Why is it that people always say that for a switch with ports in full-duplex mode you need portcount times bandwidth times two for the backplane capacity?

Let's have an example. Take a 4-port 1Gb switch. Obviously full-duplex since 1GbE doesn't do half duplex.


  • Ports 1 and 2 talk to each other at 1Gb full-duplex (Port 1 and Port 2 both send and receive at 1Gb/s)
  • Ports 3 and 4 talk to each other at 1Gb full-duplex (both send and receive at 1Gb/s)

All four ports now are using up 2Gb/s (RX+TX) (if you look at the ports). Yet, there are only four actual 1Gb/s streams, as the following illustrates:

  • Port 1 --> Port 2 @ 1Gb/s
  • Port 1 <-- Port 2 @ 1Gb/s
  • Port 3 --> Port 4 @ 1Gb/s
  • Port 3 <-- Port 4 @ 1Gb/s

Why is it that some claim you need a 8 Gb/s backplane here?? Is this a marketing trick?

Take the even more simple example of a 2-port switch? Why in the world would I need a 4Gb/s backplane then?

  • Not my downvote, but he problem is that it sounds a bit like homework. – tombull89 May 5 '13 at 14:44
  • Nobody says that. Look up oversubscription and examine the concept of wire-speed. – ewwhite May 5 '13 at 14:44
  • Well I guess I should become a professor then, if the questions I figure out myself sound like homework. – Marki May 5 '13 at 14:45
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    Off topic? Hrmm, I've answered this very question in person with numerous of my (less experienced) fellow Networking Professionals. I'm relatively new here, but it seems pretty on topic within the realm of professional networking to me. – Jeff McAdams May 6 '13 at 17:26
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    This kind-of random closing of topics for whatever reasons gets on my nerves. Probably the marketing guys trying to shut us up ;-) – Marki May 6 '13 at 19:21

Your question highlights the problem of "Marketing Math" when it deals with networking gear.

Yes, the total backplane throughput of a switch would be the sum of the port speeds. The trickiness arises when you start looking at spec sheets for chassis based switches. You have to consider the bandwidth of the backplane for streams to and from an individual switch blade. In order to ensure that your switch is non-blocking, you need the throughput of the interface for the switch blade to talk to the backplane to be double the port capacity.

Consider a hypothetical 2 slot switch...each slot populated with 2 ports of gigabit ethernet where each port on the switch is sending its data to the same numbered port of the other switch blade:

  • Slot1/Port1 -> Slot2/Port1 @ 1Gb/s
  • Slot2/Port1 -> Slot1/Port1 @ 1Gb/s
  • Slot1/Port2 -> Slot2/Port2 @ 1Gb/s
  • Slot2/Port2 -> Slot1/Port2 @ 1Gb/s

Total of 4 Gb/s, just like your example, but consider the bandwidth to and from the backplane for the same traffic pattern from the perspective of Slot 1:

  • Slot1/Port1 -> backplane @ 1Gb/s
  • backplane -> Slot1/Port1 @ 1Gb/s
  • Slot1/Port2 -> backplane @ 1Gb/s
  • backplane -> Slot1/Port2 @ 1Gb/s

Total of 4 Gb/s for the 2 port switch blade to talk to the backplane. The same applies to slot 2, so if you're measuring the sum of the bandwidths of the switch blades talking to the backplane, you come up with 8 Gb/s. While the backplane internally only has to be able to handle a total throughput equal to the port capacity, the interfaces between the backplane and the switch blades of a modular chassis will sum up to double the capacity.

This is why reading spec sheets on networking gear is not always a straightforward task and sometimes requires some digging to understand what the actual data the vendor is presenting to you is representing. Some will, for backplane bandwidth, quote the throughput of the backplane itself, while others will give you the sum of the interfaces to the backplane from the switch blades.

  • Ok, thanks. But no matter what, whether you have slots with ports or just ports (one slot), summing up the TX of the sending port with the RX of the receiving port is no correct way of specifying anything to me, since we're talking about the same data being transferred. * Slot1/Port1 -> backplane @ 1Gb/s = backplane -> Slot2/Port1 * backplane -> Slot1/Port1 @ 1Gb/s = Slot2/Port1 -> backplane * Slot1/Port2 -> backplane @ 1Gb/s = backplane -> Slot2/Port2 * backplane -> Slot1/Port2 @ 1Gb/s = Slot2/Port2 -> backplane At no point does the backplane carry 8Gb/s of data. – Marki May 5 '13 at 17:41
  • Like I said, "Marketing Math". – Jeff McAdams May 5 '13 at 18:36
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    Also, this is where you really get into needing to understand the internal architecture of the switch. If the ASIC that is making the forwarding decisions is on a blade, then the backplane really would need 8 Gb/s of throughput as the traffic would need to transit from the interface blade, across the backplane to the switching module, then back across the backplane to the output interface blade...transiting the backplane twice for each packet. – Jeff McAdams May 5 '13 at 18:38
  • Well... the backplane may be in use for the same packet more than one time, but not at the same time. ;-) – Marki May 5 '13 at 19:35
  • @Marki You're right as far as I can tell. Even if all 4 ports were to generate L2 multicast packets continuously I still count 4 Gb max for this example. Regardless where the forwarding asic is. As you say, the data gets send twice (asic on chassis backplane) or 3 times (asic on each blade), but not simultaneously. But marketing really loves to add RX and TX together to make the numbers bigger. Take a look at Powerline marketing if you want a real laugh. Those things are as half-duplex as can be and they still add RX+TX. – Tonny May 5 '13 at 22:25

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