I need help with the above problem. Any guidence would be great, thank you! A 10
ID: 3665203 • Letter: I
Question
I need help with the above problem. Any guidence would be great, thank you!
A 10 M bit (M = 10^6) file is to be transferred from a client to a server over a network in which there are two intermediate switches. All devices operate at 100 Mbps, and each link distance is 2000 m How long will it take the file to be transferred. If the fife is not segmented into packets (i.e.. sent as a whole) If the file is segmented into 1 Kbit (K-1000) packets All switches use store-and-forward processing, and you can ignore processing and queuing delays at all devices, as well as header overheads. Assume that the speed of light is 2x10^8 m/sec.Explanation / Answer
Fast Ethernet on fiber is defined under 100BASE-FX standard which provides speed up to 100 MBPS on fiber. Ethernet over fiber can be extended up to 100 meters in half-duplex mode and can reach maximum of 2000 meters in full-duplex over multimode fibers.
Example for the soultion:
When each has window size 0.5W , the two flows have combined window sizes that are sufficient to keep the pipe full. Their combined throughput will be 30 Mbps, so by symmetry each will get 15 Mbps. The round-trip delay, which is the same for A and B, can be found by:
RTT = 0.5 × 2000 × 10, 000bits 15Mbps = 666.7ms
Subtracting the delay on the reverse path, i.e., 2tp = 333.3ms, we get a forward path delay of 333.3ms, which is exactly the propagation delay 2tp, i.e., the queueing delay is tq = 0 (we ignored the transmission time 3ts from the RTT for simplicity). When A’s window is W and B’s is 0.5W , the two flows have combined window sizes that are sufficient to keep the pipe full. Their combined throughput will be 30Mbps. A will have twice as many packets in the pipeline as B, so their throughputs will have a ratio of 2:1. Thus A gets 20 Mbps and B gets 10 Mbps. The roud-trip time for A and B is the same, which is found by:
RTT = 1 × 2000 × 10, 000bits 20Mbps = 0.5 × 2000 × 10, 000bits 10Mbps = 1000ms
Subtracting the delay on the reverse path, i.e., 2tp = 333.3ms, we get a forward path delay of 666.7ms for both A and B. Their queueing delay is tq = 666.7 333.3 = 333.3ms (we ignored the transmission time 3ts from the RTT for simplicit