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).
Let I be the minimum number of bits (I) that will be required to represent the sequence numbers distinctly assuming that no time gap needs to be given between transmission of two frames.
Suppose that the sliding window protocol is used with the sender window size of 2i where is the number of bits identified in the previous question and acknowledgments are always piggybacked. After sending 2i frames, what is the minimum time the sender will have to wait before starting transmission of the next frame? (Identify the closest choice ignoring the frame processing time.)
- A.
16ms
- B.
18ms
- C.
20ms
- D.
22ms
Attempted by 29 students.
Show answer & explanation
Correct answer: C
Step 1 — how many frames fit in the link:
Bandwidth × one-way propagation = 106 bps × 25 ms = 10^6 × 0.025 = 25,000 bits = 25 frames (1000 bits each).
Step 2 — minimum sequence-number bits I:
Need smallest I with 2^I ≥ 25. 2^4 = 16 < 25, 2^5 = 32 ≥ 25, so I = 5.
Step 3 — sender window and transmission times:
Sender window = 2^I = 2^5 = 32 frames.
Frame transmission time = 1000 bits / 10^6 bps = 0.001 s = 1 ms.
Time to send 32 frames back-to-back = 32 × 1 ms = 32 ms.
Step 4 — when does the earliest acknowledgment arrive?
The first frame finishes transmission at 1 ms and arrives at the receiver after 25 ms propagation, i.e. at 26 ms.
Assuming the receiver immediately piggybacks an acknowledgment on a return frame, that return frame needs 1 ms to transmit and 25 ms to propagate back, so the ACK reaches the sender at 26 + 1 + 25 = 52 ms.
Step 5 — minimum waiting time after sending 32 frames:
Sender finished sending the 32nd frame at 32 ms. Earliest ACK arrives at 52 ms. So the sender must wait 52 − 32 = 20 ms before starting the next frame.
Answer: 20 ms (closest choice). This ignores processing time as requested and uses the piggybacking assumption.