EE555: "Broadband Networks Architecture" Professor A. Zahid, FALL Semester 2007 HW#2: Solution Stalling’s Problems (5.4=15pts, 5.5=10pts, 10.2=15pts):Extra Problem 1 (20pts, 10 per each part):Extra problem 2 (10pts): A user is connected to a FR network through a T-1 line. The Committed Information Rate (CIR) is 1 Mbps with a Committed Burst Size, Bc, of 5 Mbits each 5 seconds. The Excess Burst Size, Be, is 1 Mbits every 5 seconds. Answer the following questions: a) What is the access rate? b) What is the maximum data rate the user can use all of the time without worrying about frames being discarded? c) If the user wants to take a risk, what is the maximum data rate the user can be used with no chance of frames being discarded if there is no congestion? d) Can the user send data at a constant bit rate of 1.4 Mbps all the time? e) The user sends data at a rate of 1.4 Mbps for 2 seconds and nothing for the next three seconds. Is there a danger of discarding if there is no congestion? Is there a danger of discarding if there is congestion? Explain Solution: a) 1.544 Mbps. b) 1 Mbps (CIR). c) 1.2 Mbps = (Bc+Be)/Tc. d) No. Some frames will be discarded since the rate exceeds CIR. e) There is no risk of discarding frames (whether there is congestion or not) since the average data rate is 0.56 Mbps which is less than the CIR. Extra Problem 3 (10pts): The purpose of HEC field is to protect against errors in the header of ATM cells. Assume bit errors occur at random and that the HEC can correct all single errors and can detect all double errors only. Assume that the bit error rate p = 10^(-6). • What is the probability that the cell will be delivered to the right destination? • What is the probability that the cell is misinserted? • What is the probability that the cell is dropped? Solution:Extra Problem 4 (10pts): The IP datagram for a TCP ACK message is 40 bytes long (20 bytes of TCP header, 20 bytes of IP header). Assume this ACK is traversing an ATM network that uses AAL5 to encapsulate IP packets. How many ATM cells are needed to carry the TCP ACK? Solution: The length of the AAL/5 CS-PDU into which the ACK is encapsulated is 48 bytes (no padding is needed, just 8 bytes of trailer) and this fits in a single ATM cell. Extra problem 5 (10pts): In your opinion, what are the main problems in deploying ATM in the Internet? Suppose you have a connection between host "A" and host "B". IP packets pass through an IP network, then through an ATM network and then back through an IP network. What happens during link failures (consider a link failure inside an IP network, a failure in the link connecting the IP network to the ATM network and finally a link failure inside the core ATM network) core ATM network. Solution: ATM is usually deployed as an ATM cloud towards the backbone of the Internet (not to the desktop and not towards the edges). The main problems is that ATM is connection-oriented (meaning that it uses virtual circuit with locally significant labels, needs a setup phase, admission control, where cells follow same path). Whereas IP (which is likely to cross ATM clouds), is connection-less datagram service (meaning that it uses globally unique addresses to make route decisions, does not perform setup or admission control, and packets may follow different paths). Hence, there are problems in mapping the IP addresses into ATM addresses (through the address resolution protocol ARP). Also, there are problems in establishing the paths in the ATM cloud. Packets arrive at the ATM cloud and need to be buffered before the connection is established. Also, packets may arrive at different entry points to the ATM clouds, which may require establishing multiple virtual paths for the same IP flow. Furthermore, there is no tear-down message in IP flows in general (for TCP we may use the FIN flag for example). But in general we need to rely on timer mechanisms to tear down the ATM connections. In addition, QoS at the IP level needs to be mapped into ATM QoS, type of service and GFC fields.Finally, routing in ATM must inter-operate with routing in the Internet (they use different routing algorithms that rely on potentially different metrics to make routing decisions. ATM runs its own routing algorithm although it is considered a data-link protocol). If the failure occurs in the Internet (outside of the ATM cloud), then the dynamic Internet routing will re-route the packets around the failure. This occurs transparently (i.e., without the end points being notified). If the route change leads to a change of path such that the entry point to the ATM cloud is changed, then a new path must be setup in the ATM cloud. If the failure occurs in the ATM cloud, then the path must be torn down and re-established, which may lead to a lot of loss