Transcript Quality Management

LC Technology
Manufacturing Systems
Quality Management – Pareto Analysis
Pinpoints problems through the identification
and separation of the ‘vital few’ problems from
the trivial many.
Vilifredo Pareto: concluded that 80% of the
problems with any process are due to 20% of
the causes.
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Heat Damage
Des.Error
PCB fault
Faulty Comp
Wrong Placement
Not Soldered
Oversolder
Dry Joint
Quality Management – Pareto Analysis
Causes of poor soldering
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Quality Management – Pareto Analysis
Causes of poor soldering – descending order
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Quality Management – Pareto Analysis
Cumulative plot is made of all of the causes
80% caused by two problems
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Statistical Process Control
•Statistical procedure to verify quality
•Check manufacturing process is working correctly
→Inspect and measure manufacturing process
→Varying from target – corrective action taken
•Prevents poor quality before it occurs
Statistical Process Control
When? Manufacturing large quantities of items
•Euro coins
•Computers
•Cars etc.
Why?
•Impractical to measure each item made
•Machine/equipment/human error
How?
•Measure a small proportion of the produced items (sample)
•Use X-bar and R Charts to see if process is in control
•Conclude the quality characteristics of the whole process
Normal Distribution
Laser machine A – cutting 20mm hole
Machine A
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19.7 19.8 19.9 19.9 20 20.1 20.2 20.3 20.4
Some measurement < 20mm
Some measurements > 20mm
Natural occurrence
Normal Distribution
Machine B making same part as machine A
•Same distribution
•Skewed to right
Machine B
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19.7
19.9
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20.2
20.4
Normal Distribution
Histogram:
•Statistical information
•Column width represents a range of sizes
•Shape of histogram is proportional to spread of data
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1.875
1.85
1.825
1.8
1.775
1.75
1.725
1.7
1.675
1.65
1.625
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1.575
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1.5 1.525 1.55 1.575 1.6 1.625 1.65 1.675 1.7 1.725 1.75 1.775 1.8 1.825 1.85
Results of a survey on the heights of a group of pupils in a large school
Column width = 25mm
Normal Distribution
Larger survey – population of a town
Column width = 10mm
•Centered about mean
•Characteristic ‘Bell’ shape curve
•Number of occurrences reduce as they deviate from the mean
Normal Distribution
A very small sample interval approximates a curve as shown
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Normal Distribution
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All measurable attributes show a variation
Spread of Sizes = Normal Distribution
Normal Distribution
•The spread or ‘width’ of the curve has a
precise mathematical meaning - Variance
•The greater the variance the wider the curve
•Defined by a parameter called the standard
deviation
Normal Distribution
Calculation of standard distribution (sigma)
-measured sizes of a sample of parts

( y1  y ) 2  ( y2  y ) 2  ....( yn  y ) 2
N
y1,y2…etc are the measured values of the sample
y is the average value
N is the number of samples taken
Normal Distribution
Sharpen 5 pencils to a length of 8 mm
6.5mm
8.2mm
8.5mm
7.5mm
7.0mm
Mean average = 7.54
(each value – mean average)² +
Sigma =
number of values
(6.5 - 7.54)² + (8.2 – 7.54)² + (8.5 – 7.54)² + (7.5 – 7.54)² + (7 – 7.54)²
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SIGMA
= 0.73
Normal Distribution
If sigma is known then we know that:
•95% of parts will lie within +/- 2σ of the mean
•99.74% of parts will lie within +/- 3σ of the mean
Control Charts
• Used to establish the control limits for a process
• Used to monitor the process to show when it is
out of control
1. X-bar Chart (Mean Charts)
2. R Charts (Range Charts)
Control Charts
Process Mean = Mean of Sample Means
Upper control limit (UCL) = Process mean+3 sigma
Lower control limit (LCL) = Process mean-3 sigma
Control Charts – X-bar Charts
1. Record measurements from a number of
samples sets (4 or 5)
Control Charts – X-bar Charts
2. Calculate the mean of each sample set
Oven temperature data
Morning
Midday
Evening
Daily
Means
Monday
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Tuesday
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Wednesday
215
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Thursday
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Friday
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Saturday
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Control Charts – X-bar Charts
3. Calculate the process mean (mean of sample means)
Daily
Means
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206  207  215  214  214  210
 211 Degrees
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Control Charts – X-bar Charts
4. Calculate UCL and LCL
UCL = process mean + 3σsample
LCL = process mean - 3σsample
The standard deviation σsample of the sample means

n
where n is the sample size (3 temperature readings)
σ = process standard deviation (4.2 degrees)
σsample
4.2
 2.42 degrees
=
3
Control Charts – X-bar Charts
4. Calculate UCL and LCL
UCL = process mean + 3σsample
LCL = process mean - 3σsample
σsample
4.2
 2.42 degrees
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UCL = 211 + 3(2.42) = 218.27 degrees
LCL = 211 – 3(2.42) = 203.72 degrees
Control Charts – X-bar Charts
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UCL = 218.27
LCL = 203.72
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Control Charts – X-bar Charts
Interpreting control charts: Process out of control
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Last data point is out of control – indicates definite problem
to be addressed immediately as defective products are
being made.
Control Charts – X-bar Charts
Interpreting control charts: Process in control
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Process still in control but there is a steady increase
toward the UCL. There may be a possible problem and it
should be investigated.
Control Charts – X-bar Charts
Interpreting control charts: Process in control
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All data points are all above the process mean. This
suggests some non-random influence on the process that
should be investigated.
Control Charts – Range Charts
The range is the difference between the largest and
smallest values in a sample.
Range is used to measure the process variation
1. Record measurements from a set of samples
Oven temperature data
Morning
Midday
Evening
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Control Charts – Range Charts
2. Calculate the range = highest – lowest reading
Oven temperature data
Morning
Midday
Evening
Range
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Control Charts – Range Charts
3. Calculate UCL and LCL
UCL = D4 x Raverage
LCL = D3 x Raverage
UCL = 2.57 x 13
= 33.41 degrees
LCL = 0 x 13
= 0 degrees
Sample size (n)
D3
D4
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3.27
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2.57
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2.28
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2.11
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0.08
1.92
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0.14
1.86
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0.18
1.82
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0.22
1.78
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0.26
1.74
Control Charts – Range Charts
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UCL
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R average
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LCL
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Process Capability
Matches the natural variation in a process to
the size requirements (tolerance) imposed by
the design
Filling a box with washers:
•exact number not in all boxes
•upper limit set
•lower limit set
Process Not Capable
Process not capable: a lot of boxes will be over and
under filled
Normal distribution > specifications
Cannot achieve tolerances all of the time
Process Capable
Process capable: However there will still be a small
number of defective parts
Normal distribution is similar to specifications
Tolerances will be met most of the time
Process Capable
Process capable: No defective parts
Normal distribution < specifications
Tolerances will be met all the time
Process Capability Index
Cp 
Tolerance Range Upper Size Limit  Lower Size Limit

6
6
If Cp =1
•Process is capable
•i.e. 99.97% of the natural variation of the process will be
within the acceptable limits
Process Capability Index
If Cp > 1
•Process is capable.
•i.e. very few defects will be found – less than three per
thousand, often much less
Process Capability Index
If Cp <1
•Process is NOT capable
•i.e. the natural variation in the process will cause outputs
that are outside the acceptable limits.
Statistical Process Control
Is doing things right 99% of the time good enough?
13 major accidents at Heathrow Airport every 2 days
5000 incorrect surgical procedures per week ………………
Pharmaceutical company producing 1 000 000 tablets a week,
99% quality would mean 10 000 tablets would be defective!
Process of maintaining high quality standards is called :
Quality Assurance
Modern manufacturing companies often aim for a target of only
3 in a million defective parts.
The term six sigma is used to describe quality at this level
Sampling
Size of Sample?
Sufficient to allow accurate assessment of process
•More – does not improve accuracy
•Less – reduced confidence in result
Sampling
Size of Sample
z
s 
e
2
S = sample size
e = acceptable error - as a proportion of std. deviation
z = number relating to degree of confidence in the result
Confidence
Value for z
99%
2.58
95%
1.96
90%
1.64
80%
1.28
Sampling
Example – find mean value for weight of a
packet of sugar
•with a confidence of 95%
•acceptable error of 10%
•Weight of packet of sugar = 1000g
•Process standard deviation = 10g
Sampling
z = 1.96 from the table
e = 0.1 (i.e.10%)
Therefore the sample size s = (1.96 / 0.1)2
Therefore s = 384.16
Sample size = 385
Sampling
Assume mean weight = 1005g
Sigma = 10g
Therefore error = 10g + 10% = 11g
Result:
95% confident average weight of all
packets of sugar  994 g  1016 g
QC, QA and TQM
Quality Control:
• emerged during the 1940s and 1950s
• increase profit and reduce cost by the inspection of product
quality.
• inspect components after manufacture
• reject or rework any defective components
Disadvantages:
• just detects non-conforming products
• does not prevent defects happening
• wastage of material and time on scrapped and reworked
parts inspection process not foolproof
• possibility of non-conforming parts being shipped by
mistake
QC, QA and TQM
Quality Assurance
• Set up a quality system
• documented approach to all procedures and processes that
affect quality
• prevention and inspection is a large part of the process
• all aspects of the production process are involved
• system accredited using international standards
QC, QA and TQM
Total Quality Management
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International competition during the 1980s and 1990s
Everybody in the organisation is involved
Focussed on needs of the customer through teamwork
The aim is ‘zero defect’ production
Just-In-Time Manufacture
Modern products – shortened life cycle
Manufacturer – pressure for quick response
Quick turnaround - hold inventory of stock
Holding inventory costs money for storage
Inventory items obsolete before use
New approach: Just-In-Time Manufacture
Just-In-Time Manufacture
Underlying concept: Eliminate waste.
Minimum amount:
•Materials
•Parts
•Space
•Tools
•Time
Suppliers are coordinated with manufacturing
company.
Just-In-Time Manufacture
Kanbans – Japanese word for card
Order form for components
Passed from one station to another
Initiates the production or movement of parts
Return
Container
to
Delivery
address
Next
process
Previous
process
Number of
components