Question

1. For a 1Gbps link, 10 ms prop. delay, 1000-bit packet, compute the utilization for: a. Stop and wait protocol
b. Sliding window (window size=10)

Answer 1

For the link it is given,

Packet size =1000 bit

Bandwidth= 1 Gbps

propagation delay=10 ms

Transmission delay = $\frac{packetsize}{Bandwidth}$ =$\frac{1000}{10^{6}}$=1 ms

so total delay = transmission delay+ 2*propagation delay= 1+20=21 ms  [assuming processing and queuing delay negligible]

so now bandwidth delay product = bandwidth* delay =$10^{6}$ bps * $21 *$$10^{-3}$ s =$21000$ bit

so the system can send 21000 bits during the time it takes for the data to go from sender to receiver and the acknowledgement to come back.

However the system can send only 1000 bits.

so utilization = $\frac{1000}{21000}=\frac{1}{21}$=$4.76$%.

Part 2 :-

The above calculation will remain same just here we are sending window of 10 packets together .

so with 10 packets the amount of data we can send =$10*1000 =10000$ bits

but the bandwidth delay product will remain same which is 21000 bits .

so we can send 21000 bits but we are only sending 10000 bits .

so Utilization =$\frac{10000}{21000}=\frac{10}{21}$=$47.61$ % .