Customers enter the camera department of a store with an average of 14.46 minutes between customers....

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Customers enter the camera department of a store with an average of 14.46 minutes between customers....

Customers enter the camera department of a store with an average of 14.46 minutes between customers.

The department is staffed by one employee, who can handle an average of 11.76 customers per hour.

Assume this is a simple Poisson arrival, exponentially distributed service time situation.

Find the following information to help the manager decide if a second employee should be added:

1) The average number of customers waiting.

Please keep 4 decimals.

2)The average time a customer waits (in minutes).

3)The average time a customer is in the department (in minutes).

business Operations-Management

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Answer 1

Answer #1

λ= 60/14.46 =4.149 customer per hour

μ= 11.76 customers per hour

1) The average number of customers waiting.

Lg=λ² /μ*(μ-λ) =(4.149*4.149)/(11.76*(11.76-4.149))=0.1923 hours

2)The average time a customer waits =Lg/λ =0.1923/4.149 =0.0463 hour =2.78 min

3)The average time a customer is in the department (in minutes)

Ls= λ/(μ-λ) =4.149/(11.76-4.149)=0.545 hour

The average time a customer is in the department = Ls/λ =0.545/4.149=0.1314 hour or 7.88 min

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Answer 2

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