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 = $\frac{E[Z_j]}{E[Z_j] + E[Y_j]} = \frac{12}{12+5/2} = \frac{12}{14.5}$