Curry G.L., Feldman R.M. Manufacturing Systems Modeling and Analysis

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4.2 Random Breakdowns and Repairs During Processing 113

4.2 Random Breakdowns and Repairs During Processing

A major source of processing time variability is due to the breakdown of an operat-

ing machine and the subsequent delay while the machine is being repaired. Several

courses of action might result from the breakdown of a machine. The job undergoing

processing at the time of the breakdown might not be recoverable (i.e., lost), the job

might require additional processing before resumption of “normal” processing, or

the job might not be effected by the breakdown and normal processing can resume

immediately after the repair is complete (as if the breakdown never occurred). Only

the latter case is considered herein, although for the second case, the additional pro-

cessing time needed to resume service can be included in the machine repair time

so that the second and third situations become equivalent.

The assumption being made is that once a machine has been repaired after a

breakdown, the job that was processing at the time of the breakdown is continued

as if the breakdown never occurred. Thus, a breakdown merely extends the job pro-

cessing time (actually job residence time with the ”normal” processing time being

unaffected). When breakdowns occur, they obviously impact the job residence time

on the machine and the resulting job residence time distribution needs to be devel-

oped. This is called the effective processing time to distinguish it from the normal

processing time.

Deﬁnition 4.1. The effective processing time, T

, refers to the time that a job ﬁrst

has control of the processor until the time at which the job releases the processor so

that it is available to begin work on another job.

Only the mean and variance parameters of the effective processing time random

variable are needed and not the distribution itself. For this development, a given

job has several possibilities. The job can complete processing without a breakdown

interruption, the machine could breakdown once during service, the machine could

breakdown twice during service, etc. So the effective processing time is also a ran-

dom variable given by

= T +

∑

i=1

, (4.2)

where T is the normal (uninterrupted) processing time random variable, the R

’s

are the (i.i.d.) repair time random variables, and N is the random number of fail-

ures during the service time T . The number of failures N is a function of the time

between failures random variables, F

, for the machine in question and is assumed

independent of the actual time that it takes to do the repairs, R

A key parameter needed for expressing the effect of failures and repairs on ser-

vice times is the availability of the processor.

Deﬁnition 4.2. The availability, a, of a processor that is subject to failures is the

long-run average fraction of time that the processor is available for processing jobs.

114 4 Processing Time Variability

Using the notation above, it is not difﬁcult to express availability in terms of the

mean time to failure and the mean repair time.

Property 4.1. Processor availability is determined by

a =

E[F

]

E[ F

]+E[R

]

where F

,··· and R

,··· are i.i.d. random variables representing suc-

cessive failure times and successive repair times, respectively, for the proces-

sor.

Hopp and Spearman [2] developed an expression for the mean and variance of the

effective service time for processors that are less than 100% reliable under the as-

sumption that failures are exponentially distributed:

E[ T

E[T

]

, and (4.3)

= C

]=C

(1 +C

])a(1 −a)E[R

]

E[ T

]

. (4.4)

They show that when T

and C

are used in place of T

and C

in (4.1) the formula

gives an exact expression for the mean waiting time in the queue for a workstation

described by an M /G/1 system subject to exponential failures. (Notice that when

replaces T

, the utilization factor must be adjusted as well.) For other G/G/c

systems, it serves as an approximation.

Example 4.2. Consider a single workstation with jobs arriving according to a Pois-

son process (i.e., exponential inter-arrival times) with an average time between ar-

rivals of 75 minutes. Initially we ignore the fact that the machine at the workstation

is not 100% reliable and observe that the normal processing time is described by an

Erlang type-3 distribution with mean of 58 minutes; thus, C

= 1, E[T

]=58 min,

= 1/3, and u = 58/75 = 0.7733. These parameters used in (4.1) yield CT

= 132

min.

After presenting these results, we are told that the processing machine is not com-

pletely reliable. The time between machine breakdowns is exponentially distributed

with a mean time of 3 hours measured according to machine processing time and

does not include idle time. The repair time is distributed according to a lognormal

distribution with a mean time of 30 min and a standard deviation of 15 min yielding

a squared coefﬁcient of variation of 0.25 for the repair time. The availability is thus

given by

a =

E[ F

]

E[ F

]+E[R

]

3 + 1/2

= 0.85714 .

4.3 Operator-Machine Interactions 115

The mean of the effective processing time ( 4.3)is

E[ T

E[T ]

= 67.67 min ,

and the squared coefﬁcient of variation for the effective processing time (4.4)is

(1 + 0.25)(0.85714)(1 −0.85714)(30)

= 0.4125 .

The effective mean service time yields an effective utilization of u = 67.67/75 =

0.9023 so that the application of (4.1) results in a revised value for the expected

waiting time in the queue

(1 + 0.4125)



0.9023

1 −0.9023



67.67 min = 441 min .

Notice that the inclusion of machine failure in the model results in over a three-

fold increase in the mean waiting time; thus, to ignore failures can create signiﬁcant

errors in performance measures. This increase is due to two factors: (1) machine

failures cause an increase the effective utilization factor and (2) machine failures

cause an increase in the service variability. As the utilization factor approaches one,

small changes in the factor will have major changes in waiting times, and in this

case, the majority of the increase in waiting times is due to the utilization factor

increase; only about 5%–6% of the increase is due to the increase i n service vari-

ability. 

• Suggestion: Do Problems 4.4–4.9 and 4.13–4.14.

4.3 Operator-Machine Interactions

Operators are frequently required to setup a machine for each job. Machine prepa-

ration time usually takes signiﬁcantly more time than the job unloading operation.

If the machining operation requires a dedicated operator, then most likely the model

of that situation would require only one resource (either the machine or the oper-

ator). When an operator is used only part time during processing and the operator

is then free to perform other tasks, operators and machines can no longer be mod-

eled as one. In this situation, an operator is frequently assigned control of more than

one machine and, thus, is responsible for setting up jobs on several machines. If an

operator is assigned to cover too many machines then system performance can be

signiﬁcantly degraded because of delays resulting from waiting for the operator to

become available to perform the necessary job setups. Even when the operator is

assigned to cover only two machines, some delays will be encountered due to the

Section 4.3 can be omitted without affecting the continuity of the remainder of the text.

116 4 Processing Time Variability

timing of job completions. If a system has reasonable capacity, then the operator ma-

chine interaction problem does not signiﬁcantly impact system performance. Thus,

this level of detail is frequently omitted in system models. This interaction can, how-

ever, degrade system performance signiﬁcantly if overlooked. The operator-machine

interaction problem also offers an opportunity to illustrate how multiple resource in-

teractions can be quantiﬁed and evaluated.

In the modeling assumptions, only one job class is treated with two identical

machines and one operator. In addition, to simplify the analysis as much as possible,

exponentially distributed times are assumed for job inter-arrival times, job setup

times, and job processing times. Since we need to keep track of two resources,

namely the operator and the machines, a state space that only keeps a record of the

number of jobs in the s ystem does not carry enough information to appropriately

establish the true system state. Speciﬁcally, in addition to the number of jobs in the

system, the status of each machine-job combination must be known; that is, the state

of the system must include whether the job is “in setup” or “in processing”. If two

jobs are in the “in setup” status, then only one of them can be actually proceeding

with setup because there there is only one operator.

There is often more than one way to deﬁne a state space, so that the particular

deﬁnition chosen is up to the modeler. It is good practice to choose a state space

deﬁnition that is descriptive so that the individual deﬁning equations for the steady-

state probabilities will be easy to read. One descriptive state deﬁnition is to use a

three-tuple for the s tates. Each state is represented as (n , i, j), where n denotes the

number of jobs in the system and i and j indicate the status of the two machines.

There are three possible values for i and j: 0 indicates a machine has no job as-

sociated with it, s indicates that a machine has a job “in setup”, and p indicates a

machine has a job “in process”. For example, the state (1,s,0) indicates that there

is one job in the system and the operator is setting it up on a machine, state ( 5,s,s)

indicates that there are 5 jobs in the system with one job being set-up on a machine,

another job waiting at a machine for the operator, and 3 jobs waiting in the queue for

a machine, and state (7, p, p) indicates 7 jobs in the system with both machines busy

processing, 5 jobs queued, and the operator idle. Because the machines are identi-

cal, it is not necessary to know which machine is processing and which machine is

begin setup.

The state space representation for n ≥ 2 is made up of three individual states:

(n,s,s), (n, s, p), and (n, p, p).Forn = 0, there is no need for all three indices, but

for consistency this state is denoted as (0,0,0).Forn = 1, the possible states are

(1,s,0) and (1, p,0). The states of the system, grouped by number of jobs in the

system, are

{(0,0,0), (1,s,0),(1, p,0), (2,s,s), (2,s, p), (2, p, p), (3,s,s), (3,s, p), (3, p, p), ···}.

The inter-arrival time, setup time, and service time distributions are all assumed

to be exponentially distributed. The mean rates for these three processes are denoted

, and

, respectively. Note that if both machines are processing (indepen-

dently), the mean output rate for the system is 2

. If both machines are being setup,

4.3 Operator-Machine Interactions 117

the mean setup rate is

, not 2

, because there is only one operator. The equations

relating the steady-state probabilities for this system are:

(0,0,0)

(1,p,0)

(

(1,s,0)

(0,0,0)

(2,s,p)

(

(1,p,0)

(1,s,0)

+ 2

(2,p,p)

(

+ 2

(2,p,p)

(2,s,p)

(

(2,s,s)

(1,s,0)

(3,s,p)

(

(2,s,p)

(1,p,0)

(2,s,s)

+ 2

(3,p,p)

(

+ 2

(3,p,p)

(2,p,p)

(3,s,p)

(

(3,s,s)

(2,s,s)

(4,s,p)

(4.5)

(

(3,s,p)

(2,s,p)

(3,s,s)

+ 2

(4,p,p)

(4.6)

(

+ 2

(4,p,p)

(3,p,p)

(4,s,p)

(4.7)

(

(n,s,s)

(n−1,s,s)

(n+1,s,p)

(

(n,s,p)

(n−1,s,p)

(n,s,s)

+ 2

(n+1,p,p)

(

+ 2

(n+1,p,p)

(n,p,p)

(n+1,s,p)

plus the norming equation, which i s the sum of all probabilities equal to one. The

set of three numbered equations (4.5–4.7) are repeated with increasing indices. The

last three listed equations represent these equations for the index n (where n ≥ 3).

So the s ystem has an inﬁnite number of deﬁning equations with the ﬁrst seven being

special and all others being one of three possible general forms.

The numerical solution scheme employed is rather straightforward, but unfortu-

nately, the solution cannot be represented nicely in closed form. For speciﬁed values

of the parameter set (

), the system is solved in the following fashion. The un-

known p

(0,0,0)

is set 1.0, then all other probabilities can be solved recursively for

numerical values according to the procedure described in the next paragraph. Since

we have an inﬁnite system, it will be truncated to a ﬁnite set of probabilities at the

point that the probabilities become very small. Thus, we continue to ﬁnd proba-

bilities until the individual probability terms become very small. At that point, we

stop and determine the sum of all probabilities that have been calculated. The ﬁnal

answer then becomes the individual terms divided by this sum.

The process for evaluating the individual probabilities is actually rather straight-

forward. First given p

(0,0,0)

the ﬁrst equation is used to obtain p

(1,p,0)

. Then, us-

ing three equations at a time, the probabilities groups solved in turn are: the

group (p

(1,s,0),

(2,s,p)

, p

(2,p,p)

), then the group (p

(2,s,s),

(3,s,p)

, p

(3,p,p)

) and ﬁnally

(3,s,s),

(4,s,p)

, p

(4,p,p)

). Each of these sets is found from the solution of three linear

equations. This last set of three equations is repeated solved for (p

(n−1,s,s),

(n,s,p)

118 4 Processing Time Variability

(n,p,p)

) for increasing values of n until the sum of the three values are less that

some limit. The solution of each of these three distinct sets of equations needs only

values for previously obtained probabilities. Thus, starting with one assumed value,

(0,0,0)

, as many probabilities as necessary can be obtained; after the relative values

for these probabilities have been determined, they are normed to sum to one. One

further observation is that only one 3 ×3 matrix inverse is needed since the same

matrix reoccurs as the coefﬁcients of the unknowns for all three forms of the three

equations groups.

To be more speciﬁc, we ﬁrst observe that p

(1,p,0)

(0,0,0)

. Then the sec-

ond through fourth equations can be rewritten in matrix form as

⎡

⎣

−(

)

−(

+ 2

)

⎤

⎦

⎡

⎣

(1,s,0)

(2,s,p)

(2,p,p)

⎤

⎦

⎡

⎣

−

(0,0,0)

(

(1,p,0)

⎤

⎦

with its solution given by

⎡

⎣

(1,s,0)

(2,s,p)

(2,p,p)

⎤

⎦

⎡

⎣

−(

)

−(

+ 2

)

⎤

⎦

−1

⎡

⎣

−

(0,0,0)

(

(1,p,0)

⎤

⎦

. (4.8)

Once the values of the probabilities (p

(1,s,0)

, p

(2,s,p)

, p

(2,p,p)

) have been obtained, the

vector (p

(2,s,s)

, p

(3,s,p)

, p

(3,p,p)

) is solved similarly using the ﬁfth through seventh

equations in the system. This solution is written as

⎡

⎣

(2,s,s)

(3,s,p)

(3,p,p)

⎤

⎦

⎡

⎣

−(

)

−(

+ 2

)

⎤

⎦

−1

⎡

⎣

−

(1,s,0)

−

(1,p,0)

(2,s,p)

−

(2,p,p)

⎤

⎦

. (4.9)

Equations (4.5–4.7) can now be used to yield the general form of the solution for

n ≥ 3; namely,

⎡

⎣

(n,s,s)

(n+1,s,p)

(n+1,p,p)

⎤

⎦

⎡

⎣

−(

)

−(

+ 2

)

⎤

⎦

−1

⎡

⎣

−

(n−1,s,s)

−

(n−1,s,p)

(n,s,p)

−

(n,p,p)

⎤

⎦

. (4.10)

Notice that the solution to each system always involves the same inverse which

greatly simpliﬁes the computational burden of the process.

Not all values for the three parameters will yield a system that can be solved. If

the operator sets up too slowly or if the arrival rates are too fast for the processing

times, the queues will build up continually and no steady-state is possible. Although

developing the steady-state conditions is outside the scope of this text, they are

given in [1] and, for completeness, we state them below. Steady-state probabilities

will exist if and only if the three parameter values are such that

)

μγ

+ 2

μγ

4.3 Operator-Machine Interactions 119

Example 4.3. To illustrate the methodology and computations, consider a two-

machine system with one server. Let the mean arrival rate of jobs be 1 per hour,

the mean time to perform a setup by 15 minutes, and let the mean processing time

be 90 minutes. Recall that all the times are exponentially distributed. Thus,

= 1,

= 4, and

= 2/3. The matrix that needs to inverted, and its inverse, are

⎡

⎣

−(

)

−(

+ 2

)

⎤

⎦

−1

⎡

⎣

−0.1622 0.0473 0.0270

0.2838 0.3547 0.2027

0.4865 0.6081 −0.0811

⎤

⎦

Now setting p

(0,0,0)

to 1.0 yields p

(1,p,0)

= 1.5. Using (4.8), the ﬁrst set of three

probabilities are

(1,s,0),

(2,s,p)

, p

(2,p,p)

)=(0.2804,0.6030,1.0338) .

From these values, (4.9) is used to evaluate the next three probabilities

(2,s,s)

, p

(3,s,p),

(3,p,p)

)=(0.1082,0.3910,1.1133) .

The probabilities (p

(3,s,s)

, p

(4,s,p),

(4,p,p)

) are obtained based on these previous val-

ues using (4.10 ) to yield

(3,s,s)

, p

(4,s,p),

(4,p,p)

)=(0.0637,0.3156,1.0182) .

Repeating the use of (4.10), we obtain

(4,s,s)

, p

(5,s,p),

(5,p,p)

)=(0.0489, 0.2713,0.9014) ,

(5,s,s)

, p

(6,s,p),

(6,p,p)

)=(0.0413, 0.2367,0.7921) ,

(14,s,s)

, p

(15,s,p),

(15,p,p)

)=(0.0110, 0.0635,0.2129) .

Stopping at this point, these probabilities sum to 15.288. Dividing all of these

probabilities by 15.288 yields an approximate solution to this system. It is obvious

that since the probability p

(15,p,p)

is not very close to zero, that this truncated so-

lution will not be very close to the unlimited system solution. In fact using these

probability values, the estimate for the mean number of jobs, N

, in the system is

WIP= E[N

]=5.606 .

As the number of probabilities obtained is increased, the expected system WIP,

converges. These iteration results are displayed below where n denotes the number

of probabilities obtained. Note that it is not much work to increase the number of

probabilities obtained since they are found iteratively three at a time using (4.10)

repeatedly:

120 4 Processing Time Variability

n = 20, WIP = 6.399,

n = 30, WIP = 7.263,

n = 40, WIP = 7.603,

n = 50, WIP = 7.725,

n = 60, WIP = 7.658,

n = 70, WIP = 7.779,

n = 80, WIP = 7.783,

n = 90, WIP = 7.785,

n = 100, WIP= 7.785.

The truncated system solution changes very little as more probabilities are added

beyond the ﬁrst 80 probabilities. Thus, a reasonable solution to the unlimited system

has been obtained. The expected cycle time in the system from Little’s Law is

CT = WIP/

= 7.785 hr .

The expected number of jobs in the operator system is

1 ×



(1,s,0)

∞

∑

n=2

(n,s,p)



+ 2 ×

∞

∑

n=2

(n,s,s)

= 0.2819 ,

with the probability that the operator is idle being

(0,0,0)

+ p

(1,p,0)

∞

∑

n=2

(n,p,p)

= 0.75 ,

and the machine utilization factor being



(1,p,0)

∞

∑

n=2

(n,s,p)



+ 1 ×

∞

∑

n=2

(n,p,p)

= 0.8909 .

The approach of using a truncated system to approximate the unlimited capacity

system leads to the problem of ﬁnding the norming constant by iteratively increas-

ing the number allowed in the system until the total non-normed probability sum

stabilizes. Using the rate-generator form of the problem for Markov processes, one

can develop a closed form representation of this sum and ﬁnd the non-normed prob-

abilities total sum with the truncation mechanism (see [1]). This approach, however,

requires mechanics that will not be developed in this text. 

• Suggestion: Do Problems 4.10–4.12.

Problems 121

Problems

4.1. Consider a processing time, T , with measured parameters E[T ]=6 and C

[T ]=

2, that has four i.i.d. sub-tasks, T

for i = 1, 2,3,4.

(a) Determine E[T

] and C

] for the sub-tasks.

(b) Assume that the variability of each sub-task can be reduced (identically) so that

]=2. Determine the squared coefﬁcient of variation of the total processing

time and the percentage improvement over the “old” processing time variability.

4.2. Consider a processing time, T , with measured parameters E[T ]=8 and C

[T ]=

3, that has ﬁve i.i.d. sub-tasks, T

for i = 1, 2,3,4, 5.

(a) Determine E[T

] and C

] for the sub-tasks.

(b) Assume that the variability of each sub-task can be reduced (identically) so that

]=2. Determine the squared coefﬁcient of variation of the total processing

time and the percentage improvement over the “old” processing time variability.

4.3. Consider a processing time that has three independent sub-tasks. These are:

the job setup time S, normal processing time P, and job removal time R from the

machine. The distributional parameters for these sub-tasks are:

E[ S]=10 min, C

[S]=3 ,

E[ P]=1hr, C

[P]=

E[ R]=5min, C

[R]=1 .

(a) Determine the mean and squared coefﬁcient of variation for the job residence

time ( total processing time).

(b) After careful study the engineering department has come up with a jig for per-

forming a sizeable proportion of the job setup time off line (while the machine is

busy processing another job). The result is that the “on-line” machine setup time is

reduced, with resulting parameters E[S]=1 min and C

[S]=1. In addition, due to

an operator suggestion, the job removal time variability was reduced to C

[R]=1/3.

Note that no improvement was made in the actual machine processing time. Deter-

mine the mean of the new total processing time (job residence time) and its squared

coefﬁcient of variation. What are the percentage improvements over the “old” job

residence time parameters?

4.4. Consider a job with processing time distribution parameters E[T ]=3 hours and

[T ]=2. The machine breakdown and repair time characteristic parameters are:

E[F]=7 hr and C

[F]=1 ,(C

[F] is required to be 1)

E[ R]=1 hr and C

[R]=1 .

Find the parameters of the effective processing time: E[T

], V [T

], and C

122 4 Processing Time Variability

4.5. Consider a job with processing time distribution parameters E[T ]=3.5 hours

and C

[T ]=1.25. The machine breakdown and repair time characteristic parameters

are:

E[F]=9 hr and C

[F]=1 ,(C

[F] is required to be 1)

E[ R]=2 hr and C

[R]=1.5 .

Find the parameters of the effective processing time: E[T

] and C

4.6. Compute the percentage increase in the cycle time for a system without ma-

chine breakdowns and the s ame system with breakdowns and repairs. Job arrivals

are according to a Poisson process with a mean rate of 4 per hour. The service time

distribution parameters are E[S]=0.2 hours and C

= 1. The mean time between

breakdowns is 2 hours and the repair time distribution parameters are E[R]=1/3

hour and C

= 2.

4.7. Consider an M/M/1/3 system with a server that has exponential time between

breakdowns and exponential repair times. Develop the rate-node diagram that con-

nects the states of this system. Given an arrival rate of 5 jobs per hour, a service rate

of 4 jobs per hour, a breakdown rate of once per hour, and a mean repair time of

10 minutes, determine the steady-state probabilities for the system states and com-

pute the system performance measures of WIP

, CT

, and th

. In addition, compute

the proportion of the time that the machine is idle, down (i.e., under repair), and

processing.

4.8. Consider M/M/1/∞ system with a server that has exponential time between

breakdowns and exponential repair times. Develop the rate-node diagram that con-

nects the states of this system. Given an arrival rate of 5 jobs per hour, a service

rate of 7 jobs per hour, a breakdown rate of once per hour and a mean repair time

of 10 minutes, determine the s teady-state probabilities for the system states and

compute the system performance measures of WIP

, CT

, and th

. Compare these

performance results with those obtained by applying the breakdown adjustments of

Eqs. (4.3) and (4.4).

4.9. Consider M/M/1/∞ system with a server that has exponential time between

breakdowns and Erlang-2 repair times. Develop the rate-node diagram that connects

the states of this system. Given an arrival rate of 5 jobs per hour, a service rate

of 7 jobs per hour, a breakdown rate of once per hour, and a mean repair time

of 10 minutes, determine the s teady-state probabilities for the system states and

compute the system performance measures of WIP

, CT

, and th

. Compare these

performance results with those obtained by applying the breakdown adjustments of

Eqs. (4.3) and (4.4).

4.10. Consider a two-machine one-operator system. Let all times be exponentially

distributed with mean rates:

(

= 1,

= 3,

) .