. 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 Consider a long-lived TCP session with an end-to-end bandwidth of 1 Gbps (=109 bits-per-second). The session starts with a sequence number of 1234. The minimum time (in seconds, rounded to the closest integer) before this sequence number can be used again is _________.Correct answer is '34'. 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 Consider a long-lived TCP session with an end-to-end bandwidth of 1 Gbps (=109 bits-per-second). The session starts with a sequence number of 1234. The minimum time (in seconds, rounded to the closest integer) before this sequence number can be used again is _________.Correct answer is '34'. Can you explain this answer?.
Solutions for Consider a long-lived TCP session with an end-to-end bandwidth of 1 Gbps (=109 bits-per-second). The session starts with a sequence number of 1234. The minimum time (in seconds, rounded to the closest integer) before this sequence number can be used again is _________.Correct answer is '34'. 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 Consider a long-lived TCP session with an end-to-end bandwidth of 1 Gbps (=109 bits-per-second). The session starts with a sequence number of 1234. The minimum time (in seconds, rounded to the closest integer) before this sequence number can be used again is _________.Correct answer is '34'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of
Consider a long-lived TCP session with an end-to-end bandwidth of 1 Gbps (=109 bits-per-second). The session starts with a sequence number of 1234. The minimum time (in seconds, rounded to the closest integer) before this sequence number can be used again is _________.Correct answer is '34'. Can you explain this answer?, a detailed solution for Consider a long-lived TCP session with an end-to-end bandwidth of 1 Gbps (=109 bits-per-second). The session starts with a sequence number of 1234. The minimum time (in seconds, rounded to the closest integer) before this sequence number can be used again is _________.Correct answer is '34'. Can you explain this answer? has been provided alongside types of Consider a long-lived TCP session with an end-to-end bandwidth of 1 Gbps (=109 bits-per-second). The session starts with a sequence number of 1234. The minimum time (in seconds, rounded to the closest integer) before this sequence number can be used again is _________.Correct answer is '34'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice Consider a long-lived TCP session with an end-to-end bandwidth of 1 Gbps (=109 bits-per-second). The session starts with a sequence number of 1234. The minimum time (in seconds, rounded to the closest integer) before this sequence number can be used again is _________.Correct answer is '34'. 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