(Basic Assumption is that all packets are of same size)
The key here to solve this problem is the fact that packet transmissions from intermidiate switches (routers here) can be overlapped.
Case(1): Message is not at all splitted and no. of packets (of 54B each) is 1...
Transmission Time @ source = 54/bandwidth
Transmission Time @ switch R1 = 54/bandwidth
Transmission Time @ switch R2 = 54/bandwidth
Hence total time= 3*54/bandwidth = 162/bandwidth
Case(2): Message is split in 2 packets (30B each)
Transmission Time @ source = 2*30/bandwidth
Transmission Time @ switch R1 = 30/bandwidth ......{note 2 packets are transmitted on the line in parallel}
Transmission Time @ switch R2 = 30/bandwidth
Hence total time = 120/bandwidth
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Similarly perform with 3 packets of 22B each : Total Time = 110/bandwidth
With 4 packets of 18B each: Total time = 72/b + 18/b + 18/b = 108/bandwidth
With 6 packets of 14B each : Total time = 84/b + 14/b + 14/b = 112/bandwidth
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Thus for sending entire file in minimum time with error-free channel (to save retransmissions of bigger packet sizes) one should go for a split of 4 packets where each packet is having the size of 18B.
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TIP - There was a question in Gate-14 (Set 1) which is based on this very concept :)