- How Are Piconets Spaced, and Are They Using Power Control?
- Saturated Radios
- How Often Will They Collide?
How Often Will They Collide?
Bluetooth devices frequency-hop in a pseudo-random sequence around 79 channels. Each piconet is synchronized to a frequency-hopping sequence defined by its master, so all the devices on the piconet are on the same channel at any particular moment. The piconet is arranged so that only one device is transmitting at any particular moment.
Because only one device in a piconet can transmit at any moment, it doesn't matter how many devices are in the area. The important question is, how many piconets are there?
Suppose that one piconet transmits on channel 0. The second piconet has a 1 in 79 chance of colliding on the same frequency, so it's probability of not colliding is this:
Pno collision 2 piconets = 1 — 1/79
Now if we have add a third piconet, there's a (1-1/79)2 probability that neither the second nor third piconet will collide with the first piconet. If we have n piconets transmitting at the same time, the probability of our first piconet not colliding on the same frequency as any of them is this:
Pno collision n piconets= (1 — 1/79)n — 1
We have the probability that we will be able to transmit in one slot, but for a successful transmission, we need to both send and receive. In the receive slot, the packet is acknowledged, so if the receive slot is lost to interference, we will have to retransmit just as if the transmit slot was lost. The probability of our first piconet not colliding with any of the other piconets in both transmit and receive slots is this:
Pno collision n piconets in 2 slots= (1 — 1/79)2n — 2
Say that we want at least half our packets to get through. Collision probability is 1/2, and we can solve for n to work out how many piconets we can have before half the slot pairs are lost to interference.
0.5 = (1 — 1/79)2n — -2
log (0.5) = (2n — 2) log (1 — 1/79)
n = 1 + log(0.5) / 2 log (1 — 1/79) = 28.2
So, you could get 28 piconets coexisting and still manage to get half your packets through. Figure 2 shows the probability that a slot pair in the first piconet will escape interference through collision with another piconet when 1 to 100 piconets are coexisting.
){kind=link}
Probability of interference-free transmission against a number of coexisting piconets.
These calculations contain simplifications: We've ignored the effects of multislot packets, and we've assumed that all the piconets are transmitting at the same time. In practice, two piconets could have timing offset so that they could send single-slot packets after one another and avoid colliding in time, even though they were on the same frequency.
We have also assumed that all our piconets transmit in every slot. In practice, activity will depend upon the profile. The LAN access point profile is likely to provide bursts of activity as data is transferred, but most of the time a link to an access point will not be transferring data. An active headset link will use between 1/3 and all of the slots in a piconet, depending on whether HV1, HV2, or HV3 packets are chosen. Devices using the object push profile are likely to be active only occasionally and then for short periods. So, in practice, most Bluetooth piconets will not be continuously active. This means that, in real life, piconets are likely to coexist far better than our calculations suggest.
Still even though it is an approximate result, our calculations show that Bluetooth wireless technology fulfills its goal of robustly tolerating interference so that a large number of Bluetooth devices can interoperate in a small space.