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Two peer processes A(sender) and B(receiver) use 'stop-and—wait ARQ to send packets over a Single link with capacity C. All packets have the same length of
100 bits. The round-trip time (which is the time until A receives an acknowledgment for a sent packet) is equal to 2 seconds. Assume that no packets or Ack's are dropped and that all packets and ACK's arrive error-free. Furthermore,assume that the capacity C is
equal to 100,000 bits per second.
Find the average (transmission) rate (in bits per
seconds) with which process A sends data to
process B? ‘

(a) 34.47 bps (b) 49.97 bps
c) 51.45 bps (d) 67.75 bps
in Computer Networks by
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b) 49.97bps ??
yeah Right.
Please do explain.

2 Answers

+3 votes
Best answer
transmission rate = data size / (TT + 2 PT)

here data size = 100 bits

TT = data size / bandwidth

=100 / 10000 = 1ms


RTT = 2 * PT =2 sec= 2000ms

then transmission rate = 100 / (1+ 2000)

=0.04997 bits/msec

=49.77 bits/sec
by Boss
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Can you please also let me know the Link utilisation in this case?

I think the formula is %LU = (Throughput/Bandwidth)*100  right? I am getting 0.049 which is not matching any options that i have.

The options are
a) 0.0005     b) 0.0002     c) 0.0001     d) 0.0050

Got it, i was finding the link utilisation %, while the question asked link utilisation got ans as A.
0 votes

dont get confused by the by transmission rate means the throughput of system= datasize / total time

100 / (TT+ 2PT)

TT=data size / bandwidth

   = 100/ 100000= 1ms=0.001 sec

2PT=round trip time=2 sec

transmission rate= 100/(2+0.001) = 49.97 bps

this means even the bandwidth provided is 100000 bps, but we can transmit with the rate of 49.97 bps.

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