Question

# Suppose that customers arrive to a ATM in a POisson arrival with 3 person/hour. Assume that...

Suppose that customers arrive to a ATM in a POisson arrival with 3 person/hour. Assume that it takes on average 5 minutes with variance = 4 for the ATM to process each transaction. Answer the following questions

a. On average how many people are waiting in the line?

b. On average how much time will people need to wait in line?

Suppose there are two identical ATM machines. Answer (a) and (b) for Q2.

This is an M/G/1 queuing situation with the following parameters:

Average arrival rate, λ = 3 per hour = 0.05 per minute
Average service rate, µ = 1 in 5 minutes = 0.2 per minute
Variance of service time, σs2 = 4

Utilization, ρ = λ / µ = 0.05/0.2 = 0.25

(a)

The average number of people witing, Lq = (λ2σs2 + ρ2)/ 2(1 − ρ) = (0.0025*4 + 0.25^2) / (2*(1 - 0.25)) = 0.0483

(b)

The average time spent in waiting line Wq = Lq / λ = 0.0483 / 0.05 = 0.967 minutes

----------------------

For the second part, the queuing system is M/G/2

SCVs = Square of coefficient of variation for servcie time = σs2 / ts2 = 4 / 25 = 0.16

First compute Lq for M/M/2 queue using the following formulae:

Computing will give us P0 = 0.78 and Lq = 0.004

So,

(a)

Lq, M/G/2 = Lq, M/M/2 * (1 + SCVs) / 2 = 0.004*(1+0.16)/2 = 0.0023

(b)

Wq, M/G/2 = Lq, M/G/2 / λ = 0.0023 / 0.05 = 0.046 minutes

#### Earn Coins

Coins can be redeemed for fabulous gifts.