CrystalBall in New Product Development
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Transcript of CrystalBall in New Product Development
04/08/2023R Balisnomo 1
Name: Ramon (Ray) BalisnomoTitle/Position: Design for Six Sigma Instructor/Engr.Company: Ingersoll Rand Securities Technologies
CONTACT INFORMATION:
E-mail: [email protected]@gmail.com
Telephone: 719-896-3604
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ANALYZING TEST LAB DATAEXERCISE 1 OF 3
Establishing Product Performance Targets
1
D-bolt
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Right-click your mouse button
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40.5 41.0 41.5 42.0 42.5 43.0 43.53950
4000
4050
4100
4150
4200
4250
f(x) = − 113.2296879306 x² + 9510.080581804 x − 195485.7849684R² = 0.891471798557738
ROCKWELL HARDNESS C
PULL
STR
ENG
TH (
FT-L
BS)
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Strength Frame machine was used for measuring the pull strength.
Objectives:
1.How hard should the face plate be for maximum strength?
2.What is certainty that the deadbolt will tolerate 4,150 ft-lbs of force?
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40.5 41.0 41.5 42.0 42.5 43.0 43.53950
4000
4050
4100
4150
4200
4250
f(x) = − 113.229687930558 x² + 9510.08058180449 x − 195485.78496844R² = 0.891471798557738
ROCKWELL HARDNESS C
PULL
STR
ENG
TH (
FT-L
BS)
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Let:T = Target Value for HardnessT = Min + AdjustmentAdjustment < RangeY = predicted pull strength @ T
Then, Y = -113.23T2 + 9510.1T - 195486
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40.5 41.0 41.5 42.0 42.5 43.0 43.53950
4000
4050
4100
4150
4200
4250
f(x) = − 113.229687930558 x² + 9510.08058180449 x − 195485.78496844R² = 0.891471798557738
ROCKWELL HARDNESS C
PULL
STR
ENG
TH (
FT-L
BS)
40.9
4065
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40.5 41.0 41.5 42.0 42.5 43.0 43.53950
4000
4050
4100
4150
4200
4250
f(x) = − 113.229687930558 x² + 9510.08058180449 x − 195485.78496844R² = 0.891471798557738
ROCKWELL HARDNESS C
PULL
STR
ENG
TH (
FT-L
BS)
Min X Max X
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40.5 41.0 41.5 42.0 42.5 43.0 43.53950
4000
4050
4100
4150
4200
4250
ROCKWELL HARDNESS C
PULL
STR
ENG
TH (
FT-L
BS)
Min X Max X
Optimal Y
Target Value
Adjustment <[Max-Min]
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40.5 41.0 41.5 42.0 42.5 43.0 43.53950
4000
4050
4100
4150
4200
4250
ROCKWELL HARDNESS C
PULL
STR
ENG
TH (
FT-L
BS)
Min X Max X
Optimal Y
Adjustment <[Max-Min]
Target Value
Performance Goal
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Strength Frame machine was used for measuring the pull strength.
Answers:1. How hard should the face plate be for maximum
strength? 42.0 Rockwell C for an expected pull strength of 4,200 ft-lbs.
2. What is certainty that the deadbolt will tolerate 4,150 ft-lbs of force? 99.84% @ 42.0 Rockwell C
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ORACLE Crystal Ball is needed in new product development because:
1. Errors are inherent in all predictive models (e.g., Linear Regression or Design of Experiments). We need a tool for modeling them (risks) so that our new products perform as expected.
2. Often times we are searching for that “Sweet Spot” or Optimal Point that lets our product perform at its peak best.
3. We want to know the probability of success (or failure) associated with our design or performance goals.
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LINEAR STATISTICAL STACK-UPSEXERCISE 2 OF 3
Robust Tolerance Design
2
ND lock
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Schlage ND-Series Heavy Duty Commercial Cylindrical Lever Lock
Based on the survey response of 107 customers in July 2009
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TechnicalDefinitionof Wobble
USL = 0
USL = 0.02
Vertical
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TechnicalDefinitionof Wobble
USL = 0
Vertical
HorizontalUSL = 0.02
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TechnicalDefinitionof Wobble
USL = 0
Vertical
Horizontal
AxialUSL = 0.03
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USL = 0
USL = 0.02
VerticalTechnicalDefinitionof Wobble
Use Copy & Paste for Horizontal & Axial, but don’t forget to change USL=0.03 for Axial
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USL = 0
USL = 0.02
VerticalTechnicalDefinitionof Wobble
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USL = 0
VerticalTechnicalDefinitionof Wobble
HorizontalUSL = 0.02
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USL = 0
VerticalTechnicalDefinitionof Wobble
Horizontal
AxialUSL = 0.03
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Financial Analysis of Wobble Problem
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USL = 0
USL = 0.02
VerticalTechnicalDefinitionof Wobble
Before After OptQuest2% improvement
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USL = 0
VerticalTechnicalDefinitionof Wobble
HorizontalUSL = 0.02
Before After OptQuest2% improvement
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USL = 0
VerticalTechnicalDefinitionof Wobble
Horizontal
AxialUSL = 0.03
Before After OptQuest73% improvement
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Financial Analysis of Wobble Problem
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Optional Step: If you only want to run 1 out of 4 possible Forecast (Outputs)
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99% @ $394
95% @ $634
88% @ $900
82% @ $106290% @
$835
86% @ $957
A
B
C
D
E
F
WhichHorseWould
YouBet On?
-$380
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99% Certain we can save at least $394K by changing the nominal drawing dimensions.
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ORACLE Crystal Ball® is needed in new product development because:
1. When it comes to tolerance stack-ups1 there is no better tool for optimizing fit and function. Humans typically cannot simultaneously adjust +40 interdependent variables at the same time.
2. It’s exponentially easier to fix the problem in the Design Phase then it is to fix the problem once the product has launched.
3. It’s a good tool for translating quality and engineering terms into language finance and management can understand.
1. note: Cartesian -- not GD&T
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EFFICIENT FRONTIEREXERCISE 3 OF 3
Cost Versus Quality3
Rekey
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3. Extract Current Security Control Key 2. Rotate 30 Degrees CCW
4. Insert New Security Control Key
NE
W
5. Rotate back to home position
1. Insert Current Security Control Key
NE
W
The New Way to Rekey a Lock
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E
B
A
D C
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1 = choose this supplier to make this part
0 = do NOT choose this supplierKey:
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“Best” SupplierCpk = 1.75 (99.99%)Unit Cost = $3.49
“Cheap” SupplierCpk = 0.59 (60.64%)Unit Cost = $0.99
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Objective:
1. If the requirement is 0.5<Cpk < 0.9, find the lowest cost supplier(s)
2. If the requirement is 0.9<Cpk < 1.3, find the lowest cost supplier(s)
3. If the requirement is 1.3<Cpk < 1.7, find the lowest cost supplier(s)
4. If the requirement is Cpk > 1.7, find the lowest cost supplier(s)
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The model does not know that only one of the four suppliers can produce each part. We have to specify this constraint into the analysis.
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This constraint only allows one supplier out of four to produce each part.
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Due to time constraints, we may be running the optimization for a much-less time (only 3-5 minutes); thus, your actual results may vary from this presentation.
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1,135 simulations @ 10,000 runs per simulation (11,350,000 data points) in 10 minutes, or 18,917 data points per second.
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(0.59, $0.99)
(0.92, $1.62)
(1.35, $2.55)
(1.72, $3.44)
Trade-off Curve between Quality and Unit Cost.
Due to a shorter number of runs, your results may vary from what is being shown here.
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ORACLE Crystal Ball® is needed in new product development because:
1. The trade-offs between RISK and REWARD are difficult to quantify. Efficient Frontiers is a tool that specializes in the above.
2. Simple YES/NO or GO/STOP decisions become complex in the face of so many combinations and permutations. Sometimes it’s better to let the a Computer make the long-term business decisions.
3. We can make a lot of money by solving big and difficult problems in a short amount of time.
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Name: Ramon (Ray) BalisnomoTitle/Position: Design for Six Sigma Instructor/Engr.Company: Ingersoll Rand Securities Technologies
CONTACT INFORMATION:
E-mail: [email protected]@gmail.com
Telephone: 719-896-3604
The EndDo you have any questions?