Individual HomeworkQueueing Theory (Diagrams in Attachment)Service Facility DesignA service organization has the following process design. There are two separate but identical service centers (with one employee each) that serve two distinct geographic areas. Each center has an activity time with a mean of 10 minutes and a standard deviation of 5 minutes. Customers arrive at each center with an inter-arrival time of 12 minutes. The standard deviation of this inter-arrival time is also 12 minutes.Line 1p = 10 minutesa = 12 minutesLine 2p = 10 minutesa = 12 minutesThe organization is considering strategies to reduce wait times. In particular, it is comparing the following strategies: (A) Reduce p from 10 minutes to 5 minutes. In this case, the standard deviation would be reduced from 5 minutes to 2.5 minutes (B) Standardize the activity, which would reduce the standard deviation from 5 two 2.5 minutes. p = 10 minutes will remain unchanged. (C) Merge the two centers as shown below (in this case, the standard deviation of the inter-arrival times will be 6 minutes).Combined1. What is the current average wait time in minutes?2. What is the average wait time in minutes if strategy A is implemented?3. What is the average wait time in minutes if strategy B is implemented?4. What is the average wait time in minutes if strategy C is implemented?Individual HomeworkQueueing Theory (Diagrams in Attachment)Service Facility DesignA service organization has the following process design. There are two separate but identical service centers (with one employee each) that serve two distinct geographic areas. Each center has an activity time with a mean of 10 minutes and a standard deviation of 5 minutes. Customers arrive at each center with an inter-arrival time of 12 minutes. The standard deviation of this inter-arrival time is also 12 minutes.Line 1p = 10 minutesa = 12 minutesLine 2p = 10 minutesa = 12 minutesThe organization is considering strategies to reduce wait times. In particular, it is comparing the following strategies: (A) Reduce p from 10 minutes to 5 minutes. In this case, the standard deviation would be reduced from 5 minutes to 2.5 minutes (B) Standardize the activity, which would reduce the standard deviation from 5 two 2.5 minutes. p = 10 minutes will remain unchanged. (C) Merge the two centers as shown below (in this case, the standard deviation of the inter-arrival times will be 6 minutes).Combined1. What is the current average wait time in minutes?2. What is the average wait time in minutes if strategy A is implemented?3. What is the average wait time in minutes if strategy B is implemented?4. What is the average wait time in minutes if strategy C is implemented?