ERLPM WorkshopERLPM WorkshopStatistical AnalysisStatistical Analysis

Carl SteinhauerCarl Steinhauer

ConsultantConsultant

34th Eastern Region 34th Eastern Region Annual Airports ConferenceAnnual Airports Conference

AnalysisAnalysis

% Taller than 5’-5”% Taller than 5’-5”

% between 5’-5” and 6’-5”% between 5’-5” and 6’-5”

Average HeightAverage Height

Limit # of samplesLimit # of samples

Statistical AnalysisStatistical Analysis

Estimate average Estimate average and % within limits and % within limits

Population

0

10

20

30

40

50

60

70

80

4-7 4-9 5-1 5-3 5-5 5-7 5-9 5-11

6-1 6-3 6-5 6-7 6-9 6-11

Organize Date

Fre

qu

en

cy

OverviewOverview

P 401 Test ResultsP 401 Test Results

Statistical Analysis; PWL EstimateStatistical Analysis; PWL Estimate

Verify Production Process: PaymentVerify Production Process: Payment

TheoryTheory

1.1. AssumptionsAssumptions

2.2. Normal DistributionNormal Distribution

3.3. Tools: Average and Standard Tools: Average and Standard DeviationDeviation

4.4. Percent Within Limits Percent Within Limits (PWL) Concept(PWL) Concept

AssumptionsAssumptions

1.1. Limited # of test resultsLimited # of test results Statistical Analysis Statistical Analysis Quality characteristics of large amount of materialQuality characteristics of large amount of material

2.2. Test result variabilityTest result variability

Components:Components:

materialsmaterials

sampling-ERLPMsampling-ERLPM

testing-ERLPMtesting-ERLPM

3.3. Same ProcessSame Process

4.4. Random sampling-Lot, SublotRandom sampling-Lot, Sublot

5.5. Normal DistributionNormal Distribution

Specific ProceduresSpecific Procedures

1.1. Sublots, Lots, Partial LotsSublots, Lots, Partial Lots

2.2. CalculationsCalculations

3.3. RetestingRetesting

4.4. OutliersOutliers

Figure 1

0

5

10

15

20

255.

45

5.55

5.65

5.75

5.85

5.95

6.05

6.15

6.25

6.35

6.45

6.55

Asphalt Content %

Fre

qu

ency n=100

96 tests96 tests

0

10

20

30

40

50

60

2400

2375

2350

2325

2300

2275

2250

2225

2200

2175

2150

2125

2100

2075

2050

2025

2000

Marshall Stability

Freq

uenc

y

x+/- 1Sn=68%x+/- 1Sn=68%

x+/-2Sn+95%x+/-2Sn+95%

x+/-3Sn+99.7%x+/-3Sn+99.7%

xx

PWL Concept

LL=spec lower tolerance limitL=spec lower tolerance limit

eg. Mat density 96.3 for P401eg. Mat density 96.3 for P401

PWL-% of PWL-% of test result test result exceeding Lexceeding L

PWL Calculation ProceduresPWL Calculation ProceduresERLPM-page 47ERLPM-page 47

Section 110-AC 1505370-10CSection 110-AC 1505370-10C

Method for Computing PWL and ExamplesMethod for Computing PWL and Examples

Section 110-02Section 110-02

Spec. P401 Table 5: L and U Spec. Limits (page 24)Spec. P401 Table 5: L and U Spec. Limits (page 24)

TEST PROPERTY

Pavements Designed for Aircraft Gross Weights of

60,000 Lbs. or More or Tire Pressures of 100 Psi or More

Pavements Designed for Aircraft Gross Weights Less

Than 60,000 Lbs. or Tire Pressures Less Than 100 Psi

Number of Blows 75 50

Specification Tolerance Limits

Specification Tolerance Limits

L U L U

Stability, minimum, pounds

1800 -- 1000 --

Flow, 0.01‑inch 8 16 8 20

Air Voids Total Mix, percent

2 5 2 5

Surface Course Mat Density, percent

96.3 [101.3] 96.3 [101.3]

Base Course Mat Density, percent

95.5 101.3]-- 95.5 [101.3]

Joint density, percent 93.3 -- 93.3 --

Table 5: Marshall Acceptance LimitsTable 5: Marshall Acceptance Limits

GivenGiven

xx11=2=2

xx22=4=4

xx33=6=6

xx44=8=8

  

X = X = 2+4+6+82+4+6+8 = 5 = 5

44

X=5X=5

dd11=2-5=-3=2-5=-3 dd11²=9²=9

dd22=4-5=-1=4-5=-1 dd22²=1²=1

dd33=6-5=1=6-5=1 dd33²=1²=1

dd44=8-5=3=8-5=3 dd44²=9²=9

SnSn = = dd11²+d²+d22²+d²+d33²=d²=d44²²

n-1n-1

Sn Sn == 9+1+1+99+1+1+9 = = 2020

4-14-1 3 3

Sn Sn =2.58=2.58

(calculator (calculator nn-1)-1)

Roundout RulesRoundout RulesERLPM-page 47ERLPM-page 47

Example-last digit to be kept-nearest 10Example-last digit to be kept-nearest 10thth

4.614.614.624.624.644.644.65004.65004.664.664.674.674.684.684.694.69

Even Digit-sameEven Digit-sameOdd Digit-increase by 1Odd Digit-increase by 1This case becomes 4.6This case becomes 4.6If it was 4.7500 it would become 4.8If it was 4.7500 it would become 4.8

becomes 4.7

MAT Density –One side density MAT Density –One side density acceptance (acceptance (Manual Appendix E, page 4)Manual Appendix E, page 4)

SublotSublot1. = 96.01. = 96.02. = 97.02. = 97.03. = 99.03. = 99.04. = 100.04. = 100.0

x=98.0x=98.0SnSn=1.8=1.8

QQII==x-Lx-L

SnSn

QQII==98.0-96.398.0-96.3 = .9444 = .9444

1.81.8Section 110-Table l, N=4Section 110-Table l, N=4

PPLL=82=82

Quality Index. See Section Quality Index. See Section 110-02f and Section 8  page 48 110-02f and Section 8  page 48 of ERLPM par 8.4.1 of ERLPM par 8.4.1

Spec Tables 5Spec Tables 5

page 50 ERLPMpage 50 ERLPM

Mat Density -Two sided acceptance Mat Density -Two sided acceptance for densityfor densityx=98.0x=98.0SnSn=1.8=1.8QQuu= = U-xU-x

SnSnQQuu==101.3-98.0101.3-98.0 = 1.833 = 1.833

1.81.8Section 110 (Section 110 (ERLPM page 48 par 8.4.2. and page 50 ERLPM page 48 par 8.4.2. and page 50 ) ) --Table l, Table l,

N=4N=4PPLL= 100= 100

PWL calculation for two sided PWL calculation for two sided specificationspecification

PWL= PL + PU-100 PWL= PL + PU-100

(ERLPM page 49  par 8.5.2(ERLPM page 49  par 8.5.2 )

PWL= 82 + 100 – 100 = 82PWL= 82 + 100 – 100 = 82

Target Density 98.0 AchievedTarget Density 98.0 Achieved

SnSn=1.8 versus 1.3=1.8 versus 1.3

Acceptable QC Value as per P-401 spec. pg 24 last par

18%18%

PPL L = 82%= 82%

MAT DensityMAT Density

98.098.0

96.396.3

Target Density 98.0 AchievedTarget Density 98.0 Achieved

SnSn=1.8 versus 1.3=1.8 versus 1.3

Acceptable QC Value

18%18%

PPL L = 82%= 82%

MAT DensityMAT Density

98.098.0

96.396.3

101.3%101.3%

Effect of Quality ControlEffect of Quality Control

90 PWL90 PWL

82 PWL82 PWL

SnSn = 1.8 = 1.8

SnSn = 1.3 = 1.3

Spec. pg 24Spec. pg 24

96.3%96.3% 98% 98% x x

Air VoidsAir VoidsApp. D, page 3App. D, page 3

Sublot Sublot

1= 2.11= 2.1

2= 3.22= 3.2

3=2.53=2.5

4=6.04=6.0

X= 3.4X= 3.4

SnSn= 1.76= 1.76

Spec. page 5 - 0.65Spec. page 5 - 0.65

QL = x-L = 3.4-2.0 = .7955

Sn 1.76

PPLL((table 1table 1) = 77%) = 77%

n=4n=4

QQUU= = U-xU-x = = 5-3.4 5-3.4 = .909= .909

SnSn 1.76 1.76

PPUU((table 1table 1) = 81%) = 81%

n=4n=4

PWL= PPWL= PLL + P + PUU-100-100

PWL= 77+81-100 = 58PWL= 77+81-100 = 58

page 50  of page 50  of ERLPM ERLPM

Air VoidsAir Voids

23%23%

58%58%

19%19%

22 3.43.4 55

PWL= PPWL= PLL + P + PUU-100-100

PWL= 77 + 81-100 =58PWL= 77 + 81-100 =58

TABLE 6. PRICE ADJUSTMENT SCHEDULE 1

Percentage of Material WithinSpecification Limits (PWL)

Lot Pay Factor(Percent of Contract Unit Price)

96 – 100 106

90 – 95 PWL + 10

75 – 89 0.5 PWL + 55

55 – 74 1.4PWL – 12

Below 55 Reject 2

Payment – One side Payment – One side for densityfor density

TABLE 6. PRICE ADJUSTMENT SCHEDULE 1

Percentage of Material Within Specification Limits (PWL) Lot Pay Factor (Percent of Contract Unit Price)

93 – 100 103

90 – 93 PWL + 10

70 – 89 0.125PWL + 88.75

40 – 69 0.75PWL +45

Below 40 Reject 2

Payment – two sided for densityPayment – two sided for density

PaymentPayment

Spec-par 401-8.1a-page 29

MAT Density PWL=82

Air Voids PWL= 58

Lot Pay Factor

Air Voids- 1.4 x 58-12= 69.2%

Mat Density- 0.5 x 82+55= 96%

Use lower of 2 values- 69.2%

Lower valueLower value

Joint DensityJoint DensityAppendix E, page 5Appendix E, page 5

93.3

95.0

97.0

96.0

X= 95.3

Sn= 1.58

QL= (95.3-93.3) = 1.2658

1.58

PL= 93

Spec. page 21 par. 401-5.2(b)(3) if < 71% there is a 5% penalty

Table 5Table 5

Partial LotsPartial Lotsspec page 20spec page 20

Section P-401-5.1cSection P-401-5.1c

Sample ProblemSample Problem

Flow-Appendix D, page 5Flow-Appendix D, page 5Partial lot situation-6 sublotsPartial lot situation-6 sublots

8.0, 8.2, 8.5, 8.2, 8.9, 9.18.0, 8.2, 8.5, 8.2, 8.9, 9.1X= 8.5X= 8.5

SnSn= 0.44= 0.44

QL= QL= x-Lx-L = = 8.5-8.08.5-8.0 = 1.1364; PL= 88 ( = 1.1364; PL= 88 (table 1 ERLPM n=6)table 1 ERLPM n=6)

Sn Sn 0.44 0.44QU= QU= U-XU-X = = 16-8.516-8.5 = 18.75; PL= 100 = 18.75; PL= 100

SnSn 0.44 0.44PWL= 88 + 100-100= 88<90PWL= 88 + 100-100= 88<90

Corrective Action! 401-5.2(b)(2)Corrective Action! 401-5.2(b)(2)

OutliersOutliers

Spec -pg 23Spec -pg 23 401-5.2d401-5.2d

-pg 25-pg 25 401-5.3c401-5.3c

MAT Density and Air VoidsMAT Density and Air Voids

Test for OutliersTest for Outliers

MAT DensityMAT Density

94.094.0

96.096.0

97.097.0

98.098.0

x= 96.2x= 96.2

Sn= 1.71Sn= 1.71

QL= QL= 96.2-96.396.2-96.3 = -.0585 = -.0585

1.711.71

PL= <50%PL= <50%

ASTM E 178, par. 4ASTM E 178, par. 4

TT11= (x-x= (x-x11)/)/SnSn

TT11= = 96.2-94 96.2-94 = 1.286= 1.286

1.711.71

• Table 1-ASTM E 178Table 1-ASTM E 178

• N=4N=4

• Upper 5% significance level 1.463Upper 5% significance level 1.463

• Since 1.286<1.463 the 94.0 test value is Since 1.286<1.463 the 94.0 test value is not considered an outlier and is retained!not considered an outlier and is retained!

Sample Problem-OutliersSample Problem-Outliers

Air VoidsAir Voids2.0, 4.8, 4.9, 5.02.0, 4.8, 4.9, 5.0X=4.2X=4.2Sn=1.45Sn=1.45

QQLL= = 4.2-2.04.2-2.0 = 1.5172; PL= 100 = 1.5172; PL= 100

1.451.45

QQUU= = 5-4.25-4.2 = 0.5517; PU= 69 = 0.5517; PU= 69

1.451.45PWL= (100+69)-100=69PWL= (100+69)-100=69ASTM E 178 par. 4ASTM E 178 par. 4

TTnn= (x-x= (x-x11)/)/SnSn = = 4.2-2.04.2-2.0 = 1.517 = 1.517

1.451.45Table 1, ASTM E 178, N=4, 5% significanceTable 1, ASTM E 178, N=4, 5% significanceT= 1.463<1.517 T= 1.463<1.517 therefore 2.0 is the outlier and it is discardedtherefore 2.0 is the outlier and it is discarded

ResamplingResampling401-5.3 page 25401-5.3 page 25

MAT Density ONLYMAT Density ONLY(Appendix E-pg 4)(Appendix E-pg 4)Prior MAT Density- 96, 97, 99, 100Prior MAT Density- 96, 97, 99, 100PWL 82PWL 82

4 new cores 96, 96, 97, 984 new cores 96, 96, 97, 98AVG-all 8, 97.4AVG-all 8, 97.4Sn= Sn= 1.51 1.51

QQLL= = x-Lx-L = = 97.4-96.397.4-96.3 = .7337 = .7337

SnSn 1.51 1.51Table 1, N= 8Table 1, N= 8PWL= 77PWL= 77

Questions?Questions?

Thanks!Thanks!