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.Each medium has a specificfixed-size bandwidth pipe associated with it, and each one may or may not define a minimumand maximum valid frame size.The minimum and maximum frame sizes are important becausemost good applications written for workstations or PCs make efficient use of bandwidthavailable and use maximum-sized frames.The smaller the frame size, the higher the percentageof overhead relative to user data; in other words, smaller frame sizes mean less effectivebandwidth utilization as illustrated in Figure 5-9.Number of Packets per SecondDE53XE04 CH01.book Page 179 Friday, January 7, 2000 5:35 PMSelecting the WAN Hardware 179Figure 5-9 Bandwidth Efficiency for Small Versus Large FramesBandwidth + Packet Size = Theoretical PerformanceSmaller Packets (Less Efficient, Not Real)Larger Packets (Better Utilization)An understanding of real traffic patterns is important when designing networks.At least sometypical applications should be known so that the average packet sizes on the network can bedetermined.Sniffer traces to look at typical packet sizes for varying applications are helpful;some of the more common ones include:" Hypertext Transfer Protocol (HTTP) (World Wide Web) 400 to 1518bytes" Network File System (NFS) 64 to 1518 bytes" Telnet 64 to 1518 bytes" NetWare 500 to 1518 bytes" Multimedia 400 to 700 bytesFor optimal network designs, an understanding of the kinds of applications that will be used isnecessary to determine the typical packet sizes that will be traversing your network.Thefollowing example, taken from a real network, shows how to optimize your network design.ExampleConsider a very simple network, depicted in Figure 5-10.DE53XE05 CH01.book Page 180 Friday, January 7, 2000 5:35 PM180 Chapter 5: WAN DesignFigure 5-10 Sample NetworkOffice OfficeBuilding BuildingRouter A Router BFDDIThe network consists of six Ethernet networks that are interconnected via an FDDI backbone.Router A interconnects the Ethernet networks to the FDDI backbone.For simplicity, we assumethat all the Ethernets have traffic characteristics similar to those shown in Figure 5-11.Figure 5-11 Graph of Typical Ethernet NetworkReal Network TrafficEthernet Frame DistributionAcknowledgements DataNumber ofFrames0 64 128 256 512.1518Bytes of a FrameMost of the traffic falls between 256-byte and 1280-byte packets, with numerous 64-bytepackets that are typically acknowledgment packets.Our calculations assume that the Ethernetnetwork is fairly busy with average utilization at 40 percent; in other words, 4 Mbps of EthernetDE53XE06DE53XE07 CH01.book Page 181 Friday, January 7, 2000 5:35 PMSelecting the WAN Hardware 181bandwidth is utilized.For average traffic rates, 40 percent utilization of Ethernet bandwidth isa rather heavily utilized network because collisions are very probable and most of the traffic onthe network is retransmission traffic.However, the example is intended to show a worst-case,real-world performance scenario.For simplicity, we assume that the following traffic is on the Ethernet:" 768-byte packets, 35%" 1280-byte packets, 20%" 512-byte packets, 15%" 64-byte packets, 30%To calculate the total packets per second that would be on the Ethernet, we need to apply thefollowing formula for each of the different packet sizes: (Bandwidth Percent Media Used)/(Packet Size bits/byte) = Packets per Second.Using this formula yields:" (4 Mbps 35%)/(768 bytes 8 bits/byte) = 228 pps" (4 Mbps 20%)/(1280 bytes 8 bits/byte) = 79 pps" (4 Mbps 15%)/(512 bytes 8 bits/byte) = 147 pps" (4 Mbps 30%)/(64 bytes 8 bits/byte) = 2344 ppsThe total, 2798 pps, is not the pps rate that goes through the router.If it is, the network designis not optimal and should be changed.Rather, the 80/20 rule applies to most nonswitchednetworks, where 80 percent of the traffic stays on the local network and 20 percent goes to adifferent destination.Then we have 2798 20% = 560 pps that the router must deal with fromthat single Ethernet network.If we take six Ethernets with similar characteristics, we get anaggregate of 3360 pps that the router must support.Now consider a scenario with central servers and assume that the 80/20 rule does not apply;only 10 percent of the traffic stays local and 90 percent goes through the router to the serversthat are off the backbone.In this scenario, the router must support 6 (2798 90%) = 15,110pps for our example of six Ethernets.The appropriate router platform must be chosen that willmeet the traffic requirements.This example shows how the packets-per-second requirement for varying networks iscomputed.As will be shown in subsequent sections, all Cisco router platforms meet and greatlyexceed the pure packets-per-second requirements of real networks.Router Platform Switching PathsThis section will list the switching paths that the various router platforms support. CH01.book Page 182 Friday, January 7, 2000 5:35 PM182 Chapter 5: WAN DesignLow-End/Midrange RoutersThis category of routers includes the Cisco 2500, 4000, 4500, and 4700 series routers.Theswitching paths supported for these routers are process switching and fast switching.Fastswitching is on by default for all protocols.The aggregate performance numbers in packets per second are listed in Table 5-6.Table 5-6 Aggregate Maximum Performance for Low-End/Midrange Routers (in Packets-per-Second)Switching Paths 2500 Series 4000 4500 4700Process Switching 1,000 1,800 10,000 11,000Fast Switching 6,000 14,000 45,000 50,000Features Affecting PerformanceUnderstanding how a given feature will affect the router s switching paths is critical when designingnetworks.Many new features are initially incorporated into the process switching path and, insubsequent releases, incorporated into faster switching paths
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