The parts department of a large automobile dealership has a counter used exclusively for mechan - ics' requests for parts. The time between requests can be modeled by a negative exponential distri - bution that has a mean of five minutes. A clerk can handle requests at a rate of 15 per hour, and this can be modeled by a Poisson distribution that has a mean of 15. Suppose there are two clerks at the counter. a. On average, how many mechanics would be at the counter, including those being served? b. What is the probability that a mechanic would have to wait for service? c. If a mechanic has to wait, how long would the average wait be? d. What percentage of time are the clerks idle? e. If clerks represent a cost of $20 per hour and mechanics a cost of $30 per hour, what number of clerks would be optimal in terms of minimizing total cost?

a. The Number of mechanics at the counter would be 0.952 requests

b. The probability that a mechanic would have to wait for service is 0.2278

c. The average wait would be 0.0556 hours

d. The percentage of time are the clerks idle is 60%

e. The number of clerks would be optimal in terms of minimizing total cost is 2

Explanation:

a. According to the given data we have the following:

One request comes in average 5 minutes; therefore in one hour number of requests will be 12

Or arrival rate λ = 12 requests per hour

Service rate µ = 15 request per hour

Number of servers M = 2 (2 clerks on counter)

The probability that no mechanics is waiting in line P0 = [ (λ/µ) ^0/0! + (λ/µ) ^1 / 1! + (λ/µ) ^2/2! * (1 / 1-ρ) ]^-1

= [1 + 0.8 + 0.533]^-1 = 2.33^-1 = 0.4286

The average number of mechanics waiting in line Lq = P0 * (λ/µ) ^M * ρ / M! * (1-ρ) ^2

= (0.4286 * (0.8) ^2 * 0.40) / (2! * (1-.40) ^2)

= (0.4286 * 0.64 * 0.40) / (2 * 0.36) = 0.152

Therefore, the Number of mechanics at the counter, including those being served = Lq + λ/µ

= 0.152 + 0.8 = 0.952 requests

The Number of mechanics at the counter would be 0.952 requests

b. To calculate the probability that a mechanic would have to wait for service we would have to make the following calculations:

Average waiting time of machine Wa = 1 / (Mµ - λ)

= 1 / (2 * 15 - 12)

= 1 / (30 - 12)

= 1/18 = 0.0556 hours

Average waiting time of machine in queue Wq = Lq / λ

= 0.152 / 12

= 0.01267 hours

Therefore the probability that a mechanic would have to wait for service Pw = Wq/Wa

= 0.01267/0.0556

= 0.2278

The probability that a mechanic would have to wait for service is 0.2278

c. To calculate how long would the average wait be If a mechanic has to wait we would have to calculate the following formula:

Average waiting time of machine Wa = 1 / (Mµ - λ)

= 1 / (2 * 15 - 12)

= 1 / (30 - 12)

= 1/18 = 0.0556 hours

The average wait would be 0.0556 hours

d. To calculate the percentage of time are the clerks idle we would have to calculate the following formula:

Utility ratio ρ = λ / Mµ

= 12 / 2*15 = 12/30 = 0.4 or 40%

Therefore idle time = 1 - ρ

= 1 - 0.4 = 0.6 or 60%

The percentage of time are the clerks idle is 60%

e. The number of clerks would be optimal in terms of minimizing total cost is 2