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(a) The distance from earth to a distant planet is approximately 9 × 1010 m. What is the
channel utilization if a stop-and-wait protocol is used for frame transmission on a 64
Mbps point-to-point link? Assume that the frame size is 32 KB and the speed of light
is 3 × 108 m/s.

(b 28)In the previous problem, 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.
in Computer Networks by Boss (11k points)
edited by | 59 views

1 Answer

+1 vote
(a) Length of frame=32 KB=32*1024*8 bits

Bandwidth=64 Mbits/sec

Transmission delay($T_{t}$)=$\frac{32*1024*8}{64*10^6}$ sec

Distance=$9*10^9$ m

Propagation speed=$3*10^8$ m/s

Propagation Delay($T_{p}$)=$\frac{9*10^{10}}{3*10^8}$ sec


Efficiency=$\frac{1}{1+2a}*100$ %

=0.00068 %

(b) Let window size be N

for 100% utilization $\frac{N}{1+2a}*100=100$

i.e. N=1+2a=$1+2*\frac{T_{p}}{T_{t}}$
by Loyal (5.2k points)
edited by

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