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Compute the fraction of the bandwidth that is wasted on overhead (headers and retransmissions)
for protocol selective repeat on a heavily loaded 50-kbps satellite channel with data
frames consisting of 40 header and 3960 data bits. Assume that the signal propagation
time from the earth to the satellite is 270 msec. ACK frames never occur. NAK frames
are 40 bits. The error rate for data frames is 1%, and the error rate for NAK frames is
negligible. The sequence numbers are 8 bits.
in Computer Networks by Boss (10.5k points) | 57 views

2 Answers

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With a 50-kbps channel and 8-bit sequence numbers, the pipe is always full. The number of retransmissions per frame is about 0.01. Each good frame wastes 40 header bits, plus 1% of 4000 bits (retransmission), plus a 40-bit NAK once every 100 frames. The total overhead is 80.4 bits per 3960 data bits, giving 80.4/(3960 + 80.4) = 1.99%.
by (11 points)
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Here no protocol is mentioned. Frame Sequence No. bits=8

Assume that window size=100

Error rate of data frame=1%

Which means out of every 100 frames transmitted 100*1%=1 frame will be corrupted

There is no ack but there is NACK frame whose overhead is 40 bits

For this corrupted frame we've to send 1 NACK frame

Efficiency=$\frac{100*3960}{100*(3960+40)+40+(3960+40)}$=98.01%

Bandwidth utilization=98.01%

Bandwidth wasted=100-98.01=1.99%
by Active (5k points)

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