Probability of TimeOverlap between two Transceiversby Meshy Tags: probability, timeoverlap, transceivers 

#1
Nov1512, 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? 



#2
Nov1512, 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? 



#3
Nov1512, 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 ********************* ************************** 



#4
Nov1512, 09:22 PM

P: 5

Probability of TimeOverlap between two Transceivers
Looks like every 40 secs in that specific scenario.




#5
Nov1512, 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 :) 



#6
Nov1512, 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.




#7
Nov1612, 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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