The distance from the earth to a distant planet is approximately 9 x 10 10 meter
ID: 665351 • Letter: T
Question
The distance from the earth to a distant planet is approximately 9 x 1010 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 108 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?
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?
Explanation / Answer
In the above question you mentioned header size and processing delays at the sender and received are negligible. I assume that they are both of size 0.
a) What is the channel utilization of the stop-and-wait protocol?
each period is Tx + propagation + ack propagation = Tx + RTT
**Channel utilization = the fraction of the transmission capacity of a communication channel that contains
data (frames) transmissions
this is very poor utilization since the medium is very long and there is a lot of time wasted for waiting for the ackownoldgement.
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?
for sliding window protocol since there is no error probability I assume is 0. to get maximum 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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