Question

Calculate the total time required to transfer a 500KB file in the following cases. Assume an RTT of 50ms, a packet size of 2KB, and an initial 2xRTT of handshaking before data is sent. A. If the bandwidth is 2Mbps, what is the Bandwidth-Delay Product (BDP)

Answer 1

i) Total time = 2457.73 secs

ii) BDP = 0.01 Mbyte

Step-by-step explanation:

Packet size : 2 Kb = ( 2 * 8192 ) bits = 16384 bits

Bandwidth : 2Mbps = 2,000,000 bits/sec

RTT = 50 ms = 0.050 secs

i) Determine total time required to transfer 500 kB

= Initial handshake + Time for transmission of all files + propagation delay

= 2 x RTT + Time for transmission + RTT /2

Total time = ( 2 * 0.05 ) + (1500 * 1.6384 )+ ( 0.050 / 2 ) = 2457.73 secs

ii) Determine the Bandwidth-delay product

= data link capacity * RTT

= 2 Mbps * 50 ms = 100,000 = 0.01 MByte