Question

4. (a) 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 the order in which they

come in, and if arrival of sets follows a Poission distribution approximately with an

average rate of 10 per 8 hours day, what is the repairman’s expected idle time each

day, How many jobs are ahead of the average set just brought in?

Answer 1

Here$\lambda=\frac{10}{8×60}=\frac{1}{48}set per minute$

But $\mu=\frac{1}{30} set/minute$

Prob.that their is no unit in the system P_O=1-$\frac{\lambda}{\mu}$

=$1-\frac{5}{8}=\frac{3}{8}$

Repairmans's expected idle time in 8 hours a day:

=nP_O=$8×\frac{3}{8}$ =3hours

Expected average n.o. of jobs or average n.o.of Tv sets in the system

L_S=$\frac{\lambda}{\mu-\lambda}$

But $\mu-\lambda$

=$\frac{1}{30}-\frac{1}{48}=\frac{1}{80}$

Substuting into the formula

$=\frac {1}{48}×\frac{80}{1}=\frac{5}{3}jobs$