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In sliding window protocol, assume a 3-bit sequence number field. A and B have windows, which has 3-bit sequence number. If A sends 3 frames and waits for all the 3 acknowledgments until the timer expires. Which of the following "could not be" the sender's "window size" after the timer expires. Assume receiver receives all the three frames correctly. But ACK's send by receivers may be lost.

a. 4

b. 7

c. 6

d. 5

1 comment

explain???

With 3 bit sequence no max window size=8

So with every ack window size reduces by 1

So if 1 ack comes window size=7

If 2 ack come window size=6

If 3 comes window size=5

So 4 is not possible

@kapil ,tauhin,manoj how window size reduces here

0,1,2,3,4,5,6,7 now 3 frames are send & every acknowledment get then windowsize decrease
suppose 0,1,2 sent.
0,1,2,3,4,5,6,7 now 0 frame acknowledgement arrive
1,2,3,4,5,6,7 window size =7
now 1 frame acknowledgement arrive
2,3,4,5,6,7 window = 6
now 1 frame acknowledgement arrive
3,4,5,6,7 window size =5
Ans(A)

by
With 3 bit we will have 2^3 sequence number =8(0-7)

I think its a hint A sends 3 frame and wait for all frame means it using Selective Repeat protocol coz it doesn't support Cumulative ACK it support individual ACK for each frame which is lost or NAK if no frame is lost.

So in Selective repeat protocol Max SWS = Max sequence number +1/2= 7+1/2 = 4

Here ACK is lost so after time expires it again sends 3 frame since in first attempt he sent 3 frame and it was expecting for ACK but didn't get!

So Sender Window Size couldn't be = 4

1 comment

by this logic why window size can’t be 5 ,6, or 7?