Applied Mock Test

Question

What will be the efficiency of the channel in (% )in CSMA/CD when bandwidth is 20 Mbps. The speed of the signal is  4000 Km/s. The frame size is 1000 bits and the length of the cable 200 m.[ Note: round off the final answer in upto one decimal place]

 

Answer 1

Given

• $Bw = 20 \text { Mbps}$
• $v = 4000 \ km/s$
• $fs  = 1000 \ bits$
• $l = 200 \ m$

Transmission delay ($T_t$) $ = \frac {fs}{Bw} = \frac {1000}{20 \times 10^6} = 50 \ \mu s$

Propagation delay ($T_p$) $ = \frac {l}{v} = \frac {200\ m}{4000 \times 10^3 \ m/s} = 50 \ \mu s$

efficiency of CSMA/CD = $\frac {T_t}{T_t \ + \  6.44 \ T_p}$

$= \frac {50}{50 \ + \  6.44 \times 50}$

$= \frac {50}{50 \ + \  322}$

$ = 0.1344$

$= 13.44 \text{ %}$

Comments

Good explaination @Yaman Sahu :)

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@Chhaatra Thanks :)