## Example

Consider a very simple network, depicted in Figure 5-10.

Figure 5-10 Sample Network

The 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 assume that all the Ethernets have traffic characteristics similar to those shown in Figure 5-11.

Figure 5-11 Graph of Typical Ethernet Network

Real Network Traffic Ethernet Frame Distribution

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Bytes of a Frame

Most of the traffic falls between 256-byte and 1280-byte packets, with numerous 64-byte packets that are typically acknowledgment packets. Our calculations assume that the Ethernet network is fairly busy with average utilization at 40 percent; in other words, 4 Mbps of Ethernet bandwidth is utilized. For average traffic rates, 40 percent utilization of Ethernet bandwidth is a rather heavily utilized network because collisions are very probable and most of the traffic on the 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:

To calculate the total packets per second that would be on the Ethernet, we need to apply the following formula for each of the different packet sizes: (Bandwidth x Percent Media Used)/ (Packet Size x bits/byte) = Packets per Second.

Using this formula yields:

• (4 Mbps x 35%)/(768 bytes x 8 bits/byte) = 228 pps

• (4 Mbps x 20%)/(1280 bytes x 8 bits/byte) = 79 pps

• (4 Mbps x 15%)/(512 bytes x 8 bits/byte) = 147 pps

• (4 Mbps x 30%)/(64 bytes x 8 bits/byte) = 2344 pps

The total, 2798 pps, is not the pps rate that goes through the router. If it is, the network design is not optimal and should be changed. Rather, the 80/20 rule applies to most nonswitched networks, where 80 percent of the traffic stays on the local network and 20 percent goes to a different destination. Then we have 2798 x 20% = 560 pps that the router must deal with from that single Ethernet network. If we take six Ethernets with similar characteristics, we get an aggregate 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 servers that are off the backbone. In this scenario, the router must support 6 x (2798 x 90%) = 15,110 pps for our example of six Ethernets. The appropriate router platform must be chosen that will meet the traffic requirements.

This example shows how the packets-per-second requirement for varying networks is computed. As will be shown in subsequent sections, all Cisco router platforms meet and greatly exceed the pure packets-per-second requirements of real networks.

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