Transcript Introduction to Assembly Lines Active Learning – Module

Approximate Three-Stage Model:
Active Learning – Module 3
Dr. Cesar Malave
Texas A & M University
Background Material
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Any Manufacturing systems book has a
chapter that covers the introduction about the
transfer lines and general serial systems.
Suggested Books:
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Chapter 3(Section 3.4) of Modeling and Analysis of
Manufacturing Systems, by Ronald G.Askin and Charles
R.Stanridge, John Wiley & Sons, 1993.
Chapter 3 of Manufacturing Systems Engineering, by
Stanley B.Gershwin, Prentice Hall, 1994.
Lecture Objectives
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At the end of the lecture, each student should
be able to
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Evaluate the effectiveness (availability) of a
three-stage transfer line given the
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Buffer capacities
Failure rates for the work stations
Repair rates for the work stations
Time Management
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Introduction - 5 minutes
Readiness Assessment Test (RAT) - 5 minutes
Lecture on Three Stage Model - 15 minutes
Team Exercise - 15 minutes
Homework Discussion - 5 minutes
Conclusion - 5 minutes
Total Lecture Time - 50 minutes
Approximate Three-Stage Model

Introduction
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Markov chains can be used to model transfer
lines with any number of stages
The number of states to be considered increases
with the number of stages, say M stages with
intermediate buffers of capacity Z require
2M(Z+1)M-1 states
Readiness Assessment Test (RAT)

Consider a three-stage line with two buffers
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Assume that a maximum of one station is down at
a time.
Determine the probability for station i to be down
Pi 
xi
M
1   xm
m 1
where xi = αi / bi
Three-Stage Model (Contd..)
Deeper analysis into the model:

Consider a line without buffers
 For every unit produced, station i is down for xi cycles
 xi is the ratio of average repair time to uptime
 From stations i = 1,…,M, all the other stations are
operational except station i
 Considering the pseudo workstation 0 with cycle failure
and repair rates α0 and β0 ,we have
M
E 0   Pi  1
i 1
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Model Analysis:
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1
1
#
Workstation #
2
2
3
#
Buffer #
Let us consider the station 2 and the three types of
states it can produce
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Production is there when all stations are up
Production is there when station 1 is down, but station 2
operates because of storage utilization from the buffer 1
Production is there when station 3 is down, but station 2
operates because of storage utilization from the buffer 2
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Model Analysis (contd..):

Let us define hij(Z1,Z2) as the proportion of time
station j operates when i is under repair for the
specified buffer limits
EZ1Z2 = E00 + P1h12(Z1,Z2) + P3h32(Z1,Z2)
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- Eq 1
Effectiveness of the line can be calculated by
converting the three stage model into a two-stage
model with the help of a pseudo work station
Case 1: From buffer 1, there are two possibilities:
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Line is down when station 2 is down with failure rateα2
when station 3 is down and buffer 2 is full - with a
failure rate {α3[1 – h32(Z1,Z2)]}
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Model Analysis (contd..):
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Stations 2 and 3 along with the connecting buffer are
replaced by a pseudo station 2’ with a failure rate
α2’ = α2 +α3[1 – h32(Z1Z2)]
- Eq 2
~ α2 +α3[1 – h32(Z2)] as h32() will not
depend on Z1
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Hence for a two-stage line, effectiveness can be
written as
EZ = E0 + P1h12(Z)
- Eq 3
= E0 + P2h21(Z) where
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Pi is the probability that station i is down as
referred before
h12(Z) is nothing but P1h12(Z1Z2)
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Model Analysis (contd..):
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Had we known h32(Z1Z2), we could have solved the
two pseudo station line using the equations defined for
estimating the effectiveness of two-staged lines with
buffers and calculated the effectiveness of the threestage line by substituting the values obtained in Eq 1.
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The question is do we know the value of h32(Z1Z2) ?
The answer is no !
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Case 2: From buffer 2, there are two possibilities –
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Line is down when station 2 is down with a failure rate α2
when station 1 is down and buffer 1 is empty - with a failure
rate {α1[1 – h12(Z1,Z2)]}
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Model Analysis (contd..):
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Stations 1, 2 and the connecting buffer can be
replaced by a pseudo workstation 1’with a failure rate
α1’ = α2 +α1[1 – h12(Z1Z2)]
- Eq 4
~ α2 +α1[1 – h12(Z1)] as h12() will not
depend on Z2
Station 3 will have a failure rate α3
The two-stage pseudo line can be solved by estimating
h32(Z1Z2) from h21(Z) of Eq 3
Solving for Case 1, i.e. estimating hij() factor is
involved as an input for Case 2 and vice versa. Thus
by utilizing these two cases, the effectiveness of the
three-stage model can be found.
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Solution Procedure:
1. Initialize h12(Z1,Z2) at, say, 0.5. Denote stages 1and 2 in any
pseudo two-stage approximation as 1’ and 2’, respectively.
Calculate E00 , the effectiveness for the unbuffered line.
2. Solve the two-stage line with α1’ given by - Eq 4. Estimate
h32(Z1,Z2) = h2’1’(Z) from Eq 3. α2’ = α3
3. Solve the two-stage line with α2’ given by - Eq 2. Estimate
h12(Z1,Z2) = h1’2’(Z) from Eq 3. α1’ = α1
If suitable convergence criteria is satisfied, go to step 4,
otherwise go to step 2.
4. Finally, effectiveness for a three-stage line is estimated by
EZ1Z2 = E00 + P1h12(Z1,Z2) + P3h32(Z1,Z2)
Team Exercise
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A 20-stage transfer line with two buffers is being
considered. Tentative plans place buffers of size 15 after
workstations 10 and 15. The first 10 workstations have a
cumulative failure rate of α = 0.005. Workstations 11
through 15 have a cumulative failure rate of α = 0.01 and
workstations 16 through 20 together yield an α = 0.005.
Repair of any station would average 10 cycles in length.
Estimate the effectiveness of this line design.
Solution
Step 1:
1
E 00 
1 b
1

3

1
 0.83333
1  10(0.005 0.01 0.005)
i
i 1
Set h12(15,15) = 0.5
Step 2: Combine stations 1 and 2
α1’ = α2 +α1[1 – h12(15,15)] = 0.01 + 0.005[.5] = 0.0125
α2’ = α3 = 0.005. Hence, we find that x1’ = 0.125, x2’ = 0.05,
s = x2’/ x1’ = 0.4. Using Buzacott’s expression with s ≠ 1,we find
C = 0.951898 and E15 = 0.8751. Now, using Eq 3 with
P2 = x2’/ (1+x1’+x2’), we find h32(15,15) ≈ (E15 – E0)/P2 = 0.564
Solution (contd..)
Step 3: Combine stations 2 and 3
α2’ = α2 +α3[1 – h32(15,15)] = 0.01 + 0.005[.436] = 0.01218
α1’ = α1 = 0.005. Hence, we find that x1’ = 0.05, x2’ = 0.1218,
s = x2’/ x1’ = 2.436. Using Buzacott’s expression with s ≠ 1,we find
C = 1.04916 and E15 = 0.87740. Now, using the result E15 = E0 + P1h12 ,
estimate h12(15,15). Now P1 = x1’/ (1+x1’+x2’) = 0.04267,
we find h12(15,15) ≈ (E15 – E0)/P1 = 0.563
As our new estimate of 0.563 differs from our initial guess of 0.5, we
return to step 2. As we continue the process, we find that
h12(15,15) = h12(15,15) = 0.563
Step 4: Estimate 3-stage effectiveness
E15 15 = E00 + P1h12(15,15) + P3h32(15,15)
≈ 0.08333 + [0.05/(1+0.05+0.1+0.05)]*0.563
+ [0.05/(1+0.05+0.1+0.05)]*0.563
= 0.88
Homework
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Consider a three-stage transfer line with
buffers between each pair of stages. Stage I
has a failure rate αi and repair rate bi. The
maximum buffer sizes are Z1 and Z2 ,
respectively. Assume geometric failure and
repair rates and ample repair workers.
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How many states are there for the system?
Consider state (RWWz10) where 0<z1< Z1. Write
the balance equation for this state.
Conclusion
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Markov chain models can be used to
determine the increase in output for a single
buffer.
Accurate output determination for a general
line with many buffers is a difficult problem.