Frames of 1000 bits are sent over a 106 bps duplex link between two hosts. The…
2009
Frames of 1000 bits are sent over a 106 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.
i = 2
- B.
i = 3
- C.
i = 4
- D.
i = 5
Attempted by 47 students.
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Correct answer: D
Answer: i = 5 (minimum number of bits required).
Reasoning:
Frame transmission time = frame size / link rate = 1000 bits / 10^6 bps = 0.001 s = 1 ms.
Propagation time = 25 ms (given).
Number of frames that can be on the link simultaneously = propagation time / transmission time = 25 ms / 1 ms = 25 frames.
We need at least as many distinct sequence numbers as frames in transit, so choose i such that 2^i >= 25.
Since 2^4 = 16 < 25 <= 32 = 2^5, the minimum i satisfying this is i = 5.
Note: This uses the one-way propagation time as given. If the problem intended round-trip packing, the required window and thus i would be larger.