logistics systems engineering maintainability/serviceability/human factors

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Logistics Systems EngineeringMaintainability/Serviceability/Human Factors

NTUSY-521-N

SMUSYS 7340

Dr. Jerrell T. Stracener, SAE Fellow

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Maintainability

• Maintainability is- an engineering and management function

spanning the product or service life cycle

- a characteristic of equipment design andinstallation which is expressed in terms of ease and economy of maintenance, availability of the equipment, safety and accuracy in the performance of maintenance actions.

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Maintainability

• Objective of maintainability- to design and develop systems and

equipment which can be maintained in the least time, at the least cost, and with a minimum expenditure of support resources, without adversely affecting the item performance or safety characteristics

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Product/Service Support Resources

• logistics personnel utilization

• spare parts

• tools and test equipment

• support services

• support facilities

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What is Maintainability?

Converters for driving factory belts1. Motor Burn-out2. Wire replacements3. Torque Adjustments4. Lubrication

– What are its associated cost?Down time: StaffingProduction: Product to marketHuman factors: Stress, Leaning CurveReliability: Service performance and

guarantees

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What is Maintainability?

• Maintainability greatly influences reliability and availability of a system or subsystem.

• Maintainability must be addressed early in the design stage to prevent or reduce failure or down times of the system.

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Why is Maintainability Required?1

• Infinite Reliability is not achievable• When a system is discarded, it must be

discarded or it must be repaired• Cost usually dictates that a faulty system must

be repaired• In addition to repair, most systems must be

serviced (Consumables replaced - fuel, oil, coolant, etc.)

• Incipient failures must be detected

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• To verify that equipment has not deteriorated, its overall capability to perform must be reviewed

• Maintenance is the repair, servicing, and inspection of equipment

Why is Maintainability Required?1

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Maintenance Concept

• Maintenance defines all those activities performed on an item to retain it in or to restore it to s specified state.4

• Can be divided into two categories:1. Preventive Maintenance

Prescribe procedures to reduce the probability of failure or degradation

2. Corrective MaintenanceInitiated after fault recognitionRegain state of system for

performing required function

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Maintenance Concept

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Detection

Preparation forMaintenance

Location andIsolation

Disassembly(Access)

Removal ofFault Item

Re-assembly

Alignmentand Adjustment

ConditionVerification

or

Repair ofEquipment

Installation ofSpare/Repair Part

Failure Occurs

Failure Confirmed

Active Maintenance Commences

Faulty Item Identified

Disassembly Complete

Re-assembly Complete

Repair Completed

Maintenance Concept

Corr

ecti

ve M

ain

ten

an

ce C

ycle

6

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Maintenance is Conducted:7

• On equipment repair– Remove and replace faulty item– Adjust of align an item that has drifted out of

specification• Off equipment repair

– In a local shop– In and industrial facility

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Achieving Maintainability

• Achieving maintainability is done through planning and realizing maintenance concepts:– Fault Detection and isolation– Partitioning equipment or systems into LRUs– User documentation– Training– Logistical Support

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• Fault Detection and isolation– Goal is to localize faults down to LRU’s (last

repairable unit / line replacement unit) by performing the following:BIT (Built-in test):1. Degree of fault2. Degree of isolation3. Correctness of the fault isolation4. Test duration


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BITE (Built-in test equipment):1. Simplicity2. Standardization3. Reliability4. Maintenance

• Equipment and System Partitioning– Partition complex equipment and systems

into LRUs: PCB– Accessibility: Ease of LRU– Adjustment: Digital reduces need– Exchange: Careful of obsolescence


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• User Documentation– General Description– Operating Manual– Preventive Maintenance– Corrective Maintenance– Illustrated Spare Parts Catalog– Logistical Support

• Training of Operating & Maintenance Personnel– Well trained and motivated– Human Errors


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• User Logistical Support– Four Levels

1. Operating personnel2. First line maintenance personnel3. Maintenance personnel4. Specialist from arsenal or industry


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• Specify– Specifications, Contracts, Warranties– Program Plan

• Design– Equipment Arrangement– Equipment Location– Servicing Locations– Weapon Location– Turnaround Arena– Accessibility– Fault and Servicing Cues


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• Plan– Predesign Homework– By Analysis– Mock Ups

• Demonstrate Supportability– Verify Operation Environment


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Bottoms Up Models

• Provide output to monitor design progress vs. requirements

• Provide input data for life cycle cost• Provide trade-off capability

– Design features vs. maintainability requirements

– Performance vs. maintainability requirements• Provide Justification for maintenance

improvements perceived as the design progresses

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Bottoms Up Models

• Provide the basis for maintainability guarantees/demonstration

• Provide inputs to warranty requirements• Provide maintenance data for the logistic

support analysis record• Support post delivery design changes• Inputs

– Task Time (MH)– Task Frequency (MTBM)

Number of Personnel-Elapsed Time (hours)For each repairable item

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Bottoms Up Models

• Input Data Sources– Task Frequency

Reliability predictions de-rated to account for non-relevant failures

Because many failures are repaired on equipment, the off equipment task frequency will be less than the task frequency for on equipment

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Bottoms Up Models

• Input Data Sources (Continued)– Task Time

Touch time vs. total time That time expended by the technician to

effect the repair Touch time is design controllable

Total Time Includes the time that the technician

expends in “Overhead” functions such as part procurement and paper work

Are developed from industrial engineering data and analyst’s estimates

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Task Analysis Model

• Task analysis modeling estimates repair time– MIL-HDK-472 method V– Spreadsheet template

Allow parallel and multi-person tasks estimation

Calculates elapsed time and staff hoursReports each task element and total repair

timeSums staff hours by repairmen typeEstimates impact of hard to reach/see

tasks

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Why Do Maintainability Modeling?

• To identify the important issues• To quantify and prioritize these issues• To build better design and support systems

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Design Guidelines for Maintainability9

• General Guidelines– Plan and Implement a concept for automatic

fault detection down to the last LRU– Partition the equipment– Aim for standardization of parts, tools, and

testing equipment– Conceive operation and maintenance

procedures to be as simple as possible– Consider environmental conditions

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• Testability– Degrees of failure detection and isolation– The correctness of test results– Test duration

• Accessibility and Exchangeability– Provide self-latching access flaps– Plan for accessibility– Use preferably indirect plug connectors– Provide for speedy replaceability– Prevent faulty installation or connection


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• Operation and Adjustment– Use high standardization in selecting

operational tools– Consider human aspects– Order all steps of a procedure in a logical

sequence– Describe system status– Avoid any form of hardware adjustments


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Elements & Terminology of Maintainability

• MTTR: Mean Time to Repair

• T0.5: Median Time to Repair

• TMAX: Maximum Time to Repair)

usually the 95th percentile• MTTPM: Mean Time to Preventive Maintenance• MTBPM: Mean Time Between Preventive

Maintenance• MDT: Mean Down Time• MTBM: Mean Time Between Maintenance

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Maintainability Prediction

• System Mean Time to Repair, MTTRS

System without redundancy

E1 E2 En

n

1ii

n

1iii

n

1i i

n

1i i

i

λ

MTTRλ

MTBF1

MTBFMTTR

MTTRs

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• Example 1: Compute the mean time to repair at the system level for the following system.

• Solution:

MTTF = 500 hMTTR = 2 h

MTTF = 400 hMTTR = 2.5 h



hh

h

hh

h

hh

h

hh

h

hh

MTTRs 04.10185.0

01925.0

1001

1005.0

2501

2501

4001

4005.2

5001

5002

1

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• Example 2: How does the MTTRs of the system in the previous example change if an active redundancy is introduced to the element with MTTF = 100h?

• Solution:

hh

h

hh

h

hh

h

hh

h

hh

h

hh

MTTRs 85.00285.0

02425.0

1001

1005.0

1001

1005.0

2501

2501

4001

4005.2

5001

5002

1






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MTTF and MTBF

Mean Time to Failure (or Between Failures) MTTF (or MTBF) is the expected Time to Failure (or Between Failures)

Remarks:MTBF provides a reliability figure of merit for expected failure free operation MTBF provides the basis for estimating the number of failures in a given period of time Even though an item may be discarded after failure and its mean life characterized by MTTF, it may be meaningful to characterize the system reliability in terms of MTBF if the system is restored after item failure.

00

)()( dttRdtttfMTBF

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Logistics Systems EngineeringModeling & Analysis of Time to Repair

NTUSY-521-N

SMUSYS 7340

Dr. Jerrell T. Stracener, SAE Fellow

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Definition

• Maintainability is an inherent design characteristic of a system or product and it pertains to the ease, accuracy, safety, and economy in the performance of maintenance actions.2

• Maintainability can be created into a four-part definition:3 1. Maintainability is the probability that a failed system2. will be restored to specified performance3. within a stated period of time4. when maintained under specified conditions.

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Definition

• Maintainability is a characteristic of an item, expressed by the probability that preventive maintenance (serviceability) or repair (repairability) of the item will be performed within a stated time interval by given procedures and resources (number and skill level of the personnel, spare parts, test facilities, etc.).4

• Maintainability is the ability of an item to be retained in, or restored to, a specified condition when maintenance is performed by people having specified skill levels, using prescribed procedures and resources.5

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Maintenance and Design8

• The system’s design determines its requirements for maintenance– Reliability (How often maintenance)– Configuration (How much time for access)– Built in Test (Fault Isolation Time)– Subassembly life span (Inspection/forced

replacement)– Adjustment/alignment requirements

(Inspection)– Capacity/fill rate (Servicing)– Corrosion susceptibility (Inspection/repair)

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Normal Distribution:

A random variable X is said to have a normal (orGaussian) distribution with parameters and ,where - < < and > 0, with probability density function

- < x <

where = 3.14159… and e = 2.7183...

22

x2

1

e2

1)x(f

f(x)

x

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• Mean or expected value of X

Mean = E(X) =

• Median value of X

X0.5 =

• Standard deviation

)(XVar

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Standard Normal Distribution

If X ~ N(, ) and if , then Z ~ N(0, 1).

A normal distribution with = 0 and = 1, is calledthe standard normal distribution.

X

Z

42

x

0

z

x

Z


f(x) f(z)

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Standard Normal Distribution Table of Probabilities

http://www.smu.edu/~christ/stracener/cse7370/normaltable.html

Enter table with

and find thevalue of

z0

z

f(z)

x

Z

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Normal Distribution - example

The following example illustrates every possible case of application of the normal distribution.

Let X ~ N(100, 10)

Find:a. P(X < 105.3)b. P(X 91.7)c. P(87.1 < X 115.7)d. the value of x for which P(X x) = 0.05

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Normal Distribution - example solution

a. P(X < 105.3) =

= P(Z < 0.53) = 0.7019

10

1003.105P

x

100

x

0

z

f(x) f(z)

105.3 0.53

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b. P(X 91.7) =

= P(Z > - 0.83)= 1 - P(Z -0.83) = 1 - 0.2033= 0.7967

10

1007.91

x

P

100

x

0

z

f(x) f(z)

91.7 -0.83

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c. P(87.1 < X 115.7) =

= P(-1.29 < Z < 1.57)= F(1.57) - F(-1.29)= 0.9418 - 0.0985 = 0.8433

7.115

10

1001.87

x

P

100

x

f(x)

87.1 115.7

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d. P(X x) = 0.05P(Z z) = 0.05 implies that z = 1.64P(X x) =

therefore

x - 100 = 16.4x = 116.4

10

100

10

100 xZP

xxP

64.110

100

x

100

x

f(x)

116.4

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Normal Distribution - Example:

The time it takes a field engineer to restore a function in a logistics system can be modeled with a normal distribution having mean value 1.25 hours and standard deviation 0.46 hours. What is the probability that the time is between 1.00 and 1.75 hours? If we view 2 hours as a critically time, what is the probability that actual time to restore the function will exceed this value?

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Normal Distribution - Example Solution:

75.100.1 XP

46.0

25.175.1

46.0

25.100.1XP

09.154.0 XP

54.009.1

5675.02946.08621.0

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Normal Distribution - Example Solution:

46.0

25.122 ZPXP

63.1163.1 ZP

0516.0

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The Lognormal Model:

Definition - A random variable X is said to have the Lognormal Distribution with parameters and , where > 0 and > 0, if the probability density function of X is:

, for x > 0

, for x 0

22

xln2

1

e2x

1 )x(f

0

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Properties of the Lognormal Distribution

Probability Distribution Function

where (z) is the cumulative probability distribution function of N(0,1)

Rule: If T ~ LN(,), then Y = lnT ~ N(,)

xln

)x(F

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Properties of the Lognormal Model:

• Mean or Expected Value

2

2

1

e)X(E

1ee)X(Var222

• Variance

• Median

ex 5.0

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Lognormal Model example

The elapsed time (hours) to repair an item is a random variable. Based on analysis of data, elapsedtime to repair can be modeled by a lognormal distribution with parameters = 0.25 and = 0.50.

a. What is the probability that an elapsed time to repair will exceed 0.50 hours?b. What is the probability that an elapsed time torepair will be less than 1.2 hours?c. What is the median elapsed time to repair?d. What is the probability that an elapsed time torepair will exceed the mean elapsed time to repair?e. Sketch the cumulative probability distributionfunction.

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Lognormal Model example - solution

a. What is the probability that an elapsed time to repair will exceed 0.50 hours?

X ~ LN(, ) where = 0.25 and = 0.50

note that:Y = lnX ~ N(, )

P(X > 0.50) = P(lnX > -0.693)

0.50

0.250.693

σ

μlnXP

89.1P Z

9716.0

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b. What is the probability that an elapsed time torepair will be less than 1.2 hours?

P(X < 1.20) = P(lnX < ln1.20)

0.50

0.25182.0

σ

μlnXP

136.0P Z

4404.0

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c. What is the median elapsed time to repair?

P(X < x0.5) = 0.5

therefore

5.0lnP

xZ

0P Z

5.0

0ln 5.0

x

25.0ln 5.0 x

284.125.05.0 eex

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d. What is the probability that an elapsed time torepair will exceed the mean elapsed time to repair?

2

σμ

2

eMTTR

2

50.025.0

2

e

375.0e

455.1

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P(X > MTTR) = P(X > 1.455)

= P(lnX > 0.375)

0.50

0.250.375

σ

μlnXP

0.25ZP

4013.0

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e. Sketch the cumulative probability distributionfunction.

Cumulative Probability Distribution Function

0

0.2

0.4

0.6

0.8

1

0 1 2 3 4 5 6

time to repair

P(t

<x)

tmax

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1 Clint Van Pelt, “Maintainability and Modeling Analysis”, March 31, 19922 Benjamin S. Blanchard and Wolter J. Fabrychy, “Systems Engineering

and Analysis,” Second Edition (Englewood Cliffs, New Jersey: Prentice-Hall, Inc., 1990), pp. 389-390.

3 Daniel L. Babcock, “Managing Engineering and Technology,” Second Edition (New Jersey: Prentice-Hall, Inc., 1996), p. 209.

4 Prof. Dr. Alessandro Birolini, “Reliability Engineering: Theory and Practice,” Third Edition (Germany: Springer-Verlag Berlin Heidelberg, 1999), p.115.

5 USAF R&M 20006 Benjamin S. Blanchard and Wolter J. Fabrychy, “Systems Engineering

and Analysis,” Second Edition (Englewood Cliffs, New Jersey: Prentice-Hall, Inc., 1990), p. 394.

7 Clint Van Pelt, “Maintainability and Modeling Analysis”, March 31, 19928 Ibid

References

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9 Prof. Dr. Alessandro Birolini, “Reliability Engineering: Theory and Practice,” Third Edition (Germany: Springer-Verlag Berlin Heidelberg, 1999), pp. 145-148.

References

logistics systems engineering maintainability/serviceability/human factors

Documents

test equipment

equipment repairremove

item performance

faulty system

maintenance concepts

economy of maintenance

service performance

design stage