Computer Architecture 2010 – Advanced Topics 1 Computer Architecture Advanced Topics.
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Analysis and Design of Asynchronous Transfer Lines as
a series of G/G/m queues:Overview and Examples
Topics
• Modeling the Asynchronous Transfer Line as a series of G/G/m queues
• Modeling the impact of preemptive, non-destructive operational detractors
• Employing the derived models in line diagnosis • Employing the derived models in line design• The role of batching in the considered manufacturing
systems• An analysis of a workstation involving parallel batching
Asynchronous Transfer Lines (ATL)W2 W3
THTHB2 B3
W1
TH THB1 M1 M2 M3
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?
The G/G/1 model:A single-station
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
THB1 M1
An Important Stability Condition
•Average workload brought to station per unit time:
TH·te
• It must hold:
• Otherwise, an infinite amount of WIP will pile up in front of the station.
TH te 1.0
THB1 M1
Performance measures for a stable G/G/1 station
• Server utilization:
• Expected cycle time in the buffer: (Kingman’s approx.)
• Expected cycle time in the station:
• Average WIP in the buffer: (by Little’s law)
• Average WIP in the station:
• Squared CV of the inter-departure times:
u TH te
CTq ca
2 ce2
2
u
1 ute
CT CTq te
WIPq TH CTq
WIP TH CT WIPq u
cd2 (1 u2)ca
2 u2ce2
THB1 M1
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 u1.0.
• 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 inter-arrival 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
• Server utilization:
• Expected cycle time in the buffer:
• Expected cycle time in the station:
• Average WIP in the buffer:
• Average WIP in the station:
• Squared CV of the inter-departure times:
M1
BTH TH
M2
Mm
u (TH te ) m
CTq ca
2 ce2
2
u 2(m1) 1
m(1 u)te
CT CTq te
WIPq TH CTq
WIP TH CT WIPq mu
cd2 1 (1 u2)(ca
2 1) u2
m(ce
2 1)
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 inter-arrival 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 non-preemptive. Each of these categories requires a different analytical treatment.
Preemptive non-destructive 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 (i.e., without the delays due to the detractors), following a general distribution.
• to = E[X]; o2=Var[X]; co=o / to .
• T = random variable modeling the effective processing time = where
• 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
X U ii1
N
e2 Var[T] ( o
2 A2) to((mr2 r
2) m f )
ce2 e
2 / te2 co
2 (1 cr2)A(1 A)(mr / to)
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 variabilityco = 1/15 = 0.067 (which is very low)
Example Continued
• Effective variability:
41.4715
3600)888.01)(888.0(2)067.0()1(2
875.16888.0/15/
888.018
8
222
o
roe
oe
rf
f
t
mAAcc
Att
mm
mA
Which is very high!
Example Continued
• Suppose through a preventive maintenance program, we can reduce mf to 8 min and mr to 1 min
79.015
60)888.01)(888.0(2)067.0()1(2
875.16888.0/15/
888.018
8
222
o
roe
oe
rf
f
t
mAAcc
Att
mm
mA
Which is low!
(the same as before)
Example:employing the developed theory for diagnostic purposes
M1 B M2
to1 =19 min
co12=0.25
mf1=48 hrsmr1=8 hrs MTTR ~ expon.
to2 =22 min
co22=1.0
mf2=3.3 hrsmr2=10 min MTTR ~ expon.
Ca2=1.0
20parts
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 B M2
to1 =19 min
co12=0.25
mf1=48 hrsmr1=8 hrs MTTR ~ expon.
to2 =22 min
co22=1.0
mf2=3.3 hrsmr2=10 min MTTR ~ expon.
Ca2=1.0
20parts
A1 m f 1 /(m f 1 mr1) 48/(48 8) 0.857
te1 to1 / A1 19/0.857 22.17min
A2 m f 2 /(m f 2 mr2) 3.3/(3.310/60) 0.952
te 2 to2 / A2 22/0.952 23.11min Bottleneck machine
Bottleneck utilization: u2 TH te2 2.4 (23.11/60) 0.9244 1.0 (!)
Diagnostics example continued:An analysis based on the G/G/m model
)(! 20>>72.3560/94.8924.2
min94.89211.239244.01
9244.0
2
04.128.5
12
9244.060/11.234.2
04.122/10)952.01(952.0)11(1/)1()1(
28.51)8868.01(442.68868.0)1(
85.2560/256.6464.2
min256.64617.228868.01
8868.0
2
442.61
12
8868.060/17.224.2
442.619/608)857.01(857.0)11(25.0/)1()1(
22
22
222
22
2
22
222222
22
22
2221
21
21
21
21
22
11
11
121
21
1
11
111121
21
21
eea
q
e
orroe
aeda
eea
q
e
orroe
CTTHWIP
tu
uccCT
tTHu
tmAAccc
cucucc
CTTHWIP
tu
uccCT
tTHu
tmAAccc
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 Option 2 Option 31 42, 2.0, 50 42, 1.0, 85 10, 2.0, 110.52 42, 2.0, 50 42, 1.0, 853 25, 1.0, 100 25, 0.7, 1204 50, 0.75, 20 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 42,2.0,50 25,1.0,100 50,0.75,2050 42,1.0,85 42,1.0,85 25, 0.7,120 6,0.75,248 10,2.0,110.5
(1000 / 50) / 8 2.5
Station 1 Station 2 Station 3 Station 41/te 42 42 25 50Ce 2 2 1 0.75P 50 50 100 20m 3 3 6 3
te 0.0238 0.0238 0.04 0.02tb=B*te 1.1905 1.1905 2 1
Cb^2=Ce^2/B 0.08 0.08 0.02 0.0113u=TH*tb/m 0.9921 0.9921 0.8333 0.8333
Ca^2 0 0.4615 0.4687 0.5598Cd^2 = 1+(1-u^2)(Ca^2-1)+(u^2/sqrt(m))*(Cb^2-1) 0.4615 0.4687 0.5598 0.4691
CT = [(Ca^2+Cb^2)/2]*[u^(sqrt(2*(m+1))-1)/(m*(1-u))]*tb+tb 3.17 14.6 2.3 1.41CT1+CT2+CT3+CT4 21.48
WIPq 4.9 33.5 0.8 1
m*P 150 150 600 60960
Reducing the line cycle time by adding capacity to Station 2
1000 42,2.0,50 42,2.0,50 25,1.0,100 50,0.75,2050 42,1.0,85 42,1.0,85 25, 0.7,120 6,0.75,248 10,2.0,110.5
(1000 / 50) / 8 2.5
Station 1 Station 2 Station 3 Station 41/te 42 42 25 50Ce 2 2 1 0.75P 50 50 100 20m 3 4 6 3
te 0.0238 0.0238 0.04 0.02tb=B*te 1.1905 1.1905 2 1
Cb^2=Ce^2/B 0.08 0.08 0.02 0.0113u=TH*tb/m 0.9921 0.7441 0.8333 0.8333
Ca^2 0 0.4615 0.505 0.5709Cd^2 = 1+(1-u^2)(Ca^2-1)+(u^2/sqrt(m))*(Cb^2-1) 0.4615 0.505 0.5709 0.4725
CT = [(Ca^2+Cb^2)/2]*[u^(sqrt(2*(m+1))-1)/(m*(1-u))]*tb+tb 3.17 1.36 2.32 1.42CT1+CT2+CT3+CT4 8.27
WIPq 4.9 0.4 0.8 1.1
m*P 150 200 600 601010
Adding capacity at Station 1, the new bottleneck
1000 42,2.0,50 42,2.0,50 25,1.0,100 50,0.75,2050 42,1.0,85 42,1.0,85 25, 0.7,120 6,0.75,248 10,2.0,110.5
(1000 / 50) / 8 2.5
Station 1 Station 2 Station 3 Station 41/te 42 42 25 50Ce 2 2 1 0.75P 50 50 100 20m 4 4 6 3
te 0.0238 0.0238 0.04 0.02tb=B*te 1.1905 1.1905 2 1
Cb^2=Ce^2/B 0.08 0.08 0.02 0.0113u=TH*tb/m 0.7441 0.7441 0.8333 0.8333
Ca^2 0 0.299 0.4324 0.5487Cd^2 = 1+(1-u^2)(Ca^2-1)+(u^2/sqrt(m))*(Cb^2-1) 0.299 0.4324 0.5487 0.4657
CT = [(Ca^2+Cb^2)/2]*[u^(sqrt(2*(m+1))-1)/(m*(1-u))]*tb+tb 1.22 1.31 2.27 1.4CT1+CT2+CT3+CT4 6.2
WIPq 0.1 0.3 0.7 1
m*P 200 200 600 601060
An alternative option:Employ less variable machines at Station 1
1000 42,2.0,50 42,2.0,50 25,1.0,100 50,0.75,2050 42,1.0,85 42,1.0,85 25, 0.7,120 6,0.75,248 10,2.0,110.5
(1000 / 50) / 8 2.5
Station 1 Station 2 Station 3 Station 41/te 42 42 25 50Ce 1 2 1 0.75P 85 50 100 20m 3 4 6 3
te 0.0238 0.0238 0.04 0.02tb=B*te 1.1905 1.1905 2 1
Cb^2=Ce^2/B 0.02 0.08 0.02 0.0113u=TH*tb/m 0.9921 0.7441 0.8333 0.8333
Ca^2 0 0.4274 0.4897 0.5662Cd^2 = 1+(1-u^2)(Ca^2-1)+(u^2/sqrt(m))*(Cb^2-1) 0.4274 0.4897 0.5662 0.4711
CT = [(Ca^2+Cb^2)/2]*[u^(sqrt(2*(m+1))-1)/(m*(1-u))]*tb+tb 1.69 1.35 2.31 1.41CT1+CT2+CT3+CT4 6.76
WIPq 1.2 0.4 0.8 1
m*P 255 200 600 601115
This option is dominated by the previous one since it presents a higher CT andalso a higher deployment cost. However, final selection(s) must be assessed and validated through simulation.
Lot Sizing• If affordable, a lot-for-lot (L4L) policy will incur the lowest inventory holding
costs and it will maintain a smoother production flow.• Possible reasons for departure from a L4L policy:
– High set up times and costs => need for serial process batching to control the capacity losses
– Processes that require a large production volume in order to maintain a high utilization (e.g., fermentors, furnaces, etc.) => need for parallel process batching
• Selection of a pertinent process batch size– It must be large enough to maintain feasibility of the production
requirements– It must control the incurred
• inventory holding costs, and/or• part delays (this is a measure of disruption to the production flow
caused by batching)• Move or transfer batches: The quantities in which parts are moved between
the successive processing stations.– They should be as small as possible to maintain a smooth process flow
Optimal Parallel Batching: A factory physics approach
Model Parameters:k: (parallel) batch size B: maximum batch sizera: arrival rate (parts/hr) ca: CV of inter-arrival timest: batch processing time (hrs) ce: CV for effective batch processing timeThen CT = WTBT + CTq+t
aaaa r
kkk
krr
k
rkWTBT
2
1
2
)1(1]
1...
10[
1
trktk
ru
k
c
kt
kct
u
uccCT a
aa
a
aa
ea
q b
b
1 ;
)( ;
12
2
2
22
22
From the above,
tu
uckc
ku
ktt
u
uckc
r
kCT eaea
a
]112
/
2
1[
12
/
2
1 2222
Remark: Notice that CT as u1 but also as u0 !
Determining an optimized batch sizeLet um rat . Then u = um / k k = um / u . Substituting this expression for k in the expression for CT, we get:
tu
ucuuc
u
uut
u
uckc
ku
kCT ema
m
mea ]112
/
2
1/[]1
12
/
2
1[
2222
Recognizing that 022
ka
m
ak
cu
uc, we set 0
2
m
au
uc and we get
tu
uyt
u
uc
uuCT
m
e
m
]12
1
2
)([]1
122
1
2
1[
2
where
u
uc
uuy e
1
)(1
)( 2
To minimize CT, it suffices to minimize y(u). This can be achieved as follows:
22
2*22
22
222
1
1
1
1012)1(0
)1(
)()1()(
ee
ee
e
cc
cuuuc
uu
ucu
du
udy
and 1c
1u 0
e
*
which further implies that trctrk aea )1(*
Remark: If ce2 0, the term in the original expression for u* will significant. In that
case, we can set em
a
cu
c
1
12* and obtain u* and k* as before.