GO-BACK-N problem

Question

Answer 1

RTT = Tp + Tack + 2PD

PD = 1000km/(2 * 10^8) = 5 ms

Tack = Tp = 0 (transmission time is ignored in question).

So, RTT = 2 * 5ms = 10ms.

In RTT, in Go-Back-N protocol, N frames are transmitted. N = 8 here.

So, time to transmit 8 frames = 10 ms.

Each frame of of size 1024.

So, time to transmit 8 * 1024 = 8Kb (2^13 bits) = 10 ms.

Time taken to send 1M (2^20) bits = 10 ms * 2^7 = 1280 ms = 1.28 s.

Ref: http://web.mit.edu/modiano/www/6.263/lec3-4.pdf

Comments

"In RTT, in Go-Back-N protocol, N frames are transmitted."

How is this statement true??

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@Higgs

Since Go-Back-N flow control mechanism follows cumulative acknowledgments.

Cumulative acknowledgment means we will send one ACK for many packets. but here the question is for how many packets, we are going to send one ACK, So its depends, some people say that after sending full window, we will send one ACK but the problem here in the last transmission we might not need a full window(maybe some packets less). So window size is not good implementation for sending 1 ACK.

Some people say we will use ACK timer at receiver.whenver receiver receives any packet, start a timer called "ACK timer" which is fixed in size. By the time when ACK timer expires, how many packets have ever received in between that time, we will be going to send 1 ACK.

ACK timer is a good implementation for cumulative acks but still, there is some drawback that if ACK timer time is too big, then the advantage is we can send as many packets as possible, we need to send only 1 ack but the disadvantage is at the sender side there might be time out of Timeout timer.

Note: Timeout timer > ACk timer

here in this question, Arjun sir is using first one implementation because no information is giving about ACK timer.

after sending a full window of 8 packets, we will send one ACK according to the nature of GBN, Since transmission time is negligible, and propagation delay in solving we get 5 ms. So in one RTT, we will get one ACK for one full window.

One RTT = 2 * propagation delay

                = 2 *5 = 10 ms

So in 10 ms, we can send 8* 1024 bits of data.

10 ms -----------2¹³ bits

order to send 2²⁰ bits, it will require 1.28 seconds.

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Hi thanks for you explanation. I have some doubt on following,
Can you please explain......for Go-Back N protocol or other sliding window protocol also (i.e Stop-Wait, ARQ) at a particular time 't', the frame sequence number in sender if 'K', then acknowledgement sequence number from receiver to sender should be k+1. Is this correct if not please explain.