ProblemConsider a packet switched network with two intermediate nodes between sender A and receiver B (that is, A x y B). Define Cas the capacity of the links in bps. Define Das the bit size of application data. Define Has the header size in bits. Define Pas the number of packets for this application data. Assume no processing and propagation delays. Nodes receive the whole packet and then transmit to the next node.a)Derive a formula for the end-to-end delay for that particular application data.b)Given that C = 5,000 bps, D = 50,000 bits, H = 500 bits, what would be the optimum number of packets that minimizes the total end-to-end delay? Calculate the delay too.Solutiona) Tend-to-end = (Total #of bits transmitted) / (Channel Capacity)Since D is the total size of application data, and P is the packet number, each packet consists of D/P bits.The first packet arrives with a three packets delay. The total number of packets on the wire will be (P+2)Time necessary to send one packet is :(H + (D/P)) / CThen total time necessary to send (P+2) packets will be:Tend-to-end = (P+2)*(H + (D/P)) / Cb)
This is the end of the preview.
access the rest of the document.