NATIONAL PRODUCTIVITY COUNCIL
WELCOMES YOU TO A
PRESENTATION
ON
CONTROL CHARTS
By B.Girish
Dy. Director
Three SQC Categories

Traditional descriptive statistics


Acceptance sampling used to randomly inspect a batch
of goods to determine acceptance/rejection


e.g. the mean, standard deviation, and range
Does not help to catch in-process problems
Statistical process control (SPC)

Involves inspecting the output from a process

Quality characteristics are measured and charted

Helpful in identifying in-process variations
Statistical Process Control (SPC)


A methodology for monitoring a
process to identify special causes of
variation and signal the need to take
corrective action when appropriate
SPC relies on control charts
SPC Implementation
Requirements





Top management commitment
Project champion
Initial workable project
Employee education and training
Accurate measurement system
Traditional Statistical Tools

The Mean- measure of central
tendency
n
x


x
i 1
i
n
The Range- difference between
largest/smallest observations in a
set of data
Standard Deviation measures
the amount of data dispersion
around mean
 x
n
σ
i 1
i
X
n 1

2
Distribution of Data

Normal distributions

Skewed distribution
Sources of Variation

Common causes of variation

Random causes that we cannot identify

Unavoidable


e.g. slight differences in process variables like diameter, weight,
service time, temperature
Assignable causes of variation

Causes can be identified and eliminated

e.g. poor employee training, worn tool, machine needing repair
Common
Causes
Special
Causes
Histograms do not
take into account
changes over time.
Control charts
can tell us
when a process
changes
Introduction to Control Charts
Important uses of the control chart
1. Most processes do not operate in a state of statistical
control
2. Consequently, the routine and attentive use of control
charts will identify assignable causes. If these causes
can be eliminated from the process, variability will be
reduced and the process will be improved
3. The control chart only detects assignable causes.
Management, operator, and engineering action will be
necessary to eliminate the assignable causes.
Introduction to Control Charts

Monitor Variation in Data


Exhibit trend - make correction before
process is out of control
A Process - A Repeatable Series of Steps
Leading to a Specific Goal
Introduction to Control Charts

(continued)
Show When Changes in Data are Due to:

Special or assignable causes




Fluctuations not inherent to a process
Represent problems to be corrected
Data outside control limits or trend
Chance or common causes


Inherent random variations
Consist of numerous small causes of random variability
Introduction to Control Charts
X
UCL
Mean
11
Time
9
7
5
LCL
3
Common
Cause
Variation
80
60
40
20
0
1
time
Special
Cause
Variation
Graph of sample data plotted over
Process
Average

Commonly Used Control Charts

Variables data




x-bar and R-charts
x-bar and s-charts
Charts for individuals (x-charts)
Attribute data


For “defectives” (p-chart, np-chart)
For “defects” (c-chart, u-chart)
Introduction to Control Charts
Popularity of control charts
1) Control charts are a proven technique for improving
productivity.
2) Control charts are effective in defect prevention.
3) Control charts prevent unnecessary process adjustment.
4) Control charts provide diagnostic information.
5) Control charts provide information about process
capability.
Control Chart Selection
Quality Characteristic
variable
attribute
defective
n>1?
no
x and MR
yes
n>=10 or no
computer?
yes
x and s
defect
x and R
constant
sample
size?
yes
no
p-chart with
variable sample
size
constant
sampling
unit?
p or
np
yes
no
c
u
SPC Methods-Control Charts



Control Charts show sample data plotted on a graph with CL, UCL,
and LCL
Control chart for variables are used to monitor characteristics that
can be measured, e.g. length, weight, diameter, time
Control charts for attributes are used to monitor characteristics
that have discrete values and can be counted, e.g. % defective,
number of flaws in a shirt, number of broken eggs in a box
Control Charts for Attributes –
P-Charts & C-Charts

Use P-Charts for quality characteristics that
are discrete and involve yes/no or
good/bad decisions



Number of leaking caulking tubes in a box of 48
Number of broken eggs in a carton
Use C-Charts for discrete defects when
there can be more than one defect per unit


Number of flaws or stains in a carpet sample cut from
a production run
Number of complaints per customer at a hotel
Control Chart Design Issues




Basis for sampling
Sample size
Frequency of sampling
Location of control limits
Developing Control Charts
1.
Prepare



2.
Choose measurement
Determine how to collect data, sample
size, and frequency of sampling
Set up an initial control chart
Collect Data



Record data
Calculate appropriate statistics
Plot statistics on chart
Pre-Control
LTL
Red
Zone
UTL
Green Zone
nominal
value
Yellow Zones
Red
Zone
Control Limits
UCL = Process Average + 3 Standard
Deviations
X = Process Average - 3 Standard Deviations
LCL
UCL
+ 3
Process
Average
- 3
LCL
TIME
Next Steps
3.
Determine trial control limits


4.
Center line (process average)
Compute UCL, LCL
Analyze and interpret results



Determine if in control
Eliminate out-of-control points
Recompute control limits as
necessary
Setting Control Limits

Percentage of values
under normal curve

Control limits balance
risks like Type I error
Comparing Control Chart
Patterns
X
X
Common Cause
Variation: No Points
Outside Control
Limits
X
Special Cause
Variation: 2 Points
Outside Control
Limits
Downward Pattern:
No Points Outside
Control Limits but
Trend Exists
Typical Out-of-Control Patterns







Point outside control limits
Sudden shift in process average
Cycles
Trends
Hugging the center line
Hugging the control limits
Instability
Control Charts for Variables




Use x-bar and Rbar charts together
Used to monitor
different variables
X-bar & R-bar
Charts reveal
different problems
In statistical control
on one chart, out
of control on the
other chart? OK?
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Processes In
Control
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Process Out of Control
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Shift in Process Average
Identifying Potential Shifts
Cycles
Trend
PROCESS STREAMS
UCL
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Sudden stability
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Mixtures
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Center
line
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LCL
Interpreting Control Charts
Chart
Description
Example # 1
Chart points do
Process not form a parti- UCL 10
cular pattern and
In
x 19
Control lie within the
upper and lower lcl 30
chart limits
Chart points
UCL 10
form a particular
Process pattern OR one
x 19
Out of or more points
Control lie beyond the
lcl 30
uppor or lower
chart limits
Run
Chart points are
UCL 10
on one side of the
center line. The
x 19
number of points
in a run is called lcl 30
the “length of the
run”
Example # 2
UCL 10
x 19
lcl 30
UCL 10
x 19
lcl 30
UCL 10
x 19
lcl 30
Interpretation
The process is stable, not
changing. Doesn’t
necesarily mean to leave
the process alone. May
be opportunities to improve
the process and enjoy
substantial benefits
Alerts us that the process
is changing. Doesn’t mean
you need to take a
corrective action. May be
relate to a change you have
made. Be sureto identify
the reason\(s) before taking
any constructive actions(w)
Suggest the process has
undergone a permanent
change (positive or
negative) and is now
becoming stable. Often
requires tha t you
recompute the control
lines for future interpretation efforts.
Interpreting Control Charts
Chart
Trend
Cycle
Description
Example # 1
A continued rise
or fall in a series UCL 10
of points (7 or
more consecutivex 19
points direction) lcl 30
1
3
UCL 10
2
3
5
UCL 10
x 19
lcl 30
CL 10
1/2
1/2
1/2
1
4
x 19
lcl 30
Chart ponts show
UCL 10
the same pattern
changes (e.g.rise
x 19
or fall) over equal
periods of time lcl 30
Chart points are UCL 10
close to the center
Hugging line or to a
x 19
control limit line
lcl 30
(2 out of 3, 3 out of
4, or 4 out of 10.)
2
4
5
67
Example # 2
x 19
lcl 30
1/2
6
Interpretation
Often seen after some change
has been made. Helps tell
you if the change(s) had a
positive or negative effect.
may also be part of a
learning curve associated
with some form of training
often relates to factors that
influence the process in a
predictable manner. Factors
occur over a set time period
and a positive/negative effect
Helps determine future work
load and staffing levels
Suggests a different type of
data has been mixed into the
sub-group being sampled.
Often need to change the
sub-group, reassemble the
data, redraw the control
chart
When to Take Corrective
Action

Corrective Action Should Be Taken
When Observing Points Outside the
Control Limits or when a Trend Has
Been Detected


Eight consecutive points above the center
line (or eight below)
Eight consecutive points that are increasing
(decreasing)
Out-of-Control Processes

If the Control Chart Indicates an Out-of-Control
Condition (a Point Outside the Control Limits or
Exhibiting Trend)


Contains both common causes of variation and
assignable causes of variation
The assignable causes of variation must be identified


If detrimental to quality, assignable causes of variation
must be removed
If increases quality, assignable causes must be incorporated
into the process design
In-Control Process

If the Control Chart is Not Indicating
Any Out-of-Control Condition, then


Only common causes of variation exist
It is sometimes said to be in a state of
statistical control


If the common-cause variation is small, then
control chart can be used to monitor the
process
If the common-cause variation is too large, the
process needs to be altered
Types of Error

First Type:


Belief that observed value represents
special cause when, in fact, it is due to
common cause
Second Type:

Treating special cause variation as if it is
common cause variation
Remember


Control does not mean that the product
or service will meet the needs. It only
means that the process is consistent
(may be consistently bad).
Capability of meeting the specification.
How to use the results

By eliminating the special causes first
and then reducing the common causes,
quality can be improved.
Final Steps
5.
Use as a problem-solving tool


6.
Continue to collect and plot data
Take corrective action when
necessary
Compute process capability
Process Capability


Product Specifications

Preset product or service dimensions, tolerances

e.g. bottle fill might be 16 oz. ±.2 oz. (15.8oz.-16.2oz.)

Based on how product is to be used or what the customer expects
Process Capability – Cp and Cpk


Assessing capability involves evaluating process variability relative to
preset product or service specifications
Cp assumes that the process is centered in the specification range
spe cificat
ion width USL  LSL
Cp

proce ss width
6σ

Cpk helps to address a possible lack of centering of the process
 USL  μ μ  LSL 
C pk  min
,

3σ 
 3σ
Relationship between Process
Variability and Specification Width

Three possible ranges for Cp





Cp = 1, as in Fig. (a), process
variability just meets
specifications
Cp ≤ 1, as in Fig. (b), process not
capable of producing within
specifications
Cp ≥ 1, as in Fig. (c), process
exceeds minimal specifications
One shortcoming, Cp assumes
that the process is centered on
the specification range
Cp=Cpk when process is centered
Computing the Cp Value at Cocoa Fizz: three bottling
machines are being evaluated for possible use at the Fizz plant.
The machines must be capable of meeting the design
specification of 15.8-16.2 oz. with at least a process
capability index of 1.0 (Cp≥1)

The table below shows the information
gathered from production runs on each
machine. Are they all acceptable?

Solution:

Machine A
Cp
Machine
σ
USL-LSL
6σ

A
.05
.4
USL  LSL
.4

 1.33
6σ
6(.05)
Machine B
.3
B
.1
.4
.6
C
.2
.4
1.2
Cp=

Machine C
Cp=
Computing the Cpk Value at Cocoa Fizz


Design specifications call for a
target value of 16.0 ±0.2 OZ.
(USL = 16.2 & LSL = 15.8)
Observed process output has now
shifted and has a µ of 15.9 and a
σ of 0.1 oz.
 16.2 15.9 15.9 15.8

C pk  min
,
3(.1)
3(.1)


.1
C pk   .33
.3

Cpk is less than 1, revealing that
the process is not capable
Process Capability
(2)
11.0  10.7171
UTL  
C pu 
3
  LTL
C pl 
3
C pk  minC pl , C pu 
C pu 
3  0.0868
C pl 
C pk  C p 1  K  where K 
 1.086
10 .7171  10 .5
 0.834
3  0.0868
2  T
Tolerance
Example: same as above,but assume processis centeredat 10.7171mm
Cp
C pm 
1
  T 
2
T is theT arget
2
0.960
C pm 
10.7171 10.75
2
1
0.8682
 0.8977
Capability Versus Control
Control
Capability
Capable
Not Capable
In Control
IDEAL
Out of Control
Excel Template
Special Variables Control Charts
x-bar and s charts
 x-chart for individuals

Charts for Attributes

Fraction nonconforming (p-chart)


Fixed sample size
Variable sample size

np-chart for number nonconforming

Charts for defects


c-chart
u-chart
p Chart

Control Chart for Proportions


Is an attribute chart
Shows Proportion of Nonconforming (Success )
Items

E.g., Count # of nonconforming chairs & divide by
total chairs inspected


Chair is either conforming or nonconforming
Used with Equal or Unequal Sample Sizes Over
Time

Unequal sizes should not differ by more than ±25%
from average sample size
p Chart
Control Limits

p(1  p)
p (1  p ) 
LCLp  max  0, p  3
 UCLp  p  3
n 
n

Average Group Size
k
n
n
i 1
k
i
Average Proportion of
Nonconforming Items
k
# Defective
Items in
Xi
Sample i
i 1
p
# of Samples

k
n
i 1
i
Size of
Sample i
p Chart
Example
You’re manager of a
500-room hotel. You
want to achieve the
highest level of
service. For 7 days,
you collect data on the
readiness of 200
rooms. Is the process
in control?
p Chart
Hotel Data
Day
1
2
3
4
5
6
7
# Rooms
200
200
200
200
200
200
200
# Not
Ready Proportion
16
0.080
7
0.035
21
0.105
17
0.085
25
0.125
19
0.095
16
0.080
p Chart
Control Limits Solution
n
n
i 1
k
i
16 + 7 +...+ 16
k
k
1400

 200
7
p
X
i 1
k
n
i 1

  .0864  3
i
121

 .0864
1400
i
.0864 1  .0864 
p3
200
n
 .0864  .0596 or .0268,.1460 
p 1 p
p Chart
Control Chart Solution
0.15
P
UCL
0.10
Mean p
0.05
LCL
0.00
1
2
3
4
Day
5
6
7
Individual points are distributed around p without any pattern.
Any improvement in the process must come from reduction of
common-cause variation, which is the responsibility of the
management.
p Chart in PHStat


PHStat | Control Charts | p Chart …
Excel Spreadsheet for the Hotel Room
Example
Understanding Process Variability:
Red Bead Example
Four workers (A, B, C, D) spend 3 days to collect beads,
at 50 beads per day. The expected number of red
beads to be collected per day per worker is 10 or 20%.
Worker
Day 1
Day 2
Day 3
A
9 (18%)
11 (12%)
6 (12%)
26 (17.33%)
B
12 (24%)
12 (24%)
8 (16%)
32 (21.33%)
C
13 (26%)
6 (12%)
12 (24%)
31(20.67%)
D
7 (14%)
9 (18%)
8 (16%)
24 (16.0%)
Totals
41
38
34
All Days
113
Understanding Process Variability:
Example Calculations
Average
Day 1
Day 2
Day 3
X
10.25
9.5
8.5
9.42
p
20.5%
19%
17%
18.83%
_
113
p
 .1883
50(12)
All Days
p(1  p)
.1883(1  .1883)
p 3
 .1883  3
n
50
 .1883  .1659
LCL  .1883  .1659  .0224
UCL  .1883 +.1659  .3542
Understanding Process Variability:
Example Control Chart
UCL
.30
_
p
.20
.10
LCL
0
A1
B1
C1
D1
A2
B2 C2
D2
A3
B3
C3
D3
Morals of the Example
Variation is an inherent part
of any process.
 The system is primarily
responsible for worker
performance.
 Only management can change the system.
 Some workers will always be above average,
and some will be below.

The c Chart

Control Chart for Number of Nonconformities
(Occurrences) in a Unit (an Area of
Opportunity)


Shows Total Number of Nonconforming Items
in a Unit


Is an attribute chart
E.g., Count # of defective chairs manufactured per
day
Assume that the Size of Each Subgroup Unit
Remains Constant
c Chart Control Limits
LCLc  c  3 c
UCLc  c  3 c
Average Number of
Occurrences
k
c
c
i 1
k
i
# of Occurrences in Sample i
# of Samples
c Chart: Example
You’re manager of a
500-room hotel. You
want to achieve the
highest level of
service. For 7 days,
you collect data on the
readiness of 200
rooms. Is the process
in control?
c Chart: Hotel Data
Day
1
2
3
4
5
6
7
# Rooms
200
200
200
200
200
200
200
# Not
Ready
16
7
21
17
25
19
16
c Chart:
Control Limits Solution
k
c
c
i 1
k
i

16  7 
 19  16
7
 17.286
LCLc  c  3 c  17.286  3 17.285  4.813
UCLc  c  3 c  29.759
c Chart:
Control Chart Solution
30
c
UCL
20
c
10
LCL
0
1
2
3
4
Day
5
6
7
Individual points are distributed around c without any pattern.
Any improvement in the process must come from reduction of
common-cause variation, which is the responsibility of the
management.
Variables Control Charts:
R Chart

Monitors Variability in Process



Characteristic of interest is measured on
numerical scale
Is a variables control chart
Shows Sample Range Over Time


Difference between smallest & largest
values in inspection sample
E.g., Amount of time required for luggage
to be delivered to hotel room
R Chart
Control Limits
UCLR  D4 R
From
Table
LCLR  D3 R
k
R
R
i 1
k
i
Sample Range at
Time i or
Subgroup i
# Samples
R Chart
Example
You’re manager of a
500-room hotel. You
want to analyze the
time it takes to deliver
luggage to the room.
For 7 days, you collect
data on 5 deliveries
per day. Is the process
in control?
R Chart and Mean Chart
Hotel Data
Day
1
2
3
4
5
6
7
Sample
Average
5.32
6.59
4.88
5.70
4.07
7.34
6.79
Sample
Range
3.85
4.27
3.28
2.99
3.61
5.04
4.22
R Chart
Control Limits Solution
k
R
R
i 1
k
i
3.85  4.27 

7
 4.22
 3.894
UCLR  D4  R  2.114  3.894  8.232
LCLR  D3  R  0  3.894  0
From Table
(n = 5)
R Chart
Control Chart Solution
Minutes
8
6
4
2
0
1
2
UCL
_
R
LCL
3
4
Day
5
6
7
Variables Control Charts:
Mean Chart (The X Chart)

Shows Sample Means Over Time



Compute mean of inspection sample over
time
E.g., Average luggage delivery time in
hotel
Monitors Process Average

Must be preceded by examination of the R
chart to make sure that the process is in
control
Mean Chart
Computed
From
Table
Sample
Mean at
Time i
UCLX  X  A2 R
LCLX  X  A2 R
k
X
X
i 1
k
k
i
and
R
R
i 1
k
i
Sample
Range
at Time i
# Samples
Mean Chart Example
You’re manager of a 500room hotel. You want to
analyze the time it takes
to deliver luggage to the
room. For 7 days, you
collect data on 5
deliveries per day. Is the
process in control?
R Chart and Mean Chart
Hotel Data
Day
1
2
3
4
5
6
7
Sample
Average
5.32
6.59
4.88
5.70
4.07
7.34
6.79
Sample
Range
3.85
4.27
3.28
2.99
3.61
5.04
4.22
Mean Chart
Control Limits Solution
k
X 
X
i 1
k
i
5.32  6.59 

7
 6.79
k
R
R
i 1
k
i
3.85  4.27 

7
 4.22
 5.813
From
Table E.9
(n = 5)
 3.894
UCLX  X  A2  R  5.813  0.577  3.894  8.060
LCLX  X  A2  R  5.813  0.577  3.894  3.566
Mean Chart
Control Chart Solution
Minutes
8
6
4
2
0
1
2
UCL
LCL
3
4
Day
5
6
7
__
X
R Chart and Mean Chart
in PHStat


PHStat | Control Charts | R & Xbar
Charts …
Excel Spreadsheet for the Hotel Room
Example
Process Capability




Process Capability is the Ability of a Process to
Consistently Meet Specified Customer-Driven
Requirements
Specification Limits are Set by Management in
Response to Customer’s Expectations
The Upper Specification Limit (USL) is the
Largest Value that Can Be Obtained and Still
Conform to Customer’s Expectation
The Lower Specification Limit (LSL) is the
Smallest Value that is Still Conforming
Estimating Process Capability



Must Have an In-Control Process First
Estimate the Percentage of Product or Service
Within Specification
Assume the Population of X Values is
Approximately Normally Distributed with
Mean Estimated by X and Standard Deviation
Estimated by R / d
2
(continued)
Estimating Process Capability

For a Characteristic with an LSL and a USL
P(an
outcome will be within specification)

 P(LSL  X  USL)
 LSL  X
USL  X
= P
Z
 R / d2
R / d2






where Z is a standardized normal random variable
(continued)
Estimating Process Capability

For a Characteristic with Only a LSL
P(an outcome will be within specification)
 P(LSL  X )

 LSL  X

= P
Z
 R / d2




where Z is a standardized normal random variable
(continued)
Estimating Process Capability

For a Characteristic with Only a USL
P(an outcome will be within specification)
 P(X  USL)


USL  X
= P Z 

R / d2






where Z is a standardized normal random variable
Process Capability Example
You’re manager of a
500-room hotel. You
have instituted a policy
that 99% of all
luggage deliveries
must be completed
within 10 minutes or
less. For 7 days, you
collect data
on 5 deliveries per
Process Capability:
Hotel Data
Day
1
2
3
4
5
6
7
Sample
Average
5.32
6.59
4.88
5.70
4.07
7.34
6.79
Sample
Range
3.85
4.27
3.28
2.99
3.61
5.04
4.22
Process Capability:
Hotel Example Solution
n5
R  3.894 and d2  2.326
P(A delivery is made within specification)
= P(X  10)
X  5.813
10  5.813 

= P Z 

3.894 / 2.326 

= P( Z  2.50)  .9938
Therefore, we estimate that 99.38% of the
luggage deliveries will be made within the 10
minutes or less specification. The process is
capable of meeting the 99% goal.
Capability Indices

Aggregate Measures of a Process’ Ability to
Meet Specification Limits


The larger (>1) the values, the more capable a
process is of meeting requirements
Measure of Process Potential Performance
USL  LSL specification spread
Cp 

process spread
6  R / d2 


Cp>1 implies that a process has the potential of
having more than 99.73% of outcomes within
specifications
Capability Indices

(continued)
Measures of Actual Process Performance

For one-sided specification limits
X  LSL

CPL 
3  R / d2 
USL  X
CPU 
3  R / d2 


CPL (CPU) >1 implies that the process mean is more
than 3 standard deviations away from the lower (upper)
specification limit
Capability Indices

(continued)
For two-sided specification limits
Cpk  min CPL, CPU 



Cpk = 1 indicates that the process average is 3
standard deviations away from the closest
specification limit
Larger Cpk indicates larger capability of meeting
the requirements
Process Capability Example
You’re manager of a
500-room hotel. You
have instituted a policy
that all luggage
deliveries must be
completed within 10
minutes or less. For 7
days, you collect data
on 5 deliveries per day.
Compute an appropriate
capability index for the
Process Capability:
Hotel Data
Day
1
2
3
4
5
6
7
Sample
Average
5.32
6.59
4.88
5.70
4.07
7.34
6.79
Sample
Range
3.85
4.27
3.28
2.99
3.61
5.04
4.22
Process Capability:
Hotel Example Solution
n5
X  5.813
R  3.894
and d2  2.326
USL  X
10  5.813
CPU 

 0.833672
3  R / d 2  3  3.894 / 2.326 
Since there is only the upper specification limit, we
need to only compute CPU. The capability index for
the luggage delivery process is .8337, which is less
than 1. The upper specification limit is less than 3
standard deviations above the mean.