Question

A television repairman finds that the time spent on his jobs has an exponential distribution with a mean of 30 minutes . If he repairs sets in order in which they come in , and if arrival of sets follows a Poisson distribution approximately with an average rate of 10 per 8 hours day , what is the repairman's expected ideal time each day, how many jobs are ahead of the average set just brought in?

Answer 1

$\lambda=\dfrac{10}{8}=1.25$ sets per hour, and $\mu=\dfrac{1}{30}(60)=2$ sets per hour

(a) Expected idle time of repairman each day = Number of hours for which the repairman

remains busy in an 8-hour day (traffic intensity) is given by

              

        $8(\dfrac{\lambda}{\mu})=8(\dfrac{1.25}{2})=5 hours$

Hence, the idle time for a repairman in an 8-hour day will be: (8 – 5) = 3 hours.

(b)Expected (or average) number of TV sets in the system

    $E(L_s)=\dfrac{\lambda}{\mu-\lambda}=\dfrac{1.25}{2-1.25}=\dfrac{1.25}{0.75}=1.66 \approx 2( Approx)$.