LECTURE 1 Q. Why don't all TE1 Devices need to connect to NT2 equipment? A. NT2 equipment provides higher layer functionality such as that found in PABXs, LANs and terminal controllers. If a TE1 device is connected to the network in a simple configuration, then it does not need the functionality provided by an NT2 interface and so can connect through an NT1 interface. Q. The BRI provides 2 B channel and 1 D channel, total 144 kbps. However a BRI interface is defined at 192 kbps. Why? A. The basic rate interface includes additional information such as D channel echo bits, framing bits and DC balancing bits. Q. In what way might a carrier treat a 64 kbps 8kHz structured speech bearer service differently to a 64 kbps unrestricted, 8kHz structured bearer service? A. If the traffic being carried is known to be voice then schemes can be implemented to either improve the quality of the voice transmission using the same bit rate, or to compress the voice so that more voice traffic can be carried than could be otherwise. Q. Which bearer services might be used for a G4 fax? A. Any unstructured service. That is, any service where it is not assumed the traffic is voice. LECTURE 2 Q. The BRI D channel contention algorithm would fail if any TE1 sent more than 8 consecutive '1's as data over the D channel. Why does this never happen? A. An examination of the NISDN protocol stack shows that data is transmitted over the D channel through the LAPD protocol. The LAPD protocol deals in units of frames. Every frame begins and ends with a flag field consisting of the bit sequence 01111110. To avoid confusion between data and flag fields, any data consisting of five consecutive '1's has a '0' inserted after it. This process is called bit stuffing. Consequently, the physical layer never sees more than six consecutive '1's. Q. Why is the overhead for the BRI so much greater than for the PRI? A. The main reason is that the BRI needs a contention algorithm for D channel access. This contention algorithm is implemented through echoing back the D channel bits which adds to the overhead. Q. Why is there no contention mechanism for PRI? A. The PRI is only ever point to point. There is only ever one device able to use the D channel and so there is no contention for it. Q. How would data consisting of the bit sequence 0111110 be coded within a LAPD frame? A. 01111100 Q. Why does LAPD define supervisory frames for flow and error control when Information frames can piggyback the same information? A. Supervisory frames are necessary when information flow is mainly in the one direction. LECTURE 3 Q. Routing is a layer 3 function, but is not supported by Q.931. Why not? A. Although Q.931 has end to end significance and is, consequently a layer 3 protocol, it makes no decisions regarding routing. This is carried out by the Signalling System 7. Q. Why is there a distinction between the control and information planes within ISDN? A. The control plane is mostly concerned with reliability. Consequently, it defines a mesh network to ensure massive redundancy within the network. The information plane is concerned with performance as well as reliability so does not require the same level of redundancy. LECTURE 4 Q. Why is connection admission control in B-ISDN more difficult than in N-ISDN? A. B-ISDN applications have much more complicated traffic descriptions and Quality of Service requirements than N-ISDN. Consequently, switches must make much more difficult decisions when deciding whether or not to accept a connection. Q. Is ATM a layer 2 (data link) or layer 3 (network) protocol? A. ATM has full routing capabilities and so is a network protocol. However, in classical IP over ATM it is treated as a subnet technology (layer 2). Q. Why not overcome the time delay propagation problem by installing switches with large buffers? A. Large buffers add to average delay and increase variability of delay within a switch, making it difficult to conform to quality of service guarantees. Q. Suppose we have a variable bit rate encoded video sequence whose mean bandwidth is 1 Mbps and standard deviation is 0.5 Mbps. Assuming independence between the video streams, what will be the mean and standard deviation of an aggregate stream consisting of 10 such sequences? 100 sequences? 1000 sequences? A. This question is an application of the Central Limit Theorem. The mean of the aggregate traffic is proportional to the number of sequences, while the standard deviation is proportional to the square root of the number of sequence. 10 sequences, mean 10, sd 1.5 100 sequences, mean 100, sd 5 1000 sequences, mean 1000 sd 16. LECTURE 5 Q. ATM has been attacked as inefficient since it has a large cell header. What is the transmission efficiency of ATM? A. An ATM cell has 48 octets of data within a 53 octet cell. So its transmission efficiency is 48 / 53 = 90.6 % Q. Write pseudo-code describing an algorithm to implement early packet discard. A. If switch buffer space less than threshold then Repeat Drop cells belonging to VPI/VCI Until cell with PTI set found Q. Why does the header CRC (the HEC field) need to be recalculated at each hop in an ATM network? A. The HEC is calculated over the full header. If the header changes then the HEC will have to recalculated. The VPI/VCI fields within the header have local significance only and so will usually change. Consequently the HEC will have to be recalculated. Other fields such as the PTI and CLP may also change. LECTURE 6 Q. What cell delay variation tolerance is needed to define a peak cell rate of 80 Mbps over a line rate of 150 Mbps? A. By working through the Generic Cell Rate Algorithm it can be calculated that a CDVT of one cell time is necessary to define any peak cell rate. Q. If cell delay variation tolerance is zero and the highest possible peak cell rate is 150 Mbps, the next peak cell rate is 75 Mbps, the next is 50 Mbps. What is the next lowest peak cell rate that can be defined? A. 37.5 Mbps. LECTURE 7 Q. Why is there no contention algorithm for access to the signalling channel in ATM? A. The signalling channel is defined as being a permanent virtual channel with VPI/VCI of 0/5. That is, cells with this VPI/VCI value contain signalling information. When a client wishes to send signalling information it sends them on this channel. There is no need for exclusive access to the channel. Q. In the worked example of PNNI routing what is the next DTL to be pushed onto the stack? A. We have the following stack: A.2.2, A.2.3, ptr=2 A.1, A.2, A.3, ptr=2 A, B, ptr= 1 The top DTL is exhausted so the stack becomes A.1, A.2, A.3, ptr=2 A, B, ptr= 1 From A.2 there is a path to A.3, so the stack becomes A.1, A.2, A.3, ptr=3 A, B, ptr= 1 We now push a new DTL onto the stack to traverse through A.3 A.3.1, A.3.2, ptr=1 A.1, A.2, A.3, ptr=3 A, B, ptr= 1 LECTURE 8 Q. An ATM switch is functioning normally until a video server and video client are connected to it. When video is played back from the server through the switch, other (low bandwidth) applications connected to the switch fail. The video delivered is jerky. What are some possible explanations for this? A. The most likely explanation is that the switch is a 'blocking' switch. That is, cells can be lost within the switch itself. Blocking switches are usually low cost, bus based switches.