# computer network

1 vote
308 views

A Go-Back-5 ARQ scheme is employed on a 200 meters cable between two nodes A and B to send frames of size 10,000 bits with bit rate 1 Mbps and bit error probability of 10$^{-5}$. Headers and acknowledgements are 20 bytes each. The Speed of propagation of signals on the cable is 200,000 km/s, and the processing delay is 11ms. Assume large m (sequencing its) for parts 1,2, and 3

1. On the above figure, label (write number of) the ACKs for the frames depicted by solid arrows. On the same figure, label the frame numbers depicted in dashed lines. On the same figure, plot and label ACKs (IF NEEDED) that node B would send as response to the frames depicted by dashed arrows.
2. What is the minimum number of bits that can be used for sequence numbers in this Go-Back-5 ARQ scheme to avoid confusion at node B?
3. What is the efficiency of this Go-Back-5 system ?
4. What is the minimum value of timeout time for this Go-Back-5 to avoid early timeouts ?
5. What is the minimum window size that could be employed for the above system to avoid early window exhaustion ?
6. Given the above parameter (except for the cable length), what is the maximum distance between the two nodes for which S&W will be as efficient as a GBN.

edited
0
why you are not interested to type the question ?
0
fracture.
0

finding duplicates is tough through images !

fracture.

is your hand is fracture ?

0
yes

## Related questions

1
540 views
Consider two computers A and B are connected through a network of 30 Mbps. Assume the distance between them is 3000km and the signal propagation speed is same as the speed of light and the packet size is 12 KB. What is the minimum number of bits required for window to achieve 100% of utilization during GoBack-N and selective repeat protocol? A . 5 and 6 B. 6 and 7 C. 6 and 6 D. 7 and 8
2
481 views
Consider a network connecting two systems, ‘A’ and ‘B’ located 6000 km apart. The propagation speed of media is 2 × 106 mps. It is needed to design a Go-Back-7 sliding window protocol for this network. The average packet size is 107 bits. If network used as its full capacity, then the bandwidth of network is__________ Mbps. Here full capacity means efficiency is 100%.
1 vote
3
216 views
I am unable to understand their explanation,can anyone explain it in a better way?