Transcript Basic Factory Dynamics

Basic Factory Dynamics
1
HAL Case
Large Panel Line: produces unpopulated printed circuit boards
Line runs 24 hr/day (but 19.5 hrs of productive time)
Recent Performance:
•
•
•
•
throughput = 1,400 panels per day (71.8 panels/hr)
WIP = 47,600 panels
CT = 34 days (663 hr at 19.5 hr/day)
customer service = 75% on-time delivery
Is HAL lean?
What data do we need to decide?
2
HAL - Large Panel Line Processes
Lamination (Cores): press copper and prepreg into core blanks
Machining: trim cores to size
Internal Circuitize: etch circuitry into copper of cores
Optical Test and Repair (Internal): scan panels optically for defects
Lamination (Composites): press cores into multiple layer boards
External Circuitize: etch circuitry into copper on outside of composites
Optical Test and Repair (External): scan composites optically for defects
Drilling: holes to provide connections between layers
Copper Plate: deposits copper in holes to establish connections
Procoat: apply plastic coating to protect boards
Sizing: cut panels into boards
End of Line Test: final electrical test
3
HAL Case - Science?
External Benchmarking
• but other plants may not be comparable
Internal Benchmarking
• capacity data: what is utilization?
• but this ignores WIP effects
Need relationships between WIP, TH, CT, service!
4
Definitions
Workstations: a collection of one or more identical machines.
Parts: a component, sub-assembly, or an assembly that moves through
the workstations.
End Items: parts sold directly to customers; relationship to constituent
parts defined in bill of material.
Consumables: bits, chemicals, gasses, etc., used in process but do not
become part of the product that is sold.
Routing: sequence of workstations needed to make a part.
Order: request from customer.
Job: transfer quantity on the line.
5
Definitions (cont.)
Throughput (TH): for a line, throughput is the average quantity of
good (non-defective) parts produced per unit time.
Work in Process (WIP): inventory between the start and endpoints of
a product routing.
Raw Material Inventory (RMI): material stocked at beginning of
routing.
Crib and Finished Goods Inventory (FGI): crib inventory is
material held in a stockpoint at the end of a routing; FGI is material
held in inventory prior to shipping to the customer.
Cycle Time (CT): time between release of the job at the beginning of
the routing until it reaches an inventory point at the end of the
routing.
6
Factory Physics
Definition: A manufacturing system is a goal-oriented network of
processes through which parts flow.
Structure: Plant is made up of routings (lines), which in turn are
made up of processes.
Focus: Factory Physics is concerned with the network and flows
at the routing (line) level.
7
Parameters
Descriptors of a Line:
1) Bottleneck Rate (rb): Rate (parts/unit time or jobs/unit time)
of the process center having the highest long-term utilization.
2) Raw Process Time (T0): Sum of the long-term average
process times of each station in the line.
3) Congestion Coefficient (): A unitless measure of
congestion.
• Zero variability case,  = 0.
• “Practical worst case,”  = 1.
• “Worst possible case,”  = W0.
Note: we won’t use  quantitatively,
but point it out to recognize that lines
with same rb and T0 can behave very
differently.
8
Parameters (cont.)
Relationship:
Critical WIP (W0): WIP level in which a line having no
congestion would achieve maximum throughput (i.e., rb)
with minimum cycle time (i.e., T0).
W0 = rb T0
9
The Penny Fab
Characteristics:
•
•
•
•
Four identical tools in series.
Each takes 2 hours per piece (penny).
No variability(Best Case).
CONWIP job releases.
Parameters:
rb
= 0.5 pennies/hour (2 hrs. per penny)
T0
=
8 hours
W0
=
0.5  8 = 4 pennies

=
0 (no variability, best case conditions)
10
The Penny Fab
11
The Penny Fab (WIP=1)
Time = 0 hours
12
The Penny Fab (WIP=1)
Time = 2 hours
13
The Penny Fab (WIP=1)
Time = 4 hours
14
The Penny Fab (WIP=1)
Time = 6 hours
15
The Penny Fab (WIP=1)
Time = 8 hours
16
The Penny Fab (WIP=1)
Time = 10 hours
17
The Penny Fab (WIP=1)
Time = 12 hours
18
The Penny Fab (WIP=1)
Time = 14 hours
19
The Penny Fab (WIP=1)
Time = 16 hours
Cycle Time = 8 hours
20
Penny Fab Performance
WIP
1
2
3
4
5
6
TH
0.125
CT
8
THCT
1
21
The Penny Fab (WIP=2)
Time = 0 hours
22
The Penny Fab (WIP=2)
Time = 2 hours
23
The Penny Fab (WIP=2)
Time = 4 hours
24
The Penny Fab (WIP=2)
Time = 6 hours
25
The Penny Fab (WIP=2)
Time = 8 hours
26
The Penny Fab (WIP=2)
Time = 10 hours
System Warmed Up ….. Start Keeping Track
27
The Penny Fab (WIP=2)
Time = 12 hours
28
The Penny Fab (WIP=2)
Time = 14 hours
29
The Penny Fab (WIP=2)
Time = 16 hours
1 Item Complete
30
The Penny Fab (WIP=2)
Time = 18 hours
2 Items Complete, so Throughput = 2/(18-10) = .25
Cycle Time = 8 hours
31
Penny Fab Performance
WIP
1
2
3
4
5
6
TH
0.125
0.250
CT
8
8
THCT
1
2
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The Penny Fab (WIP=4)
Time = 0 hours
33
The Penny Fab (WIP=4)
Time = 2 hours
34
The Penny Fab (WIP=4)
Time = 4 hours
35
The Penny Fab (WIP=4)
Time = 6 hours
36
The Penny Fab (WIP=4)
Time = 8 hours
System Warmed Up, Start Keeping Track
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The Penny Fab (WIP=4)
Time = 10 hours
1 Unit Complete
38
The Penny Fab (WIP=4)
Time = 12 hours
2 Units Complete
39
The Penny Fab (WIP=4)
Time = 14 hours
3 Units Complete, so Throughput = 3/(14-8) = .5
Cycle Time = 8 hours
40
Penny Fab Performance
WIP
1
2
3
4
5
6
TH
0.125
0.250
0.375
0.500
CT
8
8
8
8
THCT
1
2
3
4
41
The Penny Fab (WIP=5)
Time = 0 hours
42
The Penny Fab (WIP=5)
Time = 2 hours
43
The Penny Fab (WIP=5)
Time = 4 hours
44
The Penny Fab (WIP=5)
Time = 6 hours
45
The Penny Fab (WIP=5)
Time = 8 hours
System Warmed Up, Start Keeping Track
46
The Penny Fab (WIP=5)
Time = 10 hours
1 Unit Complete
47
The Penny Fab (WIP=5)
Time = 12 hours
2 Units Complete, so Throughput = 2/(12-8) = .5
Cycle Time = 10 Hours
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Penny Fab Performance
WIP
1
2
3
4
5
6
TH
0.125
0.250
0.375
0.500
0.500
0.500
CT
8
8
8
8
10
12
THCT
1
2
3
4
5
6
49
TH vs. WIP: Best Case
0.6
rb
0.5
TH
0.4
0.3
1/T0
0.2
0.1
0
0
1
2
3
4
W0
5
6
7
8
9 10 11 12
WIP
50
CT vs. WIP: Best Case
26
24
22
20
18
16
14
12
10
T0 86
4
2
0
CT
1/rb
0 1 2 3 4 5 6 7 8 9 10 11 12
W0 WIP
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Best Case Performance (No Variability)
Best Case Law: The minimum cycle time (CTbest) for a given
WIP level, w, is given by
CTbest
if w  W0
T0 ,

w / rb , otherwise.
The maximum throughput (THbest) for a given WIP level, w is
given by,
TH best
w / T0 , if w  W0

otherwise.
 rb ,
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Best Case Performance (cont.)
Example: For Penny Fab, rb = 0.5 and T0 = 8, so W0 = 0.5  8 = 4,
if w  4
 8,
CTbest  
w / .5, otherwise.
T Hbest
w / 8, if w  4

0.5, otherwise.
which are exactly the curves we plotted.
53
A Manufacturing Law
Little's Law: The fundamental relation between WIP, CT, and
TH over the long-term is:
WIP  TH  CT
parts
parts
 hr
hr
Insights:
• Fundamental relationship
• Simple units transformation
• Definition of cycle time (CT = WIP/TH)
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Penny Fab Two
2 hr
5 hr
3 hr
10 hr
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Penny Fab Two
Station
Number
1
Number of
Machines
1
Process
Time
2 hr
Station
Rate
0.5 j/hr
2
2
5 hr
3
6
10 hr
0.4 j/hr
0.6 j/hr
4
2
3 hr
0.67 j/hr
0.4 p/hr
20 hr
8 pennies
rb = ____________
T0 = ____________
W0 = ____________
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Penny Fab Two Simulation (Time=0)
2
2 hr
5 hr
3 hr
10 hr
57
Penny Fab Two Simulation (Time=2)
7
4
2 hr
5 hr
3 hr
10 hr
58
Penny Fab Two Simulation (Time=4)
7
6
9
2 hr
5 hr
3 hr
10 hr
59
Penny Fab Two Simulation (Time=6)
7
8
9
2 hr
5 hr
3 hr
10 hr
60
Penny Fab Two Simulation (Time=7)
17
12
8
9
2 hr
5 hr
3 hr
10 hr
61
Penny Fab Two Simulation (Time=8)
17
12
10
9
2 hr
5 hr
3 hr
10 hr
62
Penny Fab Two Simulation (Time=9)
17
19
12
10
14
2 hr
5 hr
3 hr
10 hr
63
Penny Fab Two Simulation (Time=10)
17
19
12
12
14
2 hr
5 hr
3 hr
10 hr
64
Penny Fab Two Simulation (Time=12)
17
19
17
22
14
14
2 hr
5 hr
3 hr
10 hr
65
Penny Fab Two Simulation (Time=14)
17
19
17
22
19
24
16
2 hr
5 hr
3 hr
10 hr
66
Penny Fab Two Simulation (Time=16)
17
19
17
22
19
24
2 hr
5 hr
3 hr
10 hr
67
Penny Fab Two Simulation (Time=17)
27
19
22
22
19
24
20
2 hr
5 hr
3 hr
10 hr
68
Penny Fab Two Simulation (Time=19)
27
29
22
22
20
24
24
22
2 hr
5 hr
3 hr
10 hr
69
Penny Fab Two Simulation (Time=20)
27
Note: job will arrive at
bottleneck just in time
to prevent starvation.
29
22
22
24
24
22
22
2 hr
5 hr
3 hr
10 hr
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Penny Fab Two Simulation (Time=22)
27
29
27
32
24
24
25
24
2 hr
5 hr
Note: job will arrive at
bottleneck just in time
to prevent starvation.
3 hr
10 hr
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Penny Fab Two Simulation (Time=24)
27
29
27
32
25
29
34
27
2 hr
5 hr
3 hr
10 hr
And so on….
Bottleneck will just
stay busy; all others
will starve periodically
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Worst Case
Observation: The Best Case yields the minimum cycle time and
maximum throughput for each WIP level.
Question: What conditions would cause the maximum cycle time
and minimum throughput?
Experiment:
• set average process times same as Best Case (so rb and T0
unchanged, so we are examining the same line)
• follow a marked job through system
• imagine marked job experiences maximum queueing (this would
occur if the first job took all the time and the other jobs took no
time)
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Worst Case Penny Fab
Time = 0 hours
Job 1 takes 8 hours, jobs 2,3,4 take 0 hours
rb = 4/(8 + 0 + 0 + 0) = .5, To= 32/4 = 8
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Worst Case Penny Fab
Time = 8 hours
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Worst Case Penny Fab
Time = 16 hours
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Worst Case Penny Fab
Time = 24 hours
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Worst Case Penny Fab
Time = 32 hours
Note:
CT = 32 hours
= 4 8 = wT0
TH = 4/32 = 1/8 = 1/T0
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TH vs. WIP: Worst Case
0.6
rb
Best Case
0.5
TH
0.4
0.3
0.2
1/T0
Worst Case
0.1
0
0
1
2
3
4
5
6
7
8
9 10 11 12
W0 WIP
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CT vs. WIP: Worst Case
Worst Case
CT
32
28
24
20
16
12
T0 8
4
0
Best Case
0 1 2 3 4 5 6 7 8 9 10 11 12
W0 WIP
80
Worst Case Performance
Worst Case Law: The worst case cycle time for a given WIP
level, w, is given by,
CTworst = w T0
The worst case throughput for a given WIP level, w, is given
by,
THworst = 1 / T0
Randomness? None - perfectly predictable, but bad!
81
Practical Worst Case
Observation: There is a BIG GAP between the Best Case and
Worst Case performance. Both cases are highly unlikely in
practice.
Question: Can we find an intermediate case that:
• divides “good” and “bad” lines, and
• is computable?
Experiment: consider a line with a given rb and T0 and:
• single machine stations
• balanced lines
• variability such that all WIP configurations (states) are equally
likely(worst case) … if actual system performs worse than this
improvements are certainly possible
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PWC Example – 3 jobs, 4 stations
clumped
up states
State
1
2
3
4
5
6
7
8
9
10
Vector
(3,0,0,0)
(0,3,0,0)
(0,0,3,0)
(0,0,0,3)
(2,1,0,0)
(2,0,1,0)
(2,0,0,1)
(1,2,0,0)
(0,2,1,0)
(0,2,0,1)
State
11
12
13
14
15
16
17
18
19
20
Vector
(1,0,2,0)
(0,1,2,0)
(0,0,2,1)
(1,0,0,2)
(0,1,0,2)
(0,0,1,2)
(1,1,1,0)
(1,1,0,1)
(1,0,1,1)
(0,1,1,1)
Note: average WIP at any station is 15/20 = 0.75,
so jobs are spread evenly between stations.
spread
out states
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Practical Worst Case
Let w = jobs in system, N = no. stations in line, and t =
process time at all stations:
CT(single): suppose you are a job, upon arrival to a machine
the time it takes to get through will be the time necessary to
process the other (w-1) jobs plus the time to process your job.
So,
CT(single) = t + ((w-1)/N)t
= (1 + (w-1)/N) t
CT(line)
TH
= N [1 + (w-1)/N] t
= Nt + (w-1)t
= T0 + (w-1)/rb, since t = 1/rb
From Little’s Law
= WIP/CT
= [w/(w+W0-1)]rb, since Wo = rb x To
84
Practical Worst Case Performance
Practical Worst Case Definition: The practical worst case
(PWC) cycle time for a given WIP level, w, is given by,
CTPWC  T0 
w 1
rb
The PWC throughput for a given WIP level, w, is given by,
TH PWC 
w
rb ,
W0  w  1
where W0 is the critical WIP.
85
TH vs. WIP: Practical Worst Case
0.6
rb
Best Case
0.5
TH
0.4
0.3
0.2
1/T0
PWC
Good (lean)
Bad (fat)
Worst Case
0.1
0
0
1
2
3
4
5
6
7
8
9 10 11 12
W0 WIP
86
CT vs. WIP: Practical Worst Case
CT
32
28
24
20
16
12
T0 8
4
0
Worst Case
PWC
Bad (fat)
Best Case
Good
(lean)
0 1 2 3 4 5 6 7 8 9 10 11 12
W0 WIP
87
Penny Fab Two Performance
0.5
Note: process
times in PF2
have var equal
to PWC.
Best Case
rb 0.4
0.3
But… unlike
PWC, it has
unbalanced
line and multi
machine
stations.
TH
0.2
0.1
1/T0
Worst Case
0
0
2
4
6
8
W0
10
12
14
16
18
20
22
24
26
WIP
88
Penny Fab Two Performance (cont.)
80
70
Worst Case
60
50
CT
40
1/rb
30
T0 20
Best Case
10
0
0
2
4
6
8
W0
10
12
14
16
18
20
22
24
26
WIP
89
Back to the HAL Case - Capacity Data
Process
Lamination
Machining
Internal Circuitize
Optical Test/Repair - Int
Lamination – Composites
External Circuitize
Optical Test/Repair - Ext
Drilling
Copper Plate
Procoat
Sizing
EOL Test
rb, T0
Rate (p/hr)
191.5
186.2
114.0
150.5
158.7
159.9
150.5
185.9
136.4
117.3
126.5
169.5
114.0
Time (hr)
4.7
0.5
3.6
1.0
2.0
4.3
1.0
10.2
1.0
4.1
1.1
0.5
33.9
90
HAL Case - Situation
Critical WIP: rbT0 = 114  33.9 = 3,869
Actual Values:
• CT = 34 days = 663 hours (at 19.5 hr/day)
• WIP = 47,600 panels
• TH = 71.8 panels/hour
Conclusions:
• Throughput is 63% of capacity (bottleneck rate)
• WIP is 12.3 times critical WIP
• CT is 24.1 times raw process time
91
HAL Case - Analysis
TH Resulting from PWC with WIP = 47,600?
TH 
w
47,600
rb 
114  105.4
w  W0  1
47,600  3,869  1
Much higher
than actual TH!
WIP Required for PWC to Achieve TH = 0.63rb?
TH 
w
rb  0.63rb
w  W0  1
0.63
0.36
w
(W0  1) 
(3,869 1)  6,586
0.37
0.37
Much lower than
actual WIP!
Conclusion: actual system is much worse than PWC!
92
HAL Internal Benchmarking Outcome
Throughput (panels/hour)
120.0
“Lean" Region
100.0
Current
TH = 71.8
WIP = 47,600
80.0
60.0
“Fat" Region
40.0
Best
Worst
PWC
20.0
0.0
0
10,000 20,000 30,000 40,000 50,000
WIP
93
Labor Constrained Systems
Motivation: performance of some systems are limited by labor or
a combination of labor and equipment.
Full Flexibility with Workers Tied to Jobs:
•
•
•
•
WIP limited by number of workers (n)
capacity of line is n/T0
Best case achieves capacity and has workers in “zones”
ample capacity case also achieves full capacity with “pick and run”
policy
94
Labor Constrained Systems (cont.)
Full Flexibility with Workers Not Tied to Jobs:
• TH depends on WIP levels
• THCW(n)  TH(w)  THCW(w)
• need policy to direct workers to jobs (focus on downstream is
effective)
Agile Workforce Systems
•
•
•
•
bucket brigades
kanban with shared tasks
worksharing with overlapping zones
many others
95
Factory Dynamics Takeaways
Performance Measures:
•
•
•
•
throughput
WIP
cycle time
service
Range of Cases:
• best case
• practical worst case
• worst case
Diagnostics:
• simple assessment based on rb, T0, actual WIP,actual TH
• evaluate relative to practical worst case
96