in Computer Networks
514 views
0 votes
0 votes
Consider two links, (A,B) and (B,C), with propagation delays of d1 and d2, respectively. Assume that host A sends M packets to host C using a sliding window flow control protocol with a window of size W.

What is the minimum time it take to send all packets from A to C when the flow control protocol is implemented end-to-end between A and C ?

Note: Transmission delays are negligible

(A) 2(d1+d2) Ceil(M/W)

(B) (d1+d2) MW

(C) 2(d1+d2)ceil(W/M)

(D) 2(d1+d2)W
in Computer Networks
514 views

1 Answer

2 votes
2 votes

(A) 2(d1+d2) $\left \lceil M/W \right \rceil$

As transmission delays are negligible, so at T=0 first window will be on link and start propagating. It will take (d1+d2) time to reach C.Then an ACK from C to A will take (d1+d2) time again. So total time for one window is 2(d1+d2).

Total number of windows would be $\left \lceil M/W \right \rceil$

$\therefore$ Total time =  2(d1+d2) $\left \lceil M/W \right \rceil$

Here transmission delays are negligible, so it is working like Stop and Wait protocol.

Related questions