SMU
EMIS 7364
NTU
TO-570-N
Statistical Quality Control
Dr. Jerrell T. Stracener,
SAE Fellow
More Control Charts Material
Updated: 3/24/04
1
Operating Characteristic (OC) Function
for the x - Chart
• The OC curve describes the ability of the
x-chart to detect shifts in process quality.
• For an x-chart with s known & constant mean
m shifts from in-control value, m0 to another
value m1, where
m1 = m0 + Ks
2
Operating Characteristic (OC) Function
for the x – Chart (continued)
OCμ   βμ   P(not detecting shift on the first
subsequent sample | μ)
 PLCL  X  UCL | μ  μ1  μ 0  Kσ 




 UCL - μ 0  Kσ  
 LCL - μ 0  Kσ  
 Φ
  Φ

σ
σ




n
n




 μ 0 n  Lσ  μ 0  Kσ  n 

 Φ

σ


 μ 0 n  Lσ  μ 0  Kσ  n 

 Φ

σ


3
Operating Characteristic (OC) Function
for the x – Chart (continued)

 

 Φ Lk n Φ Lk n ,
where
Φ y  
y


and
1
e
2π
z2

2
dz
L is usually 3, the three-sigma limits
4
Example
If n=5 & L=3, determine & plot the OC function vs
K, where m1= m0 + Ks.
5
Example - Solution

 

 Φ3  k 5   Φ 3  k 5 
OCK   βK   Φ L  k n  Φ  L  k n
k
-3.0
-2.5
-2.0
-1.5
-1.0
-0.5
0.0
0.5
1.0
1.5
2.0
2.5
3.0
b
0.000104397
0.00479646
0.070492119
0.361631295
0.777546112
0.970060633
0.997300066
0.970060633
0.777546112
0.361631295
0.070492119
0.00479646
0.000104397
1.2
1
0.8
0.6
0.4
0.2
0
-4.0
-2.0
-0.2
0.0
2.0
4.0
6.0
6
OC Function of the Fraction Nonconforming
Control Chart
OCp  βp  P(acceptin g the hypothesis that
a process is in statistica l control | p)
 Pp̂  UCL | p   Pp̂  LCL | p 
 PD  nUCL | p   PD  nLCL | p 
nUCL  n 
d


p

 d  1  p  
d 0  
 FnUCL   F nLCL

n d
nLCL  n 


d 0
 p d 1  p n d
d
7
OC Function of the Fraction Nonconforming
Control Chart
Where
and
[nUCL] denotes the largest integer  nUCL
<nLCL> denotes the smallest integer  nLCL
Note: The OC curve provides a measure of the
sensitivity of the control chart – i.e., its ability to
detect a shift in the process fraction nonconforming
from the nominal value p to some other value p.
8
Example
For a fraction nonconforming control chart with
parameters
and
n = 50,
LCL = 0.0303,
UCL = 0.3697,
Determine & plot the OC curve.
9
Example - Solution
OCp  βp  PD  500.3697  | p   PD  50 0.0303 | p 
 PD  18.49 | p   PD  1.52 | p 
p
0.01
0.03
0.05
0.10
0.15
0.20
0.25
0.30
0.35
0.40
0.45
0.50
0.55
P(D<=18|p)
1.0000
1.0000
1.0000
1.0000
0.9999
0.9975
0.9713
0.8594
0.6216
0.3356
0.1273
0.0325
0.0053
P(D<=1|p) P(D<=18|p) - P(D<=1|p)
0.9106
0.0894
0.5553
0.4447
0.2794
0.7206
0.0338
0.9662
0.0291
0.9708
0.0002
0.9973
0.0001
0.9712
0.0000
0.8594
0.0000
0.6216
0.0000
0.3356
0.0000
0.1273
0.0000
0.0325
0.0000
0.0053
10
Example - Solution
OC(p)
1.0
0.8
0.6
0.4
0.2
0.0
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
-0.2
p
11
OC Function for c-charts and u-charts
• For the c-chart
OCc  βc  P(acceptin g the hypothesis that
a process is in statistica l control | c)
 PX  UCL | c   PX  LCL | c 

UCL 

d 0
e c c x LCL  e c c x

x!
x!
d 0
 FUCL   F LCL

12
OC Function for c-charts and u-charts
• For the u-chart
OCu   βu   P(acceptin g the hypothesis that
a process is in statistica l control | u)
 PX  nUCL | u   PX  nLCL | u 

nUCL 

d 0
x


nu
e  nu
x!
where x  nLCL
13
Example
Determine & plot the OC function for a u-chart with
parameter.
and
LCL = 6.48,
UCL = 32.22.
14
Example - Solution
OCu   βu   Px  UCL | u   Px  UCL | u 
 Pc  nUCL | u   Pc  nUCL | u 
 PnUCL  x  nUCL | u 

nUCL

x


nu
e  nu
d 0
u
0.01
0.03
0.05
0.10
0.15
0.20
0.25
0.30
0.35
0.40
0.45
0.50
0.55
P(D<=33|c)
1.000
1.000
1.000
1.000
1.000
0.999
0.997
0.950
0.744
0.546
0.410
0.151
0.038
x!
P(D<=6|c)
0.999
0.996
0.762
0.450
0.220
0.008
0.000
0.000
0.000
0.000
0.000
0.000
0.000
P(D<=33|c) - P(D<=6|c)
0.001
0.004
0.238
0.550
0.780
0.991
0.997
0.950
0.744
0.546
0.410
0.151
0.038
15
Example - Solution
OC(u)
1.0
0.8
0.6
0.4
0.2
0.0
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
-0.2
u
16
Average Run Length for x-Charts
• Performance of Control Charts can be
characterized by their run length distribution.
• Run Length (RL) of a control chart is defined
to be the number of samples until the process
characteristic exceeds the control limits for
the first time.
• Run Length, RL, is a random variable and
therefore has a probability distribution
17
Average Run Length for x-Charts
Let
p = P(x falls outside control limits)
Then
P(RL = 1)=
P(RL = 2)=
=
P(RL = 3)=
=
P(x1 falls outside CL)=p
P(x1 falls inside CL & x2 falls outside of CL)
(1-p)p
P(x1, x2 fall inside CL & x3 falls outside of CL)
(1-p)(1-p)p
P(RL = i) = P(x1, x2, … xi-1 fall inside CL & xi falls outside of
18
CL)
i-1
Average Run Length for x-Charts
Therefore, the probability mass function of RL is
PRL  k   1  p K 1 p for K  1,2,...
The mean or expected value of RL is

μ  ERL    K 1  p K 1 p
K 1
 p  2p1 - p   3p1 - p   4p1 - p   ...
1

2
3

 p 1  21 - p   31 - p   41 - p   ...
2

3
 p a 1  p 
a 1
a 1
19
Average Run Length for x-Charts


1

 p
2 
 1 - 1 - p  
1

p
• The Average Run Length, ARL, indicates the
number of samples needed, on the average
before x will exceed the control chart limits.
20
Probability of Out-Of-Control Signal and ARL
• Process in control with mean m0
• p = 1 – P(LCL  x  UCL)
= 0.0027
• ARL 
1
p
1

0.0027
 370,
i.e., one the average we would expect 1 outof-control signal out of 370 samples.
21
Probability of Out-Of-Control Signal and ARL
• Process in control with mean mm1m0+ds with
constant s
• What happens if the process goes out of
control?
• How long does it take until the control
charts detects the shift?
• Probability of detecting shift
s
s 

pδ   1  P μ 0  3
 x  μ0  3

n
n

22
Probability of Out-Of-Control Signal and ARL
σ
σ


 μ 0  δσ
μ0  3
 μ 0  δσ 
 μ0  3
n
n

 1  P
Z
σ
σ




n
n



 1 P  3  δ n  Z  3  δ n

 

 P Z  3  δ n  P Z  3  δ n

23
Example
For example, if n = 5, and d = 1,

 
p1  P Z  3  5  P Z  3  5
   5.236  1   0.764
 1  0.0000  0.7775
 0.2225

and
1
ARL 
p1
1

0.2225
 4.495
24