8/3/2019 sol reti

1/2

8/3/2019 sol reti

2/2

For L = 2250 U = 0.718

For L = 2500 U = 0.718

For L = 3000 U = 0.717

For L = 3500 U = 0.716

Hence the maximum data rate will occur for frames of around length 2250 bits.

5. To improve network throughput and efficiency in a cost-effective manner, what would

you do first?

(i) Double the bandwidth of the communication channel.

(ii) Improve the routing algorithm to reduce the processing and queuing delays.

(iii) Increase the window size of the flow control algorithm.

What would you do next? Assume a sliding-window protocol is being used.

Solution

(i) Double the bandwidth would halve the transmission time Tframe. This would also double a

(anew = 2aold). If W is large enough (i.e. W 1 + 2anew) then this would double the datathroughput, but with more packets in the network this can be offset by increased queuing

delays. But if W was initially chosen optimally (i.e. W = 1 + 2a old) then this would yield little

improvement on the data throughput and halve the link utilisation (although the data

throughput would actually increase slightly). Not very cost-effective solution.

(ii) The net Tprop would decrease slightly, but more importantly a would decrease yielding both

improved link utilisation and the need for a smaller W. The latter would make the network

more robust to congestion. Very cost-effective solution.

(iii) Increasing W is a brute-force to improve link utilisation since it would increase the

number of frames in the network, increase end-to-end delays and even create congestion.

Option (ii) first then (i) in tandem with (iii).