Multipoint to Multipoint Mesh Capacity Math

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In the this case we assume

  • All APs are using the same 5 GHz channel
  • All APs can "hear" all of the other devices on the same channel
  • All traffic uses all hops to get off of the network
  • The APs are distributed in a linear fashion as would be the case for a street deployment
  • The AP which is hardwired is at one of the mesh
  • The propagation environment does not generate a lot of reflection
    • With Aruba's unique Multi-polar omnis the solution provides the 11N dual stream throughput rates
    • Allowing for some derating this analysis assumes 80 Mbps goodput in a 20 MHz channel


1 HOP SYSTEM

200px-One_Hop_Shared.png

In this case there is one hop.

The leftmost unit will exchange its traffic directly on the wire.

The first mesh unit will exchange its traffic over the 80 Mbps connection to the wired unit.


2 HOP SYSTEM

300px-Two_Hop_Shared.png

In this case there is one hop to the first meshed unit and two hops to the second meshed unit.

If only one unit is busy at a given time bandwidth available to the first meshed unit is 80/1 and the bandwidth available to the second meshed unit is 80/2. This is referred to as the peak rate available at each unit.

When both units are busy and loaded equally you now have to carry the total number of hops in determining the bandwidth that both devices can carry

For example assume that the first and second meshed units have a sustained load of 1 Mbps.

The traffic from the first meshed unit has to use the 5 Ghz channel one time to send the 1 Mbps back to the wire.

The traffic from the second meshed unit has to use the 5 GHz channel two times to send the 1 Mbps back to the wire.

So, 1 Mbps of traffic on each unit generates 3 Mbps of 5 GHz traffic.

In this fashion we say that the average sustained date rate at each AP is equivalent to 80/3 or 26.6 Mbps.

Total system capacity of the meshed units is 53.2 Mbps so the mesh penalty for a two hop system is 33%.


3 HOP SYSTEM

400px-Three_Hop_Shared.png

In this case there is one hop to the first meshed unit, two hops to the second meshed unit. and three hops to the third meshed unit.

If only one unit is busy at a given time, bandwidth available to the first meshed unit is 80/1, the bandwidth available to the second meshed unit is 80/2, and the bandwidth available to the third meshed unit is 80/3. This is referred to as the peak rate available at each unit.

When both units are busy and loaded equally you now have to carry the total number of hops in determining the bandwidth that both devices can carry

For example assume that the first, second, and third meshed units have a sustained load of 1 Mbps.

The traffic from the first meshed unit has to use the 5 Ghz channel one time to send the 1 Mbps back to the wire.

The traffic from the second meshed unit has to use the 5 GHz channel two times to send the 1 Mbps back to the wire.

The traffic from the third meshed unit has to use the 5 GHz channel three times to send the 1 Mbps back to the wire.

So, 1 Mbps of traffic on each unit generates 6 Mbps of 5 GHz traffic.

In this fashion we say that the average sustained date rate at each AP is equivalent to 80/6 or 13.3 Mbps.

Total system capacity of the meshed units is 40 Mbps so the mesh penalty for a three hop system is 50%.


4 HOP SYSTEM

500px-Four_Hop_Shared.png

The logic follows for higher order systems. (here N=hops)

We now represent channel capacity of C

The peak rate that can ideally be delivered to the end of a linear mesh decreases as C/N

The sustained average that can be delivered to each AP can be shown to be equal to 2C/(N*(N+1))

The mesh penalty can similarly be calculated as (100 - 200/(N+1))%

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Last update:
‎09-25-2014 01:15 PM
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