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In selective repeat, assume the sender is using the ambiguity window constraint of is W = 2b-1, with b bits in the sequence number. Let b = 4, so frames are numbered cyclically, from 0 t 15. Make the assumptions that signals can’t jump over each other in time, and that error detection is perfect. Suppose the receiver has correctly received all frames numbered up through 10, a gap from 11, and also has received frames 12 and 13 immediately after the gap.
a) What is the number of the oldest frame that could come to the receiver when the receiver is in this state? Note the receiver doesn’t know what acknowledgments have successfully returned to the sender, other than what can be deduced based on knowing the sender’s window rule.
b) In this state, what is the most advanced frame number that the sender would be allowed to send without violating the condition to prevent ambiguity?
Expert Answer
Sender window size is 2b-1 or 8
We know that sender has sent frame 13.
and we also know the client has not sent ack for frame 10.
a)If we consider that 13 is the last frame in the window, the oldest frmae in the window is 6 (13 – 8 + 1). So the oldest frame that client can receive is 6.
b) Since we know ack for frame 10 has not been sent, sender window must not cross frame 10. In best case sender has recived ack for frame 9 and the window start from frame 10. So in that case, the most advance frame would be 17 (10 + 8 -1)
