Question 8013 – Operating-Systems
March 11, 2024IP-Header
March 12, 2024Question 9137 – Sliding-Window-Protocol
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).
Suppose that the sliding window protocol is used with the sender window size of 2l, where l is the number of bits identified in the earlier part and acknowledgements are always piggy backed. After sending 2l 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.)
Correct Answer: C
Question 11 Explanation:
From above diagram we can say that Total RTT will be
1ms (transmitting time for first frame)
+ 25ms (propagation delay from S to R)
+ 1ms (transmitting time for piggybacked ACK)
+ 25ms (propagation delay from R to S)
= 52 ms
Also total time for sender needed to transmit 32 frames (22 = 25 = 32) is 32×1ms = 32ms
So sender has to wait for,
(52 – 32)ms
= 20ms
16ms
18ms
20ms
22ms
Subscribe
Login
0 Comments