Mod10-NVN QA QC

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MODULE NO -10

Introduction to Control charts

Statistical process control

• Statistical process control is a collection of tools that when used together canresult in process stability and variability reduction.

• A stable process is a process that exhibits only common variation, or variationresulting from inherent system limitations.

• A stable process is a basic requirement for process improvement efforts.

Advantage of a stable process

• Management knows the process capability and can predict performance, costs,and quality levels.

• Productivity will be at a maximum, and costs will be minimized.

• Management will be able to measure the effects of changes in the system withgreater speed and reliability.

• If management wants to alter specification limits, it will have the data to back upits decision.

Categories of variation in piece part production

• Within-piece variation

• Piece-to-piece variation

• Time-to-time variation

Source of variation

Variation is present in every process due to a combination of the equipment,

materials, environment, and operator.

The first source of variation is the equipment. This source includes tool wear,

machine vibration, work holding-device positioning, and hydraulic and electrical

fluctuations. When all these variations are put together, there is a certain capability orprecision within which the equipment operates.

The second source of variation is the material. Since variation occurs in the finished

product, it must also occur in the raw material (which was someone else's finishedproduct). Such quality characteristics as tensile strength, ductility, thickness, porosity,

and moisture content can be expected to contribute to the overall variation in the final

product.

A third source of variation is the environment. Temperature, light, radiation,

electrostatic discharge, particle size, pressure, and humidity can all contribute to variationin the product. In order to control this source, products are sometimes manufactured in



white rooms. Experiments are conducted in outer space to learn more about the effect of

the environment on product variation.

A fourth source is the operator This source of variation includes the method by which

the operator performs the operation. The operator's physical and emotional well-being

also contribute to the variation. A cut finger, a twisted ankle, a personal problem, or aheadache can make an operator's quality performance vary. An operator's lack of

understanding of equipment and material variations due to lack of training may lead tofrequent machine adjustments, thereby compounding the variability.

The above four sources account for the true variation. There is also a reportedvariation, which is due to the inspection activity. Faulty inspection equipment, the

incorrect application of a quality standard, or too heavy a pressure on a micrometer can

be the cause of the incorrect reporting of variation. In general, variation due to inspection

should be one-tenth of the four other sources of variations. It should be noted that three ofthese sources are present in the inspection activity-an inspector, inspection equipment,

and the environment.

Chance and Assignable Causes of Quality Variation

As long as these sources of variation fluctuate in a natural or expected manner, astable pattern of many chance causes (random causes) of variation develops. Chance

causes of variation are inevitable. Because they are numerous and individually of

relatively small importance, they are difficult to detect or identify.

When only chance causes are present in a process, the process is considered to be in a

state of statistical control. It is stable and predictable. However, when an assignable cause

of variation is also present, the variation will be excessive, and the process is classified asout of control or beyond the expected natural variation.

• A process that is operating with only chance causes of variation present is said tobe in statistical control.

• A process that is operating in the presence of assignable causes is said to be out ofcontrol.

• The eventual goal of SPC is reduction or elimination of variability in the processby identification of assignable causes.

Control chart

• Control chart was developed to recognize constant patterns of variation.

• When observed variation fails to satisfy criteria for controlled patterns, the chartindicate this.

• Control chart allow us to distinguish between controlled and uncontrolled

processes



Statistical Basis of the Control Chart

Basic Principles

A typical control chart has control limits set at values such that if the process is incontrol, nearly all points will lie between the upper control limit (UCL) and the lower

control limit (LCL).

Definition :

A control chart is defined as a statistical tool used to detect the presence of assignablecauses in any manufacturing systems and it will be influenced by the pure system of

chance causes only

Control charts are of two types : Variable control charts and attribute control charts

Variable Control charts : A variable control chart is one by which it is possible tomeasure the quality characteristics of a product. The variable control charts are

(i) - chart

(ii) R – chart

(iii) – chart

Attribute Control chart : An attribute control chart is one in which iti is not possible to

measure the quality characteristics of a product i.e., it is based on visual inspection only

like good or bad success or failure, accepted or rejected. The attribute control charts are.

(i) p - chart

(ii) np – chart(iii) c – chart

(iv) u - chart

Objectives of control charts

• Control charts are used as one source of information to help whether an item oritems should be released to the customer.

• Control charts are used to decide when a normal pattern of variation occurs, theprocess should be left alone when an unstable pattern of variable occurs which

indicates the presence of assignable causes it requires an action to eliminate it.

• Control charts can be used to establish the product specification.

• To provide a method of instructing to the operating and supervisory personnel(employees) in the technique of quality control.

x



Notations

: Mean of the sample

: Standard deviation of the sample

: Mean of the population or universe: Standard deviation of the population

Central Limit Theorem

Irrespective of the shape of the distribution of the universe, the average value of a

sample size ‘n’ ( X bar1, X bar2, X bar3 --------- n ) drawn from the population will

tend towards a normal distribution as n tends to infinity.

Relation between R bar and -

= Mean Range

d2 = Depends upon sample size from the tables

x

xσ

1

xσ

1

σ 1

2

1

d

R=σ

R



Control Limits for R chart

Interpretation of Control Charts

After plotting the points on the X bar - R charts, it shows two possible states of

control. They are1. State of statistical control and

2. State of lack of control.



State of Statistical ControlA manufacturing process is said to be in a state of statistical control whenever it is

operated upon by a pure system of chance causes. The display of points in the X bar

chart and R chart will be distributed evenly and randomly around the center line and all

the points should fall between the UCL and LCL.

Control Charts - in Control VS Chance Variation

State of Lack of Control

A process is said to be in a state of lack of control whenever the state of statisticalcontrol does not hold good. In such a state we interpret the presence of assignable causes,

the reason for lack of control are

• Points violating the control limits

• Run• Trend

• Clustering

• Cycle pattern

Control Charts Interpretation

• Special: Any point above UCL or below LCL

• Run : > 7 consecutive points above or below centerline

• 1-in-20: more than 1 point in 20 consecutive points close to UCL or LCL

• Trend: 5-7 consecutive points in one direction (up or down)



Control Charts - Lack of Variability

Control Charts – Lack of Variability





Control Charts shifts in Process Levels



Control Charts Recurring Cycles



Control Charts points near or outside limits



MODULE NO -11

Applications of X bar – R Chart with Real life data

Problem 3.

The following are the X bar - R values of 20 subgroup of 5 readings each

S.G No X bar R1 34.0 4

2 31.6 23 30.8 3

4 33.8 5

5 31.6 26 33.0 5

7 28.2 13

8 33.8 19

9 37.8 610 35.8 4

11 38.4 4

12 34.0 14

13 35.0 4

14 33.8 7

15 31.6 516 33.0 7

17 32.6 3

18 31.8 9

19 35.6 6

20 33.0 4

(a) Determine the control limits for X bar and R chart.(b) Construct the and R chart and interpreter the result.

(c) What is process capability?

(d) Does it appear that the process is capable of meeting the specification limits.(e) Determine the percentage age of rejection if any

The specification limits are =33±5.

0.126

2.669

=

=

R

x

0.126

2.669

=

=

R

x

46.3320

2.669==

=

k

x x

6320

126===

K

R R



For a subgroup size of 5 from tables

A2 = 0.58

d2 = 2.326

D3 = 0.0D4 = 2.11

Control limits for R- chartUCL= D4 = 2.11 x 6.3

LCL = D3 = 0.0

CL = = 6.3

It is seen from the data two subgroup are crossing the UCL which indicates the presenceof assignable causes. So the homogenization is necessary.

Again control limits for R-chart

UCL= = 2.11 x 5.17

LCL= = 0.0

CL = = 5.17

Again one more subgroup is crossing the UCL

Again control limits for R-chart

UCL= = 2.11 x 4.17 = 9.917

LCL= = 0 x 4.7 = 0.0

CL = = 4.7Now all the points are falling with the control limits. The final values are

UCL = 9.917

LCL = 0.0CL = 4.7

Control limits for - chart

UCL =

220

19140.126

−

−−= R

17.518

93==

7.4320

1319141262 =

−

−−−= R

X

22 R A X +

R

R

R

14 R D

13 R D

1 R

24 R D

23 R D

2 R



= 33.46 + 0.58 x 4.7= 36.186

LCL =

= 30.734

CL = = 33.46

It is seen from the data that three subgroup are crossing the control limits. Which

indicates the presence of assignable causes. So homogenization is necessary

Again control limits for X-bar -chart

UCL =

= 33.22+0.58 x 4.7= 35.946

LCL == 33.22 -0.58 x 4.7

= 30.494

CL = = 33.22

Now all the points are falling within the control limits. The final value areUCL = 35.946

LCL = 30.494

CL = 33.22

The Charts are plotted for the final values

22 R A X −

X

22.33

320

2.284.388.372.6691

=

−

−−== X

221 R A X −

1 X

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

UCL=35.946

CL=33.22

LCL=30.494

SubgroupNumber

X

221 R A X +



(b) Interpretation R-chart is not in control. Some points are crossing the UCL, - chart is

not in control. Points are crossing the control limits. So process is not in a of statisticalcontrol

1 =

The process capability = 61

= 6 x 2.02

= 12.12

(d) UCL – LSL =10

Since 61 > (UCL – LCL), the process is not capable of meeting the specifications

limits.

(e) UNTL = +31

= 33.22 + 3 x 2.02

= 39.28

LNTL = -31

= 33.22 - 3 x 2.02

= 27.16

CL = = 33.22

UCL = 38

LSL = 28

Probability = 0.0052 = 0.52%

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

UCL=9.917

CL=4.7

LCL=0.0

SubgroupNumber

02.2326.2

7.4

2

2==

d

R

1

X

1

X

1 X

58.202.2

22.3328−=

−= Z Below



Probability = 0.9909 = 99.09%

Therefore 100 – 99.09 = 0.91%

Total Rejection = .052 + 0.91 = 1.43%

Problem - 4

A Control charts has been used to monitor a certain characteristic. The process is

sampled in a subgroup size of 4 at an interval of 2 hours. - chart has 3 control limits of

121 and 129 with the target value of = 125.

(a) If the product is sold to a user who has a specification of 127 ±8. What percentage of

the product will not meet the specification assuming normally distributed output.

(b) If the target value of the process can be shifted without effect on the process standard

deviation, what target value would minimise the amount of product being outside the

specifications.

(c) At this new target value what percentage of the product will not meet the specification

requirements.

Solution. Subgroup size ‘n’ = 4

UCL = 129

LCL = 121

CL = = = 125.

Specification limits = 127 ±8USL = 135

LSL = 119

From tables, for a subgroup size of 4.

A2 = 0.73

d2 = 2.059

D3 = 0.0

D4 = 2.28

UCL = + A2

Process capability = 6 1

= 6 x 2.6

36.202.2

22.3338−=

−= Z Above

X

1

X

X 1

X

X R

A

X UCL R

2

−=

48.573.0

125129=

−=



= 15.96

USL – LSL = 16

Science 61 < (USL – LSL). The process in capable of meeting the specification limits.

UNTL = +3 1

= 125 + 3 x 2.66= 132.98

LNTL = - 3 1

= 125 – 3 x 2.66

= 117.02

USL = 135

LSL = 119

Percentage of rejection

Z =

= -2.25

(a) Probability from table 0.0122 = 1.22%

(b) In order to minimize the percentage of rejection

change the process target from 125 to 127.

The percentage of rejection

Z =

= -3.00

Probability from tables = 0.00135

Percentage of rejection = 0.135

1

X

1

X

UNTL= 132..98

USL= 135

CL= 125

LNTL= 117.02

LSL= 119

σ 1

66.2

125119 CL LSL −=

−

66.2

127119−



(c) Since it is symmetric the total percentage of rejection = 0.135 x 2 = 0.27%.

Problem -5

For a certain characteristic of a product of sample size2 after 25 sub-groups. R =

0.81 and X bar = 27.635. The specification limits are 1.12 ± 0.087.

(a) In the process harmonized to the specifications.(b) What are the rejections percentages if any?

(c) Is the process capable of meeting the specifications.

(d) Harmonise the process to the specifications and obtain the control limits for

X bar-R chart after harmonizing the process to specification.

Solution. n = 2, K = 25, R = 0.81, X bar = 27.635.

Specification limits = 1.12 ± 0.087

USL = 1.207

LSL = 1.033

From tables for a subgroup size of 2.

d2 = 1.128A2 = 1.88

D3 = 0.0

D4 = 3.27

UNTL =

= 1.1054 + 3 x 0.0287

= 1.1915

LNTL =

= 1.1054 – 3 x 0.0287

= 1.0193

CL = = 1.1054

11054.1

25

635.27 X

K

X X ===

=

0324.025

81.0==

=

K

R R

0287.0128.1

0324.0

2

1===

d

Rσ

11

3 X +

11

3 X −

1

X

UNTL= 1.1915

USL=1.207

CL= 1.1054

LSL= 1.033



(a) It is clear from the figure that the process is not harmonised with the specifications

(LNTL is below LSL) (For a process to be harmonised, LNTL, UNTL must fall well

With in the USL and LSL or must be just equal to them.)

(b) The percentage of rejections

Probability = 0.0059

Percentage of rejection = 0.59%

(c) USL – LSL = 1.207 – 1.033

= 0.1746 1 = 6 x 0.0287

= 0.17226 1 (USL-LSL) i.e., 0.1722 < 0.174 the process is capable of meeting the specificationlimits.

(d) In order to harmonise the process to the specifications change the process centre to

the specifications mean

i.e., = 1.12

The control limits for X-bar-chart

UCL =

= 1.12 + 1.88 x 0.0324= 1.1809

LCL == 1.059

CL = X bar = 1.12

Control limits for R-chart

5.20287.0

1054.1033.1−=

−= Z

X

R A X 2+

R A X 2−

1059.0

0324.027.34

=

== x R DUCL

.0324.0

0.00324.00.03

==

=×==

RCL

R D LCL



MODULE NO -12

Development and use of X bar – R Chart

In order to establish a pair of control charts for the average ( X bar ) and the range

(R), it is desirable to follow a set procedure. The steps in this procedure are as follows:

1. Select the quality characteristic.

2. Choose the rational subgroup.

3. Collect the data (20 to 25 samples).4. Calculate the mean ( X bar) and R for each sample.

5. Determine the trial control limits.

6. Establish the revised control limits.7. Construction of X bar - R – Chart.

8. Interpretation of the Results.

Equations for computing 3-sigma limits on Shewhart control charts for variables



Problem 1. Control charts for X bar and R are maintained on a certain dimension of a

manufactured part which is specified as 2.05 ± 0.02 cms. Subgroup size is 4. The values

of X bar and R are computed for each subgroup. After 20 subgroups. = 41.283 andR = 0.280. If the dimensions fall above USL, rework is required, if below LSL, the part

must be scrapped. If the process is in statistical control and normally distributed.

(a) Determine the 3 control limits for X bar and R chart.

(b) What is process capability

(c) What can you conclude regarding its ability to meet specifications(d) Determine the percentage of scrap and rework

(e) What are your suggestions for improvement.

Solution. = 41.283

R = 0.280n = Sample size = 04Number of subgroup (K) = 20

The specification limits are 2.05 ± 0.02

Upper specification limit USL = 2.07 cmLower specification limit LSL = 2.03 cm

From the tables, for a subgroup size 4

A2 = 0.73 d2 = 2.059D3 = 0.0 D4 = 2.28

X

X



(c) USL – LSL = 2.07 – 2.03 = 0.04

Since the 6

1 is greater than USL – LSL , the process is not capable of meeting thespecification limit i.e., 0.0407 > 0.04.

Note: 1. If 6 1 is less than (USL – LSL). The process is capable of meeting the

specification. There should not be any rejection. If rejection occurs we can

conclude that, the process is not centered properly.

2. If 61 is equal to (USL – LSL), the process is exactly capably of meeting thespecification limits. But tight tolerances are provided. We have to prefer a skilled

operator for operating the machine.

3. If 6 1 is greater than to (USL – LSL), the process is not capable of meeting the

specifications limits. The rejections are inevitable.



(e) Since the percentage of rework is 19.49%, to minimize this, the possible ways are

(i) Change the process centre to the specification mean i.e., from 2.06415 to 2.05.

The calculations are shown below:

Z =

Probability from Normal tables is 0.9984

That is 1 – 0.9984 = 0.0016 i.e. 0.16%

94.200679.0

05.207.2=

−



The percentage of rework is 0.16%

Since it is symmetric the percentage of scrap is also 0.16%.

(ii) Widening the specification limits, for this we have to consult the design engineer,

whether the product performs its function satisfactorily or not.

(iii) Decrease the dispersion, for this we have to prefer a skilled operator and very goodraw material and a new machine, practically which is difficult.

(iv) Leave the process alone and do the 100% Inspection.(v) Calculate the cost of scrap and rework, whichever is costly make it zero, accordingly

change the process centre.

Problem 2.

Subgroup of 5 item each are taken from a manufacturing process at regular intervals.

A certain quality characteristic is measured and X bar , R values computed for eachsubgroup. After 25 subgroup

= 357.5,

R = 8.8. Assume that all the points are within the control limits onboth the charts. The specifications are 14.4 ± 0.4

(a) Compute the control limits for X bar and R chart

(b) What is the process capability

(c) Determine the percentage of rejections if any(d) What can you conclude regarding its ability to meet the specifications.

(e) Suggest the possible scrap for improving the situation. (note: n=5 from tables

A2=0.5, d2=2.236, D3 = 0, D4 = 2.11)

X



MODULE NO -13

Development and use of X bar– S Chart With Real life data

Note :Although X- bar and R charts are widely used, it is a occasionally desirable to

estimate the process standard deviation directly instead of indirectly through the use ofthe range R. This leads to control charts for X-bas and S, where S is the sample standard

deviation. Generally X-bar and s charts are preferable to their more familiar counter

parts, X – bar and R charts when either

1. The sample size n is moderately large ---say n>10 or 12.

2. The sample size n is variable

Problem – 6

The following data presents the inside diameter measurements on the piston rings toillustrate the construction and the operation of X bar and S chart. The subgroup size isfive.

Sample no Si

1 74.010 0.0148

2 74.001 0.0075

3 74.008 0.0147

4 74.003 0.0091

5 74.003 0.0122

6 73.996 0.0087

7 74.00 0.0055

8 73.997 0.0123

9 74.004 0.0055

10 73.998 0.0063

11 73.994 0.0029

12 74.001 0.0042

13 73.998 0.010514 73.990 0.0153

15 74.006 0.0073

16 73.997 0.0078

17 74.001 0.0106

18 74.007 0.0070

i X



19 73.998 0.0085

20 74.009 0.0080

21 74.000 0.0122

22 74.002 0.0074

23 74.002 0.0119

24 74.005 0.0087

25 73.998 0.0162



Note:

The control limits for the x bar chart based on S bar are identical to the X bar chartcontrol limits, where the limits were based on R bar They will not always be the same,

and in general, the X bar chart control limits based on S bar will be slightly differentthan limits based on R bar.

We can estimate the process standard deviation using the fact that S/c4 is an unbiased

estimate of . Therefore, since c4 = 0.9400 for samples of size five, our estimate of the

process standard deviation is

This estimate is very similar to that of obtained via the range method.

Problem -7

A certain product has a specification of 120 ±5. At present the estimated process

average is120 and 1 = 1.5(a) Compute the 31limits for X bar , R chart based on a subgroup size of 4

(b) If there is a shift in the process average by 2%, What percentage of product

will fail to meet the specification.(c) What is the probability of detecting the shift by X bar - chart

01.09400.0

0094.0ˆ

4

===

c

S σ



120

7454.117

0885.373.01202

==

=

−=

−=

X CL

x

R A X LCL



Below Z =

= -1.73


= 4.18%

Above: Z =

= 1.73


= 95.82%i.e. 100-95.82 = 4.18%

(c) With respect to X bar - chart

n = 0.75

122.4

USL= 125

CL=120

117.6

LSL= 115

5.1

6.117115−

5.1

4.122125−

4

5.11

==

n x

σ σ



Below Z


= 57.14%

Above Z =

= 0.1866

Probability = 0.4287 = 42.86%

i.e. =100 – 42.86 = 57.14%

Problem - 8 Subgroup of 4 items each are taken from a manufacturing process at regular intervals.

A certain quality characteristic is measured and X bar , R values are computed for each

subgroup. After 25 subgroup. X bar = 15350, R = 411.1.

(a) Compute the control limits for X bar, R chart.

(b) Assume all the points are falling within the control limits on both the charts.The specification limits are 610 ± 15. If the quality characteristic is Normally

distributed what percentage of product would fail to meet the specifications.

(c) Any product that falls below L will be scrapped and above U must be

reworked. It is suggested that the process can be centered at a level so that not

more than 0.1% of the product will be scrapped. What should be the aimed

value of to make the scrap exactly 0.1%.

(d) What percentage of rework can be expected with this centering.

122.4USL= 122 . 26

CL= 120

117.6LSL= 117.74

122.4

1866.075.0

4.12226.122=

−=

1866.075.0

4.12226.122−=

−

1

X



From tables, for a subgroup size of 4

A2 = 0.73d2 = 2.059

D3 = 0.0

D4 = 2.28Control limits for X bar - chart

(b) Specification limits are

610 ± 15

USL = 625LCL = 595

012.626

456.1673.0614

2

=

×+=

+= R A X UCL

0.0

450.160

3

=

×=

= R D LCL

456.16== RCL

6141

== X X



Probability from tables = 0.0089 = 0.89%Percentage of rework

Probability from tables = 0.947 i.e 91.47%

99.7059.2

456.16

2

1===

d

Rσ

UNTL = 637 .97

USL = 625

CL = 614

LNTL = 590.03

LSL = 595

37.199.7

614625

1

1

=−

=

−=

σ

X USL Z



Rework = 100 – 91.47%

= 8.53%For the probability 0.001 the Z value from the normal table is -3

The percentage of rework now is

For Z 0.75 the probability from normal table is 0.7734

i.e 77.34%

Percentage of rework = 100 - 77.34 = 22.66%.

99.7

5953

1

new X −=−

97.61899.735951

=×+=new X

75.099.7

97.618625=

−= Z



MODULE NO -14

Introduction to Six Sigma Concepts

What is six sigma

Six sigma is several things. First, it is a statistical measurement. It tells us howgood our products, services and processes really are. The Six sigma method allows us to

draw comparisons to other similar or dissimilar products, services and processes. In this

manner, we can see how far ahead or behind we are. Most importantly, we see where weneed to go and what we must do to get there. In other words, Six sigma helps us to

establish, our course and gauge our pace in the race for total customer satisfaction

For example, when we say a process is 6 sigma, we are saying it is Best-in-Class.Such a level of capability will only yield about 3 instances of nonconformance out of

every million opportunities for nonconformance. On the other hand, when we say that

some other process is 4 Sigma, we are saying it is average. This translates to about 6200nonconformities per million opportunities for nonconformance. In this sense, the sigma

scale of measure provides us with, a “goodness micrometer” for gauging the adequacy of

our products, services and processes. Six Sigma as a business strategy can greatly helpus to gain competitive edge. The reason for this is very simple – as you improve the

sigma rating of a process, the product quality improves and costs go down. Naturally, the

customer becomes more satisfied as a result. Let us remember there is no economics of

quality. It is always cheaper to do “Right Things, “Right First Time”.

WHAT DOES “METRICS” STAND FOR ?

M MeasureE Everything

T ThatR Results

I In

C CustomerS Satisfaction

Applicability's of six sigma

The first step toward improving the sigma capability of a process is defining what the

‘customer’ expectations are. Next, you “map” the process by which you get the workdone to meet those expectations. This means that you create a ‘box diagram’ of theprocess flow; I.e.; identifying the steps within the process. With this done, you can now

affix success criteria to each of the steps. Next, you would, want to record the number of

times each of the given success criteria is not met and calculate the total defects-per-opportunity (TDPO). Following this, the TDPO information is converted to defects-per-

opportunity (DPO) which in turn, is translated into, a sigma value (). Now, you are

ready to make direct comparisons – even apples and Oranges if you want.



Three Sigma vs Six Sigma

Three Sigma would be equivalent to one misspelled word per 15 pages of text . Six

sigma would be equivalent to one misspelled over 300000 pages, quite a differenceindeed. Now, let’s put this in real world terms. Some corporations are already running

Six Sigma. It is self-evident that they’re going to perform better over the long haul. Forexample, several prestigious Japanese Companies(which are doing so well in the World

market place) are currently running at or near the 6 sigma Level.

SIGMA - ()

Sigma is a letter in Greek alphabet.

The term “sigma” is used to designate the distribution or spread about the me an(average) of any process or procedure.

The Sigma rating indicates how often defects are likely to occur. The higher the sigmarating, the less likely a process will produce defects. As sigma rating increases, costs godown, cycle time goes down and customer satisfaction goes up.

QUALITY IMPROVEMENT

=

PRODUCTIVITY IMPROVEMENT

=

COST REDUCTION

RIGHT FIRST TIME AND EVERY TIME

What is a defect A defect is any variation of a required characteristic of the product (or its parts) or

services which is far enough from its target value to prevent the product from fulfilling

the physical and functional requirements of the customer, as viewed through the eyes of

your customer.

A defect is also anything that causes the processor or the customer to make adjustments.

Anything That Dissatisfies Your Customer

The Common Metric: Defects per Unit (DPU)

DPU is the best measure of the overall quality of the process.

DPU is the independent variable.

Process yields are dependent upon DPU



Example:We checked 500 Purchase Orders (PO) and PO had 10 defects then,

d.p.u = d/u = 10/500 = 0.02 In a P.O. we check for the following :

a) Supplier address/approval.

b) Quantity as per the indentc) Specifications as per the indent.

d) Delivery requirementse) Commercial requirements.

Then there are 5 opportunities for the defects to occur. Then, the total no.

opportunities =m u = 5x500 = 2500.Defects per opportunity, d.p.o = d/(m u) = 10/2500 = 0.004

If expressed in terms of d.p.m.o. (defects per million

opportunities) it becomes. d.p.m.o. = d.p.o x 106 = 4000 PPMFrom d.p.o., we go to the normal distribution tables and

calculate ZLT and corrected to ZST by adjusting for shift (1.5

) then.ZLT = 2.65; andZST = 2.65 + 1.5 = 4.15

No. of opportunities = No. of points checked.

If you don’t check some points then it becomes a passive opportunity. We should takeonly active opportunities into our calculation of d.p.o., and Sigma level.

Cost / QualitySix Sigma has shown that the Highest Quality Producer Is the Lowest Cost Producer

Process capability process potential index (Cp) The greater the design margin, the lower the DPU.

Design Margin is measured by Capability Index (Cp),

Where :The numerator is controlled by Design Engineer

Cp = Maximum allowable Range of Characteristic

Normal Variation of Process

The denominator is controlled by

process Engineering.

If Ford says,

Cp should be more than 1.33 for regular production.

Cp should be more than or equal to 1.67 for new jobs.

Motorola says, Cp should be more than 2.0 for all jobs.



That implies, (U – L) / 6 = 2.0 or (U – L) = 12 i.e; (U – L) = ± 6

Hence the name Six Sigma.

± 3

Process capability means 0.27%. I.e., 2700 PPM shall be out of specification.

± 6 Process capability shall mean 2.5 Parts per Billion, shall be outside the specificationlimits.

The six Sigma Methodology is a five phase improvement cycle that are employed in a

project oriented fashion through the1. Define

2. Measure

3. Analyze4. Improve

5. Control

Step 1 : Define :

Define The Customer, Critical to quality (CTQ) issues, And the Core business

Process involved. Define who customers are, what their requirements areand what their expectations are. Define Project boundaries – the start and stop of the

process. Define the process to be improved by mapping the process flow.

Step 2 : Measure :

Develop a data collection plan for the process. Collect data, to determine types ofdefects and metrics.

Measure the current performance of the core business process involved.

Step 3 : Analyze :

Analyze the data collected to determine the root causes of defects and opportunities

for improvement. Identify the gaps between current performance and goalperformance. Prioritize opportunities to improve. Identify sources of variation.

Step 4 : Improve :

Improve target solutions by designing creative solutions to fix and prevent problems.

Create innovative solutions using technology and discipline. Develop and deployimplementation plan.



Step 5 : Control :

Control the improvements to keep the process on the new course. Prevent revertingback to the “old way” Control the development, documentation and Implementation of

an ongoing monitoring plan.

Step 1 : Define:

The Problem definition has five major elements. The Business Case.

Identifying the Customers of the project, their needs & requirements.

The problem statement. Project Scope. Goals & Objectives.

Step 2 : Measure

Calculating Sigma Value for discrete data

The data being collected for this project is discrete, to calculate sigma using thediscrete method, there are three items being measured. They are :

1. Unit : The item produced.

2. Defect : Any event that does not meet customer’s requirement.

3. Opportunity : A chance for a defect to occur.

Step 2 : Measure

The Formula to Calculate DPO.

Number of Defects

DPO =

Number of opportunities x Number of units produced

DPMO = DPO x 1000000

This calculation is called defects per Million Opportunities.

Step 2 : Measure

Performance Measures

For the Month of December :Total Number of rings produced = 86,702

Number of Defective Rings = 47

Number of Opportunities = 4 opportunities

Defects per Opportunity = 47/ 86,702 * 4= 1.355 * 10 – 4

Defects per Opportunity = 47/ 86,702 * 4

= 1.355 * 10 - 4

Defects / Million Opportunities = 1.355 * 10 – 4 * 106

= 135.55 DPMOConverting DPMO to “” value = 5.1



Performance Measure

Month Number of Ringsproduced

Number ofDefective Rings DPO DPMO

SigmaValue

December 2000 86,702 47 0.0001355 135.55 5.1January 2001 1,13,345 100 0.0002145 214.5 5.0

February 2001 1,14,368 123 0.0002688 268.88 4.9

March 2001 1,14,404 451 0.0009855 984.54 4.5

The Practical Meaning of



SIX SIGMA AS A GOAL (Distribution shifted to ± 1,5 )

Sigma Level Defects in PPM Yield in %

2

3 4 5 6

308,538

66,80762102333.4

69.1462

98.319899.379099.976799.99966

Legends

“m” : Number of Opportunities.“N” : Number of parts.“d” : Number of defects.

“dpu” : Defects per unit.“dpo” : Defects per opportunity.“Yft” : First time yield.“Yrt” : Rolled thru put yield.“dpmo” : Defects per million opportunities.“TDPU” : Total defects per unit.“Zlt” : Long term sigma level“Zst” : Short term sigma level

Formulae“dpu” = d / N“dpo” = dpu/m“Yft” = e-dpu“dpmo” = dpo X 106“TDPU” = sum of dpu“Yrt” = e-TDPU“YPO” = Yrt 1/m = e-dpo“dpo” of the over all process = (1-Ypo)Cpk = Zlt / 3Z = (USL – X bar)/ Cp = Zst/3



Process Capability It is a measure of the inherent uniformity of the process. Before examining the sourcesand causes of variation and their reduction we must measure the variation.YARD-STICKS OF PROCESS CONTROL:Cp - measuring capability of a processCpk - Capability of process, but corrected for non-centering

Process Capability Indices:Cp is a measure of spread.Cp = Specification (S) / Process width (P)Cpk is a measure of centering the process and its spread.

Cpk is minimum of Cpu= (USL - µ) / 3σ and Cpl = (µ - LSL) / 3σ The relationship between Cp and Cpk is Cpk = (1 – k) Cp

where : k : Correction factor and is the minimum of (T -µ ) / S / 2 or (µ - T ) / S / 2

Reducing Variation is the key to Reducing Defects



Cp - Measure of variation

Process Capability



CASE STUDY TO A REAL LIFE PROBLEM

In a company manufacturing and assembling of Printed Circuit Boards(PCB’s),the

rejection rate was found to be very high. Upon study it was noticed that there are 16stages in the assembly process of Printed Wiring Boards(PWB’s),out of which therejection rate was more during the wave soldering process compared to the other stages.

This wave soldering process stage is a critical stage of assembly.



Hence Wave Soldering Process Stage was selected for the study, in order to reduce theprocess variability and to minimize the rejection rate. The PCB’s are classified as singlelayered, bi-layered and multi-layered boards. In the assembly section of this company,two types of PCB’s are being assembled.

On-line inspection data for the wave soldering process were collected and the attribute

control charts (p and c) were plotted which showed that the process was not in a state ofstatistical control.

The fraction rejection was found to be 0.2 (i.e., 20%) and the average defects per unitwere 1.67 for multi-layered boards and 0.5 for bi-layered boards respectively.

The on-line inspection data for wave soldering process with the existing process

parameter values was collected and sigma(σ) was calculated.

For bi-layered boards the sigma level was 3.39 and for multi-layered boards the sigmawas 3.33,which are given below.

Table 1. Product Type: Bi-layered Boards

Standard Process Parameters

Baking Temperature 75˚C

Preheat Temperature 300˚C

Hot-air Temperature 320˚C

Solder Temperature 245˚C

Solder wave height 11mm



On-line Data for Bi-layered Boards- calculationNumber of defects = 71Total no. of soldering points = 233287Defects per opportunity =

Total no. of defects = 71 ….(1)Total no. of soldering points 233287

= 0.000304From the Normal Tables ,the value of sigma is 3.39

Table 2. Product Type: Multi-layered Boards

Standard Process Parameters

Baking Temperature 75˚C

Preheat Temperature 320˚C

Hot-air Temperature 340˚C

Solder Temperature 255˚C

Solder wave height 12.5mm

On-line Data for Multi-layered Boards- calculation

Number of defects = 38Total no. of soldering points = 106828Defects per opportunity =Total no. of defects = 38 ….(2)Total no. of soldering points 106828

= 0.00036

From the Normal Tables, the value of sigma is 3.33

After conducting the Brain storming session with the operators, foreman and themanager, the causes for the rejection of the PWA’s were traced out. During theinspection of the wave soldered PWA’s, it was found that the rejections were due to thefollowing causes.



MeaslesBlow holesSolder bridgeSolder splash andIcicles

The necessary calculations are made for both Bi-layered and Multi-layered boardsand analysis was carried out for the collected data with the help of Pareto Diagram whichshowed that blow holes and solder bridges constituted for majority of rejection.

After discussions with the operators, foreman and manager, the causes for the blowholes and solder bridges were identified.

The cause and effect diagrams were drawn and critical process parameters (controlfactors) that influence the wave soldering process were identified as:

Baking TemperaturePre-heat TemperatureHot-air TemperatureSolder TemperatureSolder wave height

Two noise factors viz., ambient temperature and humidity, each with two levels areconsidered for the experimentation. In order to optimize the above identified wavesoldering process parameters, the Orthogonal Array Approach of DOE was applied.Three levels were fixed for each of the above five critical factors which are shown inTables 3 and 4 for both bi-layered and multi-layered boards respectively. With theapplication of the Linear Graphs the number of experiments to be conducted are 27 forthe factors. OA Table and physical layout for the Bi-layered and Multi-layered boards areprepared. 27 experiments were carried out for both Bi-layered and Multi-layered PCB’sseparately with a sample size of two each.

Table 3.Factors and Levels for Bi-layered Board

FACTORS LEVEL 1 LEVEL 2 LEVEL 3

Banking Temperature 75˚C 80˚C 85˚C

Preheat Temperature 300˚C 305˚C 310˚C

Hot air Temperature 320˚C 325˚C 330˚C

Solder Temperature 240˚C 245˚C 250˚C

Solder wave height 10mm 11mm 12mm



Table 3. Factors and Levels for Bi-layered Board

FACTORS LEVEL 1 LEVEL 2 LEVEL 3

Banking Temperature 75˚C 80˚C 85˚C

Preheat Temperature 320˚C 325˚C 330˚CHot air Temperature 340˚C 345˚C 350˚C

Solder Temperature 250˚C 255˚C 260˚C

Solder wave height 11.5mm 12.5mm 13.5mm

Analysis of Data and ResultsThe experimental results were analyzed to establish the optimum process parameter

values for Baking Temperature, Pre-heat Temperature, Hot-air Temperature, SolderTemperature and Solder wave height. The responses were calculated for each of theexperiments for both bi-layered and multi-layered PCB’s. From the response matrices,

the SIGNAL-TO-NOISE (S/N) ratios were calculated using the formula :

=10 log ((1/p) - 1) ……(3)where p =1 - % good …….(4)

For example ,p= 1 – (25/100) = 75=10 log ((1/0.3) – 1) = - 4.7712

The S/N ratios for each of the experiments were calculated for bi-layered and multi-layered boards. The analysis of variance (ANOVA) was carried out for bi-layered and

multi-layered boards. The optimal levels of parameters were established based on thehighest value of S/N ratios. Optimized Factor Level for the wave soldering process forboth bi-layered and multi-layered PCB’s are given below

FACTORS Bi-layered Multi-layered

Banking Temperature Level-3 85˚C Level-3 85˚C

Preheat Temperature Level-3 310˚C Level-3 330˚C

Hot air Temperature Level-3 330˚C Level-3 260˚C

Solder Temperature Level-3 250˚C Level-3 260˚C

Solder wave height Level-2 1.5mm Level-3 2.5mm

Confirmation Run Further experiments were carried out with the optimized levels of the above

parameters for both bi-layered and multi-layered PCB’s taking a sample size of 8 each tocheck the validity of the levels of the optimized parameters.



The sigma levels were calculated again for the data collected and were found to be4.1 for bi-layered and 4.125 for multi-layered PCB’s which shows that the processvariability is decreased and process capability(cp and cpk) is increased. The percentageof rejections was again calculated which is reduced to 0.2% from 20%.

ConclusionsIn this case study, both on-line quality control techniques like control charts, paretodiagram and cause and effect diagram as well as off-line quality control techniques areapplied before the manufacture of the product to control the process. The sigma level forbi-layered PCB was improved from 3.39sigma to 4.1sigma and the multi-layered PCBfrom 3.33sigma to 4.125sigma.

Since the sigma levels were increased considerably using the Orthogonal approach ofDOE, it is evident that the application of the DOE technique (during the early stagesitself) is very effective in improving the quality of any process or product by optimizingthe parameters in order to yield a product which can be produced with minimum cost and

with minimum variation. The optimal levels for the factors obtained using OA approachand levels of sigma for both bi-layered and multi-layered PCB’s are summarized below:

Table 5: Comparison of the levels of five parameters for bi-layered PWS and multi-layered PWS

FACTORS Bi-layered PWA’s Multi-layered PWA’s

Present Optimum Present Optimum

Baking Temperature 75˚C 85˚C 75˚C 85˚C

Pre-heat Temperature 300˚C 310˚C 320˚C 330˚C

Hot-air Temperature 320˚C 330˚C 340˚C 345˚CSolder Temperature 245˚C 250˚C 255˚C 260˚C

Solder Wave height 11mm 11mm 12.5mm 13.5mm

Table 6. Comparison of the Sigma Levels

Type of Printed WiringAssembly

Present Level ofSigma

Improved Levelof Sigma

Bi-layered 3.39 4.1

Multi-layered 3.33 4.125

With lesser number of experiments in Orthogonal Array approach of DOE, it ispossible to achieve the same effective results as compared to other techniques of DOElike Full Factorial, Fractional Factorial, Randomized Block Design etc.



Orthogonal Array Table

ExperimentNumber

BakingTempt.

PreheatTempt.

Hot-airTempt.

SolderTempt.

SolderWaveHeight

1 1 1 1 1 1

2 1 1 1 1 2

3 1 1 1 1 3

4 1 2 2 2 1

5 1 2 2 2 2

6 1 2 2 2 3

7 1 3 3 3 1

8 1 3 3 3 2

9 1 3 3 3 3

10 2 1 2 3 1

11 2 1 2 3 2

12 2 1 2 3 3

13 2 2 3 1 1

14 2 2 3 1 2

15 2 2 3 1 3

16 2 3 1 2 1

17 2 3 1 2 2

18 2 3 1 2 3

19 3 1 3 2 1

20 3 1 3 2 2

21 3 1 3 2 3

22 3 2 1 1 1

23 3 2 1 1 224 3 2 1 1 3

25 3 3 2 3 1

26 3 3 2 3 2

27 3 3 2 3 3



Physical Layout For Bi-layered Boards

ExperimentNumber

BakingTempt.

PreheatTempt.

Hot-airTempt.

SolderTempt.

SolderWaveHeight

1 75˚C 300˚C 320˚C 240˚C 10mm2 75˚C 300˚C 320˚C 240˚C 11mm

3 75˚C 300˚C 320˚C 240˚C 12mm

4 75˚C 305˚C 325˚C 245˚C 10mm

5 75˚C 305˚C 325˚C 245˚C 11mm

6 75˚C 305˚C 325˚C 245˚C 12mm

7 75˚C 310˚C 330˚C 250˚C 10mm

8 75˚C 310˚C 330˚C 250˚C 11mm

9 75˚C 310˚C 330˚C 250˚C 12mm10 80˚C 300˚C 325˚C 250˚C 10mm

11 80˚C 300˚C 325˚C 250˚C 11mm

12 80˚C 300˚C 325˚C 250˚C 12mm

13 80˚C 305˚C 330˚C 240˚C 10mm

14 80˚C 305˚C 330˚C 240˚C 11mm

15 80˚C 305˚C 330˚C 245˚C 12mm

16 80˚C 310˚C 320˚C 245˚C 10mm

17 80˚C 310˚C 320˚C 245˚C 11mm

18 85˚C 310˚C 320˚C 245˚C 12mm

19 85˚C 300˚C 330˚C 245˚C 10mm

20 85˚C 300˚C 330˚C 245˚C 11mm

21 85˚C 300˚C 330˚C 245˚C 12mm

22 85˚C 305˚C 320˚C 250˚C 10mm

23 85˚C 305˚C 320˚C 250˚C 11mm

24 85˚C 305˚C 320˚C 250˚C 12mm

25 85˚C 310˚C 325˚C 240˚C 10mm

26 85˚C 310˚C 325˚C 240˚C 11mm

27 85˚C 310˚C 325˚C 240˚C 12mm



Conclusion

In most of the Indian industries, the acceptance criterion is only on the basis ofspecification limits specified by the designer. If any characteristic of a product / processfalls between the specified limits, it is taken for granted that the product is uniformly

good. But as per Taguchi’s QLF, as the functional characteristic of a product deviatesfrom the target value, it causes loss to the society.

The more the deviation, the more is the loss, even if it is within the specified limits.Robust engineering methods are recommended at the early stages of product design toachieve the higher sigma levels. Robust engineering also reduces the time to market withthe help of two step optimization.

The results obtained from small scale laboratory experiments can be repeated underthe large scale manufacturing conditions if the output characteristics are selectedappropriately using S / N ratio.

Introduction to six sigma

Problem 1A press brake is set up to produce a formed part to a dimension of 3 ± 0.005. A

process study reveals that the process limits are at 3.002” ± 0.006, i.e., at a minimum of2.996 and a maximum of 3.008. After corrective action, the process limits are broughtunder control to 3.001 ± 002.

Question:1

Question 1. Calculate the Cp and Cpk of the old process.Question 2. Calculate the Cp and Cpk of the corrected process

Answers:Question 1.

specification with (s) = 0.010; process width (p) = 0.012

So Cp= S/P = 0.10/0.012 = 0.833= 3.002; design center (D) = 3.000

4.0005.0

002.0

005.0

000.3002.3

2==

−=

−=

S

D X SoK



Question 2.

Specification width (s) = 0.010; process width (p) = 0.004.

Therefore Cpk = (1-0.4)0.833 = 0.5

So Cp= S/P = 0.10/0.004 = 2.5= 3.001"; design center (D) = 3.000"

Therefore Cpk = (1-k) Cp = (1-0.2)2.5 = 2.0Using the simpler and alternate formula for Cpk:

5.0006.0003.0

006.0002.3005.3:1 ==−= pk C question In

0.2002.0

004.0

002.0

001.3005.3:2 ==

−= pk C question In

2.0005.0

001.0

005.0

000.3001.3

2==

−=

−=

S

D X SoK



MODULE NO -15

Introduction to Robust Design and its applications

Out line

• Introduction.• Quality Control• Quality Engineering• Taxonomy of Quality• Evaluation of Quality loss• Tools used in Robust Design• Process Capability• Conclusions.

Introduction

Dr. G. Taguchi:

Dr. Taguchi, is born in 1924. He started his career from Naval institute of Japanbetween 1942 – 45 and then with ministry of public health and welfare. Later he joinedministry of education subsequently moved to Nippen telephone at japan.

Dr. Taguchi is the inventor of the famous orthogonal array OA techniques for thedesign of experimentation. He published his first book on OA in 1951 Taguchi alsovisited Indian statistical institute between 1954 – 55. He wrote a book on design ofexperiments.

Dr. Taguchi philosophy is Robust Engineering Design. He blended statistics withengineering applications and pioneered work in Industrial Experimentation. He is also theinnovator of the quality loss function concept and promoted robust design related to thishe propagated signal to noise ratio phenomenon in SPC. He developed a three stage offLine QC methods viz., system design, parameter design and Tolerance design.

Genichi Taguchi, a Japanese statistician, is in the forefront of the pioneers of theQuality Control. His major contribution is the concept of Robust Design, which isacclaimed as the most significant one throughout the world. His concepts haverevolutionized the very idea of quality control and hence these techniques are widely

applied by the manufacturing and service industries successfully in the advancedcountries like Japan, US, UK, etc. In the early 1970’s, Taguchi developed the conceptof the Quality Loss Function.



Quality

• Quality is defined as “fitness for use”.• As per G. Taguchi, the quality of a product is the (minimum) loss imparted by

the product to the society from the time the product is shipped.

• Quality plays a vital role in all walks of life, starting from the household to bigengineering and service industries.

Quality Control

• It is an activity of ensuring the manufacturing of good quality products whichsatisfy the customers’ needs.

• Quality control techniques are broadly classified into On-line and Off-line.

• The On-line quality control techniques are applied to monitor a manufacturingprocess to verify the levels of quality of goods already produced.

• The Off-line quality control techniques are applied to improve the quality of aproduct/process in the design stage itself i.e., before the products aremanufactured and made available to customers.

For any company to compete in the world class market scenario, its leaders mustunderstand, digest, disseminate and guide the implementation of simple and powerfultools that go well beyond the traditional quality control techniques. They are

• Design of Experiments (DOE)

• Multiple Environment Over Stress Test (MEOST)

• Quality Function Deployment (QFD)

• Total Productive Maintenance (TPM)• Benchmarking

• Poka-Yoke

• Next Operation As a Customer (NOAC)

• Supply Chain Management (SCM)

• Failure Mode and Effect Analysis (FMEA) and

• Cycle Time Reduction

The Design of Experiments (DOE) is one of the most powerful techniques that helpsto achieve the world-class quality.

Quality EngineeringIt consists of the activities directed at reducing the variability and thereby reducing

the loss. The fundamental principle of robust design is to improve the quality ofproduct/process by minimizing the effect of the causes of variation without eliminatingthe causes. This is achieved by optimizing the product/process design to make theperformance minimum sensitive to the cause of variation. The robust design processencompasses three stages namely



System Design – It is the process of applying scientificand engineering knowledge to produce a basic functionalprototype design.

• Development of a system to function under an initial set of nominal conditions.• Requires technical knowledge from science and engineering.

• Originality / Invention / Marketing strategy.

Parameter Design - It is the process of investigation towards identifying the settings ofdesign parameters that optimizes the performance characteristics and reduces thesensitivity of engineering design to the source of variation (noise factors).

• Determination of control factor levels so that the system is least sensitive tonoise.

• Involves use of orthogonal arrays and signal – to Noise Ratio.• Improves quality at minimal cost.

Tolerance Design – It is the process of determining the tolerances around the nominal

settings identified in the parameter design process.• Specification of allowable ranges for deviations in parameter values.• Involves cause detection and removal of causes.• Typically increases product cost. However, cost may be minimized by

experimenting to find tolerances that can be relaxed without adversely affectingquality.

Taxonomy Of QualityThere are three fundamental issues regarding quality:· To evaluate the quality· To improve quality cost-effectively and· To monitor and maintain quality cost-effectively. Quality characteristics are

classified into two:· Variable characteristic and· Attribute characteristic

Variable characteristics can be classified into three types:

· Nominal-The-Best : A characteristic with a specific target value.Examples: Dimension, Clearance, Viscosity etc.

· Smaller-The-Better : Here the ideal target value is zero.Examples: Wear, Shrinkage, Deterioration etc.

· Larger-The-Better : The ideal target value is infinityExamples: Strength, Life, Fuel efficiency etc.

· Attribute Characteristics : Based on visual inspectionExamples : Appearance, Taste, Good/bad, etc.



Evaluation Of Quality Loss

• Traditional interpretation of quality loss• Taguchi’s interpretation of quality loss.

Traditional interpretation of quality loss Step Function

Taguchi emphasizes that the loss incurred by a product which falls close to LSLand which falls just below LSL is almost same. The problem with most of the traditionalmeasures of quality (rework rate, scrap rate, Cp, Cpk, etc.) is that by the time we getthese figures, the product is already either in production or in the hands of customer.

Taguchi’s interpretation of Quality Loss Function (TQLF)

• The objective of TQLF is the quantitative evaluation of loss resulting from thefunctional variation of the output quality characteristic from the target value.

• The two important points to be considered to establish Taguchi’s QLF are

* Consumer tolerance and* the customer loss

Taguchi’s Quadratic Representation of the QLF

There are three cases of TQLF namely :

• TQLF for Nominal-The-Best

• TQLF for Smaller-The-Better and

• TQLF for Larger-The-Better.



To establish the characteristics of Taguchi’s QLF, the two important aspects to be

considered are consumer tolerance and consumer loss.

Tqlf For Nominal-The-Best (NTB):

Taguchi’s QLF for the case of NTB is given byL(y) = K * (y-T) 2

Where L(y) = loss is rupees per unit of product for the output characteristic of ‘y’

T = Target value of ‘y’K = Constant of proportionality that depends upon financial importance of the outputcharacteristic

Taguchi recognized the loss as a continuous function and it does not occur suddenly.

The quadratic representation of QLF i.e., L (y) = K * (y-T) 2 .

Where Loss L(y) is minimum at y =T and L(y) increases as ‘y’ deviates from thetarget value T.

∴ K = L(y) = Ao

(y-T)2 ∆ 02

where ‘∆0’ is the consumer tolerance and ‘Ao’ is the consumer loss which areshown in Figure below.

TQLF for Nominal-The-Best

TQLF For Smaller-The-Better (STB)

When the out-put characteristic is to be a minimum value, the loss function ischaracterized as “Smaller-The-Better”. The examples for STB are shrinkage,pollution, radiation leakage etc.



The ideal value for this is zero. The Loss function is slightly different but the procedureis same as the Nominal–The–Best.

For STB, the loss function is given byL(y) = K * y 2 and K = A0 / y0 2

TQLF For Larger-The-Better (LTB)

The loss function for LTB is the reciprocal of the Smaller-The-Better case and isgiven by L (y) = K * (1 / y2) and K = Ao * yo2 . This is shown in figure below.Some examples of LTB are strength of a permanent adhesive, strength of a welded joint,fuel efficiency, corrosion resistance etc. The ideal value for LTB is infinity.

TQLF : Larger-The-Better



y = Output Characteristic (nominal-the-Best)m = target Value

• Same product, same specifications• All are 100% inspected• Cost to you is the same from all four sources.

Which factory would you choose to be your vendor? Why?

USLS

m

32

mmm



It is, therefore, very much essential to analyze and quantify the losses to the societyusing TQLF. This will help in identifying the level of one’s own quality in comparisonwith the competitors’ quality to take remedial actions for improvements, if necessary, tocompete in the present day global competition. The TQLF can be used to determine theoptimum tolerances for the levels of optimized parameters determined by the parameterdesign technique.

Tools Used in Robust Design

• Signal- To - Noise Ratio (S/N) - which measures quality

• Orthogonal Arrays - which are used to study many design parameterssimultaneously

1USLS

m

4

mm

$0.73/pc

m

$0.15/pc$1.23/pc$0.27/pc



S/N ratios for Static ProblemsS/N ratio for Smaller-The-Better

When the output characteristic can be classified as Smaller-The-Better, the standardS/N ratio is

With n observations, y1, y2, y3 ……. Yn

L(y) = K(y2)K(MSD)

whereK= A0 / Yo2MSD = y12 + y22 + y32 + ….. Yn2

n

S/N (dB) = -10 log10 (MSD)

S/N Ratio for Nominal-The-Best

When the output characteristic is classified as Nominal-The-Best the standard S/N ratiois:With n observations, y1, y2, y3 ……. yn

The Signal to Noise ratio is

S/N ratio for Larger-The-Better

When the output characteristic can be classified asLarger-The-Better , the standard S/N is:

L(y) = K(1 / y2)

K(MSD)K = Aoyo2n

MSD = 1/n yi2i=1

Signal to Noise Ratio isS/N (dB) = -10 log10 (MSD)

and n

Y n

ii

Y =

=1

1 2

1

2 )1

1( Y

n

n

ii

Y −−

= =

σ

2

2

10log10σ

Y n =



S/N ratio for Fraction Défective

The Quality Characteristic is denoted by p, a fraction assuming values between 0 and1. When the fraction defective is p, on an average we have to manufacture 1 / (1-p) to

produce one good piece.For every good piece produced there is a waste and hence a loss,equivalent to the cost of the processing {1/(1-p) - 1} pieces.

Thus, the Quality Loss Q is given byQ = K (p / 1-p)

Where K is the cost of processing one piece Ignoring K, we obtain the objective functionto be maximized in the decibel scale as

Ideal Function Realityy = Output Response y = Output ``

y y

M = Input Energy M = Input Energy

The most common way of expressing a design’s Ideal Function is :y = MWhere y = Output response

m = Input signal

−−=

p

pn

1log10 10



Classification of Parameters

Signal factors – these are the parameters set by the user of the product to express theintended value for the response of the product. The signal factors are selected by thedesign engineer based on the engineering knowledge of the product being developed.

Control factors – these are any design parameters of a system that engineers canspecify by nominal values and maintain cost effectively.

Noise factors – these are the variables that affect the system function and are eitheruncontrollable or too expensive to control.

The Engineered System consists of four (4) components:

Product/ProcessSignal factors

Control factors

Noise factors

Block Diagram of a Product/Process

Response

VOICE OF CUSTOMER

System

(Subsystem)

Engineered System

Control Factors

YOutput

Response

RESULT

MInput

Signal

INTENT

Noise Factors



The Relationship Between Loss and Noise Factors

Relationship between Control Factors Vs Noise Factors is represented graphicallybelow.

ControlFactors

NoiseFactors

R&D Advanced Product Design MFG. Users/RecycleEngineering Planning Process Design

Numberof

Factors

Product Development Cycle

Loss to the Society

Deviations of Functional Characteristics fromTarget Value

Noise Factors

Cause Deviations

InnerNoise

OuterNoise

Between-ProductNoise

Variation inOperating

Environments

Human Errors

DeteriorationManufacturingImperfections



Put a lot of thought intowhat you are going To measureas Data Selection of output

Characteristics

Put a lot of thoughtinto selecting controlFactors and levels

Put a lot of thought

into how you are goingto treat noise factors.

Make predictions

Is predictionConfirmed ?

Is predictionConfirmed ?

No

Yes

Yes

No

Put a lot of thought intowhat you are goingTo measure as dataSelection of outputCharacteristics

Put a lot of thoughtinto selecting controlFactors and levels

Put a lot of thoughtinto how you aregoing to treat noisefactors.

Design the experiment with anorthogonal array without assigningInteractions among control factors

Use L12, L18, L36 as muchAs possible

Analyze data using Signal-to-NoiseRatio response table

Make predictions

Conduct Experiment



Six Sigma Process CapabilityIt is a measure of the inherent uniformity of the process. Before examining the sourcesand causes of variation and their reduction we must measure the variation.

Yard-Sticks Of Process Control:Cp - measuring capability of a processCpk - Capability of process, but corrected for non-centering

Process Capability Indices:

Cp is a measure of spread.Cp = Specification (S) / Process width (P)Cpk is a measure of centering the process and its spread.

Cpk is minimum of Cpu= (USL - µ) / 3σ and Cpl = (µ - LSL) / 3σ The relationship between Cp and Cpk is Cpk = (1 – k) Cp

where :k : Correction factor and is the minimum of

(T -µ ) / S / 2 or (µ - T ) / S / 2

Raw Data from the Experiment

S/N Analysis

•Smaller–the-Better(db) = - 10 log

•Larger–the-Better(db) = - 10 log

•Nominal–the-Best (Type I)(db) = 10 log

•Nominal–the-Best (Type II)(db) = 10 log

Response Table & Graphs

•Control factors only

CharacteristicType

Nominal-the-Best Analysis

Smaller-the-BetterAnalysis

Larger-the-BetterAnalysis



Cp - Measure of variation

Process Capability



Sigma Levels and the Associated CP & Defects

Flow Chart Showing Stages Of Quality Engineering.

± 5

± 2

± 4

± 6

± 3

308537 PPM C = 0.67 1970’S

3.4 PPM Cp = 2

2000’S

World Class

233 PPM Cp= 1.66 Mid– to-

late

1990’S

6210 PPM Cp= 1.33 Early

1990’S

66807 PPM Cp = 1 1980’

QUALITY ENGINEERING

OFF-LINE QUALITY ENGINEERINGENGINEERING OPTIMIZATION

USING DESIGN OF EXPERIMENTS

PRODUCTDESIGN

PROCESSDESIGN

SYSTEM DESIGNINNOVATION



The Basic Steps In Parameter Design

1. Define project scope / objectives

– Define project objectives– Identify the system or subsystems

– Select team leader and members

– Establish overall strategies

2. Identify Ideal Function / Response to be Measured

– Establish intent, desired results

– Define input signal and output response– Define ideal function.

– Determine measurement feasibility

3. Develop Signal and Noise Factor Strategies

– Define Signal levels and ranges.

– Identify all noise factors– Select critical noise factors and set levels

– Determine Noise Strategies.

4. Establish control factors and levels– Identify all control factors

– Select critical control factors and set levels– Select orthogonal array– Assign control factors to orthogonal array

5. Control experiments

– Plan / prepare for experiment– Conduct experiment

– Collect data

ON-LINE QUALITY ENGINEERINGPROCESS CONTROL

PARAMETER DESIGN

TOLERANCE DESIGNOPTIMIZAZTION



6. Conduct data analysis

– Calculate S/N ratios and ’s– Complete /interpret response tables / graphs

– Perform two step optimization

– Make predictions.

7. Conduct confirmation run

8. Implement and document results

Contributions of DOE to Business Excellence: A Spider Chart

Applications of Computer based Robust Engineering

• Reduction of R&D cycle time using Simulation based Robust Engineering.• Software testing and Algorithm Optimization.

• Design of Information Systems for Pattern Analysis.

ChronicProblem Solving

90%

60%

50%

70%

60%

50%

50%

30%

70%

40%

30%30%

Profit /R.O.IImprovement

Customer

Loyalty

Cycle – Time

Reduction

Employee

Morale

DesignImprovement

Cost reduction

ReliabilityImprovement

Space

Reduction

Supplier

Improvement



MODULE NO -25

Measurement Systems Analysis and Gage R&R

Measurement System Analysis (MSA)

Introduction

The manufacturing environment, by its very nature, relies on two types of

measurements to verify quality and to quantify performance: (1) measurement of itsproducts, and (2)measurement of its processes. Therefore, product evaluation and process

improvement require accurate and precise measurement techniques. Due to the fact that

all measurements contain error, and in keeping with the basic mathematical expression:Observed value = True value + Measurement Error, understanding and managing

"measurement error," generally called Measurement Systems Analysis (MSA), it is an

extremely important function in process improvement (Montgomery, 2005).

MSA is a comprehensive set of tools for the measurement, acceptance, and analysis

of data and errors, and includes such topics as statistical process control, capabilityanalysis, and gauge repeatability and reproducibility, among others

(Besterfield,2004).MSA recognizes that measurements are made on both simple and

complex products, using both physical devices and visual inspection devices that rely

heavily on human judgment ofproduct attributes. Purpose of MSA is to statistically verify that current measurement

systems provide:

– Representative values of the characteristic being measured– Unbiased results

– Minimal variability

Organizational Uses of MSA are:

• Mandatory requirement for QS 9000 certification.

• Identify potential source of process variation.• Minimize defects.

• Increase product quality.

All measurement processes will contain some amount of variation. The variation can

come from one of two sources; 1) the difference between parts made by any process, and

2) the method of obtaining the measurements are imperfect. Measurement System Errorsare of two types, they are as follows

• Accuracy: difference between the observed measurement and the actualmeasurement.

• Precision: variation that occurs when measuring the same part with the same

instrument.



Any measurement system can have any of these problems. One could have ameasurement device that measures parts with very little variation, but is not accurate.

One could also have an instrument where the average of the measurements are very close

to the actual value, but has a large variance (not precise). Finally, one could have a device

that is neither accurate nor precise.

Measurements are said to be accurate if their tendency is to center around the actual

value of the entity being measured. Measurements are precise if they differ from oneanother by a small amount.

Measured Value = ƒ (TV + Ac + Rep + Rpr)

TV = true value

Ac = gauge accuracyRep = gauge repeatability

Rpr = gauge reproducibility

Measurement system components:

• Equipment or gage– Type of gage

• Attribute: go-no go, Vision systems (part present or not present)

• Variable: calipers, probe, coordinate measurement machines– Unit of measurement - usually at least 1/10 of tolerance

• Operator and operating instructions

222

gage product observed σ σ +=



Measurement error is considered to be the difference between a value measured and

the true value.

Types of gage variation:

• Systematic variation– Accuracy - improper calibration

– Reproducibility - different persons using same equipment with differenttechniques

• Periodic variation

– Stability - wear, deterioration, environment• Random variation

– Repeatability (unable to locate part to be measured)

Types of measurement variation

• Accuracy• Stability• Reproducibility

• Repeatability

Accuracy :

Difference between the true average and the observed average. (True average may beobtained by using a more precise measuring tool)

Stability:

The difference in the average of at least 2 sets of measurements obtained with a gage

over time.



Reproducibility:

Variation in average of measurements made by different operators using the same

gage measuring the same part.

Repeatability:The random variation in measurements when one operator uses the same gage to

measure the same part several times.

How do we improve gage capability?

• Reproducibility

– operator training, or

– more clearly define measurement scale available to the operator• Repeatability

– gage maintenance

– gage redesign to better fit application



Accuracy of Measurement

• Broken down into three components:

1. Stability: the consistency of measurements over time.2. Accuracy: a measure of the amount of bias in the system.

3. Linearity: a measure of the bias values through the expected range ofmeasurements.

Precision of Measurement

• Precision, Measurement Variation, can be broken down into two components:

1. Repeatability (Equipment variation): variation in measurements under exact

conditions.2. Reproducibility (Appraiser variation): variation in the average of measurements

when different operators measure the same part.

MSA Process Flow

1. Preparation for study2. Evaluate stability

3. Evaluate resolution

4. Determine accuracy

5. Calibration6. Evaluate linearity

7. Determine repeatability and reproducibility

Preparation for Study • Objective: establish process parameter for the study.

• Process:1. Determine which measurement system will be studied.

2. Establish test procedure.

3. Establish the number of sample parts, the number of repeated readings, and the

number of operators that will be used.4. Choose operators and sample parts

Evaluate Stability • Objective: evaluate measurement system to determine if the system is in statistical

control.

• Procedure:1. Choose sample standards.

2. Measure sample standards three to five times.

3. Plot data on a x-bar and R chart.



• Analysis:

1. Determine if process is in control.2. If process is unstable determine and correct the cause

Evaluate Resolution

• Objective: determine if the measurement system can identify and differentiatebetween small changes in the given characteristic.

• Process:

1. Choose a sample standard.

2. Measure the sample standard three to five times.3. Repeat the process 10 to 25 times.

4. Plot data on a R chart.

• Analysis

1. The resolution is inadequate if:- There are only one, two, or three possible values for the range, or

- There are only four possible values for the range when n >= 3.

Determine Accuracy • Objective: determine the variation between the observed measurement and the

actual measurement of a part.• Process:

1. Choose sample standards.

2. Measure sample standards 15 to 25 times using the same measuring device, thesame operator, and the same setup.

3. Calculate x-bar

4. Calculate bias- Bias = Average – Reference Value

5. Calculate the upper and lower 95% confidence limit (CL).

• Analysis

1. If reference value is within the 95% CL then the bias is insignificant.

2. If reference value is outside the 95% CL then the bias is significant andmeasurement system must be recalibrated.

Calibration • Objective: to ensure the instrument is accurate, and measurement bias is

minimized.

• Process: calibrate instrument .

Evaluate Linearity • Objective: determine the difference between the obtained value and a reference

value using the same instrument over the entire measurement space.

• Process:



1. Choose three to five sample standards that cover the measurement space.2. Measure sample standards 15 to 25 times.

3. Calculate the average of the readings.

4. Calculate bias.

5. Plot reference values on x-y graph.6. Calculate slope of the linear regression line.

7. Calculate linearity and percent linearity.8. Calculate R2.

• Analysis

1. The closer the slope is to zero, the better the instrument.

2. R2 gives indication of how well the “best-fit” line accounts for variability in the

x-y graph.

Determine Repeatability and Reproducibility

• Objective: determine variation in a set of measurement using a single instrument

that can be credited to the instrument itself, and to the entire measurement system.

• Process

1. Generate random order for operators and parts to complete the run.

2. Repeat process for subsequent runs.

3. Have operators take measurements.

• Analysis:

1. Plot data2. Run ANOVA (analysis of variance) on data.

3. Calculate total variance.

4. Calculate % Contribution and determine if acceptable.5. Calculate % Contribution (R&R)

6. Calculate Process to Tolerance ratio (P/T) for repeatability.

7. Determine if P/T is acceptable.

Real World Application

- Gauge R&R study of automobile radiator manufacturer.

- After studying four characteristics of radiator components the following resultswere obtained:



• Any system having greater than 30% gauge R&R is considered inadequate. As

seen in Table 1, all four characteristics’ %R&R is inadequate.

• Investigation of the measurement system led to a subsequent reduction of %R&Rin three of the four characteristics to between 12% and 23%.

• Further investigation of the fourth characteristic, inlet hole diameter, led theexaminers to a manufacturing problem. The team discovered high ovality in the

inlet hole, which was caused by the cutting tool. The tool was modified to reduce

ovality.

• Benefits of the study.

1. Reduced measurement variation.

2. Increased operator confidence regarding their aptitude for conducting gauge R&Rstudies.

3. Paved the way for further studies within the firm.

An Exercise – Calculating EV, AV, R&R, and TV

• Given: EV = 5.15(s0) , AV = 5.25(s1)

R&R = √ (EV2 + AV2)

TV = √ (EV2 + AV2 + PV2)

• Where: s0 = gauge standard deviation = 0.05

s1 = true appraiser standard deviation = 0.1

PV = part-to-part variation = 0.02

• Calculate R&R and TV

• Is the calculated R&R acceptable?



Gage R&R Analysis – Analysis of Repeatability and

Reproducibility

This is a technique to measure the precision of gages and other measurement systems.

The name of this technique originated from the operation of a gage by different operators

for measuring a collection of parts. The precision of the measurements using this gageinvolves at least two major components: the systematic difference among operators and

the differences among parts. The Gage R&R analysis is a technique to quantify each

component of the variation so that we will be able to determine what proportion of

variability is due to operators, and what is due to the parts.

A typical gage R&R study is conducted as the following: A quality characteristic of

an object of interest (could be parts, or any well defined experimental units for the study)is selected for the study. A gage or a certain instrument is chosen as the measuring

device. J operators are randomly selected. I parts are randomly chosen and prepared for

the study.

Each of the J operators is asked to measure the characteristic of each of the I parts for

r times (repeatedly measure the same part r times). The variation among the mreplications of the given parts measured by the same operation is the Repeatability of the

gage. The variability among operators is the Reproducibility.

Gage repeatability and reproducibility studies determine how much of your observedprocess variation is due to measurement system variation. The overall variation is broken

down into three categories: part-to-part, repeatability, and reproducibility. The

reproducibility component can be further broken down into its operator, and operator bypart, components.

Gage R&R Studies

Gage repeatability and reproducibility (R&R) studies involve breaking the total gage

variability into two portions:

repeatability which is the basic inherent precision of the gagereproducibility is the variability due to different operators using the gage.



• Gage variability can be broken down as

More than one operator (or different conditions) would be needed to conduct the

gage R&R study.

Statistics for Gage R&R Studies (The Tabular Method)

• Say there are p operators in the study

• The standard deviation due to repeatability can be found as

Where

And d2 is based on the # of observations per part per operator.

Statistics for Gage R&R Studies (the Tabular Method)

• The standard deviation for reproducibility is given as

Where

d2 is based on the number of operators, p

Basic Terms

• EV= Equipment Variation (Repeatability)

• AV= Appraiser Variation (Reproducibility)

• R&R= Repeatability & Reproducibility

• PV= Part Variation

• TV= Total Variation of R&R and PV

• K1-Trial, K2-Operator, & K3-Part Constants

2222

ityrepeatabililityreproducibgageerror t measuremen σ σ σ σ +==

2

ˆd

Rityrepeatabil =σ

p

R R R R

p+++

=

21

2

ˆd

R x

ilityreproducib =σ

),,min(

),,max(

21min

21max

minmax

p

p

x

x x x x

x x x x x x R

=

=−=



• Generally two or three operators

• Generally 10 units to measure

• Each unit is measured 2-3 times by each operator

• Determine if reproducibility is an issue. If it is, select the number of operators toparticipate.

• Operators selected should normally use the measurement system.• Select samples that represent the entire operating range.

• Gage must have graduations that allow at least one-tenth of the expected processvariation.

• Insure defined gaging procedures are followed.

• Measurements should be made in random order.

• Study must be observed by someone who recognizes the importance of conducting

a reliable study.

Procedure for Performing R&R Study

• Calibrate the gage, or assure that it has been calibrated.• Have the first operator measure all the samples once in random order.

• Have the second operator measure all the samples once in random order.

• Continue until all operators have measured the samples once (this is Trial 1).• Repeat above steps for the required number of trials.

• Use GR&R form to determine the statistics of the study.

– Repeatability, Reproducibility & %GR&R

– Standard deviations of each of the above– % Tolerance analysis

• Analyze results and determine action, if any.

Variable Gage R&R

% R&R Results

<5% No issues

10%≤ Gage is OK

10% – 30% Maybe acceptable based upon importance of application, and

cost factorOver 30% Gage system needs improvement/corrective action

Gasket Thickness Study

PT1 PT2 PT3 PT4 PT5 PT6 PT7 PT8 PT9 PT10 OP/TRIAL0.65 1.00 0.85 0.85 0.55 1.00 0.95 0.85 1.00 0.60 A1

0.60 1.00 0.80 0.95 0.45 1.00 0.95 0.80 1.00 0.70 A20.55 1.05 0.80 0.80 0.40 1.00 0.95 0.75 1.00 0.55 B1

0.55 0.95 0.75 0.75 0.40 1.05 0.90 0.70 0.95 0.50 B2

0.50 1.05 0.80 0.80 0.45 1.00 0.95 0.80 1.05 0.85 C10.55 1.00 0.80 0.80 0.50 1.05 0.95 0.80 1.05 0.80 C2



X bar& R Minitab Example Using Aiag49:mtw Data File Specification: 0.6 - 1.0 mm

Process Variation: 1.6 mm Reference QS Measurement System Analysis Manual

Gage R&R Study for Thickness – XBar/R Method

Source Variance StdDev 5.15*Sigma

Total Gage R&R 2.08E-03 0.045650 0.235099Repeatability 1.15E-03 0.033983 0.175015

Reproducibility 9.29E-04 0.030481 0.156975

Part-to-Part 3.08E-02 0.175577 0.904219

Total Variation 3.29E-02 0.181414 0.934282

Gage R&R Study for Thickness – XBar/R Method

Source %Contribution %Study Var %Tol %Process

Total Gage R&R 6.332 25.164 58.77 14.69

Repeatability 3.509 18.733 43.75 10.94Reproducibility 2.823 16.802 39.24 9.81

Misc:

Tolerance:

Reported by :

Date of s tudy:

Gage name:

1.11.0

0.90.8

0.70.60.50.40.3

321

Xbar Chart by Operator

S a m p l e M e a n

X=0.80753.0SL=0.8796

-3.0SL=0.7354

0.15

0.10

0.05

0.00

321

R Chart by Operator

S a m p l e R

a n g e

R=0.03833

3.0SL=0.1252

-3.0SL=0.000

10987654321

1.1

1.0

0.9

0.8

0.7

0.6

0.5

0.4

Gasket

OperatorOperator*Gasket Interaction

A v e r a g e

1

2

3

321

1.1

1.0

0.9

0.8

0.7

0.6

0.5

0.4

Operator

Response by Operator

10987654321

1.1

1.0

0.9

0.8

0.7

0.6

0.5

0.4

Gasket

Response by Gasket

%Total Var

%Study Var

%Process

%Toler

Part-to-PartReprodRepeatGage R&R

200

100

0

Components of Variation

P e r c e n t

Gage R&R (Xbar/R) for Thickness



Part-to-Part 93.668 96.782 226.05 56.51

Total Variation 100.000 100.000 233.57 58.39

Number of distinct categories = 5

Calculation Explanation

• 5.15 Sigma = the factor standard deviation. 5.15 was developed empirically to

approximate the gage population distribution variation.×5.15

• % Contribution = repeatability variance/ total variation variance.×Percentcontribution of each factor based upon the variance. Repeatability = 100

• % Study Variation = 5.15 repeatability standard deviation/ 5.15 total variation

standard deviation.× the total variation standard deviation. Repeatability = 100 ×

the factor standard deviation divided by 5.15 ×5.15

• % Tolerance 5.15 repeatability standard deviation/tolerance.× the factor

standard deviation divided by the tolerance. Repeatability = 100 ×= 5.15

• % Process Variation = 5.15 x the factor standard deviation divided by theprocess variation. Repeatability = 100 x 5.15 repeatability standard deviation/

process variation.

• Number of Distinct Categories = part standard deviation divided by the total

gage R&R standard deviation times 1.41.



MODULE NO -26

Statistical Theory of Tolerances

Purpose of Specification .

A specification is a definition of a design. The design remains a concept in the mindof the designer until he define it through verbal description, sample, drawing, writing etc.

It defines in advance what the manufacturer expects to make. It defines what the

consumer can expect to get. The specification serves as an agreement between

manufacturer and consumer on the nature of characteristics of the product.

It is helpful to recognize the distinction between a design specification and an

inspection specification. The design specification deals with what is desired in amanufactured article, i.e. it deals with the specification function. In contrast, the

inspection specification deals with means of judging whether what is desired is actually

attained, in other words it deals with inspection function (quality of conformance).

Specifying the tolerance. It is practically impossible to manufacture one article exactly

like another or one batch like another. Variability is one of the fundamental concepts ofmodern quality control. Therefore, the range of permissible difference in dimension have

been standardized under the name limits. The limit of size for dimension or a part are two

extreme permissible sizes for that dimension (high limit and low limit).

Design engineer have a tendency to specify tight tolerance for the following reason.

1. Lack of time. Tolerance are to be set up on many dimension, therefore, the designersmay not have sufficient time to give much attention to tolerances on all dimension.

Therefore, to be on the safer side the designers are tempted to specify much closertolerance

2. The concept of factor of safety. Designers have been taught to allow for the

unexpected or the unusual i.e. overloading of the machine, use of unintended purpose,

change in the condition of use . The designers may assume more factor of safety toanticipate failure of conformance by the shop.

3. Setting tolerances assuming ideal conditions. Design engineering seems to specifytolerance with reference to some what idea l conditions, assuming good machine, well

trained operators, skilled supervision and good working condition. Or they use reference

tables which may tacitly assume such factor. In actual practice, nearly ideal conditionmay be obtained during some part of the process, but almost never for say, extended

period of time.

4. Lack of knowledge of the production process. The designers may not have sufficient

knowledge about the production process. Therefore, they may design the product with

little or no critical consideration of the various production problems involved to meet the

tolerance.



5. Lack of information about the process capability. In some case the designers do not

have information regarding the production facilities available in the plant, their conditionand process capability.

6. Lack of awareness of the quantitative effect of tolerance decision on factory

economy

7. Tendency of shop personnel to loose up the tolerances. The design may beconscious of the difference between the blue print tolerance and those which are actually

enforced. Therefore, in order to get what they think they need, they tend to specify loser

tolerance then they believe necessary.

Definition: It is not possible to manufacture each & every item identically. It is a customary to

allow a certain variability in the measured quality characteristic called the tolerance.Generally in any industry design section specifies what is to be produced &sets the

dimension & tolerance of the characteristic. The responsibility of the manufacturing dept,is to manufacture the items according to the specification laid down by the designdepartment. The inspection dept. checks whether the product is meeting the specification

given by the design dept, unless there is a proper co-ordination it is difficult to

manufacture the item exactly.

while establishing the specification limits the fall pts must be considered

1. Functional utility of the product.

2. Capability of the product and process.3. Inspection procedures.

Tolerance spread: T = (U-L)It is set by the engg. design section to design the mini & max values available for the

product to work properly.

Theorems in statistical Tolerance

• Addition theorem

When the components are added together linearly



mean of the assembly = µ assembly = µ A + µ B

std deviation assembly = assembly =

If n components are assembled together linearly, then the mean of the wholeassembly = µ assembly = µ A + µ B + - - -µ n

std. deviation of assembly = assembly =

1. Difference theorem

When the components are mating together, eg: shaft and the bearing.

Assumption to be made for the above formula in statistical tolerance

1. The component dimension are independent of each other & are assembled

randomly.

2. The component dimension must follow a normal distribution3. A control chart has to be maintained for each of the component dimension /

characteristic.4. The actual average of each component is equal to the nominal value stated in the

specification.

22

B A σ σ +

222

n B A σ σ σ −−−++



Conventional tolerance Statistical tolerance

1. 100% of the interchangeability of thecomponents is possible for assembly

2. The tolerance of the interacting

dimension are smaller than the

necessary.

1 . A small % of the assembly will falloutside the tolerance limit but this can be

corrected with a selective assembly

2. This method permits a larger tolerance

On the interacting dimension

3. No assumption are necessary

4. No special process control procedures

are required

3. The interacting dimension must be

independent of each other & each

characteristic must follow a normaldistribution

4. The process average of each

components must be maintained at anominal dimension value (target value)


Statistical Tolerance

Use of statistical method of tolerance can lead to economic production, when we are

dealing with interacting dimensions. Interacting dimensions are those which mate ormerge with other dimensions to create a final result.

A dimension of an assembled product may be the sum of the dimensions of severalparts. Or an electrical resistance may be the sum of several electrical resistances of parts.Or a weight may be the sum of number of weights of parts. In such situations it is

necessary to determine the relationship of the tolerances of the sum.

The statistical theory of tolerance results in larger component tolerances with no

change in manufacturing process and no change in the assembly tolerance. Larger

tolerances, increase the production

Output, minimize waste of material and productive effort and are generally

responsible for reduction in manufacturing costs. This is the effect of statistical approach.

If an overall tolerance is fixed but not being met, then the problem is which

component tolerances should be reduced to meet overall tolerance. The statistical

theorem can help to determine which of the component tolerances have the greatest effecton the overall tolerances. This information, when coupled with economic considerations

on achieving a smaller tolerance, can form basis for a decision.



A risk involved in the use of statistical theorem is that : it is possible that an assembly

will result which falls outside of the assembly tolerance. However, the chance can becalculated and a judgment made on whether or not to accept risk. The probability that

assembly length will fall outside the tolerance limits can be found out by analyzing the

area under the normal curve for assembly lengths.

1. The component dimensions are independent and the components are assembled

randomly.2. Each component dimension should be normally distributed. Some departure from this

assumption is permissible.

3. The actual average for each component is equal to the nominal value stated in thespecification.

Problem 1.

Manufacturer A produces a metal piece the dimension of which is normally

distributed with based on a subgroup size of 4. Themanufacturer B produces a 2nd metal pieces which is also normally distributed withbased on a subgroup size 9 of a company C purchases

these 2 parts & assembles them together to obtain a combined dimension of 15 cm. What

% of the combined assembly.Would you expect to have the dimension is excess of 15.006 cm.

Solution :

from the table, for a subgroup size of 4, d2=2.059

from the table, for a subgroup size of 9, d2=2.97

cm Rcm X A A 004.0&5.8

1

==

cm Rcm X B B 005.0&5.61

==

4004.0&5.81

=== A A A ncm Rcm X

9005.0&5.61

=== B B B ncm Rcm X



The % of assembled items which have dimensions in excess of 15.006 is

The tables, for Z=2.33, the prob. Is 0.9901 i.e.. 99.01%

i.e. 100 -99.01 = 0.99%

There fore 0.99% assembled items will have dimensions more than 15.006.

Problem-2

Two Parts A&B are received in an assembly operation, where each part is

permanently attached to the other . When the combined width of the parts dose not meet

the specification of 10 ±0.02 the assembled product must be scrapped. If the width ofthe part B is normally distributed with

and the width of the part A is also normally distributed with

33.2002568.0

15006.15

=−

=

−=

−=

Z

x Z

x Z

C

c

σ

µ

σ

µ

012.0&5.6 11 == B B

X σ

008.0&5.3 11 == A A

X σ



The assembly is at random. Determine the %of the assembled product that has to be

scrapped

Since it is symmetric, the % of items of that are scrapped (The scrap & rework) is2x0.0823 = 0.1646

I.e. = 16.46%



MODULE NO -27


Hanna Varnum Diagram

A Hanna-Varnum Diagram is used to determine the probability of interference when

2 normal distributions overlap. This diagram is plotted for the ratio of differences

between the 2 means to a smaller standard deviation V/s ratio of standard deviations.

Steps involved

Step 1: Divide the larger standard deviation by smaller standard deviation.

Step 2: Locate this value on the lower scale (x-axis)Step 3: subtract the average ie difference between the 2 means

Step 4: Divide this difference by a smaller standard deviation.

Step 5: Indicate this value on the vertical scale.

Step 6: Find the point which is above the lower scale value to the right of the verticalscale value.

Step 7: Determine the interference risk from the % curves passing nearest to this pointwhich is shown in the graph.

Interference Tolerance

It is defined as a negative clearance. Interference exists in a situation where the shaftdiameter > than the bearing diameter. If the negative clearance is present, considered it

as zero.

Problem-3

Two mating parts X & Y have an average clearance of 0.015mm. control chartanalysis indicates that the standard deviations of X & Y are 0.025 and 0.075mmrespectively. Find the probability of interference between the 2 distributions and also

probability of clearance being greater than 0.0175 assume normal distribution and

random assembly



The Probability of interference between the Two distribution is

Since it is an interference, it is –ve clearance i.e 0

From the normal tables, value of z = 0.4286

ie 42.86%

42.86% assembled items have probability of interference between 2 distribution

Problem-4

Dimensions of Two mating parts E and F are normally distributed with averages of251.0mm and 250.0mm and standard deviations of 0.1mm and o.3mm respectively. If the

parts are assembled randomly, what percent of the assemblies will have (a) clearance

greater than 1.2mm and (b) no defective parts if the specifications of E and F are 251.0±0.2mm and 250.0±0.5mm respectively

18.0

01899.0079.0

015.00.

−=

−=−

= zei

Clearance



Solution:

given(a) Average clearance

To Find the area under the curve

Corresponding area from table =0.7357(a) For part E, Xmax = 251.2 mm;

Xmin = 250.8 mmXbarE = 251mm

and E = 0.1 mm

To find the area under the curve

Corresponding area = 0.9773

mm X X F E 1=−=

22 )()( F E assembly σ σ σ +=

3162.0)3.0()1.0( 22 =+=

mmmm X

mmmm X

F F

E E

3.00.250

1.00.251

==

==

σ

σ

σ

X X F Factor i

−=

1

21.0

2512.251+=

−=

21.0

2518.2502

−=−

=F Factor




Therefore shaded area = 0.9773 – 0.0228

= 0.9545

Therefore, Probability that the part E will be non- defective = 0.9545.

For part F, Xmax = 250.5

Xmin = 249.5

XbarF = 250F = 0.3


Therefore, shaded area = 0.9522 -0.0478 = 0.9044.

[ Probability that the part F will be non-defective] The probability that assembly of parts

E and F will be non-defective= PE x PF

= 0.9545 x 0.9044 = 0.86324

i.e. 86.324% assemblies will have no defective part

667.03.0

2505.2501

−=−

=F



Problem-5

Control chart analysis indicates that the standard deviation of the 2 mating parts C &

D are 0.0016 & 0.004 cm respectively. It is desired that the probability of smaller

clearance than 0.004 cm should be 0.005. What distribution between the average

dimensions of C & D should be specified by the designer. Assume the data followsnormal distribution & random assembly with this distribution specified what is the

probability that the parts assembled at random will have a greater clearance than 0.024

cm.

Solution :

For the probability of 0.005

The Z value from normal tables is -2.57

(b) % of items that have clearance > 0.024

004.0

0016.0

=

=

D

C

σ

σ

cm

x Z

C

C

C

C

015.0

0043.0

004.057.2

=

−=−

−=∴

µ

µ

σ

µ



From the tables, the probability less than 0.024 is 0.9817

There fore % of assembled items having clearance > 0.024

Is 1-0.9817

=0.0183

i.e., 1.83%.

Problem 6.

Control chart analysis indicates that the standard deviation of the 2 mating parts willhave identical values of 0.0013cm. It is desired that the probability of clearance less than

0.003 cm should be 0.002. What distribution between the average values of these

dimensions should be specified by the designer. Assume normal distribution & random

assembly with this distribution specified, what is the probability that the assembled items

will have clearance > 0.009cm.

Solution :

(a) For the probability of 0.02, the Z value from the normal tables is -2.05

(b) % of assembled items having clearance greater than 0.009 cm

0013.021

==σ σ

( )

cm

comb

00184.0

0013.022

2

2

2

1

=

=

+= σ σ σ



From the normal tables, Z=1.21,

the probability = 0.8869 i.e. 88.69%

There fore % of items having clearance > 0.009 cm is 100 – 88.69 = 11.31%

Problem -7

2 mating parts A&B have the dimension of 2.61&2.615cm respectively , control chart

analysis indicates that the standard duration of A&B are 0.0012&0.0015cm respectively.

With distribution of A&B are normal & centered about the specified dimension &the

parts are assembled at random, find the probability of interference both the 2 distribution

& also calculate the probability of clearance >0.01.

solution :

probability of interference better the 2distribution is

Since it is an –ve clearance interference =0

From the tables probability of interference =0.0047

I.e. 0.47% items have the probability of interference.

C

x Z

σ

µ −=

60.200192.0

005.00−=

−= Z



(b) Probability of clearance > 0.01

From the tables, probability of clearance > 0.01 is 0.0047

i.e. 0.47%

Probability of assembled items having clearance >0.01is

= 1-0.0047

=0.4953

i.e. 99.53%

Problem -8Control chart analysis indicates that the SD of the 2 mating points are 0.008&0.02cm

resp. It is desired that the prob, of smaller clearance than 0.002cm should be 0.005what

distributions, between average dimension of C&D should be specified by the designer

with this distributions , specified, what is the prob, that the 2parts assembled will have

clearance > 0.12

Solution:

a) For probability of 0.005 the value of Z= -2.57

cm

B Acomb B

A

0215.0

)02.0()008.0(

02.0008.0

22

22

=

+=

+===

σ σ σ σ σ



(b) % of items having clearance >0.12

From the tables, the probability is 0.9982

Therefore % of items having clearance >0.12

= 1-0.9982

= 0.0018i.e., 0.18%

Problem 9.The gross weight of cement bags with cement at the terminal dispatch stage at a

cement factory was known to follow a ND with mean =51 kg & SD = 400 gms. It is

known that the weight of cement bags before filling follows a ND with mean = 500 gms

& SD = 20 gms. If the net weight of the specimen is 50 kg minimum can it be assumedthat all the bags have the minimum net weight. If not, what % of the bags have

underweight. What should be the minimum mean gross weight required in order to have

no defect net weight.Solution :

22

E N g

E N g

σ σ σ µ µ µ

+=+=

kgweight net imum

kgkg

kgkg

E g

E g

50min

02.04.0

5.051

=

==

==

σ σ

µ µ

kg

E g N

E N g

5.505.051 =−=−=

+=

µ µ µ

µ µ µ

( ) ( )

kg N

E g N

E g N

E N g

399.0

02.04.02222

222

222

=∴

−=−=

−=

+=

σ

σ σ σ

σ σ σ

σ σ σ



(a) % of cement bags falling below 50 kg net weight

From the normal tables for Zo-1.25, the probability is 0.1056 ie 10.56% of the bags

are falling below 50

(b) When LSL=LNTL, there are no rejections LSL=LNTL

The minimum mean gross weight required in order to have no deficit net weight.

51.197+0.5 = 51.697kg

11

3σ −= X

( )

197.51

399.0350

350

350

=

+=

+=

−=

n

N N

N n

µ

σ µ

σ µ



MODULE NO -28


Application of STT in Other Areas

Setting Specification Limits on Discrete Components

It is often necessary to use information from a process capability study to set

specifications on discrete parts or components that interact with other components to

form the final product. This is particularly important in complex assemblies, or to preventtolerance stack-up where there are many interacting dimensions. This section discusses

some aspects of setting specifications. on components to ensure that the final product

meets specifications.

Linear Combinations:

In many cases, the dimension of an item is a linear combination of the dimensions ofthe component parts. That is, if the dimensions of the components are x 1, x2, ……, xn,

then the dimension of the final assembly is y = a1x1 + a2x2 + …. + anxn

If the xi are normally and independently distributed with mean µ i and variance i2,

then y is normally distributed with mean µ y=in =1and

variance 2y= i

n=1 ai2 I

2. There fore, if µ I and i2 are known for each component, the

fraction of assembled items falling outside the specifications can be determined.

Problem-1.A linkage consists of four components as shown in Fig. The lengths of x 1, x2, x3 and

x4 are known to be x1~ N(2.0, 0.0004), x2 ~ N(4.5, 0.0009), x3 ~ N(3.0, 0.0004), and x4 ~

N(2.5, 0.0001). The lengths of the components can be assumed independent, becausethey are produced on

Different machines. All lengths are in inches

The design specifications on the length of the assembled linkage are 12.00 ± 0.10. Tofind the fraction of linkages that fall within these specification limits, note that y is

normally distributed with mean µ y =2.0 + 4.5 + 3.0 + 2.5 = 12.0 and variance

x1 x2 x3 x4

y

Alinkage assembly with four components.



2y =0.0004 + 0.0009 + 0.0004 + 0.0001= 0.0018

To find the fraction of linkages that are within specification, we must evaluate

P{11.90 y 12.10} =P {y 12.10} – P {y 11.90}

= (2.36) - (-2.36)

= 0.99086 – 0.00914= 0.98172

There fore, we conclude that 98.172% of the assembled linkages will fall within the

specification limits.

Problem-2Consider the assembly shown in Fig. Suppose that the specifications on this assemblyare 6.00 ± 0.06 in. Let each component x1, x2, and x3 be normally and independently

distributed with means µ 1=1.00 in., µ 2=3.00 in., and µ 3=2.00 in., respectively. Suppose

that we want the specification limits to fall inside the natural tolerance limits of theprocess for the final assembly so that Cp= 1.50, approximately, for the final assembly,

this implies that about 7 ppm defective is allowable.

The length of the final assembly is normally distributed. Furthermore, if the allowable

number of assemblies nonconforming to specifications is 7 ppm, this implies that the

natural tolerance limits must be located at µ ± 4.449 y. Now µ y = µ 1 + µ 2 + µ 3 =1.00 +3.00 + 2.00 = 6.00, so the process is centered at the nominal value. Therefore, the

maximum possible value of y that would yield the desired value of Cp is

That is if y 0.0134, then the number of nonconforming assemblies produced will

be less than or equal to 7 ppm. Now let us see how this affects the specifications on the

individual components. The variance of the length of the final assembly is

−Φ−

−Φ=0018.0

00.129.11

0018.0

00.1210.12

x1 x2 x3

y

Assemble for Example

0134.049.4

06.0

== yσ

( ) 00018.00134.022

3

2

2

2

1

2 =≤++= σ σ σ σ y



Suppose that the variances of the component lengths are all equal; that is ,

(say). Then

And the maximum possible value for the variance of the length of any component is

Effectively, if 2 0.00006 for each component, then the natural tolerance limits for

the final assembly will be inside the specification limits such that Cp = 1.50.

This can be translated in to specification limits on the individual components. If weassume that the natural tolerance limits and the

Specification limits for the components are to coincide exactly, then the specification

limits for each component are as follows

Problem-3.

A shaft is to be assembled into a bearing. The international diameter of the bearing isa normal random variable – say, x1 – with mean µ 1 = 1.500 in. and standard deviation 1 = 0.0020 in. The external diameter of the shaft – say, x2 – is normally distributed with

mean µ 2 = 1.480 in. and standard deviation 2 = 0.0040 in. The assembly is shown in

figure. When the two parts are assembled. Interference will occur if the shaft diameter is

larger than the bearing diameter – that is, if y = x1 – x2 < 0 Note that the distribution of yis normal with mean

µ y= µ 1 - µ 2 = 1.500 - 1.480 = 0.020

and variance

Therefore, the probability of interference is

P {interference} = P{y,0}

22

3

2

2

2

1 σ σ σ σ === 22 3σ σ = y

00006.03

00018.0

3

2

2 === yσ

σ

0232.000.200006.000.300.2:

0232.000.300006.000.300.3:

0232.000.100006.000.300.1:

3

2

1

±=±

±=±

±=±

x

x

x

( ) ( ) 00002.00040.00020.0222

2

2

1

2

=+=+= σ σ σ y

( )

( ) ppm4000004.0

47.4

00002.0

020.00

=

−Φ=

−Φ=



Which indicates that very few assemblies will have interference. In problems of this

type, we occasionally define a minimum clearance – say, C – such thatP {interference < C} =

Thus, C becomes the natural tolerance for the assembly and can be compared with the

design specification. In our example, if we establish = 0.0001 ( i.e., only 1 out of

10,000 assemblies or 100 ppm will have clearance less than or equal to C), then have

Which implies that C = 0.020 – (3.77) 0.00002 = 0.0034. That is, only 1 out of 10,000

assemblies will have clearance less than 0.0034 in.

Problem-4.Figure below shows an assembly consisting of four components. The lengths of x1,

x2, x3 and x4 are known to be x1~ N(2.5, 0.032), x2 ~ N(2.4, 0.022), x3 ~ N(2.4, 0.042),and x4 ~ N(3.0, 0.012). The lengths of the components can be assumed independent,

because they are produced on different machines. All lengths are in cm.

The design specifications are 10.25 ± 0.15. To find the fraction of linkages that fall

within these specification limits, note that y is normally distributed with mean

µ y =2.5 + 2.4 + 2.4 + 3.0= 10.3 cm and variance

2y = 0.03

2 + 0.02

2+ 0.04

2+ 0.01

2= 0.003 cm

2

To find the fraction of components that are within specification, we must evaluate

P{10.10 y 10.40} =P {y 10.40} – P {y 10.10}

= (1.818) - (-3.636)= 0.9655 – 0.0001

= 0.9654

71.300002.0

020.0

0001.0

−=−

−=−

C

Z C

y

y

σ

µ

x2 x1

Assembly of shaft and a bearing

X

X

X

−Φ−

−Φ=

055.0

30.1010.10

055.0

30.1040.10



There fore, we conclude that 96.54% of the assembled components will fall within the

specification limits.


Problem 5.Control chart analysis indicates that the standard deviation of the 2 mating parts will

have identical values of 0.0013cm. It is desired that the probability of clearance less than0.003 cm should be 0.002. What distribution between the average values of these

dimensions should be specified by the designer. Assume normal distribution & random

assembly with this distribution specified, what is the probability that the assembled items

will have clearance > 0.009cm.

Solution :

(a) For the probability of 0.02, the Z value from the normal tables is -2.05

(b) % of assembled items having clearance greater than 0.009 cm

From the normal tables, Z=1.21, the probability = 0.8869 i.e. 88.69%

There fore % of items having clearance > 0.009 cm is 100 – 88.69 = 11.31%

0013.021

==σ σ

( )

cm

comb

00184.0

0013.02 2

2

2

2

1

=

=

+= σ σ σ

cmC

C

006772.0

00184.0

003.005.2

=

−=−

µ

µ

0.003

2 %

00184.0

?

=

=

C

C

σ

µ

21.1

00184.0

006772.0009.0

=

−=

−=

Z

x Z

C

C

σ

µ

0..009

00184.0

006772.0

=

=

C

C

σ

µ



problem -7 2 mating parts A&B have the dimension of 2.61&2.615cm respectively, control chart

analysis indicates that the standard duration of A&B are 0.0012&0.0015cm respectively.

With distribution of A&B are normal & centered about the specified dimension &the

parts are assembled at random, find the probability of interference both the 2 distribution

& also calculate the probability of clearance >0.01.

Probability of interference better the 2distribution is

Since it is an –ve clearance interference =0

From the tables probability of interference =0.0047 I.e. 0.47% items have theprobability of interference.

From the tables, probability of clearance > 0.01 is0.0047i.e. 0.47%

Probability of assembled items having clearance >0.01is= 1-0.0047

=0.4953

i.e. 99.53%

C

x Z

σ

µ −=

60.200192.0

005.00−=

−= Z



problem -8 Control chart analysis indicates that the SD of the 2 mating points are 0.008&0.02cm

resp. It is desired that the prob, of smaller clearance than 0.002cm should be 0.005what

distributions, between average dimension of C&D should be specified by the designer

with this distributions , specified, what is the prob, that the 2parts assembled will have

clearance > 0.12

From the normal tables for Zo-1.25, the probability is 0.1056 ie 10.56% of the bags arefalling below 50

(b) When LSL=LNTL, there are no rejections LSL=LNTL

The minimum mean gross weight regds. In order to have No deficit net weight.

51.197+0.5 = 51.697kg

11

3σ −= X

( )

197.51

399.0350

350350

=

+=

+=−=

n

N N

N n

µ

σ µ σ µ



MODULE NO -29

Reliability

Introduction:The concept of reliability has been known for a number of years, but it has assumed

greater significance ,and importance during the past decade, particularly due to impact ofautomation, development in complex missile and space programmers. The manufacture

of highly complex equipment has served to focus greater attention on reliability. The

complex products, equipments are made up of hundreds or thousands of components

whose individual reliability determines the reliability of the entire equipment. Usingvarious types of materials and fabricating operations, the industry has to build reliable

performance into equipment and the products manufactured.

As regards to the Indian industry, the reliability concept is yet to find a footing. The

solutions to many of the problems of quality and economy remain handicapped because

of inadequate appreciation of the reliability principles and techniques. However,

reliability is only one of the tools of the management which must be supplemented byother tools like quality control and design of experiments for the solution of problems of

quality and cost.

Quality Control and Reliability

Quality control maintains the consistency of the product and thus affects reliability.

But it is entirely a separate function. Reliability is associated with quality over the longterm whereas quality control is associated with the relatively short period of time,

required for manufacture of the products. The task of reliability is to see that in a product

design, full account has been taken of every contingency which may cause a breakdownin use and to forecast the components or assemblies that are likely to become defective in

service. However, the equipment is designed, still it may be unreliable, if somecomponent has not been fully evaluated under all service conditions, even if the

production standards have been maintained by quality control during manufacture.

Need for a reliable product

The reliability of a system, equipment or product is very important aspect of qualityfor its consistent performance over its expected life span. In fact, Uninterrupted service

and hazard free operation is the essential requirement of large complex systems like

electric power generation and distribution plants or communication network such asrailways, aero plane, automobile vehicles etc. In these cases a sudden failure of even a

single component, assembly or system results in a health hazard, accident, or interruption

in continuity of service.

Thermal power plants provide electric power for domestic, commercial, industrial and

agricultural use. Reliability problems may cause shut down or reduced generation of

power resulting in load shedding and many other problems including loss of productiveactivities. Failure of anyone system of an air-craft may result in forced landing or an

accident. Sudden stoppage of suburban railway train due to fault in the single system

faulty carriage, interruption in the power supply or faulty track, sets up a chain of events



leading to disruption of service or accidents. Similarly, sudden failure of a car break

system while it is running may cause severe accident. Unpredicted failure of a singlecritical component may be cause of anyone of the above.

What is true of power plants, air-crafts, railways etc. is also true for other products

like washing machine, mixer grinder, T.V. sets, Refrigerators etc. though failure of suchproducts may cause inconvenience on a smaller scale. The problem of assuring and

maintaining has many responsible factors, including original equipment design, control ofquality during manufacture, acceptance inspection, field trials, life testing and design

modifications. Therefore, deficiencies in design and manufacture of products which go to

build such complex systems needs to be detected by elaborate testing at the development

stage and later corrected by a planned programme of maintenance.

Definitions of Reliability

Reliability is ordinarily associated with the performance of the product. However,there would be little point in having an electric lamp which may light at the time of

purchase but which may burn off after 200 hours of use. Reliability is the probabilitythat a device will perform its intended function satisfactorily without failure for the statedperiod of time under the specified operating conditions.

In the above definition there are 4 factors which are essential to the concept of

reliability. Whenever the customer purchases a product he expects that it should givesatisfactory performance over a reasonably long period. Hence, what is important is that a

product should function and continue to function for a reasonable time. In practice, in

majority of the cases, it may not be possible to test each and every product for its life orother performance requirements. Nevertheless, it is a well known experience that each

individual unit of product varies from the other units; some may have relatively long life.

In view of the existence of this variation,

Reliability is the probability of a product functioning in the intended manner over

its intended life under the environmental conditions encountered.

From this definition, there are, four factors associated with reliability. These are :

1) Numerical value 2) Intended function 3) Life 4) Environmental condition

The introduction of this element of probability really makes the quantitative

measurement of reliability possible. In other words, such measurements help to make

reliability a number-a probability-that can be expressed as a standard.The second consideration for a product to be reliable is that it must perform a certain

function or do a certain job when called upon. The phrase 'functioning in the intended

manner' (satisfactory performance) implies that the device is intended for certainapplication.

For example, in the case of electric iron, the intended application is that of applying

intended degree of heat to the various types of fabrics. If instead it is used to keep a room



at a certain temperature, the electric iron might be inadequate because of the greater area

to be heated and the change in environment.

The third consideration for a product to be reliable is that it of time which ensures that

the product is capable of working satisfactorily throughout the expected life. The fourth

consideration for a product to be reliable is that of the environment conditions whichhave to be viewed broadly so as to include storage and transport conditions. Since these

conditions too have significant effect on product reliability. When an equipment workswell, and works whenever called upon to do the job for which it is designed, such

equipment is said to be reliable. Failure is defined as the inability of an equipment not to

breakdown in operation.

The causes of unreliability of the product are many: one of the major causes is the

increasing complexity of product. The multiplication law of probability Illustrates this

very simply. Given an assembly made up of five components, each of which has areliability of function of 0.95, the reliability of function of assembly is (0.95)5 or about

0.78. Many assemblies which are electronic in nature involve thousands of parts (aballistic missile has more than 40,000 parts). Therefore, to have reasonable chance ofsurvival for such assemblies the component reliability is of prime importance.

Basic Elements of Reliability

The basic elements required for an adequate specification or definition of reliabilityare as follows:

1. Numerical value of probability.

2. Statement defining successful product performance.3. Statement defining the environment in which the equipment must operate.

4 Statement of the required operating time.

5. The type of distribution likely to be encountered in reliability measurement.Reliability follows the

distribution of Poisson

where mean life

T required life

Failure Pattern for Complex Product Complex products often follow a familiarpattern of failure. When the failure rate (number of failures per unit time) is plotted

against a continuous time scale, the resulting chart is known as "bath tub curve" (because

of its shape).

θ T

e_

θ

ZONE-1

randomfailure

ZONE-2 ZONE-3

wear out failureearlyfailure F

a

i

l

u

r

e

/

F

a

i

l

u

r

e

r

a t

e



This curve exhibits three distinct zones. These zones differ from each other in

frequency of failure and in the cause of failure pattern. These are as follows:

1.Early failure period: (or burn in or the debugging period). This is characterized by

high failure rates. It begins at the first point during manufacture

That total equipment operation is possible and continues for such a period of time as

permits (through maintenance and repairs), the elimination of marginal parts initiallydefective though not inoperative and unrecognizable as such until premature failure

Commonly, these are early failures resulting from defect in manufacturing, or other

deficiencies which can be detected by debugging, running on or extended testing.Failures in this zone are due to one or more of the following causes: ( i.e., assignable

causes)

• Design deficiency• Manufacturing error

2. Random failure period: The constant failure rate period. It is characterized by a moreor less constant failure rate. This is the rate at which the normal usage of the product

occurs without any expectation of failures. Failures in this zone are due to chance causes.

These are chance failures which may result from the limitations inherent in the design

plus accidents caused by usage or poor maintenance or hidden defects which escapeinspection. The period from A to B is the normal operating period in which the average

failure rate remains fairly constant.

3.Wear out period: These are failures due to abrasion fatigue, creep, corrosion,

vibration etc., e.g., the metal becomes embrittled, the insulation dries out. A reduction in

failure rate requires preventive replacement of these dying components before they resultin catastrophic failure Failures in this zone are due to one or more of the following

causes:

• Ageing• Reduction

• Wear & tear.

Achievement of Reliability. There are five effective areas for the achievement ofreliability of the product. They are

(I) Design

(ii) Production

(iii) Measurement and testing

(iv) Maintenance and

(v) Field operation.

Design is the main cause of unreliability and a greater percentage of causes of

unreliability can be traced out in this area.



Designing for Reliability

The following factors should be considered for achieving a reliable design:

1. Simplicity of product. The design should be as simple as possible. Error rate is

directly proportional to complexity. The greater the number of components the greater

the chance of failure. Increased reliability is a natural by-product ofequipment/simplification.

2. Derating : Derating means providing a large safety margin. It is also used as a method

of achieving design reliability. For example, a material with tensile strength of 10,000

kg/cm2 might be prescribed where only 7,000 kg/cm2 is required.

3 . Redundancy. Redundancy is the provision of stand-by or parallel components or

assemblies to take over in the event of failure of the primary item.Even though we use

most reliable components and keep their number a minimum, there may be one or twosuch components which may have lesser reliability.

To overcome this, more number of such components are included and so arranged thatthe whole equipment will continue to survive so long as at least one of these components

survives. This technique is known as redundancy. Auxiliary power generators are

examples of redundant items. They are put in service when the primary system fails.

4 . Safe operation: Part should be designed with fail safety in mind. How the component

fails is of importance. If possible, failure should occur in non catastrophic manner and

should do not harm to operator.Auxiliary power generators are examples of redundantitems. They are put in service when the primary system fails.

5. Protection from extreme environmental conditions. An item protected fromextremes of environmental conditions will have increased reliability. For example, pilots

of supersonic space-craft are protected from the effects of extremes of heat and cold.

Electric motors of common household appliances are rubber mounted to protect themfrom vibration.

6. "Maintainability" and "serviceability" are important considerations in designing for

reliability. Ease of maintenance and service contributes to higher field reliability. It isevident that an item which is easy to maintain naturally receives better maintenance and

service.

Reliability Tests. Reliability testing means the tests conducted to verify that a product

will work satisfactorily for a given time period. Reliability testing therefore consists of

functional test, environmental test and life testing.

Functional Test. Functional testing involves a test to determine if the product will

function at time zero.

Environmental Test. Environmental conditions (temperature, humidity, vibration, etc.)are critical to many products.



Environmental test consists of determining the expected environmental levels and then

carrying the functional test under the environments under which the product has tooperate.

Relationship between the failure rate and the mean time between failures (MTBF) &

MTTF & MTTR :MTBF ( repairable systems only) is defined as mean time intervalbetween the successive failure. It is denoted by . Related to the MTBF is the failure rate

which is denoted by .

The failure rate is the reciprocal of the MTBF

If a large number of items are placed in the test of the same type & operated until each

one fails.

The mean time to failure ( irreparable systems)

Where n= number for items failedT = test duration

Prove that the failure rate is the reciprocal of the MTBF:

Proof: Let us test ‘n’ items each ‘t’ hours & items which fails are repairable.suppose that there are ‘r’ failures, that failure rate is

Hence proved

MTBF

1=λ

n

Ti

MTTF

n

i

== 1

λ

λ

t nr

nt

r

durationtest total

failed itemsof no

=

==,

λ λ

λ

1

.

1

==

===

nt

nt MTBF

r

nt

failed itemsof no

durationtest Total MTBF



Prove that the failure rate is the reciprocal of the MTBF:

Proof: Let us test ‘n’ items each ‘t’ hours & items which fails are repairable.suppose that there are ‘r’ failures, that failure rate is

The reliability function of any system is given by

R=probability of survival (therefore it may be denoted as P also ie probability)

Problem 1:

Determine the reliability of a system for a period of 100 hrs if the MTBF is 500 hrs.

Solution : Reliability of a system

MTBF = 500 hrs, t =100 hrs

= 0.8187 is the probability of survival of the system. Ie 81.87%

Problem 2: A device as a failure rate of 5x10-6 failures / hrs

a) What is the reliability for an operating period of 100hrs.

b) If 10,000items are placed in the test, how many failure are expected in 100hrs© What is the MTBF.

(d) What is the reliability of a systems for an operating time equal to MTBF.

e) If the useful life is 1,00,000hrs what is the reliability for operating over its useful life.

λ

λ

t nr

nt

r

durationtest total

failed itemsof no

=

==,

λ λ

λ

1

.

1

==

===

nt

nt MTBF

r

nt

failed itemsof no

durationtest Total MTBF

t e R λ −=

t e R λ −=

hr failures MTBF

rateFailure / 002.0500

11===λ

)100(002.0−= e R



Solution :

Failure rate of the system = failures / hrs(a)

(b) n = 10,000 , t = 100hrs

Expected no of failures= (10,000) (100) ( 5X10-6 )

r = 5 failures

( c)

d)

e)

Problem 3: A piece of ground support equipment has a specified mean time between

failure of 100hrs, what is its reliability for a mission period time of 1hrs , 10hrs, 50hrs,

100hrs, 200hrs, & 300hrs graph these answer by plotting the mission time V/Sreliability. Assume exponent tonal distribue.

Solution : MTBF = 100hrs

9995.0

)100105(6

==

=−−

−

Re

e R

X X

t λ

6105 − Χ=λ

hrs X

MTBF

000,00,2105

116

=

==−

λ

6065.0

000,00,1

3679.0

000,00,2

000,00,2

)000,00,1105(

)200000105(

6

6

=

=

=

=

=

==

==

Χ Χ−

Χ−

−

−

−

e R

hrst

R

e R

t MTBF

hrs MTBF wheree R

X

t λ

9048.0

10)(

9900.0

1)(

/ 01.0100

11

1001.0

101.0

==

=

==

==

===

Χ−

Χ−

−

e R

hrst b

e

hr t e Ra

hr failures MTBF

t λ

λ



( c ) t = 50hrs , R = e-0.01X50 = 0.6065

(d) t = 100hrs, R = e-0.01X100 = 0.3679

(e) t = 200hrs, R = e-0.01X200 = 0.1353

(f) t = 300hrs, R = e-0.01X300 = 0.0498

As the duration increases the reliability of the systems decreases

Systems Reliability

1. Systems connected in series

2. Systems connected in parallel

System connected in series : follow a multiplication law of probability Consider 3

components A, B, C

If a system consists of 3 components A,B,C connected in series , then the reliability of

systems

If a system consists of 3 components A,B,C connected in series , then the reliability of

systems

A C

I / PO / P

B

Mission time

R e

l i a

b i l i t y

RS = RA. R B. RC.



Systems connected in parallelHere the function of A can bedone by B vice versa.If the systems consists of the

Components A&B connected In Parallel with there reliability

RA & RB , then the reliability of the systems is,Rs = 1- (1-RA) (1-RB)

If n components are connected in parallel then

RB

A

B

I / P O / P

RA

[ ]( )

[ ][ ]n

S

n

S

n B A

n B AS

R R

R R

then R R R If

R R R R

)1(1

)1(1

.....................

)1......().........1()1(1

,

−=−

−−=

==

−−−−=



MODULE NO -30

Reliability

Prove that the failure rate of the systems is equal to the sum of the failure rates of the

components of the system.

orProve that for a series system, the failure rates are additive

Proof : Let RS , RA , RB , RC……………… be the reliability of the system & its

component parts A , B , C, ……………….

For independent component in series, & exponentially distributed failure rate,

RS = RA . RB . RC . ……………..__(1)

If are the failure rate of the system & its components parts

for a mean time tmFrom equation (1) RS = RA . RB . RC

Problem 1:

A systems has 3 units with a failure rate of 1.5, 4 &3.8 failure for 10+6

hrs

(a) Find the MTBF of the systems

(b) Determine the reliability of the system for 10 hrs if the components are connected

in series.

(c) If the components are connected in parallel

Solution:

Failure rate is always per hr, But here it is for 106 hrs

therefore 106 hrs – 1.5

1 hr = 1.5 X 10-6

= 1.5 failure for 106 hrs =1.5 x 10

-6 failures / hr

= 4 x 10-6

failures /

n B AS ei λ λ λ λ .................... ++=

...................,,, C B AS λ λ λ λ

mC m Bm Am t t t t eeee

λ λ λ λ −−−− = ..

proved Hence

RHS LHS for samearebasetheSince

ee

ee

C B AS

s

t t

C B A

C B Amms

................

&

.......)..........(

......)..........(

+++=

=

=+++

+++−−

λ λ λ λ

λ λ λ λ

λ λ λ λ

C

B

A

λ

λ

λ



= 3.8 x 10-6

failures / hr

Assuming that components are in series,

(a) MTBF of the systems = MTBFs

(b)

(c) If the components are connected in parallel,

Problem 2 :

A series system has 3 independent parts A,B,&C which have an MTBF of 100, 400

& 800 hrs respectively find

(a) MTBF of the system

(b) Failure rate of the system in failures per million hrs

(c) Failure of the systems in percent failure for 1000hrs

(d) Reliability of the systems for 30 hrs

Solution:

hr failures X s

s

C B As

/ 103.9

)8.345.1(6

10

6

−=++=

++=−

λ

λ

λ λ λ λ

9999.0

88.107526103.9

11

)10103.9(

6

6

=

==

= Χ

==

Χ Χ−

−

−

−

s

t

s

S

s

R

ee R

seriesinconnected component

hrs MTBF

sλ

λ

[ ]

9999.0

9999.0

)1)(1)(1(1

10104

10105.1

6

6

===

===

−−−−=

−

−

−−

−−

X X t

B

X X t

A

C B As

ee R

ee R

R R R R

λ

λ

( )

9999.0

)9999.01)(9999.01)(9999.01(1

9999.010108.3 6

=

−−−−=

=== −−−

s

s

X X t

C

R

R

ee R λ

hrs MTBF

hrs MTBF

hrs MTBF

C

B

A

800

400

100

=

=

=

hr failures

hr failures

hr failures

C

B

A

/ 00125.0800

1

/ 0025.0400

1

/ 01.0100

1

==

==

==

λ

λ

λ



(a) MTBFs

= 0.01+ 0.0025 + 0.00125

= 0.01375 failures/hr

(b)

1 million = 106

= 0.01375 X 106 failures / million hrs

= 13750 failures / million hrs

(c) = 0.01375 failures / hrs

= 0.01375 X 103 failures / 1000hrs

= 0.01375X103 / 100 percent failures / 1000hrs

= 1375 percent failures / 1000hrs(d)

Problem 3 :

A systems is composed of 10,000 parts, what average failure rate / part must be

achieved to get a system MTBF of 250 hrs . Assume series with independent.

Solution : MTBF = 250 hrs

Problem 4 :

Determine the reliability of an equipment having an MTBF of 50 hrs for an operating

period of 45 hrs. If the reliability has to be improved by 20% what % change in the

MTBF is required

solution : MTBF = 50 hrs

t = 45 hrs

C B AS λ λ λ λ ++=

hrs MTBF S

S 727.7201375.0

11===

λ hr failuresS / 01375.0=λ

S λ

66199.0

)3001375.0(

=

=

=−

−

X

t

e

e R S λ

part per hr failures

for hr failures

MTBF

S

S

/ 004.0

000,10 / 40

004.0....................

004.01

000,101

=

=

==

==

λ

λ

λ λ

λ

40657.0

/ 02.050

11

4502.0

=

=

=

===

−

−

X

t

e

e R

hr failures MTBF

λ

λ



Change in reliability by 20 % i.e. By 20%

RN = (0.20) 0.40657) + (0.40657)

= 0.487884

- 0.7176 = - ie loge=1

e X L

e

e R

N n

X

t

N

N

S

log45)487850.0(

487884.0)45(

λ

λ

λ

−=

=

=−

−

e N log*45*λ

%78.25

2578.050

5089.62%

89.12

89.6250

)0041.00159.002.0(

89.620159.0

1 / 0159.0

=

=−

=

=

+−==−=

==∴=

increase

MTBF int improvemen

int improvemen

MTBF hr failures N N

λ

λ



MODULE NO -31

Reliability

Problem 5:

The MTBF of a certain unit is 50hrs. Calculate the reliability for 75hrs of operatingperiod. If the reliability of the unit is increased by 10 % , 20%, 30%, 40%&, & 50%,

calculate.

(a) The% changes in the MTBF that is necessary.(b) Plot a graph , % change in reliability V/S % change in MTBF.

Solution :

MTBF = 50 hrs, t = 75 hrs

R increased by 10% R10 = 1.10X0.223 = 0.2453

R increased by 20 % R20 = 1.20 X 0.223 = 0.2676

R increased by 30 % R30 1.30 X 0.223 = 0.2899

R increased by 40 % R40 = 1.4 X 0.223 = 0.3122

223.0

/ 02.0

50

11

)75(02.0

=

==

===

−−e Re R

hr failures

MTBF t λ

λ

hrs MTBF

e X

ee R t

37.531

0187.0

log752453.0ln

2453.0

10

)75(

10

==∴

=

−=

== −−

λ

λ

λ

λ λ

hrs

MTBF

ee

89.56

01758.0

101758.0

log752676.0ln2676.0

20

)75(

=

=∴=

Χ−== −

λ

λ λ

hrs MTBF

ee

57.60

0165.0

log75)2899.0(ln2899.0

20

)75(

=

=

Χ−== −

λ

λ λ

hrs MTBF

ee

43.64

01552.0

log75)3122.0(ln3122.0

40

)75(

=

=

Χ−== −

λ

λ λ



R increased by 50 % R50 = 1.50 X 0.223 = 0.3345

Reliability MTBF (hrs) % Change in MTBF

0.223

0.2453

02676

0.2899

50

53.37

56.89

60.57

0.3122

0.3345

64.43

68.486

hrs MTBF

ee

486.6801460.0

1

01460.0

log75)3345.0(ln3345.0

50

)75(

==

=

Χ−== −

λ

λ λ

%14.2110050

5057.60

%78.1310050

5089.56

%74.610050

5037.53

= Χ

−

= Χ

−

= Χ

−

%972.3610050

50486.68

%86.2810050

5043.64

= Χ

−

= Χ

−

GRAPH

0

10

20

30

40

1 2 3 4 5 6

% Change in Reliability

% C

h a n g e i n

M T B F



Problem 6 An electronic system consist of a power supply whose failure rate is 30 failures / 106

hrs, a receiver whose failure rate is 25 failures / 106 & an amplifier whose failure rate is

20 failures / 106 hrs in series . The equipment is to operate for 200 hrs. Determine its

reliability.

Solution:

Since the components are connected in series

Problem-7A cassette player has got 4 subsystems, namely spool control systems, magazine

pickup head, amplification & sound system & other systems, all the 4 systems must

perform satisfactorily. The MTBF of the various subsystems are :

spool ctrl systems = 2000 hrs

Mag. Pickup head = 2500 hrs

Amplification & sound systems = 4000hrsOther systems = 3000 hrs

Calculate the reliability of the cassette player for 1500 hrs reception time. What is the

mean time between failures of a cassette player?

Solution:

Pλ r λ aλ

hr failures X

hr failures X

hr failures X

P

r

P

/ 1020

/ 1025

/ 1030

6

6

6

=

=

=

λ

λ

λ

9851.0

Re

/ 1075

10)202530(

2001075

6

6

6

=

=

=

=

++=

++==

−

−

R

e R

e Rsystemtheof liability

hr failure X

X

systemof rate failure

X X

t

ar PS

S λ

λ λ λ λ

SCS mPH S A& Os



Since the comp. are connected in series,

S = 0.0005+0.0004+0.00025+0.00033

= 1.483 x 10-3

failures / hr.

Problem-8:

An item is required to have a failure rate not greater than 0.1% for 1000 hrs of

operation.

(a) Assuming a constant failure rate what is the probability that one of these units will

survive for a required 2000 hrs of service.

(b) Determine the minimum acceptable failure rate where the probability of survival

for a required 2000 hrs of operation is 0.99

Solution: = 0.1% for 1000 hrs

= 0.1 / 100 for 1000 hrs

= 0.001 x 10-3

fail / hrs

= 0.1 x 10-5

fail / hr

(b)

hr failures

hr failures

hr failures

hr failures

Os

S A

MPH

SCS

/ 00033.03000

1

/ 00025.04000

1

/ 0004.02500

1

/ 0005.02000

1

&

==

==

==

==

λ

λ

λ

λ

10807.0

Re)(

150010483.13

=

=

=××−

−

−

e

e Rliabilitya t sλ

hrs

systemtheof MTBF bS

68.675

10483.1

1

1)(

3

=×

=

=

−

λ

998.0

)(200101.0 5

=

== ××−− −

ee Ra t λ

6

)200(

10025.5

log200099.0ln

99.0

−

−

−

×=

−=

=

=

λ

λ

λ

λ

e

e

e R t



Problem-9

A participant in a motor rally is required. to complete a mission time of 2500 km. his

vehicle can be imagined to have 3 subsystems namely fuel, ignition & other systems. The

mean time to repair ( MTTR) of the 3 sub systems are known to be 6000, 8000, 10,000

kms respectively. Find the reliability of his completing the mission without repair.

Solution:

Since the comp. are connected in series

Problem-10

What is the failure rate for a piece of equipment if the probability of survival is 88%

for 900 hrs of operating period. Express the failure rate in terms of percent failures for

1000 hrs.

Solution:

R = 0.88, t = 900

km failureshrs MTTR

km failureshrs MTTR

km failureshrs MTTR

OO

I I

F F

/ 10110000

1ie10000

/ 1025.18000

1ie8000

/ 1066.16000

1ie6000

4

4

4

−

−

−

×===

×===

×===

λ

λ

λ

I λ Oλ F λ

km failures

S I F S

/ 1091.3

10)125.166.1(

4

4

−

−

×=

++=

++= λ λ λ λ

376.0

)25001091.3( 4

=

=

=××−

−

−

e

e R t S λ

hr failures

e

e R

e

t

/ 1042.1

log90088.0ln

88.0

4

)900(

−

−

−

×=

×−=

=

=

λ

λ

λ

λ

hrs failures percent

hrs failures percent

hr failures percent

1000 / 42.1

1000 / 10001001042.1

/ 1001042.1

4

4

=

××=

×=

−

−λ



Problem-11:

A 750 hrs life test is performed on 6 components. One component fails after 350 hrs

of operation. All others survive the test. Compute the failure rate

.Solution:

Redundancy

One of the methods for improving the reliability of the system is by initializing the

concept of redundancy. To enhance the reliability of the system, quite often additionalunits are built in to the system to perform the same functions. In such a system, one

component failure will not necessarily cause the system failure.

Since additional components are available to perform the same function. Redundancy

is defined as the characteristic of the system by virtue of which marginal component

failures are prevented from causing system failures due to the presence of additional

components.

In order to increase the reliability of the system select the component which has the

least reliability & then arrange it parallely?

For example:

RS = (RA). (RB). (RC)

= (0.8). (0.5). (0.7)

= 0.28

In order to increase the above systems reliability, select the component which has the

least reliability & arrange it parallely.

hr failure

duratriontest total

failured itemsof norateFailure

/ 10439.2

)3501()7505(

1

.

4−×=

×+×=

=λ

A R B R C R

0.8 0.7 O/PI/P 0.5



RB’ = 1-[(1-RB) (1-RB)]

= 1-(1-0.5)2

RB’ = 0.75

RS = (0.8) (0.75) (0.7)

= 0.42

It is clear from the system that reliability of the system is increased from 0.28 to 0.42.

Derivation

Definition of improvement factor (IF)

If in the case of ‘n’ parallel redundancies, the improvement factor (IF) =1-R / 1-R S

Where (1-R) = unreliability of each component(1-RS) = unreliability of the system

If ‘n’ components are connected in parallel with the same reliability as shown in the

fig below.

We know that, if n component are connected in parallel, then the reliability of the

systems is

C R

O/PI/P0.7

B R

B R

A R

0.8

0.5

0.5

A R ' B R C R

0.8 0.7 0.75I/P O/P



RS = 1-[(1-RA) (1-RB) - - - (1-Rn)

Since the reliabilities RA = RB = - - - Rn, then

RS = 1- (1-R)n

There fore 1-RS = (1-R)n

Problem-1

A system consists of 3 components A, B & C. The configuration of the system and

the reliabilities of the elements are given below. Calculate the reliability of the system.

Solution:

RAB = 1 - [ (1-RA)(1-RB) ]

= 1 – [ (1-0.8) (1-0.7) ] = 0.94

RS = (RAB) (Rc) = (0.94) (0.9) = 0.846.

Problem-2

Calculate the reliability of the configuration given below.

n

S

n

Q IF

Q Rif R

R

R

R IF

−=

=−−−=

−−=

1

)1(1

1

)1(

1



Solution:

RAB = 1 - [ (1-RA)(1-RB) ]

= 1 – [ (1-0.7) (1-0.7) ] = 0.91

RABC = (RAB) (RC)

= (0.91) (0.9) = 0.819

RABCD = 1 - [ (1-RABC )(1-RD) ]

= 1 – [ (1-0.819) (1-0.8) ] = 0.9638

RABCDE = (0.9638) (0.9)

= 0.86742

Problem-3:What is the reliability of the systems shown below. P(A)=P(B)=P(C)=0.8 P(D) =

0.95, P(E) = 0.85



How would the reliability be improved further if the sub system E is also made

parallel redundant. Show the configuration of the system.

Solution:

Case 1:

RABC = 1 - [ (1-RA)(1-RB) (1-RC) ]

= 1 – [ (1-0.8)3 ] = 0.992

RABCDE = (RABC) (RD) (RE)

= (0.992) (0.95) (0.85)

= 0.80104

RABC = 1 - [ (1-RA)(1-RB) (1-RC) ]

= 1 – [ (1-0.8)3 ] = 0.992

RE’=1-[(1-RE)2]

=1-[(1-0.85)2=0.9775



RABCDE’ = (RABC) (RD) (RE’)= (0.992) (0.95) (0.9775)

= 0.9212

% improvement in reliability

= 0.9212-0.80104 / 0.80104 = 15%

Problem-4:Determine the probability of success for the system with all units operating

P(A) = 90%, P(A) =85%, P(c) =75% & P(D) = 80%

Solution:

P(A) = 0.9, P(B) =0.85, P(C) = 0.75, P(D) =0.80

RD’ = 1 - [ (1-RD)(1-RD) (1-RD) ]

= 1 – [ (1-0.8)3 ] = 0.992

RABCD’A = (RA) (RBC) (RD’) (RA)

= (0.9) (0.9625) (0.992) (0.9)

= 0.773388

Therefore probability of success for the system with all units operating ( i.e.

reliability of the system) = 0.77388.



Problem-5

An electronic system consists of 5 subsystems with the following MTBF’s. (SS =

subsystem)

SS A = 12,500 hrs SS D = 9,850 hrs

SS B = 2,830 hrs SS E = 15,500 hrs

SS C = 11,000 hrs

These 5 subsystems are arranged in series configuration. What is the probability of

survival for an operating period of 800 hrs.

Solution:

RS = (RA) (RB) (RC) (RD) (RE)

= (0.938) (0.7146) (0.9299) (0.95008)

= 0.5458

938.0

/ 108500,12

11

800108

5

5

===

×===

××−−

−

−

ee R

hr failures MTBF

t

A

A

A

λ

λ

7146.0

/ 108380,2

11

800102.4

4

4

===

×===

××−−

−

−

ee R

hr failures MTBF

t

B

B

B

λ

λ

9299.0

/ 1009.9000,11

11

8001009.9

5

5

===

×===

××−−

−

−

ee R

hr failures MTBF

t

C

C

C

λ

λ

9216.0

/ 1002.1850,911

8001002.1

4

4

===

×===

××−−

−

−

ee R

hr failures MTBF

t

D

D

D

λ

λ

95008.0

/ 104.6500,15

11

800104.6

5

5

===

×===

××−−

−

−

ee R

hr failures MTBF

t

E

E

E

λ

λ



Problem-6

A step down transformer, rectifier, filter comprises a series of systems. The following

are the failure rates of these components.

Transformer =1.56 percent failures / 1000 hrs

Amplifier = 2 percent failures / 1000 hrsFilter = 1.7 percent failures / 1000 hrs

The equipment is operate for 1500 hrs. What is the probability of survival of the system.

Solution:

RS = (RT) (RA) (RF)

= (0.97687) (0.97044) (0.97482)

= 0.92412

Problem 7:

Determine the reliability of the systems for 20 hrs of operating period. The

configuration is given below. The failure rate per hr are also given.

hr fail

hr failures percent T

/ 1056.1

1000 / 56.1

5−×=

=λ

hr failures

hr failures percent A

/ 102

1000 / 2

5−×=

=λ

hr failures

hr failures percent F

/ 107.1

1000 / 7.15−×=

=λ

97687.015001056.15

=== ××−− −

ee R Tt

T

λ

97044.015001025

=== ××−− −

ee R At

A

λ

97482.01500107.1 5

=== ××−− −

ee R Ft

F

λ

,025.0,02.0,02.0,015.0,01.0 ===== A A A A A λ λ λ λ λ



RBC = 1 – [ (1-RB ) (1-Rc )]

= 1- [(1 – 0.7408) (1 – 0.6703)]

= 0.9145

RABC = ( RA ) ( RBC )

= (0.8187) (0.9145)

= 0.7487

6065.0

6703.0

6703.0

7408.0

8187.0

20025.0

2002.0

2002.0

20015.0

2001.0

===

===

===

===

===

Χ−−

Χ−−

Χ−−

Χ−−

Χ−−

ee R

ee R

ee R

ee R

ee R

t E E

t D

D

t C

C

t B

B

t A

A

λ

λ

λ

λ

λ



RABCD = 1 – [ ( 1- RABCD ) (1- RD ) ]

= 1 – ([ ( 1 – 0.7487) ( 1 – 0.6703)]

= 0.917146

RABCDE = (RABCD ) ( RC )

= (0.917146) ( 0.6065)

= 0.5562

Mod10-NVN QA QC

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