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Design and Optimize Sample Plans atThe Dow Chemical Company
Swee-Teng Chin and Leo Chiang
Analytical Technology CenterThe Dow Chemical Company
September 12, 2011
Introduction
• In 2010 Dow had annual sales of $53.7 billion and employed approximately 50,000 people world wide. The Company’s more than 5,000 products are manufactured at 188 sites in 35 countries across the globe.
The Dow Chemical Company
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The Manufacturing Environment
• Most of the sample are made in pounds not parts• Quality release in the manufacturing is based on
analytical inspection of grab samples against specifications and mostly one sample per lot
• This presentation if focus on large sample size that is applied under special circumstances
• 2 topics:1. Optimize Sampling Plan to reduce number of samples2. Design a new sample plan to understand the physical
properties of catalyst
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1. Optimize a New Sample Plan
• The catalyst was made by the vendor and Dow performed QC testing on them before loading it up to the plant
• The current sampling plan use more than half of the existing resources
• Based on historical data, the catalyst have been consistently well performed
Background
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Objective
• To eliminate any unnecessary catalyst QC testing for cost and labor reduction
1. Optimize a New Sample Plan
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Current Sampling Plan
1000 kg (2205 lb)
Bag 1
1000 kg (2205 lb)
Bag 2
1000 kg (2205 lb)
Bag 18
1 25
1 lot
1 million lb of catalyst in a plant unit
1 liter
Dow QC labVendor QC lab
1000 kg (2205 lb)
Bag 1
1000 kg (2205 lb)
Bag 2
1000 kg (2205 lb)
Bag 18
1 lot
1 liter
2
100 mls 100 mls
Dow QC labVendor QC lab
1. Optimize a New Sample Plan
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Analysis and Results
• The current sampling plan is based on composite sample and no data available for each sampling stage.
• Use acceptance sampling plan approach to find the optimum sample size with respect to the risk involved– The uncertainty of testing the catalyst is modeled using a
binomial probability distribution. – The approach was found in the Remund et al.1 where it is
applied on seed purity analysis
1Remund et al. (2001), “Statistical considerations in seed purity testing for transgenic traits,” Seed Science Research.
1. Optimize a New Sample Plan
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The Binomial model1
• Let p = the proportion of defected bag of catalystn = number of bags, N = number of segments or groups, So total number of bags = N*n
• Prob(no defected bag in a group of n bags) = (1 - p)n
• Prob(at least one defected bag in a given bag group) = P = 1 - (1 - p)n
• c = the maximum number of the N groups that can test defect and result in acceptance of all the groups in a reactor
• Probability of acceptance lot
• Probability of rejection
)|(Prob1)|(Prob PcXPcX ≤−=>
1Remund et al. (2001), “Statistical considerations in seed purity testing for transgenic traits,” Seed Science Research.
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( ) jNjc
jPP
jN
PcX −
=
−
=≤ ∑ 1)|(Prob
0
1. Optimize a New Sample Plan
Assumptions
• The sample is random sample• The analysis is solely based on the information of the
sampled product and no information of the process is used (process is unchanged)
• The measurement data can be approximated by a normal distribution
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1. Optimize a New Sample Plan
•AQL (Acceptable Quality Level) = the lowest level of defected bag in a 1 million lb catalyst that the current production practices can support
•LQL (Lower Quality Limit) = the lowest level of defected bag in a 1 million lb catalyst that is considered acceptable to consumer
•Producer’s risk (or Type I error): probability of rejecting a good lot (1 million lb catalyst) or 100 – Prob(AQL)
•Consumer’s risk (or Type II error): probability of accepting a bad lot (1 million lb catalyst) or Prob(LQL)
•c = the maximum number of the N groups that can test defect and result in acceptance of all the groups in a reactor
Operating Characteristic (OC) Curve1
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1Remund et al. (2001), “Statistical considerations in seed purity testing for transgenic traits,” Seed Science Research.
1. Optimize a New Sample Plan
Hypothesis Testing 590:H 590:H a0 >= µµ
( )( )( )( )
5.3193%or 05319.09468.01
14.393.5855911
3.14 585.93,N~X when 59113.14 585.93,N~X when 591
2
2
=−=
−
<−=
<−=
≥=
ZP
XPXPα
( )( )
2.801%or 028.014.3597591
3.14 597,N~X when 591 2
=
−
<=
<=
ZP
XPβ
Producer Risk = Type I error, α = P(Ho is rejected when it is true)
Consumer Risk = Type II error, β = P(Not rejecting Ho when Ho is false)
580 585 590 595
Normal(585.928,3.13649)
Calculating the producer and consumer risks for current sampling plan
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1. Optimize a New Sample Plan
Possible sample sizes for the QC test
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1. Optimize a New Sample Plan
N n Producer Consumer(number of segments) (number of bags) Risk (%) Risk (%)
450 1 5.50 2.56225 2 5.49 2.57150 3 5.48 2.5890 5 5.46 2.6175 6 5.45 2.6250 9 5.42 2.6745 10 5.41 2.6830 15 5.35 2.7525 18 5.32 2.8018 25 5.24 2.9115 30 5.19 2.9910 45 5.02 3.279 50 4.97 3.376 75 4.69 3.933 150 3.83 6.632 225 2.93 12.07
c 1AQL 0.08LQL 1.23
Summary Results
• In the possible ranges of sample sizes, the producer and consumer risks do not seem to have a big change in values for both tests
• Optimized sample plan has 60% reduction in sample size – From 25 samples to 9 samples!
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1. Optimize a New Sample Plan
2. Design a New Sample Plan
Background
• Dow buys a catalyst from a vendor. – Catalyst is shipped in drums.
• The current sampling plan is not sufficient to distinguish between good and bad catalyst based on the specification – Sometimes batches pass the spec but they performed poorly in
the plant. This costs $$!
• The current plan calls for pulling a small sample which is a composite across multiple drums
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Objective
• Develop a sampling plan for the fresh catalyst that we received from the vendor to understand the underlying distribution of its physical properties.
2. Design a New Sample Plan
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Analysis
• The quality test on the catalyst are usually done by the vendor
• Need to determine an appropriate sample size:– Use information from the ASTM method to estimate
the measurement variation. • Sample size calculation is based Power analysis on
one sample meanStandard deviation 1.06 2.12Difference to Detect 1.78 2.11Sample Size 5 10Type I error (probability of rejecting a good lot), α = 0.05Type II error (probability of accepting a bad lot), β = 0.2
2. Design a New Sample Plan
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AnalysisTotal number of pellets per lot
1 location sampling Number of drums 91Number of pellets per drum 10Total pellets 910
3 location samplingNumber of drums 5Number of pellets per location within a drum 10Number of pellets per drum 30Total pellets 150
Grand total pellets 1570
2. Design a New Sample Plan
• But … the estimated sample size is for one location in a drum rather than the whole lot− This is to ensure that we fully
understand the variance structure of the product.
• There might be a significant drum to drum variation or within drum variation that we might not be aware– 3 location (top, middle and
bottom of a drum) samples to estimate within drum variation.
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Results
• Results on the data analysis shows that 1. There is no significant drum to drum variation and
within drum variation. 2. New specification was defined based on the
results
2. Design a New Sample Plan
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Specification analysis
• Old specification is based on a lower specification on average value of 15 on one of the critical physical test
• Data is skewed to the left, hence a spec based on average is not sufficient especially it is based on small composite sampling plan
• New specification is based on– Median not mean– Some small fraction of lower right end of the same physical
property
10 20 30 40
100.0%
99.5%
97.5%
90.0%
75.0%
50.0%
25.0%
10.0%
2.5%
0.5%
0.0%
maximum
quartile
median
quartile
minimum
46.8
41.9175
34.5288
24.3
19.85
15.35
12.075
10.01
7.76
6.252
5.2
Quantiles
Mean
Std Dev
Std Err Mean
Upper 95% Mean
Lower 95% Mean
N
16.6
6.50
0.23
17.0
16.1
750
Moments
2. Design a New Sample Plan
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Results
• New batch is made and has been tested by the vendor as good batch based on the proposed sampling plan
• Dow need to validate this batch using a reduce sampling plan with the same level of protection but with less number of sample– Variable sampling plan – Double sampling plan
2. Design a New Sample Plan
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Variables sampling plan2
• Requires that the measurement to be normally distributed
• Assuming the standard deviation is unknown
If USL andX ≤ USL – k S
Take a representative sample of n units, measure each unit and calculate
average (X) and standard deviation (S)
Yes
No
ACCEPT
REJECT
START
If LSL andX ≥ LSL + k S
No
REJECT
2. Design a New Sample Plan
2Wayne Taylor, “Successful Acceptance Sampling“, Taylor Enterprise, Inc., www. variation.com
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Double sampling plan
• Characterized by five parameters:– n1 = first sample size– a1 = first accept number– r1 = first reject number– n2 = second sample size– a2 = second accept number
2. Design a New Sample Plan
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Double sampling plan2
Count ≤ a1 Count ≥ r1Compare
count
ACCEPT REJECT
Take a representativesample of n1 units and
count number of defectives
Count ≤ a2 Count > a2Compare
count
ACCEPT REJECT
Take second representativesample of n2 units and
count the total number ofdefectives in both samples
START
Continue To Second Stage
2. Design a New Sample Plan
2Wayne Taylor, “Successful Acceptance Sampling“, Taylor Enterprise, Inc. www. variation.com
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Comparison between the Sampling Plans
Both of the variable and double sampling plan requires less sample to test but the execution procedure
is more completed.
2. Design a New Sample Plan
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Conclusion
2. Design a New Sample Plan
• The original spec based on the sample plan from the vendor did not work– Performance didn’t always match the predicted
performance based on quality analysis.• Vendor will be using a new sample plan and Dow will
be using a new reduced sample plan to validate the results.
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General Conclusions
• At The Dow Chemical Company, we use statistical methods for designing and optimizing sampling plans.– Need to take into account the risk and costs.
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