7 During the workday the one candy machine stays busy and the boss believes that

Question

7 During the workday the one candy machine stays busy and the boss believes that people hang around in line just to visit. He has information that during the 10-hour work day 216 purchases are made. You know that the machine take 1.5 minutes on average to serve a customer. You decide to do a type one-model analysis so you can brief the boss on the real situation (Assume Poisson distribution). a. What is the average machine utilization? b. What is the average time in line? c. What is the average time in system? d. What is average number of people in line? e. What is average number of people in system? f. What should you tell the boss?

Explanation / Answer

We will address the problem as an M/M/1 Queue System.

Back-up Theory

An M/M/1 queue system is characterized by arrivals following Poisson pattern with average rate , [this is also the same as exponential arrival with average inter-arrival time = 1/ ] service time following Exponential Distribution with average service time of (1/µ) [this is also the same as Poisson service with average service rate = 1/µ] and single 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 n customers in the system is given by Pn = n(1 - ) ……(1)

The steady-state probability of no customers in the system is given by P0 = (1 - ) ……(2)

Average queue length = E(m) = (2)/{µ(µ - )} ………………………………………..(3)

Average number of customers in the system = E(n) = ()/(µ - )………………………..(4)

Average waiting time = E(w) = ()/{ µ(µ - )} ……………………………………………..(5)

Average time spent in the system = E(v) = {1/(µ - )}……………………………………..(6)

Percentage idle time of service channel = P0 = (1 - ) …………………………………….(7)

Machine utilization = 1 - P0 = …………………..…………………………………….(8)

Preparatory Calculations:

During 10 hour day 216 purchases are made => average number of purchases (arrivals) per hour = 21.6 = ………………………………………………………………………(9)

Machine takes on an average 1.5 minutes to serve a customer => µ = 60/1.5 = 40…….(10)

And hence = 21.6/40 = 0.54 …………………………………………………………… (11)

(a) average machine utilization

     = = 0.54 or 54% or 5.4 hours per day ANSWER [vide (8) and (11)]

(b) average time in line

     = E(w) = ()/{ µ(µ - )}

     = 21.6/(40 x 18.4) = 0.0293 hour = 1.76 minutes ANSWER [vide (5), (9) and (10)]

(c) average time in system

     = E(v) = 1/(µ - )

     = 1/18.4 = 0.0543 hour = 3.26 minutes ANSWER [vide (6), (9) and (10)]

(d) average number of people in line

     E(m) = (2)/{µ(µ - )}

     = (21.6)2/(40 x 18.4) = 0.63 ANSWER [vide (3), (9) and (10)]

(e) average number of people in system

     E(n) = ()/(µ - )

     = (21.6)/(18.4) = 1.17 ANSWER [vide (4), (9) and (10)]

(f) Report to boss

Boss’s contention that the machine is busy is not quite correct. Machine utilization is only 54% i.e., almost half the time the machine is idle.

However, his belief that people are only hanging around to visit is well founded because both average waiting time and time spent in the system are pretty low. ANSWER

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