The 4M company has a work center with a single turret lathe. Jobs arrive at this

Question

The 4M company has a work center with a single turret lathe. Jobs arrive at this work center an exponential distribution with a mean of 0.25 days per job. a.) On average, how many jobs are waiting in the work center? b.) On average, how long will ajob stay in the center? bs per day. The lathe p c.) Since each drop 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 of waiting near the lathe, how big should be the storage space near the lathe? - Queuing Theory Problem - Please solve parts a.), b.), and c.) and show all the steps to get to the final solution(s.

Explanation / Answer

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 , [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 = 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)

Probability of waiting = 1 - P0 = .……..……………………………………………………………….(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 =>

= 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

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.