Question # 1 Assume a network with a bandwidth of 2000Mbits/sec. It has asending
ID: 3608352 • Letter: Q
Question
Question # 1
Assume a network with a bandwidth of 2000Mbits/sec. It has asending overhead of 100sec and a receiving overhead of150sec. Assume two machines connected together. It isrequired to send a 20,000 byte message from one machine to theother (including header), and the message format allows 15, 00bytes in a single message. Calculate the total latency to send themessage from one machine to another assuming they are 20m apart (asin a SAN). Next, perform the same calculation but assume themachines are 800m apart (as in a LAN). Finally, assume they are2000Km apart (as in a WAN).
Assume that signals propagate at 66% of the speed of lightin a conductor, and that the speed of light is 300,000Km/sec.
Explanation / Answer
By using the assumption, we get: Distance between two machines in Km Time of flight =-------------------------------------------------- 2/3 x 300,000Km/sec Total Latency = Sender overhead + Time of flight + Messagesize/bandwidth + Receiver overhead For SAN: Total latency = 100sec + (0.020Km/(2/3 x 300,000Km/sec)) + 20,000bytes/ 2000Mbits/sec + 150sec = 100sec + 0.1sec + 80sec + 150sec = 330.1sec For LAN Total latency = 100sec + (0.8Km/(2/3 x 300,000Km/sec)) + 20,000bytes/ 2000Mbits/sec + 150sec = 100sec + 4sec + 80sec + 150sec = 334sec For WAN Total latency = 100sec + (2000Km/(2/3 x 300,000Km/sec)) + 20,000bytes/ 2000Mbits/sec + 120sec = 100sec + 10,000sec + 80sec + 120sec = 10300sec