Chapter 9: Managing Flow Variability

1

• Chapter 9 Managing Flow Variability
• Sections 9.3.4 to End
• Team 10
• Alex Ichiroku
• Vivian Ramos
• Hamid Orandi
• (andShehzad Khan)

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9.3.4 Control Charts

• 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
• Dont make any adjustments to the process
• If the observed values are outside this range
• The process is out of control
• Need to investigate whats causing the problems
  the assignable cause

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9.3.4 Control Charts Continued

• Let ? be the expected value of the performance
• Set up a control band around ?
• UCL Upper Control Limit
• LCL Lower Control Limit
• Calculate the standard deviation, ?
• Decide how tightly we want to monitor and
  control the process
• The smaller the value of z, the tighter the
  control

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9.3.4 Control Charts Continued

• The Upper and Lower Control Bands
• LCL ? - z? UCL ? z?
• Process Control Chart

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9.3.4 Control Charts Continued

• If observed data within the control band
• Performance variability is normal
• If observed data outside the control band
• Process is out of control
• Data Misinterpretation
• Type I error, ? Process is in control, but
  data outside the Control Band
• Type II error, ? Process is out of control,
  but data inside the Control Band

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9.3.4 Control Charts Continued

• Optimal Degree of Control
• Depends on 2 things
•      1. How much variability in the performance
  measure we consider acceptable
• 2. How frequently we monitor the process
  performance.

• Optimal frequency of monitoring is a balance
  between the costs and benefits

The value of z determines how tightly we
control the process LCL ? - z?
UCL ? z? A small value of z
Narrower Control Band -- Tighter Control
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9.3.4 Control Charts Continued

• Optimal Degree of Control
• If we set z to be too small
•      ? Well end up doing unnecessary
  investigation
• ? Incur additional costs

• If we set z to be too large
• ? Well accept a lot more variations as
  normal
• ? We wouldnt look for problems in the
  process less costly

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9.3.4 Control Charts Continued

• Optimal Degree of Control

• In practice, a value of z 3 is used
• ? 99.73 of all measurements will fall
  within the normal range

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9.3.4 Control Charts Continued

• Average and Variation Control Charts
• Monitor process performance by taking random
  samples
• For each Sample
•      ? Calculate the average value, A1,
  A2.AN
• ? Calculate the variance of each sample,
  V1, V2.VN

Sample Averages ? Normally
distributed ? Mean of ?A ? Standard
Deviation of ?A
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9.3.4 Control Charts Continued

• Average and Variation Control Charts
• ?A ?/?n (n sample size)
• LCL ? - z?/?n and UCL ?
  z?/?n
•     

• Take it one step further

Estimate ? by the overall average of all
the sample averages, ?A ?A
(A1 A2AN) / N (N of samples)
Also estimate ? by the standard deviation of
all N x n observations, S
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9.3.4 Control Charts Continued

• Average and Variation Control Charts
• Our New, Improved equations for Upper and Lower
  Control Limits are
• LCL ?A - zs/?n and UCL ?A
  zs/?n
•     

• We can do the same calculations with the Sample
  Variances

Calculate?V -- the average variance of the
sample variances ?V (V1
V2VN) / N (N of samples)
Also calculate SV -- the standard deviation
of the variances
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9.3.4 Control Charts Continued

• Average and Variation Control Charts
• The New equations for Variance Control Limits
  are
• LCL ?V - z sV and UCL ?V z sV

If observed variations fall within this
range Process Variability is stable
If not Need to investigate the
cause of abnormal variations
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9.3.4 Control Charts Continued

• Average and Variation Control Charts
• Garage Door Example revisited
•     

Ex A1 (81 73 85 90 80) / 5 81.8
kg Ex V1 (90 - 73) 17 kg Std. Dev.
of Door Weights s 4.2 kg Std. Dev. of
Sample Variances sV 3.5 kg
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9.3.4 Control Charts Continued

• Average and Variation Control Charts
• Average Weights of Garage Door Samples
•     

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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
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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
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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 Observations
Plot Control Charts
Abnormal 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 didnt 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
Plot the results on a graph
Scatter Plot
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9.4 Process Capability

• Ease of external product measures (door
  operations and durability) and internal 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 s (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 s
  1, we can use the standard normal table in
  Appendix II to compute
• AT US
• Prob (W 85) in terms of
• Z (W-µ)/ s
• 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,t
  rue) .037073)
• THEN
• Prob (75W85)
• .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 theyll 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 3s 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)/3s (.1984 as calculated later)
• and (µ -LS)/3s (.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)/3s .1984 and define a single measure of
  process capabilities as
• Cpk min(US-µ/)3s, (µ -LS)/3s (.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)/3s or (µ -LS)/3s
• 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)/6s (.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 -3s
  from the µ.
• Consequently, 6s is sometimes referred to as the
  natural tolerance of the process.
• Ex 9.8
• Cpk min(US- µ)/3s , (µ -LS)/3s
• 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)/6s
• (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.
  Its 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
  Cp1.3 (see Table 9.4) Test s
  (85-75)/(6)(1.282) 1.300kg

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Table 9.4
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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 /s), (µ -LS)/s (
  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)/2s
• Ex 9.9
• Currently the sigma capability of door making
  process is
• Smin(85-82.5)/(2)(4.2) .5952
• By centering the process correctly, its sigma
  capability increases to
• Smin(85-75)/(2)(4.2) 1.19
• THUS, with a 3s that is correctly centered, the
  US and LS are 3s away from the mean, which
  corresponds to Cp1, 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 6s from the mean each.
• Extraordinary high degree of precision.
• Corresponds to Cp2 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
• s (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
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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)-zs 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 customers 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 lt W lt 85)
• Prob (W lt 85) Prob (W lt 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
  .8830
• Z -1.19 (1 - .8830)
  .1170
• Prob (W lt 85) Prob (W lt 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 (75 lt W lt 85)
• Prob (W lt 85) Prob (W lt 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
  .9772
• Z -2.0 (1 - .9772)
  .0228
• Prob (W lt 85) Prob (W lt 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 customers
  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

• ???