Question

The 4M company has a work center with a single turret lathe. Jobs arrive at this work center according to a Poisson process at a mean rate of 2 jobs per day, The lathe processing time has an exponential distribution with a mean of 0.25 day per job.

a) On average, how many jobs are waiting.in the work center?

b) On average, how long will a job stay in the center?

c).Since each job takes a big space, the waiting jobs are currently waiting in the warehouse. The production manager is proposing to add a storage space near the lathe. If an arriving job will have at least 90% chance waiting near the lathe, how big should be the storage space near the lathe?

Answer 1

SOLUTION

This is a direct application 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.]

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

Jobs arrive at the work center according to Poisson process at mean rate of 2 jobs per day =>

λ = 2/day ……............................................................................................................................…. (9)

And lathe processing time has exponential distribution with mean of 0.25 days per job =>

(1/µ) = 0.25 or µ = 4/day ..........................................................................................................…. (10)   

Part (a)

Average number of jobs waiting at the center = E(m) = 4/(4 x 2) = 0.5 [vide (3) above]

So, Average number of jobs waiting at the center = 0.5 ANSWER

Part (b)

Average stay time of jobs at the center

= waiting time + service time

= E(v) = ½ [vide (3) above]

= ½ day ANSWER [Note: Unit of time is day.]

Part (c)

Since on an average 2 jobs arrive per day and given 90% chance of waiting, (2 x 0.9) = 1.8 arriving jobs are expected to be waiting. So, size of the space near the lathe must be sufficient to accommodate 2 jobs. ANSWER