Question

consider the bandwidth of the line is 1.2mbps; packet size is 1KB; and round trip time is 50 msec. What is the link utilization in stop and wait?

Answer 1

Answer:

Explanation:

given,

Bandwidth of line = 1.2 Mbps

packet size = 1 Kb

round trip time =  50 msec

transition delay = $\dfrac{Packet\ size}{bandwidth}$

$t_t = \dfrac{1}{1.2}$

$t_t = \dfrac{2^{10}\times 8\ bits}{1.2 \times 10^6\ bits/sec}$

$t_t = 6.862 msec$

propagation delay

$T_p = \dfrac{round\ trip}{2}$

$T_p = \dfrac{50}{2}$

$T_p = 25\ msec$

now calculation of 'a'

$a = \dfrac{T_p}{T_t}$

$a = \dfrac{25}{6.862}$

a = 3.643

Calculating Link Utilization-

 Link Utilization or Efficiency (η)

= $\dfrac{1}{1+2a}$

= $\dfrac{1}{1+2\times 3.643}$

= $\dfrac{1}{8.286}$

= 0.1206

= 12.06%