The distance between two stations M and N is L km. All frames

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The distance between two stations M and N is L km. All frames are K bits long. The propagation delay per kilometer is t s. Let R b/s be the channel capacity. Assuming that processing delay is negligible, the minimum number of bits for the sequence number field in a frame for maximum utilization, when the sliding window protocol is used, is

1. log₂ 2LTR + 2K
  K
2. log₂ 2LTR
  K
3. log₂ 2LTR + K
  K
4. log₂ 2LTR + K
  2K

Correct Option: C

We can send (RTT /Transmission Time) number of packets for maximum utilization of the channel, as in this time we get the first ACK back and till that time, we can continue sending packets.
So, ((Transmission Time + 2*Propagation Time)/ transmission Time)
Number of packets should be sent.
Therefore, bits required for the sequence number field :

[log₂ {(K/R +2Lt) / (K / R)}] =		log₂	2LTR + K
			K