Question

4. Customers at a Post Office line up to transact business with a single clerk at the Postal Agency. Customers arrive at the rate of four per minute, following a Poisson distribution. The single Postal clerk takes 12 seconds to service a customer, following an exponential distribution.
   1. What is the probability that there are more than two customers in the system?
   2. What is the probability that the system is empty?
   3. How long will the average customer have to wait before reaching the Postal clerk?
   4. What is the expected number of customers in the queue?
   5. What is the average number in the system?
   6. If a second Postal Clerk is added (who works at the same pace), how will the operating characteristics computed in (b), (c), (d), and (e) change?

Answer 1

Solution

Parts (a) to (d) are cases of M/M/1 queue system and Parts (e) is a case of M/M/2 queue system

Back-up Theory

An M/M/1 queue system is characterized by arrivals following Poisson pattern with average rate λ, [this is also the same as exponential arrival with average inter-arrival time = 1/λ] service time following Exponential Distribution with average service time of (1/µ) [this is also the same as Poisson service with average service rate = 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 (λ/µ) = ρ

The steady-state probability of n customers in the system is given by P_n = ρⁿ(1 - ρ) …………(1)

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

Average queue length = E(m) = (λ²)/{µ(µ - λ)} …………………………………………..........…..(3)

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

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

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

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

Probability of waiting = 1 - P₀ = ρ …………………..………………………………........………..(8)

Now to work out the solution,

Given,

customers arrive at the rate of four per minute, following a Poisson distribution, => λ = 4/minute ..... (9)

Postal clerk takes 12 seconds to service a customer, following an exponential distribution

=> µ = 5/minute [1 minute = 60 seconds and so 60/12] ……………………………………….. (10)

And hence, ρ = 4/5 = 0.8 …………………………………………………………….....…………. (11)

Part (a)

Probability that there are more than two customers in the system

= Σ(n = 3, 4, …..)P_n

= 1 – (P₀ + P₁ + P₂)

= 1 - P₀(1+ ρ¹+ ρ²) [vide (1) and (2)]

= 1 - (1 - ρ)(1+ ρ¹+ ρ²) [vide (2)]

= 1 – 0.2(1+ 0.8+ 0.64) [vide (11]

= 1 – 0.488

= 0.512   Answer 1

Part (b)

Probability that the system is empty

= P₀ [vide (8)]

= 1 – 0.8 [vide (8)]

= 0.2 Answer 2

Part (c)

Time the average customer have to wait before reaching the Postal clerk

= E(w)

= (λ)/{µ(µ - λ)} [vide (5)]

= 4/(5 x 1) [vide (9) and (10)]

= 0.67 minutes

= 40 seconds Answer 3

Part (d)

Expected number of customers in the queue

= E(m)

= (λ²)/{µ(µ - λ)} [vide (3)]

= 16/(5 x 1) [vide (9) and (10)]

= 3.2 Answer 4

Part (e)

Average number of customers in the system

= E(n)

= (λ)/(µ - λ) [vide (4)]

= 5/1 [vide (9) and (10)]

= 5 Answer 5

Part (f)

If a second Postal Clerk is added (who works at the same pace), it becomes M/M/2.

Back-up Theory

An M/M/2 queue system is characterized by arrivals following Poisson pattern with average rate λ, service time following Exponential Distribution with average service rate of µ and 2 service channels.

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 (λ/µ) = ρ

[Note: The formulae used here are the simplified forms of the general M/M/C formulae]

The steady-state probability of no customers in the system is given by

P₀ = (2 - ρ)/(2 + ρ)………………………………………………………….…………...………(1)

The steady-state probability of n customers in the system is given by

P_n = P₀ x ρⁿ/n! for n < 2 ………………………………………………………………………(2a)

P_n = P₀ x ρⁿ/{c!(c^n-c) for n ≥ 2 ..………………………………………………………………(2b)

In particular, P₁ = ρP₀;   P₂ = (P₀ρ²)/2;   P₃ = (P₀ρ³)/12;   P₄ = (P₀ρ⁴)/96………………(2c)

Average queue length = E(m) = (λ³)/{µ(2µ - λ)²}……………….…………………………..(3)

Average number of customers in the system = E(n) = E(m) + ρ …..……………………..(4)

Average waiting time = E(w) = E(m)/(λ)…… ……………………………………..………..(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, λ = 4/minute µ = 5/minute and hence, ρ = 4/5 = 0.8 ……………………………. (8)

P₀ = (2 - ρ)/(2 + ρ) [vide (1)]

= 1.2/2.8 [vide (8)]

= 0.4286 ………………………………………………………………..…………………. (9)

P₁ = 0.4286 x 0.8 [vide (2)]

= 0.34288 …………………………………………………………………………………. (10)

P₂ = 0.4286 x 0.8²/2 [vide (2)]

= 0.137152 …………………………………………………………………………………. (11)

Probability that there are more than two customers in the system

= Σ(n = 3, 4, …..)P_n

= 1 – (P₀ + P₁ + P₂)

= 1- 0.9086 [vide (9) to (11)

= 0.0914   Answer 6

Probability that the system is empty

= P₀ [vide (7)]

= 0.4286 [vide (9)] Answer 7

Expected number of customers in the queue

= E(m)

= (λ³)/{µ(2µ - λ)²} [vide (3)]

= 64/(5 x 36) [vide (8)]

= 0.3556 Answer 8

Time the average customer have to wait before reaching the Postal clerk

= E(w)

= E(m)/(λ) [vide (5)]

= 0.3556/4 [vide Answer 8 and (8)]

= 0.0889 minutes

= 5.33 seconds Answer 9

Average number of customers in the system

= E(n)

= E(m) + ρ [vide (4)]

= 0.3556 + 0.8 [vide Answer 8 and (8)]

= 1.1556 Answer 10

DONE