Router System Capacity Vs Packet Forwarding Capacity

When I read a spec document of Juniper Mx104 3D router, they mention 2 things. This router has 4 X 10GBE ports.

1.System capacity 80 gpbs 2.Packet forwarding capacity 55 to 60 Mpps

Questions:

1. What is the difference between those technically.
2. Additionally, this router has, 4 line card slots. Assuming if I plug in 4 line cards of 2 X 10 GBE ports, I exceed the capacity of the system theoretically. Is that ok?

How to balance between these capacity specification?

If the question is relevant, please close it.

• Did any answer help you? If so, you should accept the answer so that the question doesn't keep popping up forever, looking for an answer. Alternatively, you can provide your own answer and accept it. Aug 6, 2017 at 20:12

We cannot know for sure what Juniper refers to when stating "system capacity 80gps"

If we take the embedded 4 x 10G interface and we use them at full duplex then we get 80gbps. So it could refer to this fact. In this case adding line card would increase this number.

MPPS is much more informative as it refers to the actual capacity of the switch of processing packets.

From Cisco support forum:

For example, 2960-48PST-S is 13.3 Mpps.

The figure MPPS expresses the maximum number of frames per second that can be processed by the device. It is not dependent on frame size but clearly small frames require higher packet rates.

To give you an idea of what this number says:

smallest frames in Ethernet are 64 bytes in size, taking in account the preamble (8 bytes) and the minimum inter-frame gap (the last two counts roughly for 20.2 bytes) to fill a GE port in one direction you need 1484560 frame per second.

10^9 / [(64+20.2)*8] where 8 is bits/byte.

So a number of 13.3 MPPS is equivalent to [((13.3 M * (64+20.2) * 8)) / 10^9 = 8.95 / 2=4.47] 4.47 GE ports filled with smallest frames bidirectional.

On the other hand frames of max size 1518 bytes require 81264 fps to fill a GE port in one direction.

So this number expresses the forwarding capability of the device.

A non blocking device with 48 GE ports would require 2 * 1484560 * 48 as MPPS or higher.

Note that the MX104 support frames up to 9192 bytes.

60mpps of 9192 bytes frames give you a max bandwidth of 4420 / 2210gbps (bidirectional / unidirectional)

60Mpps of 64 bytes frames give only 37/18,5 gbps.

If we are stuck with standard Ethernet MTU of 1500 we get something around 730/365gbps

• The post you quoted is from a on-time-only poster and should be taken with a grain of salt. I have no idea where this "roughly 20.2 bytes" comes from. The preamble is 8 bytes and the IPG is 12 bytes. Total on-the-wire minimum for an Ethernet frame is 84 bytes or 672 bits. GbE tops out at 10^9/672 = 1,488,095 frames/s.
– Zac67
May 6, 2018 at 12:36

Specifications from most vendors follow the same rules:

Bandwidth from non-blocking ports is simply added and counted twice for full-duplex: 40 GbE ports => "80 Gbit/s" - this is a theoretical in-flight maximum

The packet forwarding rate is a bit more real-life. It's the (best case) forwarding capacity of the backplane. Unlike bandwidth it's only counted once. A 40-port GbE device has a physical limit of 10^9/672*40 = 59.5 Mp/s. (672 bits is the minimum Ethernet packet size.) A lower forwarding rate means you won't get full wire speed with small packets.

Note that the packet forwarding rate is best case - any additional processing (ACLs, NAT, firewall rules, ...) may eat into the forwarding rate, exactly how depends on the device and its architecture.