Question

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?

Answer 1

Answer:

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