1
$\begingroup$

I am an undergraduate studying Data Coms and Networking and this question about go-back-N ARQ popped up that I am struggling to get my head around and cant find anything similar online. Any help at all would be greatly appreciated.

The only thing I have at all is that throughput could be defined by (Packet bits + overhead)/N. Is this at all on the right path?

QUESTION Consider the Go-Back-N ARQ protocol with N=5 and 4 bits used for sequence numbering. The user data frame is 3000 bits long where 100 bits are overhead for header and error check, etc. The frame transmission rate is 2Mbps. Assume that the medium does not reorder the frames. The outstanding frames will be retransmitted when the sending window becomes empty. The probability that a data frame would be received without any errors is 0.3.

i) What is the effective data transmission rate of this protocol?

ii) Suppose that at time t, the receiver is waiting for an information frame with binary sequence number 1001. What is the possible sequence number Slast in binary inside the sender’s window at time t? Justify your answer.

$\endgroup$
  • 2
    $\begingroup$ What did you try? Where did you get stuck? We're happy to help you understand the concepts but just solving exercises for you is unlikely to achieve that. You might find this page helpful in improving your question. Once you understand the concepts, this question should be a mere matter of arithmetic. Why don't you start by solving a simpler version of this question? For instance, suppose the probability was 1.0 instead of 0.3? Then could you answer it? $\endgroup$ – D.W. Apr 25 '17 at 7:02
  • $\begingroup$ I'm trying to get my head around this concept also.. it's so confusing. looking at part ii), if R is waiting for frame 9, the last sent frame could be 8. but the sender could also have: received ACK 9; AND sent 9,10,11,12,13? alternatively, if the sender had timed out and not received the ACK for 9, the sender's window may be different. the sender may have sent; 8,7,6,5,4 this means the sender's window could be: 4,5,6,7,8,9,10,11,12,13 could it not? $\endgroup$ – Exprise Apr 25 '17 at 9:06
0
$\begingroup$

Window size Ws is 5. the time to transmit one frame is: tf=3000/2 Mbps=1.5ms As the frame may be received in error with probability 0.7, each frame need to be transmitted 10/3 times on average. Then the time needed for successfully transmitting one frame is

tf+7/3*Wstf=

38/3*tf=19ms.

The effective rate is (3000-100)/19ms=152.6kbps. ii) 1001 in binary is 9 in decimal. Frame 8 must have be received by the receiver and the ACK has been sent back to the sender. Therefore, frame 3 must not be included in sender’s window. As frame 9 hasn’t been received yet, 9 must be greater than or equal to Slast. Therefore, Slast can be any of these: 4 to 9 in decimal. Orin binary, 0100, 0101, 0110, 0111, 1000, 1001.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.