Question

How long does it take a packet of length 1,000 bytes to propagate over a link of distance 2,500 km, propagation speed 2.5 · 108 m/s, and transmission rate 2 Mbps?

Answer 1

Answer:

10 ms      

Explanation:

Length of packet = 1000 bytes

Distance of link = 2500 km

Propagation Speed = 2.5 x 10^8 m/s

Transmission Rate = 2 Mpbs

To find: Time a packet takes to propagate over a link

Propagation Time = ?

Solution:

Propagation Time = Distance of Link/Propagation Speed

                      tprop = d/s=

Distance is in kilometers. First convert it into meters.

2500*1000 = 2500000 meters

Distance of link = 2500000 m

Propagation Speed = 2.5 x 10^8 = 2.5 x 100000000 = 250000000

Propagation Time = (2500*)/(2.5*10^8)

                                = 2500000/250000000

                                =0.01 sec

Convert 0.01 second into millisecond:

                               0.01 x 1000 = 10 ms

So

                      Propagation Time = 10 ms

Answer 2

Corrected Question:

How long does it take a packet of length 1,000 bytes to propagate over a link of distance 2,500 km, propagation speed 2.5 x $10^{8}$ m/s, and transmission rate 2 Mbps?

Answer:

4.01 ms

Explanation:

The total time(T) taken for the packet to propagate over the link is the sum of the delay in transmission ($T_{d}$) and the delay in propagation ($T_{p}$).

i.e T = $T_{d}$ + $T_{p}$    -----------------------(i)

Where

(1) $T_{d}$ is the quotient of the packet length (L) in bits and the transmission rate (R) in bits per second.

=> $T_{d}$ = L / R    ---------------------- (ii)

=> L = 1000 bytes = 1000 x 8 bits = 8 000 bits

=> R = 2 Mbps = 2 x 1 000 000 = 2 000 000 bps

Substituting the values of L and R into equation (ii) we have,

=>  $T_{d}$ = 8000/2000000

=>  $T_{d}$  = 4 ms

(2)$T_{p}$ is the quotient of the distance (d = 2500km) of the link and the propagation speed of the link (s =  2.5 x $10^{8}$ m/s)

=> $T_{p}$ = d / s

=> $T_{p}$ = 2500 /  2.5 x $10^{8}$

=> $T_{p}$ = 0.01 ms

(3) Recall that the time taken (T) for the packet to propagate is ;

=> T = $T_{d}$ + $T_{p}$  

Substituting the values of  $T_{d}$ and $T_{p}$ into the equation gives;

=> T = 4 + 0.01

=> T = 4.01 ms

Therefore the time taken is 4.01 ms