Given the maximum lifetime of a segment is 30 sec and link capacity is…
Given the maximum lifetime of a segment is 30 sec and link capacity is 500Mbps, find the no. of bits required to avoid wraparound during this time?
Answer: A. 34 — Explanation: Given time = 30 sec, Bandwidth B = 500 Mbps In 1 sec ------ 500 Mb Therefore, 30 sec ----- 30 * 500 * 106= 15 * 109 bits No. of bits required to…
- A.
34
- B.
36
- C.
38
- D.
40
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Correct answer: A
Explanation:
Given time = 30 sec, Bandwidth B = 500 Mbps
In 1 sec ------ 500 Mb
Therefore, 30 sec ----- 30 * 500 * 106= 15 * 109
bits
No. of bits required to avoid wrap around = ceil(log2 (15*109)) bits= ceil(33.804) bits = 34 bits 9)) bits=ceil(33.804) bits = 34 bits
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