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marin [14]
4 years ago
13

Consider an M/M/1 queue in which the expected number of customers in the system is 4, and the expected waiting time in the syste

m is 80 minutes. What is the probability that a customer’s service time is less than 40 minutes?
Mathematics
1 answer:
boyakko [2]4 years ago
6 0

Answer:

Step-by-step explanation:

Given

Expected no of customer in the system is 4 i.e. length of queue is L=4

Expected waiting time in the system W=80\ min

Length of queue is given by

L=\frac{\rho }{1-\rho }

4=\frac{\rho }{1-\rho }

thus \rho =\frac{4}{5}

where \rho =\frac{mean\ arrival\ rate}{mean\ service\ rate}

L is also given by

L=\lambda \times W

therefore

\lambda =\frac{4}{80}=\frac{1}{20}

where \lambda=mean arrival rate

\mu=mean service rate

Probability that customer service rate is less than 40 minutes is

P=1-\rho e^{-\mu (1-\rho )\cdot t}

P=1-0.8\times e^{-\frac{1}{16}\times 0.2\times 40}

P=1-0.4852

P=0.5147

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