Question

Machine X sends a 4KB (kilobyte = 1000 bytes) file over a 8Mb/s (mega bit per second) single-hop link to machine Y which lies 150Km away. Someone starts a stopwatch when machine X starts sending the first bit on the link towards machine Y . The stopwatch is stopped when last bit is received by machine Y . What is the minimum time the stopwatch may be showing? [Assume propagation at speed of light c = 3 × 108 m/s]; [1 byte = 8 bits]

Answer 1

File Size (L) = 4KB = 4 x 1000 = 4000 bytes = 4000 x 8 bits

Bandwidth (B) = 8Mb/s = 8 x 10⁶ bits/s

Propagation speed (c) = 3 x 10⁸ m /s

Distance between Machine X and Y = 150Km (D) = 150 x 10³ m

Total time = Transmission Delay time + Propagation delay time

Total time = (L/B) + ( D/ c)

Therefore, minimum time stop watch will be showing is 4.5 milliseconds i.e, 4.5ms.