Question

A 20 Kbps satellite link has a propagation delay of 400 ms. The transmitter employs to "go back n ARQ" scheme with n set to 10. Assuming that each frame is 100 bytes long, what is the maximum data rate possible?

• A. 10 Kbps
• B. 15 Kbps
• C. 5 Kbps
• D. 20 Kbps

Answer 1

The correct option is A 10 Kbps

Tx = 1000*8 bits/20 Kbps = 40 ms
Tp = 400 ms.
a = Tp / Tx = 400/40 =10
Efficiency of GBN = W/(1 + 2a)
where w = window size
$10=10/(1+20)=10/21$
BW utilization or throughput or max data rate = efficiency * BW =(10/21)*20
It is nearly 10 Kbps