a) Customers arrive at ABC bank at an average rate of 50 per hour, following Poi
ID: 3216127 • Letter: A
Question
a) Customers arrive at ABC bank at an average rate of 50 per hour, following Poisson distribution. There are 6 tellers and each of them can serve 10 customers in an average per hour. Interpret your answers. i. What are the average number of customers in the line and in the system? ii. What are the average time for customer to wait in the line and in the system? iii. Calculate the utilization rate & the ideal rate iv. What is the probability that more than 5 customers in the system? quantitative analsis
Explanation / Answer
Solution
This is a M/M/C Queue System application.
Back-up Theory
An M/M/C queue system is characterized by arrivals following Poisson pattern with average rate ?, service time following Exponential Distribution with average service rate of µ and multiple (> 1) service channel.
Let n = number of customers in the system and m = number of customers in the queue.
[Trivially, n = m + number of customers under service.]
Let (?/µ) = ?
The steady-state probability of no customers in the system is given by
P0 = 1/(S1 + S2), where S1 = ?[0,c - 1](?n/n!) and S2 = (?c)/[c!{1 - (?/c)}]………………(1)
The steady-state probability of n customers in the system is given by
Pn = P0 x ?n/n! for n < c ……………………………………………………………………(2a)
Pn = P0 x ?n/{c!(cn-c) for n ? c ..……………………………………………………………(2b)
Average queue length = E(m) = P0{(?µ)(?/µ)c}/{(c - 1)!(cµ - ?)2}………………………..(3)
Average number of customers in the system = E(n) = E(m) + (?/µ)………………………..(4)
Average waiting time = E(w) = E(m)/(?)…… ……………………………………………..(5)
Average time spent in the system = E(v) = E(w) + (1/µ)..…………………………………..(6)
Percentage idle time of service channel = P0 ………. …………………………………….(7)
Now, to work out the solution,
Given
?/hr =
50
µ/hr =
10
? =
5
c =
6
i). Average number of customers in the line = E(m) = 2.94 [vide (3)] ANSWER 1
Average number of customers in the system = E(n) = 7.94 [vide (4)] ANSWER 2
ii). Average time for customer to wait in the line = E(w)
= 0.0588 hr = 3.53 minutes [vide (5)] ANSWER 3
Average time for customer to wait in the system = E(v)
= 0.1588 hr = 9.53 minutes [vide (6)] ANSWER 3
iii). Utilization rate = 1 - P0 = 1 – 0.004512 [vide (1)]
= 0.9955 or 99.55% ANSWER 5
Idle rate = P0 = 0.004512 or 0.45% [vide (1)] ANSWER 6
iv). Probability that more than 5 customers in the system
= 1 – P(5 or less customers in the system)
= 1 – ?[n = 0, 5]Pn
= 1 – 0.412484 [Details of calculations given below.]
= 0.5875 ANSWER 7
n
Pn (n < c)
0
0.004512
1
0.022561
2
0.056402
3
0.094003
4
0.117503
5
0.117503
sum
0.412484
DONE
?/hr =
50
µ/hr =
10
? =
5
c =
6
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