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_{s}^{2} = 4

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

(a)

The average number of people witing,
** L_{q} =
(λ^{2}σ_{s}^{2} + ρ^{2})/ 2(1 −
ρ)** = (0.0025*4 + 0.25^2) / (2*(1 - 0.25)) =

(b)

The average time spent in waiting line
** W_{q} = L_{q} /
λ** = 0.0483 / 0.05 =

**----------------------**

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

SCV_{s} = Square of coefficient of variation for servcie
time = σ_{s}^{2} / t_{s}^{2} = 4 /
25 = 0.16

First compute *L*_{q} for *M/M/2* queue
using the following formulae:

Computing will give us P_{0} = 0.78 and
*L*_{q} = 0.004

So,

(a)

*L _{q, M/G/2}* =

(b)

*W _{q, M/G/2}* =

*please use excel and provide the formulas
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