Transcript Introduction to Assembly Lines Active Learning – Module

Introduction to Transfer Lines
Active Learning – Module 1
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(Sections 3.1 and 3.2) 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
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At the end of the lecture, every student should be
able to
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Understand and analyze about serial production systems
that are subject to machine failures and random
processing times
Evaluate the effectiveness (availability) of a transfer line
given
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Failure rates for the work stations
Repair rates for the work stations
Time Management
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Readiness Assessment Test (RAT) - 5 minutes
Introduction - 5 minutes
Lecture on Paced Lines without buffers - 20 minutes
Spot Exercise - 5 minutes
Team Exercise - 5 minutes
Homework Discussion - 5 minutes
Conclusion - 5 minutes
Total Lecture Time - 50 minutes
Readiness Assessment Test (RAT)
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Solve the following problem on probability:
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Six red discs, numbered from 1 to 6, and 4 green discs,
numbered from 7 to 10, are placed in box A. Ten blue discs,
numbered from 1 to 10, are placed in box B. Two discs are
drawn from Box A and two discs are drawn from Box B. The
four discs are drawn at random and without replacement.
Find the probability that the discs drawn are:
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one red disc, one green disc and two blue discs with all
four discs odd numbered.
Solution
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Probability: The total number of occurrences of a selection/event
"a", divided by the total number of occurrences (a) plus the number of
failures of those occurrences "b" (i.e.. total possible outcomes)
Hence, probability is given by P = a /(a+b)
The problem can be solved the following way:
Probability = P (red-odd)*P (green-odd)*P (blue-odd)*P (blue-odd) +
P (green-odd)*P (red-odd)*P (blue-odd)*P (blue-odd)
= ( 3/10 * 2/9 * 5/10 * 4/9 ) + ( 2/10 * 3/9 * 5/10 * 4/9 )
= 4/135
Introduction
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Transfer Line: A set of serial, automatic
machine and/or inspection stations linked by a
common material transfer and control system.
Buffers: Means for protecting workstations
from failures elsewhere in the line, thereby
improving station utilization.
Introduction (Contd..)
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1
1
#
Workstation #
2
2
3
3
#
Buffer #
Sample transfer line with intermediate buffers
Reasons for a station to be non-functional
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Station failure
Total line failure
Station blocked
Station starved
Introduction (Contd..)
Classification of Failures
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Time Dependent Failures
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Occur with a
chronological frequency
Time between failures is
measured in units like
hours
E.g.: Daily Maintenance
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Operation
Dependent Failures
Occur while the
system is running
Time between
failures is
measured in cycles
E.g.: Tool Wear
Introduction (contd..)
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Effectiveness (Availability) of a line can be defined as:
E (uptime)
E=
E (uptime + downtime)
E (uptime): Productive cycle time ( expected value )
E (uptime + downtime): Total cycle time ( expected )
Paced Lines without Buffers
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Operation Dependent Failures: Assumptions
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Mean time to failure(MTTF) follows a geometric
distribution with failure rate i and the density
t 1
function is given by fi (t ) = i (1 - i)
Mean cycles to failure(MCTF) is 1/ i
The number of cycles for repair at station i is
geometric with mean 1/bi cycles.
Assume bi = b, i = 1,2,…M
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Assumptions (contd..)
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All uptime and downtime random variables are
independent.
Idle stations do not fail.
Failures occur at the end of the cycle, they do not
destroy the product.
A maximum of one station can fail on any cycle.
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Deeper insight into the assumptions:
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Geometric distribution(GD) application in transfer lines.
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Every workstation will have two states – failure and working
Probability that a station has failed =
and working = 1 Probability that the station fails at the 1st cycle is
Probability that the station fails at the 2nd cycle = (1 - )
Probability that the station fails at the 3rd cycle = (1 - )2
Probability that the station fails at the t th cycle = (1 - )t-1
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GD assumption allows the use of discrete time and discrete state
Markow chain model to solve the above.
At most one station failure assumption allows us to ignore the
2nd order terms.
Idle station failure distinguishes between the two types of
failures.
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Model Analysis:
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Probability that the transfer line (M stations) first
fails at the end of the tth cycle :
t 1
M

 
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P(T = t ) =  (1   i ) 1   (1   i )
 i =1
  i =1

M
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Let us define
 M
 M
b = 1   (1  ai )   ai
 i =1
 i =1
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Hence the above probability equation becomes
P(T = t) = b(1 - b)t-1
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M station line behaves like a single station but with
failure parameter b replacing i ..
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Model Analysis (contd..):
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By expanding the terms in b and noting that i
being small, higher order terms approach zero and
hence
M
b   i
i =1
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Effectiveness of a line can be calculated as the
ratio of the expected productive cycles between
failures(uptime) divide by expected total cycles
between failures(uptime + downtime).
b 1
1
E = 1 1 =
b +b
1 + bb 1
where b-1 is the mean number of cycles for repair to
a station.
Spot Exercise
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Find the line availability for a three-station line in
the respective cases 1 and 2: Also analyze the
variation in the results obtained.
 The first, second and third stations average a
failure every 10, 20 and 30 cycles respectively.
Average repair time is 2 cycles.
 The first, second and third stations average a
failure every 5, 10 and 15 seconds respectively.
Average repair time is 2 seconds.
Solution
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Case 1:
1 = 1/ 10, 2 = 1/ 20,3 = 1/ 30 and b = ½
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1
E=
= 0.775
1 + (1/10+ 1/20 + 1 / 30) * 2
Case 2:
E = (1 + 2 / 10) 1 * (1 + 2 / 20) 1 * (1 + 2 / 30) 1
= 0.7102
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Analysis: Effectiveness has been reduced from 0.75 to
0.71 because stations 1 and 2 continue the aging process
while idle as a result of a failure at the other station
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Time Dependent Failures: Assumptions
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Time to failure for station i is exponential with rate
parameter i
Repair times are exponential with parameter bi
For an operating time of t,
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Expected number of failures for a station i : it
Expected repair time
: it/bi
Effectiveness of a station i is given by
:
Ei = lim
E ( uptime)
t 
E (uptime + downtime)
=
As stations are independent,
E = E1*E2*E3*…
t
t + it / bi
=
1
1 + i / bi
Team Exercise
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Repairmen can fix a failed workstation in 10
minutes on an average. Cycle time is 20
seconds. Stations seem to operate 250 cycles
on average before failing. Estimate the daily
(8-hour shift) production for a six-station line.
What is the average daily workload of the
operator. ( problem 3.16 – Askin & Stanridge)
Solution
Given:
Average Repair time = 10 min, Cycle time C = 20sec
αi = 1/250, b-1 = 30
(a) Estimate the output for 8 hours = 8*60*3 = 1440 cycles
E = [1 + b-1Σ αi ]-1 = [1 + 30(0.024)]-1 = 0.581
Units/shift = 0.581 * 1440 = 837
(b) Average daily workload of the repairmen
The % of time repairmen are busy = (1 – E) * 100
= 41.9 % of shift
= 3.35 hrs
Homework
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A 15 stage transfer line is being considered. Stations
1 through 10 should have the same reliability. Each
station is expected to operate approximately 1000
cycles between failures. The last 5 stations are
expected to break down about once every 600
cycles. Repair times will vary but should average
about the equivalent of 12 cycles in duration.
Estimate the availability of the line.
Conclusion
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Transfer Lines : Serial production systems that are
very costly and subjected to random failures
Failures are of two types – Time Dependent and
Operation Dependent
They can affect the entire line or just a single station
Reliability models help in estimating system
effectiveness when buffers are not used