Question

Suppose two hosts, A and B, are separated by 7,500 kilometers and are connected by a direct link of R = 10 Mbps. Suppose the propagation speed over the link is 2.5 x 10 8 meters/sec. Consider sending a large packet of 500,000 bits from Host A to Host B. How many milliseconds (ms) does it take before the receiver has received the entire 500,000-bit file?

Answer 1

Answer:

50 ms (milliseconds) will be taken in total to take the entire file by user.

Explanation:

In Computer networks, propagation delay is defined as the time in which a packet is sent from sender to receiver completely. It is computed by taking the ratio of link length and propagation speed.

We know that:

BDP(in bits) = total bandwidth (in bits/sec) * trip time(in sec)

Now according to given condition we have:

Bandwidth = R = 10 Mbps = 10,000,000 bps

BDP = 500,000 bits

For finding Propagation Delay:

Propagation Delay = BDP/ R

Propagation Delay = 500,000/10,000,000 sec

Propagation Delay = 0.05 sec

Converting in milliseconds:

Propagation Delay = 50 ms

Hence. the delay would be 0.05 seconds and in milliseconds they will be equal to 50 ms