Exam Revisit (I)
•
Diagonal term vs. Off-diagonal term
•
Matrix rearrangement
⎧ FR1 ⎫ ⎡ X
⎪ FR 2⎪ ⎢ X
⎪ ⎢
⎪
⎬=
⎨
⎪ FR3 ⎪ ⎢ O
⎪⎩ FR 4⎪⎭ ⎢⎣ O
O
X
X
X
O
X
X
X
O ⎤ ⎧ DP1 ⎫
O ⎥⎥ ⎪⎪ DP 2⎪⎪
⎬
⎨
O ⎥ ⎪ DP3 ⎪
⎥
X ⎦ ⎪⎩ DP 4⎪⎭
Taesik Lee © 2005
Exam Revisit (II)
•
Allowable tolerance / Probability of Success
⎧ DPa ⎫
⎪ DPb ⎪
⎪
2 0 2 2 0 ⎤⎪
⎧ FR1 ⎫ ⎧2⎫ ⎡ 1
⎪⎪ DPc ⎪⎪
⎪
⎪ ⎪ ⎪ ⎢
⎥
⎨ FR 2⎬ = ⎨1 ⎬ = ⎢0.5 1 0 0 1 0 ⎥ ⎨
⎬
DPd
⎪ FR3 ⎪ ⎪3 ⎪ ⎢ 0.1 0.2 0 0 3 0.5⎥ ⎪
⎪
⎩
⎭ ⎩ ⎭ ⎣
⎦ ⎪ DPe ⎪
⎪
⎪
⎪⎩ DPf ⎪⎭
⎧ FR 2⎫ ⎧2⎫ ⎡0.5 0 0 ⎤ ⎧ DPa ⎫
⎪
⎪ ⎪ ⎪ ⎢
⎪
⎥⎪
⎨ FR1 ⎬ = ⎨1 ⎬ = ⎢ 1 2 0 ⎥ ⎨ DPd ⎬
⎪ FR3 ⎪ ⎪3 ⎪ ⎢ 0.1 0 0.5⎥ ⎪ DPf ⎪
⎭
⎦⎩
⎭ ⎩ ⎭ ⎣
⎩
∆DP 2 + = 2∆FR 2 + = 0.2
∆DP1+ = −0.5∆DP 2 + + 0.5∆FR1+ = −0.05
∆DP3 = −0.2∆DP 2 + 2∆FR3 = 0.16
Taesik Lee © 2005
Exam Revisit (III)
DP1
A
DP1
0.2
0.19
0.16
0.05
0.1
-0.2
0.2
-0.1
DP2
0.1
-0.2
-0.1 -0.05
0.05
0.2
DP2
-0.05
-0.19
-0.2
DP1
DP3
0.2
Joint p.d.f. for (DP1,DP2)
0.16
0.1
0.1
0.1
-0.2
0.2
-0.1
DP2
0.1
-0.2
0.2
-0.1
DP2
-0.1
-0.1
Joint p.d.f. for (DP2,DP3)
A
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-0.2
Design of Manufacturing Systems
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•
What is a manufacturing system?
Photo removed for copyright reasons.
Photo removed for copyright reasons.
1910… Ford Motor Company
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2010… Semiconductor Fab
Design of fixed manufacturing systems
for discrete identical parts
Small Scale Problems
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Ι. Simple deterministic scheduling problem
C
D
E
B
F
A
Robot
G
IN OUT
Part
Process module for X
X
Design a manufacturing
system
to eliminate the root cause
of a problem (symptom)
In/Out buffer
Taesik Lee © 2005
Photoresist processing
Vapor
Prime
VP chill
Adhesion promoter
Coating
Photoresist film
Soft Bake
SB chill
Solvent evaporation
Substrate
Developed image
(negative resist)
HB chill
Hard Bake
Developing
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Chemical reaction
in exposed area
PEB chill
PEB
(Post Exposure
Bake)
Exposure
Deterministic scheduling problem
Machine diagram removed for
copyright reasons.
D
C
E
B
F
Robot
A
G
IN OUT
Part
X
Process module for X
In/Out buffer
Taesik Lee © 2005
Process
Time (sec)
# of
modules
A
40
2
B
20
1
C
17
1
D
60
2
E
15
1
F
40
2
G
35
2
Level 1
FRs
#.1
#.2
Perform process steps
with desirable quality
Satisfy process flow
and throughput
DPs
Process modules
System configuration
⎡ FR1⎤ ⎡X
⎢FR2⎥ = ⎢
⎣
⎦ ⎣X
X ⎤ ⎡ DP1⎤
⎥⎢
⎥
X ⎦ ⎣DP2⎦
Level 2
FRs
#.1
Manage the recipe
#.2
Support the system
physically
Move wafer when
process is over
#.3
DPs
Recipe handling
module
System layout
Transport system
Taesik Lee © 2005
⎡ FR2.1⎤ ⎡X
⎢FR2.2⎥ = ⎢O
⎢
⎥ ⎢
⎢⎣ FR2.3⎥⎦ ⎢⎣X
O
X
X
O ⎤ ⎡ DP2.1⎤
X ⎥⎥ ⎢⎢DP2.2⎥⎥
X ⎥⎦ ⎢⎣ DP2.3⎥⎦
Level 3 - Sub FRs/DPs of FR2.1
FRs
#.1
#.2
#.3
Keep TAKTprocess below
TAKTsystem
Maintain # of moves by main
robot not to degrade target
throughput
Locate process modules into
200-APS frame
DPs
Number of each process
module
Number of IBTA
Layout (module
arrangement)
⎡ FR2.2.1⎤ ⎡X
⎢FR2.2.2⎥ = ⎢O
⎢
⎥ ⎢
⎢⎣ FR2.2.3⎥⎦ ⎢⎣X
O
X
X
O ⎤ ⎡ DP2.2.1⎤
X ⎥⎥ ⎢⎢DP2.2.2⎥⎥
X ⎥⎦ ⎢⎣ DP2.2.3⎥⎦
Level 3 - Sub FRs/DPs of FR2.2
FRs
#.1
#.2
#.3
#.4
#.
#.
Coordinate transport
function
Move wafer from CES
to VP
From VP to VPC
From VPC to CT
M
From HB to HBC
From HBC to CES
DPs)
Command and control
algorithm
CES handler
IBTA
Central handler
M
Central handler
SI handler
Taesik Lee © 2005
* Design matrix depends on
a process plan and selection
of DPs.
•
•
•
•
FR1: move wafer from process 1 to 2
FR2: move wafer from process 2 to 3
:
FR5: move wafer from process 5 to 6
•
•
DP1: robot 1
DP2: robot 2
•
•
•
t=0
t = t1
t = t2
FR = {FR1}
FR = {FR4}
FR = {FR2, FR3, FR5}
DP = {DP1}
DP = {DP2}
DP = {DP1, DP2}
Coupling due to an insufficient number of DPs
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D
C
•
Problem definition
B
– Conflict : more than one modules
competing for a robot
– The conflicts make the waiting time of
wafers inconsistent, which degrades onwafer result variation.
Robot
A
SP
T1
P2
P1
SP
T2
T1
P1
SP
SP
P1
SP
T2
T1
In/Out buffer P3
P2
P1
T2
T1
P3
P2
T2
P3
Delay time
T1
P1
P3
P2
Conflict
P1
T3
T2
T1
Process module for X
X
P3
P2
G
IN OUT
Part
P1
E
P2
T2
T1
P2
P1
SP
P3
T2
P3
T1
P2
P1
SP
T3
T2
T1
P1
T3
P3
P2
T1
T2
P2
P3
T2
Example : Process timing diagram with a sending period(6 unit)
Taesik Lee © 2005
T3
P3
Deterministic scheduling problem
i
i −1
i
j =1
j =0
j =1
t i = ∑ Pj + ∑ MvPk j + ∑ MvPl j + n ⋅ SP , n = 0,1,2,...
Dividing both sides by its SP yields
′
i
′
i −1
i
′
t i = ∑ Pj + ∑ MvPk j + ∑ MvPl j + n , n = 0,1,2,...
j =1
′
j =0
j =1
Original pick-up time
Taking only the decimal,
G
τ i = ti ′ − int(ti ′ )
moment of ith transport task
within a period
τD
E
Process
τi indicates the (normalized)
F
τC
D
C
B
A
IN
Conflict
0
0.5
1
1.5
Time (normalized by sending period)
Taesik Lee © 2005
2
Solution
•
Basic concept
– Break the conflicts among number of transport requests from process
modules
– Use predetermined “queue” as a decoupler between process and
transport
– Insert optimum queue at possible process steps
i
i −1
i
i
j =1
j =0
j =1
j =1
t = ∑ Pj + ∑ MvPk j + ∑ MvPl j + n ⋅ SP + ∑ q j , n = 0,1,2,...
*
i
Taesik Lee © 2005
Solution
Condition for no-conflict:
~ ≤ τ * −τ * ≤ − ~
tmax
1 tmax
i
j
Where
*
*
τ i −τ j = τ i −τ j +
i
∑
k =1
for i = 1,2,... , N ; j = 1,2,… , (i − 1)
qk ′ −
~
tmax : longest transport time
j
∑
qk ′ = τ i − τ j +
k =1
N
∑
(aik − a jk ) ⋅ qk'
k =1
Optimize values of qk along with sending period, subject to no-conflict
condition and process constraint (qcritical = 0 sec)
N
min
∑
q 'j
j =1
Taesik Lee © 2005
Solution
Process
Time (sec)
Delay (sec)
A
40
2
B
20
8
C
17
0
D
60
5
E
15
9
A’
40
9
F
35
3
Adjusted pick-up time
A
D
B
IN
0
0.5
1
1.5
2
Transforming a potentially combinatorial
complexity problem to a periodic problem
Solution is obtained for one (and repeating) period
Taesik Lee © 2005
Manufacturing Systems Design
Large Scale Problems
Taesik Lee © 2005
Customer’s view on Toyota products
Models rated
at or below average
GM
Chrysler
28
7
3
19
10
Ford
10
1
Nissan
Honda
Models rated
above average
*The Wall Street Journal
May 4, 2000
Toyota
5
1
8
1
10
• World’s No.2 Automaker
• $12B profit (2003)
• No1. JD Power Initial Quality Prize
• Market capitalization of Toyota ($104B) >
GM ($24B) + Ford ($23B) + DC ($37B) (2003.11.1)
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TPS / Lean manufacturing system
Set of 19 slides removed for copyright reasons.
Source: Production System Design presentation by Dr. David Cochran
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Conclusion
Cartoon removed for copyright reasons.
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