Station A uses 32 byte packets to transmit messages to Station

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 ms and the bottleneck bandwidth on the path between A and B is 12 kbps. What is the optimal window size that A should use?

Bandwidth delay production = Roundtrip delay × Bottle neck bandwidth = 80 × 10^–3 × 128 × 1024 bit

=	80 × 10^–3 × 128 × 1024	bit
	8

[because 1 byte = 8 bit]

Now, Optimal size window =	Bandwidth Delay Production	bit
	Packet size

=	80 × 10^–1 × 128 × 1024	bits = 40
	8 × 32