Question

Question 2 Customers arrive at the checkout counter (shown in the figure below) at random from 1 to 8 minutes apart. Each possible value of inter-arrival time has the same probability of occurrence, as shown in Table 2.6. The service times vary from 1 to 6 minutes with the probabilities shown in Table 2.7. Departure Arrival Checkout Counter Table 26 Distribution of TIme Between Amivals Time baweerm Arrivals Table 27 Service-Time Distribution Minutesy) Prohablity Service Tme 0.125 0.125 0.125 125 0.125 0125 0.125 0125 Minutes)Probability 0.10 020 030 0.10 Assuming that the single channet queue and one checkout counter to attend the customers, simulate the arrival of the first 20 customers and calculate the following (a) Average inter-arrival time. (b) Average service time. (c) Average waiting time. (d) Average idle time. Use the three digit random numbers from 1-999 generated from excel for arrival time data and use the two digit random numbers from 1-99 generated from excel for the service time data. Use Excel spread sheet to find the answer

Answer 1

a) Average inter-arrival time = sum of product of arrival times and its probability = 0.125*(1+2+3+4+5+6+7+8)=0.125*36

=4.5 minutes

b) Average service time = sum of product of service times and its probability = (0.1*1)+(0.2*2)+(0.3*3)+(0.25*4)+(0.1*5)+(0.05*6)

=3.2 minutes

c)$\lambda$=1/Average inter-arrival time=1/4.5

$\mu$=1/ Average service time= 1/3.2

$\rho$=$\lambda$/$\mu$ = 3.2/4.5 = 0.711

Length of the queue=L_q = $\frac{\rho^{2}}{1 - \rho}$ =$\frac{0.5057}{1 - 0.711}$=1.75=2(approximately)

Average waiting time = $\frac{L_{q}}{\lambda}$ =2/(1/4.5) = 9 minutes (approximately)

d) Average Idle time proportion = $1-\rho$=1-0.711 = 0.289=28.9%

Therefore, 28.9% of the time the counter will be idle.