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Lilit [14]
4 years ago
10

in a waiting line system, there is only one server. Customer arrivals follow Poisson distribution with mean inter-arrival time 5

minutes. Service time follow exponential distribution and a server can serve 15 customers per hour on average. What is the average waiting time?a. 2 minb. none of the answerc. 20 mind. 30 mine.16 min
Business
1 answer:
nignag [31]4 years ago
4 0

Answer:

The problem can be represented by single queue model.  

Inter-arrival time = 5 min

Arrival rate = 60 min / 5 min

Arrival rate =12 customers per hour

Service rate = µ =15 customers per hour  

Utilization of server

P = λ/µ

P = 12 / 15

P = 0.80

Number of customers in queue

L = (12^2)/15(15 - 12)

L = 3.2 customers per hour  

Waiting time in queue

W = 3.2 / 12  

W = 0.266 hour

Convert hours into minutes as follows:

Waiting time in queue = 0.2667 x 60

Waiting time in queue = 16 minutes

Waiting time in system

Ws = 02667 + 1/15

Ws = 0.333 hour

Waiting time in system = 0.333 x 60

Waiting time in system = 20 mins

Average waiting time in queue = 16 mins  

Average waiting time in system = 20 mins

Therefore, average waiting time in the system is 20mins.  

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Explanation:

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Answer:

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Journal Entry

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Explanation:

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Explanation:

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Social Movements are marked by 4 stages: Emergence, coalescence, bureaucratization, and decline

Stage 1: Emergence or Incipient Stage - This stage is where individuals in isolated areas begin to see or believe there is a problem with a policy or an area of society. There is disappointment with certain social or political issues. It is the stage of realisation of a problem and dissatisfaction with the problem and this is the stage identified in the question.

Stage 2: Coalescence - This stage goes beyond the dissatisfaction to people, groups coming together to discuss the issue and then campaigns and protests begin to emerge

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_____ involves interpreting the meaning of the received messages
mr Goodwill [35]

Answer:

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