Chapter 13
MultiSegment Configuration Guidelines
The individual media chapters you've just read covered the basic configuration guidelines for a single segment media system. However, when it comes to building a more complex halfduplex Ethernet system based on repeater hubs, you need to know what the multisegment guidelines have to say.
The official configuration guidelines provide two approaches for verifying the configuration of a halfduplex shared Ethernet channel: Transmission System Model 1 and Transmission System Model 2. Model 1 provides a set of "canned" configuration rules. As long as your halfduplex network system meets these basic rules, it will function correctly in terms of the essential timing specifications. Model 2 provides a set of calculation aids that make it possible for you to evaluate more complex network topologies that aren't covered under the Model 1 configuration rules.
This chapter describes the rules for combining multiple segments with repeater hubs to build complex halfduplex Ethernet systems operating at 10, 100 and 1000Mbps. We begin by looking at the scope of the configuration guidelines. To help make it clear how the guidelines apply to a single Ethernet system, we then need to describe the function of a collision domain. Following that, we describe the Model 1 and Model 2 rules as they apply to each Ethernet system.
Scope of the Configuration Guidelines
The configuration guidelines apply to the Ethernet equipment described in the IEEE 802.3 standard. Further, the Ethernet media segments must be built according to the recommendations in each media system standard. If your halfduplex network system includes Ethernet equipment or media segments that are not described in the standard, you may not be able to use the configuration guidelines to verify its operation.
The engineers in the IEEE committee developed the configuration rules based on the known signal timing and electrical performance specifications of Ethernet equipment that fully conforms to the published standard. That way, they could predict what the behavior of the Ethernet equipment would be, and how the signal timing would function across multiple segments.
Using noncompliant equipment and media segments makes it impossible to evaluate the timing of the network. Linking media segments together with equipment not described in the standard also makes it impossible to evaluate the timing. In both cases, there is no way for the design engineers to know how such equipment and media segments will behave. While such an Ethernet may function perfectly well, it will be "outside the standard," and you will not be able to use the official IEEE configuration guidelines to verify that such a system meets the halfduplex timing specifications.
When it comes to evaluating the configuration of your network and for later troubleshooting, you should document each network link in your system when it is installed. The documentation should include the length of each cable segment in the link, including any transceiver cables and patch cables. Also included should be the cable type used in each segment and any information you can collect on the cable manufacturer, the cable ID numbers printed on the outer sheath, and the cable delay in bit times provided by the manufacturer. The standard recommends that you create your own forms based on Table 131 for use in collecting information and documenting your network.
Table 131: Sample Cable Segment Documentation Form

Horizontal Cabling 
Transceiver Cables 
Wiring Closet Patch Cord(s) 
Station Patch Cords 
Length





Type (e.g., Category 5) 




Cable Manufacturer 




Cable Code/ID





Collision Domain
The multisegment configuration guidelines apply to a halfduplex Ethernet collision domain described in Chapter 3, The Media Access Control Protocol. A collision domain is formally defined as a single Carrier Sense Multiple Access with Collision Detection (CSMA/CD) network in which there will be a collision if two computers attached to the system transmit at the same time.
An Ethernet system composed of a single segment or of multiple segments linked with repeater hubs constitutes a single collision domain. Figure 131 shows two repeater hubs connecting three computers. Since only repeater connections are used between the segments in this network, all of the segments and computers are in the same collision domain.
Figure 131. Repeater hubs create a single collision domain

Another important point is that all segments within a given collision domain must operate at the same speed. That's because repeater hubs assume that all segments connected to the repeater are operating at the same speed and have the same roundtrip timing constraints. This is also why there are three sets of halfduplex configuration guidelines, one each for 10, 100, and 1000 Mbps Ethernet. Each of the three Ethernet speeds has its own roundtrip timing constraints and its own set of configuration guidelines.
The configuration guidelines described in this chapter are taken directly from the IEEE 802.3 standard, which describes the standards for the operation of a single halfduplex Ethernet local area network (LAN). Therefore, these guidelines only apply to a single collision domain, and have nothing to say about combining multiple Ethernet collision domains with packet switching devices such as switching hubs or routers. Switching hubs enable you to create new collision domains on each port, allowing you to link many networks together. You can also link segments operating at different speeds with switching hubs. The operation and configuration of switching hubs are described in Chapter 18, Ethernet Switching Hubs.
Model 1 Configuration
Guidelines for 10 Mbps
The first configuration model provided in the 802.3 standard describes a set of multisegment configuration rules for combining various 10 Mbps Ethernet segments. Bold text is taken directly from the IEEE standard.
 Repeater sets are required for all segment interconnection. A "repeater set" is a repeater and its associated transceivers (i.e., medium attachment units, or MAUs) and attachment unit interface (AUI) cables, if any. Repeaters must comply with all IEEE specifications in clause 9 of the 802.3 standard, and are used for signal retiming and reshaping, preamble regeneration, etc.
 MAUs that are part of repeater sets count toward the maximum number of MAUs on a segment. Twistedpair, fiber optic and thin coax repeater hubs typically use internal MAUs located inside each port of the repeater. Thick Ethernet repeaters use an outboard MAU to connect to the thick coax.
 The transmission path permitted between any two DTEs may consist of up to five segments, four repeater sets (including optional AUIs), two MAUs, and two AUIs. The repeater sets are assumed to have their own MAUs, which are not counted in this rule.
 AUI cables for 10BASEFP and 10BASEFL shall not exceed 25 m. (Since two MAUs per segment are required, 25 m per MAU results in a total AUI cable length of 50 m per segment.)
 When a transmission path consists of four repeaters and five segments, up to three of the segments may be mixing and the remainder must be link segments. When five segments are present, each fiber optic link segment (FOIRL, 10BASEFB, or 10BASEFL) shall not exceed 500 m, and each 10BASEFP segment shall not exceed 300 m. A mixing segment is defined in the standard as one that may have more than two medium dependent interfaces attached to it (e.g., a coaxial cable segment). A link segment is defined as a pointtopoint fullduplex medium that connects two and only two MAUs.
 When a transmission path consists of three repeater sets and four segments, the following restrictions apply:
 The maximum allowable length of any interrepeater fiber segment shall not exceed 1000 m for FOIRL, 10BASEFB, and 10BASEFL segments and shall not exceed 700 m for 10BASEFP segments.
 The maximum allowable length of any repeater to DTE fiber segment shall not exceed 400 m for 10BASEFL segments and shall not exceed 300 m for 10BASEFP segments and 400 m for segments terminated in a 10BASEFL MAU.
 There is no restriction on the number of mixing segments in this case. In other words, when using three repeater sets and four segments, all segments may be mixing segments if desired.
Figure 132 shows an example of one possible maximum Ethernet configuration that meets the canned configuration rules. The maximum packet transmission path in this system is between station 1 and station 2, since there are four repeaters and five media segments in that particular path. Two of the segments in the path are mixing segments, and the other three are link segments.
Figure 132. A maximum Model 1 10 Mbps configuration

While the canned configuration rules are based on conservative timing calculations, you shouldn't let that lead you to believe that you can bend these rules and get away with it. Despite the allowances made in the standards for manufacturing tolerances and equipment variances, there isn't a lot of engineering margin left in maximumsized Ethernets. If you want maximum performance and reliability, then you need to stick to the published guidelines.
In addition, while the configuration guidelines emphasize the maximum limits of the system, you should beware of stretching things as far as they can go. Ethernets, like many other systems, work best when they aren't being pushed to their limits.
An oversimplified version of the 10 Mbps Model 1 rules, called the "543" rule, has been circulating for some years. Various forms of the 543 rule have been published, and some of them include misleading terms that are incorrect. To quote from one widely distributed configuration guide, the 543 rule means that there may be as many as five segments connected in series in a network. This guide further states that up to four repeaters may be used, and up to three "populated segments." A populated segment is defined as a segment that is "attached to PCs."
While this may sound like an easy to remember rule of thumb, the "543" rule is an oversimplification of the actual configuration rules described above. Worse, the use of the term "populated segment" is misleading. This definition means that a coax segment could be regarded as an "unpopulated" segment in a network system as long as two conditions were met. First, the coax segment was not used to support PCs and, second, the segment was only used as a link segment to connect to a repeater at each end. However, this is incorrect.
A link segment is specifically defined in the 802.3 standard as a segment based on a pointtopoint fullduplex media type that connects twoand only twoMAUs. A fullduplex medium means that the medium provides separate transmit and receive data paths. This is important, since collision detection occurs faster on a fullduplex medium than it does on coaxial segments. This difference in timing is factored into the total roundtrip timing delays that are incorporated in the Model 1 configuration guidelines. That's why the notion of an "unpopulated" coax segment that could be used as a link segment is misleading and incorrect.
To recast the 543 rule into something closer to reality, we can define it to mean that you can have up to five segments in series, with up to four repeaters, and no more than three "mixing" segments. If three mixing segments are used, then the remaining two segments must be link segments as defined in 802.3. Actually, you can have up to four mixing segments under some circumstances as described in the real 802.3 rules above, so even our corrected 543 rule is still an oversimplification. 
Model 2 Configuration
Guidelines for 10 Mbps
The second configuration model provided by the IEEE provides a set of calculation aids that make it possible for you to check the validity of more complex Ethernet systems. We will be describing the network models and segment timing values provided in the standard for making the Model 2 calculations.
While the detailed description of this calculation method may seem complex, in reality the calculation method is a very straightforward process based on simple multiplication and addition. You may find the following description of the network models and timing values confusing at first glance. If so, you may wish to skip ahead to the section, Simple 10 Mbps Model 2 Configuration, to see how easy the actual calculations can be.
There are two sets of calculations provided in the standard that must be performed for each Ethernet system you evaluate. The first set of calculations verifies the roundtrip signal delay time, while the second set verifies that the amount of interframe gap shrinkage is within normal limits. Both calculations are based on network models that evaluate the worstcase path through the network.
Network Models and Delay Values
The network models and the delay values provided in the Model 2 guidelines were deliberately designed to hide a lot of complexity while still making it possible for you to calculate the timing values for any Ethernet system. Each component in an Ethernet system provides a certain amount of delay, all of which are listed in the 802.3 standard in excruciating detail.
As the Ethernet signal moves through the system, it also encounters startup delays that vary depending on which kind of equipment is involved. If you were an expert, you could use a calculator and a copy of the 802.3 standard to calculate the total of all of the bit time delays. You could also calculate the complex timing delays involved in detecting and signaling collisions, all of which differ depending on the media type involved and even the direction in which the signal travels. Fortunately, the IEEE standard has provided a better way to do things.
In the Model 2 configuration guidelines, the standard provides a set of network models and segment delay values that incorporate all of the complex delay calculations and other considerations we've just mentioned. All you need to do is to use the network models provided and follow the rules for the calculations involved. Although the network models and the rules for using them may seem arbitrary at first glance, by following the rules you can quickly and simply evaluate the roundtrip timing for a complex Ethernet system.
If you are curious as to exactly what goes into the segment delay values and the path delay calculations, you can refer to the IEEE 802.3 standard. The standard lists every delay component used, and explains why the calculation rules are set up the way they are.
Figure 133 shows the network model which is used in the standard for calculating the roundtrip timing of the worstcase path. The worstcase path is the path through your network system that has the longest segments and the most repeaters between any two stations. The calculation model includes a left and right end segment, and as many middle segments as needed.
Figure 133. Network model for roundtrip timing

To check the roundtrip timing on your network, you make a similar model of the worstcase path in your system. We will show how the roundtrip timing model is used by evaluating two sample networks later in this chapter. The network model used for interframe gap shrinkage is very similar to the roundtrip timing model, as you will see in the section on calculating interframe gap shrinkage.
Finding the WorstCase Path
You begin the process of checking an Ethernet system by finding the path in the network with the maximum delay. This is the path with the longest roundtrip time and largest number of repeaters between two stations. In some cases, you may decide that you have more than one candidate for worstcase path in your system. If that's the case, then you should identify all the paths through your network that look like they meet the worstcase definition. Following this, you can do the calculations for each worstcase path you have found, and if any path exceeds the limits for roundtrip timing or interframe gap, then the network system does not pass the test.
You should have a complete and uptodate map of your network on hand that you can use to find the worstcase path between two stations. However, if your system is not well documented then you will have to investigate and map the network yourself. The information you need includes:
 The type of segments being used (twisted pair, fiber optic, coax).
 How long the segments are.
 The location of all repeaters in the system.
 How the segments and repeaters in the system are laid out.
Once you have this information then you can determine what the maximum path between any two stations is, and what kinds of segments are used in the maximum path.
After you've found your worstcase path(s), the next thing you need to do is make a model of your path based on the network model shown in Figure 133. You do this by assigning the segment at one end of your worstcase path to be a left end segment, leaving a right end segment and possibly one or more middle segments.
To help you do this, draw a sketch of your worstcase path, noting the segment types and lengths. Then arbitrarily assign one of the end segments to be the left end; it doesn't matter which kind of segment it is. This leaves you with a right end segment; all other segments in the worstcase path become middle segments.
Calculating RoundTrip Delay Time
One goal of the configuration guidelines is to make sure that any two stations on a halfduplex Ethernet LAN can contend fairly for access to the shared Ethernet channel if they happen to transmit at the same time. When this happens, each station attempting to transmit must be notified of channel contention (collision) by receiving a collision signal within the correct collision timing window.
The way to verify whether your Ethernet system meets the limits is by calculating the total path delay, or roundtrip timing, of the worstcase path in your system. This is done using segment delay values, which are provided in terms of bit time values for each Ethernet media type. A bit time is the amount of time required to send one data bit on the network, which is 100 nanoseconds (ns) for a 10 Mbps Ethernet system. Table 132 shows the segment delay values provided in the standard for use in calculating the total worstcase path delay.
Table 132: Roundtrip Delay Values in Bit Times
Segment Type 
Max Length (in meters) 
Left End 
Middle Segment 
Right End 
RT Delay/
meter 
Base 
Max 
Base 
Max 
Base 
Max 
10BASE5 
500 
11.75 
55.05 
46.5 
89.8 
169.5 
212.8 
0.0866 
10BASE2 
185 
11.75 
30.731 
46.5 
65.48 
169.5 
188.48 
0.1026 
FOIRL 
1000 
7.75 
107.75 
29 
129 
152 
252 
0.1 
10BASET 
100 
15.25 
26.55 
42 
53.3 
165 
176.3 
0.113 
10BASEFL 
2000 
12.25 
212.25 
33.5 
233.5 
156.5 
356.5 
0.1 
Excess AUI 
48 
0 
4.88 
0 
4.88 
0 
4.88 
0.1026 
Calculating the total roundtrip delay is a matter of adding up the delay values found on the worst case path in your network. Once you have calculated the segment delay values for each segment in the worstcase path on your LAN, you then add the segment delay values together to find the total path delay. The standard recommends that you add a margin of 5 bit times to this total. If the result is less than or equal to 575 bit times, the path passes the test.
This value ensures that a station at the end of a worstcase path will be notified of a collision and stop transmitting within 575 bit times. This includes 511 bits of the frame plus the 64 bits of frame preamble and start frame delimiter (511 + 64 = 575). Once you know that the roundtrip timing for the worstcase path in your network is okay, then you can be sure that all other paths must be okay as well.
There is one more item to check in the calculation for total path delay. If the path you are checking has left and right end segments of different segment types, then you must check the path twice. The first time through, you must use the left end path delay values of one of the segment types, and the second time through you must use the left end path delay values of the other segment type. The total path delay must pass the delay calculations no matter which set of path delay values are used. We will show how this is done in the complex network example later in this chapter.
Calculating the Interframe Gap Shrinkage
The interframe gap is a 96 bit time delay provided between frame transmissions to allow the network interfaces and other components some recovery time between frames. As frames travel through an Ethernet system, the variable timing delays in network components combined with the effects of signal reconstruction circuits in the repeaters can result in an apparent shrinkage of the interframe gap. Too small a gap between frames can overrun the frame reception capability of network interfaces, leading to lost frames. Therefore, it's important to ensure that a minimum interframe gap is maintained at all receivers (stations).
Figure 134. Network model for interframe gap shrinkage

The network model for checking the interframe gap shrinkage is shown in Figure 134. As you can see, it looks a lot like the roundtrip path delay model, except that it includes a transmitting end segment. When you are doing the calculations for interframe gap shrinkage, only the transmitting end and the middle segments are of interest, since only signals on these segments must travel through a repeater to reach the receiving end station. The final segment connected to the receiving end station does not contribute any gap shrinkage and is therefore not included in the interframe gap calculations. Table 133 provides the values used for calculating interframe gap shrinkage.
Table 133: Interframe Gap Shrinkage in Bit Times
Segment Type 
Transmitting End 
MidSegment 
Coax 
16 
11 
Link segment 
10.5 
8 
When the receive and transmit end segments are not the same media type, the standard says to use the end segment with the largest number of shrinkage bit times as the transmitting end for the purposes of this calculation. This will provide the worstcase value for interframe gap shrinkage. If the total is less than or equal to 49 bit times, then the worstcase path passes the shrinkage test.
Model 1 Configuration Guidelines for Fast Ethernet
Transmission System Model 1 of the Fast Ethernet standard provides simplified configuration guidelines. The goal of the configuration guidelines is to make sure that the important Fast Ethernet timing requirements are met, so that the medium access control (MAC) protocol will function correctly. The basic rules for Fast Ethernet configuration include:
 All copper (twistedpair) segments must be less than or equal to 100 meters in length.
 Fiber segments must be less than or equal to 412 meters in length.
 If medium independent interface (MII) cables are used, they must not exceed 0.5 meters each.
When it comes to evaluating network timing, delays attributable to the MII do not need to be accounted for separately, since these delays are incorporated into station and repeater delays.
With these rules in mind, Table 134 shows the maximum collision domain diameter for segments using Class I and Class II repeaters. These repeaters are described in Chapter 17, Ethernet Repeater Hubs. The maximum collision domain diameter in a given Fast Ethernet system is the longest distance between any two stations (DTEs) in the collision domain.
Table 134: Model 1Maximum Fast Ethernet Collision Domain in Meters
Repeater Type 
All Copper 
All Fiber 
Copper and Fiber Mix (e.g., T4 and FX) 
Copper and Fiber Mix (TX and FX) 
DTEDTE Single Segment 
100 
412 
N/A 
N/A 
One Class I Repeater 
200 
272 
231 
260.8 
One Class II Repeater 
200 
320 
N/A 
308.8 
Two Class II Repeaters 
205 
228 
N/A 
216.2 
The first row in Table 134 shows that a DTEtoDTE (stationtostation) link with no intervening repeater may be made up of a maximum of 100 meters of copper, or 412 meters of fiber optic cable. The next row provides the maximum collision domain diameter when using a Class I repeater, including the case of all twistedpair and fiber optic cables, or a network with a mix of twistedpair and fiber cables. The third row shows the maximum collision domain length with a single Class II repeater in the link.
The last row of Table 134 shows the maximum collision domain allowed when two Class II repeaters are used in a link. In this last configuration, the total twistedpair segment length is assumed to be 105 meters on the mixed fiber and twistedpair segment. This includes 100 meters for the segment length from the repeater port to the station, and five meters for a short segment that links the two repeaters together in a wiring closet.
Figure 135 shows an example of a maximum configuration based on the 100 Mbps simplified guidelines we've just seen. Note that the maximum collision domain diameter includes the distance:
A (100 m) + B (5 m) + C (100 m)
Figure 135.
One possible maximum 100 Mbps configuration

The interrepeater segment length can be longer than 5 m as long as the maximum diameter of the collision domain does not exceed the guidelines for the segment types and repeaters being used. Segment B in Figure 135 could be 10 meters in length, for instance, as long as other segment lengths are adjusted to keep the maximum collision diameter to 205 meters. While it's possible to vary the length of the interrepeater segment in this fashion, you should be wary of doing so and carefully consider the consequences. Next we'll describe the major reason why you want to avoid using interrepeater segments longer than 5 meters.
Longer InterRepeater Links
The chief problem with using longer interrepeater links is that this makes your network timing rely on the use of shorter than standard segments from the repeater ports to the stations, which could cause confusion and problems later on. These days everyone assumes that twistedpair segment lengths can be up to 100 meters long. Because of that, a new segment that's 100 meters long could be attached to a system with a long interrepeater link later. In this case, the maximum diameter between some stations could then become 210 meters. If the signal delay on this long path exceeds 512 bit times, then the network may experience problems, such as late collisions. You can avoid this by keeping the length of interrepeater segments to five meters or less.
A switching hub is just another station (DTE) as far as the guidelines for a collision domain are concerned. The switching hub shown in Figure 135 provides a way to link separate network technologiesin this case, a standard 100BASET segment and a fullduplex Ethernet link. The switching hub is shown linked to a campus router with a fullduplex fiber link that spans up to two kilometers. This makes it possible to provide a 100 Mbps Ethernet connection to the rest of a campus network using a router port located in a central section of the network.
Figure 136 shows an example of a maximum configuration based on a mixture of fiber optic and copper segments. Note that there are two paths representing the maximum collision domain diameter. This includes the distance A (100 m) + C (208.8 m), or the distance B (100 m) + C (208.8 m), for a total of 308.8 meters in both cases.
Figure 136. Mixed fiber and copper 100 Mbps configuration

A Class II repeater can be used to link the copper (TX) and fiber (FX) segments, since these segments both use the same encoding scheme.
Model 2 Configuration Guidelines for Fast Ethernet
Transmission System Model 2 for Fast Ethernet segments provides a set of calculations for verifying the signal timing budget of more complex halfduplex Fast Ethernet LANs. These calculations are much simpler than the Model 2 calculations used in the original 10 Mbps system, since the Fast Ethernet system uses only link segments.
The maximum diameter and the number of segments and repeaters in a halfduplex 100BASET system are limited by the roundtrip signal timing required to ensure that the collision detect mechanism will work correctly. The Model 2 configuration calculations provide the information you need to verify the timing budget of a set of standard 100BASET segments and repeaters. This ensures that their combined signal delays fit within the timing budget required by the standard.
You may notice that these calculations appear to have a different roundtrip timing budget than the timing budget provided in the 10 Mbps media system. This is because media segments in the Fast Ethernet system are based on different signaling systems than 10 Mbps Ethernet, and because the conversion of signals between the Ethernet interface and the media segments consumes a number of bit times.
You may also notice that there is no calculation for interframe gap shrinkage, unlike the one found in the 10 Mbps Model 2 calculations. That's because the maximum number of repeaters allowed in a Fast Ethernet system is limited, thus eliminating the risk of excessive interframe gap shrinkage.
Calculating RoundTrip Delay Time
Once you have determined the worstcase path (see Finding the WorstCase Path in the Model 2 Configuration Guidelines for 10 Mbps section), your next step is to calculate the total roundtrip delay. This can be accomplished by taking the sum of all the delay values for the individual segment in the path, plus the station delays and repeater delays. The calculation model in the standard provides a set of delay values measured in bit times, as shown in Table 135. Note that the RoundTrip Delay in Bit Times per Meter only applies to the cable types in the table. The device types in the table (DTE, repeater) have only a maximum roundtrip delay through each device listed.
Table 135: 100BASET Component Delays
Component 
RoundTrip Delay in Bit Times per Meter 
Maximum RoundTrip Delay in Bit Times 
Two TX/FX DTEs 
N/A 
100 
Two T4 DTEs 
N/A 
138 
One T4 and one TX/FX DTE 
N/A 
127 
Category 3 Cable 
1.14 
114 (100 meters) 
Category 4 Cable 
1.14 
114 (100 meters) 
Category 5 Cable 
1.112 
111.2 (100 meters) 
Shielded TwistedPair Cable 
1.112 
111.2 (100 meters) 
Fiber Optic Cable 
1.0 
412 (412 meters) 
Class I Repeater 
N/A 
140 
Class II Repeater with all ports TX/FX 
N/A 
92 
Class II Repeater with any T4 port 
N/A 
67 
To calculate the roundtrip delay value, you multiply the length of the segment (in meters) times the RoundTrip Delay in Bit Times per Meter listed in the table for the segment type. This results in the roundtrip delay in bit times for that segment. If your segment is at the maximum length you can use the Maximum RoundTrip Delay in Bit Times value listed in the table for that segment type. If you're not sure of the segment length, you can also use the maximum length in your calculations just to be safe.
Once you have calculated the segment delay values for each segment in the worstcase path, you then add the segment delay values together. You also add the delay values for two stations (DTEs), and the delay for any repeaters in the path, to find the total path delay. Your vendor may provide values for cable, station, and repeater timing, which you can use instead of the ones in the table.
To this total path delay value, you should add a safety margin of zero to four bit times, with four bit times of margin recommended in the standard. This helps account for unexpected delays, such as those caused by long patch cables between a wall jack in the office and the computer. If the result is less than or equal to 512 bit times, the path passes the test.
Calculating Your Own Segment Delay Values
The segment delay value varies depending on the kind of segment used, and on the quality of cable in the segment if it is a copper segment. More accurate cable delay values may be provided by the manufacturer of the cable. If you know the propagation delay of the cable you are using, you can also look up the delay for that cable in Table 136.
Table 136: Conversion Table for Cable Propagation Times
Speed Relative to c 
Bit Time/Meter 
Nanoseconds/Meter 
100 Mbps Fast Ethernet 
1000 Mbps Gigabit Ethernet 
0.4 
8.34 
0.834 
8.34 
0.5 
6.67 
0.667 
6.67 
0.51 
6.54 
0.654 
6.54 
0.52 
6.41 
0.641 
6.41 
0.53 
6.29 
0.629 
6.29 
0.54 
6.18 
0.618 
6.18 
0.55 
6.06 
0.606 
6.06 
0.56 
5.96 
0.596 
5.96 
0.57 
5.85 
0.585 
5.85 
0.58 
5.75 
0.575 
5.75 
0.5852 
5.70 
0.570 
5.70 
0.59 
5.65 
0.565 
5.65 
0.6 
5.56 
0.556 
5.56 
0.61 
5.47 
0.547 
5.47 
0.62 
5.38 
0.538 
5.38 
0.63 
5.29 
0.529 
5.29 
0.64 
5.21 
0.521 
5.21 
0.65 
5.13 
0.513 
5.13 
0.654 
5.10 
0.510 
5.10 
0.66 
5.05 
0.505 
5.05 
0.666 
5.01 
0.501 
5.01 
0.67 
4.98 
0.498 
4.98 
0.68 
4.91 
0.491 
4.91 
0.69 
4.83 
0.483 
4.83 
0.7 
4.77 
0.477 
4.77 
0.8 
4.17 
0.417 
4.17 
0.9 
3.71 
0.371 
3.71 
Table 136 is taken from the standard and provides a set of delay values in bit times per meter, listed in terms of the speed of signal propagation on the cable. The speed (propagation time) is provided as a percentage of the speed of light; this is also called the Nominal Velocity of Propagation, or NVP, in vendor literature.
If you know the NVP of the cable you are using, then this table can provide the delay value in bit times per meter for that cable. Calculate the total delay value for your cable by multiplying the Bit Time/Meter value by the length of your segment. The result of this calculation must be multiplied by two to get the total roundtrip delay value for your segment. The only difference between 100 Mbps Fast Ethernet and 1000 Mbps Gigabit Ethernet in the above table is that the bit time in Fast Ethernet is ten times longer than the bit time in Gigabit Ethernet. For example, since the bit time is one nanosecond in Gigabit Ethernet, a propagation time of 8.34 nanoseconds per meter translates to 8.34 bit times.
Typical Propagation Values for Cables
As an example of vendor NVP specifications, Table 137 lists some typical propagation rates for Category 5 cable provided by two major vendors. These values apply to both 100 Mbps Fast Ethernet and 1000 Mbps Gigabit Ethernet systems.
Table 137: Typical VendorSupplied Cable Propagation Times
Vendor 
Part Number 
Jacket 
NVP 
AT&T 
1061 
nonplenum 
70% 
AT&T 
2061 
plenum 
75% 
Belden 
1583A 
nonplenum 
72% 
Belden 
1585A 
plenum 
75% 
Model 1 Configuration Guidelines for Gigabit Ethernet
Transmission System Model 1 of the Gigabit Ethernet standard provides simplified configuration guidelines. The goal of the configuration guidelines is to ensure that the timing requirements for Gigabit Ethernet are met so that the medium access control (MAC) protocol will function correctly. The rules for halfduplex Gigabit Ethernet configuration are:
 The system is limited to a single repeater.
 Segment lengths are limited to the lesser of 316 meters (1,036.7 feet) or the maximum transmission distance of the segment media type.
The maximum length in terms of bit times for a single segment is 316 meters. However, any media signaling limitations which reduce the maximum transmission distance of the link to below 316 meters take precedence. The fiber optic media signaling limitations can be found in Chapter 16, Fiber Optic Cables and Connectors. Table 138 shows the maximum collision domain diameter for a Gigabit Ethernet system for the segment types shown. The maximum diameter of the collision domain is the longest distance between any two stations (DTEs) in the collision domain.
Table 138: Model 1Maximum Gigabit Ethernet Collision Domain in Meters
Configuration 
Category 5 UTP 
1000BASECX 
Fiber Optic 1000BASESX/LX 
Category 5 and Fiber Optic 
1000BASECX and 1000BASESX/LX 
DTEDTE Single Segment 
100 
25 
316 
N/A 
N/A 
One Repeater 
200 
50 
220 
210 
220 
The first row in Table 138 shows the maximum lengths for a DTEtoDTE (stationtostation) link. With no intervening repeater the link may be up of a maximum of 100 m of copper, 25 m of 1000BASECX cable, or 316 m of fiber optic cable. Some of the Gigabit Ethernet fiber optic links described in Chapter 12, Gigabit Ethernet Fiber Optic Media System (1000BASEX), are limited to quite a bit less than 316 m due to signal transmission considerations. In those cases, you will not be able to reach the 316 m maximum allowed by the bit timing budget of the system.
The row labeled One Repeater provides the maximum collision domain diameter when using the single repeater allowed in a halfduplex Gigabit Ethernet system. This includes the case of all twistedpair cable (200 m), all fiber optic cable (220 m) or a mix of fiber optic and copper cables.
Model 2 Configuration Guidelines for Gigabit Ethernet
Transmission System Model 2 for Gigabit Ethernet segments provides a set of calculations for verifying the signal timing budget of more complex halfduplex Gigabit Ethernet LANs. These calculations are much simpler than the Model 2 calculations for either the 10 Mbps or 100 Mbps Ethernet systems, since Gigabit Ethernet only uses link segments and only allows one repeater. Therefore, the only calculation needed is the worstcase path delay value (PDV).
Calculating the Path Delay Value
Once you have determined the worstcase path (see "Finding the WorstCase Path" in the "Model 2 Configuration Guidelines for 10 Mbps" section), you then calculate the total roundtrip delay value for the path, or PDV. The PDV is made up of the sum of segment delay values, repeater delay, DTE delays, and a safety margin.
Segment Delay Value
The calculation model in the standard provides a set of delay values measured in bit times, as shown in Table 139. To calculate the roundtrip delay value, you multiply the length of the segment (in meters) times the RoundTrip Delay in Bit Times per Meter listed in the table for the segment type. This results in the roundtrip delay in bit times for that segment.
Table 139: 1000BASET Component Delays
Component 
RoundTrip Delay in Bit Times per Meter 
Maximum RoundTrip Delay in Bit Times 
Two DTEs 
N/A 
864 
Category 5 UTP Cable Segment 
11.12 
1112 (100 m) 
Shielded Jumper Cable (CX) 
10.10 
253 (25 m) 
Fiber Optic Cable Segment 
10.10 
1111 (110 m) 
Repeater 
N/A 
976 
The result of this calculation is the roundtrip delay in bit times for that segment. You can use the Maximum RoundTrip Delay in Bit Times value listed in the table for that segment type if your segment is at the maximum length. The max delay values can also be used if you're not sure of the segment length and want to use the maximum length in your calculations just to be safe. To calculate cable delays, you can use the conversion values provided in righthand column of Table 136.
To complete the PDV calculation, you add the entire set of segment delay values together, along with the delay values for two stations (DTEs), and the delay for any repeaters in the path. Your vendor may provide values for cable, station and repeater timing, which you can use instead of the ones in the tables provide here.
To this total path delay value, you add a safety margin of from zero to 40 bit times, with 32 bit times of margin recommended in the standard. This helps account for any unexpected delays, such as those caused by extra long patch cords between a wall jack in the office and the computer. If the result is less than or equal to 4,096 bit times, the path passes the test.
Sample Network Configurations
Next, we'll look at a few sample network configurations to show how the configuration rules work in the real world. The 10 Mbps examples will be the most complex, since the 10 Mbps system has the most complex set of segments and timing rules. Following that we will show a single example for the 100 Mbps system, since the configuration rules are much simpler for Fast Ethernet. There is no need for a Gigabit Ethernet example, as the configuration rules are extremely simple, allowing for only a single repeater hub. In addition, all Gigabit Ethernet equipment being sold today only supports fullduplex mode, which means there are no halfduplex Gigabit Ethernet systems.
Simple 10 Mbps Model 2 Configuration
Let's look at how the 10 Mbps Model 2 calculations work with a very simple example first. Figure 137 shows a network with three 10BASEFL segments connected to a fiber optic repeater. Two of the segments are 2 km (2,000 m) in length, and one is 1.5 km in length.
Figure 137. Simple 10 Mbps configuration example

Although this is a simple network, it is a configuration that is not described in the Model 1 configuration rules. Therefore, the only way to verify its operation is to perform the Model 2 calculations. By looking at Figure 137, we see that the worstcase delay path is between Station 1 and Station 2, as this path has the longest distance between any two stations. Next, we will evaluate this worstcase path for total roundtrip delay and interframe gap shrinkage.
Roundtrip delay
Since there are only two media segments in the worstcase path, the network model for roundtrip delay only has a left and right end segment. There are no middle segments to deal with. We'll assume for the purposes of this example that the fiber optic transceivers are connected directly to the stations and repeater, which eliminates the need to add extra bit times for transceiver cable length. Both segments in the worstcase path are the maximum allowable length, which means we can simply use the "Max" values from Table 132.
According to the table, the Max left end segment delay value for a 2 km 10BASEFL link is 212.25 bit times. For the 2 km right end segment, the Max delay value is 356.5 bit times. Add them together, plus the five bit times margin recommended in the standard, and the total is: 573.75 bit times. This is less than the 575 maximum bit time budget allowed for a 10 Mbps network, which means that the worstcase path is okay. All shorter paths will have smaller delay values, so all paths in this Ethernet system meet the requirements of the standard as far as roundtrip timing is concerned. To complete the interframe gap calculation, we need to compute the gap shrinkage in this network system.
Interframe gap shrinkage
Since there are only two segments, we only have to look at a single transmitting end segment when calculating the interframe gap shrinkage. There are no middle segments to deal with, and the receive end segment does not count in the calculations for interframe gap. Since both segments are of the same media type, finding the worstcase value is easy. According to Table 133, the interframe gap value for the link segments is 10.5 bit times, and that becomes our total shrinkage value for this worstcase path. This is well under the 49 bit times of interframe shrinkage allowed for a 10 Mbps network.
As you can see, the example network meets both the roundtrip delay requirements and the interframe shrinkage requirements, thus it qualifies as a valid network according to the Model 2 configuration method.
Complex 10 Mbps Model 2 Configuration
The next example is more difficult, comprised of many different segment types, extra transceiver cables, etc. All these extra bits and pieces also make the example more complicated to explain, although the basic process of looking up the bit times and adding them together is still quite simple.
For this complex configuration example, please refer back to Figure 132 earlier in the chapter. This figure shows one possible maximumlength system using four repeaters and five segments. According to the Model 1 rulebased configuration method, we've already seen that this network complies with the standards. To check that, we'll evaluate this network again, this time using the calculation method provided for Model 2.
As usual, we start the process by finding the worstcase path in the sample network. By examination, you can see that the path between Stations 1 and 2 in Figure 132 is the maximum delay path. It contains the largest number of segments and repeaters in the path between any two stations in the network. Next, we make a network model out of the worstcase path. Let's start the process by arbitrarily designating the thin Ethernet end segment as the left end segment. That leaves us with three middle segments composed of a 10BASE5 segment and two fiber optic link segments, and a right end segment comprised of a 10BASET link segment.
Next, we need to calculate the segment delay value for the 10BASE2 left end segment. This can be accomplished by adding the left end base value for 10BASE2 coax (11.75) to the product of the roundtrip delay times the length in meters (185 × 0.1026 = 18.981) results in a total segment delay value of 30.731 for the thin coax segment. However, since 185 m is the maximum segment length allowed for 10BASE2 segments, we can simply look up the Max left hand segment value from Table 132, which, not surprisingly, is 30.731. The 10BASE2 thin Ethernet segment is shown attached directly to the DTE and repeater, and there is no transceiver cable in use. Therefore, we don't have to add any excess AUI cable length timing to the value for this segment.
Calculating separate left end values
Since the left and right end segments in our worstcase path are different media types, the standard notes that we need to do the path delay calculations twice. First calculate the total path delay using the 10BASE2 segment as the left end segment and the 10BASET segment as the right end. Then swap their places and make the calculation again, using the 10BASET segment as the left end segment this time, and the 10BASE2 segment as the right end segment. The largest value that results from the two calculations is the one we must use in verifying the network.
AUI delay value
The segment delay values provided in the table include allowances for a transceiver cable (AUI) of up to two meters in length at each end of the segment. This allowance helps takes care of any timing delays that may occur due to wires inside the ports of a repeater.
Media systems with external transceivers connected with transceiver cables require that we account for the timing delay in these transceiver cables. You can find out how long the transceiver cables are, and use that length multiplied by the roundtrip delay per meter to develop an extra transceiver cable delay time which is then added to the total path delay calculation. If you're not sure how long the transceiver cables are in your network, you can use the maximum delay shown for an transceiver cable, which is 4.88 for all segment locations, left end, middle, or right end.
Calculating middle segment values
Let's continue the process of finding the total roundtrip delay time by doing the calculations for the middle segments. In the worstcase path for the network in Figure 132, there are three middle segments composed of a maximum length 10BASE5 segment, and two 500 m long 10BASEFL fiber optic segments. By looking in Table 132 under the Middle Segments column, we find that the 10BASE5 segment has a Max delay value of 89.8.
Note that the repeaters are connected to the 10BASE5 segment with transceiver cables and outboard MAUs. That means we need to add the delay for two transceiver cables. Let's assume that we don't know how long the transceiver cables are. Therefore, we'll use the value for two maximumlength transceiver cables in the segment, one at each connection to a repeater. That gives us a transceiver cable delay of 9.76 to add to the total path delay.
We can calculate the segment delay value for the 10BASEFL middle segments by multiplying the 500 meter length of each segment by the RT Delay/meter value, which is 0.1, giving us a result of 50. We then add 50 to the middle segment base value for a 10BASEFL segment, which is 33.5, for a total segment delay of 83.5.
Although it's not shown in Figure 132, fiber optic links often use outboard fiber optic transceivers and transceiver cables to make a connection to a station. Just to make things a little harder, let's assume we used two transceiver cables, each being 25 m in length, to make a connection from the repeaters to outboard fiber optic transceivers on the 10BASEFL segments. That gives us a total of 50 m of transceiver cable on each 10BASEFL segment. Since we have two such middle segments, we can represent the total transceiver cable length for both segments by adding 9.76 extra bit times to the total path delay.
Completing the roundtrip timing calculation
We started our calculations with the 10BASE2 segment assigned to the left end segment, which leaves us with a 10BASET right end segment. This segment is 100 m long, which is the length provided in the "Max" column for a 10BASET segment. Depending on the cable quality, a 10BASET segment can be longer than 100 m, but we'll assume that the link in our example is 100 m. That makes the Max value for the 10BASET right end segment 176.3. Adding all the segment delay values together, we get the result shown in Table 1310.
Table 1310: RoundTrip Path Delay with 10BASE2 Left End Segment
Link 
Media 
BitTime Delay 
Left End 
10BASE2 
30.731 
Midsegment 
10BASE5 
89.8 
Midsegment 
10BASEFL 
83.5 
Midsegment 
10BASEFL 
83.5 
Right End 
10BASET 
176.3 
Excess Length AUI 
Quan. Four 
19.52 

Total Path Delay = 
483.351 
To complete the process, we need to perform a second set of calculations with the left and right segments swapped. In this case, the left end becomes a maximum length 10BASET segment, with a value of 26.55, and the right end becomes a maximum length 10BASE2 segment with a value of 188.48. Note that the excess length AUI values do not change. As shown in Table 132, the bit time values for AUI cables are the same no matter where the cables are used. Adding the bit time values again, we get the following result in Table 1311.
Table 1311: RoundTrip Path Delay with 10BASET Left End Segment
Link 
Media 
BitTime 
Left End 
10BASET 
26.55 
Midsegment 
10BASE5 
89.8 
Midsegment 
10BASEFL 
83.5 
Midsegment 
10BASEFL 
83.5 
Right End 
10BASE2 
188.48 
Excess Length AUI 
Quan. Three 
19.52 

Total Path Delay = 
491.35 
The second set of calculations shown in Table 1311 produced a larger value than the total from Table 1310. According to the standard we must use this value for the worstcase roundtrip delay for this Ethernet. The standard also recommends adding a margin of five bit times to form the total path delay value. We are allowed to add anywhere from zero to five bits margin, but five bit times is recommended.
Adding five bit times for margin brings us up to a total delay value of 496.35 bit times, which is less than the maximum of 575 bit times allowed by the standard. Therefore, our complex network is qualified in terms of the worstcase roundtrip timing delay. All shorter paths will have smaller delay values, which means that all paths in the Ethernet system shown in Figure 132 meet the requirements of the standard as far as roundtrip timing is concerned.
Interframe gap shrinkage
We finish the evaluation of the complex network example shown in Figure 132 by calculating the worstcase interframe gap shrinkage for that network. This is done by evaluating the same worstcase path we used in the path delay calculations. However, for the purposes of calculating gap shrinkage we only evaluate the transmitting and midsegments.
Once again we start by applying a network model to the worstcase path, in this case the network model for interframe gap shrinkage. To calculate interframe gap shrinkage, the transmitting segment should be assigned the end segment in the worstcase path of your network system that has the largest shrinkage value. As shown in Table 133, the coax media segment has the largest value, so for the purposes of evaluating our sample network we will assign the 10BASE2 thin coax segment to the role of transmitting end segment. That leaves us with middle segments consisting of one coax and two link segments, and a 10BASET receive end segment which is simply ignored. The totals are seen in Table 1312.
Table 1312: Total Interframe Gap Shrinkage
Media/Link 
Shrinkage in Bit Times 
Transmitting End Coax 
16 
Midsegment Coax 
11 
Midsegment Link 
8 
Midsegment Link 
8 
Total PVV = 
43 
As you can see, the total path variability value for our sample network equals 43. This is less than the 49 bit time maximum allowed in the standard, which means that this network meets the requirements for interframe gap shrinkage.
100 Mbps Model 2 Configuration
For this example, you'll need to refer back to Figure 135, which shows one possible maximum length network. As we've seen, the Model 1 rulebased configuration method shows that this system is okay. To check that, we'll evaluate the same system using the calculation method provided in Model 2.
Worstcase path
In the sample network, the two longest paths are between Station 1 and Station 2, and between Station 1 and the switching hub. Signals from Station 1 must go through two repeaters and two 100 m segments, as well as a 5 m interrepeater segment to reach either Station 2 or the switching hub. As far as the configuration guidelines are concerned, the switching hub is considered as another station.
Both of these paths in the network include the same segment lengths and number of repeaters, so we will evaluate one of them as the worstcase path. Let's assume that all three segments are 100BASETX segments, based on Category 5 cables. By looking up the Max Delay value in Table 135 for a Category 5 segment, we find 111.2 bit times.
The delay of a 5 m interrepeater segment can be found by multiplying the roundtrip Delay per Meter for Category 5 cable (1.112) times the length of the segment in meters (5). This results in 5.56 bit times for the roundtrip delay on that segment. Now that we know the segment roundtrip delay values, we can complete the evaluation by following the steps for calculating the total roundtrip delay for the worstcase path.
To calculate the total roundtrip delay, we use the delay times for stations and repeaters found in Table 135. As shown in Table 1313, the total roundtrip path delay value for the sample network is 511.96 bit times when using Category 5 cable. This is less than the maximum of 512 bit times, which means that the network passes the test for roundtrip delay.
Table 1313: RoundTrip Delay in Sample Network, Default Timing Values
Two TX DTEs 
100 
100 meter Cat 5 segment 
111.2 
100 meter Cat 5 segment 
111.2 
5 meter Cat 5 segment 
5.56 
Class II repeater delay 
92 
Class II repeater delaySee All repeater ports TX or FX. 
92 
Total Delay = 
511.96 
Note, however, that there is no margin of up to 4 bit times provided in this calculation. There are no spare bit times to use for margin, because the bit time values shown in Table 135 are all worstcase maximums. This table provides worstcase values that you can use if you don't know what the actual cable bit times, repeater timing, or station timing values are.
For a more realistic look, let's see what happens if we work this example again, using actual cable specifications provided by a vendor. In Table 1314, let's assume that the Category 5 cable is AT&T type 1061 cable, a nonplenum cable which has an NVP of 70 percent as shown in Table 137. If we look up that speed in Table 136, we find that a cable with a speed of 0.7 is rated at 0.477 bit times per meter. The roundtrip bit time will be twice that, or 0.954 bit times. Therefore, timing for 100 m will be 95.4 bit times, and for 5 m it will be 4.77 bit times. Again, refer to Table 1314 to see how things add up using these different cable values.
Table 1314: RoundTrip Delay Using Vendor Timing for Cable
Two TX DTEs 
100 
100 m Cat 5 segment 
95.4 
100 m Cat 5 segment 
95.4 
5 m Cat 5 segment 
4.77 
Class II repeater delay 
92 
Class II repeater delay 
92 
Margin 
4 
Total Delay = 
483.57 
When realworld cable values are used instead of the worstcase default values in Table 135, there is enough timing left to provide for 4 bit times of margin. This meets the goal of 512 bit times, with bit times to spare.
Working with bit time values
Some vendors note that their repeater delay values are smaller than the value listed in Table 135, which will make it easier to meet the 512 bit time maximum. While these extra bit times could theoretically be used to provide an interrepeater segment longer than five meters, this approach could lead to problems.
While providing a longer interrepeater link might appear to be a useful feature, you should consider what would happen if that vendor's repeater failed and had to be replaced with another vendor's repeater whose delay time was larger. If that were to occur, then the worstcase path in your network might end up with excessive delays due to the bit times consumed by the longer interrepeater segment you had implemented. You can avoid this problem by designing your network conservatively and not pushing things to the edge of the timing budget.
It is possible to use more than one Class I or two Class II repeaters in a given collision domain. This can be done if the segment lengths are kept short enough to provide the extra bit time budget required by the repeaters. However, the majority of network installations are based on building cabling systems with 100 m segment lengths (typically implemented as 90 m "in the walls" and 10 m for patch cables). A network design with so many repeaters that the segments must be kept very short to meet the timing specifications is not going to be useful in most situations.
Footnotes
