Question

1. Clients arrive at a law office with a single lawyer at a rate of 2.5 per hour.

       The lawyer takes an average of 10 minutes with each client. Assuming that

       arrivals are Poisson distributed and the legal service time is exponentially

       distributed, answer the following questions:

1. Compute that answers for this problem using the first six queuing equations.
2. Service goals for this law office dictate that an arriving client should not have to wait more than an average of 5 minutes to see the lawyer. Is this goal being met? Explain. If the goal is not being met, what do you recommend?
3. If the lawyer can reduce the average time spent per customer to 8 minutes, what is the mean service rate? Will the service goal be met?

Answer 1

Solution

The solution is based of the theory of M/M/1 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.]

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

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

Average waiting time = E(w) = (λ)/{µ(µ - λ)} = (1/µ)ρ/(1 – ρ) = E(n)/µ ………………..................................……..(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 λ = 2.5 per hour, µ = 6 per hour [10 minutes per client => 6 clients per 60 minutes.], ρ = 2.5/6 = 5/12 .... (9)

Q1

Vide (3), average queue length = 6.25/(6 x 3.5) = 0.3 client Answer 1

Vide (4), Average number of customers in the system = 2.5/3.5 = 0.7 client Answer 2

Vide (5), Average waiting time = 2.5/(6 x 3.5) = 0.12 hour = 7.2 minutes Answer 3

Vide (6), Average time spent in the system = 1/3.5 = 0.29 hour = 17 minutes Answer 4

Vide (7), Percentage idle time of service channel = 100(1 – 5/12) = 700/12 = 58.33% Answer 5

Vide (8), Probability of waiting = 5/12 = 0.42 Answer 6

Q2

Vide Answer 3, service goal of maximum waiting time of 5 minutes is NOT met. Answer 7

Recommendation:

Service time needs to be reduced, may be by appointing an assisitant to the lawyer or improving service mechanism.  Answer 8

Q3

When service time per client is reduced to 8 minutes, service rate = 60/8 = 7.5 clients per hour. Answer 9

With µ = 7.5, Vide (5), Average waiting time = 2.5/(7.5 x 5) = 1/15 hour = 4 minutes Answer 10

Clearly, service goal of maximum waiting time of 5 minutes is met. Answer 11

DONE