Frame size (jamming signal)

Question

Answer 1

125000 Bytes (or 125 KB) should be the right answer.

Jamming signal is like a negative acknowledgement used to inform all the nodes connected to a link that collision occurred.

Suppose a node P is transmitting frames to other node Q which is 200 km away from P.

Now suppose the frame sent by P to Q collides with another frame, at a point very near to Q after travelling 199.99 km, then to inform node P about collision, the jamming signal has to travel almost 200 km.

So in the worst case, P will get informed about collision only after two way propagation delay.

& if P receives jamming signal after it has completed its frame transmission, then will never know that which frame caused the collision & that the frame that it sent to Q has been collided & lost.

So the frame that P is going to transmit should be large enough to be transmitted until a longest two way propagation time.

That is frame transmission time must be at least as much as 2*Propagation time.

Here 2 * Propagation time = 2 * 10^-3 seconds,

& Transmission rate = 500 Mega bits per second.

Let frame size be L bits.

Then (L / 500 * 10⁶) >= 2 * 10^-3

Hence L >= 10⁶ bits or L >= 1,25,000 Bytes or L >= 125 KB.

Comments

Yes, it will be 125007 Bytes as we have to give answer in bytes, we must consider jamming signal transmission time.I am sorry, My bad.

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But answer in test series given as 125000 , so, here my question is in Gate what should we consider - 125000 or 125006 ?

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If the size of jamming signal is given then, then we must consider its transmission time & 125006 Bytes will be right answer to this question.

Answer 2

TT ≦ 2PT  
 
   x/500*10^6 = 2*(200*10^3 / 2*10^8)

x= 2000 bits
x= 250 bytes