Ethernet is a Carrier Sense Multiple Access with Collision Detection (CSMA/CD) network. Each station on the network listens for a carrier and attempts to transmit data when it senses the absence of that signal. Unfortunately, two stations may attempt to simultaneously transmit data, resulting in the occurrence of a collision. Even when one station thinks there is no carrier, it is quite possible that a carrier signal is propagating down the transmission path . Thus, a station transmitting data when its sampling of the line indicates the absence of a carrier may also result in a collision.
Because of the random nature of collisions, Ethernet bus performance is not deterministic and performance characteristics and message transmission delays are not predictable. However, over a period of time, you can determine average and peak utilization, data elements that you can use to split one Ethernet LAN into two or more LANs via the use of bridges or other network devices to increase individual network performance.
In this section we compute the frame rate on Ethernet, 100BASE-TX Fast Ethernet, and Gigabit Ethernet networks. Because the frame flow on 100BASE-TX is ten times that of the 10BASE-T Ethernet network operating at 10 Mbps, we will first focus our computations on the lower-speed network. Once we compute the frame rate on 10 Mbps Ethernet, we will simply multiply the result by 10 to determine the frame rate on Fast Ethernet.
Although we will perform a similar operation for Gigabit Ethernet by multiplying Ethernet results by 100, we will also note that this process is only applicable to certain types of frames . That is, Gigabit Ethernet requires carrier extension technology to insure the transmission of a minimum length frame, which was described earlier in this book and which will be reviewed in this chapter when we consider the Gigabit Ethernet frame rate.
The top portion of Table 7.1 illustrates the IEEE 802.3 (Ethernet) frame format. In this illustration, the seven-byte preamble field and the one-byte start of frame delimiter field were combined into a common eight-byte preamble field for simplicity. In actuality, the preamble field used by Ethernet is an eight-byte sequence of alternating 1's and 0's, while the IEEE 802.3 frame format uses a seven-byte preamble field of alternating 1's and 0's. The start of frame delimiter one-byte field used in IEEE 802.3 frames follows a seven-byte preamble field and the sequence of alternating 1's and 0's, but ends with two set bits instead of the 1 and 0 used in the Ethernet preamble field. Because computations required to estimate Ethernet network and bridge and router performance are based on frame length and not frame composition, the use of a common eight-byte preamble field, while not technically correct from a frame composition basis, does not affect our computations. As indicated by the tabulation of frame field lengths in the lower portion of Table 7.1, the frame size can vary from a minimum of 72 to a maximum of 1526 bytes.
Preamble | Destination Address | Source Address | Length or Type | Data | Frame Check Sequence |
8 | 6 | 6 | 2 | 46 ‰ n ‰ 1500 | 4 |
Frame Size (bytes) | ||
---|---|---|
Field | Minimum Size Frame | Maximum Size Frame |
Preamble | 8 | 8 |
Destination Address | 6 | 6 |
Source Address | 6 | 6 |
Length or Type | 2 | 2 |
Data | 46 | 1500 |
Frame Check Sequence | 4 | 4 |
Total Size | 72 | 1526 |
Under Ethernet and IEEE 802.3 standards, there is a dead time of 9.6 microseconds ( ¼ s) between frames. Using the frame size and dead time between frames, you can compute the maximum number of frames per second that can flow on an Ethernet network. For our example, let us assume we have a 10-Mbps LAN, such as a 10BASE-2, 10BASE-5, or 10BASE-T network. Here, the bit time then becomes 1/10 7 or 100 nanoseconds (ns).
Now let us assume that all frames are at the maximum length of 1526 bytes. Then, the time per frame becomes:
Because one 1526-byte frame requires 1.23 ms, then in one second there can be 1/1.23 ms or approximately 812 maximum sized frames. Thus, the maximum transmission rate on an Ethernet network is 812 frames per second when information is transferred in 1500-byte units within a sequence of frames. One example of a situation in which data would be transferred in 1500-byte units is when a workstation downloads a file from a server or transfers a file to another workstation or to a server. When this type of data transfer occurs, the data field of a large number of sequential frames would be filled to their maximum size of 1500 bytes. If the last portion of the file being transferred is less than 1500 bytes, then the data field of the last frame used to transport the file would be less than 1500 bytes in length.
Now that the maximum number of frames that can traverse an Ethernet network when the data field is at its maximum size has been determined, let us compute the frame rate when the data field is at its minimum length.
When that situation occurs, the data field contains up to 46 characters of information, because PAD characters are required to fill the data field to a minimum length of 46 characters. This results in a minimum size Ethernet frame being 72 characters in length.
For a minimum frame length of 72 bytes, the time per frame is: 9.6 ¼ s + 72 bytes * 8 bits/byte * 100 ms/bit, or 67.2 * 10 ˆ’ 6 seconds. Thus, in one second there can be a maximum of 1/67.2 * 10 ˆ’ 6 , or 14,880 minimum-size 72-byte frames.
Table 7.2 summarizes the frame processing requirements for a 10-Mbps Ethernet network under 50 percent and 100 percent load conditions based upon minimum and maximum frame sizes. Note that those frame processing requirements define the frame examination (filtering) operating rate of a bridge, switch, or router port connected to an Ethernet network operating at 10 Mbps. That rate indicates the number of frames per second a device connected to a 10-Mbps Ethernet LAN must be capable of examining under heavy (50 percent load) and full (100 percent load) traffic conditions. Those frame processing requirements also define the maximum frame forwarding rate for a bridge, switch, or router connected to a single network, because if all frames were routed off the network, the forwarding rate would equal the filtering rate.
Frames per Second | |||
---|---|---|---|
Network Type | Average Frame Size (bytes) | 50% load | 100% load |
Ethernet | 1526 | 406 | 812 |
72 | 7440 | 14880 | |
Fast Ethernet | 1526 | 4060 | 8120 |
72 | 74400 | 148800 |
As an example of the potential utilization of information contained in Table 7.2, assume you were considering the acquisition of a two-port 10BASE-T bridge. That bridge must have a filtering capability at or above 29,780 72-byte frames per second to ensure it is capable of examining every frame that can flow on a 10BASE-T network connected to each port. Similarly, under a worst-case operational scenario, the bridge must be capable of forwarding 29,780 72-byte frames per second through the bridge to ensure that no frames requiring forwarding are lost.
As discussed in Chapter 2, Gigabit Ethernet uses carrier extension technology to ensure that the minimum length frame is 512 bytes, not including its preamble and start of frame delimiter (SFD) fields. Figure 7.1 provides a review of the format of a Gigabit Ethernet frame, to include its carrier extension variable field.
Concerning the latter, the carrier extension field will vary in length from a maximum of 448 bytes when a minimum-length 64-byte frame is formed , to 0 bytes when a frame with an information field equal to or greater than 494 bytes in length is transmitted. Note that the actual frame length flowing on the media includes preamble and SFD fields, requiring a minimum length frame of 520 bytes for Gigabit Ethernet.
For a Gigabit Ethernet minimum length frame of 520 bytes, the time per frame computations use a dead time of 0.096 ¼ s between frames and a bit duration of 1 ns. Thus, for a minimum frame length of 520 bytes, the time per frame becomes:
Then, in one second, there can be a maximum of 1/4.256 ¼ s or 234,962 minimum-size 520-byte frames. To compute the maximum number of maximum length frames that can flow on a Gigabit Ethernet network, we would use a frame length of 1526 bytes, to include the preamble and start of frame delimiter fields. Doing so, the time required to transmit a maximum-length Gigabit Ethernet frame becomes:
Thus, in one second, there can be a maximum of 1/12.304 10 ˆ’ 6 or 81,200 maximum-length frames per second. Table 7.3 summarizes the Gigabit Ethernet frame processing capability.
Frames per Second | ||
---|---|---|
Average Frame Size (bytes) | 50% load | 100% load |
520 | 117481 | 234962 |
1526 | 40600 | 81200 |
If you compare the entries in Table 7.2 and Table 7.3, you will note that Ethernet supports a data flow of 14,880 minimum-length frames per second, while Gigabit Ethernet's support for 520-byte minimum-length frames, which in effect could be 72 byte frames with a 448-byte carrier extension, is limited to 234,962 frames per second. This is 15.79 times the capability of a 10-Mbps Ethernet. If we compare Fast Ethernet's minimum-length frame per second rate of 14,800 to Gigabit Ethernet's rate of 234,962, the ratio decreases to 1.579:1. Thus, instead of obtaining a 100:1 and 10:1 ratio, we obtain 15.79:1 and 1.579:1 ratios, which indicates that for interactive query response applications the use of Gigabit Ethernet can be expected to provide less than a 60 percent improvement over Fast Ethernet instead of a ten-fold improvement!
Perhaps because of the previously developed Gigabit Ethernet computations, Alteon Networks introduced a Gigabit Ethernet extended frame transmission capability for its network adapters and Gigabit switch ports. This technology is being used under license or via original equipment manufacturer (OEM) agreements by other vendors and represents a technology that is gaining acceptance as "a mechanism to enhance the performance of Gigabit Ethernet." Whether or not jumbo frame technology provides a significant improvement in Gigabit performance can be determined by putting pen to paper to perform a series of computations.
Under Alteon Networks' jumbo frames technology, Ethernet frames are extended to permit an information field up to 9,000 bytes in length. This extension provides the ability to transport up to six times the maximum information transported by one standardized maximum-length frame while only requiring one set of overhead bytes for framing of the information field. To obtain an appreciation of the performance capability of jumbo frames, let us first compute the maximum number of maximum-length frames that can be transmitted on a Gigabit network.
The time required to transmit a jumbo Gigabit Ethernet frames is:
Then, in one second, 1/72.096 10 ˆ’ 6 , or 13870 frames can be transmitted.
We can compare Gigabit Ethernet maximum-length frames to jumbo frames by examining their data transfer capability. Because 81,274 maximum-length Gigabit frames can be transmitted per second and each frame can transport 1500 bytes in its information field, we obtain a maximum data transfer capability of 81,274 frames/second * 1500 bytes/frame, or 121,911,000 bytes/second. At 8 bits per byte, we obtain a maximum data transfer of 121,911,000 bytes/second * 8 bits/byte, or 0.975288 Gbps. When jumbo frames are used, we can transmit 13,870 frames per second. Because the information field is now 9,000 bytes, this results in a maximum data transfer capability of 13,870 frames/second * 9000 bytes/frame * 8 bits/byte, or 0.99864 Gbps. Based upon the preceding , the use of jumbo frames can be expected to provide an increase in data transfer capability of approximately 23 Kbps (998.640 Mbps ˆ’ 975.288 Mbps) for sustained file transfers and other applications that can fill jumbo frames. Whether or not the 23-Kbps increase in data transfer is worth a proprietary solution is left up to the reader to determine. However, it should be noted that the use of jumbo frames is only applicable to full-duplex Gigabit Ethernet where one end of the connection is a switch port that also supports jumbo frame technology. When we turn our attention to LAN switches later in this book, we will also discuss the use of jumbo frame support by switches. Now that we have an appreciation for Ethernet, Fast Ethernet, and Gigabit Ethernet frame processing requirements of bridges and routers, let us turn our attention to several BASIC programs to compute a series of bridge, switch, and router frame processing requirements.
From the two sets of computations previously performed in this section, Ethernet bridge, switch, and router processing requirements and 50- and 100-percent network load conditions were determined. To facilitate performing additional computations, a general model of Ethernet network device processing requirements will be developed that will be incorporated into a BASIC language program. That program, which we will appropriately name EPERFORM.BAS, will be exercised by varying the Ethernet frame size from its minimum frame length of 72 bytes to its maximum frame length of 1526 bytes. Once again, it is important to note that the frame length will be varied from 72 bytes to 1526 bytes as we must include the preamble and start of frame delimiter fields, which collectively add 8 bytes to the MAC frame length. This addition is necessary because both fields are included with each frame that flows on the network and affects the maximum number of frames that can be transmitted per unit time. Similar to other BASIC language programs, EPERFORM.BAS can be located in the directory BASIC at the Web URL previously noted in this book.
To develop a general model that provides the maximum number of frames that can be transmitted on an Ethernet network, we can simply replace the specific frame length used in prior computations by the variable FLENGTH. Then, we obtain the maximum frame rate in frames per second (FPS) under a 100-percent network load by exercising the following equation:
Table 7.4 lists the contents of the program EPERFORM.BAS developed to exercise the previously developed Ethernet frame rate model for frame lengths varying from 72 to 1526 bytes in length under 50- and 100-percent load conditions. In this program, the FOR J loop was used to define four sets of variables for use by the FOR FLENGTH loop, which performs the computations required to determine Ethernet frame processing requirements based on different frame lengths. The variable FPS computes the frame per second rate based on a bit duration of 100 ms (0.0000001), which represents the bit duration of a 10-Mbps Ethernet network. You can either lower the bit duration by a factor of 10 and change the interframe gap from 0.0000096 to 0.00000096 to compute a table of frame rates for Fast Ethernet, or multiply the results obtained from the execution of the unmodified program by 10. To facilitate determining the frame processing requirements of a Fast Ethernet network, the previously described modifications were made, with the modified program stored as the file FSTPRFM.BAS in the BASIC directory at the previously noted Web URL.
REM PROGRAM EPERFORM.BAS LPRINT "THIS PROGRAM COMPUTES ETHERNET BRIDGE FRAME PROCESSING REQUIREMENTS" LPRINT " BASED UPON VARYING AVERAGE ETHERNET FRAME LENGTHS" LPRINT LPRINT "AVERAGE FRAME LENGTH FRAME PROCESSING REQUIREMENT" LPRINT " 50% LOAD 100% LOAD" LPRINT FOR J = 1 TO 12 STEP 3 READ A, B, C DATA 72,72,1,80,100,20,125,1500,25,1526,1526,1 FOR FLENGTH = A TO B STEP C FPS = 1/(.0000096 + FLENGTH * 8 *.0000001) LPRINT USING " ##### "; FLENGTH; LPRINT USING " ######.## ######.##"; FPS/2; FPS NEXT FLENGTH NEXT J END |
The first iteration of the J loop simply sets FLENGTH to compute the frame processing requirements for a frame length of 72 bytes. The second iteration of the J loop results in the FLENGTH loop computing the frame processing requirement for frame lengths from 80 to 100 bytes in increments of 20 bytes. The third iteration of the J loop results in the FLENGTH loop computing the frame processing requirements for frame lengths from 125 to 1500 bytes in increments of 25 bytes. Finally, the fourth iteration completes the computations for the maximum frame length of 1526 bytes.
Table 7.5 lists the results obtained from the execution of the program EPERFORM.BAS. Using monitoring equipment, such as a protocol analyzer, you can determine the average frame length transmitted on your network. Then you can use that data in conjunction with the frame processing requirements columns listed in Table 7.5 to determine the frame processing requirements for your specific network environment.
THIS PROGRAM COMPUTES ETHERNET BRIDGE FRAME PROCESSING REQUIREMENTS BASED UPON VARYING AVERAGE ETHERNET FRAME LENGTHS AVERAGE FRAME LENGTH FRAME PROCESSING REQUIREMENT 50% LOAD 100% LOAD 72 7440.48 14880.95 80 6793.48 13586.96 100 5580.36 11160.71 125 4562.04 9124.09 150 3858.02 7716.05 175 3342.25 6684.49 200 2948.11 5896.23 225 2637.13 5274.26 250 2385.50 4770.99 275 2177.70 4355.40 300 2003.21 4006.41 325 1854.60 3709.20 350 1726.52 3453.04 375 1614.99 3229.97 400 1516.99 3033.98 425 1430.21 2860.41 450 1352.81 2705.63 475 1283.37 2566.74 500 1220.70 2441.41 525 1163.87 2327.75 550 1112.10 2224.20 575 1064.74 2129.47 600 1021.24 2042.48 625 981.16 1962.32 650 944.11 1888.22 675 909.75 1819.51 700 877.81 1755.62 725 848.03 1696.07 750 820.21 1640.42 775 794.16 1588.31 800 769.70 1539.41 825 746.71 1493.43 850 725.06 1450.12 875 704.62 1409.24 900 685.31 1370.61 925 667.02 1334.04 950 649.69 1299.38 975 633.23 1266.46 1000 617.59 1235.18 1025 602.70 1205.40 1050 588.51 1177.02 1075 574.98 1149.95 1100 562.05 1124.10 1125 549.69 1099.38 1150 537.87 1075.73 1175 526.54 1053.07 1200 515.68 1031.35 1225 505.25 1010.51 1250 495.25 990.49 1275 485.63 971.25 1300 476.37 952.74 1325 467.46 934.93 1350 458.88 917.77 1375 450.61 901.23 1400 442.63 885.27 1425 434.93 869.87 1450 427.50 854.99 1475 420.31 840.62 1500 413.36 826.72 1526 406.37 812.74 |
As indicated in the footnote in Table 7.5, by multiplying an entry in either frame processing column by 10 you can obtain the frame processing requirement for a 100BASE-TX Fast Ethernet network. As an alternative, you can execute the program FSTPRFM.EXE to generate a display similar to the one shown in Table 7.5 for the Fast Ethernet bridge frame processing requirements.
Due to improvements in bridge, switch, and router frame processing resulting from the incorporation of low-cost, high-performance microprocessors, such as the Intel Pentium series, today you should be able to obtain networking devices with a processing capability that exceeds the frame processing requirements for all Ethernet frame lengths. Thus, the data listed in Table 7.5 is primarily of value for deciding whether or not you should continue to use older bridges and routers as your network operation changes. If you are using or considering the installation of Fast Ethernet, the frame processing requirements listed in Table 7.5 increase by a factor of ten. Under such circumstances, the ability of some networking devices may not be capable of supporting a 100-percent load requirement. This is especially true if you anticipate adding a new adapter card and driver to support 100BASE-T operations in an existing PC-based bridge. Thus, you should carefully examine the frame processing capability of bridges, switches, and routers prior to using them for Fast Ethernet applications.
To provide you with the ability to determine the frame processing requirements of bridges, switches, and routers that have a Gigabit Ethernet interface, the program GBITPFRM.BAS was developed. Table 7.6 lists the statements in the program. Note that there are two key changes when comparing this program to the program EPERFORM.BAS listed in Table 7.4. First, for any frame length under 520 bytes, the frame length variable GLENGTH is set to 520 bytes for computing the FPS value. Second, the FPS computation uses a dead time of 9.6 * 10 ˆ’ 8 sec and a bit duration of 100 ns associated with 10-Mbps Ethernet. Table 7.7 contains the results of the execution of the previously described program.
REM PROGRAM GBITPFRM.BAS LPRINT "THIS PROGRAM COMPUTES GIGABIT ETHERNET BRIDGE FRAME PROCESSING" LPRINT "REQUIREMENTS BASED UPON VARYING AVERAGE ETHERNET FRAME LENGTHS" LPRINT LPRINT "AVERAGE FRAME LENGTH FRAME PROCESSING REQUIREMENT" LPRINT " 50% LOAD 100% LOAD" LPRINT FOR J = 1 TO 12 STEP 3 READ A, B, C DATA 72,72,1,80,100,20,125,1500,25,1526,1526,1 FOR FLENGTH = A TO B STEP C IF FLENGTH < 520 THEN GLENGTH = 520 ELSE GLENGTH = FLENGTH FPS = 1/(9.599999999999999D-08 + GLENGTH * 8 *.000000001#) LPRINT USING " ##### "; FLENGTH; LPRINT USING " ######.## ######.##"; FPS/2; FPS NEXT FLENGTH NEXT J END |
Another area in which the data presented in Table 7.5 and Table 7.7 can be extremely valuable is in evaluating bridges, switches, and routers. Suppose you were evaluating two bridges manufactured by the well-known vendors "X" and "Y." Let us assume vendor X's sales literature lists an Ethernet frame processing capability of 15,000 frames per second without specifying the average frame length, while vendor Y's sales literature lists an Ethernet frame processing capability of 20,000 frames per second, also without specifying the average frame length. Which vendor product provides a higher level of capability?
THIS PROGRAM COMPUTES GIGABIT ETHERNET BRIDGE FRAME PROCESSING REQUIREMENTS BASED UPON VARYING AVERAGE ETHERNET FRAME LENGTHS AVERAGE FRAME LENGTH FRAME PROCESSING REQUIREMENT 50% LOAD 100% LOAD 72 117481.20 234962.41 80 117481.20 234962.41 100 117481.20 234962.41 125 117481.20 234962.41 150 117481.20 234962.41 175 117481.20 234962.41 200 117481.20 234962.41 225 117481.20 234962.41 250 117481.20 234962.41 275 117481.20 234962.41 300 117481.20 234962.41 325 117481.20 234962.41 350 117481.20 234962.41 375 117481.20 234962.41 400 117481.20 234962.41 425 117481.20 234962.41 450 117481.20 234962.41 475 117481.20 234962.41 500 117481.20 234962.41 525 116387.34 232774.67 550 111209.96 222419.92 575 106473.59 212947.19 600 102124.18 204248.36 625 98116.17 196232.34 650 94410.88 188821.75 675 90975.26 181950.52 700 87780.90 175561.80 725 84803.26 169606.52 750 82021.00 164042.00 775 79415.50 158831.00 800 76970.45 153940.89 825 74671.45 149342.89 850 72505.80 145011.59 875 70462.23 140924.47 900 68530.70 137061.41 925 66702.24 133404.48 950 64968.82 129937.63 975 63323.20 126646.41 1000 61758.89 123517.79 1025 60270.01 120540.02 1050 58851.22 117702.45 1075 57497.70 114995.40 1100 56205.04 112410.07 1125 54969.22 109938.44 1150 53786.57 107573.15 1175 52653.75 105307.50 1200 51567.66 103135.31 1225 50525.46 101050.93 1250 49524.56 99049.13 1275 48562.55 97125.09 1300 47637.20 95274.39 1325 46746.45 93492.90 1350 45888.40 91776.80 1375 45061.29 90122.57 1400 44263.46 88526.91 1425 43493.39 86986.78 1450 42749.66 85499.31 1475 42030.93 84061.87 1500 41335.98 82671.96 1526 40637.19 81274.38 |
Because each vendor's frame processing rate exceeds 14,880 frames per second, both have the ability to process a fully loaded Ethernet network. Thus, the higher processing capability of vendor Y is irrelevant and you should consider both products to be equivalent with respect to their frame processing capability.
Now let us assume that instead of working with legacy Ethernet, your organization is considering installing a Gigabit Ethernet backbone. Let us further assume that you are evaluating the capability of several switch vendors concerning their ability to support a Gigabit Ethernet uplink. Based on the evaluation of two switch vendor products, you noted that vendor X indicates a maximum frame processing capability of 200,000 frames per second while vendor Y, whose product costs 40 percent more, claims a frame processing capability of 500,000 frames per second. Without referring to Table 7.7, your first impression might be that you are obtaining a 150-percent performance increase from vendor Y for a 40-percent increase in cost. However, upon examining the frame rate obtainable on a Gigabit Ethernet network, you now realize it is limited to a maximum of 234,962 frames per second. Thus, the achievable performance difference between the two products is actually 17.48 percent, whereas the cost difference is 40 percent. This means that vendor Y's product is not the bargain it appears to be!
Now that the maximum number of frames that can be carried on 10-Mbps, 100-Mbps, and 1-Gbps Ethernet LANs has been determined, that information can be used to determine the utilization of the LAN. To do so we must recognize that the actual number of bits that can be carried on an Ethernet LAN will always be less than its operating rate due to the dead time between frames. For example, to compute the actual number of bits transmitted in one second using the maximum-length frame, you must subtract the number of bits that cannot be transmitted during the 812 slots of dead time (9.6 ¼ s for a 10-Mbps Ethernet) from the LAN operating rate. Then, when 1526-byte frames are transmitted, the actual maximum Ethernet network data transfer operating rate becomes:
Thus, for 100-percent utilization of a 10-Mbps Ethernet when a maximum frame size of 1526 bytes is used, 9.922 Mbps must be transmitted in one second. For a Fast Ethernet LAN, idle characters are transmitted between frames, which in effect results in a dead time between frames. That dead time is one tenth that of a 10-Mbps Ethernet, or 0.96 ¼ s, while the bit duration is reduced to 10 ns. Thus, the actual maximum 100-Mbps Fast Ethernet data transfer operating rate when 1526-byte frames are transmitted becomes:
We can perform a similar computation for Gigabit Ethernet, adjusting the dead time between frames, its bit duration, and the number of frames transportable per second as follows for maximum length frames:
Because many inexpensive test devices are capable of counting frames, bits, or both frames and bits, using such equipment with knowledge of the maximum achievable data transfer rate on the network provides us with the ability to compute its utilization. In addition, you may be able to defer the purchase of a more expensive LAN performance analyzer, which, due to the prices of such equipment, may result in considerable savings.
To illustrate the computations required to determine the level of Ethernet network utilization, let us assume that the monitoring of an Ethernet LAN indicates that during 10 minutes of monitoring, a total of 280,000 frames with an average data field length of 100 bytes were counted. The average frame length, including 100 data bytes, would be 126 bytes due to the 26 overhead bytes required to transport each frame. Then the average number of frames per second would be computed as follows:
The number of bits flowing on the network is computed by multiplying the frame size by 8 bits/byte and then multiplying the result by the frame size. Thus, 126 bytes/frame * 8 bits/byte * 466.67 frames/second is 470,403 bits. Then, the utilization in percent would be 470,403/9,922,048 * 100, or 4.74.
Based on readily available performance statistics, a 100-node Ethernet can normally be expected to have an average utilization under 2 percent, with worst second, minute, and hour percentages of 40, 15 to 20, and 3 to 5, respectively. Similar utilization levels may not be applicable to 100-Mbps Fast Ethernet networks because such networks operate at ten times the rate of 10BASE-T LANs. This means that they can support a significant increase in traffic prior to reaching a higher level of utilization.
These preceding performance statistics represent the activity on a typical Ethernet network in that, at any particular time, many network users are performing local processing, such as composing a memorandum or an electronic message. Other network users may be reading a manual, talking on the telephone, or performing an activity completely unrelated to network usage. Thus, only a few people are actually transmitting or receiving information using the network. Concerning those people, one network user may be transmitting a short electronic mail message of a few hundred characters while another network user might be downloading a file from the server or accessing a server facility. Thus, a typical 2-percent level of network utilization on a 10-Mbps Ethernet network equates to a data transfer of 9,922,048 * 0.02, or approximately 198 Kbps. At this data transfer rate, many people can be sending electronic messages, interacting with the file server, and performing file transfer operations.
On a Fast Ethernet network, a 2-percent level of network utilization equates to a data transfer rate ten times that of a 10-Mbps Ethernet network, or approximately 1.98 Mbps. To put this number in perspective, let us assume that the typical length of an electronic mail message is 1000 characters, or 8000 bits. This means that at a 2-percent level of network utilization, a 100-Mbps Fast Ethernet network could support the transfer of almost 250 1000-character electronic mail messages per second!
When a number of network users initiate file transfers, you can expect a short peak level of utilization to approach or surpass 40 percent on a 10-Mbps Ethernet network. Because a 640-Kbyte file transfer will require less than 0.07 seconds at 10 Mbps, many file transfers will rapidly be completed, which eliminates the potential for one file transfer overlapping another file transfer operation if two people initiate a file transfer just a second or two apart from one another. This explains why the worst minute utilization of a 10-Mbps Ethernet network is typically reduced to a range between 15 and 20 percent in comparison to a worst second utilization of 40 percent. Because our previous computation is much better than the typical worst minute utilization, it would appear that the monitored LAN is not overloaded. However, an extension of monitoring of several hours of activity during peak periods should be considered to ensure that utilization peaks were not inadvertently missed.
When we consider the level of utilization on a Gigabit Ethernet backbone network, it is important to remember that all frames with a length below 520 bytes are extended to a length of 520 bytes via carrier extension technology. This means that the performance level of Gigabit Ethernet can be taxed by a large amount of query-response traffic flowing onto a backbone Gigabit Ethernet LAN. You can see this by noting from Table 7.7 that the maximum number of frames a Gigabit Ethernet can transport is limited to 234,962 per second for all frames less than or equal to 520 bytes in length. In comparison, a Fast Ethernet LAN can support the transmission of 148,800 72-byte frames per second. This means that it becomes theoretically possible for two Fast Ethernet LANs joined via a Gigabit Ethernet backbone to saturate the backbone, although the Gigabit network theoretically should be capable of supporting ten Fast Ethernet LANs. When we examine the use of LAN switches later in this book, we will also turn our attention to the use of Gigabit Ethernet as a backbone network.
Although knowledge concerning the average frame length and frame rate is important, by themselves they do not provide definitive information concerning the rate at which information can be transferred on a network. This is because a portion of an Ethernet frame represents overhead and does not carry actual data. Thus, to obtain a more realistic indication of the ability of an Ethernet network to transfer information, you must compute the information transfer rate in bits per second (bps). This calculation is performed by first subtracting 26 bytes from the frame length for frames with a data field of 46 or more bytes, as there are 26 overhead bytes in each frame. Next, you would multiply the frame rate by the adjusted frame length and then multiply the result by 8 to obtain the information transfer rate in bits per second.
Table 7.8 lists the statements contained in a program named EITR.BAS. This program was developed to compute the information transfer rate in bps based on 16 average frame lengths and their corresponding frame transfer rates. Readers will note that the program EITR.BAS represents a simple modification to the previously developed program EPERFORM.BAS. Similar to the method noted for modifying the program EPERFORM.BAS for computing Fast Ethernet statistics, you can modify the program listing contained in Table 7.8. That is, you would change the computation for the variable FPS (frames per second) by modifying the interframe gap time and bit duration as indicated earlier in this chapter when the coding for the program EPERFORM.BAS was described. The previously described changes are included on the file FEITR.BAS, located in the BASIC directory at the noted Web URL, while the executable version of that file is named FEITR.EXE in that directory.
CLS REM PROGRAM EITR.BAS PRINT "INFORMATION TRANSFER RATE VERSUS AVERAGE FRAME LENGTH" PRINT PRINT "AVERAGE FRAME 100% LOAD INFORMATION TRANSFER" PRINT " LENGTH FRAMES/SEC RATE IN BPS " FOR J = 1 TO 12 STEP 3 READ A, B, C DATA 72,72,1,80,100,20,125,1500,125,1526,1526,1 FOR FLENGTH = A TO B STEP C FPS = 1/(.0000096 + FLENGTH * 8 *.0000001) PRINT USING "##### ######"; FLENGTH; FPS; PRINT USING " ########"; FPS * (FLENGTH - 26) * 8 NEXT FLENGTH NEXT J END |
The results of the execution of EITR.BAS are listed in Table 7.9. To obtain the frames per second and information transfer rate for Fast Ethernet, you can multiply the entries in the second and third columns of Table 7.9 by 10 or execute the program FEITR.
Average Frame Length | 100% Load Frames/Sec | Information Transfer Rate (bps) |
---|---|---|
72 | 14881 | 5476191 |
80 | 13587 | 5869565 |
100 | 11161 | 6607143 |
125 | 9124 | 7226278 |
250 | 4771 | 8549618 |
375 | 3230 | 9018088 |
500 | 2441 | 9257812 |
625 | 1962 | 9403454 |
750 | 1640 | 9501312 |
875 | 1409 | 9571590 |
1000 | 1235 | 9624506 |
1125 | 1099 | 9665787 |
1250 | 990 | 9698891 |
1375 | 901 | 9726027 |
1500 | 827 | 9748677 |
1526 | 813 | 9752926 |
In examining the data contained in Table 7.9, let us focus attention on the information transfer rate in the third column. Note that at an average frame length of 72 bytes, the information transfer rate is approximately 5.48 Mbps, or slightly more than half the Ethernet 10-Mbps operating rate. At an average frame length of 1526 bytes in which all frames are the maximum length, the information transfer rate increases to approximately 9.75 Mbps. This explains why a large 10-Mbps Ethernet network can safely handle many simultaneous file transfer operations without degradation. A file transfer increases the average frame length, which increases the ability of an Ethernet network to transport information.
To illustrate how the information transfer rate depends on the average frame length, the results obtained from the execution of EITR.BAS were plotted as a line graph in Figure 7.2. In examining the entries on the y-axis of Figure 7.2, note that they range up to 10 Mbps, representing the information transfer rate on a 10-Mbps Ethernet network. Because Figure 7.2 is based on the plot of column 3 versus column 1 from Table 7.9, you can simply multiply the y-axis values by 10 to obtain a plot of the frame length versus the information transfer rate for 100-Mbps Fast Ethernet.
One of the more interesting aspects of networking is its resemblance, on occasion, to a magician, with a comparison between products many times appearing as an illusion when you probe deeper. We can vividly note one illusion by comparing the information transfer rate of Gigabit Ethernet to Fast Ethernet and conventional or legacy 10-Mbps Ethernet.
As previously noted, Gigabit frames less than 512 bytes in length, not including the preamble and start of frame delimiter fields, are extended to a length of 512 bytes through carrier extension technology. Thus, the actual information transfer capacity of a Gigabit frame depends on its length.
Table 7.10 contains the program listing of GITR.BAS, which was developed to compute the information transfer rate of Gigabit Ethernet based on a range of frame lengths. Note that when the frame length is less than 500, that value plus 8 is subtracted from 512 to determine the number of carrier extensions. Because we are working with frames that include the preamble and start of frame delimiter fields, an additional 8 bytes is subtracted to determine the number of carrier extensions. Next, the LPRINT statement subtracts the number of carrier extensions plus 26 overhead Ethernet bytes from the fixed Gigabit frame length of 520 bytes for all frames less than or equal to 520 and multiplies that amount by 8 to compute the number of information transporting bits in a frame. Multiplying that number by the variable FPS results in the information transfer rate for all frames up to 520 bytes in length. When frames exceed 520 bytes in length, the second series of LPRINT statements is invoked. This results in the frame length being decremented by the 26 overhead bytes in order to compute the information transfer rate. Table 7.11 illustrates the results obtained from the execution of the program GITR.BAS.
REM PROGRAM GITR.BAS LPRINT "THIS PROGRAM COMPUTES GIGABIT ETHERNET INFORMATION TRANSFER RATE" LPRINT LPRINT LPRINT "AVERAGE FRAME 100% LOAD INFORMATION TRANSFER" LPRINT " LENGTH FRAMES/SEC RATE IN BPS" LPRINT FOR J = 1 TO 12 STEP 3 READ A, B, C DATA 72,72,1,80,100,20,125,1500,25,1526,1526,1 FOR FLENGTH = A TO B STEP C IF FLENGTH < 520 THEN GLENGTH = 520 ELSE GLENGTH = FLENGTH FPS = 1/(9.599999999999999D-08 + GLENGTH * 8 *.000000001#) IF FLENGTH > 520 GOTO SKIP CARRIEREXT = 512 - (FLENGTH + 8) LPRINT USING " ##### ########## "; FLENGTH; FPS; LPRINT USING " ########## "; FPS * (520 - (CARRIEREXT + 26)) * 8 GOTO SKIP1 SKIP: LPRINT USING " ##### ######### "; FLENGTH; FPS; LPRINT USING " ########## "; FPS * (FLENGTH - 26) * 8 SKIP1: NEXT FLENGTH NEXT J END |
THIS PROGRAM COMPUTES GIGABIT ETHERNET INFORMATION TRANSFER RATE AVERAGE FRAME 100% LOAD INFORMATION TRANSFER LENGTH FRAMES/SEC RATE IN BPS 72 234962 116541352 80 234962 131578944 100 234962 169172928 125 234962 216165408 150 234962 263157888 175 234962 310150368 200 234962 357142848 225 234962 404135328 250 234962 451127808 275 234962 498120288 300 234962 545112768 325 234962 592105280 350 234962 639097728 375 234962 686090240 400 234962 733082688 425 234962 780075200 450 234962 827067648 475 234962 874060160 500 234962 921052608 525 232775 929236480 550 222420 932384320 575 212947 935264064 600 204248 937908480 625 196232 940345408 650 188822 942598144 675 181951 944687104 700 175562 946629184 725 169607 948439616 750 164042 950131264 775 158831 951715328 800 153941 953201984 825 149343 954599744 850 145012 955916416 875 140924 957158976 900 137061 958333376 925 133404 959445056 950 129938 960499008 975 126646 961499520 1000 123518 962450624 1025 120540 963355776 1050 117702 964218432 1075 114995 965041408 1100 112410 965827328 1125 109938 966578752 1150 107573 967297728 1175 105308 967986560 1200 103135 968646848 1225 101051 969280512 1250 99049 969889024 1275 97125 970473920 1300 95274 971036608 1325 93493 971578176 1350 91777 972099840 1375 90123 972602752 1400 88527 973087808 1425 86987 973556032 1450 85499 974008192 1475 84062 974445184 1500 82672 974867776 1526 81274 975292608 |
In examining the entries in Table 7.11, note that the information transfer rate of Gigabit Ethernet is only approximately 20 times that of 10-Mbps Ethernet for minimal-length frames. Because Fast Ethernet has ten times the information transfer rate of legacy Ethernet, this also means that Gigabit Ethernet only provides approximately twice the information transfer capability of Fast Ethernet when the average frame length is relatively small. Because many people might assume that Gigabit Ethernet always provides a ten-fold increase in information transfer in comparison to Fast Ethernet, this is a relatively good example of an illusion of networking!
In our examination of the overhead associated with the composition of Ethernet frames in Chapter 2, we noted that relatively short frames have a relatively large overhead, owing to the necessity to use pad characters to fill a data field to a minimum of 46 characters. At that time, we noted that by composing a client screen to accept several items of information rather than perform separate queries, we could enhance the efficiency of Ethernet frames, as they would transport larger data fields. At that time we did not notice an optimum data field size other than the fact that a minimum data field of 1500 characters is the most efficient.
In examining Figure 7.2, note that for a frame length between 375 and 625 characters, including frame overhead characters, you can achieve an information transfer rate between 9 and 9.5 Mbps. A similar graph is applicable for Fast Ethernet, with the y-axis increased in value by a factor of 10. Thus, for both Ethernet and Fast Ethernet, you can achieve a very high level of information transfer when frames are between 375 and 625 characters in length. This indicates that by attempting to keep your client queries to that range, you can significantly increase the network information transfer rate to over 90 percent of that obtainable by a maximum-length frame. Thus, a frame length between 375 and 625 characters would be an appropriate range to make programmers developing client/server applications aware of as a design goal to increase the transmission efficiency of client/server applications.
Because Gigabit Ethernet primarily represents a backbone network technology, your hard efforts involved in developing client queries that flow in frames between 375 and 625 characters in length can cause a degree of harm if data is uplinked onto a Gigabit network. This is because you will note from Table 7.11 that the information transfer rate of Gigabit Ethernet begins to exceed 900 Mbps only when the frame length exceeds 500 bytes. Thus, the development of client/server applications may require you to investigate both the efficiency of frames on the network that clients use as well as a backbone network that may be used to connect clients to enterprise servers.