Question

Station A uses 32 byte packets to transmit messages to Station B using a sliding window protocol. The round trip delay between A and B is 80 milliseconds and the bottleneck bandwidth on the path between A and B is 128 kbps. What is the optimal size that A should use ?

• A. 40
• B. 160
• C. 20
• D. 320

Answer 1

The correct option is A 40

Given round trip delay t = 80 ms
$=80\times {10}^{-3}$ sec
$R=128kbps=128\times {10}^{3}bps$
L = Rt
$=128\times {10}^3\times 80\times {10}^{-3}=128\times 80$
So, optional window size = n
$=\frac{128\times 80}{32\times 8}=\frac{10240}{256}$
$n=40$