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Probability of Time-Overlap between two Transceivers

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Meshy
#1
Nov15-12, 07:23 PM
P: 5
Ok, so I had this thought for low power transistors to only broadcast certain data at specific time increments. I cannot figure out how to answer this for the heck of me.

Two transceivers, and they only broadcast SEND requests when they have data to send.

Transceiver A) Broadcasts every 8 seconds for 2 seconds. If it doesn't get a SEND request from another transceiver in those 2 seconds it goes back to sleep.

Transceiver B) Broadcasts every 8 seconds for 2 seconds. If it doesn't get a SEND request from another transceiver in those 2 seconds it goes back to sleep.

Transceiver X) Broadcasts every X seconds for Y seconds. If it doesn't get a SEND request from another transceiver in those Y seconds it goes back to sleep. ::

If they were started a random times, lets say 6 second differential or w/e, just as long as they weren't started at the same time, how long till I', guaranteed one finds the other? Is there a mathematical model to calculate this?
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haruspex
#2
Nov15-12, 07:51 PM
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Clearly A and B are not guaranteed ever to get in touch.
For simplicity, to start with, I would assume each broadcasts/listens for a time period of 1, and take the cycle lengths to be whole multiples of that. What then would be the relationship between two cycle lengths to ensure they coincide sometime?
Meshy
#3
Nov15-12, 09:18 PM
P: 5
Now that I think of it, if they were both 8 seconds they aren't guaranteed to ever communicate. But at 6 seconds, they should. Lemme give it another shot:


-, * = 2 sec increments


----*----*----*----*----*----*----*----*----*----*----*----*----*----*----*----*----*----*----*----*----*
---*---*---*---*---*---*---*---*---*---*---*---*---*---*---*---*---*---*---*---*---*---*---*---*---*---*

Meshy
#4
Nov15-12, 09:22 PM
P: 5
Probability of Time-Overlap between two Transceivers

Looks like every 40 secs in that specific scenario.
Meshy
#5
Nov15-12, 09:30 PM
P: 5
Set - or * to t. x1 = 5, x2 = 4.

f(x,y,t)=>

f(5,4,2)=40;
f(5,4,1)=20;

f(x,y,t)=x*y*t

So X*Y*T it is I guess.

Lemme try, (2,3,1):

2X3X1 = 6

-*-*-*-*-*-*-*-*-*-*
--*--*--*--*--*--*--*


Yup that's six increments. Wow. Figured it out myself :)
Meshy
#6
Nov15-12, 09:36 PM
P: 5
So I just realized it's just the multiplication of the 2 time increments. Obviously had too much on my mind today. lol. SOLVED.
haruspex
#7
Nov16-12, 02:55 AM
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I'm glad you're happy, but if that's the answer then I clearly did not understand the question in the first place.
If A sends for 2 seconds starting at time 6t, and B does 2 seconds starting at time 6t+3, t=0, 1, ..., they will never be awake together. Likewise with 1 second at 10t, 15t+2. What have I misunderstood?


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