Question

Problem 1 REGIONAL AIRLINES Regional Airlines is establishing a new telephone system for handling flight reservations. During the 10:00 A.M. to 11:00 A.M. time period, calls to the reservation agent occur ran- domly at an average of one call every 3.75 minutes. Historical service time data show that a reservation agent spends an average of 3 minutes with each customer. The waiting line model assumptions of Poisson arrivals and exponential service times appear reasonable for the telephone reservation system. Regional Airlines' management believes that offering an efficient telephone reserva- tion system is an important part of establishing an image as a service-oriented airline. If the system is properly implemented, Regional Airlines will establish good customer relations, which in the long run will increase business. However, if the telephone reservation system is frequently overloaded and customers have difficulty contacting an agent, a negative customer reaction may lead to an eventual loss of business. The cost of a ticket reservation agent is $20 per hour. Thus, management wants to provide good service, but it does not want to incur the cost of overstaffing the telephone reservation operation by using mor agents than necessary

Answer 1

Solution

The solution is based on M/M/1 Queueing 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

Traffic Intensity = ρ = (λ/µ)……………………………..............................……………………………………..(A)

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 number of customers in the waiting line, L_q = (λ²)/{µ(µ - λ)} …………....……………………..…..(3)

Average number of customers in the system = L = (λ)/(µ - λ) = ρ/(1 – ρ)……….…….........................…..(4)

Average time a customer spends in the waiting line, W_q = (λ)/{µ(µ - λ)} = (1/µ)ρ/(1 – ρ…………………..(5)

Average time a customer spends in the system W = {1/(µ - λ)}………………….......……………………..(6)

Now, to work out the solution,

In the given problem, λ = 60/3.75 = 16 per hour [one arrival per 3.75 minutes => 60/3.75 per h............(7)

µ = 20 per hour [3 minutes per customer => 20 customers per hour]..................................................... (8)

(7), (8) and (A) => ρ = 0.8 ....................................................................................................................... (9)

Part (a)

Vide (2), and (9)

Probability that no customer are in the system, P₀ = 1 – 0.8

= 0.2 Answer 1

Part (b)

Vide (3),

Average number of customers in the waiting line, L_q = (16²)/{20 x 4)

= 3.2 calls Answer 2

Part (c)

Vide (4),

Average number of customers in the system, L = ρ/(1 – ρ)

= 0.8/0.2

= 4 calls Answer 3

Part (d)

Vide (5),

Average time a customer spends in the waiting line, W_q = (λ)/{µ(µ - λ)}

= 16/(20 x 4)

= 0.2 hour

= 12 minutes per hour Answer 4

Part (e)

Vide (6),

Average time a customer spends in the system, W = 1/(µ - λ)

= ¼ hour

= 15 minutes per hour Answer 5

Part (f)

If there is no customer in the system, an arriving customer does not have to wait and hence

Probability an arriving customer has to wait, P_W

= 1 – Probability that there is no customer in the system

= 1 - P₀

= 1 – 0.2 [Vide Answer 1]

= 0.8 Answer 6

DONE