A channel has bandwidth $B = 10 Mbps$ and round-trip time RTT = 80 ms. Frame size is 1000 bytes (8000 bits). Calculate the MINIMUM sender window size required to achieve 100% channel utilization.
$Tx = 8000 bits / 10^7 bps = 0.8 ms. RTT = 2 * Tp = 80 ms. Efficiency eta = N * Tx / (Tx + RTT) = 1 => N = (Tx + RTT) / Tx = (0.8 + 80) / 0.8 = 80.8 / 0.8 = 101 frames.$
In a Stop-and-Wait protocol, transmission delay Tx = 2 ms and propagation delay Tp = 48 ms. Calculate the channel utilization percentage (efficiency * 100).
eta = Tx / (Tx + 2*Tp) = 2 / (2 + 2*48) = 2 / 100 = 0.02 = 2%.
A 1 km Ethernet network operates at 100 Mbps with signal speed 2 * 10^8 m/s. What is the MINIMUM required frame size in bits for CSMA/CD collision detection?
$Tp = 1000 / (2 * 10^8) = 5 microseconds. L_{min} = 2 * Tp * B = 2 * (5 * 10^{-6} s) * (100 * 10^6 bps) = 1000 bits.$
In Go-Back-N ARQ using 5-bit sequence numbers, what is the MAXIMUM sender window size Ws?
$Ws_max = 2^k - 1 = 2^5 - 1 = 31.$
In Selective Repeat ARQ using 6-bit sequence numbers, what is the MAXIMUM sender window size Ws?
$Ws_max = 2^{k-1} = 2^{6-1} = 32.$