Transcript Analysis and Design of Asynchronous Transfer Lines as a

Analysis and Design of
Asynchronous Transfer Lines as
a series of G/G/m queues
Topics
• The negative impact of variability in the operation
of Asynchronous Transfer Lines
• Modeling the Asynchronous Transfer Line as a
series of G/G/m queues
• Modeling the impact of various operational
detractors
• Employing the derived models in line diagnosis
• Employing the derived models in line design
Asynchronous Transfer Lines (ATL)
W1
TH
B1
M1
W2
TH B2
M2
W3
TH B3
M3
TH
Some important issues:
• What is the maximum throughput that is sustainable through this
line?
• What is the expected cycle time through the line?
• What is the expected WIP at the different stations of the line?
• What is the expected utilization of the different machines?
• How does the adopted batch size affect the performance of the
line?
• How do different detractors, like machine breakdowns, setups,
and maintenance, affect the performance of the line?
Analyzing a single workstation with
deterministic inter-arrival and
processing times
Case I: ta = tp = 1.0
B1
M1
TH
WIP
1
t
1
Arrival
2
3
4
Departure
5
TH = 1 part / time unit
Expected CT = tp
Analyzing a single workstation with
deterministic inter-arrival and
processing times
Case II: tp = 1.0; ta = 1.5 > tp
B1
WIP
M1
TH
Starvation!
1
t
1
Arrival
2
3
4
Departure
5
TH = 2/3 part / time unit
Expected CT = tp
Analyzing a single workstation with
deterministic inter-arrival and
processing times
Case III: tp = 1.0; ta = 0.5
B1
M1
TH
WIP
3
Congestion!
2
1
t
1
Arrival
2
3
4
Departure
5
TH = 1 part / time unit
Expected CT  
A single workstation with
variable inter-arrival times
Case I: tp=1; taN(1,0.12) (ca=a / ta = 0.1)
B1
M1
TH
WIP
3
2
TH < 1 part / time unit
Expected CT  
1
t
1
Arrival
2
3
4
Departure
5
A single workstation with
variable inter-arrival times
Case II: tp=1; taN(1,1.02) (ca=a / ta = 1.0)
B1
M1
TH
WIP
3
2
TH < 1 part / time unit
Expected CT  
1
t
1
Arrival
2
3
4
Departure
5
A single workstation with
variable processing times
Case I: ta=1; tpN(1,1.02)
B1
M1
TH
WIP
3
2
TH < 1 part / time unit
Expected CT  
1
t
1
Arrival
2
3
4
Departure
5
Remarks
• Synchronization of job arrivals and completions
maximizes throughput and minimizes experienced cycle
times.
• Variability in job inter-arrival or processing times
causes starvation and congestion, which respectively
reduce the station throughput and increase the job cycle
times.
• In general, the higher the variability in the inter-arrival
and/or processing times, the more intense its disruptive
effects on the performance of the station.
• The coefficient of variation (CV) defines a natural
measure of the variability in a certain random variable.
The propagation of variability
W1
B1
W2
M1
B2
Case I: tp=1; taN(1,1.02)
Case II: ta=1; tpN(1,1.02)
WIP
WIP
3
3
2
2
1
1
1
W1 arrivals
2
3
4
5
W1 departures
M2 TH
t
1
W2 arrivals
2
3
4
5
t
Remarks
• The variability experienced at a certain station
propagates to the downstream part of the line due to the
fact that the arrivals at a downstream station are
determined by the departures of its neighboring upstream
station.
• The intensity of the propagated variability is modulated
by the utilization of the station under consideration.
• In general, a highly utilized station propagates the
variability experienced in the job processing times, but
attenuates the variability experienced in the job interarrival times.
• A station with very low utilization has the opposite
effects.
The G/G/1 model:
A single-station
B1
M1
TH
Modeling Assumptions:
• Part release rate = Target throughput rate = TH
• Infinite Buffering Capacity
• one server
• Server mean processing time = te
• St. deviation of processing time = e
• Coefficient of variation (CV) of processing time: ce = e / te
• Coefficient of variation of inter-arrival times = ca
An Important Stability Condition
B1
M1
TH
•Average workload brought to station per unit time:
TH·te
• It must hold:
TH  t e  1.0
• Otherwise, an infinite amount of WIP will pile up in
front of the station.

Performance measures for
a stable G/G/1 station
B1
M1
TH
• Server utilization: u  TH  t e
• Expected cycle time in the buffer:
c a2  c e2 u
CTq 
t e (Kingman’s approx.)
2 1 u
• Expectedcycle time in the station: CT  CTq  te
• Average WIP in the buffer: WIPq  TH  CTq (by Little’s law)
• Average WIP in the station:

WIP  TH  CT  WIPq  u
• Squared CV of the
inter-departure times:
c d2  (1 u2 )c a2  u2c e2
Remarks
• For a station with variable job inter-arrival and/or processing times,
utilization must be strictly less than one in order to attain stable
operation.
• Furthermore, expected cycle times and WIP grow to very large
values as u1.0.
• Expected cycle times and WIP can also grow large due to high
values of ca and/or ce; i.e., extensive variability in the job interarrival and/or processing times has a negative impact on the
performance of the line.
• In case that the job inter-arrival times are exponentially distributed,
ca=1.0, and the resulting expression for CTq is exact (a result
known as the Pollaczek-Kintchine formula).
• The expression for cd2 characterizes the propagation of the station
variability to the downstream part of the line, and it quantifies the
dependence of this propagation upon the station utilization.
Performance measures for
a stable G/G/m station
M1
TH
B
M2
TH
Mm
• Server utilization: u  (TH  t e ) m
c a2  c e2 u 2(m 1)1
te
• Expected cycle time in the buffer: CTq 
2
m(1 u)
• Expectedcycle time in the station: CT  CTq  te
• Average WIP in the buffer: WIPq  TH  CTq

• Average WIP in the station:

WIP  TH  CT  WIPq  m  u
2
u
2
2
2
2
c

1
(1
u
)(c
1)

(c
• Squared CV of the
inter-departure
times:
d
a
e 1)

m
Analyzing a multi-station ATL
TH
Key observations:
• A target production rate TH is achievable only if each station satisfies the stability
requirement u < 1.0.
• For a stable system, the average production rate of every station will be equal to TH.
• For every pair of stations, the inter-departure times of the first constitute the interarrival times of the second.
• Then, the entire line can be evaluated on a station by station basis, working from the
first station to the last, and using the equations for the basic G/G/m model.
Operational detractors:
A primal source for the line variability
• Effective processing time = time that the part occupies
the server
• Effective processing time = Actual processing time +
any additional non-processing time
• Actual processing time typically presents fairly low
variability ( SCV < 1.0).
• Non-processing time is due to detractors like machine
breakdowns, setups, operator unavailability, lack of
consumables, etc.
• Detractors are distinguished to preemptive and nonpreemptive. Each of these categories requires a different
analytical treatment.
Preemptive operational detractors
• Outages that take place while the part is
being processed.
• Some typical examples:
– machine breakdowns
– lack of consumables
– operator unavailability
Modeling the impact of
preemptive detractors
•
•
•
•
•
•
•
•
•
X = random variable modeling the natural processing time, following a general
distribution.
to = E[X]; o2=Var[X]; co=o / to .
N
T = random variable modeling the effective processing time = X 
U where
i1 i
Ui = random variable modeling the duration of the i-th outage, following a
general distribution, and
N = random variable modeling the number of outages during a the processing of
a single part.

mr=E[Ui]; r2=Var[Ui]; cr = r / mr
Time between outages is exponentially distributed with mean mf.
Availability A = mf / (mf+mr) = percentage of time the system is up.
Then,
te = E[T] = to / A or equivalently re = 1/te = A (1/to) = A ro

 e2  Var[T]  ( o2 A2 )  to ((mr2   r2 ) m f )
c e2   e2 /te2  c o2  (1 c r2 )A(1 A)(mr /t o )
Breakdown Example
•
•
•
•
•
•
Data: Injection molding machine has:
15 second stroke (to = 15 sec)
1 second standard deviation (so = 1 sec)
8 hour mean time to failure (mf = 28800 sec)
1 hour repair time (mr = 3600 sec)
Natural variability
co = 1/15 = 0.067 (which is very low)
Example Continued
• Effective variability:
mf
8
A

 0.888
m f  mr 8  1
te  to / A  15 / 0.888  16.875
mr
3600
2
c  c  2 A(1  A)
 (0.067)  2(0.888)(1  0.888)
 47.41
to
15
2
e
2
o
Which is very high!
Example Continued
• Suppose through a preventive maintenance
program, we can reduce mf to 8 min and mr to 1
min
mf
8
A

 0.888
m f  mr 8  1
te  to / A  15 / 0.888  16.875
(the same as before)
mr
60
2
c  c  2 A(1  A)
 (0.067)  2(0.888)(1  0.888)
 0.79
to
15
2
e
2
o
Which is low!
Non-preemptive operational
detractors
• Activities that may take place between the
processing of two consecutive parts.
• Some typical examples:
– setups
– preventive maintenance
– operator breaks
Modeling the impact of
non-preemptive detractors
•
•
•
•
•
•
•
•
•
X = random variable modeling the natural processing time, following a general
distribution.
to = E[X]; o2=Var[X]; co=o / to .
NS = average number of parts processed between two consecutive setups
It is also assumed that the number of parts between two consecutive setups
follows a geometric distribution, which when combined with the previous bullet, it
implies that probability for a setup after any given job = 1/ NS.
Z = random variable modeling the duration of a setup
tS = E[Z]; S2 = Var[Z]
S = random variable modeling the setup time experienced by any given job =
 Z with probability1/N s
 
- 1/Ns
0 with probability 1
T = random variable modeling the effective processing time = X+S
Then,
E[S] = tS / NS ; Var[S] = (S2 / NS) + tS2((NS-1) / NS2);
te = E[T] = to+tS / NS ;  e2  Var[T]   o2   S2 NS  tS2 (NS 1) NS2 ;

 

c e2   e2 /te2
Setup Example
•
Data:
– Fast, inflexible machine:
(2 hr setup every 10 jobs)
No difference!
– Slower, flexible machine:
(no setups)
 t0  1 hour
   0.25 hours
 0
 N s  10 jobs/setup

 t s  2 hours

 t  t  t s  1  2  1.2 hours
0
 e
Ns
10

 r  1  1  0.83 jobs/hour
 e t
1.2
e

 t0  1.2 hours
  0.6 hours
 0

 r  1  1  0.83 jobs/hour
e

t0 1.2

Setup Example (cont.)
•
Compare mean and variance
–Fast, inflexible machine – 2 hr setup every 10 jobs
 0  0.25 hours
c0 
0
t0
 0.25
cs  0.25
0.25 


2
2
2
2 Ns 1
2 10  1
σe   0 
 ts

0.25


2
 0.4475


2
2
Ns
Ns
10
10
2
2
s
ce2 
 e2
t
2
e

0.4475
1.2 
2
 0.3107638
–Slower, flexible machine – no setups
0.6 

2
2
ce  c0  2 
 0.25
2
t0
1.2 
Conclusion: Flexibility can reduce variability.
 02
•
2
Setup Example (cont.)
•
New Machine: Consider a third machine same as previous
machine with setups, but with shorter, more frequent
setups
N s  5 jobs/setup
ts  1 hour
re 
•
Analysis:
1
1

 0.83 jobs/hr
te 1.2
0.25 

Ns 1
2
2 5 1
σ  
t
  0.25  
1
 0.235
2
2
Ns
Ns
5
5
2
e
c 
2
e
•
2
0
 e2
t
2
e


2
s
2
2
s
0.235
1.2 
2
 0.163194
Conclusion: Shorter, more frequent setups induce less variability.
Example:employing the developed
theory for diagnostic purposes
M1
Ca2=1.0
to1 =19 min
co12=0.25
mf1=48 hrs
mr1=8 hrs
MTTR ~ expon.
B
20
parts
M2
to2 =22 min
co22=1.0
mf2=3.3 hrs
mr2=10 min
MTTR ~ expon.
Desired throughput is TH = 2.4 jobs / hr but practical experience has shown
that it is not attainable by this line. We need to understand why this is not
possible.
Diagnostics example continued:
Capacity analysis based on mean values
M1
Ca2=1.0
to1 =19 min
co12=0.25
mf1=48 hrs
mr1=8 hrs
MTTR ~ expon.
B
20
parts
M2
to2 =22 min
co22=1.0
mf2=3.3 hrs
mr2=10 min
MTTR ~ expon.
A1  m f 1 /(m f 1  m r1 )  48/(48  8)  0.857
t e1  t o1 / A1  19/0.857 22.17min
A2  m f 2 /(m f 2  m r2 )  3.3/(3.3  10/60)  0.952
t e 2  t o2 / A2  22/0.952 23.11min  Bottleneck machine
Bottleneck ut ilization
: u2  TH  t e 2  2.4  (23.11/60)  0.9244 1.0 (!)
Diagnostics example continued:
An analysis based on the G/G/m model
c e12  c o12  (1 c r12)A1(1 A1)mr1 /t o1  0.25 (1 1)  0.857 (1 0.857)  8  60/19  6.442
u1  TH  t e1  2.4  22.17/60  0.8868
c a12  c e12 u1
1 6.442 0.8868
CTq1 
t e1 
22.17  646.256min
2 1 u1
2 1 0.8868
WIPq1  TH  CTq1  2.4  646.256/60  25.85
2
c a22  c d1
 u12c e12  (1 u12 )c a12  0.88682  6.442 (1 0.88682 ) 1  5.28
2
c e2
 c2  (1 c r22 )A2 (1 A2 )mr2 /t o2  1 (1 1)  0.952 (1 0.952) 10/22  1.04
u2  TH  t e2  2.4  23.11/60  0.9244
2
c a2 2  c e2
u2
5.28 1.04 0.9244
CTq 2 
t e2 
23.11 892.94min
2 1 u2
2
1 0.9244
WIPq 2  TH  CTq 2  2.4  892.94/60  35.72 >> 20 (!)
i.e., the long outages of M1, combined with the inadequate capacity of the
interconnecting buffer, starve the bottleneck!
Example: ATL Design
• Need to design a new 4-station assembly line for circuit board
assembly.
• The technology options for the four stations are tabulated below
(each option defines the processing rate in pieces per hour, the CV
of the effective processing time, and the cost per equipment unit in
thousands of dollars).
•
Station Option 1
1 42, 2.0, 50
2 42, 2.0, 50
3 25, 1.0, 100
4 50, 0.75, 20
Option 2
Option 3
42, 1.0, 85 10, 2.0, 110.5
42, 1.0, 85
25, 0.7, 120
6, 0.75, 24
Example: ATL Design (cont.)
• Each station can employ only one technology option.
• The maximum production rate to be supported by the line
is 1000 panels / day.
• The desired average cycle time through the line is one day.
• One day is equivalent to an 8-hour shift.
• Workpieces will go through the line in totes of 50 panels
each, which will be released into the line at a constant rate
determined by the target production rate.
A baseline design:Meeting the desired
prod. rate with a low cost
1000 42,2.0,50
50 42,1.0,85
8 10,2.0,110.5
(1000 / 50) / 8
42,2.0,50
42,1.0,85
25,1.0,100 50,0.75,20
25, 0.7,120 6,0.75,24
2.5
Station 1
42
2
50
3
Station 2
42
2
50
3
Station 3
25
1
100
6
Station 4
50
0.75
20
3
te
tb=B*te
Cb^2=Ce^2/B
u=TH*tb/m
0.0238
1.1905
0.08
0.9921
0.0238
1.1905
0.08
0.9921
0.04
2
0.02
0.8333
0.02
1
0.0113
0.8333
Ca^2
Cd^2 = 1+(1-u^2)(Ca^2-1)+(u^2/sqrt(m))*(Cb^2-1)
0
0.4615
0.4615
0.4687
0.4687
0.5598
0.5598
0.4691
3.17
14.6
2.3
4.9
33.5
0.8
1.41
21.48
1
150
150
600
1/te
Ce
P
m
CT = [(Ca^2+Cb^2)/2]*[u^(sqrt(2*(m+1))-1)/(m*(1-u))]*tb+tb
CT1+CT2+CT3+CT4
WIPq
m*P
60
960
Reducing the line cycle time by
adding capacity to Station 2
1000 42,2.0,50
50 42,1.0,85
8 10,2.0,110.5
(1000 / 50) / 8
42,2.0,50
42,1.0,85
25,1.0,100 50,0.75,20
25, 0.7,120 6,0.75,24
2.5
Station 1
42
2
50
3
Station 2
42
2
50
4
Station 3
25
1
100
6
Station 4
50
0.75
20
3
te
tb=B*te
Cb^2=Ce^2/B
u=TH*tb/m
0.0238
1.1905
0.08
0.9921
0.0238
1.1905
0.08
0.7441
0.04
2
0.02
0.8333
0.02
1
0.0113
0.8333
Ca^2
Cd^2 = 1+(1-u^2)(Ca^2-1)+(u^2/sqrt(m))*(Cb^2-1)
0
0.4615
0.4615
0.505
0.505
0.5709
0.5709
0.4725
3.17
1.36
2.32
4.9
0.4
0.8
1.42
8.27
1.1
150
200
600
1/te
Ce
P
m
CT = [(Ca^2+Cb^2)/2]*[u^(sqrt(2*(m+1))-1)/(m*(1-u))]*tb+tb
CT1+CT2+CT3+CT4
WIPq
m*P
60
1010
Adding capacity at Station 1,
the new bottleneck
1000 42,2.0,50
50 42,1.0,85
8 10,2.0,110.5
(1000 / 50) / 8
42,2.0,50
42,1.0,85
25,1.0,100 50,0.75,20
25, 0.7,120 6,0.75,24
2.5
Station 1
42
2
50
4
Station 2
42
2
50
4
Station 3
25
1
100
6
Station 4
50
0.75
20
3
te
tb=B*te
Cb^2=Ce^2/B
u=TH*tb/m
0.0238
1.1905
0.08
0.7441
0.0238
1.1905
0.08
0.7441
0.04
2
0.02
0.8333
0.02
1
0.0113
0.8333
Ca^2
Cd^2 = 1+(1-u^2)(Ca^2-1)+(u^2/sqrt(m))*(Cb^2-1)
0
0.299
0.299
0.4324
0.4324
0.5487
0.5487
0.4657
1.22
1.31
2.27
0.1
0.3
0.7
1.4
6.2
1
200
200
600
1/te
Ce
P
m
CT = [(Ca^2+Cb^2)/2]*[u^(sqrt(2*(m+1))-1)/(m*(1-u))]*tb+tb
CT1+CT2+CT3+CT4
WIPq
m*P
60
1060
An alternative option:Employ less
variable machines at Station 1
1000 42,2.0,50
50 42,1.0,85
8 10,2.0,110.5
(1000 / 50) / 8
42,2.0,50
42,1.0,85
25,1.0,100 50,0.75,20
25, 0.7,120 6,0.75,24
2.5
Station 1
42
1
85
3
Station 2
42
2
50
4
Station 3
25
1
100
6
Station 4
50
0.75
20
3
te
tb=B*te
Cb^2=Ce^2/B
u=TH*tb/m
0.0238
1.1905
0.02
0.9921
0.0238
1.1905
0.08
0.7441
0.04
2
0.02
0.8333
0.02
1
0.0113
0.8333
Ca^2
Cd^2 = 1+(1-u^2)(Ca^2-1)+(u^2/sqrt(m))*(Cb^2-1)
0
0.4274
0.4274
0.4897
0.4897
0.5662
0.5662
0.4711
1.69
1.35
2.31
1.2
0.4
0.8
1.41
6.76
1
255
200
600
1/te
Ce
P
m
CT = [(Ca^2+Cb^2)/2]*[u^(sqrt(2*(m+1))-1)/(m*(1-u))]*tb+tb
CT1+CT2+CT3+CT4
WIPq
m*P
This option is dominated by the previous one since it presents a higher CT and
also a higher deployment cost. However, final selection(s) must be assessed and
validated through simulation.
60
1115