### Introduction to Assembly Lines Active Learning – Module

```Two-Stage Paced Lines
Active Learning – Module 2
Dr. Cesar Malave
Texas A & M University
Background Material


Any Manufacturing systems book has a
chapter that covers the introduction about the
transfer lines and general serial systems.
Suggested Books:


Chapter 3(Section 3.3) of Modeling and Analysis of
Manufacturing Systems, by Ronald G.Askin and Charles
R.Stanridge, John Wiley & Sons, Inc, 1993.
Chapter 3 of Manufacturing Systems Engineering, by
Stanley B.Gershwin, Prentice Hall, 1994.
Lecture Objectives

At the end of the lecture, every student should be
able to

Evaluate the effectiveness (availability) of a transfer line
given




Buffer capacity
Failure rates for the work stations
Repair rates for the work stations
Determine the optimal location of the buffer in any N
stage transfer line.
Time Management

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
Readiness Assessment Test (RAT) - 5 minutes
Lecture on Paced Lines with buffers - 10 minutes
Spot Exercise - 5 minutes
Paced Lines with buffers (contd..) – 10 minutes
Team Exercise - 10 minutes
Homework Discussion - 5 minutes
Conclusion - 5 minutes
Total Lecture Time - 50 minutes

Here are the possible reasons for a station to
be down. Analyze each of these reasons and
the importance of buffers in every case.




Station Failure
Total Line Failure
Station Blocked
Station Starved
RAT Analysis




Station Failure: Fractured tool, quality out-of-control signal, missing/defective
part program, or jammed mechanism. Although this station is down, other
stations can operate if they are fed product by the buffer, and have space for
sending completed product.
Total Line Failure: All the stations are inoperative. A power outage or error in
the central line controller would cause a total line failure.
Station Blocked: On completion of a cycle, if station i is not able to pass the
part to station i+1, station i is blocked. Failure of the handling system, failure of
a downstream station prior to the next buffer, or failure of a downstream station
with the intermediate buffer between these stations currently being full. Suppose
station i+1 is down, and its input buffer is filled, then station i must remain idle
while it waits for downstream space for the just completed part.
Station Starved: Station i is starved if an upstream failure has halted the flow of
parts into station i. Even if operational, the starved station will sit idle.
Two-Stage Paced Lines with Buffers




Two serial stages are separated by an inventory
buffer.
Buffer reduces the dependence between the stations
– blocking/starving effects are reduced.
Buffer state should be taken into account while
calculating line effectiveness.
Markow Chain Model is used.
Serial Stages
Buffer
Assumptions and Conventions









Markow Chain Environment (s1 ,s2 ,z)
si is the status for station i and z is the number of items
in the buffer
W – station in working condition (operational)
R – station in repair
Failures and Repairs occur at the end of a cycle
When a cycle starts, if both stations are working, station 2
receives its next part from station 1.
If station 1 is down, station 2 grabs a part from the buffer
If buffer is empty, station 2 is starved
If station 2 is down and station 1 is up, part from station 1 is
sent to the buffer
If the buffer is full, station 1 becomes blocked

Chapman – Kolmogorov result : steady-state
balance equations



S be the set of states of the system
P(s) be the probability that the system is in state s
p(u,v) is the transition probability from state u
(beginning) to state v (ending)
P(s1) =  P(s)*p(s,s1)
The steady-state balance equations can be
obtained by applying the above equation.
Transitions for Two-stage line with buffer
At time t, both stations are working, the transitions include





Station 1 is being repaired while station 2 is working

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WW  WW: probability = (1 - α1)(1 - α2)
WW  WR: probability = (1 - α1)α2
WW  RW: probability = α1(1 - α2)
Z, does not change
RW  WW:

if Z = 0, then probability = b1

if Z = x > 0, then probability = b1(1 - α2), Z = x - 1
RW  RW:

if Z = 0, then probability = (1 - b1)

if Z = x > 0, then probability = (1 - b1)(1 - α2), Z = x - 1
RW  RR:

if Z = x > 0, probability = (1 - b1)α2, Z = x – 1
Example: P(WW0) = (1 - α1 - α2) P(WW0) + b1 P(RW0) + b1 P(RW0)

Steady-state equation obtained by the Chapman-Kolmogorov Result

Station 1 is working, while station 2 is repaired

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WR  WW:

if Z = 0, then probability = b2

if Z > x = 0, then probability = (1 – α1) b2, Z = x + 1
WR  WR:

if Z = 0, then probability = 1 – b2

if Z > x = 0, then probability = (1 – α1)(1 - b2 ), Z = x + 1
WR  RR:

if Z > x = 0, probability = α1 (1 - b2), Z = x + 1
Both stations are being repaired



RR  WR:

if Z = x = 0, probability = b1 (1 - b2), Z = x + 1
RR  RW:

if Z = x = 0, probability = (1 – b1) b2, Z = x + 1
RR  RR:

if Z = x = 0, probability = (1 – b1) (1 - b2), Z = x + 1

System Effectiveness for a buffer of maximum size z can be calculated as
Z
Z
x 0
x 1
Ez   P(WWx )   P( RWx )

Buzacott’s closed-form expression for the effectiveness for a buffer of
maximum size z :
where

1  sC Z


1  x1  (1  x2 ) sC Z
EZ  
1  r  b1 (1  x)  Zb2 (1  x)

2

(
1

2
x
)[
1

r

b
(
1

x
)

Zb
(
1

x
)
2
2

C
s 1
s 1
(1   2 )(b1  b2 )  1b2 (1   2  b1  b2 )
(1   2 )(b1  b2 )   2b1 (1   2  b1  b2 )
and xi = αi / bi (ratio of average repair time to uptime),
s = x2 / x1, r = a2 / a1
Spot Exercise

A paced assembly line has a cycle time of 3 minutes.
The line has eight workstations and a buffer of
capacity 10 is placed between the fourth and fifth
workstations. Each station has a 1 percent chance of
breaking down in any cycle. Repairs average 12
minutes. Estimate the number of good parts made
per hour.
Solution
Given:
Cycle time C = 3 min (paced), m = 8, αi = 0.01, b = 0.25
Thus, Cycles/hr = (60 min/hr)/(3 min/cycle) = 20 cycles/hr
Now, α1 = 0.4, Z =10 and α2 = 0.4
Effectiveness can be found from Buzacott’s closed form
expression,
E10 
1  r  b1 (1  x)  Zb2 (1  x)
 0.82
2
(1  2 x)[1  r  b2 (1  x)]  Zb2 (1  x)
where x = 0.04/0.25 = 0.16 and s = 1, r = 1
Hence, the number of parts/hr = E10 * Cycles/hr = 16.40
System Reduction

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Aggregation of a set of stations that must be jointly
active or idle and which have a common repair rate,
into a single station
Holds effective for all serial stations as well as stations
that act as feeder stations for the main line
Aggregated failure rate is obtained by summing the
individual failure rates
System Reduction Rules:

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Combination Rule
Median Buffer Location
Reversibility

Combination Rule

A set of stations without any intermittent buffers can be replaced with
a single station
m
 ' j  i and bj = b, provided all stations must stop if
i 
α1
any individual station stops
α2
α3
α4
α5
Single Line
α1+α2
α1
α2
α3+α4+α5
=
α3
α4
α6
Feeder Line (unbuffered)
α 1+ α 2+ α 3+ α 6
α 4+ α 5
α5

Combination Rule(Contd..)
Feeder Line (buffered)
α2
α3
α4
α1
=
α2+α3+α4
α1

Median Buffer Location

If only one buffer is to be inserted, it should be placed in the middle
of the line. The upper bound on availability is given by
  
min1  i 
1i  M
b

1

Median Buffer Location(Contd..)

A median location is the one for which, if r is the last station
before buffer and the two conditions below are satisfied
M
M
r

 r 1

max i ,  i   max i ,  i 
 i 1 i r 1 
 i 1 i r  2 
M
M
r

 r 1

max i ,  i   max i ,  i 
i r
 i 1 i r 1 
 i 1


Reversibility

If the direction of flow is reversed in a serial line, production
rate remains the same.
Team Exercise

Consider a paced assembly system with four
workstations. Mean cycles to failure are estimated to
be 100, 200, 100 and 50 cycles, respectively. Repair
times should average 8 cycles.
 Find line availability assuming no buffers.
 A buffer of size 5 would be profitable if
availability increased by at least 0.04. The buffer
could be located after station 2 or 3. Which
location is the best? Should the buffer be
included?
Solution
Given:
α1 = 0.01, α2 = 0.005, α3 = 0.01, α4 = 0.02, b = 0.125 for all i
(a) Thus, E0 = [1 + b-1Σ αi ]-1 = [1 + 8(0.045)]-1 = 0.735
(b) According to the median location rule, the buffer should be
located after station 3.
Now, α1 = 0.025 (0.01+0.005+0.01), Z = 5 and α2 = 0.02
Effectiveness can be found from Buzacott’s closed form
expression,
1  sC Z
E5 
 0.764
Z
1  x1  (1  x2 )sC
where s = 0.8, x1 = 0.2, x2 = 0.16 and C = 0.9153
Now, E5 – E0 = 0.029 < 0.04, and do not include the buffer
Homework

A 10 stage transfer line is being considered with failure rates
i = 0.004 ( i =1,2,..,10) and bj = 0.2 ( i =1,2,..,10). Estimate
the following
 Effectiveness of the line without buffers
 Effectiveness of the line with a buffer size of 5 after the
station 5
 Effectiveness of the line with a buffer size of 5 after the
station 3
 Optimal location of the buffer
Conclusion



Buffers allow for partial independence between the
stages in the line, thereby improving line
effectiveness against station failures
The size of the buffer may have an extreme influence
on a flow production system. The importance of
buffers increases with the amount of variability
inherent in the system.
Although buffer involvement leads to considerable
investment and factory space, it is crucial to find the
optimal number and distribution of the buffers within
a flow production system.
```