Statisztikai folyamatszabályozás

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Transcript Statisztikai folyamatszabályozás

Statistical Process Control
Production and Process
Management
Where to Inspect in the Process
• Raw materials and purchased parts – supplier
certification programs can eliminate the need for
inspection
• Finished goods – for customer satisfaction, quality at the
source can eliminate the need for inspection
• Before a costly operation – not to waste costly labor or
machine time on items that are already defective
• Before an irreversible process – in many cases items
can be reworked up to a certain point, beyond that point
• Before a covering process – painting can mask deffects
Process stability and process
capability
• Statistical process control (SPC) is used to
evaluate process output to decide if a
process is „in control” or if corrective
action is needed.
• Quality Conformance: does the output of
a process conform to specifications
• These are independent
Variation of the process
• Random variation (or chance) – natural variation
in the output of a process, created by countless
minor factors, we can not affect these factors
• Assignable variation – in process output a
variation whose cause can be identified.
• In control processes – contains random
variations
• Out of control processes – contains assigneable
variations
Sampling distribution vs. Process
distribution
• Both distribution have the
same mean
• The variability of the sampling
distribution is less than the
variability of the process
• The sampling distribution is
normal even if the profess
distribution is not normal
• Central limit theorem: states
thet the sample size increase
the distribution of the sample
averages approaches a normal
distribution regardless of the
shape of the sampled
distribution
• In the case of normal
distribution
– 99,74% of the datas fall
into m± 3 σ
– 95,44% of the datas fall
into m± 2 σ
– 68,26% of the datas fall
into m± 1 σ
– If all of the measured datas
fall into the m± 3 σ
intervall, that means, the
process is in control.
Sampling
• Random sampling
– Each itemhas the same probability to be selected
– Most common
– Hard to realise
• Systhematic sampling
– According to time or pieces
• Rational subgoup
– Logically homogeneous
– If variation among different subgroups is not accounted fo, then
an unawanted source of nonrandom variation is being introduced
– Morning and evening measurement in hospitals (body
temperature)
• Variables – generate data that are
measured (continuus scale, for example
length of a part)
• Attributes – generate data that are
counted (number of defective parts,
number of calls per day)
Control limits
• The dividing lines between random and
nonrandom deviation from the mean of the
distribution
• UCL – Upper Control limit
• CL – Central line
• LCL – lower Control limit
• This is counted from the process itself. It is
not the same as specification limits!
Specification limits
• USL – Upper specification limit
• LCL – lower specification limit
• These reflect external specifications, and
determined in advance, it is not counted
from the process.
Control chart
Hypothesis test
• H0 = the process is stable
Decision
Stable
Reality
not stable
Stable
OK
Type I error (risk
of the producer)
not stable
Type II error risk OK
of the costumer)
• Type I error – concluding a process is not in
control when it is actually is – producers risk – it
takes unnecessary burden on the producer who
must searh fot something is not there
• Type II error – concluding a process is in control
when it is actually not – customers risk –
because the producer didn’t realise something is
wrong and passes it on to the costumer
Control charts
x and R – mean and range chart
• Sample size – n=4 or n=5 can be handled
well, with short itervals,
• Sampling freuency – to reflec every affects
as chenges of shifts, operators etc.
• Number of samples – 25 or more
• mean
• range
• n is the sample size
• Means of samples’ means
• Means of ranges
• m is the number of samples
x
x1  x2  ......  xn
n
R  xmax  xmin
x1  x 2  ....  x m
x
m
R
R1  R2  ......  R3
m
Control limits
UCLR  D4 R
UCLx  x  A2 R
LCLR  D3 R
LCLx  x  A2 R
A2, D3, D4 are constants and depends on the sample size
Exercise
6
5
7
day2 8
day3 7
day4 6
6
6
7
6
6
5
7
6
4
8
6
4
2
0
1
2
3
4
day
Means
Cl x-bar
LCL x-bar
UCL x-bar
Rchart
Centimeter
day1 6
centimeter
x-bar chart
6
4
2
0
1
2
3
4
Day
Sample Range
R-bar
UCL R
Control charts for attributes
• When the process charasterictic is
counted rather than measured
• p-chart – fraction of defective items in a
sample
• c-chart – number of defects per unit
p-chart
• p-average fraction
defective in the
population
• P and σ can be counted
from the samples
• min 25 samples – m
• Big samlpe size is
needed (50-200 pieces) –
n
• Number of defective item
–np
• If the LCL is negativ,
lower limit will be 0.
UCLp  p  z p
LCL p  p  z p
p 
p (1  p )
n
np
p
nm
Exercise
• z=3,00
• p=220/(20*100)=0,11
• σ=(0,11(10,11)/100)1/2=0,03
• UCL=0,11+3*0,03=0,2
• LCL=0,11-3*0,3=0,02
c-chart
• To control the occurrences (defects) per unit
• c1, c2 a number of defects per unit, k is the number of units
UCLc  c  3 c
LCLc  c  3 c
Exercise
Solution
c
45
 2,5
18
UCLc  2,5  3 2,5  7,24
LCLc  2,5  3 2,5  2,24  0
• Determine
Run and trend tests
– Runs up and down (u/d)
– Above and below median (med)
• Count the number of runs and compared with the number of runs
that would be expected in a completely random series.
E (r ) med 
N
2  N 1
 1 E (r )U / D 
2
3
– N number of observations or data points,
– E(r) expected number of runs
•
•
•
•
16  N  29  med  N  1
U / D 
4
90
Determine the standard deviation
Too few or too maní runs can be an indication of nonrandomness
Determine z score using the following formula:
obs  E (r )
z

counted z must be fall into the interval of (-2;2) to accept
nonrandomness (this means that the 95,5% of the time random
process will produce an observed number of runs within 2σ of the
expected number)
It can be (-1,96;1.96) 95% of time
Or (-2,33;2,33) 98% of time
Example
Solution
•
•
•
•
•
•
•
E(r)med=N/2+1=20/2+1=11
E(r)u/d=(2N-1)/3=(2*20-1)/3=13
σmed=[(N-1)/4]1/2=[(20-1)/4]1/2=2,18
σu/d= =[(16N-29)/90]1/2 =[(16*20-29)/90]1/2=1,80
zmed=(10-11)/2,18=-0,46
Zu/d=(17-13)/1,8=2,22
Although the median test doesn’t reveal any
pattern, the up down test does.
Index of process capability
• CP (capability process) – it refers to the inherent variability of
process output relative to the variation allowed by designed
specifications
• the higher CP the best capablity
• I the case of CP>1 the process can fulfill the requirements
• It make sense when the process is centered
USL  LSL
Cp 
6ˆ
Process capability when process is
not centered
•  - estimated process average (using
̂grand mean of the samples)
•
- estimated standard deviation
(USL  ˆ )
Cpu 
3ˆ
( ˆ  LSL)
Cpl 
3ˆ
Cpk  min{ Cpu; Cpl )
R
ˆ 
d2
̂  x
Process capability when process is
not centered II
• When sampling is not achievable, than for
the total population
Pp 
(USL  LSL)
6
(   LSL)
Ppl 
3
Ppk  min{ Ppu; Ppl}
(USL   )
Ppu 
3

2
(
x

x
)
 i
(n  1)
USL
LSL
Cp=1
Cpk=1
6σ
• When the process is not centered the is the fault
of operator but when standard deviation is
higher than the tolerance limit, managers must
interfer in  a new machine is needed ,
Cp>1
Cpk>1
Cpk<1
Cp<1
process capacity is It can’t occure
proper
Process capacity is Managers
not proper it is the responsible for
workers fault
Exercise
• For an overheat projector, the thickness of a
component is specified to be between 30 and
40 millimeters. Thirty samples of components
yielded a grand mean ( x ) 34 mm, with a
standard deviation (̂ ) 3,5 mm. Calculate the
process capability index by following the steps
previously outlined. If the process is not highly
capable, what proportion of product will not
conform?
Solution
USL  LSL 40  30
Cp 

6ˆ
3  3,5
Cpu 
(USL  ˆ ) 40  34

 0,57
3ˆ
3  3,5
( ˆ  LSL) 34  30
Cpl 

 0,38
3ˆ
3  3,5
• Process is out of control
• To determine number of products use table of
normal distribution
USL  x 40  34
zu 

 1,71
ˆ
3,5
LSL  x 30  34
zl 

 1,14
ˆ
3,5
• 0,1271+0,0436=0,1707 17,07% of products
doesn’t meet specification
Thank you for your attention