ATM is a network mechanism intended to accommodate real-time traffic as well as bulk data transfer. We present ATM here as a LAN layer, for which it is still sometimes used, but it was originally proposed as a replacement for the IP layer as well, and, to an extent, the Transport layer. These broader plans were not greeted with universal enthusiasm within the IETF. When used as a LAN layer, IP packets are transmitted over ATM as in5.5.1 ATM Segmentation and Reassembly.
A distinctive feature of ATM is its small packet size. ATM has its roots in the telephone industry, and was therefore particularly intended to support voice. A significant source of delay in voice traffic is the packet
fill time: at DS0 speeds (64 kbps), voice data accumulates at 8 bytes/ms. If we are sending 1 kB packets, this means voice is delayed by about 1/8 second, meaning in turn that when one person stops speaking, the earliest they can hear the other’s response is 1/4 second later. Slightly smaller levels of voice delay can introduce an annoying echo. Smaller packets reduce the fill time and thus the delay: when voice is sent over IP (VoIP), one common method is to send 160 bytes every 20 ms.
ATM took this small-packet strategy even further: packets have 48 bytes of data, plus 5 bytes of header. Such small packets are often calledcells. To manage such a small header, virtual-circuit routing is a necessity. IP packets of such small size would likely consume more than 50% of the bandwidth on headers, if the LAN header were included.
Aside from reduced voice fill-time, other benefits to small cells are reduced store-and-forward delay and minimal queuing delay, at least for high-priority traffic. Prioritizing traffic and giving precedence to high- priority traffic is standard, but high-priority traffic is never allowed tointerrupttransmission already begun of a low-priority packet. If you have a high-priority voice cell, and someone else has a 1500-byte packet just started, your cell has to wait about 30 cell times, because 1500 bytes is about 30 cells. However, if their low-priority traffic is instead made up of 30 cells, you have only to wait for their first cell to finish; the delay is 1/30 as much.
ATM also made the decision to requirefixed-sizecells. The penalty for one partially used cell among many is small. Having a fixed cell size simplifies hardware design, and, in theory, allows it easier to design for parallelism.
Unfortunately, the designers of ATM also chose to mandateno cell reordering. This means cells can use a smaller sequence-number field, but also makes parallel switches much harder to build. A typical parallel switch design might involve distributing incoming cells among any of several input queues; the queues would then handle the VCI lookups in parallel and forward the cells to the appropriate output queues. With such an architecture, avoiding reordering is difficult. It is not clear to what extent the no-reordering decision was related to the later decline of ATM in the marketplace.
ATM cells have 48 bytes of data and a 5-byte header. The header contains up to 28 bits of VCI information, three “type” bits, one cell-loss priority, or CLP, bit, and an 8-bit checksum over the header only. The VCI is divided into 8-12 bits of Virtual Path Identifier and 16 bits of Virtual Channel Identifier, the latter supposedly for customer use to separate out multiple connections between two endpoints. Forwarding is by full switching only, and there is no mechanism for physical (LAN) broadcast.
5.5.1 ATM Segmentation and Reassembly
Due to the small packet size, ATM defines its own mechanisms for segmentation and reassembly of larger packets. Thus, individual ATM links in an IP network are quite practical. These mechanisms are called
ATM Adaptation Layers, and there are four of them: AALs 1, 2, 3/4 and 5 (AAL 3 and AAL 4 were once separate layers, which merged). AALs 1 and 2 are used only for voice-type traffic; we will not consider them further.
The ATM segmentation-and-reassembly mechanism defined here is intended to apply only to largedata; no cells are ever further subdivided. Furthermore, segmentation is always applied at the point where the data enters the network; reassembly is done at exit from the ATM path. IPv4 fragmentation, on the other hand, applies conceptually to IP packets, and may be performed by routers within the network.
For AAL 3/4, we first define a high-level “wrapper” for an IP packet, called the CS-PDU (Convergence Sublayer - Protocol Data Unit). This prefixes 32 bits on the front and another 32 bits (plus padding) on the rear. We then chop this into as many 44-byte chunks as are needed; each chunk goes into a 48-byte ATM payload, along with the following 32 bits worth of additional header/trailer:
• 2-bittypefield:
– 10: begin new CS-PDU
– 00: continue CS-PDU
– 01: end of CS-PDU
– 11: single-segment CS-PDU
• 4-bit sequence number, 0-15, good for catching up to 15 dropped cells • 10-bit MessageID field
• CRC-10 checksum.
We now have a total of 9 bytes of header for 44 bytes of data; this is more than 20% overhead. This did not sit well with the IP-over-ATM community (such as it was), and so AAL 5 was developed.
AAL 5 moved the checksum to the CS-PDU and increased it to 32 bits from 10 bits. The MID field was discarded, as no one used it, anyway (if you wanted to send several different types of messages, you simply created several virtual circuits). A bit from the ATM header was taken over and used to indicate:
• 1: start of new CS-PDU
• 0: continuation of an existing CS-PDU
The CS-PDU is now chopped into 48-byte chunks, which are then used as the entire body of each ATM cell. With 5 bytes of header for 48 bytes of data, overhead is down to 10%. Errors are detected by the
CS-PDU CRC-32. This also detects lost cells (impossible with a per-cell CRC!), as we no longer have any cell sequence number.
For both AAL3/4 and AAL5,reassemblyis simply a matter of stringing together consecutive cellsin order of arrival, starting a new CS-PDU whenever the appropriate bits indicate this. For AAL3/4 the receiver has to strip off the 4-byte AAL3/4 headers; for AAL5 the receiver has to verify the CRC-32 checksum once all cells are received. Different cells from different virtual circuits can be jumbled together on the ATM “backbone”, but on any one virtual circuit the cells from one higher-level packet must be sent one right after the other.
A typical IP packet divides into about 20 cells. For AAL 3/4, this means a total of 200 bits devoted to CRC codes, versus only 32 bits for AAL 5. It might seem that AAL 3/4 would be more reliable because of this, but, paradoxically, it was not! The reason for this is that errors arerare, and so we typically have one or at most two per CS-PDU. Suppose we have only a single error,iea single cluster of corrupted bits small enough that it is likely confined to a single cell. In AAL 3/4 the CRC-10 checksum will fail to detect that error (that is, the checksum of the corrupted packet will by chance happen to equal the checksum of the original packet) with probability 1/210. The AAL 5 CRC-32 checksum, however, will fail to detect the error with probability 1/232. Even if there are enough errors that two cells are corrupted, the two CRC-10s together will fail to detect the error with probability 1/220; the CRC-32 is better. AAL 3/4 is more reliable only when we have errors in at least four cells, at which point we might do better to switch to an error-correctingcode.
Moral: one checksum over the entire message is often better than multiple shorter checksums over parts of the message.
5.6 Epilog
There are not many wired LANs that are not called “Ethernet”.
While it is sometimes tempting (in the IP world at least) to write off ATM as a niche technology, virtual circuits are a serious conceptual alternative to datagram forwarding. As we shall see in 25 Quality of Service, IP has problems handling real-time traffic, and virtual circuits offer a solution. The Internet has so far embraced only small steps towards virtual circuits (such as MPLS, 25.12 Multi-Protocol Label Switching (MPLS)), but they remain a tantalizing strategy.
5.7 Exercises
Exercises are given fractional (floating point) numbers, to allow for interpolation of new exercises.Exercises marked with a♢have solutions or hints at24.5 Solutions for Other LANs.
1.0. Suppose remote host A uses a VPN connection to connect to host B, with IP address 200.0.0.7. A’s normal Internet connection is via deviceeth0with IP address 12.1.2.3; A’s VPN connection is via device
ppp0 with IP address 10.0.0.44. Whenever A wants to send a packet via ppp0, it is encapsulated and forwarded over the connection to B at 200.0.0.7.
(a). Suppose A’s IP forwarding table is set up so that all traffic to 200.0.0.7 useseth0and all traffic to anywhere else usesppp0. What happens if an intruder M attempts to open a connection to A at 12.1.2.3?
What route will packets from A to M take?
(b). Suppose A’s IP forwarding table is (mis)configured so thatalloutbound traffic usesppp0. Describe what will happen when A tries to send a packet.
2.0. Suppose remote host A wishes to use a TCP-based VPN connection to connect to host B, with IP address 200.0.0.7. However, the VPN software is not available for host A. Host A is, however, able to run that software on a virtual machine Vhosted byA; A and V have respective IP addresses 10.0.0.1 and 10.0.0.2 on the virtual network connecting them. V reaches the outside world through network address translation (9.7 Network Address Translation), with A acting as V’s NAT router. When V runs the VPN software, it forwards packets addressed to B the usual way, through A using NAT. Traffic to any other destination it encapsulates over the VPN.
Can A configure its IP forwarding table so that it can make use of the VPN? If not, why not? If so, how? (If you prefer, you may assume V is a physical host connecting to a second interface on A; A still acts as V’s NAT router.)
3.0. Token Bus was a proprietary Ethernet-based network. It worked like Token Ring in that a small token packet was sent from one station to the next in agreed-upon order, and a station could transmit only when it had just received the token.
(a). If the data rate is 10 Mbps and the token is 64 bytes long (the 10-Mbps Ethernet minimum packet size), what is the average wait to receive the token on an idle network with 40 stations? (The average number of stations the token must pass through is 40/2 = 20.) Ignore the propagation delay and the gap Ethernet requires between packets.
(b)♢. Sketch a protocol by which stations can sort themselves out to decide the order of token transmission; that is, an order of the stations S0. . . Sn-1where station Sisends the token to station S(i+1) mod n.
4.0.[SM90]contained a proposal for sending IP packets over ATM as N cells as in AAL-5, followed by one cell containing the XOR of all the previous cells. This way, the receiver can recover from the loss of any one cell. Suppose N=20 here; with the SM90 mechanism, each packet would require 21 cells to transmit; that is, we always send 5% more. Suppose thecellloss-rate is p (presumably very small). If we send 20 cells without the SM90 mechanism, we have a probability of about 20p that any one cell will be lost, and we will have to retransmit the entire 20 again. This gives an average retransmission amount of about 20p extra packets. For what value of p do the with-SM90 and the without-SM90 approaches involve about the same total number of cell transmissions?
5.0. In the example in5.4 Virtual Circuits, give the VCI table for switch S5. 6.0. Suppose we have the following network:
A S1 S2 B
C S3 S4 D
The virtual-circuit switching tables are below. Ports are identified by the node at the other end. Identify all the connections. Give the path for each connection and the VCI on each link of the path.
SwitchS1:
VCIin portin VCIout portout
1 A 2 S3
2 A 2 S2
3 A 3 S2
SwitchS2:
VCIin portin VCIout portout
2 S4 1 B
2 S1 3 S4
3 S1 4 S4
SwitchS3:
VCIin portin VCIout portout
2 S1 2 S4
3 S4 2 C
SwitchS4:
VCIin portin VCIout portout
2 S3 2 S2
3 S2 3 S3
4 S2 1 D
7.0.♢We have the same network as the previous exercise:
A S1 S2 B
C S3 S4 D
The virtual-circuit switching tables are below. Ports are identified by the node at the other end. Identify all the connections. Give the path for each connection and the VCI on each link of the path.
SwitchS1:
VCIin portin VCIout portout
1 A 2 S2
SwitchS2:
VCIin portin VCIout portout
2 S1 3 S4
1 B 2 S4
SwitchS3:
VCIin portin VCIout portout
2 S1 2 S4
1 S4 3 S1
SwitchS4:
VCIin portin VCIout portout
3 S2 2 D
2 S2 1 S3
8.0. Suppose we have the following network:
A S1 S2 B
C S3 S4 D
Give virtual-circuit switching tables for the following connections. Route via a shortest path.
(a). A–D (b). C–B, via S4 (c). B–D
(d). A–D, via whichever of S2 or S3 wasnotused in part (a)
9.0. Below is a set of switches S1 through S4. Define VCI-table entries so the virtual circuit from A to B follows the path
AÝÑS1ÝÑS2ÝÑS4ÝÑS3ÝÑS1ÝÑS2ÝÑS4ÝÑS3ÝÑB That is, each switch is visitedtwice.
A S1 S2
B S3 S4
At the lowest (logical) level, network links look like serial lines. In this chapter we address how packet structures are built on top of serial lines, via encoding and framing. Encoding determines how bits and bytes are represented on a serial line; framing allows the receiver to identify the beginnings and endings of packets.
We then conclude with the high-speed serial lines offered by the telecommunications industry, T-carrier and SONET, upon which almost all long-haulpoint-to-pointlinks that tie the Internet together are based.
6.1 Encoding and Framing
A typical serial line is ultimately a stream ofbits, not bytes. How do we identify byte boundaries? This is made slightly more complicated by the fact that, beneath the logical level of the serial line, we generally have to avoid transmitting long runs of identical bits, because the receiver may simply lose count; this is theclock synchronizationproblem (sometimes called the clock recovery problem). This means that, one way or another, we cannot always just send the desired bits sequentially; for example, extra bits are often inserted to break up long runs. Exactly how we do this is theencodingmechanism.
Once we have settled the transmission of bits, the next step is to determine how the receiver identifies the start of each new packet. Ethernet packets are separated by physical gaps, but for most other link mechanisms packets are sent end-to-end, with no breaks. How we tell when one packet stops and the next begins is the
framingproblem. To summarize:
• encoding: correctly recognizing all the bits in a stream • framing: correctly recognizing packet boundaries These are related, though not the same.
For long (multi-kilometer) electrical serial lines, in addition to the clock-related serial-line requirements we also want the average voltage to be zero; that is, we want no DC component. We will mostly concern ourselves here, however, only with lines short enough for this not to be a major concern.
6.1.1 NRZ
NRZ (Non-Return to Zero) is perhaps the simplest encoding; it corresponds to direct bit-by-bit transmission of the 0’s and 1’s in the data. We have two signal levels,loandhi, we set the signal to one or the other of these depending on whether the data bit is 0 or 1, as in the diagram below. Note that in the diagram the signal bits have been aligned with thestartof the pulse representing that signal value.
0 0 1 0 1 1 0 1 1 1 1 0 0 0 0 0 1 NRZ Encoding: 1 = hi, 0 = lo
NRZ replaces an earlier RZ (Return to Zero) encoding, in which hi and lo corresponded to +1 and -1, and between each pair of pulses corresponding to consecutive bits there was a brief return to the 0 level. One drawback to NRZ is that we cannot distinguish between 0-bits and a signal that is simply idle. However, the more serious problem is the lack ofsynchronization: during long runs of 0’s or long runs of 1’s, the receiver can “lose count”,egif the receiver’s clock is running a little fast or slow. The receiver’s clock can and does resynchronize whenever there is atransitionfrom one level to the other. However, suppose bits are sent at one per µs, the sender sends five 1-bits in a row, and the receiver’s clock is running 10% fast. The signal sent is a 5-µs hi pulse, but when the pulse ends the receiver’s clock reads 5.5 µs due to the clock speedup. Should this represent five 1-bits or six 1-bits?
6.1.2 NRZI
An alternative that helps here (though not obviously at first) isNRZI, or NRZ Inverted. In this encoding, we represent a 0-bit as no change, and a 1-bit as atransitionfrom lo to hi or hi to lo:
0 0 1 0 1 1 0 1 1 1 1 0 0 0 0 0 1
NRZI Encoding: 1 = transition, 0 = no transition
Now there is a signal transition aligned above every 1-bit; a 0-bit is represented by the lack of a transition. This solves the synchronization problem for runs of 1-bits, but does nothing to address runs of 0-bits. However, NRZI can be combined with techniques to minimize runs of 0-bits, such as 4B/5B (below).
6.1.3 Manchester
Manchester encoding sends the data stream using NRZI, with the addition of a clock transitionbetween each pair of consecutive data bits. This means that the signaling rate is now double the data rate, eg 20 MHz for 10Mbps Ethernet (which does use Manchester encoding). The signaling is as if we doubled the bandwidth and inserted a 1-bit between each pair of consecutive data bits, removing this extra bit at the receiver:
clock clock clock clock clock clock clock clock clock clock clock clock clock clock clock clock clock