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EE555-USC-Spring08-Silvester-HW1_solutions

EE555-USC-Spring08-Silvester-HW1_solutions - EE555...

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EE555 – Broadband Network Architectures (Spring 2009) Homework 1 Solutions: Networking Basics and Modeling Question 1 . (10 points) i. In datagram packet switching, a routing decision needs to be made for each packet at every intermediate node, causing increased routing load and additional delay. However, in virtual circuit packet switching, no routing decision needs to be made once the circuit is set up, only a simple (hardware based) table look-up is needed. ii. In addition, if not all packets follow the same path (as is possible in datagram) there will be increased jitter and the possibility of out of order delivery iii. In datagram packet switching, there is no delay in setting up a circuit, whereas in virtual packet switching, there is usually a delay in connection establishment, typically it has to wait a full RTT before sending out the first packet. iv. In virtual circuit switching, while the connection request contains the full address for the destination, each data packet contains only a small identifier, making the per-packet header overhead small. On the contrary, in datagram packet switching, there is significant control information and processing overhead. However, in many implementations of VC, the VC label is added as extra overhead in front of the IP header. v. In virtual circuit switching, if a switch or a link fails, the connection is broken and a new path needs to be established. This is not the case in datagram switching. vi. Datagram packet switching has the potential to route around congestion resulting in greater line efficiency. In practice this is difficult to achieve (over short time horizons). vii. Connection setup in virtual circuit switching provides an opportunity to reserve resources; this is not the case for datagram packet switching. Virtual circuit packet switching is commonly implemented in a WAN either by permanent virtual circuits (PVC) or switched virtual circuits (SVC) often using MPLS in today’s networks. Question 2. (10 points) Assuming there is no other traffic in the network and that packets can be sent continuously (i.e. no flow or congestion control), we can consider that the file is transmitted in a continuous stream of packets back to back. (Note we are assuming that there are 7 routers and 6 links) Number of packets = 10 7 /10 4 = 10 3 Transmission time for one packet = 11000/10 6 s = 11ms Total transmission time = 11s The ETE propagation delay is 18ms Total processing time = 7*0.1ms = 0.7ms (processing will also be pipelined) In addition, we must add the store and forward delay (for the first packet) – it must be transmitted 6 times
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