Suppose two hosts, A and B, are connected by a 10 Mbps link. The length of a packet is 12 Kb (Kilobits, i.e., 12 × 103 bits). The length of the link is 40 km. Assume that signals propagate at the speed of light (i.e., the ideal speed of 3 × 108 m/s (meters per second)).

Answer is provided in the explanation section

Explanation:

Given data:

Bandwidth of link = 10* 106 bps

Length of packet = 12* 103 bits

Distance of link = 40 * 103m

Transmission Speed = 3 * 108 meters per second  

Formulas:

Transmission Delay = data size / bandwidth = (L /B) second

Propagation Delay = distance/transmission speed = d/s

Solution:

Transmission Delay = (12* 103 bits) / (10* 106 bps) = 0.0012 s = 1.2 millisecond

Propagation Delay = (40 * 103 meters)/ (3 * 108mps) = 0.000133 = 0.13 millisecond

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