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Suppose the length of a 10Base5 cable is $2500$ m. If the speed of propagation in a thick coaxial cable is $200,000,000$ m/s, how long does it take for a bit to travel from the beginning to the end of the network? Assume that there is a $10$ μs delay in the equipment due to repeaters and there is a repeater for every $500$ meters?

  1. $26.5$ micro seconds
  2. $52.5$ microseconds
  3. $47.5$ micro seconds
  4. $37.5$ micro seconds
asked in Computer Networks by Boss (13.7k points)
edited by | 98 views

2 Answers

+2 votes

Propagation delay ($T_{p}$)=$\frac{500m}{2*10^8 m/sec}$= $2.5{\mu }$sec

Total time taken= $2.5*5+10*4$= $52.5{\mu }$sec

Hence Option B) is correct

 

answered by Boss (12.8k points)
0

 why we are not taking transmission time on each  repeter......for first repeater it will zero 

2.5+4*10+(0.1+2.5)*4=======52.9

 

@Arjun sir @GATEBOOK

0
It is negligible.
0 votes

The answer is 52.5 microseconds 

40  microseconds  due to the router and 12.5 due to the propagation delay

 

answered by Active (1.6k points)
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