Transcript Quality Management

Quality Management
09. lecture
Statistical process control
Variability of process
• Random variation – uncontrollable, caused
by chance, centered around a mean with a
consistent amount of dispersion
• Non-random variation – has a systematic
cause, shift in process mean
• Process stability – only random variation
exist
Sampling methods
• Less expensive
• Take less time
• Less intrusive
• 100% sampling – during acceptance sampling or workin-process inspection
• Random sample: equal chance to be inspected,
independence among observations
• Systematic samples: according to time or sequence
• Rational subgroup: logically homogeneous, if we not
separate these groups, non-random variation can biased
results
Process control chart
• Tools for monitoring process variation
• Continuous variable
• Attribute – either or situation
• Weight will be variable, while number of
defective items will be attributes
Steps
•
•
•
•
•
Identify critical operations
Identify critical product characteristics
Determine whether variables or attributes
Select the proper control charts
Determine control limits and improve the
process
• Update the limits
Control limits
•
•
•
•
UCL – Upper Control limit
CL – Central line
LCL – lower Control limit
Control limits comes from the process and
are very different from specification limits.
Distribution
• Central limit
theorem:
• If the samples
number is high
(above 30) than
the mean of the
samples will
follow normal
distribution
Hypothesis test
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•
•
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•
H0: μ=11 cm
H1: μ≠11 cm
95% (z=1,96) rejection limit
If σ=0,001 (n=10), than the rejection limits:
11+1,96*0,001 and 11-1,96*0,001
(11,00196;10,99804)
The sample mean μ=10,998 falls between the
rejection limits, we accept the null hypothesis
• Then we accept that a process is in control
Errors during hypothesis test
Decision
In control
Reality
In control
Out of
control
OK
Out of
control
Error type of
one (risk of
supplier)
Error type of two OK
(risk of
customer)
X mean and R control chart
• Mean chart monitor the average of the
process
• Range chart monitor the dispersion of the
process
• K>25, n=4 or 5
x1  x 2  ...... x n
n
• Sample mean
• Range of sample
R  xmax  xmin
• n is number of observations
x
• Average of sample means
• Average of ranges
• k is number of samples
x
R
x1  x 2  .... x k
k
R 1  R 2  ...... R k
k
Counting of control limits
UCLR  D4 R
UCL x  x  A2 R
LCLR  D3 R
LCLx  x  A2 R
A2, D3, D4 comes from factor for control limits table
Exercise
Obs1 Obs2
day1
6
6
day2
8
6
day3
day4
7
6
Obs3
5
6
Obs4
7
7
6
5
6
4
6
7
Rchart
8
6
4
2
0
Centimeter
centimeter
x-bar chart
6
4
2
0
1
2
3
1
4
2
day
Means
Cl x-bar
LCL x-bar
3
4
Day
UCL x-bar
Sample Range
R-bar
UCL R
X and MR (moving range) chart
• If it is not possible to draw samples
• Only one or two units per day are produced
• Central limit theorem doesn’t apply  you make
sure that the datas normally distributed. If the
distribution is not normal
– Use g chart of h chart
• X – individual observation from a population  3
std dev limit is a natural variation  X chart
limits ate not control limits. They are natural
limits.
LCLx  x  E2  (MR )
UCLx  x  E2  (MR )
UCLMR  D4  (MR )
LCLMR  0
Exercise
• The following table shows
the daily trips. The trucks
generally take 6,5 hours
to make the daily trip. The
owner want to know
whether there is any
other reason of the
increasing delivery time,
or it just depend on the
traffic.
• Use X chart and MR chart
to determine.
Travel Time
MR
6,2
-
6,1
0,1
6,5
0,4
7,2
0,7
6,8
0,4
7,7
0,9
Solution
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Xmean=6,75
MRmean=0,5
UCLx=6,75+2,66*0,5=8,08
CLx=6,75
LCLx=6,75-2,66*0,5=5,42
UCLMR=3,268*0,5=1,634
CLMR=0,5
LCLMR=0
1,7
1,5
1,3
1,1
0,9
0,7
0,5
0,3
0,1
-0,1 1
1,5
2
2,5
3
3,5
4
4,5
5
Median chart
• If counting average takes too much time or
effort
• Number of observations (n) is better to be
odd number, (3,5,7)
• 20<k<25
• In sum the number of observations must
reach 100
~
~
LCLx  x  A2  R
~
~
UCLx  x  A2  R
Example
• The table below contains observations of a
process. Use median chart and determine,
whether the process is in control.
Obs 1
Obs 2
Obs 3
Obs 4
Obs 5
1,1
1,2
1,4
1,5
1,6
1
1,02
1,5
1,6
1,6
1,2
1,4
1,4
1,4
1,5
1,3
1,3
1,3
1,5
1,6
Solution
• CLx=1,4
• LCLx=1,4-0,691*0,425=1,1063
• UCLx=1,4+0,691*0,425=1,693
Obs 1
1,1
Obs 2
1,2
Obs 3
1,4
Obs 4
1,5
Obs 5
1,6
Median
1,4
Range
0,5
1
1,2
1,3
1,02
1,4
1,3
1,5
1,4
1,3
1,6
1,4
1,5
1,6
1,5
1,6
1,5
1,4
1,3
0,6
0,3
0,3
s and X mean chart
• Concerned about the dispersion of a process,
than R chart is not sufficently precise
• Use std dev chart, when variationis small (high
tech industry)
• New formula must be used for compute limits of
x mean chart
• Si – the dtd dev for sample i
• K number of samples
• B3 and A3 factors
UCLs  B4  s
LCLs  B3  s
s

s
k
LCLx  x  A3  (s )
i
UCLx  x  A3  (s )
Example
• Determine useing s chart whether the process is
in control, we have 4 samples with n=3.
Obs1
Obs2
Obs3
Sample
2,0000
1,9995
2,0002
1
1,9998
2,0003
2,0002
2
1,9997
2,0000
2,0004
3
2,0003
1,9998
1,9997
4
2,0004
2,0001
2,0000
Sample
sum
Mean
Std.dev
1
1,99990
0,000361
2
2,00010
0,000265
3
2,00003
0,000351
4
1,99993
0,000321
5
2,00017
0,000208
10,00013
0,00151
• UCLs=2,568*(0,00151/5)=0,00
0775
• CLs=0,000302
• LCLs=0
• UCLx=2,000026+1,954*0,0003
02=2,00061
• CLx=2,000026
• LCLx=2,0000261,954*0,000302=1,99943
2,00020
2,00015
2,00010
2,00005
2,00000
1,99995
1,99990
1,99985
1
1,5
2
2,5
1,5
2
2,5
3
3,5
4
4,5
5
0,0008
0,0007
0,0006
0,0005
0,0004
0,0003
0,0002
0,0001
0
1
3
3,5
4
4,5
5
Process capability
• If the process is in control, than there is
only non-random variation in the process.
But it doesn’t mean that the products
produced by the process meet the
specifications or defect-free.
• Process capability refers to the ability of a
process to produce a product that meet
the specifications.
Specification limit
• USL – Upper specification limit
• LSL – lower specification limit
• Specification limit comes from outside,
determined by engineers or administration,
and not calculated from the process.
Population capability
• If there are no subgroups, calculate population
capability, where
•  - population mean
( xi  x ) 2
•  - population process std.dev

(USL   )
Ppu 
3
Ppl 
(   LSL )
3
Ppk  min{Ppu; Ppl)
n 1
Capability index
• 1. select critical operation
• 2. select k sample of size n
– 19<k<26
– n>50 (if n binomial)
– 1<n<11 (measurement)
• Use control chart whether it is
stable
• Compare process natural
tolerance limit with
specification limits
• Compute capability indexes:
Cpl, Cpu, Cpk
• ˆ - computed population
process mean
• ˆ - estimated process std.dev
R
ˆ 
d2
(USL  ˆ )
Cpu 
3ˆ
( ˆ  LSL )
Cpl 
3ˆ
Cpk  min{Cpu; Cpl)
Cp 
USL  LSL
6ˆ
USL
LSL
Cp=1
Cpk=1
6σ
Exercise
• For an overhead projector, the thickness
of component is specified to be between
30 and 40 millimeters. Thirty samples of
components yield a grand mean ( x ) of
34 millimeters with a standard deviation
(ˆ ) of 3,5. Calculate process capability
index. If the process is not capable, what
proportion of a product will not conform?
Solution
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•
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Cpu=(40-34)/3*3,5=0,57
Cpl=(34-30)/3*3,5=0,38
Cpk=0,38
The process is not capable.
To determine the proportion of product that not
conform, we need to use normal distribution table.
• Z=(LSL- x )/ ˆ =(30-34)/3,5=-1,14
• Z=(USL- x )/ˆ =(40-34)/3,5=1,71
• 0,1271+0,0436=0,1707 17,07% will not conform
Thank you for your attention!