How to find transmission rate ,stop and wait ARQ protocol,help please?


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


is it b?? 49.97bps??


Just calculate throughput for this question:

Length of packet (L)= 100bits
2Tp = Round trip time = 2s
Bandwidth(B)= 100,000 bits/second
Transmission time (Tt) = L/B

Thoughput = Length of packet/(Transmission time + Propagation Delay)

or you can use,
Throughput = Efficiency x Bandwidth

where efficiency = 1/(1+2Tp/Tt)

Option (B) would be the answer.