Question

There are two reservation counters for all ticket booking for customers, who arrive in a Poisson fashion at an average rate of 10 per hour. The service time for booking clerks at both the counters are exponentially distributed with mean of 5 minutes. These countrrs remain open for 12 hours per day.
i) Find the hours of the day for which all the clerks are busy.
ii) Find the expected waiting time of customers in the queue.

Answer 1

Answer to Question #84513 – Math – Statistics and Probability

There are two reservation counters for all ticket booking for customers, who arrive in a Poisson fashion at an average rate of 10 per hour. The service time for booking clerks at both the counters are exponentially distributed with mean of 5 minutes. These counters remain open for 12 hours per day.

Question

i) Find the hours of the day for which all the clerks are busy.

Solution

Two counters can serve:

$$\frac{2}{5} \frac{\text{customers}}{\text{minutes}} = \frac{10}{25} \frac{\text{customers}}{\text{minutes}}$$

So clerks are busy 25 minutes in one hour.

Then clerks are busy per day:

$$25 \cdot 12 = 300 \text{ min} = 5 \text{ hours}$$

Question

ii) Find the expected waiting time of customers in the queue.

Solution

Arrival rate:

$$\lambda = \frac{10}{1 \text{ hour}}$$

Service rate:

$$\mu = \frac{1}{5 \text{ min}} = \frac{12}{1 \text{ hour}}$$

The expected waiting time of customers in the queue:

$$W _ {q} = \frac {\lambda}{2 \mu (\mu - \lambda)} = \frac {1 0}{2 \cdot 1 2 \cdot (1 2 - 1 0)} = \frac {1 0}{4 8} = \frac {5}{2 4} \mathrm {h o u r s} = 1 2. 5 \mathrm {m i n}$$

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