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 dа ta:

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