Question 41
What will be the efficiency of a Stop and Wait protocol, if the transmission time for a frame is 20ns and the propagation time is 30ns?
Step-by-Step Solution
Detailed Explanation:
$Efficiency η = Ttt / (Ttt + 2 \times Tprop).$
Ttt = 20 ns, Tprop = 30 ns ⇒ Ttt + 2 Tprop = 20 + 60 = 80 ns.
η = 20 / 80 = 0.25 = 25%.