Statistics

Question

A single-server queuing system with an infinite calling population and a first-come, first-served queue discipline has the following arrival and service rates:

λ = 36 customers per hour
µ = 50 customers per hour

Determine P0, P4, L, Lq, W, Wq, and U.

Answer 1

In this queuing system, we are dealing with a M/M/1 queue, which is characterized by:

1. Arrival rate (λ\lambda) = 36 customers per hour
2. Service rate (μ\mu) = 50 customers per hour
3. Queue discipline = First-Come, First-Served (FCFS)

The formulas for the various performance metrics in an M/M/1 queue are as follows:

1. Utilization (UU):

The utilization factor represents the proportion of time the server is busy.

U=λμU = \frac{\lambda}{\mu}Substitute the given values:

U=3650=0.72U = \frac{36}{50} = 0.722. Probability that the system is empty (P0P_0):

The probability that there are zero customers in the system (i.e., the server is idle) is:

P0=1−U=1−0.72=0.28P_0 = 1 - U = 1 - 0.72 = 0.283. Probability that there are 4 customers in the system (P4P_4):

The probability that there are exactly nn customers in the system is given by:

Pn=(1−U)⋅UnP_n = (1 - U) \cdot U^nFor n=4n = 4:

P4=(1−0.72)⋅0.724=0.28⋅0.724≈0.28⋅0.2687≈0.0753P_4 = (1 - 0.72) \cdot 0.72^4 = 0.28 \cdot 0.72^4 \approx 0.28 \cdot 0.2687 \approx 0.07534. Average number of customers in the system (LL):

The average number of customers in the system (both in the queue and being served) is:

L=U1−U=0.721−0.72=0.720.28≈2.57L = \frac{U}{1 - U} = \frac{0.72}{1 - 0.72} = \frac{0.72}{0.28} \approx 2.575. Average number of customers in the queue (LqL_q):

The average number of customers waiting in the queue is:

Lq=U21−U=0.7221−0.72=0.51840.28≈1.85L_q = \frac{U^2}{1 - U} = \frac{0.72^2}{1 - 0.72} = \frac{0.5184}{0.28} \approx 1.856. Average time spent in the system (WW):

The average time a customer spends in the system is the average number of customers in the system divided by the arrival rate:

W=Lλ=2.5736≈0.0714 hours≈4.29 minutesW = \frac{L}{\lambda} = \frac{2.57}{36} \approx 0.0714 \text{ hours} \approx 4.29 \text{ minutes}7. Average time spent in the queue (WqW_q):

The average time a customer spends in the queue is the average number of customers in the queue divided by the arrival rate:

Wq=Lqλ=1.8536≈0.0514 hours≈3.09 minutesW_q = \frac{L_q}{\lambda} = \frac{1.85}{36} \approx 0.0514 \text{ hours} \approx 3.09 \text{ minutes} Summary of Results:

1. P0P_0 = 0.28
2. P4P_4 = 0.0753
3. LL = 2.57 customers
4. LqL_q = 1.85 customers
5. WW = 0.0714 hours ≈ 4.29 minutes
6. WqW_q = 0.0514 hours ≈ 3.09 minutes
7. UU = 0.72 (server utilization)

These are the key performance metrics for the given M/M/1 queuing system.