Question

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.

Answer 1

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, σ_s² = 4

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

(a)

The average number of people witing, L_q = (λ²σ_s² + ρ²)/ 2(1 − ρ) = (0.0025*4 + 0.25^2) / (2*(1 - 0.25)) = 0.0483

(b)

The average time spent in waiting line W_q = L_q / λ = 0.0483 / 0.05 = 0.967 minutes

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

SCV_s = Square of coefficient of variation for servcie time = σ_s² / t_s² = 4 / 25 = 0.16

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

Computing will give us P₀ = 0.78 and L_q = 0.004

So,

(a)

L_{q, M/G/2} = L_{q, M/M/2} * (1 + SCV_s) / 2 = 0.004*(1+0.16)/2 = 0.0023

(b)

W_{q, M/G/2} = L_{q, M/G/2} / λ = 0.0023 / 0.05 = 0.046 minutes