Ethernet

Question

Consider two nodes A and B on the same Ethernet, and suppose the propagation delay b/w the two nodes is 225-bit times. Suppose at the time both nodes A and B begin to transmit a frame.Both nodes transmit a 50-bit jam signal after detecting  collision.
For 10^7 bit/set find the time at which both nodes A and B sense idle channel?

Answer 1

Firstly, note that here time is given in bits not in sec. Now,

1: Both stn detect collision at t= 225.

2: after detecting collision, both will sent jam signal to inform other stn in that network about the occurrence of collision. here jam signal is 50 bits.

So both stn stop transmitting their jam signal at t= 225+50 = 275.

3: Last bit of jam signal will now propagate to other end. And Tp is given to be 225 bit time. So time at which both the jam signal will reach completely to opp end  and channel is sensed idle at t= 275+225 = 500 bits ( Final answer in bits)

Now "For 10^7 bit/sec. find the time at which both nodes A and B sense idle channel?" - Since Bandwidth is given, ques is asked in units of time (sec).

So now converting answer in bits to sec :-

time at which both nodes A and B sense idle channel = 500 bits  = 500 / BW = 500/ 10^7

     = 50 micro sec.

Comments

here is don't take 2*PT time for the collision because both are transmitting  simultaneously so that its take 225 bit times 

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Why are we not considering the transmission delays?? Could anyone explain?

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why would we consider that

in layman language, a transmission delay is a time when sender uploads a data to transmit to the packet over the link

and in question, it's asking after the packet has transferred over the link

Answer 2

Propagation delay (Tp)

= 225 bit times

= 225 bit / 10 Mbps

= 22.5 x 10-6 sec

= 22.5 μsec

At t = 0,

 

• Nodes A and B start transmitting their frame.
• Since both, the stations start simultaneously, so collision occurs in the midway.
• Time after which collision occurs = Half of the propagation delay.
• So, time after which collision occurs = 22.5 μsec / 2 = 11.25 μsec.

 

 

At t = 11.25 μsec,

• After the collision occurs at t = 11.25 μsec, collided signals start traveling back.
• Collided signals reach the respective nodes after time = Half of propagation delay
• Collided signals reach the respective nodes after time = 22.5 μsec / 2 = 11.25 μsec.
• Thus, at t = 22.5 μsec, collided signals reach the respective nodes.
• At t = 22.5 μsec,

 

• As soon as nodes discover the collision, they immediately release the jamming signal.
• Time taken to finish transmitting the jam signal = 50 bit time = 50 bits/ 10 Mbps = 5 μsec.

 

Thus,

Time at which the jamming signal is completely transmitted

= 22.5 μsec + 5 μsec

= 27.5 μsec 

Answer 3

Is it 27.5microsec.........??

Comments

No, Answer is 50 micro sec.