Question

Consider a point to point link 50 km in length. At what speed would propagation delay (at a speed 2 x 108 meters/second) equal transmit delay for 100 byte packets.

Answer 1

Answer:

The complete question is here:

Consider a point-to-point link 2 km in length. At what bandwidth would propagation delay (at a speed of 2 x 108 m/s) equal transmission delay for

(a) 100-byte packets?

(b) 512-byte packets?

Explanation:

Given:

Length of the link = 50 km

Propagation delay = distance/speed = d/s

To find:

At what speed would propagation delay (at a speed 2 x 10^8 meters/second) equal transmit delay for 100 byte packets?

Solution:

Propagation Delay = t prop

                                 = 50 * 10^3 / 2 * 10^8

                                 = 50000 / 200000000

                                 = 0.00025

                                 = 25 ms

(a) When propagation delay is equal to transmission delay for 100 packets:

Transmission Delay = packet length / data rate = L / R

So when,

Transmission Delay = Propagation Delay

100 * 8  / bits/sec = 50 * 10^3  / 2 * 10^8

Let y denotes the data rate in bit / sec

speed = bandwidth = y bits/sec

         100 * 8  / y bits/sec = 50 * 10^3  / 2 * 10^8

                        y bit/sec   =  100 * 8  / 25 * 10^ -5

                                         = 800 / 0.00025

                                     y  = 3200000 bit/sec

                                      y = 3200 Kbps

(b) 512-byte packets

512 * 8  / y bits/sec = 50 * 10^3  / 2 * 10^8

                        y bit/sec   =  512 * 8  / 25 * 10^ -5

                                         = 4096 / 0.00025

                                     y  = 16384000 bit/sec

                                      y = 16384 Kbps

Answer 2

Answer:

The bandwidth is 3200 Kbps

Explanation:

Given:

The length of the point to point link = 50 km = 50×10³ = 50000 m

Speed of transmission = 2 × 10⁸ m/s

Therefore the propagation delay is the ratio of the length of the link to speed of transmission

Therefore,  $t_{prop} = \frac{distance}{speed}$

$t_{prop} = \frac{50000}{2*10^{8} }=2.5*10^{-4}$

Therefore the propagation delay in 25 ms

100 byte = (100 × 8) bits

Transmission delay = $\frac{(100*8)bits }{(x)bits/sec}$

If propagation delay is equal to transmission delay;

$\frac{(100*8)bits }{(x)bits/sec}= \frac{50000}{2*10^{8} }$

$x=\frac{100*8*2*10^{8} }{50000}$

$x=\frac{100*8}{2.5*10 ^-4}=3200000$

x = 3200 Kbps