Chapter 7
Performance analysis of failureprone production lines
Learning objectives :
Understanding the mathematical models of production lines
Understanding the impact of machine failures
Understanding the role of buffers
Able to correctly dimension buffer capacities
Textbook :
S.B. Gershwin, Manufacturing Systems Engineering, Prentice Hall,
1994.
J. Li and S.M. Meerkov, Production Systems Engineering
1
Plan
• Basic concepts
• Failure-prone single-machine systems
• Production lines with unlimited buffers
• Production lines without buffers
• Aggregation of parallel machines and consecutive
dependent machines
• Two-machine production lines with intermediate buffer
• Long failure-prone production lines
2
Basic concepts
3
Production lines or tranfer lines
M1
B1
M2
B2
machine
M3
B3
M4
Buffer
Frequent production disruption by machine failures
Buffers are of finite capacity
Production often varies wildly
4
Production lines or tranfer lines
5
Production capability of a machine
Cycle time or processing time (t) :
• time necessary to process a part by a machine
• Constant cycle times (large assembly systems)
• Variable or random cycle time (job-shop environment).
Maximal production rate or max. capacity (U) :
• U = 1/t (parts per time unit)
6
Machine reliability model
• TBF = Time Between Failure (Tup)
• TTR = Time To Repair (Tdown)
• MTBF = Mean TBF
• MTTR = Mean TTR
• Failure rate l = 1/MTBF
• Repair rate m = 1/MTTR
l
UP
DOWN
m
7
Models of machine failures
ODF – Operation-Dependent Failure: the state of the
machine degrades ony when it produces.
Implication : an ODF cannot fail when it is not producing.
TDF – Time-Dependent Failure: the state of the
machine degrades all the time even if it is not producing.
Implication : a TDF machine can fail even if it is not
producing.
8
Models of machine failures
Cause of failures:
ODF: tool wear
TDF: electricity supply, electronic components, …
ODF failures account for 80% of disruptions in
manufacturing systems (Hanifin & Buzacott)
In this chapter, we mainly focus on ODF failures.
Example of a two machine production line to explain the
difference
9
Mathematical models of buffers
Buffer capacity : N
State of a buffer : number of parts in it varying from 0 to N
Assumptions:
• A part, produced by a machine, is immediately placed in
the downstream buffer, if it is not full.
• A part is immediately available for processing by a
machine, if the upstream buffer is not empty.
10
Mathematical models of buffers
Buffering capacity of a moving convey
l
v
 Ttravel 
N0  

 t 
N  K  N0
Ttravel 
where
l = length of the convey
v = speed of the convey
K = maximum number of carriers in the convey
11
Interaction between machines and buffers
Blocking Before Service (BBS):
A machine cannot operate and is blocked if
• it is up
• its downstream buffer is full
• no part can be removed from that buffer
Blocking After Service (BAS):
• A machine continue to produce even if the downstream buffer is full.
• The machine is blocked at the completion of the part if the buffer remains
full.
Buffer capacity convetion: NBBS = NBAS +1
Starvation : an idle up machine is starved if its upstream buffer is empty/
12
Performance measures
Throughput rate also called productivity (TH):
• Number of parts produced per time unit.
Production rate (PR):
• Number of parts produced per cycle time.
• Concept appropriate for synchronuous production systems with
all machines having identical cycle times
TH = U×PR
13
Performance measures
Work-in-process of the i-th buffer (WIPi)
• Average number of parts contained in the i-th buffer.
Total work-in-process (WIP):
• Average number of parts) in the system
• WIP = WIP1 + WIP2 + ...
Probability of blocking (BLi)
Probability of starvation (STi)
14
Failure-prone single-machine
systems
15
Throughput rate
MTBF
m
TH  U
U
MTBF  MTTR
lm
Proof:
•Average length of an UP period = MTBF
•Average length of a DOWN period = MTTR
•Production of an UP period = U. MTBF
•Length of an UP-DOWN cycle = MTBF + MTTR
•Throughput rate : TH = (U.MTBF) / (MTBF + MTTR).
UP
DOWN
16
A machine operating at a reduced speed U' < U
Operation Dependent Failure case
TH  U '
m
U'
l
m
U
Time Dependent Failure case:
TH  U '
m
lm
17
Production lines with unlimited
buffers
18
Assumptions
M1
B1
M2
B2
M3
B3
M4
• The first machine M1 is never starved
• The last machine is never blocked
• Each machine produces if its upstream buffer is not empty
19
Bottleneck machine
M1
B1
M2
B2
M3
B3
M4
A machine Mi is said to be a bottleneck if it proper
productivity (or isolated productivity) is smaller than
that of other machines, i.e.
Ui
mi
li  mi
Uj
mj
lj  m j
, j  i
20
Throughput rate
M1
B1
M2
B2
M3
B3
M4
The throughput rate of a production line with unlimited buffers
is equal to that of the bottleneck machines, i.e. (Why?)
TH  MIN
i
U i MTBFi
Um
 MIN i i
i
MTBFi  MTTRi
li  mi
The throughput rate of a machine Mi is equal to that of the
slowest upstream machine, i.e.
TH j  MIN
i j
U i MTBFi
Um
 MIN i i
i  j li  mi
MTBFi  MTTRi
21
Case of a single bottleneck machine
M1
B1
M2
B2
B3
M3
M4
The level of the input buffer of the bottleck machine grows
without limit (at which slope?)
All downstream buffers remain limited.
B2
B3
22
Case of two bottleneck machines
M1
B1
M2
B2
B3
M3
M4
The level of the input buffer of each bottleneck machines grows
without limit.
All other buffers downstream of the first bottleneck remain
limited.
B1
B3
B2
Master GI2007
23
Production lines without buffers
24
Assumptions
M1
B1
M2
B2
M3
B3
M4
If a machine breaks down or it takes longer time for an
operation,
then all other machines must wait.
(immedicate propagration of disruptions)
Impact : the productivity of the line is usually smaller than that
of the bottleneck machine.
25
Case of reliable machines with different cycle times
t1  1
t2  2
M1
M2
The progress of products in the line is synchronized to allow
the completion of all on-going operations .
The cycle time of the line is that of the slowest machine, i.e.
TH  MIN U i  MIN
i
2
i
2
1
ti
M1 wait
2
26
Failure-prone lines with identical cycle times
M1
B1
M2
B2
M3
B3
M4
Assumptions:
When a machine breaks down, all other machines must wait.
The probability of two machines failed at the same time is small
enough and can be neglected. (true in practice)
27
Failure-prone lines with identical cycle times
M1
B1
M2
B2
M3
B3
M4
Productivity:
TH 
U
li
1 
i mi

1

li 
t 1   
i mi 

28
Proof
1) Each time interval can be decomposed as follows:
UP
M4
All UP
UP
M2
UP
M1
some machine DOWN
2) tn : instant when the line produces n parts. The time interval
[0, tn) includes :
3)
•
a total duration of nt of all UP periods,
•
for each machine Mi, nt/MTBFi failures requiring with total repair
time of (nt/MTBFi) MTTRi.
tn  nt  
i
nt
MTTRi
MTBFi

l 
 nt 1   i 
i mi 

n
1

x  tn

l 
t 1   i 
i mi 

TH  lim
29
Impact of the length of the line
The longer the line is, the higher the capacity loss is.
Unlimited buffer U/(1l/m)
0,6
0,5
Throughput
0,4
lost capacity
0,3
Zero-buffers
U/(1nl/m)
0,2
0,1
0
0
5
10
15
20
Nb machines
30
Aggregation of parallel machines and
consecutive dependent machines
31
Aggregation
M5
M1
B1
M2
M3
M4
B6
B4
M7
M6
consecutive
dependent machines
M1
B1
M234
parallel
machines
B4
M56
B6
M7
32
Aggregation of parallel machines
M1
M2
Meq
MS
Identical parallel machines : ti = t = 1/U, li = l, mi = m
Ueq = S×U
leq = l
meq = m
33
Aggregation of parallel machines
M1
M2
Meq
MS
Non Identical parallel machines : ti = 1/Ui, li, mi, ei = 1/(1+ li/mi)
S
U eq   U i
leq eeq
i 1
eeq 
1
U eq

S
i 1
U i ei


1
1  leq meq
S
i 1
li ei
S
 eeq
av failure frequency
availability of Meq
34
Aggregration of consecutive dependent machines
M2
M1
...
Meq
MS
Machines of identical cycle time : ti = t = 1/U, li, mi
leq   i 1 li
S
eeq 
1
1
 E ( L) 
leq
meq
1
S li 1
  i 1
meq
leq mi
Failure rate equivalence
1
S li
1   i 1
mi
Flow rate equivalence
Average stoppage time equivalence
35
Aggregration of consecutive dependent machines
Machines of nonidentical cycle time : ti = 1/Ui, li, mi
• All machines slowed down to slowest one :
U = min{Ui, i= 1, ..., S}
• Reduced failure rate : li = Uli /Ui
• Equivalent machine cycle time : Ueq = U
• Failure rate equivalence : leq = Si li
• Flow rate equivalence : THeq = TH(L)
1
S li 1
  i 1
• Average stoppage time equivalence
meq
leq mi
36
Two-machine production lines with
intermediate buffer
37
Motivation & cost of intermediate buffers
M1
B1
M2
B2
M3
B3
M4
Motivation:
• Avoid loss of production capacity
Costs:
• Increasing WIP and production delay
• Larger factory space
• More complicated material handling
Effect of failures:
• Unlimited buffer : no upward propagation of disruptions
• No buffer: Instantaneous propagation
• Finite buffers : delayed and partial propagations
38
Motivation & cost of intermediate buffers
New phenomena:
•Blocking
Failure of M3
•Starvation
M1
M2
M3
M4
M3
M4
3t time units after the failure
where t is the cycle time
M1
M2
39
CTMC model of
reliable line with exponential processing times
M1
B
M2
Assumptions:
• The two machines are reliable and never fail
• The processing times are exponentially distributed random
variables with mean 1/p1 on M1 and 1/p2 on M2
• The buffer capacity is K
• Each machine can hold a part on it for processing.
• M1 is never starved and M2 is never blocked
40
CTMC model of
reliable line with exponential processing times
B
M1
M2
The following state variable
X(t) = number of parts in B
+ the part on M2 if any
+ the finished part blocked on M1 if any
is a continuous time Markov chain
p1
0
p1
…
1
p2
p1
p2
p1
K+1
p2
K+2
p2
41
CTMC model of
reliable line with exponential processing times
M1
B
M2
CTMC model equivalent to M/M/1/(K+2).
0 
1 r
n
,


r
 0 , if r  1
n
K 3
1 r
1
n 
, if r  1
K 3
with r = p1/p2, corresponding to the traffic intensity.
p1
0
p1
…
1
p2
p1
p2
p1
K
p2
K+2
p2
42
CTMC model of
reliable line with exponential processing times
Performance measures (case r ≠ 1)
Starving probability of M2 : 0 = (1-r)/(1-rK+3)
Blocking probability of M1: K+2 = rK+2(1-r)/(1-rK+3)
Throughput rate :
TH = p2(1-0) = p1(1-K+2) = p1(1- rK+2)/(1-rK+3)
Mean WIP
K 2
r
E  X    n n 
K 3
1

r
n 0
1  r K  2
K 2 
 ( K  2) r


 1  r

43
CTMC model of
reliable line with exponential processing times
Example : p1 = 10, p2 = 9, r = 10/9
0,4
0,35
0,3
0,25
Blocking prob.
0,2
9,5
Unlimited buffer
9
0,15
0,1
starving prob
8,5
0,05
0
0
7,5
5
10
15
20
25
30
35
Buffer capacity
7
30
Zero-buffer
6,5
25
6
20
5,5
5
0
5
10
15
20
Buffer capacity
25
30
35
Encours
Throughput
8
15
10
5
0
0
5
10
15
20
Buffer capacity
25
30
35
44
Exponential model of Failure-prone lines
M1
B
M2
Assumptions:
• Machines can break down.
• Exponentially distributed times to failures and time to repair
with TBFi = EXP(li) and TTRi = EXP (mi)
• Exponential processing times with T1 = EXP(p1) and T2 =
EXP(p2)
• Buffer of capacity K
• Each machines holds the part in process.
• M1 never starved and M2 never blocked
45
Exponential model of Failure-prone lines
M1
B
M2
The system can be described by the following state variables:
x = number of parts in B + part on M2 if any + finished part
blocked on M1 if any
ai = 1 if Mi is UP and 0 if Mi is DOWN
The state vector (a1, a2, x) is a continuous time Markov chain
46
Exponential model of Failure-prone lines
0, 0, 0
0, 0, 1
0, 0, 2
1, 0, 0
1, 0, 1
1, 0, 2
m2
0, 1, 0
m1
0, 1, 1
l1
1, 1, 0
0, 1, 2
l2
p2
1, 1, 1
1, 1, 2
p1
Exponential model of the case K = 0
• Analytical expressions of steady-state probabilities
available in the book of SB Gershwin
• Can be used to evaluate the performance measures
47
Slotted time model of a failure-prone line
Assumptions
M1
B
M2
• Synchronized line, i.e. ti = t, with a buffer of capacity N.
• All parts remain in buffers and machines do not hold parts.
• Slotted time indexed t = 1, 2, 3, …
• Machine state change at the begining of a period: machine working (W),
under repair (R), blocked (B), starved (I).
• Buffer state change at the end of a period.
• Blocking Before Service: M1 blocked if B is full, M2 starved if B is empty
• A machine Mi in state W in t breaks down in period t+1 with proba pi and, with
proba 1 - pi, moves to state W or B or I.
• A machine Mi in state R in t moves to state W in t+1 with proba ri and, with
proba 1 - ri, remains in R.
• A machine in B or I in t moves to W in t+1 if the other machine is repaired.
48
Slotted time model of a failure-prone line
Discrete Time Markov chain
M1
B
M2
The state vector (a1, a2, x) with
• ai(t) = 1/0 depending on whether Mi is UP or DOWN at the
begining of t
• x(t) = number of parts in B at the end of t
is a discrete time Markov chain.
Buffer state change :
• x(t) = x(t-1) + a1(t)×1{x(t-1)<N} - a2(t)×1{x(t-1)>0}
• 0≤x(t) ≤ N
49
Slotted time model of a failure-prone line
Discrete Time Markov chain
M1
B
M2
Transient states :
• (1, 0, 0), (1, 1, 0), (0, 0, 0), (1, 0, 1)
• (0,0,N), (0,1,N), (1,1,N), (0,1,N-1)
Flow balance equations for states (a1, a2, x) with 2 ≤ x ≤ N-2
(1,1,x) = (1-p1)(1-p2)(1,1,x) + r1(1-p2)(0,1,x) + (1-p1)r2(1,0,x) + r1r2(0,0,x)
(0,0,x) = (1-r1)(1-r2)(0,0,x) + p1(1-r2)(1,0,x) + (1-r1)p2(0,1,x) + p1p2(0,0,x)
(1,0,x) = (1-p1)p2(1,1,x-1) + (1- p1)(1-r2)(1,0,x-1) + r1p2(0,1,x-1) + r1(1-r2)(0,0,x-1)
(0,1,x) = p1(1-p2)(1,1,x+1) + (1-r1)(1-p2)(0,1,x+1) + p1r2(1,0,x+1) + (1-r1)r2(0,0,x+1)
Other boundary equations can be derived similarly.
50
Slotted time model of a failure-prone line
Discrete Time Markov chain
M1
B
M2
Performance measures :
• Efficiency of M1 : E1 = Sa1 = 1, x < N  (a1, a2, x)
• Efficiency of M2 : E2 = Sa2 = 1, x > 0 ( a1, a2, x)
• Throughput rate : TH = E1×U = E2×U
• WIP : S x( a1, a2, x)
• Probability of starvation : (0, 1, 0)
• Probability of blocking : (1, 0, N)
51
Slotted time model of a failure-prone line
Analytical results
Efficiency of the line E(N) = probability a machine is producing

1  r * N
,
if I1  I 2

* N
1  I1  (1  I 2 ) r 
E(N)  
r1  r2  r1r2  r1r2 (1  I ) N

,if I1  I 2  I
 ( r  r )(1  2 I )  r r (1  I )2 ( N  1)
1 2
 1 2
a1  p1  p2  p1 p2  r1 p2
a 2  p1  p2  p1 p2  r2 p1
1  r1  r2  r1r2  p1r2
 2  r1  r2  r1r2  p2 r1
p
a
rp 
  1 2 , r*  1 2 1 , Ii  i
 2a1
r2 p1  2
ri
WIP to be determined with expressions of steady-state
probabilities available in the book of S.B. Gerswhin
52
Continuous flow models of failure-prone lines
Assumptions
M1
B
M2
Only synchronuous lines are considered, i.e. Ui = U.
Each machine produces continuously.
When a machine Mi produces,
• a flow moves out of its upstream buffer at rate U
• a flow is injected in its downstream buffer at rate U
53
Continuous flow models of failure-prone lines
How good is continuous flow approximation
M1
B
M2
The continuous flow model is a good approximation for a high
volume production line with large enough buffer capacity.
Theorem (David, Xie, Dallery):
THContinuous(h) < THDiscret(h) < THContinuous(h+2)
where THDiscret (h) is a discrete flow line similar to the
Exponential model but with constant processing times and
buffer capacity h.
Result holds for longer lines.
54
Continuous flow models of failure-prone lines
Dynamic behavior
M1
B
M2
Model parameters:
 li: failure rate of Mi
 mi: repair rate of Mi
• U: maximum production rate of Mi
• h : buffer capacity
State variables
 ai(t) =1/0 : state of machine Mi at time t
• x(t): buffer level at t (a real variable)
Auxillary variable:
•ui(t) : production rate of Mi at t
55
Continuous flow models of failure-prone lines
Dynamic behavior
u1(t)
h
u2(t)
M1
B
M2
a1(t)
x(t)
a2(t)
u1 = U
u1 = U
u1 = 0
u1 = U
u1 = 0
u1 = 0
a1 = 1
a1 = 1
a1 = 1
a1 = 1
a1 = 0
a1 = 0
u2 = U
u2 = 0
u2 = 0
u2 = U
u2 = U
u2 = 0
a2 = 1
a2 = 0
a2 = 0
a2 = 1
a2 = 1
a2 = 1
x(t)
F2
blocking
R2
F1
Starving
R1
F2
56
Continuous flow models of failure-prone lines
Dynamic behavior
u1(t)
M1
a1(t)
u2(t)
B
x(t), h
M2
a2(t)
Blockage of M1 : a1(t) = 1, x(t) = h, a2(t) = 0
Starvation of M2 : a1(t) = 0, x(t) = 0, a2(t) = 1
In all other case : u1(t) = a1(t) U, u2(t) = a2(t) U
57
Continuous flow models of failure-prone lines
Performance measures
online proof
Efficiency of the line, E(h) that is the probability a machine
is producing
Throughput rate :
TH(h) = E(h)U
Probability of starving of M2:
ps(h) = 1 – E(h)/e2
Probability of blocking of M1:
pb(h) = 1 – E(h)/e1
Isolated efficiency of Mi :
ei = 1/(1+Ii) = mi/(li+mi)
Mean buffer level:
Q(h)
58
Continuous flow models of failure-prone lines
Analytical solution
Case I1 = l1/m1  I2 = l2/m2
E (h) 
Q (h) 
a
I 2 eah  I1
I 2 (1  I 2 ) e ah  I1 (1  I1 )
I1 I 2 m1  m2
U
1  e ah  I 2 (1  I 2 ) he ah
I 2  I1 m1m2
(
)
I 2 (1  I 2 ) eah  I1 (1  I1 )
l2 m1  l1m2 l1  l2  m1  m2
U
( l1  l2 )( m1  m2 )
59
Continuous flow models of failure-prone lines
Analytical solution
Case I1 = l1/m1  I2 = l2/m2 = I
E (h) 
l l
1 I
h 1 2 U
I
l1l2
(1  I )2 h 
I
(1  2 I )
l1  l2
U
l1l2
h I  l1  l2 
(1  I )  1 
U
2 l1 
m2 
Q (h) 
h
l1  l2
2
1

I
h

I
1

2
I
U
( )
(
)
2
l1l2
60
Continuous flow models of failure-prone lines
Analytical solution
Case U=1, l1  m1  l2  m2 = 0,1
0,55
infinite buffer
0,5
Throughput
0,45
In this case,
0,4
Q(h) = 0,5h
zero-buffer
0,35
0,3
0,25
0,2
0
50
100
150
200
250
300
350
Buffer capacity h
61
Numerical results
U = 1, l1 = 0.1, m2 = 0.1, l2 = 0.1
Throughput
m1 = 0,14
m1 = 0,12
m1 = 0,10
m1 = 0,08
m1 = 0,06
Buffer capacity
Discussions:
Why are the curves increasing?
Why do there reach an asymptote?
What is TH when N= 0?
What is the limit of TH as N tends to infinity?
Why are the curves with smaller m1 lower?
62
Numerical results
U = 1, l1 = 0.1, m2 = 0.1, l2 = 0.1
WIP
m1 = 0,14
m1 = 0,12
m1 = 0,10
m1 = 0,08
m1 = 0,06
Buffer capacity
Discussions:
• Why are the curves increasing?
• Why different asymptotes?
• What is the limit of WIP as N→?
• Why are the curves with smaller m1 lower?
63
Numerical results
U = 1, l1 = 0.1, m2 = 0.1, l2 = 0.1
Questions :
• If we want to increas production rate, which machine should
we improve?
• What would happen to production rate if we improved any
other machine?
64
Numerical results
U = 1, l1 = 0.1, m1 = 0.1, m2 = 0.1, l2 = 0.1
Throughput
Improvement to nonbottleneck machine.
Same graph for
improvement of
machine 2
Buffer capacity
65
Numerical results
U = 1, l1 = 0.1 , m1 = 0.1, m2 = 0.1, l2 = 0.1
Average inventory
Inventory increases as
the (non-bottleneck
upstream machine is
improved and as the
buffer space is
increased.
Buffer capacity
66
Numerical results
U = 1, l1 = 0.1 , m1 = 0.1, m2 = 0.1, l2 = 0.1
Average inventory
• Inventory decreases
as the (non-bottleneck)
downstream machine
is improved
•Inventory increases as
the buffer space
isincreased.
Buffer capacity
67
Numerical results
U = 1, m2 = 0.8, l2 = 0.09, h = 10
Should we prefer short and frequent disruptions or long
and infrequent disruptions?
Throughput
•
l1 and m1 vary together and
m1/(l1 + m1) = 0.9
• Answer: short and frequent
failures.
• Why?
Repair rate m1
68
Continuous flow models of failure-prone lines
Reversiblity
M1
B
M2
M1
B
M2
L
L'
Reversibility Theorem
(hold for any nb of machine and for all models)
E(L) = E(L')
Q(L) = h - Q(L')
ps(L) = pb(L')
pb(L) = ps(L')
Proof for the continuous L2
line
69
Continuous flow models of failure-prone lines
Dynamic behavior
u1(t)
M1
a1(t)
u2(t)
B
x(t), h
M2
a2(t)
A continuous time Markov process with hybrid state space
characterized by
Internal state distribution (0 < x < h):
Fa1a2(x) = P{a1(t) = a1, a2(t) = a2, 0 < x(t)  x}
fa1a2(x) = d Fa1a2(x) /dx
Boundary distribution:
Pa1a2(0), Pa1a2(h)
70
Case I1 = l1/m1  I2 = l2/m2
a
l2 m1  l1m2 l1  l2  m1  m2
U
( l1  l2 )( m1  m2 )
f10 ( x )  f 01 ( x )  Ce ax
f 00 ( x ) 
l1  l2 ax
m  m2 ax
Ce , f11 ( x )  1
Ce
m1  m2
l1  l2
P11 ( h ) 
U
l1
Ce ah , P11 ( 0 ) 
U
l2
C
l1  l2 ah
l1  l2
P10 ( h )  U
Ce , P01 ( 0 )  U
C
l1m2
l2 m1
P01 ( h )  P00 ( h )  P10 ( 0 )  P00 ( 0 )  0
1 U ( l1  l2 )  1  I 2 ah 1  I1 

e 


C l2 m1  l1m2  I1
I2 
71
Case I1 = l1/m1  I2 = l2/m2=I
f10 ( x )  f 01 ( x )  C0
C0
I
U
U
P11 ( h )  C0 , P11 ( 0 ) 
C0
f 00 ( x )  IC0 , f11 ( x ) 
l1
l2
l1  l2
l1  l2
P10 ( h )  U
C0 , P01 ( 0 )  U
C0
l1m2
l2 m1
P01 ( h )  P00 ( h )  P10 ( 0 )  P00 ( 0 )  0
l l
1
 U 1 2 (1  2 I ) 
C0
l1l2
(1  I )
I
2
h
72
Long failure-prone production lines
73
Introduction
M1
B1
M2
B2
M3
B3
M4
•
The performance evaluation of a general failure-prone line is difficult
due to the lack of analytical solution and the state space explosion.
•
The number of states for a M machines lines with buffers of capacity N
is about 2M(N+1)M-1. For M = 10 and N = 100, there are over 1021 states.
•
A so-called DDX decomposition method is capable of obtaining an
approximative but precise enough analytical estimation.
•
Other approximation methods exist but the DDX method is considered as
one of the most efficient ones and can be extended to other systems such
as assembly lines.
•
Focus on Continuous flow model but all results can be extended to
discrete flow models
74
Notation
M1
B1
M2
B2
M3
B3
M4
Given isolated machine performances:
Ii = li/mi
ei = 1/(1+Ii) : isolated efficiency of Mi
eiU : isolated productivity of Mi
Unknown system performance measures:
Ei : Probability that Mi is producing
THi = Ei U: throughput rate of Mi
psi : probability of starvation of Mi
pbi : proba of blockage of Mi
75
Aggregation method
Equivalent machine
M1
B1
M2
B2
M1
B1
M2 L12
M3
B3
M4
Replace L12 by a machine M12 of equivalent isolation
throughput rate (flow equivalence), i.e.
1
m12
eM 
 E ( L12 )
l12 
 m12
1  l12 / m12
E ( L12 )
12
Repair time of M12 = Average stoppage time of M2 in L12:
P(a 2  0)m2
ps( L12 )m1
1
1
1


m12 P(a 2  0)m2  ps( L12 )m1 m2 P(a 2  0)m2  ps( L12 )m1 m1
P(a 2  0)  1  E ( L12 )  ps( L12 )
76
Aggregation method
Equivalent machine
M1
B1
M2
M12
B2
B2
M3
B3
M4
M3
B3
M4
M123
B3
M4
M1234
Repeating the aggregation process leads to an approximated
estimation of the throughput of the line.
77
Decomposition method
Properties of a continuous line
M1
B1
M2
B2
M3
B3
M4
Flow conservation:
THi = TH1, i = 2, …, K
(1)
Ei = E1, i = 2, …, K
(2)
Flow-idle time relation:
Ei = ei (1- psi –pbi) , i = 2, …, K (3)
Proof:
Ei = P{ai(t) = 1 & Mi not blocked & Mi not starved}
= P{ai(t) = 1 | Mi not blocked & Mi not starved}. P{Mi not blocked & Mi not starved}
= ei (1- psi –pbi)
since the proba that Mi is blocked and starved simultaneously is null in continuous flow
model.
78
Decomposition method
Decomposition
Decompose a K-machine line into K-1 lines of two-machines
L:
L(1)
M1
B1
M2
l1
m1
h1
l2
m2
Mu(1)
B(1)
Md(1)
lu(1)
mu(1)
h1
ld(1)
md(1)
B2
h2
B3
M4
l3
m3
h3
l4
m4
Mu(i) = upstream subline of Bi
Md(i) = downstream subline of Bi
Mu(2)
B(2)
lu(2)
mu(2)
h2
Objective: the input/output flow of
B(i) is similar to that of Bi in L
M3
Md(2)
L(2)
ld(2)
md(2)
Mu(3)
B(3)
Md(3)
lu(3)
mu(3)
h3
ld(3)
md(3)
L(3)
79
Decomposition method
Decomposition
Notation :
Iu(i), eu(i), Id(i), ed(i)
E(i) : proba that Md(i) is producing
ps(i) : proba of starvation of Md(i)
pb(i) : proba. of blockage of Mu(i)
From the objective of decomposition:
E(i) = Ei+1, i = 1, …, K-1
(4)
ps(i) = psi+1 , i = 1, …, K-1
(5)
pb(i) = pbi , i = 1, …, K-1
(6)
80
Decomposition method
Decomposition
Apply(3) to L(i):
E(i) = eu(i)(1- pb(i)) , i = 1, …, K-1
(7)
E(i) = ed(i)(1- ps(i)) , i = 1, …, K-1 (
8)
Combine (2) & (4)
E(i) = E(1) , i = 1, …, K-1
(I)
Combine (3), (4), (5), (6),
E(i-1) = ei (1 - ps(i-1) - pb(i))
Combining with (7) & (8)
Id(i-1) + Iu(i) = 1/E(i-1) + Ii –1
(II)
81
Decomposition method
Decomposition
Repair time of Mu(i) :
A) Failure of Mu(i)
= failure of Mi, with prob. 1- a
= failure of Mu(i-1), with prob. a
B) Repair time of Mu(i)
MTTRu(i) = aMTTRu(i-1) + (1-a) MTTRi
1/mu(i) = a/mu(i-1) + (1-a) /mi
(9)
where a = percentage of stoppages of Mu(i) caused by a failure of Mu(i-1)
82
Decomposition method
Decomposition
C)
nb of flows interruptions of B(i-1)
=
a
ps ( i  1) mu ( i  1)
E ( i ) lu ( i )
nb of flow resumptions of B(i-1)
(10)
Nb of failures of Mu(i)
State-transition of Mu(i)
lu(i)
E(i) working
DOWN
mu(i)
idle
83
Decomposition method
Decomposition
Combine (9)-(10),

ps ( i  1) mu ( i  1)  1

 1 


mu ( i ) E ( i ) I u ( i ) mu ( i ) 
E ( i ) I u ( i ) mu ( i )  mi

ps ( i  1) 
ps ( i  1) mu ( i  1)
1 
 mi  mu ( i ) 
E
i
I
i
E ( i ) Iu ( i )
( ) u( )

1
ps ( i  1)
1
mu(i) = X. mu(i-1) + (1-X) mi
(III)
with X = ps(i-1) / (Iu(i).E(i)).
84
Decomposition method
Decomposition
Repair time of Md(i)
md(i) = Y. mu(i+1) + (1-Y) mi+1
(IV)
with Y = pb(i+1) / (Id(i).E(i)).
Boundary equations:
lu(1) = l1, mu(1) = m1, ld(K-1) = lK, mu(K-1) = mK, (V)
85
Decomposition method
Decomposition
Equation system (I) – (V),
(I)
E(i) = E(1) , i = 1, …, K-1
(II)
Id(i-1) + Iu(i) = 1/E(i-1) + Ii –1
(III)
mu(i) = Xmu(i-1) + (1-X) mi with
X
ps ( i  1)
Iu (i ) E (i )
pb ( i  1)
(IV)
md(i) = Ymu(i+1) + (1-Y) mi+1 with Y  I ( i ) E ( i )
d
(V)
lu(1) = l1, mu(1) = m1, ld(K-1) = lK, mu(K-1) = mK
• 4(K-1) equation
• 4(K-1) unknowns : lu(i), mu(i), ld(i), md(i)
• E(i), ps(i), pb(i) are functions of lu(i), mu(i), ld(i), md(i)
86
Decomposition method
Decomposition
DDX Algorithm:
Step 1: Initialisation lu(i) = li, mu(i) = mi, ld(i) = li+1, mu(i) = mi+1
Step 2: Forward update lu(i), mu(i) by equation (I)-(II)-(III)
For i = 2 to K-1, do
2.1 Evaluate the line L(i-1) to obtain E(i-1), ps(i-1), pb(i-1)
2.2 From (II), Iu(i) = 1/E(i-1) + Ii –1 - Id(i-1)
2.3 From (III)-(I), mu(i) = Xmu(i-1) + (1-X) mi with X = ps(i-1) / (Iu(i).E(i-1)).
Step 3: Backward update ld(i), md(i) with equations (I)-(II)-(IV)
For i = K-2 to 1, do
3.1 Evaluate the line L(i+1) to obtain E(i+1), ps(i+1), pb(i+1)
3.2 From (II), Id(i) = 1/E(i+1) + Ii+1 –1 - Iu(i+1)
3.3 From (IV)-(I), md(i) = Ymd(i+1) + (1-Y) mi+1 with Y = pb(i+1) / (Id(i).E(i+1))
Step 4: Repeat (2) – (3) till convergence, i.e. E(i) = E(1).
87
Distribution of material in a line with
average buffer
50 identical machines l = 0.01, m = 0.1, U = 1, hi = 20
88
Effect of bottleneck
average buffer
50 identical machines : l = 0.01, m = 0.1, U = 1, hi = 20
except bottleneck at M10 with l10 =0.0375
89
Increase one buffer capacity
8-machines with l = 0.09, m = 0.75, U = 1.2, hi = 30
except h6
Why buffer
increases and
which buffer
decreases?
buffer capacity h6
90
Distribution of buffer capacity
Which has a higher throughput rate?
9-machine line with two buffering options:
• 8 buffers equally sized
M1
B1
M2
B2
M3
B3
M4
B4
M5
B5
M6
B6
M7
B7
M8
B8
M9
• 2 buffers equally sized
M1
M2 M3
B3
M4
M5 M6
B6
M7
M8 M9
91
Distribution of buffer capacity
Throughput
All machines have
l = 0.001, m =
0.019, U = 1
What are the
asymptotes
Is 8 buffers always
faster?
Total buffer space
92
Distribution of buffer capacity
Throughput
Is 8 buffers always
faster?
Perhaps not, but the
difference is not
significant in
systems with very
small buffers.
Total buffer space
93
Design buffer space distribution
Design the buffers for a 20-machine production line
The machines have been selected, and the only decision
remaining is the amount of space to allocate for in-process
inventory.
The goal is to determine the smallest amount of inprocess inventory space so that the line meets a
production rate target.
94
Design buffer space distribution
The common operation time is one operation per minute.
The target production rate is 0.88 parts per minute.
95
Design buffer space distribution
Case 1 : MTBF = 200 minutes and MTTR = 10.5 minutes
for all machines (ei = 0.95 parts per minute)
Case 2 : Like Case 1 except Machine 5. For Machine 5,
MTBF = 100 and MTTR = 10.5 minutes (ei = 0.905 parts
per minute)
Case 3 : Like Case 1 except Machine 5. For Machine 5,
MTBF = 200 and MTTR = 21 minutes (ei = 0.905 parts per
minute)
96
Design buffer space distribution
Are buffers really needed?
Line
Production rate with no buffer
Case 1 0.487
Case 2 0.475
Case 3 0.475
Yes. How to compute these numbers? (homework)
97
Design buffer space distribution
Optimal buffer space distribution
Observation:
Buffer space is needed most where buffer level variability is
greatest!
98