Frames of 1000 bits are sent over a 10 ^ 6 bps duplex link between two hosts.…
2016
Frames of 1000 bits are sent over a 10 ^ 6 bps duplex link between two hosts. The propagation time is 25ms. Frames are to be transmitted into this link to maximally pack them in transit (within the link). What is the minimum number of bits (I) that will be required to represent the sequence numbers distinctly? Assume that no time gap needs to be given between transmission of two frames.
- A.
l = 2
- B.
l = 3
- C.
l = 4
- D.
l = 5
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Correct answer: D
Transmission time per frame is 1000 bits / 106 bps = 1 ms. Propagation time is 25 ms. The number of frames that can be in transit simultaneously to maximize link utilization is the ratio of propagation time to transmission time: 25 ms / 1 ms = 25 frames. To represent 25 distinct sequence numbers, we need the smallest integer n such that 2n >= 25. Since 24 = 16 and 25 = 32, the minimum number of bits required is 5.
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