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A multiple-channel queuing system with a Poisson arrival rate and exponential service time has an average arrival rate of four customers per hour and an average service time of 18 minutes per customer. The minimum number of servers required to avoid an overloaded system is

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Answer:

1.2

Explanation:

The computation of the minimum number of servers required to avoid an overloaded system is shown below:-

Average service time =  Average service time ÷ number of minutes in a hour

18 min = 18 ÷ 60 per hr

= 0.3 hr per customer

Now for four customers, it would be

min servers = 4 × 0.3

= 1.2

So, the correct answer is min servers = 2 as servers cant be in fractions

Therefore we have applied the above formula.

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