MIT 2.852
Manufacturing Systems Analysis
Lectures 18–19
Loops
Stanley B. Gershwin
Spring, 2007
c 2007 Stanley B. Gershwin.
Copyright �
Problem
Statement
B1
M2
B2
M3
B3
M1
M4
B6
M6
B5
M5
B4
• Finite buffers (0 � ni(t) � Ni).
• Single closed loop – fixed population (
�
i
ni(t) = N ).
• Focus is on the Buzacott model (deterministic processing time; geometric
up and down times). Repair probability = ri; failure probability = pi. Many
results are true for more general loops.
• Goal: calculate production rate and inventory distribution.
c 2007 Stanley B. Gershwin.
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Problem
Statement
Motivation
• Limited pallets/fixtures.
• CONWIP (or hybrid) control systems.
• Extension to more complex systems and policies.
c 2007 Stanley B. Gershwin.
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Two-Machine Loops
Special Case
Refer to MSE Section 5.6, page 205.
N
N1
r1
p1
B1
M1
M2
N2
r2
p2
=�
r1
p1
N*
M1
B
M2
r2
p2
B2
P loop(r1, p1, r2, p2, N1, N2) = P line(r1, p1, r2, p2, N �)
where
N � = min(n, N1) − max(0, n − N2)
c 2007 Stanley B. Gershwin.
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4
Two-Machine Loops
Special Case
.14
.01
.1
.01
20
N2=10
N2=15
N2=20
N2=30
N2=40
0.885
0.88
production rate
r1
p1
r2
p2
N1
0.89
0.875
0.87
0.865
0.86
0
c 2007 Stanley B. Gershwin.
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10
20
30
population
40
50
60
5
Expected
population
method
• Treat the loop as a line in which the first machine
and the last are the same.
• In the resulting decomposition, one equation is
missing.
• The missing equation is replaced by the expected
�
population constraint ( i n̄i = N ).
c 2007 Stanley B. Gershwin.
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Expected
population
method
Evaluate i, i − 1, i + 1 modulo k (ie, replace 0 by k and replace k + 1 by 1).
ps(i − 1)ru(i)
ru(i) = ru(i − 1)X(i) + ri(1 − X(i));
X(i) =
pu(i)E(i − 1)
�
�
1
1
pd(i − 1)
+
−2−
, i = 1, ..., k
pu(i) = ru(i)
E(i − 1)
ei
rd(i − 1)
pb(i + 1)rd(i)
rd(i) = rd(i + 1)Y (i + 1) + ri+1(1 − Y (i + 1)); Y (i + 1) =
.
pd(i)E(i + 1)
�
�
1
1
pu(i + 1)
+
−2−
, i = k, ..., 1
pd(i) = rd(i)
E(i + 1)
ei+1
ru(i + 1)
This is 4k equations in 4k unknowns. But only 4k − 1 of them are independent
because the derivation uses E(i) = E(i + 1) for i = 1, ..., k. The first k − 1
are E(1) = E(2), E(2) = E(3), ..., E(k − 1) = E(k). But this implies
E(k) = E(1), which is the same as the kth equation.
c 2007 Stanley B. Gershwin.
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Expected
population
method
Therefore, we need one more equation. We can use
�
n̄i = N
i
If we guess pu(1) (say), we can evaluate
�
TOTAL
n̄
= i n̄i as a function of pu(1). We search for
the value of pu(1) such that n̄ TOTAL = N .
c 2007 Stanley B. Gershwin.
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8
Expected
population
method
Behavior:
• Accuracy good for large systems, not so good for
small systems.
• Accuracy good for intermediate-size populations; not
so good for very small or very large populations.
c 2007 Stanley B. Gershwin.
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9
Expected
population
method
• Hypothesis: The reason for the accuracy behavior of the
population constraint method is the correlation in the buffers.
� The number of parts in the system is actually constant .
� The expected population method treats the population as random, with a
specified mean.
� If we know that a buffer is almost full, we know that there are fewer parts
in the rest of the network, so probabilities of blockage are reduced and
probabilities of starvation are increased. (Similarly if it is almost empty.)
� Suppose the population is smaller than the smallest buffer. Then there
will be no blockage. The expected population method does not take this
into account.
c 2007 Stanley B. Gershwin.
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10
Loop Behavior
To construct a method that deals with the invariant
(rather than the expected value of the invariant), we
investigate how buffer levels are related to one another
and to the starvation and blocking of machines.
In a line, every downstream machine could block a
given machine, and every upstream machine could
starve it. In a loop, blocking and starvation are more
complicated.
c 2007 Stanley B. Gershwin.
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Ranges
Loop Behavior
B1
M2
B2
M3
B3
M1
M4
B6
M6
B5
M5
B4
• The range of blocking of a machine is the set of all machines
that could block it if they stayed down for a long enough time.
• The range of starvation of a machine is the set of all machines
that could starve it if they stayed down for a long enough time.
c 2007 Stanley B. Gershwin.
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Ranges
Loop Behavior
Range of Blocking
10
B1
10
M2
B2
10
M3
B3
M1
M4
7
B6
0
0
M6
B5
M5
B4
• All buffer sizes are 10.
• Population is 37.
• If M4 stays down for a long time, it will block M1.
• Therefore M4 is in the range of blocking of M1.
• Similarly, M2 and M3 are in the range of blocking of M1.
c 2007 Stanley B. Gershwin.
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Ranges
Loop Behavior
Range of starvation
10
7
B1
M2
B2
10
M3
B3
M1
M4
0
B6
10
0
M6
B5
M5
B4
• If M5 stays down for a long time, it will starve M1.
• Therefore M5 is in the range of starvation of M1.
• Similarly, M6 is in the range of starvation of M1.
c 2007 Stanley B. Gershwin.
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Loop Behavior
Ranges
Line
• The range of blocking of a machine in a line is the entire
downstream part of the line.
• The range of starvation of a machine in a line is the entire
upstream part of the line.
c 2007 Stanley B. Gershwin.
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Ranges
Loop Behavior
Line
In an acyclic network, if Mj is downstream of Mi, then the range
of blocking of Mj is a subset of the range of blocking of Mi.
Mi
Mj
Range of blocking of M j
Range of blocking of M i
Similarly for the range of starvation.
c 2007 Stanley B. Gershwin.
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Ranges
Loop Behavior
Line
In an acyclic network, if Mj is downstream of Mi, any real machine whose
failure could cause Md(j) to appear to be down could also cause Md(i) to
appear to be down.
Mi
Bi
M
j
Md(i)
Md(j)
Consequently, we can express rd(i) as a function of the parameters of L(j).
This is not possible in a network with a loop because some machine that
blocks Mj does not block Mi.
c 2007 Stanley B. Gershwin.
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Ranges
Loop Behavior
Difficulty for decomposition
L(1)
Range of blocking of M6
B1
M2
B2
M3
M u(1)
B(1)
B3
M1
Range of blocking of M1
M4
B1
B6
M6
B5
M5
d
B(6)
M2
B2
M3
B3
B4
Range of starvation of M1
M (6)
d
M (1)
M4
M1
u
M (6)
B6
L(6)
M6
B5
M5
B4
Range of starvation of M2
Ranges of blocking and starvation of adjacent machines are not subsets or
supersets of one another in a loop.
c 2007 Stanley B. Gershwin.
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Ranges
Loop Behavior
Difficulty for decomposition
Range of blocking of M1
Range of blocking of M2
10
B1
10
M2
B2
10
M3
B3
M1
M4
7
B6
0
0
M6
B5
M5
B4
M5 can block M2. Therefore the parameters of M5 should directly affect the
parameters of Md(1) in a decomposition. However, M5 cannot block M1 so
the parameters of M5 should not directly affect the parameters of Md(6).
Therefore, the parameters of Md(6) cannot be functions of the parameters of
Md(1).
c 2007 Stanley B. Gershwin.
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19
Multiple Failure
Mode Line
Decomposition
• To deal with this issue, we introduce a new
decomposition.
• In this decomposition, we do not create failures of
virtual machines that are mixtures of failures of real
machines.
• Instead, we allow the virtual machines to have
distinct failure modes, each one corresponding to the
failure mode of a real machine.
c 2007 Stanley B. Gershwin.
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Multiple Failure
Mode Line
Decomposition
1,2
3
4
5,6,7
1,2,
3,4
5,6,7,
8,9,10
8
9,10
• There is an observer in each buffer who is told that
he is actually in the buffer of a two-machine line.
c 2007 Stanley B. Gershwin.
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21
Multiple Failure
Mode Line
Decomposition
1,2
3
4
5,6,7
1,2,
3,4
5,6,7,
8,9,10
8
9,10
• Each machine in the original line may and in the two-machine
lines must have multiple failure modes.
• For each failure mode downstream of a given buffer, there is a
corresponding mode in the downstream machine of its
two-machine line.
• Similarly for upstream modes.
c 2007 Stanley B. Gershwin.
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Multiple Failure
Mode Line
Decomposition
1,2
3
4
5,6,7
1,2,
3,4
5,6,7,
8,9,10
8
9,10
• The downstream failure modes appear to the observer after
propagation through blockage .
• The upstream failure modes appear to the observer after
propagation through starvation .
• The two-machine lines are more complex that in earlier
decompositions but the decomposition equations are simpler.
c 2007 Stanley B. Gershwin.
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23
Multiple Failure
Mode Line
Decomposition
Two-Machine Line
up
down
Form a Markov chain and find the steady-state probability
distribution. The solution technique is very similar to that of the
two-machine-state model. Determine the production rate,
probability of starvation and probability of blocking in each down
mode, average inventory.
c 2007 Stanley B. Gershwin.
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Multiple Failure
Mode Line
Decomposition
Line Decomposition
• A set of decomposition equations are formulated.
• They are solved by a Dallery-David-Xie-like
algorithm.
• The results are a little more accurate than earlier
methods, especially for machines with very different
failures.
c 2007 Stanley B. Gershwin.
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Multiple Failure
Mode Line
Decomposition
1,2
3
Line Decomposition
4
5,6,7
1,2,
3,4
5,6,7,
8,9,10
8
9,10
• In the upstream machine of the building block, failure mode 4 is a local
mode; modes 1, 2, and 3 are remote modes. Modes 5, 6, and 7 are local
modes of the downstream machine; 8, 9, and 10 are remote modes.
• For every mode, the repair probability is the same as the repair probability
of the corresponding mode in the real line.
• Local modes: the probability of failure into a local mode is the same as the
probability of failure in that mode of the real machine.
c 2007 Stanley B. Gershwin.
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Multiple Failure
Mode Line
Decomposition
1,2
3
Line Decomposition
4
5,6,7
1,2,
3,4
5,6,7,
8,9,10
8
9,10
• Remote modes: i is the building block number; j and f are the machine
number and mode number of a remote failure mode. Then
Ps,jf (i − 1)
Pb,jf (i)
u
d
pjf (i) =
rjf ;
pjf (i − 1) =
rjf
E(i)
E(i − 1)
where pu
jf (i) is the probability of failure of the upstream machine into mode
jf ; Ps,jf (i − 1) is the probability of starvation of line i − 1 due to mode jf ;
rjf is the probability of repair of the upstream machine from mode jf ; etc.
• Also, E(i − 1) = E(i).
d
(i),
p
• pu
jf
jf (i) are used to evaluate E(i), Ps,jf (i),Pb,jf (i) from
two-machine line i in an iterative method.
c 2007 Stanley B. Gershwin.
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Multiple Failure
Mode Line
Decomposition
Line Decomposition
Consider
pu
jf (i)
=
Ps,jf (i − 1)
rjf ;
pdjf (i
− 1) =
Pb,jf (i)
rjf
E(i)
E(i − 1)
In a line, jf refers to all modes of all upstream machines in the
first equation; and all modes of all downstream machines in the
second equation.
We can interpret the upstream machines as the range of
starvation and the downstream machines as the range of
blockage of the line.
c 2007 Stanley B. Gershwin.
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Multiple Failure
Mode Line
Decomposition
B1
Extension to Loops
M2
B2
M3
B3
M1
M4
B6
M6
B5
M5
B4
• Use the multiple-mode decomposition, but adjust the
ranges of blocking and starvation accordingly.
• However, this does not take into account the local
information that the observer has.
c 2007 Stanley B. Gershwin.
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Thresholds
10
B1
10
10
M2
B2
M3
B3
M4
M1
5
B6
2
M6
B5
0
M5
B4
5
d
M (6)
B6
u
M (6)
• The B6 observer knows how many parts there are in his buffer.
• If there are 5, he knows that the modes he sees in M d(6) could
be those corresponding to the modes of M1, M2, M3, and M4.
c 2007 Stanley B. Gershwin.
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Thresholds
10
B1
9
10
M2
B2
M3
B3
M1
M4
8
B6
0
M6
B5
0
M5
B4
8
d
M (6)
B6
u
M (6)
• However, if there are 8, he knows that the modes he sees in
M d(6) could only be those corresponding to the modes of M1,
M2, and M3; and not those of M4.
• The transition probabilities of the two-machine line therefore
depend on whether the buffer level is less than 7 or not.
c 2007 Stanley B. Gershwin.
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Thresholds
• This would require a new model of a two-machine
line.
• The same issue arises for starvation.
• In general, there can be more than one threshold in
a buffer.
• Consequently, this makes the two-machine line very
complicated.
c 2007 Stanley B. Gershwin.
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32
Transformation
• Purpose: to avoid the complexities caused by
thresholds.
• Idea: Wherever there is a threshold in a buffer,
break up the buffer into smaller buffers separated by
perfectly reliable machines.
c 2007 Stanley B. Gershwin.
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33
Transformation
buffer size
20
3
B1
18
13
M2
B2
1
• When M2 fails for a long time,
B1 fills up, and there is one
part in B4. Therefore there is
a threshold of 1 in B4.
M3
M1
1
15
B4
threshold
M4
population = 21
c 2007 Stanley B. Gershwin.
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B3
• When M1 fails for a long time,
B4 and B3 fill up, and there
is one part in B2. Therefore
there is a threshold of 1 in B2.
5
• When M3 fails for a long time,
B2 fills up, and there are 18
parts in B1. Therefore there is
a threshold of 18 in B1.
34
Transformation
buffer size
20
3
B1
18
13
M2
B2
1
M3
M1
• Note: B1 has two thresholds
and B3 has none.
1
15
B4
threshold
M4
population = 21
c 2007 Stanley B. Gershwin.
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B3
• When M4 fails for a long time,
B3 and B2 fill up, and there
are 13 parts in B1. Therefore
there is a threshold of 13 in
B1 .
5
• Note: The number of thresh­
olds equals the number of ma­
chines.
35
Transformation
20
B1
M2
B2
3
M*4
15
5
B4
M4
B3
M2
M3
M3
M1
M*1
M*3
M1
M*2
M4
• Break up each buffer into a sequence of buffers of size 1 and
reliable machines.
• Count backwards from each real machine the number of
buffers equal to the population.
• Identify the reliable machine that the count ends at.
c 2007 Stanley B. Gershwin.
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36
Transformation
M1
2
M*3
5
M*4
13
M2
B1
1
M*2
14
B4
M4
2
5
B3
M3
1
M*1
B2
• Collapse all the sequences of unmarked reliable
machines and buffers of size 1 into larger buffers.
c 2007 Stanley B. Gershwin.
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37
Transformation
• Ideally, this would be equivalent to the original
system.
• However, the reliable machines cause a delay, so
transformation is not exact for the discrete/
deterministic case.
• This transformation is exact for continuous-material
machines.
c 2007 Stanley B. Gershwin.
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38
Transformation
Small populations
• If the population is smaller than the largest buffer, at
least one machine will never be blocked.
• However, that violates the assumptions of the
two-machine lines.
• We can reduce the sizes of the larger buffers so that
no buffer is larger than the population. This does not
change performance.
c 2007 Stanley B. Gershwin.
Copyright �
39
Numerical
Results
Accuracy
• Many cases were compared with simulation:
� Three-machine cases: all throughput errors under 1%; buffer
level errors averaged 3%, but were as high as 10%.
� Six-machine cases: mean throughput error 1.1% with a
maximum of 2.7%; average buffer level error 5% with a
maximum of 21%.
� Ten-machine cases: mean throughput error 1.4% with a
maximum of 4%; average buffer level error 6% with a
maximum of 44%.
c 2007 Stanley B. Gershwin.
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40
Numerical
Results
Other algorithm attributes
• Convergence reliability: almost always.
• Speed: execution time increases rapidly with loop
size.
• Maximum size system: 18 machines. Memory
requirements grow rapidly also.
c 2007 Stanley B. Gershwin.
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41
The Batman Effect
Numerical
Results
0.815
• Error is very small, but there
are apparent discontinuities.
Decomposition
Simulation
0.81
0.805
Throughput
0.8
0.795
0.79
0.785
0.78
0.775
0.77
0
5
10
15
Population
20
c 2007 Stanley B. Gershwin.
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25
30
• This is because we cannot
deal with buffers of size 1, and
because we do not need to
introduce reliable machines in
cases where there would be
no thresholds.
42
Behavior
Numerical
Results
B1
*
M2
B2
M1
M3
B4
M4
B3
• All buffer sizes 10. Population 15. Identical machines
except for M1.
• Observe average buffer levels and production rate as
a function of r1.
c 2007 Stanley B. Gershwin.
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43
Behavior
Numerical
Results
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
• Production rate vs. r1.
• Usual saturating graph.
c 2007 Stanley B. Gershwin.
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44
Behavior
Numerical
Results
10
b1 average
b2 average
b3 average
b4 average
9
8
B1
*
M2
7
B2
6
5
M1
M3
4
3
2
B4
M4
B3
1
0
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
• When r1 is small, M1 is a bottleneck, so B4 holds 10
parts, B3 holds 5 parts, and the others are empty.
• As r1 increases, material is more evenly distributed.
When r1 = 0.1, the network is totally symmetrical.
c 2007 Stanley B. Gershwin.
Copyright �
45
Applications
• Design system with pallets/fixtures. The fixtures and
the space to hold them in are expensive.
• Design system with tokens/kanbans (CONWIP). By
limiting population, we reduce production rate, but
we also reduce inventory.
c 2007 Stanley B. Gershwin.
Copyright �
46
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2.852 Manufacturing Systems Analysis
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