Transcript Lean Thinking

Systems Thinking and the
Theory of Constraints
A Statement for Quality Goes Here
These sides and note were prepared using
1. Managing Business Flow processes. Anupindi, Chopra, Deshmukh, Van Mieghem,
and Zemel.Pearson Prentice Hall.
2. Few of the graphs of the slides of Prentice Hall for this book, originally prepared by
professor Deshmukh.
Introduction ~ The Garage Door Manufacturer
According to the sales manager of a high-tech manufacturer of
garage doors, while the company has 15% of market share,
customers are not satisfied
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Door Quality in terms of safety, durability, and ease of use
High Price compared competitors’ process
Not on-time orders
Poor After Sales Service
We can not rely of subjective statements and opinions
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Collect and analyze concrete data –facts- on performance
measures that drive customer satisfaction
Identify, correct, and prevent sources of future problems
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9.1 Performance Variability
All internal and external performance measures display vary
from tome to time.
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
External Measurements - customer satisfaction, product
rankings, customer complaints.
Internal Measurements - flow units cost, quality, and time.
No two cars rolling off an assembly line have identical cost. No
two customers for identical transaction spend the same time in
a bank. The same meal you have had in two different
occasions in a restaurant do not taste exactly the same.
 Sources of Variability
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Internal: imprecise equipment, untrained workers, and lack of
standard operating procedures.
External: inconsistent raw materials, supplier delays, consumer
taste change, and changing economic conditions.
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9.1 Performance Variability
 A discrepancy between the actual and the expected
performance often leads to cost↑, flow time↑, quality↓ 
dissatisfied customers.
 Processes with greater variability are judged less satisfactory
than those with consistent, predictable performance.
 What is the base of the customer judgment the exact unit of
product or service s/he gets, not how the average product
performs. Customers perceive any variation in their product or
service from what they expected as a loss in value.
 In general, a product is classified as defective if its cost, quality,
availability or flow time differ significantly from their expected
values, leading to dissatisfied customers.
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Quality Management Terms
 Quality of Design. How well product specifications aim to
meet customer requirements (what we promise consumers ~ in
terms of what the product can do). Quality Function
Deployment (QFD) is a conceptual framework for translating
customers’ functional requirements (such as ease of operation
of a door or its durability) into concrete design specifications
(such as the door weight should be between 75 and 85 kg.)
 Quality of Conformance. How closely the actual product
conforms to the chosen design specifications. Ex. # defects per
car, fraction of output that meets specifications. Ex. irline
conformance can be measured in terms of the percentage of
flights delayed for more than 15 minutes OR the number of
reservation errors made in a specific period of time.
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9.2 Analysis of Variability
To analyze and improve variability there are diagnostic tools to
help us:
1.
2.
3.
4.
5.
Monitor the actual process performance over time
Analyze variability in the process
Uncover root causes
Eliminate those causes
Prevent them from recurring in the future
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9.2.1 Check Sheets
 check Sheet is simply a tally of the types and frequency of
problems with a product or a service experienced by customers.
 Pareto Chart is a bar chart of frequencies of occurrences in nonincreasing order. The 80-20 Pareto principle states that 20% of
problem types account for 80% of all occurrences.
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Type of Complaint
Number of Complaints
Cost
IIII IIII
Response Time
IIII
Customization
IIII
Service Quality
IIII IIII IIII
Door Quality
IIII IIII IIII IIII IIII
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Service Quality
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Response Time
Customization
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9.2.3 Histograms
Collect data on door weight – Ex. one door, five times a day, 20
days, total of 100 door weight.
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 Histogram is a bar plot that
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Frequency
displays the frequency
distribution of an observed
performance characteristic. Ex.
14% of the doors weighed about
83 kg, 8% weighed about 81 kg,
and so forth.
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9.2.4 Run Charts
 Run chart is a plot of some measure of process performance
monitored over time.
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9.2.5 Multi-Vari Charts
Multi-vari chart is a plot of high-average-low values of
performance measurement sampled over time.
Time\Day
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Comparison
 Pareto Chart. The importance of each item. Quality was the
most important item. Quality was then defined as finish, ease
of use, and durability. Ease of use and durability which are
subjective, must be translated into some thing measurable. We
translate them into weight. If weight is high, it cannot operate
easily, if weight is low, it will not be durable. A high quality
door, based on engineering design must weight 82.5 lbs.
 Histogram. Shows the tendency (mean) and the standard
deviation. Ex. For door weight.
 Run Chart. Can show trend.
 Multi-Vari Chart. Shows average and variability inside the
samples and among the samples.
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Process Management
Two aspects to process management;
 Process planning’s goal is to produce and deliver products
that satisfy targeted customer needs.
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Structuring the process
Designing operating procedures
Developing key competencies such as process capability,
flexibility, capacity, and cost efficiency
 Process control’s goal is to ensure that actual performance
conforms to the planned performance.
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Tracking deviations between the actual and the planned
performance and taking corrective actions to identify and
eliminate sources of these variations.
There could be various reasons behind variation in performance.
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9.3.1 The Feedback Control Principle
Process performance
management is based on
the general principle of
feedback control of
dynamical systems.
Applying the feedback control principle to process control.
“involves periodically monitoring the actual process
performance (in terms of cost, quality, availability, and response
time), comparing it to the planned levels of performance,
identifying causes of the observed discrepancy between the two,
and taking corrective actions to eliminate those causes.”
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Plan-Do-Check-Act (PDCA)
 Process planning and process control are similar to the
Plan-Do-Check-Act (PDCA) cycle. Performed
continuously to monitor and improve the process
performance.
 Problems in Process Control



Performance variances are determined by comparison of
the current and previous period’s performances.
Decisions are based on results of this comparison.
Some variances may be due to factors beyond a worker’s
control.
 According to W. Edward Deming, incentives based on
factors that are beyond a worker’s control is like
rewarding or punishing workers according to a lottery.
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Two categories of performance variability
 Normal Variability. Is statistically predictable and
includes both structural variability and stochastic
variability. Cannot be removed easily. Is not in
worker’s control. Can be removed only by process redesign, more precise equipment, skilled workers, better
material, etc.
 Abnormal variability. Unpredictable and disturbs the
state of statistical equilibrium of the process by
changing parameters of its distribution in an
unexpected way. Implies that one or more performance
affecting factors may have changed due to external
causes or process tampering. Can be identified and
removed easily therefore is worker’s responsibility.
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Process Control
 If observed performance variability is
 Normal - due to random causes - process is in control
 Abnormal - due to assignable causes - process is out of
control
 The short run goal is:
1. Estimate normal stochastic variability.
2. Accept it as an inevitable and avoid tampering
3. Detect presence of abnormal variability
4. Identify and eliminate its sources
 The long run goal is to reduce normal variability by
improving process.
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9.3.3 Control Limit Policy
 How to decide whether observed variability is normal or
abnormal?
 Control Limit Policy
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
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Control band - A range within which any variation in
performance is interpreted as normal due to causes that
cannot be identified or eliminated in short run.
Variability outside this range is abnormal.
Lower limit of acceptable mileage, control band for house
temperature.
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Process Control
 Process control is useful to control any type of
process.
 Application of control limit policy
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Managing inventory, process capacity and flow time.
Cash management - liquidate some assets if cash falls
below a certain level.
Stock trading - purchase a stock if and when its price
drops to a specific level.
 Control limit policy has usage in a wide variety
of business in form of critical threshold for
taking action
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9.3.4 Statistical Process Control
 Statistical process control involves setting a “range of
acceptable variations” in the performance of the process,
around its mean.
 If the observed values are within this range:
 Accept the variations as “normal”
 Don’t make any adjustments to the process
 If the observed values are outside this range:
 The process is out of control
 Need to investigate what’s causing the problems – the
assignable cause
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9.3.4 Process Control Charts
 Let  be the expected value and  be the standard deviation
of the performance. Set up an Upper Control Limit (UCL)
and a Lower Control Limit (LCL).
LCL =  - z
UCL =  + z
Decide how tightly to monitor and control the process. The
smaller the z, the tighter the control
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9.3.4 Process Control Charts
 If observed data within the control limits and does not
show any systematic pattern  Performance variability is
normal . Otherwise  Process is out of control
 Type I error ( error). Process is in control, its statistical
parameters have not changed, but data falls outside the
limits.
 Type II error ( error) Process is out of control, its statistical
parameters have changed, but data falls inside the limits.
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9.3.4 Control Charts … Continued
 Optimal Degree of Control depends on 2 things:


How much variability in the performance measure we consider
acceptable
How frequently we monitor the process performance.
 Optimal frequency of monitoring is a balance between the
costs and benefits
 If we set ‘z’ to be too small: We’ll end up doing unnecessary
investigation. Incur additional costs.
 If we set ‘z’ to be too large: We’ll accept a lot more variations as
normal. We wouldn’t look for problems in the process – less
costly
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9.3.4 Control Charts … Continued
In practice, a value of z = 3 is used. 99.73% of all measurements will
fall within the “normal” range
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We have collected 20 samples, each of size 5, n=5, of our variable
of interest X – the door weight in our example. We have 100
pieces of data. We can simple use excel to compute the average
and standard deviation of this data.
Overall average weight X  82.5
Standard deviation s  4.2
Variance s 2  17.64
A higher value of the average indicates a shift in the entire
distribution to the right, so that all doors produced are
consistently heavier. An increase in the value of the standard
deviation means a wider spread of the distribution around the
mean, implying that many doors are much heavier or lighter than
the overall average weight.
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X Bar Chart
If we compute the average of the random variable X, in each
sample of n, in our example 5, and show it by
X
Average Door Weigh t in each sample : X 
n
X has any dostributi on with Mean  and Standard Deviation of 
X has Normal dostributi on with Mean  and Standard Deviation of 
n
Average of Average Door Weigh t : X  82.5
s
4.2
Standard Deviation of Average Door Weigh t : s X 

 1.88
n
5
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X Bar Chart
Therefore, if we compute the average weight door
68.26% of all doors will weigh within 82.5 + (1)(1.88),
95.44% of doors will weight within 82.5 + (2)(1.88), and
99.73% of door weights will be within 82.5 + (3)(1.88), or
between and 76.86 and 88.14 .
UCL
Average
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LCL
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Day
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Process Control and Improvement
Out of Control
In Control
Improved
UCL

LCL
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R Chart
Range in a Sample of Size n : R
Time\Day
Range
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Average Range in a Sample of Size n : R
Standard Deviation of R : sR
R  10.1
sR  3.5
Range
UCL = 10.1+3(3.5) = 20.6 , LCL = 10.1-3(3.5) = -0.4 = 0
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UCL
LCL
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Day
Process Is “In Control” (i.e., variation is stable)
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Number of Defects (c) Chart
Discrete Quality Measurement:
D = Number of “defects” (errors) per unit of work
Examples: Number of typos/page, errors/thousand transactions,
equipment breakdowns/shift, bags lost/thousand flown,
power outages/year, customer complaints/month,
defects/car.......
If
n = No. of opportunities for defects to occur, and
p = Probability of a defect/error occurrence in each
then
D
~

Binomial (n, p) with mean np, variance np(1-p)
Poisson (m) with m = mean = variance = np , if
n is large (≥ 20) and p is small (≤ 0.05)
With m = np = average number of defects per unit,
Control limits = m + 3 √m
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Performance Variation
Stable
Unstable
Trend Cyclical Shift
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Performance Variation
Stable
Unstable
Trend
Cyclical
Shift
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Performance Variation
Stable
Unstable
Trend
Cyclical
Shift
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Performance Variation
Stable
Unstable
Trend
Cyclical
Shift
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Performance Variation
Stable
Unstable
Trend
Cyclical
Shift
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X Bar Chart
If the door weight distribution was Normal, 68.26% of all
doors will weigh within 82.5 + (1)(4.2), 95.44% of doors
will weight within 82.5 + (2)(4.2), and 99.73% of door
weights will be within 82.5 + (3)(4.2). This would be the
distribution of the weight of each individual door.
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9.3.4 Control Charts … Continued

Average and Variation Control Charts
Let z = 3
Sample Averages
UCL = A + zs/n = 82.5 + 3 (4.2) / 5 = 88.13
LCL = A - zs/n = 82.5 – 3 (4.2) / 5 = 76.87
Average Weight Control Chart
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Average Wt. (Kg)
88
UCL = 88.13
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LCL = 76.87
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9.3.4 Control Charts … Continued

Average and Variation Control Charts
Let z = 3
Sample Variances
UCL = V + z sV = 10.1 + 3 (3.5) = 20.6
LCL = V - zs sV = 10.1 – 3 (3.5) = - 0.4
Variance (range) of Wt.
(Kg)
Variance Control Chart
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UCL = 20.6
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LCL = 0
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9.3.4 Control Charts … Continued

Extensions
Continuous Variables:
 Garage Door Weights
 Processing Costs
 Customer Waiting Time
Use Normal distribution
Discrete Variables:
 Number of Customer Complaints
 Whether a Flow Unit is Defective
 Number of Defects per Flow Unit Produced
Use Binomial or Poisson distribution
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9.3.5 Cause-Effect Diagrams

Cause-Effect Diagrams
Sample
Plot
Abnormal
Observation
s
Control
Charts
Variability !!
Now what?!!
Brainstorm Session!!
Answer 5 “WHY” Questions !
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9.3.5 Cause-Effect Diagrams … Continued

Why…? Why…? Why…? (+2)
Our famous “Garage Door” Example:
1. Why are these doors so heavy?
Because the Sheet Metal was too ‘thick’.
2. Why was the sheet metal too thick?
Because the rollers at the steel mill were set
incorrectly.
3. Why were the rollers set incorrectly?
Because the supplier is not able to meet our
specifications.
4. Why did we select this supplier?
Because our Project Supervisor was too busy
getting the product out – didn’t have time to
research other vendors.
5. Why did he get himself in this situation?
Because he gets paid by meeting the
production quotas.
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9.3.5 Cause-Effect Diagrams … Continued

Fishbone Diagram
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9.3.6 Scatter Plots

The Thickness of the Sheet Metals
 Change Settings on Rollers
 Measure the Weight of the Garage Doors
 Determine Relationship between the two
Roller Settings & Garage Door Weights
Plot the results on a graph:
Door Weight (Kg)
Scatter Plot
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9.4 Process Capability
 Ease of external product measures (door operations and durability) and internal

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measures (door weight)
Product specification limits vs. process control limits
Individual units, NOT sample averages - must meet customer specifications.
Once process is in control, then the estimates of μ (82.5kg) and σ (4.2k) are reliable.
Hence we can estimate the process capabilities.
Process capabilities - the ability of the process to meet customer specifications
Three measures of process capabilities:
 9.4.1 Fraction of Output within Specifications
 9.4.2 Process Capability Ratios (Cpk and Cp)
 9.4.3 Six-Sigma Capability
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9.4.1 Fraction of Output within Specifications
 The fraction of the process output that meets customer specifications.
 We can compute this fraction by:
- Actual observation (see Histogram, Fig 9.3)
- Using theoretical probability distribution
Ex. 9.7:
- US: 85kg; LS: 75 kg (the range of performance variation that customer is
willing to accept)
See figure 9.3 Histogram: In an observation of 100 samples, the process is 74%
capable of meeting customer requirements, and 26% defectives!!!
OR:
 Let W (door weight): normal random variable with mean = 82.5 kg and
standard deviation at 4.2 kg,
Then the proportion of door falling within the specified limits is:
Prob (75 ≤ W ≤ 85) = Prob (W ≤ 85) - Prob (W ≤ 75)
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9.4.1 Fraction of Output within Specifications cont…
 Let Z = standard normal variable with μ = 0 and σ = 1, we can use the standard
normal table in Appendix II to compute:
AT US:
Prob (W≤ 85) in terms of:
Z = (W-μ)/ σ
As Prob [Z≤ (85-82.5)/4.2] = Prob (Z≤.5952) = .724 (see Appendix II)
(In Excel: Prob (W ≤ 85) = NORMDIST (85,82.5,4.2,True) = .724158)
AT LS:
Prob (W ≤ 75)
= Prob (Z≤ (75-82.5)/4.2) = Prob (Z ≤ -1.79) = .0367 in Appendix II
(In Excel: Prob (W ≤ 75) = NORMDIST(75,82.5,4.2,true) = .037073)
THEN:
Prob (75≤W≤85)
= .724 - .0367 = .6873
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9.4.1 Fraction of Output within Specifications cont…
SO with normal approximation, the process is capable of producing 69% of doors
within the specifications, or delivering 31% defective doors!!!
Specifications refer to INDIVIDUAL doors, not AVERAGES.
We cannot comfort customer that there is a 30% chance that they’ll get doors that is
either TOO LIGHT or TOO HEAVY!!!
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9.4.2 Process Capability Ratios (C pk and Cp)
 2nd measure of process capability that is easier to compute is the process capability





ratio (Cpk)
If the mean is 3σ above the LS (or below the US), there is very little chance of a
product falling below LS (or above US).
So we use:
(US- μ)/3σ
(.1984 as calculated later)
and (μ -LS)/3σ
(.5952 as calculated later)
as measures of how well process output would fall within our specifications.
The higher the value, the more capable the process is in meeting specifications.
OR take the smaller of the two ratios [aka (US- μ)/3σ =.1984] and define a single
measure of process capabilities as:
Cpk = min[(US-μ/)3σ, (μ -LS)/3σ]
(.1984, as calculated later)
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9.4.2 Process Capability Ratios (C pk and Cp)
 Cpk of 1+- represents a capable process
 Not too high (or too low)
 Lower values = only better than expected quality
Ex: processing cost, delivery time delay, or # of error per transaction process
 If the process is properly centered
 Cpk is then either:
(US- μ)/3σ or (μ -LS)/3σ
As both are equal for a centered process.
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9.4.2 Process Capability Ratios (C pk and Cp) cont…
 Therefore, for a correctly centered process, we may simply define the process
capability ratio as:
 Cp = (US-LS)/6σ
(.3968, as calculated later)
Numerator = voice of the customer / denominator = the voice of the process
 Recall: with normal distribution:
Most process output is 99.73% falls within +-3σ from the μ.
 Consequently, 6σ is sometimes referred to as the natural tolerance of the process.
Ex: 9.8
Cpk = min[(US- μ)/3σ , (μ -LS)/3σ ]
= min {(85-82.5)/(3)(4.2)], (82.5-75)/(3)(4.2)]}
= min {.1984, .5952}
=.1984
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9.4.2 Process Capability Ratios (C pk and Cp)
 If the process is correctly centered at μ = 80kg (between 75 and 85kg), we
compute the process capability ratio as
Cp = (US-LS)/6σ
= (85-75)/[(6)(4.2)]
= .3968
 NOTE: Cpk = .1984 (or Cp = .3968) does not mean that the process is
capable of meeting customer requirements by 19.84% (or 39.68%), of the
time. It’s about 69%.
 Defects are counted in parts per million (ppm) or ppb, and the process is
assumed to be properly centered. IN THIS CASE, If we like no more than
100 defects per million (.01% defectives), we SHOULD HAVE the
probability distribution of door weighs so closely concentrated around the
mean that the standard deviation is 1.282 kg, or Cp=1.3 (see Table 9.4)
Test: σ = (85-75)/(6)(1.282)] = 1.300kg
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Table 9.4
Table 9.4 Relationship Between Process Capability Ratio and Proportion Defective
Defects (ppm)
10000
1000
100
10
1
2 ppb
Cp
0.86
1
1.3
1.47
1.63
2
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9.4.3 Six-Sigma Capability




The 3rd process capability
Known as Sigma measure, which is computed as
S = min[(US- μ /σ), (μ -LS)/σ] (= min(.5152,1.7857) = .5152 to be calculated later)
S-Sigma process
If process is correctly centered at the middle of the specifications,
S = [(US-LS)/2σ]
Ex: 9.9
Currently the sigma capability of door making process is
S=min(85-82.5)/[(2)(4.2)] = .5952
By centering the process correctly, its sigma capability increases to
S=min(85-75)/[(2)(4.2)] = 1.19
THUS, with a 3σ that is correctly centered, the US and LS are 3σ away from the
mean, which corresponds to Cp=1, and 99.73% of the output will meet the
specifications.
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9.4.3 Six-Sigma Capability cont…
 SIMILARLY, a correctly centered six-sigma process has a standard deviation so
small that the US and LS limits are 6σ from the mean each.
 Extraordinary high degree of precision.
Corresponds to Cp=2 or 2 defective units per billion produced!!! (see Table 9.5)
 In order for door making process to be a six-sigma process, its standard deviation
must be:
σ = (85-75)/(2)(6)] = .833kg
 Adjusting for Mean Shifts
Allowing for a shift in the mean of +-1.5 standard deviation from the center of
specifications.
Allowing for this shift, a six-sigma process amounts to producing an average of 3.4
defective units per million. (see table 9.5)
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Table 9.5
Table 9.5 Fraction Defective and Sigma Measure
Sigma S
3
4
5
Capability Ratio Cp
1
1.33
1.667
Defects (ppm)
66810
6210
233
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2
3.4
54
9.4.3 Six-Sigma Capability cont…
 Why Six-Sigma?




See table 9.5
Improvement in process capabilities from a 3-sigma to 4-sigma = 10-fold
reduction in the fraction defective (66810 to 6210 defects)
While 4-sigma to 5-sigma = 30-fold improvement (6210 to 232 defects)
While 5-sigma to 6-sigma = 70-fold improvement (232 to 3.4 defects, per
million!!!).
 Average companies deliver about 4-sigma quality, where best-in-class companies
aim for six-sigma.
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9.4.3 Six-Sigma Capability cont…
 Why High Standards?

-
The overall quality of the entire product/process that requires ALL of them to
work satisfactorily will be significantly lower.
Ex:
If product contains 100 parts and each part is 99% reliable, the chance that the
product (all its parts) will work is only (.99)100 = .366, or 36.6%!!!
Also, costs associated with each defects may be high
Expectations keep rising
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9.4.3 Six-Sigma Capability cont…
 Safety capability
- We may also express process capabilities in terms of the desired margin [(US-LS)zσ] as safety capability
- It represents an allowance planned for variability in supply and/or demand
- Greater process capability means less variability
- If process output is closely clustered around its mean, most of the output will fall
within the specifications
- Higher capability thus means less chance of producing defectives
- Higher capability = robustness
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9.4.4 Capability and Control
 So in Ex. 9.7: the production process is not performing well in terms of MEETING
THE CUSTOMER SPECIFICATIONS. Only 69% meets output specifications!!!
(See 9.4.1: Fraction of Output within Specifications)
 Yet in example 9.6, “the process was in control!!!”, or WITHIN US & LS LIMITS.
 Meeting customer specifics: indicates internal stability and statistical predictability
of the process performance.
 In control (aka within LS and US range): ability to meet external customer’s
requirements.
 Observation of a process in control ensures that the resulting estimates of the
process mean and standard deviation are reliable so that our measurement of the
process capability is accurate.
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9.5 Process Capability Improvement
 Shift the process mean
 Reduce the variability
 Both
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9.5.1 Mean Shift
 Examine where the current process mean lies in comparison to the specification
range (i.e. closer to the LS or the US)
 Alter the process to bring the process mean to the center of the specification range
in order to increase the proportion of outputs that fall within specification
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Ex 9.10
 MBPF garage doors (currently)
-specification range: 75 to 85 kgs
-process mean: 82.5 kgs
-proportion of output falling within specifications: .6873
 The process mean of 82.5 kgs was very close to the US of 85 kgs (i.e. too
thick/heavy)
 To lower the process mean towards the center of the specification range the
supplier could change the thickness setting on their rolling machine
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Ex 9.10 Continued
 Center of the specification range: (75 + 85)/2 = 80 kgs
 New process mean: 80 kgs
 If the door weight (W) is a normal random variable, then the proportion of doors




falling within specifications is: Prob (75 =< W =< 85)
Prob (W =< 85) – Prob (W =< 75)
Z = (weight – process mean)/standard deviation
Z = (85 – 80)/4.2 = 1.19
Z = (75 – 80)/4.2 = -1.19
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Ex 9.10 Continued
 [from table A2.1 on page 319]
Z = 1.19
Z = -1.19
(1 - .8830)
.8830
.1170
 Prob (W =< 85) – Prob (W =< 75) =
.8830 - .1170 = .7660
 By shifting the process mean from 82.5 kgs to 80 kgs, the proportion of garage
doors that falls within specifications increases from .6873 to .7660
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9.5.2 Variability Reduction
 Measured by standard deviation
 A higher standard deviation value means higher variability amongst outputs
 Lowering the standard deviation value would ultimately lead to a greater
proportion of output that falls within the specification range
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9.5.2 Variability Reduction Continued
 Possible causes for the variability MBPF experienced are:
-old equipment
-poorly maintained equipment
-improperly trained employees
 Investments to correct these problems would decrease variability however doing
so is usually time consuming and requires a lot of effort
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Ex 9.11
 Assume investments are made to decrease the standard deviation from 4.2 to 2.5





kgs
The proportion of doors falling within specifications:
Prob (W =< 85) – Prob (W =< 75)
Z = (weight – process mean)/standard deviation
Z = (85 – 80)/2.5 = 2.0
Z = (75 – 80)/2.5 = -2.0
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Ex 9.11 Continued
 [from table A2.1 on page 319]
Z = 2.0
Z = -2.0
(1 - .9772)
.9772
.0228
 Prob (W =< 85) – Prob (W =< 75) =
.9772 - .0228 = .9544
 By shifting the standard deviation from 4.2 kgs to 2.5 kgs and the process mean
from 82.5 kgs to 80 kgs, the proportion of garage doors that falls within
specifications increases from .6873 to .9544
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9.5.3 Effect of Process Improvement on Process
Control
 Changing the process mean or variability requires re-calculating the control limits
 This is required because changing the process mean or variability will also change
what is considered abnormal variability and when to look for an assignable cause
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9.6 Product and Process Design
 Reducing the variability from product and process design
-simplification
-standardization
-mistake proofing
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Simplification
 Reduce the number of parts (or stages) in a product (or process)
-less chance of confusion and error
 Use interchangeable parts and a modular design
-simplifies materials handling and inventory control
 Eliminate non-value adding steps
-reduces the opportunity for making mistakes
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Standardization
 Use standard parts and procedures
-reduces operator discretion, ambiguity, and opportunity for making mistakes
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Mistake Proofing
 Designing a product/process to eliminate the chance of human error
-ex. color coding parts to make assembly easier
-ex. designing parts that need to be connected with perfect symmetry or with
obvious asymmetry to prevent assembly errors
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9.6.2 Robust Design
 Designing the product in a way so its actual performance will not be affected by
variability in the production process or the customer’s operating environment
 The designer must identify a combination of design parameters that protect the
product from the process related and environment related factors that determine
product performance
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QUESTIONS
???
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