Alternating Renewal Process

Alternating Renewal Process

1. Definition

Consider a process X(t) with “on” periods and “off” periods
  • Let Z_j equiv ON time of cycle j
  • Let Y_j equiv OFF time of cycle j
Then Z_j and Y_j must satisfy the following property in order for X(t) to be regenerative
The pair (Z_j, Y_j) must be i.i.d; in particular, (Z_j, Y_j) independent of (Z_i, Y_i) for i neq j. That is
  • Z_j are i.i.d
  • Y_j are i.i.d
  • However, Z_j and Y_j may be independent for the same j
Cycle time X_j = Z_j + Y_j.
Then by the regenerative process, we have
the average “up” time = frac{mbox{average up time in a cycle}}{mbox{average time of one cycle}} = frac{E[Z_j]}{E[Z_j] + E[Y_j]}

2. Example

Cars pass a point on highway according to Poisson process with rate lambda = 2/min. 1/5 of cars are speeding (>10 mph over the pretend speed limit). Assume time to issue a ticket~UNIF[10,14] minutes(one officer)
Question: What fraction of speeding cars pass when the officer is busy.
Answer: The answer to the question is equivalent to the fraction of time the officer is busy.
  • Let Y_j =  time spent waiting to give a ticker
  • Let Z_j = time spent giving a ticker
Speeders arrive according to Poisson process with rate 2 cdot 1/5 = 2/5 per min.
Fraction of time busy = frac{E[Z_j]}{E[Z_j] + E[Y_j]} = frac{12}{12+5/2} = frac{12}{14.5}

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