One mechanic services five drilling machines for a steel plate manufacturer. Machines break down on an average of once every six working days, and breakdowns tend to follow a Poisson distribution....

One mechanic services five drilling machines for a steel plate manufacturer. Machines break down on an average of once every six working days, and breakdowns tend to follow a Poisson distribution. The mechanic can handle an average of one repair job per day. Repairs follow an exponential distribution.

a. How many machines are waiting for service, on average?

b. What is the average waiting time n the queue?

c. What is the average wait in the system?

Expert Answers
sciencesolve eNotes educator| Certified Educator

a. You need to evaluate how many machines are waiting for service, on average, using the following formula, such that:

`L_s = (lambda)/(mu - lambda)`

`lambda` represents the mean number of machine breakdowns

`lambda = 1/6 = 0.16`

`mu` represents the mean of repairs

`mu = 1`

`L_s = 0.16/(1 - 0.16) => L_s = 0.16/0.84 => L_s = 0.190` machines are waiting for service, on average

b. You need to evaluate the average waiting time in the queue, using the following formula, such that:

`W_q = lambda/(mu(mu - lambda)) => W_q = 0.16/(1(1 - 0.16)) => W_q = 0.190` the average waiting time in the queue

c. You need to evaluate the average wait in the system, using the following formula, such that:

`L_q = ((lambda)^2)/(mu(mu - lambda)) => L_q = (0.16^2)/(1(1 - 0.16)) =>` `L_q = 0.03 t` he average wait in the system

sciencesolve eNotes educator| Certified Educator

b. The average waiting time in the queue is evaluated using the following formula, such that:

`W_q= (lambda)/(mu(mu - lambda))`

`lambda` represents the mean number of machine breakdowns

`mu` represents the mean of repairs

`lambda` = 1 breakdown/6 days => `lambda = 1/6`

`mu` = 1 repair job per day

`W_q = (lambda)/(mu/(mu - lambda))=> W_q = 1/(6(1 - 1/6)) => W_q = 1/5`

c. The average waiting time in the system is evaluated using the following formula, such that:

`L_q = (lambda^2)/(mu(mu - lambda))=>L_q = 6/(36*5) => L_q = 1/30 `