criticality analysis for lhc cryogenic … criterions for a criticality ranking numerical input data...

CRITICALITY ANALYSIS FOR LHC CRYOGENIC PROCESS Determination of critical components

1


2 Jimmy Martin

A criticality analysis ? What for ?

• The current overall availability reached is above 90% • This value must be increased to maximise the global LHC’s efficiency

Methodology of study : 2 STEPS • Step 1 : Process functional analysis

• Step 2 : WEIBULL analysis (Reliability computations) + FMECA criterions for a criticality ranking

Numerical input data

• Statistics for 2011-2012 of ‘CM’ conditions losses (E. DURET) • Cumulated events from Logbook OA (‘failure/stop’ and ‘problem solved’ categories)

Studied components

• From 1st January 2011 to 1st August 2012 Period of study

• Valves: CV’s for helium – PV’s for nitrogen • Turbines, warm and cold compressors systems • Electric heaters

Waloddi WEIBULL


3 Jimmy Martin

STEP 1 : PROCESS FUNCTIONAL ANALYSIS

Aim of the analysis • Good comprehension of the process layout in order to make easier the next criticality analysis

tasks The chosen methodology

• Determination of elementary process functions, given the requested thermal-hydraulic conditions for cryogenic fluids

• Development into secondary and tertiary functions, as far as elementary technical solutions can be highlighted

Main function and its associated sub functions

Technical solutions

‘Warm’ compression station

Example :


4 Jimmy Martin

STEP 2 : CRITICALITY ANALYSIS

Definition of criticality • Level of criticality of an elementary component : estimated by studying the occurrence, the

severity and the probability of non-detection associated to a (combined) failure mode(s) :

C = O ∗ S ∗ D

Criticality index (from 0 to 1)

Occurrence index (from 0 to 1) • Built by using statistical data from logbook • Computation of failure rate functions associated to

hardware components (cf. WEIBULL analysis)

Severity index (from 0 to 1) • Built by using statistical data from ‘CM loss’ monitoring

files (cf. E. DURET )

Non-detection index (from 0 to 1) • Not included because of a lack of input data

(equivalent to the ‘Risk Priority Number’ or ‘RPN’ criterion used in the FMECA analysis)


5 Jimmy Martin


Occurrence index ‘O’ : determination

Studied population of components + ‘Logbook OA data’

05

10152025

Histogram of causes

02468

1012

Command systems

0

0,001

0,002

0,003

0,004

0,005

0,006

0 200 400 600

Lam

bda

(fai

lure

/day

)

time (day)

Decreasing Pareto analysis of causes (like a ‘Russian dolls’ game)

Statistical computations, using the WEIBULL’s tools

(λ(t), R(t), etc.)

⇒ However :

• λ(t) = ?? • R(t) = ?? • WEIBULL’s models = ??

EXCEL file ‘LHC CRYO – statistical study’

Extraction


Jimmy Martin


Some definitions : • R(t) = reliability probability (from 0 to 1) of a given population of components • F(t) = failure probability = 1 - R(t) • λ(t) = failure rate function associated to a population : number of failures by unit of time, as a function of time variable • MTBF = ‘Mean Time Between Failures’ of a population : estimator of the population reliability

How to compute λ(t), F(t), R(t) and MTBFs from ‘Logbook OA’ data ?

By trying to correlate statistical data to known mathematical

models (WEIBULL,…)


Under the WEIBULL model hypothesis : 𝑅 𝑡 = 𝑒−(𝑡−𝛾𝜂 )𝛽

𝐹 𝑡 = 1 − 𝑒−(𝑡−𝛾𝜂 )𝛽

𝜆 𝑡 =𝛽𝜂 ∗

𝑡 − 𝛾𝜂

𝛽−1

Where: t = time variable β = shape parameter γ = position parameter η = scale parameter


Jimmy Martin


Linear regression method to determine β, η and γ parameters :


𝑅 𝑡 = 𝑒−(𝑡−𝛾𝜂 )𝛽 𝐿𝐿 𝐿𝐿

11 − 𝐹∗ 𝑡

= 𝛽 ∗ 𝐿𝐿 𝑡 − 𝛽 ∗ 𝐿𝐿(𝜂) We can demonstrate that :

Where F*(t) is the unbiased computed failure probability from ‘Logbook’ data : 𝐹∗ 𝑡 =𝑁 𝑡 − 𝑁 0𝑁 0 + 1

• 𝑁 0 is the overall number of failures of the sample, and so, the number of survivors at time t = 0;

• 𝑁 𝑡 is the number of survivors at time t.

If a linear relation 𝒚 = 𝒂 ∗ 𝒙 + 𝒃 can be highlighted between 𝑳𝑳 𝑳𝑳 𝟏𝟏−𝑭∗ 𝒕

and Ln(t), so we can assume

the cumulated frequency of failures follows a WEIBULL law, whose parameters are :

y = 0,8757x - 3,8033 R² = 0,9876

-2,50

-2,00

-1,50

-1,00

-0,50

0,00

0,50

1,00

0,00 1,00 2,00 3,00 4,00 5,00 6,00

Ln(L

n(1/

(1-F

*(t)

)))

Ln(t)

𝛽 = 𝑎

𝜂 = 𝑒−𝑏𝛽

𝛾 = 0 (𝑖𝑖 𝑖𝑖 R2 close to 1)

In this example :

𝛽 = 0.8757

𝜂 = 76.95

𝛾 = 0


Jimmy Martin


Given the β, η and γ parameters :


• It is now possible to compute theoretical R(t) and λ(t) functions • The β value defines the trend of λ(t) :

00,0010,0020,0030,0040,0050,0060,0070,0080,009

1 6 11 16 21 26 31 36 41 46 51 56 61 66 71 76 81 86 91 96 101

106

111

116

121

λ(t)

β < 1 1 < β β = 1

• A bilinear scatter when plotting 𝐿𝐿 𝐿𝐿 11−𝐹∗ 𝑡

as a function of Ln(t) reveals a bi-Weibull distribution which currently leads to

the typical ‘bathtub’ plot for λ(t) :

0

0,0005

0,001

0,0015

0,002

0,0025

0,003

0 100 200 300 400 500 600 700

λ(t)

time t (day)

y = 0.8555x - 5.6952 R² = 0.9819

y = 3.5604x - 21.553 R² = 0.9649

-4,7

-3,7

-2,7

-1,7

-0,7

0,3

1,3

2,2 2,7 3,2 3,7 4,2 4,7 5,2 5,7 6,2 6,7

ln(ln

(1/(

1-F*

(t)))

)

ln(t)


Jimmy Martin


The Kolmogorov-Smirnov statistical test to check the WEIBULL distribution hypothesis :


• Computation of the maximum absolute value of the difference between the theoretical and observed F(t) : 𝐷 = 𝑀𝑎𝑀 𝐹𝑡𝑡𝑡𝑡 𝑡 − 𝐹∗(𝑡)

• Determination of the admissible limit value for D from a Kolmogorov-Smirnov table, as a function of the number N(0) of

events and the chosen level of risk α (what gives the level of confidence of the WEIBULL hypothesis) :

• If the condition 𝐷 ≤ 𝐷𝑙𝑙𝑙𝑙𝑡 is respected, so the WEIBULL hypothesis can be accepted with an associated level of confidence (1 –α)%

Example :

0,0000

0,1000

0,2000

0,3000

0,4000

0,5000

0,6000

0,7000

0,8000

0,9000

1,0000

0 100 200 300 400 500 600

F(t)

t (day)

theoreal


Jimmy Martin


Construction of the Occurrence index :


• Built by considering the overall number of events and the failure rate’s trend related to a studied population:

If: • β < 1 λ’s trend = ↘ • β = 1 λ’s trend = → • 1 < β ≤ 2 λ’s trend = ↗ + • 2 < β λ’s trend = ↗ ++

Code of colours: • Green 0.25; • Yellow 0.5; • Orange 0.75; • Red 1.

0

0,001

0,002

0,003

0,004

0,005

0,006

0 200 400 600

Lam

bda

(fai

lure

/day

)

time (day)

0 1 2

CV valves - MEC

Warm Compressors(without leakage)

PV valves for N2

Turbines - command

Electric Heatears

CV valves - COM

Cold Compressors -regulator

Previously computed λ(t) functions for the studied populations of components

Occurrence ranking


Jimmy Martin


Construction of the Severity index :

Severity index ‘S’ : determination

• Built by considering the overall number of losses of ‘CM’ and time before recovering these conditions, for a studied population:


Statistical data about ‘CM’ losses (cf. E. DURET’s files)

0 0,5 1

PV valves for N2

CV valves - MEC

Electric Heatears

CV valves - COM

Turbines - command

Warm Compressors (withoutleakage)

Cold Compressors - regulator

Severity ranking


Jimmy Martin


Construction of the Criticality index :

Criticality index ‘C’ : determination

0 0,1 0,2 0,3 0,4 0,5 0,6 0,7 0,8

PV valves for N2

CV valves - MEC

Electric Heatears

CV valves - COM

Turbines - command

Warm Compressors (withoutleakage)


Severity ranking

• Built by multiplying the two previous indexes of occurrence and severity:

0,0 20,0 40,0 60,0 80,0

CV valves - MEC

PV valves for N2

Electric Heatears

Warm Compressors (without leakage)

Turbines - command

CV valves - COM


0 1 2

CV valves - MEC

Warm Compressors(without leakage)

PV valves for N2

Turbines - command

Electric Heatears

CV valves - COM

Cold Compressors -regulator

Occurrence ranking

CRITICALITY RANKING

Recall : C = O ∗ S ∗ D (With D = 1)


13 Jimmy Martin

Cold compressors – global study • 25 events along the studied period

• A unique Weibull distribution: the failure rate’s trend is increasing (95% of confidence)

y = 2.683x - 15.847 R² = 0.9433

-4

-3

-2

-1

0

1

2

4,6 5,1 5,6 6,1 6,6

ln(ln

(1/(

1-F*

(t))))

ln(t)

0,0000

0,0020

0,0040

0,0060

0,0080

0,0100

0,0120

0,0140

0 100 200 300 400 500 600

λ(t)

time (day)

𝜆 𝑡 =2.683367

∗𝑡

367

1.683


14 Jimmy Martin

Cold compressors – global study

• Pareto analysis of causes:

CAUSE Quantity % COM Command 21 84.0 UTI Utilities 2 8.0 UNK Unknown 2 8.0 MEC Mechanical 0 0.0 ENV Environment 0 0.0

Total 25 100.0 0

5

10

15

20

25

Command Utilities Unknown Mechanical Environment

Global Pareto histogram for CCs

• The main failure mode is related to command about 84% of the overall • Through the ‘command’ category, the cold compressor’s regulator and frequency converter are

mainly responsible to the ‘command’ failures

CAUSE Quantity % CCR CC regulation 10 47.6 VFREQ Frequency converter 4 19.0 PLC QURC regulation 3 14.3 INSTRU Instrumentation 3 14.3 SIGN Links (profibus, etc.) 1 4.8

Total 21 100.0 0

5

10

15

CC regulation Frequencyconverter

QURC regulation Instrumentation Links (profibus,etc)

Pareto histogram for COM


15 Jimmy Martin

Cold compressors – ‘regulator’ causes

• 11 events along the studied period • A unique Weibull distribution: the failure rate’s trend is increasing (95% of confidence)

0,0000

0,0020

0,0040

0,0060

0,0080

0,0100

0,0120

0 100 200 300 400 500 600

Lam

bda

(fai

lure

/day

)

time (day)

VREQ and CCR failure distribution

CC1

CC2

CC3

CC4

VFREQ and CCR failure distribution Element Quantity % %Air Liquide %Linde CC1 6 42.9 83.3 16.7 CC2 4 28.6 0 100 CC3 2 14.3 0 100 CC4 2 14.3 - 100 Total 14 100.0 % value 35.7 64.3

Linde/A.Liquide failure%

%AirLiquide

%Linde

• CC1 and then CC2 main causes of failures • A possible correlation between thermo hydraulic conditions and ‘regulator’ failures?

𝜆 𝑡 =2.7631

406∗

𝑡406

1.7631


16 Jimmy Martin

Warm compressors – excluding leakage

• 21 events along the studied period • A unique Weibull distribution: the failure rate’s trend is decreasing (95% of confidence)

y = 0,8155x - 4,6538 R² = 0,9504

-3,5-3

-2,5-2

-1,5-1

-0,50

0,51

1,5

2 2,5 3 3,5 4 4,5 5 5,5 6 6,5 7

ln(ln

(1/(

1-F*

(t))))

ln(t)

y = 0,0078x-0,184

0,0025

0,003

0,0035

0,004

0,0045

0,005

0,0055

0 100 200 300 400 500 600

Lam

bda

(fai

lure

/day

)

time (day)

• Pareto analysis of causes: CAUSE Quantity %

COM Command 8 38.1 MEC Mechanical 6 28.6 UTI Utilities 4 19.0 UNK Unknown 2 9.5 ENV Environment 1 4.8

Total 21 100.0 0

2

4

6

8

10

Command Mechanical Utilities Unknown Environment

• Main causes: ‘command’ and ‘mechanical’ • Not enough data to follow the statistical analysis

𝜆 𝑡 =0.8155

301∗

𝑡301

−0.1845


17 Jimmy Martin

Turbines – global study

• 25 events along the studied period • A unique Weibull distribution: the failure rate’s trend is constant (95% of confidence)


• Main causes: ‘command’ and ‘mechanical’ • What are the individual failure rate’s trends for these previous causes?

y = 1,0529x - 6,3469 R² = 0,9292 -4

-3

-2

-1

0

1

2

2,2 2,7 3,2 3,7 4,2 4,7 5,2 5,7 6,2 6,7

ln(ln

(1/(

1-F*

(t))))

ln(t) y = 0,0018x0,0529

0

0,0005

0,001

0,0015

0,002

0,0025

0,003

0 100 200 300 400 500 600

Lam

bda

(fai

lure

/day

)

time (day)

CAUSE Quantity % COM Command 15 60.0 MEC Mechanical 8 32.0 UNK Unknown 2 8.0 ENV Environment 0 0.0 UTI Utilities 0 0.0

Total 25 100.0 02468

10121416

Command Mechanical Unknown Environment Utilities

𝜆 𝑡 =1.0529

415∗

𝑡415

−0.0529


18 Jimmy Martin

Turbines – ‘command’ causes

• 15 events along the studied period • A unique Weibull distribution: the failure rate’s trend is increasing (95% of confidence), so the

failure rate’s trend for ‘mechanical’ causes is necessarily decreasing


• Main causes: ‘flow switch’ and ‘TU regulator’ • Not enough data to follow the statistical analysis

y = 1,262x - 7,8751 R² = 0,8721

-3

-2,5

-2

-1,5

-1

-0,5

0

0,5

1

3,7 4,2 4,7 5,2 5,7 6,2 6,7

ln(ln

(1/(

1-F*

(t))))

ln(t)

y = 0,0005x0,262

0

0,0005

0,001

0,0015

0,002

0,0025

0,003

0 100 200 300 400 500 600

Lam

bda

(fai

lure

/day

)

time (day)

CAUSE Quantity % FS Flow Switch 6 40.0 PLC TU regulation 5 33.3 INSTRU Instrumentation 3 20.0 SETTING Settings 1 6.7

Total 15 100.0

0

1

2

3

4

5

6

7

Flow Switch TU regulation Instrumentation Settings

𝜆 𝑡 =1.262513

∗𝑡

513

0.262


19 Jimmy Martin

CVs – global study • 43 events along the studied period

• Two distinguishable Weibull distributions, i.e. two distinct failure rate’s trends difficult case !

y = 0.8555x - 5.6952 R² = 0.9819

y = 3.5604x - 21.553 R² = 0.9649

-4,7

-3,7

-2,7

-1,7

-0,7

0,3

1,3

2,2 2,7 3,2 3,7 4,2 4,7 5,2 5,7 6,2 6,7

ln(ln

(1/(

1-F*

(t))))

ln(t)

0

0,005

0,01

0,015

0,02

0,025

0,03

0,035

0,04

0 100 200 300 400 500 600 700

Lam

bda

(fai

lure

/day

)

time (day)

• Good fitting between theoretical and observed reliability probabilities (cf. K-Smirnov test) • 95% of confidence in the Weibull distribution’s hypothesis

• Use of bi-WEIBULL distribution, expressed by :

𝑅 𝑡 = �𝑒−(𝛼𝑡)𝜃 , 𝑡 < T

𝑒−[ 𝛼𝑡 𝜃+ 𝑡−𝛾𝜂

𝛽] , 𝑡 ≥ T

𝜆 𝑡 = �𝛼𝛼 ∗ (𝛼𝑡)𝜃−1, 𝑡 < T

𝛼𝛼 ∗ 𝛼𝑡 𝜃−1 + 𝛽𝜂

∗𝑡 − 𝛾𝜂

𝛽−1

, 𝑡 ≥ T

𝝀 𝒕 = �𝟎.𝟎𝟎𝟏𝟏 ∗ 𝟎.𝟎𝟎𝟏𝟎 ∗ 𝒕 −𝟎.𝟏𝟏𝟏, 𝒕 < 𝟎𝟑𝟑 𝒅𝒂𝒚𝒅

𝟎.𝟎𝟎𝟏𝟏 ∗ 𝟎.𝟎𝟎𝟏𝟎 ∗ 𝒕 −𝟎.𝟏𝟏𝟏 + 𝟗.𝟎𝟎𝟑𝟑𝟑𝟎𝟑

∗𝒕

𝟑𝟎𝟑

𝟖.𝟎𝟎𝟑𝟑, 𝒕 ≥ 𝟎𝟑𝟑 𝒅𝒂𝒚𝒅

For the complete mathematical method see technical note LHC-CRYO-OP-A-110


20 Jimmy Martin

CVs – global study


CAUSE N %

Command 21 48.8

Mechanical 11 25.6

Unknown 5 11.6

Utilities 5 11.6

Environment 1 2.3

Total 43 100 0

5

10

15

20

25

Command Mechanical Unknown Utilities Environment

Global Pareto histogram for CVs

• Two failure modes are dominant: ‘command’ and ‘mechanical’ causes both represent around 75% of the overall


21 Jimmy Martin

CVs – ‘command’ causes • 21 events along the studied period


0

0,001

0,002

0,003

0,004

0,005

0,006

0 100 200 300 400 500 600

Lam

bda

(fai

lure

/day

)

time (day)

y = 1,8996x - 11,522 R² = 0,9254

-3,7

-3,2

-2,7

-2,2

-1,7

-1,2

-0,7

-0,2

0,3

0,8

1,3

4,1 4,6 5,1 5,6 6,1 6,6

ln(ln

(1/(

1-F*

(t))))

ln(t)

• Pareto analysis of causes: CAUSE Quantity %

PROFIBUS 8 38.1 SIPART 7 33.3 SETTING 3 14.3 SAFETY COMPONENT 1 4.8 UNKNOWN 1 4.8 INSTRU 1 4.8

Total 21 100

0123456789

PROFIBUS SIPART SETTING SAFETYCOMPONENT

UNKNOWN INSTRU𝜆 𝑡 =1.8996

431∗

𝑡431

0.8996


22 Jimmy Martin

CVs – ‘mechanical’ causes • 11 events along the studied period

• A unique Weibull distribution: the failure rate’s trend is decreasing

• Good fitting between theoretical and observed failure probabilities • 95% of confidence in the Weibull distribution’s hypothesis

y = 0,6494x - 3,7642 R² = 0,968

-2,7

-2,2

-1,7

-1,2

-0,7

-0,2

0,3

0,8

1,3

2,2 2,7 3,2 3,7 4,2 4,7 5,2 5,7 6,2 6,7

ln(ln

(1/(

1-F*

(t))))

ln(t)

y = 0,0151x-0,351

0

0,001

0,002

0,003

0,004

0,005

0,006

0,007

0,008

0 100 200 300 400 500 600

Lam

bda

(fai

lure

/day

)

time (day)

𝜆 𝑡 =0.6494

329∗

𝑡329

−0.3506


23 Jimmy Martin

PVs for nitrogen: global study

• Only 5 events along the studied period but a good fitting between data and theoretical distribution! (this is due to the manifestation of a unique failure mode, i.e. leakage)


• Leakage due to the bad design of PVs (i.e. PV409) for nitrogen

y = 14.344x - 88.375 R² = 0.9562

-2

-1,8

-1,6

-1,4

-1,2

-1

-0,8

-0,6

-0,4

-0,2

06,02 6,04 6,06 6,08 6,1 6,12 6,14 6,16 6,18

ln(ln

(1/(

1-F*

(t))))

ln(t)

y = 6E-38x13,344

0

0,005

0,01

0,015

0,02

0,025

0,03

0,035

410 420 430 440 450 460 470 480

Lam

bda

(fai

lure

/day

)

time (day)

0

0,1

0,2

0,3

0,4

0,5

0,6

0,7

410 420 430 440 450 460 470 480

F(t)

t (day)

theo

real


24 Jimmy Martin

Electric heaters (from tunnel) – global study

• 8 events along the studied period, cumulated from February 2012 • A unique Weibull distribution: the failure rate’s trend is increasing (95% of confidence)


y = 8,9778x - 55,668 R² = 0,8372

-2,5

-2

-1,5

-1

-0,5

0

0,5

1

6 6,05 6,1 6,15 6,2 6,25 6,3 6,35

ln(ln

(1/(

1-F*

(t))))

ln(t)

0

0,005

0,01

0,015

0,02

0,025

0,03

0,035

0,04

400 420 440 460 480 500 520 540 560

Lam

bda

(fai

lure

/day

)

time (day)

0

1

2

3

4

5

6

7

0

0,1

0,2

0,3

0,4

0,5

0,6

0,7

0,8

0,9

1

400 420 440 460 480 500 520 540 560

F(t)

t (day)

theo

real


25 Jimmy Martin

Criticality ranking – index of occurrence • Built by considering the overall number of events and the failure rate’s trend related to a studied

population:

Population type

λ’s trend

N of events

MTBF (day)

Index O

Cold compressors – general

β = 2.68

↗ ++

25

327

1

Cold compressors –

regulator

β = 2.76

↗ ++

11

362

1

Warm compressors –

general

β = 0.89

↘

26

325

0.5

Warm compressors –

without leakage

β = 0.82

↘

21

336

0.5

Turbines – general

β = 1.05

→

25

407

0.75

Turbines – ‘command’

β = 1.26

↗ +

15

477

0.75

CVs – general

↗ ++

43

-

1

CVs – ‘command’

β = 1.9

↗ +

21

382

1

CVs – ‘mechanical’

β = 0.65

↘

11

450

0.25

Electric heaters

β = 8.98

↗ ++

8

467

0.75

PVs for N2

β = 14.3

↗ ++

5

457

0.75

If: • β < 1 λ’s trend = ↘ • β = 1 λ’s trend = → • 1 < β ≤ 2 λ’s trend = ↗ + • 2 < β λ’s trend = ↗ ++


0 0,2 0,4 0,6 0,8 1 1,2

CV valves - MEC


PV valves for N2

Turbines - command

Electric Heatears

CV valves - COM


INDEX OF OCCURENCE OF PROCESS HARDWARE COMPONENTS


26 Jimmy Martin

Criticality ranking – Index of severity • Built by considering the overall number of losses of ‘CM’ and time before recovering these

conditions, for a studied population:


0 0,1 0,2 0,3 0,4 0,5 0,6 0,7 0,8

PV valves for N2

CV valves - MEC

Electric Heatears

CV valves - COM

Turbines - command



SEVERITY OF PROCESS HARDWARE COMPONENTS


27 Jimmy Martin

Criticality ranking – index of criticality • Built by multiplying the two previous indexes of occurrence and severity:

0,0 10,0 20,0 30,0 40,0 50,0 60,0 70,0 80,0

CV valves - MEC

PV valves for N2

Electric Heatears


Turbines - command

CV valves - COM


GLOBAL CRITICALITY OF PROCESS HARDWARE COMPONENTS Recall :

C = O ∗ S ∗ D (With D = 1)


28 Jimmy Martin

CONCLUSION • For the highlighted most critical hardware components, a refined analysis of origins of failure

modes must be set up. These components are: • Cold compressors’ regulators • Profibus and Sipart devices for CVs • Flow switches and regulators for TUs

• Perhaps this kind of analysis will be extended to other components, and then will lead to a coherent ensemble of statistical data related to process’ reliability, probable use of this data as the input of a MARKOV chains computation :

THANK YOU FOR YOUR ATTENTION

• The current way to put information into ‘Logbook OA’ must be also improved to make easier a future statistical study

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It could possible to : • Predict an overall availability

for the ‘CRYO’ process • Improve the maintenance

action plans by firstly simulate them

criticality analysis for lhc cryogenic … criterions for a criticality ranking numerical input data...

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