Probability of Time-Overlap between two Transceivers

[Meshy]

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, let's 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?

 

[haruspex]

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]

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]

Looks like every 40 secs in that specific scenario.

 

[Meshy]

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]

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]

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?