Jpm Tqm Course Mat-6 T-3

7/31/2019 Jpm Tqm Course Mat-6 T-3

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SEVEN TOOLS OF QUALITY CONTROLSeven Tools for identifying Quality problems and their causes for solution &improvement :

1) Check Sheet : A fact finding tool for collection of information & data of themeasure of variations and count the number of defectivesby type (data

sheet may be in the form of SPREAD SHEETS arranging data recorded in matrixform). The purpose :

i) recording and keeping track with respect to previous lots

ii) providing a record of frequency of quality problemiii) showing where quality problem occurred,

(for quality problem identification and analysis).

2) Stratification : A technique of arranging the collected observations /measured data

in different groups, based on segregated CAUSES, like

(i) RAW MATERIALS : a) Supplier-wise

b) Supply Batch-wise

(ii) PRODUCTION : a) Machine-wiseb) Operation-wise (Work-Center wise)c) Shift-wise or batch-wise


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SEVEN TOOLS OF QUALITY CONTROL

3) Scatter Diagram :A two variables graph, plotting of measurements ofthe dependable variable of the process (quality characteristic) on Y -axis againstan independent process variable like Size / Dimension of job, Cutting Speed,

Depth of Cut, etc (on X -axis), to establish the nature of variation of the curve,showing co-relation between the two variables. The co-relation may be :

i) Positive Co-relation : increase of dependable variable (quality characteristic)on Y-axis caused by increase of independent variable on X-axis, ie Upward.

ii) Negative Co-relation : decrease of dependable variable (quality characteristic)

on Y-axis caused by increase of independent variable on X-axis, ie Downward.iii) Zero Co-relation : No change of dependable variable (quality characteristic)on Y-axis caused by any variation of independent variable on X-axis, ie II toX-axis.

.

.

. .

..

.

..

.

... . . .. .

..

Quality QualityQuality

X X X

Positive Co-relation Negativ

e Co-relation

Zero Co-relation


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SEVEN TOOLS OF QUALITY CONTROL

4) Control Charts : Showing sample measurements of various lots with respect toCentral Line (CL), Upper Control Limit (UCL) and Lower Control Limit (LCL), UNTL & USLand LNTL & LSL.

5) Histogram :To analyze (i) How well the histogram is centered,(ii) How wide the histogram is, and

(iii) Shape of the histogram : whether Normal Frequency

distribution or not.

6) Pareto Chart : To determine the most frequently occurring problems / defectsaccording to their Causative Importance (Cause of defect / rejection / failure).

The hierarchy of the Causative factors for the problem makes it easier to tackle them insame order. So, the most frequent Causative Factor/s are identified and separated from theless frequent Causative Factors.

Pareto Charts are created by plotting the cumulative frequency data of the relativefrequencies arranged in descending order.

1

23

5

4

6 7

64%

5%

3%

2%1%

16%9%

1,2,3,4,5,6,7 are causative factors or type of failure.

1 + 2 = 64% + 16% = 80%

3 + 4 + 5 + 6 +7 = 9% + 5% + 3% + 2% + 1% = 20%

Type/Cause of rejection/defect

Frequency of Failure

dimension

frequency


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Assignments :Make Cause & Effect Analysis with the help of Ishikawa Fish Bone

Diagram, for

i) House Painting Peeling.ii) Non-conformance of Shaft diameter in production shop.

iii) Delayed flight departures.


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House PaintingPeeling

Material Method

Machine /Equipment

Environment

Dirt mixed

Wrong type

Contaminated

Cheap quality

Too thin layer

Too thick layer

Paint on dirty surface

Wrong method ofpaint application

Dirty Brush

Used Brush

Brush with Bad bristle Air Pollution

High Temperature

High Humidity

Acid RainWrong Brush


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CAUSAL RELATION

1. Necessary Cause : The prime cause that MUST BE PRESENT foreffect to occur.

Without this cause the effect can not come.Example : Combustion is essentially necessary to drive an I.C. Engine.

2. Sufficient Cause :Each of the Sufficient Causes can alone produce theeffect unaided by any other cause (without the help of any other cause), thoughmore than one sufficient causes may be present there for that effect.

Any one of the Sufficient Causescan alonecause the effect.

Example : battery is dead , Faulty Spark Plugs, Empty Fuel Tank, are thesufficient Causes for the effect that a car will not start. Only one of suchsufficient causes can alone result in the effect that the car will fail to start.

On the other hand, any one of the Sufficient Causes may not be present forthe effect.

3. Contributory Cause : The cause that helps to produce the effect butcan not do so by itself without the presence of other contributory cause/s. Inother words, all the Contributory Causes are to be present to cause the effect.Example : to effect FIRE, contributory causes include the presence of(i) Oxygen,

(ii) Fuel or Inflammable materials,

(iii) Initiation of Ignition.


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BRAIN STORMINGIt is the systematic process of idea generation, with the objective :

to uncover (open) the hidden quality issues and to help coming up of new innovative and effective ideas,

(i) for solving quality problems / issues as well as(ii) for quality improvement,

from the participating employees (workers, engineers, supervisors, experts, etc)of the organization.

In Brainstorming process, an open discussion is taken up among the selectedparticipants / members (say, workers) under the leadership of theSupervisor, whose role is as a FACILITATOR.

Brainstorming process should have the following characteristics,

1. At the first step, the quality problem or quality issue is identified.2. Each individual member of the brainstorming group can suggest an

idea, one by one, relating to solve the quality problem or for qualityimprovement, related to the issue under consideration.

3. If any member has no new idea, he /she may pass.


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4. No criticism is permitted and members are encouraged to generatenew and innovative idea without any fear and hesitation.

5. The idea must be new and unexposed so far and not in use in existing

system.6. Wild ideas are not neglected and not discouraged. They are recorded,

because they may trigger other good ideas from somebody else.

7. The Facilitator writes down each and every new idea onthe blackboard for discussions on it, so that every one can see it and think over

it, and discuss on it.8. Detail & freediscussions on the idea is encouraged for its

(i) effectiveness, (ii) implementation, (iii) ability to solve theproblem under consideration, (iv) all positive & negative featuresas well as side-effects, are carried on.

9. Any modification of the idea for the betterment and more effectivenessis also encouraged in the discussion.

10. The process continues until no further new idea is coming up.

11. The best idea is recommended for consideration.


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ROOT CAUSE ANALYSISLong term relief from any quality problem can be achieved by eliminating theROOT CAUSE of the problem and not by eliminating only the symptoms of the

problem (which is temporary remedy).When any quality problem is located or identified, immediately Root CauseAnalysis should be taken up to locate the source of the quality failure or qualityproblem, ie the Root Cause leading to the quality failure, so that the RootCause can be rectified to stop / prevent the further occurrence of such quality

failure.By Root Cause Analysis, locating and eliminating the Root Causes leads to thelong term relief from generating poor quality can be achieved.

Various methods / techniques of Root Cause Analysis are

(i) Why Why or 5 Why .

(ii) Ishikawa - Fish Bone Diagram.

(iii) Failure Mode & Effect Analysis (FMEA)

(iv) Fault Tree Analysis (FTA)

5 WHY (Wh Wh ) M th d


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5 WHY (Why Why) MethodIt is one of the useful methods of Root Cause Analysis. Proceeding towards root causeby consecutively asking WHY at each stage of occurrence, atleast 5 times.In a classic example of this method used in Toyota Motors, a machine repeatedlyfailed because of a fuse blew.

Replacing the fuse would have been one solution obviously, however very temporaryone and in that case the failure might continue to repeat frequently if the root causebehind such fuse blow was not identified and rectified.So, Toyota applied the 5 Why technique to solve it and find the permanent solution.First WHY : Why did the machine fail ?

because the fuse blew.Second WHY : Why did the Fuse blow ?

because the bearing of the machine did not have adequatelubrication.

Third WHY : Why did the bearing have not adequate lubrication?because the lubrication pump was not working properly.

Fourth WHY : Why the lubrication Pump did not work properly ?because the axle of the lubrication pump was worn out.

Fifth WHY : Why was the pump axle worn ?because sludge (lubricating oil with particulate matter) seeped

into the pump axle.This was the Root Cause.Toyota attached a STRAINERto the lubrication pumpto prevent sludge seepinginto the pump axle, and obviously the problem of the machine failure was solved

instantaneously for long term.


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FMEA (Failure Mode & Effects Analysis)FMEA is a systematic approach to review the design of existing product by analyzing the

mode of failure of the product to satisfy the customer, linking with its Causes &Effects, leading to the recommendations for remedial actions.

Typical example of FMEA for Pearle Snacks Biscuits :

FailureMode

Effect of Failure Causes of Failure RPN Remedial Actions

Stale &damped

Bad taste & loss ofcrunchiness, perceived

as old ; leading to lossof sales

High moisturecontent, expiry of

true self-life, poorpackaging

550 Remove moisture byvacuum & baking, better

sealing in packaging,specify less (true) life

Broken Poor display at snackstable & perceived asold, leading to loss ofsales

Poor packaging,too brittle,too thin

670 Change the recipe,increase thickness,improve packaging withplastic guard

Too salty &too hot

Consumption isreduced, health hazardresulting in loss ofsales

Uneven mixing ofspices & salt,process control ispoor

460 Revise the productionprocess, improve process-control

Severity Ranking = S, Occurrence (repetition)= O, Failure of prior Detection = D,

[ S, O and D are estimated on 1 10 scale]

Risk Priority Number (RPN) = S x O x D


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Poka Yoke i h f i t k fi i t ti


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Poka Yoke :is an approach for mistake-proofing processes using automaticdevices or methods to avoid simple and very common human errors.The concept was developed and refined in early 1960s by Shigeo Shingo, a Japanesemanufacturing engineer who developed Toyota Production System of modern conceptOf Lean manufacturing. Baka Yoke (Fool Proof) term was modified later to Poka Yoke

(Mistake Proof) on sentimental ground.Poka Yoke is focused on two aspects :(1) Prediction : recognizing that a defect is about to occur and providing a

warning, and(2) Detection : recognizing that a defect has occurred and stopping the process

flow using defective materials, to stop continuing the generation of defectiveproducts.

Many application of Poka Yoke are very simple yet Creative, effective & helpful.Usually they are inexpensive to implement.Example :At Yamada Electric plant Of Japan, workers assembled a switch having two pushbuttons supported by two springs. Occasionally, a worker forgot to insert a spring

under each button, which led to a costly and embarrassing repair at thecustomers premises.Previously the worker used to take one by one two springs from a box containing largequantity of spring before inserting the screw during assembly.Shingo instructed first to place two spring from the box onto a small dish in frontof the parts box and then assemble the switch. Now, if a spring remains on the

small dish the worker immediately detects his mistake and rectifies.

S E l f P k Y k


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Some Examples of Poke Yoke :

1. Machines have limit switches connected to warning lightand/or soundthat tell operator in case parts are positioned improperly on the machine.

2. A device on a drill counts the number of holes drilled on a work pieceand a buzzer sounds if the work piece is removed before the correct

number of holes has been drilled.3. Cassette-covers were being frequently scratched when the screwdriver

slipped out of the screw slot and slid against the plastic covers. As PokaYoke, the screw head design were changed to head with partlyslotted at middleposition to prevent the screwdriver to slip.

4. Computer programs display awarning messageif a file that has notbeen saved is to be closed, or during deleting any file.

5. The SIM Card fitting (one corner of sim card is cut) in the cell phone.

6. If there is any manufacturing defect in a part produced, the Poka Yokedesigns thedevice so that the next machine in the process line will notaccept it or will not start to process on it.

7. One production step at Motorola involves putting alphabetic characters ona key board and the checking to make sure the each key is placedcorrectly. A group of workers designed a clear transparent templatewith the letters positioned slightly off center. Now, by holding thistemplate over the keyboard under assembly, the operator can easily spot

any mistake .

KANO M d l V i f C t


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KANO Model on Voice of CustomerNoriaki KANO was a Japanese professor who classified customer requirements/expectations in 3 categories, Dissatisfiers, Satisfiers and Delighters.1. DISSATISFIERs :CTQ (Critical to Quality)Characteristics essentially

needed by the customers, but not expressed by them.

Customers are not extra satisfied if DISSATISFIERs are met, but they areextremely dissatisfied if DISSATISFIERs are not met,because,DISSATISFIERs are assumed as obvious and basic eligibility criteriafor quality of the product (like safety features of an automobile).DISSATISFIERsact asEligibility of product for entry to market-competition.

2. SATISFIERs : (1) CTQs which the customer prefers to get in the product,(2) they express that they give importance to such CTQs, and(3) more or better such characteristics are met, customers are moresatisfied.Better grade of SATISFIERs make the customer more satisfied.Customers compare the product of different producers based on the degreeof Satisfier Factors (example : Fuel efficiency, speed pick-up rates, sun-roofinsulation, etc of the automobile). Satisfiers make product more competitive.

3. DELIGHTERs : New and innovative quality features which are beyondcustomers expectation, whosepresence leads to very high customer-perception of quality. Delighters have very high impact on customerspurchase-decision by adding extremely high value to the productwithextremely high customer satisfaction. (example : previously Camera and nowinternet facility in mobile phone, independent rear seat audio control in the car and

weather channel button on radio in the automobile). They act as Differentiator.


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TIME


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Trade-off matrix

Design

characteristics

Customerrequirements

Target values

Relationshipmatrix

Competitiveassessment

Importance

1 2

3

4

5

6

Quality Function Deployment

C i i


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CompetitiveAssessment of

Customer Requirements

Irons

well

Easyand

safetouse

Competitive Assessment

Customer Requirements 1 2 3 4 5

Presses quickly 9 B A X

Removes wrinkles 8 AB XDoesnt stick to fabric 6 X BA

Provides enough steam 8 AB X

Doesnt spot fabric 6 X AB

Doesnt scorch fabric 9 A XB

Heats quickly 6 X B AAutomatic shut-off 3 ABX

Quick cool-down 3 X A B

Doesnt break when dropped 5 AB X

Doesnt burn when touched 5 AB X

Not too heavy 8 X A B

F


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Energyneededtopress

Weightofiron

Sizeofsoleplate

Thicknessof

soleplate

Materialused

insoleplate

Numberofho

les

Sizeofholes

Flowofwaterfrom

holes

Timerequiredtoreach450F

Timetogofrom

450to100

Protectiveco

verforsoleplate

Automaticsh

utoff

Customer Requirements

Presses quickly - - + + + -

Removes wrinkles + + + + +

Doesnt stick to fabric - + + + +

Provides enough steam + + + +

Doesnt spot fabric + - - -

Doesnt scorch fabric + + + - +

Heats quickly - - + -

Automatic shut-off +

Quick cool-down - - + +

Doesnt break when dropped + + + +

Doesnt burn when touched + + + +

Not too heavy + - - - + -

Irons

well

Easyand

safetouse

FromCustomer

Requirements

to DesignCharacteristics


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Energyneededtopress

Weightofiron

Sizeofsoleplate

Thicknessofsoleplate

Materialusedin

soleplate

Numberofholes

Sizeofholes

Flowofwaterfromholes

Timerequiredto

reach450

Timetogofrom

450to100

Protectivecoverforsoleplate

Automaticshuto

ff

--

++

+

Tradeoff Matrix


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Energy

neededtopress

Weight

ofiron

Sizeof

soleplate

Thicknessofsoleplate

Materia

lusedinsoleplate

Numberofholes

Sizeof

holes

Flowof

waterfromholes

Timere

quiredtoreach450

Timeto

gofrom450to10

0

Protect

ivecoverforsoleplate

Automa

ticshutoff

Units of measure ft-lb lb in. cm ty ea mm oz/s sec sec Y/N Y/N

Iron A 3 1.4 8x4 2 SS 27 15 0.5 45 500 N Y

Iron B 4 1.2 8x4 1 MG 27 15 0.3 35 350 N YOur Iron (X) 2 1.7 9x5 4 T 35 15 0.7 50 600 N Y

Estimated impact 3 4 4 4 5 4 3 2 5 5 3 0

Estimated cost 3 3 3 3 4 3 3 3 4 4 5 2

Targets 1.2 8x5 3 SS 30 30 500

Design changes * * * * * * *

Ob

jective

me

asures

Targeted Changes inDesign


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Quality

Management

System (QMS)&

Quality Assurance

Focus : ISO - 9001 ISO : International Standard Organization


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ISO : International Standard OrganizationISO 9000 : Standard forQuality Management & Quality Assurance[Previous ISO 9000 was edited in 1993 and next updating and revision in2000] It is the standard for the Organization having the required QualityManagement System, but not the standard of the Products.

[ ISO 9000 was equivalent to ANSI / ASQC Q 90 (US version]ISO 9000 :Guide line for selection and use.ISO 9001 :Models for Quality Assurance in design, development,

Production, Installation and Services.

ISO 9002 :Models of Quality Management System for QualityAssurance in Production, Installation and Services.

ISO 9003 : Models for Final Inspection, Auditing and Testing forQuality Assurance.

ISO 9004 : Guide line to Managers to build their QMS for effective

Quality Management.[ From 2000 version, ISO 9001 and ISO 9002 have been merged tosingle one ISO 9001, which is applicable to all industries of Productionincluding manufacturing and services, with or without design anddevelopment.Now, the latest version of ISO-9001 is 2008.]


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[ Being closely related to ASQC (American Society for Quality Control)RAB (Registrar Accreditation Board) is the controlling authorities, for the

accreditation (approval) & registration of the REGISTRARs for certificationof ISO-9000.]

First Party : Supplier / producer organization seeking Certificate for Compliancewith ISO 9001, for Quality Assurance and total Customer Satisfaction.

Second Party : The REGISTRARs who are CERTIFICATE issuing authorities,after conducting appropriate assessment (through Audit by Lead Auditors)of existing Quality Management System of the certification-seekingorganization (supplier / Producer) to assure that its Quality ManagementSystem is indeed in compliance with appropriate standard (ISO-9001) topositively achieve the specified quality and reliability of their products.

Third Party : The LEAD AUDITORs , theAuditorsempanelled and registered withRAB who conduct the QMS Audit on behalf of the Registrar.

Organizations which may require to have ISO 9001 certification :a) Manufacturing Industries : Automobile, Metal, Minerals, Fabrication &

Machining, Assembly line, Process (Chemicals, Petro-chemical, Paints,Pharmaceutical, Food, Pharmaceuticals) , Electronic goods, Consumergoods, Leather, Furniture, etc.

b) Service Industries : Educational Service (Schools and colleges); MedicalUnits (Clinic, Nursing home, Hospital); IT; Accounting & Auditing; FinancialServices; Hotels; Restaurants; 3PL & 4PL; Consultancy; Retail Outlets;Communication Network Service; Postal; Courier; Transport (Roadway;Railway; Airlines; Shipping); Construction; Repair & Maintenance, Security,etc.

(Registrar Accreditation Board)


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(Registrar Accreditation Board)

R A B

REGISTRARs Certified Lead -AUDITORs

Employ (4)

Accredits & registers (1) Certifies & Empanels (2)

Supplier /Producer

Audit /Inspection (5)

Compliance Report (6)

Issue ISO-9001Certificate (7)

Consumer /Customer

Quality Assurance ofthe Products

Certification Relationship Diagram

Applies for Certification (3)

Benefits of having ISO 9001 Certification


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Benefits of having ISO 9001 Certification

1) Improves Competitiveness of the business

2) Improves the image (goodwill) and the credibility of the Organization

in the market by Assuring the Quality of production3) Improvesconfidence of the organization through QMS4) Improves profitability of the organization by improving economy of

production by reducing:i) Quality Rejectionsii) Scraps generation

iii) Wastes5) Improves Productivity by Efficient Utilization of input Resources

(materials, manpower, machines, utilities, etc)6) Easy and prompt identification & rectification of fault (if any) for

continual improvement of quality, through pin pointing the

responsibilities and accountabilitiesof all employees at all levels.7) COMPLIANCE to Quality target and specification of productis assured

Steps towards getting ISO 9001 Certification


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Steps towards getting ISO 9001 Certification1. Take decision for ISO 9001 Certification ( decide on what business,

which unit/s or branch, etc).

2. Select and employ CONSULTANT for assistance (if required).

3. Select RAB accredited REGISTRAR (who will issue the Certificate),with consultation with the employed Consultant.

4. Select MR (Management Representative) for the organization, withconsultation with the employed Consultant.

5. Conduct extensive Quality Awareness programme and QualityTraining programme among all the employees related in theorganization, particularly in that unit or branch, including themanagement & executive staff, and employees at all levels top tobottom.

6. Select RepresentativesforInternal Audit of Quality Management

System of Quality Assurance, from all the branches and departmentsof the organization, with consultation with the employed Consultant.

7. Prepare, establish and maintain the Documentations.

8. Train up all the personnel withthe Documentations and maintenanceof the Forms & Formats.

9. Maintain such Quality Management System (QMS).

10. Conduct INTERNAL AUDIT (including departmental audit and over


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10. Conduct INTERNAL AUDIT (including departmental audit and overall audit).

11. Rectify all defects and all the non-conformances / non-compliances.

12. Apply to selected Accredited Registrar for Certification andInvite the certified LEAD AUDITOR for audit.

13. Conduct a number of MOCK AUDIT, and each time rectify all the non-conformances and re-audit, before the Third Party Audit.

14. Third Party Audit (TPA) is conducted by the LEAD AUDITOR,may be in a number of times. Non-Compliances found in a ThirdParty Audit is rectified and Lead Auditor is invited for next TPA.

15. Ultimately, in the Final Third Party Audit, there should not be anynon-complianceand the Lead Auditor is satisfied to recommend tothe accredited REGISTRAR for issuing the Certificate.

16. The accredited Registrar then issues ISO-9001 certificate.

17. The Certificate is valid for some defined period of time, after whichrenewal must be applied for and there after the Third Party Audit isagain done in the same way by certified Lead Auditor beforerenewed Certificate is issued.

18. Accredited Registrar has the right to inspect and re-audit withoutnotice at any time in the validity period of the running Certificateand to cancel the Certificate if non-compliance is found.

DOCUMENTATIONS :


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DOCUMENTATIONS : for ISO-9001 Certification1. QUALITY MANUAL : Open Document containing : (1) Quality Policy

and (2) Quality Performance target with , (3) overall QualityProcedures for Quality Management System.

2. PROCEDUREs : Confidential Document containing : (1) DetailProcedures quality maintenance and appraisal and (2) Guidelines toeach individual, each work-unit and each department, to establishQuality System for achieving assured quality .

3. WORK INSTRUCTIONS : Confidential Document containing : Detail

Instructions & Procedural steps (how to work, work-proceduralsteps) to be strictly followed by each individual, each work unit andeach department, for achieving assured quality .

4. FORMS & FORMATS : Confidential Document to be filled up atevery steps and every time carefully and accurately, with provision

for cross-check, for record and reference needed to maintain full-proof quality management system as well as to pin-point theresponsibility & accountability of any fault, failure occurred orchances to occur and thereby to get prompt opportunity forrectification.

TAGUCHIs QUALITY LOSS FUNCTION


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TAGUCHI s QUALITY LOSS FUNCTIONQuality Guru GenichiTaguchi said any deviation from the target value of aquality characteristics results in extra cost to some segments of the society. Hedefined deviation from the quality target as Social Loss in terms of :(i) Loss to Producers, (ii) Loss to the Consumers, and (iii) Loss to Society.

The ideal value of performance characteristic is the Target Value.A high quality product performs very close to this Target Valueconsistentlythroughout its specified Life spanunder all operating conditions.Taguchis Quality Loss Function estimates the Social Loss from the failure of aproduct to meet the quality specification/requirement.

Taguchi expressed that such Loss (L) for the variation of the performancecharacteristic (X)from the Target Value (T) is directly proportional to thesquare of theDeviation (d). The Loss Function can be expressed as

L d2=> L = K . d2 => L= K. (X T)2

Where, K : Quality Loss Coefficient,X : Average value of quality characteristic actually achievedT : Target value of quality characteristic

The Quality Loss Function remains valid at all times during the productlife.


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Now, if at USL or at LSL (ie with permitted deviation), the Loss isA(which isthe average repair cost), for such accepted quality ,

A = K x Dp2

=> Quality Loss Coefficient,

Where Permitted Deviation, Dp = (USL - T) or (T - LSL)K = Quality Loss CoefficientA = Average Repair Cost

In case of manufacturing in big lots,Lav = K

2

=>

Where, = Standard Deviation of the populationS = Standard Deviation of the sample [ S2 is variance]T = Target set by the customer or designerXav =Mean value of measured parameter

Lav = K[S2 + (Xav T)

2 ]

K = A / Dp2

L = K. (X T)2


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Loss (Rs)

X

L

(Quality Characteristic ie Process

Performance Characteristic)TargetValue

LowerSpec.Limit

Upper

Spec.Limit

Customer Tolerance Band

QUALITY LOSS FUNCTION

Xav

Dp Dp

(Xav T)

Problem-1


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A hot-oil-bath manufacturer has set specification of temperature as 1000 60 C for theoil bath being sold by them. The average repair cost at specified limit is $ 360.

Find the Quality Loss, when the oil bath performance temperatures are 980C or 1040C.

Dp = (USL T) = (T LSL) = 60 C, T = 1000 C, A = $ 360,

K = A Dp2 = $ 360 62 = $ 10

Loss at980 C, L 98 = K (X T)2 = $ 10 x (98 100)2 = $ 40Loss at1040 C, L 104 = K (X T)2 = $ 10 x (104 100)2 = $ 160

Problem-2For a blower manufacturer the speed of the blower has been specified as 1000 3 rpm.The average repair cost for the product with specified limits USL & LSL of the speed, isRs 2300. Determine the Quality Loss if the product shows the speed at 1002 rpm.

Dp = (USL T) = (T LSL) = 3 rpm, T = 1000 rpm, A = Rs 2300,

K = A Dp2 = Rs 2300 32 = Rs 256

Loss at1002 rpm, L 1002 = K (X T)2 = Rs 256 x (1002 1000)2 = Rs 1024

Problem-3


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Problem 3

Compute the average Quality Loss for a process that produces steel shafts forautomobile, if the target value of diameter is specified as 6.40 inch. TheCoefficient of Quality Loss is estimated as $ 95000 and the measurements ofthe sample of sample size eight are 6.36, 6.40, 6.38, 6.39, 6.43, 6.39, 6.46and 6.42 inch.

Sample Size, n = 8, T = 6.40 inch, K = $ 95000

Xav = 6.36 + 6.40 + 6.38 + 6.39 + 6.43 + 6.39 + 6.46 + 6.42 = 6.40375 inch,

8S = Standard Deviation of the sample = [ { (XavXi )2} (n 1) ]

=[ (6.40375 6.36)2 + (6.40375 6.36)2+ + (6.40375 6.36)2] (8 1)= 0.0315945 inch

Lav = K[S2 + (Xav T)

2 ]

= $ 95000[(0.0315945)2 + (6.40375 6.40)2 ] = $ 96.20

QUALITY B E N C H M A R K I N G


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QUALITY B E N C H M A R K I N G1. It is the continuous and systematic process of

(a)measuring / evaluating the (i) Quality /performance of theproducts, (ii) Business process, (iii) work Process & work Procedure ,of the target organizations which are recognized as industry-leaders,

and(b) identifying the gaps between such best practices & thoseexisting in the own business, to establish the rational performance goalsand practices for this business.

2. It is the method of improving business performance continuouslyby learning from the business leaders, how to do things better inorder to be best in the class.

3. It is the on-going practice of measurementof relativeperformanceagainst industry leaders, in the key process areas.

4. It is the search for the best practices that will lead to superiorperformance.

Benchmarking focuses on any of the following :1. World class business leaders.2. Business leaders in the country.3. Leaders in the same industry.4. Potential competitors in the same industry.5. Internally (with improved target with respect to present self performance

assessment)

Benchmarking Steps :


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Benchmarking Steps :1) Planning : i) Identify the need and decide what is to be benchmarked.

ii) Identify the comparative (best-in-class) organization ie Focusorganization.

iii) Determine the scope & objectives and develop a projectPlan for benchmarking.

2) Analysis : iv) Determinedata collection method.

v) Collect dataon focus organizations as well as ownorganization.

vi) Compare data and analyze to determine the current

performance gaps, for recommendation.3) Integration : vii) Project future performance levels.

viii) Communicate Benchmark findings to gain acceptancefrom all relevant departments and from all levels.

4) Action : ix) Establish functional goals.

x) DevelopAction Plan to attain the superior performancebenchmark-level.

xi) Implement the specific action plan and monitor theprogress and improvement.

xii) Recalibrate (update) Benchmark.

AREAS OF BENCHMARKING


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AREAS OF BENCHMARKING1. COMPETITIVE BENCHMARKING on

i) Products : Pricing & Quality (Features, Performance characteristicsand other quality characteristics)

ii) Process & Technologyiii) Work-procedure in production, packing, Quality Control, Logistics& Supply Chain, maintenance, etc.

iv Business Performance

v) Customer Service

2. PROCESS BENCHMARKINGIt includes key business process or work processes like Distribution,Order-entry, Procurement, Order-processing, Training &Development, etc.

3. STRATEGIC BENCHMARKING

It includes the strategy adopted by the successful competitors to achievesuccess through their Order Winning Strategy, having competitiveadvantage.

4. INTERNAL BENCHMARKING

Benchmarking is set at higher level of performance and improvedsystem with reference to existing one, inside the organization.

Malcolm Baldrige National Quality Award (MBNQA)


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Malcolm Baldrige National Quality Award (MBNQA)The Malcolm Baldrige National Quality Award is an annual award that recognizesU.S. organizations in the field of (i) business, (ii) health care, (iii) education, and(iv) nonprofit sectors for performance excellence in Quality.

[It is is named after the quality-management champion Malcolm Baldridge (1922-87)who was the secretaryof commerce (1981-87) in Ronald Reagan administration. It ismanaged by the National Institute of Standards and Technology and conferred bythe American Society For Quality.]This prestigious quality award is being given annually from 1987 in USA, to one or twoOrganizations (both private and public), for the best achievement in the area of Total

Quality Management, in each of the following three categories of business sector,(i) Manufacturing,(ii) Services, and(iii) Small business with employee number less than 500.

Objective of the MBN Quality Award :

(i) To stimulateU.S. business companies to improve their quality level.(ii) To establish useful criteria for the business companies for evaluating theirown individual quality improvement effort.

(iii) To SET the Quality of successful organization as the Quality BENCHMARK(ideal/example in the industry as Industry Leader), for improvement of quality ofother industries.

(iv) To help other organizations to learn how to manage quality by disseminating

information about the Award Winning Program.

Malcolm Baldrige National Quality Award

O
http://www.businessdictionary.com/definition/champion.htmlhttp://www.businessdictionary.com/definition/secretary.htmlhttp://www.businessdictionary.com/definition/commerce.htmlhttp://www.businessdictionary.com/definition/administration.htmlhttp://www.businessdictionary.com/definition/standards.htmlhttp://www.businessdictionary.com/definition/technology.htmlhttp://www.businessdictionary.com/definition/technology.htmlhttp://www.businessdictionary.com/definition/standards.htmlhttp://www.businessdictionary.com/definition/administration.htmlhttp://www.businessdictionary.com/definition/commerce.htmlhttp://www.businessdictionary.com/definition/secretary.htmlhttp://www.businessdictionary.com/definition/champion.html


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(v) Award Winner organizations become confident & competent and the MODEL or Benchmarkfor other organizations to emulate in establishing their Total Quality Management program.They become the Leaders of Quality. [eg Motorola, Xerox, Cadillac, Milliken, IBM, FederalExpress, etc]

(vi) To set the standards to be used as the BASELINE for TQM, in the following sevenCRITERIA for Performance Excellence :

(1) Leadership :The organizations leadership system and senior leaders personalleadership(2) Strategic Planning :How the organization sets strategic directions and how itdevelops the critical strategies and action plans(3) Customers and Market Focus: How the company determines requirements,expectations, and preferences of customers and markets

(4) Human Resources : How the company enables employees to develop andutilize their full potential, aligned with the companys objectives(5) Information, Measurement, Analysis & Knowledge Management : Theselection, management, and effectiveness of use of information and data to supportkey companyprocesses and action plans, and the companys performance management system

(6) Process Management : How key processes are designed, implemented,managed, and improved(7) Business Result : The organizations performance and improvement in keybusiness areas like :

(a) Customer satisfaction, (b) Product & Services,(c) Financial & Market Share (d) Organizational effectiveness,

(e) Governance & Social responsibility.

Points allotted for Assessment in MBNQA


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Criteria Points

1. Leadership 100

2. Information & Analysis 60

3. Strategic Quality Planning 904. Human Resource Utilization 150

5. Quality Assurance 150

6. Quality Result 150

7. Customer Satisfaction 300

T O T A L 1000

Who can use MBNQA :

Whether the organization is,1. small or large business,

2. involved in service or manufacturing,

3. government or nonprofit making or private organizations, and

4. has one office or multiple offices/sites inside the country or across the

globe.

QUALITY CIRCLE


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QUALITY CIRCLE Quality Circles were first seen in the United States in the 1950s Quality Circles were developed by Dr. Kaoru ISHIKAWAin Japan in the 1960s Quality Circles were re-exported to the US in the early 1970s 1980s brought Total Quality Management and a reduction in the use of Quality Circles

Quality Circles can be a very useful tool if used properly.

What is a Quality Circle?QUALITY CIRCLE is the VOLUNTARY groups of employees (working onsimilar tasks or sharing the same area of responsibility) who agree to

meet on a regular basis (i) to identify & solve all the quality problemsand issues related to that work-area, and (ii) for continuous improvement.

Quality Circle operates on the philosophy of employees participationwith empowerment in decision-making and problem-solving whichimproves the quality of work.

Quality circle provides opportunities to the employees at the worker-levelto (i) identify quality problem, if any, (ii) solve the quality problemeffectively and (iii) take measure for continuous improvement and thus toproceed towards the top level of the organization.

OBJECTIVE : (i) Increase Productivity,(ii) Improve Quality

(iii) Boost Employee Morale

QUALITY CIRCLE

Ten Essential Characteristics of Quality Circle :


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Ten Essential Characteristics of Quality Circle :1. Membership is purely VOLUNTARY.2. Membership is drawn from workers doing similar job or from same work-

area.3. The size of quality circle should be appropriate (neither large nor small).

4. Q.Circle Team sets all the Regulations of the Circle and its task-Priorities.5. Q. Circle Team identifies the Quality Problems & decides (by making

Consensus of the Quality Circle team, on how to solve them.6. Leadershipof theCircleshould be selected from and within the Members

of the Circle or some times may be the Supervisor.7. Full management-support towards the functioning of Quality Circle.

8. Extensive TRAINING of the operator-members on Quality Issues, Solvingof Quality Problems and Continuous improvement

9. Members need to be EMPOWERED in decision making10. Members need to have the support, recognition, appreciation and

rewards from Senior Management for the functioning & performance ofthe Quality Circle.

# S I Z E of the Quality Circle

The Size of Quality Circle should usually be between 4 to 12 members butthey can be upto a maximum of fifteen (15). Ideal size is 8 to 10 .

[Larger size of Quality Circle, tend to lose their DYNAMISM , while

Smaller size of Quality Circle, often lose their CREATIVE CAPACITY.]

QUALITY CIRCLE


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# Common Causes of failure of Quality CircleImplementation :

1. Inappropriate SIZE of Quality Circle (ideal 8 to 10)2. Wrong Selection of Members of Quality-Circle (if not from same

work area)

3. Members are not truly Voluntary, but forced

4. Wrong Selection of the Leader of Quality-Circle (if not from and

among the members of the Circle, or if top official / managementperson)

5. Inadequate Training of Members

6. Lack of Management Interest and support

7. Not really empowering Quality Circles to make relevant decisions.

8. Infrequent (inadequate number of) Quality Circle MEETING (shouldbe atleast once in a month).

9. Quality Circles functioning & performance not recognized orappreciated or rewarded by the management.

10. Unfavorable industrial Relation with trade unions.

11. Unsure about the Purpose of Quality Circle.

TOTAL PRODUCTIVE MAINTENANCE (TPM)


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O O ( )Total Productive Maintenance (TPM) is a team-oriented andParticipation - based Equipment-maintenance program.

It is a combination of (1)Preventive & Predictive Maintenance with

(2) Autonomous (participative) Maintenance.

It emphasizes on Equipment Maintenance with:(i) Operators involvement with Team work, and(ii) Operators feeling of Ownership of Equipment & of its Performance,(iii) Preventive & Predictive Maintenance.

OBJECTIVE : To improve PRODUCTIVITY by

1) maintaining improved product QUALITY through improving workingcondition of machine, so that cost of quality-rejection comes down;

2) reducing equipment Maintenance Cost through operators selfmaintenance;3) increasing production volume through

(i) decreasing machine breakdown time (increasing equipmentavailability) and(ii) reducing the chance of processing-speed reduction of the

machine

TPM bi


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TPM combines(i) the American practiceof Preventive and Predictive Maintenance(ii) with the Japanese concepts of Total quality control and total employeeinvolvement, promoting operators maintenance through day-to-daymonitoring activities.

TPM ensures Rapid and Continuous Improvement of Productivity inmanufacturing industries by eliminating the losses like

1. Equipment Breakdown Losses (equipment Idle Time),2. Speed Losses (due to inefficiency of equipment), and3. Quality Losses(due to Poor Quality Output).

TPMeliminate the RISK of :1. Equipment Breakdowns, creating interruption in production2. Quality Defects, generated from faulty / defective equipment3. Failure of On-time delivery

4. Equipment Related Productivity Losses5. Other Manufacturing Losses (Wastes, Scraps, etc)

Measurement of EFFECTIVENESS of TPM


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Measurement of EFFECTIVENESS of TPMTPM helps to improve(i) Equipment Availability, (ii) Performance Efficiencyand (iii) Quality Rate of output.

Overall Equipment Effectiveness (OEE)

Overall Equipment Effectiveness (OEE) is a measure of effectiveness ofTPM implementation and is the product of three efficiency measures:(i) availability, (ii)productivity, and (iii) quality.OEE is the combined measure of these three improvement criteria as a resultof the success or effectiveness of implementing TPM.

The three measurables Equipment Availability (Time) Equipment Performance (Speed) Quality Yield (Quality performance)

which consider 'SIX BIG Losses

1. Equipment DowntimeLoss (Breakdowns)2. Quality RejectionsLoss (Scrap Rate)3. Reduced equipment-speed Loss4. Set-up timeLoss (Engineering Adjustment)5. Start-up loss

6. Minor Stoppages Loss (Unplanned Breaks)

EQUIPMENT PERFORMANCE MEASURES


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Overall Equipment Effectiveness (OEE)

Overall Equipment Effectiveness,

OEE = Equip. Availability x Performance Efficiency x Quality RateOEE = E A x P E x Q R

Equipment Availability = Net Operating Time x 100%

Total Available Time

[ Net Operating Time = Total Available Time Total Time Loss ]

Performance Efficiency = Design Cycle Time x Total items Run x 100%

Net Operating Time

[Total Parts Run is the Total no. of output produced including rejects]

Quality Rate (Quality Yield) = Total items Run Total Defects x 100%Total items Run

[ Typical example of very good OEE = 87% x 90% x 96% = 75%]

Example (OEE)One Metal Processing plant has the present performance data for one eek of


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One Metal Processing plant has the present performance data for one week ofoperation, as follow. Analyze the OEE and advise. Operating schedule = 120 hrs/week Non-operation time = 250 min/week Maintenance downtime = 500 min/week

Changeover downtime = 4140 min/week Ideal production rate = 9.2 pieces/min Total output = 15,906 pieces/week Rejected pieces = 558 per weekOEE =EA x PE x QRTotal available time = (120 hrs)(60 min/hr) = 7200 minTotal time loss = (250 + 500 + 4140) minutesNet Operating time = Total available time Total time loss

= 7200 (250 + 500 + 4140) = 2,310 minEA = Net Operating time Total available timeEA= 2310 7200 = 0.32PE = (Design cycle time x Total output) Net Operating time

PE = [(1/9.2)min/ pieces x (15,906)] (2,310 min)= [(0.109)(15,906)] 2,310 = 0.75

QR = (Total output rejects) Total output= (15,906 558) 15,906 = 0.96

OEE = EA x PE x QR = (0.32)x(0.75)x(0.96) = 0.23 = 23%

DEMING WHEEL / PDCA CYCLE


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G / C C CMeaning of PDCA cycle :The PDCA Cycle is a checklist of the four stages which you must go through to get from`problem-faced' to `problem solved and for continuous quality improvement.PDCA (plan-do-check-act) is an interactive four-step problem-solving process

typically used in BPI (Business Process Improvement). It is also known as the DemingCircle,, Deming cycle, Deming wheel, Shewhart cycle or plan-do-study-act.

PLAN : Plan for changes to bring about improvement. Establish theobjectives and processes necessary to deliver results in accordance with the expectedoutput. By focussing onto the expected output, it differs from other techniques in respectthat the completeness and accuracy of the specification is also part of the improvement.Customer Supplier mapping, Flow-charting, Root Cause Analysis, Brainstorming,

FMEA, FTA, Pareto Analysis,etc.DO : Do changes on a small scale first to trial them.Implement the newprocesses to test.

CHECK : Check to see if changes are workingby Measure the result of newprocesses and compare the results against the expected results to ascertain anydifferences. Analyze the differences to determine their cause.

ACT : Act to get greatest benefits from changes.

Each will be part of either one or more of the P-D-C-A steps. Determine whereto apply changes that will include improvement. When a pass through thesefour steps does not result in the need to improve, refine the scope to whichPDCA is applied until there is a plan that involves improvement.


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BATH-TUB CURVE :


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The PROBABILITY OF FAILURE or Un-Reliability of a product generally

follow a typical Hazard-Curve or Failure Curve called BATH-TUB CURVE.

The typical Hazard-Curve showing the variation of Hazard Rate or Failure Rate

(failures per time) of the product with respect to Time variation, alongproducts life, is a smooth curve in the shape of a Bath-Tub, and because of

its shape such curve is called Bath- Tub Curve.

Failure Rate

Time

Infant Failure Service FailureWear-outFailure

Infant Mortality /

U S E F U L L I F E Wear-out Life

Break-in Period

(I) Bath-Tub Curve has threeclear Failure Zones (i) Infant FailureZone, (ii) Service Failure Zone, and (iii) Wear-out Failure Zone.


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o e, ( ) Se ce a u e o e, a d ( ) ea out a u e o e(II) Bath-Tub Curve indicates that the Product LIFE is divided into two

major sectors (i) USEFUL LIFE, and (ii) WEAR-OUT LIFE.[Useful Life includes Infant Failure Zone & Service Failure Zone.]

Infant Failure Zone (Infant Mortality / Break-in Period): is short initialperiod of product life, during which Early Failure of the product or componenttakes place due to the defects in(i) Product Design (Inherent DESIGN Weakness of the part / section),(iii) Assembly (eg alignment, fitting, etc)(iv) quality of material of construction,

(v) workmanship,(vi) surface preparation / lining (eg insulation, surface hardening ,surfacefinish, etc), and due to

(vii) Improper operation (eg application of excess load; beyond specifiedoperating condition like dust, temperature, humidity, pressure, etc)

In this zone, Failure Ratedecreases rapidlywith Time to reach a steady level.

Service Failure Zone : Failure Rate or Hazard Rate becomes (i) constantlyLOWEST, (ii) failures are very Occasional & Random, (iii) fairly evenlydistributed, (iv) failure density is STEADY and exponential in nature (FailureRate is constant and parallel to Time-axis) leading to Constant and lowFailure Rate curve.Wear-out Failure Zone : The Failure Rateincreases rapidlywith Time due to

ageing, wear & tear, fatigue and corrosion, etc., (specially after specified LIFE)

RELIBILITY : of the product is the PROBABILITY that the product will


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p p

perform as per theSPECIFIED (defined) QUALITY

- duringits SPECIFIED LIFE TIME

- underSPECIFIED OPERATING & UNVIRONMENTAL

CONDITION.Reliability is the Probability of success in performance of the goods /Product .In Reliability testing /FAILURE TEST, Reliability, Hazard Rate, FailureDensity, etc. at a given time, is given by,

1)Reliability, R= Number of survivors at the end of a given timeTotal population of product at the beginning

2)Hazard Rate, Z = Number of failure during particular unit intervalAverage population during the interval

3)Failure Density, fd= Number of failure during given time intervalTotal population of product at the beginning

4)Failure Rate, :Number of Failure of the product per unit time (say, so

many failures per hour).

5)Mean Time Between Failures (MTBF) : it is the mean of the time


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gaps between two consecutive failures of the product under operation. MTBFindicates meanTime gap between n th. failure and (n+1)th. failure.

6)Mean Time To Failure,where, N = No. of specimen (sample size),

ti= Time of failure forith specimenwhere, i = 1 to N

7)RELIABILITYof an item during the particular operating period t,

MTTF = (ti ) N

MTTF = 1

Rt = e .t

For Constant Hazard Rate

[Euler's constant orNapier's constant,

e= 2 + 1/2 + 1/(2 3) + 1/(2 3 4) + 1/(2 3 4 5) +

e= 2.71828182846]

Example-1 :


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Failure Test was conducted for electric bulbs on a sample of size of 1000 withtotal duration of test of 10 days. The result Data of such failure test is givenbelow. Find at the end of each unit time-period (i) Failure density, (ii) HazardRate, (iii) Reliability and also calculate (iv) Mean Hazard Rate, .

Time (t)day

0 1 2 3 4 5 6 7 8 9 10

No. offailures(f)

0 210 143 117 96 77 65 90 135 50 17

Answer-1 :


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Time(t) days

No. offailures (ft)

Cumulativefailures (Ft)

No. ofSurvivors(St=N Ft)

Failure

Density fd(= ft / N)

HazardRate (Zt=ft/Mean of St& St-1)

Reliability

(Rt = St /N)

0 0 0 1000 - - 1.0001 210 210 790 0.210 0.235 0.790

2 143 353 647 0.143 0.199 0.647

3 117 470 530 0.117 0.199 0.530

4 96 566 434 0.096 0.199 0.434

5 77 643 357 0.077 0.199 0.357

6 65 708 292 0.065 0.200 0.292

7 90 798 202 0.090 0.364 0.202

8 135 933 67 0.135 1.0037 0.067

9 50 983 17 0.050 1.190 0.017

10 17 1000 0 0.017 2.000 0.000

fd1 = 210 / 1000 = 0.210 Total : 5.785

Z1 = 210 (1000+790) / 2 = 210/895 = 0.235 Z2= 143 (790+647) / 2 = 0.199

Rt1 = 790 / 1000 = 0.790

Mean Hazard Rate, h = 5.785 10 = 0.5785= 57.85%

Example-2 :

D t i MTTF f i i ti f 1000 h lif if th T t f F il d t


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Determine MTTF for a mission time of 1000 hours life, if the Test of Failure data ona sample of 10such items indicates the time to fail as shown below. Find alsothe Reliability of the item for the specified life of 1000 Hours

Item No. 1 2 3 4 5 6 7 8 9 10

Time toFail (Hrs)

807 820 810 875 900 837 850 790 866 815

Answer-2 :

MTTF = (ti ) N = (807+820+810+875+900+837+850+790+866+815) 10

= 8370 10 = 837 Hours

Considering constant Hazard Rate,

Failure Rate, = 1 MTTF = 1 837 = 0.0012 failures / Hour

Reliability for 1000 Hours specified life, R1000 = e .t

= e 0.0012 x 1000 =e 1.2 = 0.3012 = 30.12%

Example- 3 :


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What is the Reliability of a product item for an operating period of 200hours, if the Failure Test shows a failure rate of 0.4 x 10 5 failures / hour.

If originally 5000 such items were under this test, how many items failed

in 200 hours ?

Answer-3 :

Given, = 0.4 x 10 5 failures / hour t = 200 hours

Reliability for 200 Hours specified life, R200 = e .t

= e 0.000004 x 200 = e 0.0008 =0.9992 = 99.92%

Original no. of item under test, No = 5000

If Number of survival after 200 hours = Ns,Reliability at 200hours, R200= Ns / No

=> Ns = No x R200 = 5000 x 0.9992 = 4996

No. of failed items in 200 hours, Nf= NoNs = 5000 4996 = 4 items


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RELIABILITYOF A SYSTEM

Series Configuration System Reliability- Series Configuration


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When Reliability (Probability of successful operation) of the System depends

on successful functioning of all its components/units, simultaneously.

1 2 3

System

Configuration Block Diagram

Units : 1, 2, 3, , n

Configuration : Series

Events : Successful function of the

units X1, X2, X3, .. , Xn

Probability of successful function of the units : R(X1), R(X2), R(X3), . , R(Xn).

Condition of successful function of System : All the units to function successfullyand under the condition of,

(i) If the reliability of X1, X2, X3, , Xn are INDEPENDENT (successful functioningof any unit is not dependent on successful functioning of remaining units)

System Reliability, R(sys) = R[(X1) and (X2) and (X3) and .. (Xn)]

Reliability of System in SERIES Configuration is alwaysless than theReliability of the poorest unit.

System Reliability, R(sys) = R(X1) x R(X2) x R(X3) x .. x R(Xn)

R(sys) < R(X)minsince R < 1,

Parallel Configuration System Reliability


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Parallel Configuration System ReliabilityWhen Reliability (Probability of successful operation) of the System depends

on successful functioning of any one of its components/unit, at any time.For successful functioning of the System, atleast one of its unit has to

function satisfactorily

Configuration Block Diagram

System

Units: 1, 2, 3, , nConfiguration : ParallelEvents : Successful functioning of the

units as X1, X2, X3, .. , Xn andUnsuccessful functioning of the

units as X1, X2, X3, .. , XnReliability (Probabilityof successful function) ofthe units :

R(X1), R(X2), R(X3), . , R(Xn).Un-Reliability (Probability of failures) of the units :

_ _ _ _

P(X1), P(X2), P(X3), . , P(Xn)._Where, P(X1) = 1 R(X1)

R(sys) = R(X1 or X2 or X3 or Xn)___ _ _ _ _

Or, P(sys) = P(X1 and X2 and X3 and Xn)_ _ _ _

= P(X1) x P(X2) x P(X3) x x P(Xn)

1

2

3

System Reliability- Parallel Configuration(i) If the unit failures are INDEPENDENT of the failure of other units


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(i) If the unit failures are INDEPENDENT of the failure of other units,___ _ _ _ _

P(sys) = P(X1) x P(X2) x P(X3) x x P(Xn)___

Now, R (sys) = 1 P(sys)_ _ _ _

So, R (sys) = 1 [ P(X1) x P(X2) x P(X3) x x P(Xn) ]

(ii) If the units are identical and unit failures are INDEPENDENT of the failureof other units,

For Parallel System

For Series System

# In Parallel Config,

(i) System-Reliabilityincreases with increased no. of units &

(ii) System Reliability ismore than the maximum individual- component-Reliability

Butin Series config.

(i) System-Reliability decreases with no. of units &

(ii) System Reliabilityisless than the minimum individual- component-

Reliability]

R (sys) = 1 [ 1 R(X) ]n

R (sys) = 1 [ 1 R(X1)] x [ 1 - R(X2)] x [1 - R(X3)] x x [1 - R(Xn) ]

R (sys) = [ R(X) ]n

System Reliability Parallel

For TWO units when failures are INDEPENDENT ,


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For TWO units when failures are INDEPENDENT ,

Pf sys = Pf (X1 and X2) = Pf (x1) x Pf(X2)

__ __ __

=> R(sys) = R(X1) x R(X2)

=> 1R (sys) = [ 1 R(a)] x [ 1 - R(b)]

=> R (sys) = 1 [ 1 R(a)] x [ 1 - R(b)]

Alternatively,

R(sys) = R (a or b)

= R (a) + R (b) R (a and b)

=> R (sys) = R (a) + R (b) R (a) . R (b)

Example-1 System Reliability - SeriesIn a Control Valve system consisting Sensor, Actuator, Linkage and Valve, the


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In a Control Valve system consisting Sensor, Actuator, Linkage and Valve, theunits are independently functioning and interconnected in a Series

Configuration. Reliability of individual units are measured as 91%, 84%, 88%

and 94% respectively.

What will be the Reliability of System? [63.23%]Example-2

A system has 40 identical but independent units connected in series config.

Probability of Failure of each unit is 2%.

Estimate the Reliability of the system. [20%]

Example-3

A system has 10 identical but independent components connected in series

configuration. Probability of the system to be achieved is 95%. Estimate the

required Reliability of each component. [99.49%]

Example-4

A system has 4 identical but independent components connected in series

configuration. Each component has a constant failure rate of 0.006 failures /

hour.

Determine the Reliability of the system for operating hour of 100 hours.

[9.07%]

Example-5 System Reliability - Series

A communication system consists of 3 component a Transmitter a receiver


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A communication system consists of 3 component a Transmitter, a receiverand an encoder connected in series configuration. Failure of one unit causesystem failure.Each component operates for different period of operation during 8 hours of

communication. The operational periods of the components during 8 hours ofoperation of the system are 6, 8 & 4 hours, respectively.The failure rate of the components are 0.0238, 0.0013 & 0.0200 failure/hour,respectively.What will be the Reliability of the communication system for 8 hours ?

Answer-5

System Reliability, R(sys) = R1 x R2 x R3 =0.8669 x 0.9897 x0.9231= 79.20%

Unit Operation Period

(t) hrsFailure Rate

( )x t

Reliability

(Ri = e- t)

Transmitter 6 0.0238 0.1428 0.8669

Receiver 8 0.0013 0.0104 0.9897Encoder 4 0.0200 0.0800 0.9231

Example-6 System Reliability- Parallel

A system has 5 identical units connected in parallel configuration Each unit


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A system has 5 identical units connected in parallel configuration. Each unithas reliability of 90%. If unit-failures are independent, what is system

reliability ? [99.99%]

Example-7A system with parallel configuration has 10 identical but independent

components. If System Reliability is targeted to be 0.95, estimate how

poor can the components be ? [25.81%]

Example-8

A system has 4 identical components connected in parallel configuration and

shows a system reliability of 0.90. How many more components should be

required to be added in parallel to get a system reliability of 0.99 ? [4]

Example-9

A system has 2 components of reliability 0.95 and 0.92 , in parallelconfiguration. Calculate the system reliability. [99.60%]

# Answer to Ex-8 System Reliability - ParallelA system has 4 identical components connected in parallel configuration and


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A system has 4 identical components connected in parallel configuration andshows a system reliability of 0.90. How many more components should berequired to be added in parallel to get a system reliability of 0.99 ?

Ans : For independently functioning units in Parallel Configuration,

R(sys) = 1 [ 1 R(X) ] n=> 0.90 = 1 [ 1 R(X) ] 4

=> [1 - R(X) ] 4 = 1 0.90 = 0.10

=> 1 - R(X) = (0.10) 0.25 = 0.5623

=> R(X) = 0.4377

Now, to achieve system reliabilityR(sys) =0.99,R(sys) = 1 [ 1 R(X) ] n

=> 0.99 = 1 [ 10.4377 ] n

=> [1 0.4377 ] n = 1 0.99= 0.01

=> (0.5623) n = 0.01

=> n . log 0.5623 = log 0.01

=> n = log 0.01 log 0.5623 = 7.99 8

So, to get system reliability of 0.99, 4more components are requiredto be added in parallel. Ans.

Product (System) Reliability can be improved by :


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1) IMPROVING QUALITY & RELIABILITY (material, design, size,workmanship, additional safety features, etc) of all the componentorunit.

However, it requires increased processing Time and COST.

Required System Reliability can be targeted at the Optimizing the TotalEffective Cost, where

Total Effective Cost = (Production Cost + R & D Cost) + (MaintenanceCost + Cost of Spares)

Quality &Reliability

Cost

1.0

Optimum Cost

Optimum Reliability

Total Effective Cost

Production Cost

R & D Cost

Cost of Spares

Maintenance Cost

COST TRADE-OFF

0

Redundancy Preference Zone

2) By REDUNDANCY Reliability ImprovementRedundancy is deliberately creating / providing new (additional)


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Redundancy is deliberately creating / providing new (additional)PARALLEL path of the component or unit (sub-system), in thesystem (product), to improve the reliability of the system(product), without further improving the quality or reliability ofindividual component (unit) or sub-system, in the product.

example :

aa

b

R(s) = R (a or b) = R(a) + R(b) R(a and b) = R(a) + R(b) R(a) . R( b)

= R(a) + R(b) [ 1 R(a) ] So, R(s) > R(a) andsimilarly R(s) > R(b)

thus RELIABILITY IS IMPROVED.

If, a & b are identical, R(s) = 2 R(a) [R(a)] 2

Redundancy may be of 3 types :

1)Element Redundancy (each element is duplicated in parallel)

2) Unit Redundancy (each sub-system is duplicated in parallel)

3) Stand-by Redundancy

Original : Reliability R(a)With Redundancy : Reliability R(s)

Example : Reliability Improvement By Redundancy


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a1 b1 a1

a2

a1

a2

b1

b2

b1

b2

Original

Configuration

( a1 & b1)

Element Redundancy

(a1 or a2) & (b1 or b2)

Unit Redundancy

(a1 & b1) or (a2 & b2)

Assumption :Redundancy-elements are identical (a1 = a2 and b1 = b2)

1) Originally: Reliability, R(s) = R[a1 & b1] = R(a1) x R(b1) =

for identical elements (i.e. a = b) R(s) = R x R = R2

1) Element Redundancy: Reliability, R(s) = R[(a1 or a2) & (b1 or b2)]= R[a1 or a2] x R[b1 or b2]

= [R(a1) + R(a2) R(a1).R(a2)] x [R(b1) + R(b2) R(b1).R(b2)]

= = [R(a) x R(b)].[(2 R(a))x(2 R(b))]

R(s) >> R(a) x R(b)

for identical elements (i.e. a = b) R(s) = (2R R2)2 = R2 .(2 R)2

2) Unit Redundancy: Reliability, R(s) =R[(a1 & b1) or (a2 & b2)]= [R(a1) x R(b1)] + [R(a2) x R(b2)][R(a1) x R(b1)] x [R(a2) x R(b2)]= for identical elements (i.e. a = b)

R(s) = (2R2 R4) = R2 .(2 R2)

R(a) x R(b)

[ 2R(a) R2(a) ] x [ 2R(b) R2(b)]

2 R(a) x R(b) [R(a)2

x R(b)2

]

R R2 (2 R)2 R2 (4 + R2 4R)


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R Element Redundancy = R2 .(2 R)2 = R2 .(4 + R2 4R)

=R2 .[(2 R2) + (2 + 2R2 4R)]

=R2 . (2 R2) + R2 . (2 + 2R2 4R)

=R Unit Redundancy + 2R2 (1 + R2 2R)

R Element Redundancy =R Unit Redundancy + [ 2R2 (1 R)2]

Now, [ 2R2 (1 R)2 ] is definitely positive, as R R Unit Redundancy

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Jpm Tqm Course Mat-6 T-3