The distance from earth to a distant planet is approximately 9 × 10^10 m. (1) Wh

Question

The distance from earth to a distant planet is approximately 9 × 10^10 m. (1) What is the channel utilization of a stop-and-wait protocol is used for frame transmission on a 64 Mbits per sec point-to-point link? Assume that the frame size is 32 KBytes and the speed of light is 3 × 10^8 m/s. (2) Suppose a sliding window protocol is used instead. For what send window size will the link utilization be 100%? You may ignore the protocol processing times at the sender and the receiver.

Please answer part 2, and show / explain all work

Explanation / Answer

Distance = 9 x 10^10 m
Datarate = 64 Mbps
Size = 32 KByte = 256 Kbits
Propagation Speed = 3 x 10^8 m/s

Transmission Delay = Packet Size / Datarate = 256 Kb/ 64 Mbps = 4 ms = 0.004 s
Propagation Delay = Distance / Propagation Speed = 9×10^10 / 3×10^8 = 300 s

Assuming no processing delay on receiver and ack size is negligible, only one packet is sent in RTT ie 600 s
Utilization = 0.004 / (0.004 + 600) = 6 x 10^-6

suppose a sliding window protocol is used instead

In order to have efficiency of 100%, the window size should be equal to number of packets that can be sent in one RTT
So RTT here is 600 s, trasmission time of 1 packet is 4 ms, so we can send 1.5×10^5 packets in 1 RTT
So window size should be 150000

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