The distance from the earth to a distant planet is approximately 9 x 10^1 meters

Question

The distance from the earth to a distant planet is approximately 9 x 10^1 meters. Assume that a stop-and-wait protocol is used for frame transmission on a 64 Mbps point-to-point link, given that the link is noise-less (error-free), the frame size is 32 Kbytes, the speed of light is 3 x 10^8 meters/second, and the header size and processing delays at the sender and received are negligible. (a) What is the channel utilization of the stop-and-wait protocol? (5 points) (b) Assume that a sliding window protocol is used instead of the stop-and-wait protocol. What is the optimum window size for the sender that will achieve maximum utilization of the link? (4 points)

Explanation / Answer

Utilization is the part of time that the medium is being used with "good" byte, that is bytes of payload (no headers or ack/nack frames).

In the question mentioned above there is no header specified and ack size is not given
Theregore I am assuming that they both are of size 0.
for the S&W case:

each period is Tx + propagation + ack propagation = Tx + RTT

Tx=32KB/64Mbps = 0.004 seconds (assuming base 10 for kilo and mega)
RTT= 2 *( 9*10^10m) / (3*10^8m/s) = 600 seconds
Utilization = 0.004 / (600 + 0.004) = 6.667x10^-6 = 6.667x10^-4 %

this is very bad utilization since the medium is very long and there is a lot of time wasted for waiting for the ack.

for sliding window: since there is no error probability I assume is is 0. to get to 100% utilization you need to keep transmitting packets while you wait for the acks that means for the whole period.

period = 600.004 seconds
1 Tx = 0.004 seconds

for non stop Tx you need to transmit 600.004/0.004 packets each period => 150001 should be your window size.

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