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Suppose that the one-way propagation delay for a 100 Mbps Ethernet having 48-bit jamming signal is 1.04 micro-seconds. The minimum frame size in bits is: 

  1. 112
  2. 160
  3. 208
  4. 256
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Answer would be D. 256.

Here, we would have to consider the worst case of collision, viz near the receiver's end. So, here this may either result in an overpowered voltage signal, a 0 power or a underpowered signal depending on the collision. So in case if the signal collision results into a 0 power, i.e; the two signals cancel each other, and since the collision is near receiver, so receiver will immediately stop sending and hence the collision signal will be sensed by only receiver. The receiver will then transmit the jamming signal which would reach the sender to inform about the collision. Hence, sender needs to transmit the packet till it can receive the jamming signal by receiver.

Therefore, 

$T_{t} \geq 2*T_{p} + T_{t_{jam}}$

$\Rightarrow T_{t_{min}} = 2*1.04 *10^{-6} + \frac{48}{BW}$

$\Rightarrow frame size_{min}= 2.08 * BW + 48 = 208 +48 \left[ BW = 100 * 10^{6} bps \right ] = 256bits $

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