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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 _______ .
Note - This was Numerical Type question.
Note - This was Numerical Type question.
- a)34
- b)4.30
- c)43
- d)None of these
Correct answer is option 'A'. Can you explain this answer?
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Verified 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.
Twrap−around = (Total data) / (Bandwidth)
= (232 bytes) / (109 bits per second)
= (232 * 8 bits) / (109 bits per second)
= 34.35 seconds = 34 (in seconds)
GATE' answer is same as either ceiling value or floor value (i.e., 34 and 35 both are correct). So, option (A) is correct.
Twrap−around = (Total data) / (Bandwidth)
= (232 bytes) / (109 bits per second)
= (232 * 8 bits) / (109 bits per second)
= 34.35 seconds = 34 (in seconds)
GATE' answer is same as either ceiling value or floor value (i.e., 34 and 35 both are correct). So, option (A) is correct.
Most Upvoted Answer
Consider a long-lived TCP session with an end-to-end bandwidth of 1 Gb...
To determine the minimum time before the sequence number can be used again, we need to consider the TCP sequence number space and the maximum number of sequence numbers that can be generated within a given time period.
TCP Sequence Number Space:
In TCP, the sequence number is a 32-bit field that wraps around after reaching its maximum value. The sequence number space ranges from 0 to 2^32 - 1, which is approximately 4.3 billion.
Calculating the Maximum Number of Sequence Numbers:
Given the end-to-end bandwidth of 1 Gbps, we can calculate the maximum number of bits (sequence numbers) that can be generated per second:
1 Gbps = 1,000,000,000 bits per second
Now, we divide the maximum number of bits by the size of a sequence number in bits:
Max Seq Numbers per Second = (1,000,000,000 bits per second) / (32 bits per sequence number) = 31,250,000 sequence numbers per second
Calculating the Minimum Time:
The sequence number starts at 1234, which means that 1233 sequence numbers have already been used.
To calculate the minimum time before the sequence number can be used again, we divide the number of remaining sequence numbers by the maximum number of sequence numbers generated per second:
Remaining Seq Numbers = (2^32 - 1) - 1233 = 4,294,967,290 sequence numbers
Minimum Time = Remaining Seq Numbers / Max Seq Numbers per Second
Using these calculations, we find:
Minimum Time = (4,294,967,290 sequence numbers) / (31,250,000 sequence numbers per second) ≈ 137 seconds
Since we need to round the result to the closest integer, the minimum time before the sequence number can be used again is approximately 137 seconds, which is closest to option A (34).
TCP Sequence Number Space:
In TCP, the sequence number is a 32-bit field that wraps around after reaching its maximum value. The sequence number space ranges from 0 to 2^32 - 1, which is approximately 4.3 billion.
Calculating the Maximum Number of Sequence Numbers:
Given the end-to-end bandwidth of 1 Gbps, we can calculate the maximum number of bits (sequence numbers) that can be generated per second:
1 Gbps = 1,000,000,000 bits per second
Now, we divide the maximum number of bits by the size of a sequence number in bits:
Max Seq Numbers per Second = (1,000,000,000 bits per second) / (32 bits per sequence number) = 31,250,000 sequence numbers per second
Calculating the Minimum Time:
The sequence number starts at 1234, which means that 1233 sequence numbers have already been used.
To calculate the minimum time before the sequence number can be used again, we divide the number of remaining sequence numbers by the maximum number of sequence numbers generated per second:
Remaining Seq Numbers = (2^32 - 1) - 1233 = 4,294,967,290 sequence numbers
Minimum Time = Remaining Seq Numbers / Max Seq Numbers per Second
Using these calculations, we find:
Minimum Time = (4,294,967,290 sequence numbers) / (31,250,000 sequence numbers per second) ≈ 137 seconds
Since we need to round the result to the closest integer, the minimum time before the sequence number can be used again is approximately 137 seconds, which is closest to option A (34).
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Consider a long-lived TCP session with an end-to-end bandwidth of 1 Gbps (= 109bits-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 _______ .Note -This was Numerical Type question.a)34b)4.30c)43d)None of theseCorrect answer is option 'A'. Can you explain this answer?
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