Question

2.Consider a point-point link 50 km in length. At what bandwidth would propagation delay (at a speed of 2 x 10⁸ m/s) equal transmit delay for 100-byte packets? What about 512-byte packets?

Answer 1

So, the propagation delay = d / s

where d = distance between the two points

s = speed of propagation

And, the transmit delay = (L / B)

where, L = size of data packet

B = bandwidth

(a) In this problem, we have d = 50 km = 5 * 10⁴ m

s = 2 * 10⁸ m/s

L = 100 bytes = 100 * 8 bits = 800 bits

We have to find B for which transmit delay is equal to propagation delay.

So, here is the solved equation.

So, the required bandwidth is 3.2 Mbps.

(b) Here, we have

d = 50 km = 5 * 10⁴ m

s = 2 * 10⁸ m/s

L = 512 bytes = 512 * 8 bits = 4096 bits.

Here is the solved equation.

The reuired bandwidth is 16.384 Mbps.

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