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P7. In this problem, we consider sending real-time voice from Host A to Host B
over a packet-switched network (VoIP). Host A converts analog voice to a
digital 64 kbps bit stream on the fly. Host A then groups the bits into 56-byte
packets. There is one link between Hosts A and B; its transmission rate is 2
Mbps and its propagation delay is 10 msec. As soon as Host A gathers a
packet, it sends it to Host B. As soon as Host B receives an entire packet, it
converts the packet’s bits to an analog signal. How much time elapses from
the time a bit is created (from the original analog signal at Host A) until the
bit is decoded (as part of the analog signal at Host B)?

+1 vote
64 kbps means  in 1 sec 64 k bits converted hence time for 1 bit=1/64k

time taken to group 56 bytes=56*8*1/64*10^-3=7 m sec (encoding time)

same time will be taken to decode=7 m sec ( decoding time)

transmission time=56*8/2*10^-6=224 micro sec

propagation time=10 m sec

if we consider only one such packet in that case total time=7 m sec

hence total time taken=TT+PT+encoding time+decoding time

=224 micro sec +10 m sec +14 m sec

=24.224 m sec
edited by
0
Why 2*PT?

Why propagation delay is taken twice?
0
sorry for that..i took another thing in consideration..now it has been edited