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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).

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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

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