CSMA/CD

Question

In standard ethernet with transmission rate of 20 Mbps, the length of the cables is 2500 m and the size of frame is 512 bits. The propagation speed of a signal in a cable is 2 × 10⁸ m/s. The percentage of the time channel is idle or not used by a station is _________ (in approximate integer value).

Comments

i'm getting 24.12 %

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A short explanation would be highly appreciated :)

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@Pawan Kumar  So ,  Percentage of time channel is idle = percentage of time station are doing useful work ?  I do get efficiency as 24.12% approx for this but I am confused what is asked in question here ?

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 Pawan Kumar 2  VS    It must be 100- efficiency?

Please correct me if I am wrong!

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Yes it should be 100-efficiency

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@VS @Pawan Kumar 2 I am getting 76% please tell whether its correct

Idle channel % = 100 - efficiency

which is 100 - $\frac{Tt}{e*2Tp + Tt + Tp}$

Answer 1

1/(1+5.44*Tp/Tt) =1/(1+5.44*12.5/25.6) =0.27=27%.   (1-27%)=72%. Ans

Comments

@Saheb_spec It is $Efficiency = 1/(1+6.44Tp/Tt)$ in CSMA/CD.

Answer 2

Given: 

• BW=$20$Mbps
• Distance=$2500$m
• L=$512$m
• V=$2\times10^8$m/sec

$T_p=\frac{d}{v}\implies\frac{2500m*sec}{2\times10^8m}=12.5\mu sec$

$T_t=\frac{L}{BW}\implies\frac{512 bit*sec}{20\times10^6 bit}=25.6\mu sec$

The efficiency of Ethernet: $\eta=\frac{1}{1+6.44a},\text{where $a=T_p/T_t$}$

Now $a=\frac{T_p}{T_t}=0.488$

$\therefore \eta=\frac{1}{1+6.44*0.488}=24.12\%$

So The percentage of time the channel is idle=$100-24.12=75.88\%$

Ref:  Efficiency of Ethernet