green belt project documentation x1707700.ppt
DESCRIPTION
green beltTRANSCRIPT
PowerPoint PresentationLalta Prasad Gupta
X1707700
NA
Last manufacturing process stage where technically the problem can get generated
Reactivation Process
Reactivation and Final Inspection
33.8 %
Maximum rejection in a month - 45.4 %
Minimum rejection in a month - 28.2 %
Slide number: *
1 Press
To reduce rejection from 33.8 % to 6.76 %
Annual Savings in Rs. Lakhs if the defect is made zero and horizontally deployed to other part numbers
0.43 Lakhs
1.30±0.1
No
Average Rejection = 33.8 %
Phase – 1- Problem Definition
Number of pieces rejected last month (for the part number identified for study)
1750 Nos
Nil
1750 Nos
Scrap cost/piece
Nil
0.032 Lakhs
0.032 Lakhs
0.43 Lakhs
Input Material
Legend: PC – Paired Comparison, PPS – Product/Process search, CS – Component Search, MCS – Modified Component Search, CC – Concentration chart
MVA – Multi-Vari analysis, OBSER – Observation, VS – Variable Search, FF – Full Factorial
SSV’s
Shainin DOE tool Selection
Slide number: *
Objective: Are Identified SSV’S the cause for the problem
Technique/Tool used: Product Process Search
Slide number: *
hino (2)
Sr. No.
hino
Sr. No.
hino (2)
Sr. No.
hino
Sr. No.
CALCULATION OF COUNT FOR BOTTOM FACING THICKNESS
TOP COUNT= 7
BOTTOM COUNT=5.5
TOTAL COUNT =12.5
New Spec = 0.475-0.595mm
So less than 75% of tolerance so can be controlled.
Hence its a root cause
hino (2)
Sr. No.
hino
Sr. No.
CALCULATION OF COUNT FOR CORE THICKNESS
TOP COUNT= 2
BOTTOM COUNT=1
TOTAL COUNT =3
As count is less than 6 so it is not a cause
hino (2)
Sr. No.
56.7
hino
Sr. No.
CALCULATION OF COUNT FOR TOP FACING THICKNESS
TOP COUNT= 5
BOTTOM COUNT=2
TOTAL COUNT =7
New Spec = 0.473-0.593mm
So less than 75% of tolerance so can be controlled.
Hence its a root cause
hino (2)
Sr. No.
56.7
hino
Sr. No.
Phase – 2- Measure and Analyze
Conclusion(s): Top Facing Thickness and Bottom Facing Thickness are Roots Causes Identified.
Slide number: *
Total Analysis summary
Cause(s) identified for the problem: Top Facing Thickness And Bottom Facing Thickness Are Causes Identified
Root cause(s) identified for the problem: Top Facing Thickness and Bottom Facing Thickness
Slide number: *
Validation Of Root Cause Before Improvement
Count is 16 so root cause is validated for derived spec.
Sheet3 (2)
Sr. No.
56.7
hino
Sr. No.
Phase – 2- Measure and Analyze
Count is 16 so root cause is validated for derived spec.
Sheet3 (2)
Sr. No.
56.7
hino
Sr. No.
Conclusion
As Top Facing And Bottom Facing Shares same specification in drawing so a common specification is concluded for both the facings.
Specification:- 0.53±0.06
Slide number: *
1
2
Yes / No
Slide number: *
Piece / Lot
Better ( B )
Current ( C )
2
5
7
No
Status of implementation
Parameter :
Step 2 : Identify are there multiple Product and Process streams
Number of streams :
Type of streams :
Phase – 4- Control
Data Grouping and Sample details
Number of Groups (Number of time blocks x number of streams)
Number of Samples in Each group (It is preferable to collect 5 samples continuously from the process so that the inherent variations are captured
Checking the consistency of part to part variation (Step 7)
Average Range (R-bar) (Round off to one decimal more than the data)
3
Upper Control Limit (UCL) = D4*R-bar (Round off to the same decimal as data)
6
0
Yes
If the Part to Part variation is not consistent, STOP, do not proceed further. Plan for DOE
Is the Range Chart plotted
Yes
Yes
If the stratification level is <=3, then the Part to Part variation is very less and the parameter does not require any monitoring. STOP. Do not proceed further
Are there 7 consequtive points increasing, decreasing and one side of the Mean range
Yes
If yes write causes (if possible)
If the range contains patterns as described above, STOP. Plan for DOE
Estimating Part to Part variation (Step 9)
Sigma = R-bar/d2 (Round off to one decimal more than data)
1
6 * Sigma (Round off to the same decimal as data)
6
Overall Average (Round off to one decimal more than data)
145
Number of Groups = Sqrt (Number of data points)
Sqrt(600)=25
6
1.0
Class Inverval (C.I) after rounding off to the same decimal as data
Ist group minimum+CL
Draw the Histogram here
Does the Overall Average lie in the group having maximum frequency or in the adjacent groups
Yes
Is there a gradual decreasing trend in frequency on both sides of the group having the maximum frequency
Yes
Are there two Modes (Two groups having maximum frequency) and both the groups are distinctly seperated
No
Based on the above analysis, can you conclude that the histogram is Normal
Yes
Note: If the Histogram is non-normal, the estimated rejection which we do in the next step may not be correct. Data transformation is required to correct non-normality
Rejection Analysis Using Sigma (Only for Normal distributions) (Step 11)
Average
145
0
USL
150
0
LSL
140
1.67
Rejection Analysis if the results above show rejection on one side higher than the other side
Target (Design)
USL
LSL
6 Sigma > 75% of Tolerance and <=100% of tolerance
6 Sigma > Tolerance
6 Sigma < Tolerance
6 Sigma = Tolerance
6 Sigma > Tolerance
Actual Part to Part variation <= 50% of tolerance
Actual Part to Part variation > 50% of tolerance
Use Average and Range chart for monitoring, but it has to be done with 100% inspection
Use DOE to reduce variation
0.005
Use Average and Range chart for monitoring
Do DOE to reduce variation or to question the tolerance for that parameter
Remove 100% inspection if done on this parameter
Calculation of limits for Average if the decision taken is Average and Range monitoring
Average
UCL = Average + A2*R-bar
LCL = Average - A2* R-bar
Select which of the following DOE tools will be used if 6 Sigma > Tolerance
DOE Tools selected for Process Improvement
Multi-Vari Analysis
Concentration Chart
Paired Comparison
Component Search
Product/Process Search
Variable Search
Full Factorial
EVOP (Evolutionary Operations) for Process Optimization
Analysis Completed by
&L&"Times New Roman,Regular"Rev: 1/Jan 2003&RPage: &P/&N
Samples D3 D4 2 0 3.267 3 0 2.575 4 0 2.282 5 0 2.115 6 0 2.004
Sample A2 2 1.880 3 1.023 4 0.729 5 0.577 6 0.483
Sample d2 2 1.128 3 1.693 4 2.059 5 2.326 6 2.534
Slide number: *
Step: 7
Slide number: *
Data Grouping and Sample details
Number of Groups (Number of time blocks x number of streams)
Time block * No of stream : 25*8 =200
Number of Samples in Each group (It is preferable to collect 5 samples continuously from the process so that the inherent variations are captured
5
Checking the consistency of part to part variation (Step 7)
Average Range (R-bar) (Round off to one decimal more than the data)
Upper Control Limit (UCL) = D4*R-bar (Round off to the same decimal as data)
Lower Control Limit (LCL) = D3*R-bar
0
Is the Part to Part variation consistent
If the Part to Part variation is not consistent, STOP, do not proceed further. Plan for DOE
Is the Range Chart plotted
Are the Stratification levels more than 3
If the stratification level is <=3, then the Part to Part variation is very less and the parameter does not require any monitoring. STOP. Do not proceed further
Are there 7 consequtive points increasing, decreasing and one side of the Mean range
If yes write causes (if possible)
If the range contains patterns as described above, STOP. Plan for DOE
Estimating Part to Part variation (Step 9)
Sigma = R-bar/d2 (Round off to one decimal more than data)
1
6 * Sigma (Round off to the same decimal as data)
6
Overall Average (Round off to one decimal more than data)
145
Number of Groups = Sqrt (Number of data points)
Sqrt(600)=25
6
1.0
Class Inverval (C.I) after rounding off to the same decimal as data
Ist group minimum+CL
Draw the Histogram here
Does the Overall Average lie in the group having maximum frequency or in the adjacent groups
Yes
Is there a gradual decreasing trend in frequency on both sides of the group having the maximum frequency
Yes
Are there two Modes (Two groups having maximum frequency) and both the groups are distinctly seperated
No
Based on the above analysis, can you conclude that the histogram is Normal
Yes
Note: If the Histogram is non-normal, the estimated rejection which we do in the next step may not be correct. Data transformation is required to correct non-normality
Rejection Analysis Using Sigma (Only for Normal distributions) (Step 11)
Average
145
0
USL
150
0
LSL
140
1.67
Rejection Analysis if the results above show rejection on one side higher than the other side
Target (Design)
USL
LSL
6 Sigma > 75% of Tolerance and <=100% of tolerance
6 Sigma > Tolerance
6 Sigma < Tolerance
6 Sigma = Tolerance
6 Sigma > Tolerance
Actual Part to Part variation <= 50% of tolerance
Actual Part to Part variation > 50% of tolerance
Use Average and Range chart for monitoring, but it has to be done with 100% inspection
Use DOE to reduce variation
0.005
Use Average and Range chart for monitoring
Do DOE to reduce variation or to question the tolerance for that parameter
Remove 100% inspection if done on this parameter
Calculation of limits for Average if the decision taken is Average and Range monitoring
Average
UCL = Average + A2*R-bar
LCL = Average - A2* R-bar
Select which of the following DOE tools will be used if 6 Sigma > Tolerance
DOE Tools selected for Process Improvement
Multi-Vari Analysis
Concentration Chart
Paired Comparison
Component Search
Product/Process Search
Variable Search
Full Factorial
EVOP (Evolutionary Operations) for Process Optimization
Analysis Completed by
&L&"Times New Roman,Regular"Rev: 1/Jan 2003&RPage: &P/&N
Samples D3 D4 2 0 3.267 3 0 2.575 4 0 2.282 5 0 2.115 6 0 2.004
Sample A2 2 1.880 3 1.023 4 0.729 5 0.577 6 0.483
Sample d2 2 1.128 3 1.693 4 2.059 5 2.326 6 2.534
Slide number: *
Estimated part to part variation at 99.73% confidence level is :
Sheet1
Data Grouping and Sample details
Number of Groups (Number of time blocks x number of streams)
Time block * No of stream : 25*8 =200
Number of Samples in Each group (It is preferable to collect 5 samples continuously from the process so that the inherent variations are captured
5
Checking the consistency of part to part variation (Step 7)
Average Range (R-bar) (Round off to one decimal more than the data)
3
Upper Control Limit (UCL) = D4*R-bar (Round off to the same decimal as data)
6
0
Yes
If the Part to Part variation is not consistent, STOP, do not proceed further. Plan for DOE
Is the Range Chart plotted
Yes
Yes
If the stratification level is <=3, then the Part to Part variation is very less and the parameter does not require any monitoring. STOP. Do not proceed further
Are there 7 consequtive points increasing, decreasing and one side of the Mean range
Yes
If yes write causes (if possible)
If the range contains patterns as described above, STOP. Plan for DOE
Estimating Part to Part variation (Step 9)
Sigma = R-bar/d2 (Round off to one decimal more than data)
6 * Sigma (Round off to the same decimal as data)
Overall Average (Round off to one decimal more than data)
Construction of Histogram (Step 10)
Number of Groups = Sqrt (Number of data points)
Sqrt(600)=25
6
1.0
Class Inverval (C.I) after rounding off to the same decimal as data
Ist group minimum+CL
Draw the Histogram here
Does the Overall Average lie in the group having maximum frequency or in the adjacent groups
Yes
Is there a gradual decreasing trend in frequency on both sides of the group having the maximum frequency
Yes
Are there two Modes (Two groups having maximum frequency) and both the groups are distinctly seperated
No
Based on the above analysis, can you conclude that the histogram is Normal
Yes
Note: If the Histogram is non-normal, the estimated rejection which we do in the next step may not be correct. Data transformation is required to correct non-normality
Rejection Analysis Using Sigma (Only for Normal distributions) (Step 11)
Average
145
0
USL
150
0
LSL
140
1.67
Rejection Analysis if the results above show rejection on one side higher than the other side
Target (Design)
USL
LSL
6 Sigma > 75% of Tolerance and <=100% of tolerance
6 Sigma > Tolerance
6 Sigma < Tolerance
6 Sigma = Tolerance
6 Sigma > Tolerance
Actual Part to Part variation <= 50% of tolerance
Actual Part to Part variation > 50% of tolerance
Use Average and Range chart for monitoring, but it has to be done with 100% inspection
Use DOE to reduce variation
0.005
Use Average and Range chart for monitoring
Do DOE to reduce variation or to question the tolerance for that parameter
Remove 100% inspection if done on this parameter
Calculation of limits for Average if the decision taken is Average and Range monitoring
Average
UCL = Average + A2*R-bar
LCL = Average - A2* R-bar
Select which of the following DOE tools will be used if 6 Sigma > Tolerance
DOE Tools selected for Process Improvement
Multi-Vari Analysis
Concentration Chart
Paired Comparison
Component Search
Product/Process Search
Variable Search
Full Factorial
EVOP (Evolutionary Operations) for Process Optimization
Analysis Completed by
&L&"Times New Roman,Regular"Rev: 1/Jan 2003&RPage: &P/&N
Samples D3 D4 2 0 3.267 3 0 2.575 4 0 2.282 5 0 2.115 6 0 2.004
Sample A2 2 1.880 3 1.023 4 0.729 5 0.577 6 0.483
Sample d2 2 1.128 3 1.693 4 2.059 5 2.326 6 2.534
Slide number: *
Data Grouping and Sample details
Number of Groups (Number of time blocks x number of streams)
Time block * No of stream : 25*8 =200
Number of Samples in Each group (It is preferable to collect 5 samples continuously from the process so that the inherent variations are captured
5
Checking the consistency of part to part variation (Step 7)
Average Range (R-bar) (Round off to one decimal more than the data)
3
Upper Control Limit (UCL) = D4*R-bar (Round off to the same decimal as data)
6
0
Yes
If the Part to Part variation is not consistent, STOP, do not proceed further. Plan for DOE
Is the Range Chart plotted
Yes
Yes
If the stratification level is <=3, then the Part to Part variation is very less and the parameter does not require any monitoring. STOP. Do not proceed further
Are there 7 consequtive points increasing, decreasing and one side of the Mean range
Yes
If yes write causes (if possible)
If the range contains patterns as described above, STOP. Plan for DOE
Estimating Part to Part variation (Step 9)
Sigma = R-bar/d2 (Round off to one decimal more than data)
1
6 * Sigma (Round off to the same decimal as data)
6
Overall Average (Round off to one decimal more than data)
145
Number of Groups = Sqrt (Number of data points)
Range = Maximum Value - Minimum Value
Class Interval (C.I) = Range/Number of groups
Class Inverval (C.I) after rounding off to the same decimal as data
Draw the Histogram here
Does the Overall Average lie in the group having maximum frequency or in the adjacent groups
Is there a gradual decreasing trend in frequency on both sides of the group having the maximum frequency
Are there two Modes (Two groups having maximum frequency) and both the groups are distinctly seperated
Based on the above analysis, can you conclude that the histogram is Normal
Note: If the Histogram is non-normal, the estimated rejection which we do in the next step may not be correct. Data transformation is required to correct non-normality
Rejection Analysis Using Sigma (Only for Normal distributions) (Step 11)
Average
145
0
USL
150
0
LSL
140
1.67
Rejection Analysis if the results above show rejection on one side higher than the other side
Target (Design)
USL
LSL
6 Sigma > 75% of Tolerance and <=100% of tolerance
6 Sigma > Tolerance
6 Sigma < Tolerance
6 Sigma = Tolerance
6 Sigma > Tolerance
Actual Part to Part variation <= 50% of tolerance
Actual Part to Part variation > 50% of tolerance
Use Average and Range chart for monitoring, but it has to be done with 100% inspection
Use DOE to reduce variation
0.005
Use Average and Range chart for monitoring
Do DOE to reduce variation or to question the tolerance for that parameter
Remove 100% inspection if done on this parameter
Calculation of limits for Average if the decision taken is Average and Range monitoring
Average
UCL = Average + A2*R-bar
LCL = Average - A2* R-bar
Select which of the following DOE tools will be used if 6 Sigma > Tolerance
DOE Tools selected for Process Improvement
Multi-Vari Analysis
Concentration Chart
Paired Comparison
Component Search
Product/Process Search
Variable Search
Full Factorial
EVOP (Evolutionary Operations) for Process Optimization
Analysis Completed by
&L&"Times New Roman,Regular"Rev: 1/Jan 2003&RPage: &P/&N
Samples D3 D4 2 0 3.267 3 0 2.575 4 0 2.282 5 0 2.115 6 0 2.004
Sample A2 2 1.880 3 1.023 4 0.729 5 0.577 6 0.483
Sample d2 2 1.128 3 1.693 4 2.059 5 2.326 6 2.534
Slide number: *
Data Grouping and Sample details
Number of Groups (Number of time blocks x number of streams)
Time block * No of stream : 25*8 =200
Number of Samples in Each group (It is preferable to collect 5 samples continuously from the process so that the inherent variations are captured
5
Checking the consistency of part to part variation (Step 7)
Average Range (R-bar) (Round off to one decimal more than the data)
3
Upper Control Limit (UCL) = D4*R-bar (Round off to the same decimal as data)
6
0
Yes
If the Part to Part variation is not consistent, STOP, do not proceed further. Plan for DOE
Is the Range Chart plotted
Yes
Yes
If the stratification level is <=3, then the Part to Part variation is very less and the parameter does not require any monitoring. STOP. Do not proceed further
Are there 7 consequtive points increasing, decreasing and one side of the Mean range
Yes
If yes write causes (if possible)
If the range contains patterns as described above, STOP. Plan for DOE
Estimating Part to Part variation (Step 9)
Sigma = R-bar/d2 (Round off to one decimal more than data)
1
6 * Sigma (Round off to the same decimal as data)
6
Overall Average (Round off to one decimal more than data)
145
Number of Groups = Sqrt (Number of data points)
Sqrt(600)=25
6
1.0
Class Inverval (C.I) after rounding off to the same decimal as data
Ist group minimum+CL
Draw the Histogram here
Does the Overall Average lie in the group having maximum frequency or in the adjacent groups
Yes
Is there a gradual decreasing trend in frequency on both sides of the group having the maximum frequency
Yes
Are there two Modes (Two groups having maximum frequency) and both the groups are distinctly seperated
No
Based on the above analysis, can you conclude that the histogram is Normal
Yes
Note: If the Histogram is non-normal, the estimated rejection which we do in the next step may not be correct. Data transformation is required to correct non-normality
Rejection Analysis Using Sigma (Only for Normal distributions) (Step 11)
Average
145
0
USL
150
0
LSL
140
1.67
Rejection Analysis if the results above show rejection on one side higher than the other side
Target (Design)
USL
LSL
6 Sigma > 75% of Tolerance and <=100% of tolerance
6 Sigma > Tolerance
6 Sigma < Tolerance
6 Sigma = Tolerance
6 Sigma > Tolerance
Actual Part to Part variation <= 50% of tolerance
Actual Part to Part variation > 50% of tolerance
Use Average and Range chart for monitoring, but it has to be done with 100% inspection
Use DOE to reduce variation
Use Pre-control chart for monotoring
Use Average and Range chart for monitoring
Do DOE to reduce variation or to question the tolerance for that parameter
Remove 100% inspection if done on this parameter
Calculation of limits for Average if the decision taken is Average and Range monitoring
Average
UCL = Average + A2*R-bar
LCL = Average - A2* R-bar
Select which of the following DOE tools will be used if 6 Sigma > Tolerance
DOE Tools selected for Process Improvement
Multi-Vari Analysis
Concentration Chart
Paired Comparison
Component Search
Product/Process Search
Variable Search
Full Factorial
EVOP (Evolutionary Operations) for Process Optimization
Analysis Completed by
&L&"Times New Roman,Regular"Rev: 1/Jan 2003&RPage: &P/&N
Samples D3 D4 2 0 3.267 3 0 2.575 4 0 2.282 5 0 2.115 6 0 2.004
Sample A2 2 1.880 3 1.023 4 0.729 5 0.577 6 0.483
Sample d2 2 1.128 3 1.693 4 2.059 5 2.326 6 2.534
Slide number: *
Project Summary
Number of drill down’s (funneling) done to reach the root cause(s):
Number of predominant root cause(s) identified:
Time in months for completing the project:
Tools/Techniques used:
Slide number: *
Overall scrap reduced from 1.95% to 1.02%.
2. Crack scrap reduced from 0.62% to 0.02%.
Cycle time reduced from 460 seconds to 422 seconds.
Estimated cost benefits is Rs. 2,18,996 per year.
Slide number: *
blocks x number of streams)
Part Details
Number of Samples in Each group (It
is preferable to collect 5 samples
continuously from the process so
that the inherent variations are
captured
0
If yes
Is the Range Chart plotted
If the stratification level is <=3, then the Part to Part variation is very less and the
parameter does not require any monitoring. STOP. Do not proceed further
Are there 7 consequtive points increasing,
decreasing and one side of the Mean range
If the range contains patterns as described above, STOP. Plan for DOE
Lower Control Limit (LCL) = D3*R-bar
Is the Part to Part variation consistent
If the Part to Part variation is not consistent, STOP, do not proceed further. Plan for
DOE
Checking the consistency of part to part variation (Step 7)
Average Range (R-bar) (Round off to one
decimal more than the data)
Upper Control Limit (UCL) = D4*R-bar
(Round off to the same decimal as data)
Samples D3 D4
2 0 3.267
3 0 2.575
4 0 2.282
5 0 2.115
6 0 2.004
than data)
6 * Sigma (Round off to the same decimal as data)
Estimating Part to Part variation (Step 9)
Overall Average (Round off to one decimal more
than data)
Sample d2
2 1.128
3 1.693
4 2.059
5 2.326
6 2.534
Range = Maximum Value - Minimum Value
Construction of Histogram (Step 10)Number of Groups = Sqrt (Number of data points)
Class Interval (C.I) = Range/Number of groups
Are there two Modes (Two groups having maximum
frequency) and both the groups are distinctly seperated
Class Inverval (C.I) after rounding off to the same decimal
as data
Does the Overall Average lie in the group having maximum
frequency or in the adjacent groups
Draw the Histogram here
Based on the above analysis, can you conclude that the
histogram is Normal
Is there a gradual decreasing trend in frequency on both
sides of the group having the maximum frequency
Average
Sigma
Rejection Analysis Using Sigma (Only for Normal distributions) (Step 11)
Cpk = Zusl/3,Zlsl/3
Projected Rejection % below LSL
Cpk = Zusl/3,Zlsl/3
Projected Rejection % below LSL
(Based on the Normal table)
Rejection Analysis if the results above show rejection on one side higher than the other side
Z(USL) = (USL - Target)/Sigma
Z(LSL) = (LSL - Target)/Sigma
Actual Part to
this parameter
monitoring, but it has to be done with
100% inspection
Do DOE to reduce variation or to
question the tolerance for that parameter
6 Sigma > Tolerance
6 Sigma <= 75% of Tolerance
6 Sigma > 75% of Tolerance and <=100% of tolerance
Sr. No.
Sr. No.
Sr. No.
X1707700
NA
Last manufacturing process stage where technically the problem can get generated
Reactivation Process
Reactivation and Final Inspection
33.8 %
Maximum rejection in a month - 45.4 %
Minimum rejection in a month - 28.2 %
Slide number: *
1 Press
To reduce rejection from 33.8 % to 6.76 %
Annual Savings in Rs. Lakhs if the defect is made zero and horizontally deployed to other part numbers
0.43 Lakhs
1.30±0.1
No
Average Rejection = 33.8 %
Phase – 1- Problem Definition
Number of pieces rejected last month (for the part number identified for study)
1750 Nos
Nil
1750 Nos
Scrap cost/piece
Nil
0.032 Lakhs
0.032 Lakhs
0.43 Lakhs
Input Material
Legend: PC – Paired Comparison, PPS – Product/Process search, CS – Component Search, MCS – Modified Component Search, CC – Concentration chart
MVA – Multi-Vari analysis, OBSER – Observation, VS – Variable Search, FF – Full Factorial
SSV’s
Shainin DOE tool Selection
Slide number: *
Objective: Are Identified SSV’S the cause for the problem
Technique/Tool used: Product Process Search
Slide number: *
hino (2)
Sr. No.
hino
Sr. No.
hino (2)
Sr. No.
hino
Sr. No.
CALCULATION OF COUNT FOR BOTTOM FACING THICKNESS
TOP COUNT= 7
BOTTOM COUNT=5.5
TOTAL COUNT =12.5
New Spec = 0.475-0.595mm
So less than 75% of tolerance so can be controlled.
Hence its a root cause
hino (2)
Sr. No.
hino
Sr. No.
CALCULATION OF COUNT FOR CORE THICKNESS
TOP COUNT= 2
BOTTOM COUNT=1
TOTAL COUNT =3
As count is less than 6 so it is not a cause
hino (2)
Sr. No.
56.7
hino
Sr. No.
CALCULATION OF COUNT FOR TOP FACING THICKNESS
TOP COUNT= 5
BOTTOM COUNT=2
TOTAL COUNT =7
New Spec = 0.473-0.593mm
So less than 75% of tolerance so can be controlled.
Hence its a root cause
hino (2)
Sr. No.
56.7
hino
Sr. No.
Phase – 2- Measure and Analyze
Conclusion(s): Top Facing Thickness and Bottom Facing Thickness are Roots Causes Identified.
Slide number: *
Total Analysis summary
Cause(s) identified for the problem: Top Facing Thickness And Bottom Facing Thickness Are Causes Identified
Root cause(s) identified for the problem: Top Facing Thickness and Bottom Facing Thickness
Slide number: *
Validation Of Root Cause Before Improvement
Count is 16 so root cause is validated for derived spec.
Sheet3 (2)
Sr. No.
56.7
hino
Sr. No.
Phase – 2- Measure and Analyze
Count is 16 so root cause is validated for derived spec.
Sheet3 (2)
Sr. No.
56.7
hino
Sr. No.
Conclusion
As Top Facing And Bottom Facing Shares same specification in drawing so a common specification is concluded for both the facings.
Specification:- 0.53±0.06
Slide number: *
1
2
Yes / No
Slide number: *
Piece / Lot
Better ( B )
Current ( C )
2
5
7
No
Status of implementation
Parameter :
Step 2 : Identify are there multiple Product and Process streams
Number of streams :
Type of streams :
Phase – 4- Control
Data Grouping and Sample details
Number of Groups (Number of time blocks x number of streams)
Number of Samples in Each group (It is preferable to collect 5 samples continuously from the process so that the inherent variations are captured
Checking the consistency of part to part variation (Step 7)
Average Range (R-bar) (Round off to one decimal more than the data)
3
Upper Control Limit (UCL) = D4*R-bar (Round off to the same decimal as data)
6
0
Yes
If the Part to Part variation is not consistent, STOP, do not proceed further. Plan for DOE
Is the Range Chart plotted
Yes
Yes
If the stratification level is <=3, then the Part to Part variation is very less and the parameter does not require any monitoring. STOP. Do not proceed further
Are there 7 consequtive points increasing, decreasing and one side of the Mean range
Yes
If yes write causes (if possible)
If the range contains patterns as described above, STOP. Plan for DOE
Estimating Part to Part variation (Step 9)
Sigma = R-bar/d2 (Round off to one decimal more than data)
1
6 * Sigma (Round off to the same decimal as data)
6
Overall Average (Round off to one decimal more than data)
145
Number of Groups = Sqrt (Number of data points)
Sqrt(600)=25
6
1.0
Class Inverval (C.I) after rounding off to the same decimal as data
Ist group minimum+CL
Draw the Histogram here
Does the Overall Average lie in the group having maximum frequency or in the adjacent groups
Yes
Is there a gradual decreasing trend in frequency on both sides of the group having the maximum frequency
Yes
Are there two Modes (Two groups having maximum frequency) and both the groups are distinctly seperated
No
Based on the above analysis, can you conclude that the histogram is Normal
Yes
Note: If the Histogram is non-normal, the estimated rejection which we do in the next step may not be correct. Data transformation is required to correct non-normality
Rejection Analysis Using Sigma (Only for Normal distributions) (Step 11)
Average
145
0
USL
150
0
LSL
140
1.67
Rejection Analysis if the results above show rejection on one side higher than the other side
Target (Design)
USL
LSL
6 Sigma > 75% of Tolerance and <=100% of tolerance
6 Sigma > Tolerance
6 Sigma < Tolerance
6 Sigma = Tolerance
6 Sigma > Tolerance
Actual Part to Part variation <= 50% of tolerance
Actual Part to Part variation > 50% of tolerance
Use Average and Range chart for monitoring, but it has to be done with 100% inspection
Use DOE to reduce variation
0.005
Use Average and Range chart for monitoring
Do DOE to reduce variation or to question the tolerance for that parameter
Remove 100% inspection if done on this parameter
Calculation of limits for Average if the decision taken is Average and Range monitoring
Average
UCL = Average + A2*R-bar
LCL = Average - A2* R-bar
Select which of the following DOE tools will be used if 6 Sigma > Tolerance
DOE Tools selected for Process Improvement
Multi-Vari Analysis
Concentration Chart
Paired Comparison
Component Search
Product/Process Search
Variable Search
Full Factorial
EVOP (Evolutionary Operations) for Process Optimization
Analysis Completed by
&L&"Times New Roman,Regular"Rev: 1/Jan 2003&RPage: &P/&N
Samples D3 D4 2 0 3.267 3 0 2.575 4 0 2.282 5 0 2.115 6 0 2.004
Sample A2 2 1.880 3 1.023 4 0.729 5 0.577 6 0.483
Sample d2 2 1.128 3 1.693 4 2.059 5 2.326 6 2.534
Slide number: *
Step: 7
Slide number: *
Data Grouping and Sample details
Number of Groups (Number of time blocks x number of streams)
Time block * No of stream : 25*8 =200
Number of Samples in Each group (It is preferable to collect 5 samples continuously from the process so that the inherent variations are captured
5
Checking the consistency of part to part variation (Step 7)
Average Range (R-bar) (Round off to one decimal more than the data)
Upper Control Limit (UCL) = D4*R-bar (Round off to the same decimal as data)
Lower Control Limit (LCL) = D3*R-bar
0
Is the Part to Part variation consistent
If the Part to Part variation is not consistent, STOP, do not proceed further. Plan for DOE
Is the Range Chart plotted
Are the Stratification levels more than 3
If the stratification level is <=3, then the Part to Part variation is very less and the parameter does not require any monitoring. STOP. Do not proceed further
Are there 7 consequtive points increasing, decreasing and one side of the Mean range
If yes write causes (if possible)
If the range contains patterns as described above, STOP. Plan for DOE
Estimating Part to Part variation (Step 9)
Sigma = R-bar/d2 (Round off to one decimal more than data)
1
6 * Sigma (Round off to the same decimal as data)
6
Overall Average (Round off to one decimal more than data)
145
Number of Groups = Sqrt (Number of data points)
Sqrt(600)=25
6
1.0
Class Inverval (C.I) after rounding off to the same decimal as data
Ist group minimum+CL
Draw the Histogram here
Does the Overall Average lie in the group having maximum frequency or in the adjacent groups
Yes
Is there a gradual decreasing trend in frequency on both sides of the group having the maximum frequency
Yes
Are there two Modes (Two groups having maximum frequency) and both the groups are distinctly seperated
No
Based on the above analysis, can you conclude that the histogram is Normal
Yes
Note: If the Histogram is non-normal, the estimated rejection which we do in the next step may not be correct. Data transformation is required to correct non-normality
Rejection Analysis Using Sigma (Only for Normal distributions) (Step 11)
Average
145
0
USL
150
0
LSL
140
1.67
Rejection Analysis if the results above show rejection on one side higher than the other side
Target (Design)
USL
LSL
6 Sigma > 75% of Tolerance and <=100% of tolerance
6 Sigma > Tolerance
6 Sigma < Tolerance
6 Sigma = Tolerance
6 Sigma > Tolerance
Actual Part to Part variation <= 50% of tolerance
Actual Part to Part variation > 50% of tolerance
Use Average and Range chart for monitoring, but it has to be done with 100% inspection
Use DOE to reduce variation
0.005
Use Average and Range chart for monitoring
Do DOE to reduce variation or to question the tolerance for that parameter
Remove 100% inspection if done on this parameter
Calculation of limits for Average if the decision taken is Average and Range monitoring
Average
UCL = Average + A2*R-bar
LCL = Average - A2* R-bar
Select which of the following DOE tools will be used if 6 Sigma > Tolerance
DOE Tools selected for Process Improvement
Multi-Vari Analysis
Concentration Chart
Paired Comparison
Component Search
Product/Process Search
Variable Search
Full Factorial
EVOP (Evolutionary Operations) for Process Optimization
Analysis Completed by
&L&"Times New Roman,Regular"Rev: 1/Jan 2003&RPage: &P/&N
Samples D3 D4 2 0 3.267 3 0 2.575 4 0 2.282 5 0 2.115 6 0 2.004
Sample A2 2 1.880 3 1.023 4 0.729 5 0.577 6 0.483
Sample d2 2 1.128 3 1.693 4 2.059 5 2.326 6 2.534
Slide number: *
Estimated part to part variation at 99.73% confidence level is :
Sheet1
Data Grouping and Sample details
Number of Groups (Number of time blocks x number of streams)
Time block * No of stream : 25*8 =200
Number of Samples in Each group (It is preferable to collect 5 samples continuously from the process so that the inherent variations are captured
5
Checking the consistency of part to part variation (Step 7)
Average Range (R-bar) (Round off to one decimal more than the data)
3
Upper Control Limit (UCL) = D4*R-bar (Round off to the same decimal as data)
6
0
Yes
If the Part to Part variation is not consistent, STOP, do not proceed further. Plan for DOE
Is the Range Chart plotted
Yes
Yes
If the stratification level is <=3, then the Part to Part variation is very less and the parameter does not require any monitoring. STOP. Do not proceed further
Are there 7 consequtive points increasing, decreasing and one side of the Mean range
Yes
If yes write causes (if possible)
If the range contains patterns as described above, STOP. Plan for DOE
Estimating Part to Part variation (Step 9)
Sigma = R-bar/d2 (Round off to one decimal more than data)
6 * Sigma (Round off to the same decimal as data)
Overall Average (Round off to one decimal more than data)
Construction of Histogram (Step 10)
Number of Groups = Sqrt (Number of data points)
Sqrt(600)=25
6
1.0
Class Inverval (C.I) after rounding off to the same decimal as data
Ist group minimum+CL
Draw the Histogram here
Does the Overall Average lie in the group having maximum frequency or in the adjacent groups
Yes
Is there a gradual decreasing trend in frequency on both sides of the group having the maximum frequency
Yes
Are there two Modes (Two groups having maximum frequency) and both the groups are distinctly seperated
No
Based on the above analysis, can you conclude that the histogram is Normal
Yes
Note: If the Histogram is non-normal, the estimated rejection which we do in the next step may not be correct. Data transformation is required to correct non-normality
Rejection Analysis Using Sigma (Only for Normal distributions) (Step 11)
Average
145
0
USL
150
0
LSL
140
1.67
Rejection Analysis if the results above show rejection on one side higher than the other side
Target (Design)
USL
LSL
6 Sigma > 75% of Tolerance and <=100% of tolerance
6 Sigma > Tolerance
6 Sigma < Tolerance
6 Sigma = Tolerance
6 Sigma > Tolerance
Actual Part to Part variation <= 50% of tolerance
Actual Part to Part variation > 50% of tolerance
Use Average and Range chart for monitoring, but it has to be done with 100% inspection
Use DOE to reduce variation
0.005
Use Average and Range chart for monitoring
Do DOE to reduce variation or to question the tolerance for that parameter
Remove 100% inspection if done on this parameter
Calculation of limits for Average if the decision taken is Average and Range monitoring
Average
UCL = Average + A2*R-bar
LCL = Average - A2* R-bar
Select which of the following DOE tools will be used if 6 Sigma > Tolerance
DOE Tools selected for Process Improvement
Multi-Vari Analysis
Concentration Chart
Paired Comparison
Component Search
Product/Process Search
Variable Search
Full Factorial
EVOP (Evolutionary Operations) for Process Optimization
Analysis Completed by
&L&"Times New Roman,Regular"Rev: 1/Jan 2003&RPage: &P/&N
Samples D3 D4 2 0 3.267 3 0 2.575 4 0 2.282 5 0 2.115 6 0 2.004
Sample A2 2 1.880 3 1.023 4 0.729 5 0.577 6 0.483
Sample d2 2 1.128 3 1.693 4 2.059 5 2.326 6 2.534
Slide number: *
Data Grouping and Sample details
Number of Groups (Number of time blocks x number of streams)
Time block * No of stream : 25*8 =200
Number of Samples in Each group (It is preferable to collect 5 samples continuously from the process so that the inherent variations are captured
5
Checking the consistency of part to part variation (Step 7)
Average Range (R-bar) (Round off to one decimal more than the data)
3
Upper Control Limit (UCL) = D4*R-bar (Round off to the same decimal as data)
6
0
Yes
If the Part to Part variation is not consistent, STOP, do not proceed further. Plan for DOE
Is the Range Chart plotted
Yes
Yes
If the stratification level is <=3, then the Part to Part variation is very less and the parameter does not require any monitoring. STOP. Do not proceed further
Are there 7 consequtive points increasing, decreasing and one side of the Mean range
Yes
If yes write causes (if possible)
If the range contains patterns as described above, STOP. Plan for DOE
Estimating Part to Part variation (Step 9)
Sigma = R-bar/d2 (Round off to one decimal more than data)
1
6 * Sigma (Round off to the same decimal as data)
6
Overall Average (Round off to one decimal more than data)
145
Number of Groups = Sqrt (Number of data points)
Range = Maximum Value - Minimum Value
Class Interval (C.I) = Range/Number of groups
Class Inverval (C.I) after rounding off to the same decimal as data
Draw the Histogram here
Does the Overall Average lie in the group having maximum frequency or in the adjacent groups
Is there a gradual decreasing trend in frequency on both sides of the group having the maximum frequency
Are there two Modes (Two groups having maximum frequency) and both the groups are distinctly seperated
Based on the above analysis, can you conclude that the histogram is Normal
Note: If the Histogram is non-normal, the estimated rejection which we do in the next step may not be correct. Data transformation is required to correct non-normality
Rejection Analysis Using Sigma (Only for Normal distributions) (Step 11)
Average
145
0
USL
150
0
LSL
140
1.67
Rejection Analysis if the results above show rejection on one side higher than the other side
Target (Design)
USL
LSL
6 Sigma > 75% of Tolerance and <=100% of tolerance
6 Sigma > Tolerance
6 Sigma < Tolerance
6 Sigma = Tolerance
6 Sigma > Tolerance
Actual Part to Part variation <= 50% of tolerance
Actual Part to Part variation > 50% of tolerance
Use Average and Range chart for monitoring, but it has to be done with 100% inspection
Use DOE to reduce variation
0.005
Use Average and Range chart for monitoring
Do DOE to reduce variation or to question the tolerance for that parameter
Remove 100% inspection if done on this parameter
Calculation of limits for Average if the decision taken is Average and Range monitoring
Average
UCL = Average + A2*R-bar
LCL = Average - A2* R-bar
Select which of the following DOE tools will be used if 6 Sigma > Tolerance
DOE Tools selected for Process Improvement
Multi-Vari Analysis
Concentration Chart
Paired Comparison
Component Search
Product/Process Search
Variable Search
Full Factorial
EVOP (Evolutionary Operations) for Process Optimization
Analysis Completed by
&L&"Times New Roman,Regular"Rev: 1/Jan 2003&RPage: &P/&N
Samples D3 D4 2 0 3.267 3 0 2.575 4 0 2.282 5 0 2.115 6 0 2.004
Sample A2 2 1.880 3 1.023 4 0.729 5 0.577 6 0.483
Sample d2 2 1.128 3 1.693 4 2.059 5 2.326 6 2.534
Slide number: *
Data Grouping and Sample details
Number of Groups (Number of time blocks x number of streams)
Time block * No of stream : 25*8 =200
Number of Samples in Each group (It is preferable to collect 5 samples continuously from the process so that the inherent variations are captured
5
Checking the consistency of part to part variation (Step 7)
Average Range (R-bar) (Round off to one decimal more than the data)
3
Upper Control Limit (UCL) = D4*R-bar (Round off to the same decimal as data)
6
0
Yes
If the Part to Part variation is not consistent, STOP, do not proceed further. Plan for DOE
Is the Range Chart plotted
Yes
Yes
If the stratification level is <=3, then the Part to Part variation is very less and the parameter does not require any monitoring. STOP. Do not proceed further
Are there 7 consequtive points increasing, decreasing and one side of the Mean range
Yes
If yes write causes (if possible)
If the range contains patterns as described above, STOP. Plan for DOE
Estimating Part to Part variation (Step 9)
Sigma = R-bar/d2 (Round off to one decimal more than data)
1
6 * Sigma (Round off to the same decimal as data)
6
Overall Average (Round off to one decimal more than data)
145
Number of Groups = Sqrt (Number of data points)
Sqrt(600)=25
6
1.0
Class Inverval (C.I) after rounding off to the same decimal as data
Ist group minimum+CL
Draw the Histogram here
Does the Overall Average lie in the group having maximum frequency or in the adjacent groups
Yes
Is there a gradual decreasing trend in frequency on both sides of the group having the maximum frequency
Yes
Are there two Modes (Two groups having maximum frequency) and both the groups are distinctly seperated
No
Based on the above analysis, can you conclude that the histogram is Normal
Yes
Note: If the Histogram is non-normal, the estimated rejection which we do in the next step may not be correct. Data transformation is required to correct non-normality
Rejection Analysis Using Sigma (Only for Normal distributions) (Step 11)
Average
145
0
USL
150
0
LSL
140
1.67
Rejection Analysis if the results above show rejection on one side higher than the other side
Target (Design)
USL
LSL
6 Sigma > 75% of Tolerance and <=100% of tolerance
6 Sigma > Tolerance
6 Sigma < Tolerance
6 Sigma = Tolerance
6 Sigma > Tolerance
Actual Part to Part variation <= 50% of tolerance
Actual Part to Part variation > 50% of tolerance
Use Average and Range chart for monitoring, but it has to be done with 100% inspection
Use DOE to reduce variation
Use Pre-control chart for monotoring
Use Average and Range chart for monitoring
Do DOE to reduce variation or to question the tolerance for that parameter
Remove 100% inspection if done on this parameter
Calculation of limits for Average if the decision taken is Average and Range monitoring
Average
UCL = Average + A2*R-bar
LCL = Average - A2* R-bar
Select which of the following DOE tools will be used if 6 Sigma > Tolerance
DOE Tools selected for Process Improvement
Multi-Vari Analysis
Concentration Chart
Paired Comparison
Component Search
Product/Process Search
Variable Search
Full Factorial
EVOP (Evolutionary Operations) for Process Optimization
Analysis Completed by
&L&"Times New Roman,Regular"Rev: 1/Jan 2003&RPage: &P/&N
Samples D3 D4 2 0 3.267 3 0 2.575 4 0 2.282 5 0 2.115 6 0 2.004
Sample A2 2 1.880 3 1.023 4 0.729 5 0.577 6 0.483
Sample d2 2 1.128 3 1.693 4 2.059 5 2.326 6 2.534
Slide number: *
Project Summary
Number of drill down’s (funneling) done to reach the root cause(s):
Number of predominant root cause(s) identified:
Time in months for completing the project:
Tools/Techniques used:
Slide number: *
Overall scrap reduced from 1.95% to 1.02%.
2. Crack scrap reduced from 0.62% to 0.02%.
Cycle time reduced from 460 seconds to 422 seconds.
Estimated cost benefits is Rs. 2,18,996 per year.
Slide number: *
blocks x number of streams)
Part Details
Number of Samples in Each group (It
is preferable to collect 5 samples
continuously from the process so
that the inherent variations are
captured
0
If yes
Is the Range Chart plotted
If the stratification level is <=3, then the Part to Part variation is very less and the
parameter does not require any monitoring. STOP. Do not proceed further
Are there 7 consequtive points increasing,
decreasing and one side of the Mean range
If the range contains patterns as described above, STOP. Plan for DOE
Lower Control Limit (LCL) = D3*R-bar
Is the Part to Part variation consistent
If the Part to Part variation is not consistent, STOP, do not proceed further. Plan for
DOE
Checking the consistency of part to part variation (Step 7)
Average Range (R-bar) (Round off to one
decimal more than the data)
Upper Control Limit (UCL) = D4*R-bar
(Round off to the same decimal as data)
Samples D3 D4
2 0 3.267
3 0 2.575
4 0 2.282
5 0 2.115
6 0 2.004
than data)
6 * Sigma (Round off to the same decimal as data)
Estimating Part to Part variation (Step 9)
Overall Average (Round off to one decimal more
than data)
Sample d2
2 1.128
3 1.693
4 2.059
5 2.326
6 2.534
Range = Maximum Value - Minimum Value
Construction of Histogram (Step 10)Number of Groups = Sqrt (Number of data points)
Class Interval (C.I) = Range/Number of groups
Are there two Modes (Two groups having maximum
frequency) and both the groups are distinctly seperated
Class Inverval (C.I) after rounding off to the same decimal
as data
Does the Overall Average lie in the group having maximum
frequency or in the adjacent groups
Draw the Histogram here
Based on the above analysis, can you conclude that the
histogram is Normal
Is there a gradual decreasing trend in frequency on both
sides of the group having the maximum frequency
Average
Sigma
Rejection Analysis Using Sigma (Only for Normal distributions) (Step 11)
Cpk = Zusl/3,Zlsl/3
Projected Rejection % below LSL
Cpk = Zusl/3,Zlsl/3
Projected Rejection % below LSL
(Based on the Normal table)
Rejection Analysis if the results above show rejection on one side higher than the other side
Z(USL) = (USL - Target)/Sigma
Z(LSL) = (LSL - Target)/Sigma
Actual Part to
this parameter
monitoring, but it has to be done with
100% inspection
Do DOE to reduce variation or to
question the tolerance for that parameter
6 Sigma > Tolerance
6 Sigma <= 75% of Tolerance
6 Sigma > 75% of Tolerance and <=100% of tolerance
Sr. No.
Sr. No.
Sr. No.