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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: - PowerPoint PPT Presentation
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Two-Stage Paced LinesActive 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 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
Readiness Assessment Test (RAT)
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
WW WW: probability = (1 - α1)(1 - α2) WW WR: probability = (1 - α1)α2
WW RW: probability = α1(1 - α2) Z, does not change
Station 1 is being repaired while station 2 is working
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
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
Buzacott’s closed-form expression for the effectiveness for a buffer of
maximum size z :
where
and xi = αi / bi (ratio of average repair time to uptime), s = x2 / x1, r = a2 / a1
Z
x
Z
xz RWxPWWxPE
10
)()(
1
)1()1(1)[21(
)1()1(1
1)1(1
1
222
21
21
sxZbxbrx
xZbxbr
ssCxx
sC
EZ
Z
Z
)())((
)())((
2121122121
2121212121
bbbbb
bbbbbC
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.
SolutionGiven: 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,
where x = 0.04/0.25 = 0.16 and s = 1, r = 1
Hence, the number of parts/hr = E10 * Cycles/hr = 16.40
82.0)1()]1(1)[21(
)1()1(12
22
2110
xZbxbrx
xZbxbrE
System Reduction
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: Combination Rule Median Buffer Location Reversibility
Combination Rule A set of stations without any intermittent buffers can be replaced with
a single station
and bj = b, provided all stations must stop if
any individual station stops
Single Line
α1+α2 = α3+α4+α5
Feeder Line (unbuffered)
m
i
ij '
α1 α2 α5α4α3
α2 α3α1 α5α4
α 1+ α 2+ α 3+ α 6α 4+ α 5
α6
Combination Rule(Contd..)
Feeder Line (buffered)
=
α2+α3+α4
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
α4α3α2 α1
α1
1
11min
bi
Mi
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
Reversibility If the direction of flow is reversed in a serial line, production
rate remains the same.
M
rii
r
ii
M
rii
r
ii
2
1
111
,max,max
M
rii
r
ii
M
rii
r
ii ,max,max
1
111
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?
SolutionGiven: 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,
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
764.0)1(1
1
21
5
Z
Z
sCxx
sCE
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.