Operations Management. Chapter 8  Statistical Process Control. PowerPoint presentation to accompany Heizer/Render Principles of Operations Management, 7e Operations Management, 9e . Statistical Process Control (SPC) Control Charts for Variables The Central Limit Theorem
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Operations Management
Chapter 8 
Statistical Process Control
PowerPoint presentation to accompany
Heizer/Render
Principles of Operations Management, 7e
Operations Management, 9e
When you complete this supplement, you should be able to:
Identify or Define:
When you complete this supplement, you should be able to:
Identify or Define:
When you complete this supplement, you should be able to:
Identify or Define:
When you complete this supplement, you should be able to:
Describe or Explain:
Each of these represents one sample of five boxes of cereal
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Frequency
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Weight
To measure the process, we take samples and analyze the sample statistics following these steps
(a)Samples of the product, say five boxes of cereal taken off the filling machine line, vary from each other in weight
Figure S6.1
The solid line represents the distribution
Frequency
Weight
To measure the process, we take samples and analyze the sample statistics following these steps
(b)After enough samples are taken from a stable process, they form a pattern called a distribution
Figure S6.1
Central tendency
Variation
Shape
Frequency
Weight
Weight
Weight
To measure the process, we take samples and analyze the sample statistics following these steps
(c)There are many types of distributions, including the normal (bellshaped) distribution, but distributions do differ in terms of central tendency (mean), standard deviation or variance, and shape
Figure S6.1
Prediction
Frequency
Time
Weight
To measure the process, we take samples and analyze the sample statistics following these steps
(d)If only natural causes of variation are present, the output of a process forms a distribution that is stable over time and is predictable
Figure S6.1
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Prediction
Frequency
Time
Weight
To measure the process, we take samples and analyze the sample statistics following these steps
(e)If assignable causes are present, the process output is not stable over time and is not predicable
Figure S6.1
Constructed from historical data, the purpose of control charts is to help distinguish between natural variations and variations due to assignable causes
Variables
Attributes
The mean of the sampling distribution (x) will be the same as the population mean m
x = m
s
n
The standard deviation of the sampling distribution (sx) will equal the population standard deviation (s) divided by the square root of the sample size, n
sx =
Regardless of the distribution of the population, the distribution of sample means drawn from the population will tend to follow a normal curve
(a) In statistical control and capable of producing within control limits
Frequency
Upper control limit
Lower control limit
(b) In statistical control but not capable of producing within control limits
(c) Out of control
Size
(weight, length, speed, etc.)
Figure S6.2
Three population distributions
Mean of sample means = x
Beta
Standard deviation of the sample means
Normal
= sx =
Uniform
s
n

3sx2sx1sxx+1sx+2sx+3sx
95.45% fall within ± 2sx
99.73% of all x
fall within ± 3sx
Distribution of sample means
Figure S6.3
Sampling distribution of means
Process distribution of means
x = m
(mean)
Figure S6.4
Take samples from the population and compute the appropriate sample statistic
Use the sample statistic to calculate control limits and draw the control chart
Plot sample results on the control chart and determine the state of the process (in or out of control)
Investigate possible assignable causes and take any indicated actions
Continue sampling from the process and reset the control limits when necessary
For xCharts when we know s
Upper control limit (UCL) = x + zsx
Lower control limit (LCL) = x  zsx
wherex=mean of the sample means or a target value set for the process
z=number of normal standard deviations
sx=standard deviation of the sample means
=s/ n
s=population standard deviation
n=sample size
Hour 1
SampleWeight ofNumberOat Flakes
117
213
316
418
517
616
715
817
916
Mean16.1
s =1
HourMeanHourMean
116.1715.2
216.8816.4
315.5916.3
416.51014.8
516.51114.2
616.41217.3
n = 9
UCLx = x + zsx= 16 + 3(1/3) = 17 ozs
LCLx = x  zsx = 16  3(1/3) = 15 ozs
For 99.73% control limits, z = 3
Variation due to assignable causes
Out of control
17 = UCL
Variation due to natural causes
16 = Mean
15 = LCL
Variation due to assignable causes

123456789101112
Out of control
Sample number
Control Chart for sample of 9 boxes
For xCharts when we don’t know s
Upper control limit (UCL) = x + A2R
Lower control limit (LCL) = x  A2R
whereR=average range of the samples
A2=control chart factor found in Table S6.1
x=mean of the sample means
Sample Size Mean Factor Upper Range Lower Range
n A2D4D3
21.8803.2680
31.0232.5740
4.7292.2820
5.5772.1150
6.4832.0040
7.4191.9240.076
8.3731.8640.136
9.3371.8160.184
10.3081.7770.223
12.2661.7160.284
Table S6.1
Process average x = 16.01 ounces
Average range R = .25
Sample size n = 5
Process average x = 16.01 ounces
Average range R = .25
Sample size n = 5
UCLx = x + A2R
= 16.01 + (.577)(.25)
= 16.01 + .144
= 16.154 ounces
From Table S6.1
Process average x = 16.01 ounces
Average range R = .25
Sample size n = 5
UCLx = x + A2R
= 16.01 + (.577)(.25)
= 16.01 + .144
= 16.154 ounces
UCL = 16.154
Mean = 16.01
LCLx = x  A2R
= 16.01  .144
= 15.866 ounces
LCL = 15.866
Upper control limit (UCLR) = D4R
Lower control limit (LCLR) = D3R
where
R=average range of the samples
D3 and D4=control chart factors from Table S6.1
For RCharts
Average range R = 5.3 pounds
Sample size n = 5
From Table S6.1 D4= 2.115, D3 = 0
UCLR = D4R
= (2.115)(5.3)
= 11.2 pounds
UCL = 11.2
Mean = 5.3
LCLR = D3R
= (0)(5.3)
= 0 pounds
LCL = 0
(a)
These sampling distributions result in the charts below
(Sampling mean is shifting upward but range is consistent)
UCL
(xchart detects shift in central tendency)
xchart
LCL
UCL
(Rchart does not detect change in mean)
Rchart
LCL
Figure S6.5
(b)
These sampling distributions result in the charts below
(Sampling mean is constant but dispersion is increasing)
UCL
(xchart does not detect the increase in dispersion)
xchart
LCL
UCL
(Rchart detects increase in dispersion)
Rchart
LCL
Figure S6.5
p(1  p)
n
sp =
UCLp = p + zsp
^
^
LCLp = p  zsp
^
wherep=mean fraction defective in the sample
z=number of standard deviations
sp=standard deviation of the sampling distribution
n=sample size
^
Population will be a binomial distribution, but applying the Central Limit Theorem allows us to assume a normal distribution for the sample statistics
SampleNumberFractionSampleNumberFraction
Numberof ErrorsDefectiveNumberof ErrorsDefective
16.06116.06
25.05121.01
30.00138.08
41.01147.07
54.04155.05
62.02164.04
75.051711.11
83.03183.03
93.03190.00
102.02204.04
Total = 80
(.04)(1  .04)
100
80
(100)(20)
sp = = .02
p = = .04
^
UCLp= 0.10
.11 –
.10 –
.09 –
.08 –
.07 –
.06 –
.05 –
.04 –
.03 –
.02 –
.01 –
.00 –
LCLp= 0.00
UCLp = p + zsp= .04 + 3(.02) = .10
^
Fraction defective
p = 0.04
LCLp = p  zsp = .04  3(.02) = 0
^

2468101214161820
Sample number
UCLp= 0.10
.11 –
.10 –
.09 –
.08 –
.07 –
.06 –
.05 –
.04 –
.03 –
.02 –
.01 –
.00 –
LCLp= 0.00
UCLp = p + zsp= .04 + 3(.02) = .10
^
Fraction defective
p = 0.04
LCLp = p  zsp = .04  3(.02) = 0
^

2468101214161820
Sample number
Possible assignable causes present
UCLc = c + 3 c
LCLc = c  3 c
wherec=mean number defective in the sample
Population will be a Poisson distribution, but applying the Central Limit Theorem allows us to assume a normal distribution for the sample statistics
c = 54 complaints/9 days = 6 complaints/day
UCLc= c + 3 c
= 6 + 3 6
= 13.35
UCLc= 13.35
14 –
12 –
10 –
8 –
6 –
4 –
2 –
0 –
Number defective
LCLc= c  3 c
= 3  3 6
= 0
c = 6
LCLc= 0

1

2

3

4

5

6

7

8

9
Day
Upper control limit
Target
Lower control limit
Normal behavior. Process is “in control.”
Figure S6.7
Upper control limit
Target
Lower control limit
One plot out above (or below). Investigate for cause. Process is “out of control.”
Figure S6.7
Upper control limit
Target
Lower control limit
Trends in either direction, 5 plots. Investigate for cause of progressive change.
Figure S6.7
Upper control limit
Target
Lower control limit
Two plots very near lower (or upper) control. Investigate for cause.
Figure S6.7
Upper control limit
Target
Lower control limit
Run of 5 above (or below) central line. Investigate for cause.
Figure S6.7
Upper control limit
Target
Lower control limit
Erratic behavior. Investigate.
Figure S6.7
Variables Data
Attribute Data
Attribute Data
Upper Specification  Lower Specification
6s
Cp =
Process mean x = 210.0 minutes
Process standard deviation s = .516 minutes
Design specification = 210 ± 3 minutes
Upper Specification  Lower Specification
6s
Cp =
Insurance claims process
Process mean x = 210.0 minutes
Process standard deviation s = .516 minutes
Design specification = 210 ± 3 minutes
Upper Specification  Lower Specification
6s
Cp =
213  207
6(.516)
= = 1.938
Insurance claims process
Process mean x = 210.0 minutes
Process standard deviation s = .516 minutes
Design specification = 210 ± 3 minutes
Upper Specification  Lower Specification
6s
Cp =
213  207
6(.516)
= = 1.938
Insurance claims process
Process is capable
UpperSpecification  xLimit
3s
Lowerx SpecificationLimit
3s
Cpk = minimum of ,
New process mean x = .250 inches
Process standard deviation s = .0005 inches
Upper Specification Limit = .251 inches
Lower Specification Limit = .249 inches
New Cutting Machine
New process mean x = .250 inches
Process standard deviation s = .0005 inches
Upper Specification Limit = .251 inches
Lower Specification Limit = .249 inches
(.251)  .250
(3).0005
Cpk = minimum of ,
New Cutting Machine
New process mean x = .250 inches
Process standard deviation s = .0005 inches
Upper Specification Limit = .251 inches
Lower Specification Limit = .249 inches
(.251)  .250
(3).0005
.250  (.249)
(3).0005
Cpk = minimum of ,
.001
.0015
Cpk = = 0.67
New Cutting Machine
Both calculations result in
New machine is NOT capable
Cpk = negative number
Cpk = zero
Cpk = between 0 and 1
Cpk = 1
Cpk > 1
Figure S6.8
Keep whole shipment
100 –
75 –
50 –
25 –
0 –
Return whole shipment
P(Accept Whole Shipment)
CutOff

0102030405060708090100
% Defective in Lot
100 –
95 –
75 –
50 –
25 –
10 –
0 –
= 0.05 producer’s risk for AQL
Probability of Acceptance

012345678
Percent defective
= 0.10
AQL
LTPD
Consumer’s risk for LTPD
Good lots
Indifference zone
Bad lots
Figure S6.9
n = 50, c = 1
n = 100, c = 2
(Pd)(Pa)(N  n)
N
AOQ =
where
Pd= true percent defective of the lot
Pa= probability of accepting the lot
N= number of items in the lot
n= number of items in the sample
If a sampling plan replaces all defectives
If we know the incoming percent defective for the lot
We can compute the average outgoing quality (AOQ) in percent defective
The maximum AOQ is the highest percent defective or the lowest average quality and is called the average outgoing quality level (AOQL)
Lower specification limit
Upper specification limit
(a)Acceptance sampling (Some bad units accepted)
(b)Statistical process control (Keep the process in control)
(c)Cpk >1 (Design a process that is in control)
Process mean, m
Figure S6.10