Jobs arrive to a shop at a mean rate of three every one-half hour, according to a poisson distribution. Each job requires a number of machining operations. The total time of all operations is approximated by an exponential distribution with a mean of 7.5 minutes. After the machining operation, the job proceeds to the inspection area for quality check and packaging for distribution or scrap, in case of defects. The time to do quality check and packaging/scrap is exponential with a mean of 2 minutes. Assume there is sufficient room for parts to wait for both machining and quality inspection/packaging. 1. Prepare a flow process chart of the scenario (job shop) above using the appropriate flowcharting tools (icons). 2. Make any assumption necessary to answer the following questions related to the described job shop above. Given the steady state and that a job has just arrived: a. What is the probability of the next job arriving in less than 5 minutes? b. What is the probability of two jobs arriving in the next 5 minutes? c. What is the average number of jobs waiting for machining operations? d. What is the probability of the machining center being busy? e. What is the probability that both the machining center and the quality check area are idle? f. What is the expected waiting time of a job for the machining operation?

2.

The given waiting line model is a single-server multiple phase system.

a.

Probability of the next job arriving in-between two time periods is as follows:-

P(a, b) = e^{–}^{a} – e^{–}^{b}.

Arrival rate, = 3 / 0.5 = 6 jobs per hour.

**Probability of the next job arriving in less than 5 minutes (i.e., 1 / 12 hours) = Arrival between 0 ^{th} minute and 5^{th} minute = P(0, 1 / 12) = e^{-6×0} – e^{-6x(1/12)} = 1 – e^{-0.5} = 0.39 **

b.

Probability of exactly ‘n’ jobs arriving in a period of ‘t’ hours is as follows:-

P(n) = [e^{–}^{t} . (t)^{n}] / n!

Arrival rate, = 3 / 0.5 = 6 jobs per hour.

**Probability of two jobs arriving in the next 5 minutes = P(2)** = [**e ^{-6 x (1/12)} . (6 x (1 / 12))^{2}] / 2! = 0.08**

c.

**Machining:**

Arrival rate, = 3 / 0.5 = 6 jobs per hour.

Service rate, = 60 / 7.5 = 8 jobs per hour.

Average number of jobs waiting in line for service = L_{q} = ^{2} / [ ( – )] = 6^{2}/ [8x (8 -6)] = 2.25 jobs.

Average number of jobs being processed = r = / = 6 / 8 = 0.75

**Average number of jobs waiting in the system = L _{s} = L_{q} + r = 2.25 + 0.75 = 3**

d.

Probability of zero job in the machining center = 1 – ( / ) = 1 – 0.75 = 0.25

**Probability of machining center being busy = 1 – 0.25 = 0.75**

e.

Probability of machining center being idle = 1 – ( / ) = 1 – 0.75 = 0.25.

Probability of quality check area being idle = 1 – ( / ) = 1 – (6 / 30) = 0.8

The operations in the two centers are done independently of the other.

**Probability of both machining center and quality check area being idle = 0.25 x 0.8 = 0.2**

f.

**Machining:**

Average time jobs are waiting in line = W_{q} = L_{q} / = 2.25 / 6 = 0.375 hours.

Average time jobs are waiting in the system = W_{s} = W_{q} + (1 / ) =( L_{q} / ) + (1 / ) = 0.375 + ( 1 / 8) = 0.5 hours = 30 minutes.

3 jobs are waiting in the system for 30 minutes.

**Expected waiting time for a single job in the system = 30 / 3 = 10 minutes.**

1. The Flow Process Chart of the given system is as follows:

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