4 probabilistic design

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INTRODUCTION 2 - 1

JMBrowne July 12, 2000 11:53 am 4ProbabilisticDesign Page 1 of 52

2.1 INTRODUCTION

In the design of a product for mass-production we are faced with

the challenge that every item produced will be different

. These differ-

ences will be slight to the casual observer, but may combine in the indi-

vidual items to give vastly different performance characteristics, and

thus impact the perceived quality of the product. These differences are

caused by, among many other things, drift in machine settings, batch

variability in material properties and operator input.The value of each design parameter embodied in any item is there-

fore likely to be different from the value in any other item. If we mea-

sure the values of a design parameter (a length, say) in all the items in a

production run we will get data on the frequency of occurrence of the

values of the parameter. If there are sufficient data values we can rescale

the frequency to give a probability. Design parameters may thus be

viewed as random variables.

Most physical variables used in engineering design are in fact ran-

dom variables. Standard calculations are really calculations with their

mean values. If we are interested in the possible range of values our re-sult might have, then we must use more information in our calculation

algorithm than the mean values alone.

The classical approach to design is to apply safety factors to each

design parameter to allow for uncertainties. If the design is complex,

these safety factors can compound to cause overdesign with an uncer-

tain reliability. And in some important cases, where there is an upper


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2 - 2 PROBABILISTIC DESIGN


and lower specification or functional limit, the safety factor method can-

not be used at all.

Probabilistic design studies how to make calculations with the

probability distributions of the design parameters, instead of the nomi-

nal or mean values only. This will then allow the designer to design for

a specific reliability or specification conformance, and hence maximize

safety, quality and economy.

Design parameters are usually independent random variables

.

Each type of parameter will have a distribution. Common distributions

for design parameters are the normal, log-normal, poisson, uniform, tri-angular, exponential and weibull distributions.

2.2 TYPES OF PROBABILITY DISTRIBUTIONS

Detailed briefly below are the types of probability distributions

more commonly found in engineering.

2.2.1 Normal

The distribution is symmetric and bell-shaped

The variable may itself be the sum of a large number of individualeffects.

Example: Heights of the adult male population.

Example: Dimension of a fabricated part.

2.2.2 Lognormal

The variable can increase without bound, but is limited to a finitevalue at the lower limit

The distribution is positively skewed (most of the values beingcloser to the lower limit).

The logarithm of the variable yields a normal distribution.

Example: Real estate values, river flow rates, strengths of materi-als, fracture toughness.

CALCUWITHUNSURNUMBE

0 . 9 3 1 . 0 0 1 . 0 8

. 2 1 1 . 0 8 1 . 9 5 2 . 8 2


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TYPES OF PROBABILITY DISTRIBUTIONS 2 - 3


2.2.3 Weibull

A distribution possessing three parameters enabling it to beadjusted to cover all stages of the bathtub reliability curve.

A shape parameter of 1 gives an exponential distribution.A shape parameter of 3.25 gives an approximation to the normal.

Finds principal application in situations involving wear, fatigueand fracture.

Example: Failure rates, life-time expectancies

2.2.4 Exponential

Describes the amount of time between occurrences.

Complements the Poisson distribution (which describes thenumber of occurrences per unit time.

Example: Time between telephone calls.

Example: Mean time between failures

2.2.5 Triangular

Used when the only information known is the

minimum

, the mostlikely

, and the maximum

values of a variable.

Example: Item costs from different suppliers or future estimation.

2.2.6 Uniform

All values between the minimum and maximum are equally likely

Example: A number from a random number generator.

2.2.7 Poisson (discrete)

Describes the number of times an event occurs in a given interval.

The number of possible occurrences in the interval is not limited.

The occurrences are independent.

The average number of occurrences is fixed.

Example: Number of telephone calls per minute.

. 0 0 1 . 2 7 2 . 5 5 3 . 8 2

0 . 0 0 1 . 1 5 2 . 3 0 3 . 4 5

. 5 0 0 . 6 5 0 . 8 0 0 . 9 5 1 . 1

. 9 0 0 . 9 5 1 . 0 0 1 . 0 5 1 . 1

. 0 0 3 . 2 5 6 . 5 0 9 . 7 5


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Example: Number of errors per page in a document

Example: Number of defects per square metre in sheets of steel.

2.2.8 Binomial (discrete)

Describes the number of successes in a fixed number of trials.

For each trial only two outcomes are possible - success or failure.

The trials are independent

The probability of success remains the same from trial to trial.

Example: Number of heads in ten tosses of a coin

Example: Number of defective items in a given batch, given thatthe average rate of producing defectives is known.

2.2.9 Geometric (discrete)

Describes the number of trials until the first successful occurrence.

The number of trials is not fixed and continue until the first success

The probability of success is the same from trial to trial

Example: Number of times to spin a roulette wheel before you win.

Example: Number of wells you would dig before the next gusher.

2.2.10 Custom

Used to describe a unique situation that cannot be described by any

of the standard distributions.

The area under the curve should equal 1.

2.2.11 Comparison of distributions

Poisson: Number of times an event occurs in a given interval.

Exponential: Interval until next occurrence of event.

Binomial: Number of successes in a fixed number of trials.

Geometric: Number of trials until the next success.

Large number of trials: Binomial approaches normal.

0 4 7 1 0

1 7 1 3 1 9

0 0 3 . 2 5 5 . 5 0 7 . 7 5 1 0 . 0


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DESCRIBING PROBABILITY DISTRIBUTIONS 2 - 5


2.3 DESCRIBING PROBABILITY DISTRIBUTIONS

2.3.1 Types of description

The types of description we will use for describing probability dis-

tributions include its parameters, its probability density function (pdf),

its cumulative distribution function (cdf), and its set of moments.

Theparameters

of a given type of distribution are the mathematicalparameters in the formula for the distribution (not to be confusedwith design parameters).

The probability density function

describes the basic shape and

location of the distribution. The graphs shown in the previoussection are probability density functions.

The cumulative distribution function

allows us to read off the areaunder the probability density function in a given range. This arearepresents the probability that the random variable will lie in this

range.

It is the probability density function, the parameters which describe

it, and the first few of its moments that will be of most use to us in prob-

abilistic design.For brevity we may use the terms pdf, distribution, or density

function instead of probability density function.

Since the distribution is the main description of a random variable

we will sometimes use the terms interchangeably.

2.3.2 Properties of probability density functions

In order to be called a probability density function, a function must

have the following properties: (Any function that looks like a blob of

goo on the axis is probably a good candidate)

It is indeed a function (no undercuts)

The area between it and the axis is unity

The support

of the probability density function is the domain of the

random variable over which the function is defined.

The full definition of a probability density function comprises the

specification of its formula and its support.


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2.3.3 Functions of random variables

Functions of random variables are central to the design of products

for quality and reliability. Since the performance of a product is gener-

ally a function of its design parameters, and the design parameters are

random variables, the performance is a function of random variables,

and is thus itself a random variable.

A central tool in the design of quality products therefore, is the

ability to calculate functions of random variables.

2.3.4 Notation

Generally we will denote a probability density function of a ran-

dom variable x by f(x). However, when we are considering a function z

= g(x) we will distinguish the two probability density functions by de-

noting them f

x

(x) and f

z

(z).

2.3.5 In sum

Design parameters and the quality variables which depend on them

are most often random variables which we describe by probability den-

sity functions.

Problem 2.1 Uniformly distributed design parameter

A design parameter is a random variable uniformly distributed be-

tween 1 and 3. Sketch its probability density function and its cumulative

distribution function showing pertinent values on the axes.

2.4 GRAPHICAL FUNCTIONS OF A RANDOM

VARIABLE

In this section we shall describe the concept of a function of a ran-

dom variable in graphical terms.

The most important attribute of a function when applied to a ran-

dom variable is whether it has an inverse over the support of the density

function of the random variable. If it does, then it is straightforward to

compute the function of the random variable. If not, then the function

must be broken up into pieces so that each piece has an inverse, the


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GRAPHICAL FUNCTIONS OF A RANDOM VARIABLE 2 - 7


transformation associated with each piece applied, and the results

summed.

2.4.1 The concept

Suppose x is a random variable with probability density function

f

x

(x), and that z = g(x) is an invertible function of x over the support of

f

x

(x).

The central concept is as follows:

The probability of x being in the interval [x

1

, x

2

] is equal to the

probability that z is in the interval [z

1

, z

2

] = [f

x

(x

1

), f

x

(x

2

)].Geometrically, this is equivalent to saying that the area under the

probability density function of x in the interval [x

1

, x

2

] is equal to the

area under the probability density function of z in the interval [f

x

(x

1

),

f

x

(x

2

)].

2.4.2 The fundamental formula

Equating the two probabilities (areas) we obtain

A = | f

z

(z) dz | = | f

x

(x) dx |

=> f

z

(z) = f

x

(x) / | dz/dx |

Note that because the probability (area) is always positive, the

same relationship will exist whether the gradient of the function g(x) is

x

z


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positive or negative. Hence we always take the absolute value of the de-

rivative dz/dx.

2.4.3 Dimensional considerations

Probability is dimensionless. However, x and z may have (differ-

ent) dimensions (units) [x] and[z], say. The probability density func-

tions f

x

(x) and f

z

(z) must have dimensions 1/[x] and 1/[z] respectively.

This fact corroborates with the formula above and may be used as

a check on the correctness of any functional transformation.

2.4.4 Examples

This simple relationship between area elements of the two density

functions may be used to perform a graphical determination of a func-

tion of a random variable. It is of course generally more accurate to de-

termine the result analytically, however it is useful to be able to

visualize the process graphically.

A linear function through zero


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GRAPHICAL FUNCTIONS OF A RANDOM VARIABLE 2 - 9


A general linear function

A concave function


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A convex function

Note that the lower gradient of the transformation function leads to

a concentration

of the probability in the corresponding region of the

transformed probability density function.

Much of our success in the design of quality products will depend

on our being able to tune the design to make use of these regions of lowgradient, hence minimizing the variability of the design output distribu-

tions.

Problem 2.2 Sketching distributions

Sketch yourself a distribution and a transformation function.

Sketch the shape of the resulting transformed distribution.

Problem 2.3 Sound output

The sound output of a product has been determined to follow a tri-

angular distribution with mode 2 units, lower limit 1 unit and upper limit

3 units.

Graphically determine the probability density function for the (nat-

ural) logarithm of the sound output.


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ANALYTICAL FUNCTIONS OF A RANDOM VARIABLE 2 - 11


2.5 ANALYTICAL FUNCTIONS OF A RANDOM

VARIABLE

2.5.1 Definition of an invertible function

The process above is straightforward if, over the support of x, there

is only one value of x for each value of z. Functions with this property

are called invertible.

2.5.2 Non-invertible functions of a random variable

If the function is not invertible, the following process may be ap-

plied:

1. Break the function up into piecewise invertible pieces overintervals [xi, xj]

2. For each piece, follow the procedure below for an invertiblefunction. The result will be valid over the interval [g(xi), g(xj)](or [g(xj), g(xi)], whichever is in the correct order), and zerooutside of it.

3. Add the results.

Functions which are constant (flat) over an interval give rise to a

discrete jump in the probability density function of z.

2.5.3 Examples of invertible and non-invertible functions

Invertible functions


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Non-invertible functions

2.5.4 Invertible functions of a random variable

The procedure for calculating an invertible function of a random

variable is as follows:

Given:

A. A probability density function: fx(x), x1 x x2B. A transformation function: g(x)

1. Calculate dz/dx from z = g(x)

2. Solve for x in terms of z to get x = g -1(z) = h(z). (There should

be only one solution since the function is invertible)

3. Substitute h(z) for x in fx(x) / |dz/dx| to get fz(z)

4. Determine the new support: g(x1) z g(x2)

We apply this procedure to some simple cases below. We assume a

general (undefined) pdf fx

(x) transformed by an invertible function g(x)

which we can differentiate. The original probability density function is

shown in light grey and the result of the function (or transformation) in

darker grey.

Addition of a constant: [z = x + a]

1. dz/dx = 1

2. h(z) = z - a


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ANALYTICAL FUNCTIONS OF A RANDOM VARIABLE 2 - 13


3. fz(z) = fx(z-a)

4. x1+a z x2+a

Geometrically, the addition of a constant to a random variable sim-

ply gives another random variable all values of which are increased (dis-

placed to the right) by that constant.

Example: The conversion of a random temperature expressed in

Celsius to one expressed in Kelvin.

Multiplication by a constant: [z = a x]

1. dz/dx = a2. h(z) = z/a

3. fz(z) = fx(z/a) / |a|

4. a x1 z a x2

Geometrically, the multiplication of random variable by a constant

simply gives another random variable all values of which are multiplied

by (stretched to the right) by that constant.

Example: The conversion of a dimension expressed in metres to

one expressed in millimetres.

The general linear transformation: [z = a x + b]

1. dz/dx = a

2. h(z) = (z-b)/a

3. fz(z) = fx((z-b)/a) / |a|

4. a x1+b z a x2+b

Geometrically, a general linear function of a random variable pro-

duces both a shift and a change in scale. The form of the function re-

mains the same.

Example: The conversion of a random temperature expressed in

Celsius to one expressed in Fahrenheit.

The exponential transformation: [z = ex]

1. dz/dx = ex

2. h(z) = ln z

3. fz(z) = fx(ln z) / |z|

4. exp(x1) z exp(x2)


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Example: The conversion of a variable expressed on a logarithmic

scale back to one expressed on a linear scale.

2.5.5 Example

Exponential transformation of a Uniform distribution

A. Probability density function: fx(x) = 1/(b-a), a x b, a > 0B. Transformation function: z = c exp(k x) where c and k are con-

stants

1. Calculate dz/dx from z = c exp(k x):dz/dx = k c exp(k x)

2. Solve for x in terms of z to get x = g-1(z) = h(z):

x = h(z) = ln(z/c)/k

3. Substitute h(z) for x in fx(x) / |dz/dx| to get fz(z):

fz(z) = (1/(b-a)) (1/|k c exp(k x)|)

= (1/(b-a)) (1/|k c exp(k (ln(z/c)/k))|)

= (1/(b-a)) (1/|k z|)

4. Determine the new support for fz(z):

c exp(k a) z c exp(k b)

Problem 2.4 Sphere volume

A manufacturer makes spheres to meet a specification on the vol-

ume. The process is known to deliver spheres with their diameters nor-

mally distributed with mean 10 mm and standard deviation 1 mm.

1. Determine the formula for the probability density function of the

volume.

2. Compare the true mean volume with the approximate mean vol-ume calculated from . (Advanced exercise).

Problem 2.5 Alloy steel

The percentage x of an alloy in a steel is exponentially distributed

with probability density function

fx(x) = a exp(- a x), 0 x , a constant.

6---10

3


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MOMENTS 2 - 15


The ultimate tensile strength of the steel, z, is logarithmically relat-

ed to the percentage of alloy by z = loge(x/b).

Derive the formula for the probability density function fz(z) of the

ultimate tensile strength, and state its support.

2.6 MOMENTS

Moments of a distribution are a way of summarizing the important

characteristics of a distribution as single numbers, without having to

cope with too much detail. The first few (lower order) moments are gen-

erally of most interest to us. An analogy might be to the reduction of a

vibration trace to its first few harmonics.

More precise mechanical analogies are

1. The mean is the centre of area of the distribution - summarizing

the location properties of the distribution.

2. The variance is the second moment of area of the distribution

about the mean - summarizing the way in which the area is spread over

the object.

Because we generally lack detailed information about the probabil-

ity density functions of our design parameters, we will usually be mak-

ing our calculations with the first few moments, often just the mean and

variance.

Following are some definitions of moments and coefficients based

on them.

2.6.1 (Non-central) moments

The nth (non-central) moment of a distribution f(x) about the

origin is

The first non-central moment is called the mean.

The mean of a random variablex will be denoted x, or simply where the context is clear.

The mean is also the expectation ofx, denotedE[x].

' n( )x

' n( )x xnf x( )dx

=


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2.6.2 Central moments

The nth central moment (n)x of a distribution f(x) about themean of a distribution is

The first central moment of any distribution is zero.

The second central moment is called the variance, denoted vx.

The third central moment is called the skew, denoted sx.

The fourth central moment is called the kurtosis, denoted kx

.

Since we will be dealing mostly with central moments, we will of-

ten refer to them simply as moments.

2.6.3 Variance

The variance is, after the mean, the most important moment of a

distribution. Its unit is the square of the unit of the random variable and

hence is always positive. It measures the spread of the distribution. A

zero variance thus implies a deterministic variable.

2.6.4 Skew

The skew is the next most important moment. Its unit is the cube of

the unit of the random variable and hence may be positive or negative.

A positively skewed distribution has its longer tail to the right. A nega-

tively skewed distribution has its longer tail to the left. We will some-

times use the skew to test how valid it is to assume a given distribution

is symmetric (and hence perhaps approximatable by a Normal distribu-

tion).

2.6.5 Kurtosis

We include here the kurtosis mainly for completeness. Since the

kurtosis measures the squatness of the distribution, it is useful for dif-

ferentiating different types of symmetric distributions (for example the

Normal and the Uniform). However since most of the distributions we

will be using are bell-shaped, we will not use the kurtosis much. It is al-

ways positive.

n( )x x ( )nf x( )dx

=


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2.6.11 Terminology

There are varying definitions in the literature for skew and kurtosis

and their dimensionless ratios. It is wise to check the definition the au-

thor is using.

2.6.12 A note on notation

In situations where there are several random variables, for exam-

ple, x, y, we will use x, y, for the mean of x, y, , and x, y, for the their variance. If we dealing with a single random variable, we

will often drop the subscripts.

2.7 THE NORMAL DISTRIBUTION

The normal distribution is the most important distribution in the ap-

plication of probability theory to science and engineering. The Central

Limit Theorem (to be discussed later) tells us that the Normal distribu-

tion has an interesting involvement in the description of complex prob-

abilistic systems.

It will be worth getting a good intuitive feel for its properties.

It is symmetric

Its support is from -Infinity to +Infinity

99.7% of the distribution lies within 3 standard deviations of themean

95% of the distribution lies within 2 standard deviations of themean

68% of the distribution lies within 1 standard deviations of themean. The inflection point on the curve is at this point.

Because of its symmetry, its odd central moments are zero.

Its even central moments are given by (where is the variance):

{, 3 2, 3x5 3, 3x5x7 4, 3x5x7x9 5, }= {, 3 2, 15 3, 105 4, 945 5, }


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THE NORMAL DISTRIBUTION 2 - 19


Its probability density function is

The graph of its probability density function for = 0 and = 1 is

Its cumulative distribution function is

The graph of it cumulative distribution function = 0 and = 1 is

1

2--------------e

1

2---

x

------------

2

1

2--- 1 Erf

x

2------------

+


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Problem 2.6 Probability of a continuous random variable

1. What is the probability that a normally distributed random vari-

able has its mean value?

2. What is the probability that a normally distributed random vari-able lies between and + 2?

3. What is the probability that a normally distributed random vari-

able is greater than + 6?

Problem 2.7 Sketching a Normal distribution

Sketch carefully a normal distribution with mean 9 and variance 9.

A random variable is distributed as above. What is the probability

that it is less than zero?

2.8 MEANS FROM NOMINAL VALUES

The usual design specification on a parameter is given by a nominal

value and a tolerance. The nominal value is usually the value given as n

in the specification [n - t1, n + t2]. The question arises: Given only a

specification on a parameter in this form, what should we assume the


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STANDARD DEVIATIONS FROM TOLERANCES 2 - 21


mean value of the parameter to be?

Until more research is done in this area, we propose that the mean

be estimated as = (t1 + t2)/2.

2.9 STANDARD DEVIATIONS FROM TOLERANCES

While mean values are often easy to find from data sources, it is

usually more difficult to obtain an estimate of the variance (or standard

deviation) of a design parameter.

This section discusses some rules of thumb for estimating standarddeviations from tolerances.

2.9.1 Estimation from tolerance range

If we know that the parameter is approximately normally distrib-

uted and the proportion of product that is expected to lie within a certain

tolerance range, then a rule of thumb for estimating the random vari-

ables standard deviation from the properties of the normal distribution

is:

If expect 68% to lie within then set x

. = If expect 95% to lie within then set x. = /2If expect 99.7% to lie within then set x. = /3

2.9.2 Estimation from limited data

A rule of thumb which enables standard deviations to be estimated

from limited data is given by Haugen:

If the estimate of the tolerance that is required is obtained:From about 4 samples then set x = From about 25 samples then set x = /2From about 500 samples then set x = /3

2.9.3 Estimation from knowledge of manufacturer

A further rule of thumb given by Shooman is:

If the product is being made by:


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Commercial, early development, little known or inexperiencedmanufacturers, then set x =

Military, mature, reputable, or experienced manufacturers, then setx = /3

2.10 THE EXPECTATION OPERATOR

The expectation of a function g(x, y, ) of random variablesx, y,

with probability density function f(x, y, ) is denoted E[g(x, y, )]

and is defined as the integral:

The following properties may be proven from the definition:

2.10.1 The expectation of sums and products

Constant

The expectation of a constant c is the constant itself. That is

E[c] = c

Linear sum

Ifa, b, are constants, then

E[a g1(x, y,) + b g2(x, y,) +] = a E[g1(x, y,)] + b E[g2(x, y,)]

Product of independent random variables

Ifx,y, are independentrandom variables, then

E[g1(x) g2(y) ] = E[g1(x)] E[g2(y)]

2.10.2 Relation of moments to the Expectation

Non-central moments as Expectations

E[1] = 1

E[x] = xE[xn] =

E g x y , ,( )[ ] g x y , ,( )f x y , ,( ) xd yd

=

' n( )x


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THE EXPECTATION OPERATOR 2 - 23


Central moments as Expectations

E[x - x] = 0E[(x - x)

2] = xE[(x - x)

3] = sxE[(x - x)

4] = kxE[(x - x)

n] = (n)x

Problem 2.8 Moments as Expectations

Prove the formulae relating moments and Expectations above from

the definitions of moment and Expectation.

2.10.3 Relation of central and non-central moments

Central moments in terms of non-central moments

Non-central moments in terms of central moments

Problem 2.9 Expectation of a linear sum

From the definition of Expectation, prove the formula for the Ex-

pectation of a linear sum of functions of random variables.

Problem 2.10 First central moment

Show that E[x - x] = 0.

E x x( )n

[ ] En

i x

ix( )

n i

i 0=

n

=

E x x

( )n

[ ]n

i

x

( )n i

i 0=

n

E x

i[ ]=

E xn

[ ] E x x( ) x+( )n

[ ]=

E xn

[ ]n

i E x x( )

i[ ] x( )

n i

i 0=

n

=


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Problem 2.11 Central and non-central moments

Fill out the detail in the above derivations relating central and non-

central moments.

Problem 2.12 Variance and skew

1. Derive expressions from first principles for the variance and

skew in terms of non-central moments. Use the binomial expansion and

the properties of the Expectation operator.

2. Verify your results using the general formulae above.

Problem 2.13 Mean values of a power

1. Derive expressions from first principles for the mean values of

second, third, and fourth powers of a random variable in terms of its

central moments. Use the binomial expansion and the properties of the

Expectation operator.

2. Verify your results using the general formulae above.

Problem 2.14 Mean second moment of areaA beam of circular cross-section has a normally distributed diame-

terD with mean 100 mm and standard deviation 2 mm.

1. Calculate the mean second moment of area ( ) about a

diameter.

2. Compare this with the nominal second moment of area based on

a nominal diameter of 100 mm.

(Hint: The kurtosis of a normal distribution is 34).

Problem 2.15 Volume of sphere

The performance of a product is dependent on the volume V of a

contained steel sphere of diameter D remaining within tight specifica-

tions. The machine manufacturing the spheres is controlled by the spec-

ification on the nominal diameter.

By using the expectation operator and the identity an = ((a - b) +

I64------D

4=


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LINEAR FUNCTIONS 2 - 25


b)n, or otherwise, derive an exact formula for the mean V in terms ofD, and higher order central moments.

2.11 LINEAR FUNCTIONS

2.11.1 Introduction

In this section we begin to explore an approximate method for

computing with random variables by considering only the first few mo-

ments of a distribution (typically only the mean and variance, but some-times the skew and kurtosis). Exact methods in computation with

random variables are often exceedingly complex and insufficiently gen-

eral. However even the approximate probabilistic approaches developed

here are an order of magnitude more powerful in engineering design for

quality and reliability than the traditional factor of safety approach.

In this section we look only at linear functions. In a later section we

will look at more general function types.

2.11.2 General formulae

For the special case of a linear function of several variables, themoments may be derived exactly and take particularly simple forms.

Note that the relations below are true, independentof the types of

underlying distributions possessed by the xi.

However in the general case, z will not have a distribution of any

known standard type.

If x1, x2, x3, are independentrandom variables, a1, a2, a3, are

constants, and z = a1 x1 + a2 x2 + a3 x3 + , then the first four moments

of z are given by:

z = a1x1 + a2x2 + a3x3 + = aixi

z = a12x1 + a2

2x2 + a32x3 + = ai

2xi

sz = a13 sx1 + a2

3 sx2 + a33 sx3 + = ai

3 sxi

kz = a14 kx1 + a2

4 kx2 + a34 kx3 +

+ 6{a12x1 a2

2x2 + a22x2 a3

2x3+ a12x1 a3

2x3 + }


26/52



= ai4 kxi + 6 ai

4xi aj4xj [i


27/52



ADDITION OF TWO NORMAL RANDOM VARIABLES

2.11.5 The Central Limit Theorem

The sum of a number of independent but not necessarily identically

distributedrandom variables tends to become normally distributedas

the number increases, provided that no one random variable contributes

appreciably more than the others to the sum; that is, no type of distribu-tion dominates.

This is an important result for designers. It means, for example, that

the overall dimension of an assembly of component parts, independent

of the distribution types of each component dimension, will tendto be

normally distributed. Knowing this, the designer can work back from

the individual component tolerances to get an estimate of the proportion

of assemblies which will lie outside any given specification.

The six greyed graphs below are, sequentially, the distributions of

the average of 1, 2, 3, 4, 5, and 6 independent identically Uniformly dis-

tributed random variables on [-1, 1]. The full line is the Normal distri-bution which has the same variance as the average.

It can be seen that even the average of only three Uniform distribu-

tions gives a result surprisingly close to Normal.


28/52



Problem 2.16 Inconsistency?

If x = y in the formula for the variance of a difference, that is, z =

x - y = 0, does this imply z = 2x = 2y?

Problem 2.17 Tolerance build-up

An assembly is made up of several components whose nominal

lengths are L1, L2, L3, and L4. It is important that the distance L = L1+L2- (L3+L4) be kept within specification limits.

1. Write down a formula for the standard deviation L as a functionof1, 2, 3, and 4.

2. What can be said about the type of distribution that L has?


29/52



Problem 2.18 Springs in parallel

An assembly contains two springs of stiffness K1 and K2 connected

in parallel. Their stiffness distributions have moments of

K1: {1 = 500 N/m, 1 = 144 (N/m)2, s1 = 10 (N/m)

3}

K2: {2 = 300 N/m, 2 = 25 (N/m)2, s2 = -10 (N/m)

3}.

1. Calculate the mean, variance and skew of the distribution of the

overall stiffness K of the system.

2. Is the resulting distribution symmetric?

Problem 2.19 Counterweights

Two designs are proposed for a sensitive counterweight. Design A

utilizes 4 spheres each of mass m. Design B utilizes two spheres each of

mass 2m.

Assuming that the coefficient of variation of the mass of each of the

spheres is the same, determine the ratio of the standard deviation of the

total mass of Design A to that of Design B.

Problem 2.20 Algenon and Biggles

Bricks are manufactured with heights of a given mean and vari-

ance. Algenon Ant climbs straight up a vertical stack of N bricks (no

mortar). His brother Biggles (a little disoriented) goes straight up and

down the first brick for the same number of brick traverses (hence cov-

ering the same mean distance).

Determine the ratio of the standard deviation of Biggles' journey to

the standard deviation of Algenon's journey.

Problem 2.21 Machine supportSuppose that a machine is to be supported with a number of springs

of the same nominal stiffness, and that the overall stiffness of the spring

assembly is to be within a given tolerance of a fixed target value K. Sup-

pose also that all the springs in the assembly have the same percentage

tolerance on their stiffness no matter what size they are. That is, the stiff-

nesses of the springs have the same coefficient of variation.


30/52



Discuss the influence of the number of springs in the assembly on

the variability of its stiffness.

Problem 2.22 Moon lander

In a design analysis of a suspension system for a moon lander, it

has been determined that the overall damping coefficient of the system

is a critical quality variable, and should be held within tight specifica-

tions.

Suppose that the shock absorbers are linear over their range of ap-

plication, and that the standard deviation of the damping coefficient ofeach shock absorber is a fixed fraction of its mean value for any sizeshock absorber.

Suppose also that there are two systems proposed: System F with

4 parallel shock absorbers, and System G with 16 parallel shock absorb-

ers, where both systems have the same total mean damping coefficient.

(You may assume that the damping coefficients are additive).

Determine the ratio of the standard deviation of the overall damp-

ing coefficient of assembly G to that of assembly F.

2.12 RELIABILITY

The reliabilityR of a system is the probability that the system will

perform as expected.

The unreliabilityQ of a system is the probability that the system

will fail to perform as expected.

R + Q = 1

Remark on terminology:

The term reliability is commonly used to refer to the probability offailure of one item due to degradation over time. It is not generally used

for the general conformance to specification of the product coming off

the end of a production line. For simplicity however we will often use

the term reliability to mean probability of conforming to specifica-

tion.

Thus the reliability of a mass-produced productis equivalent to the

proportion of the product within specification.


31/52

RELIABILITY 2 - 31


2.12.1 Operating windows

Many systems (products, designs) depend for their correct perfor-

mance on the values of their quality variables (design parameters or

functions of design parameters) remaining within given bounds, limits,

or tolerances. This leads to viewing these bounds as the frame of an op-

erating window.

A design specification might read something like:

The parameterx must lie in the rangexL toxU.

Sincex will usually have a distribution of values it is most likelythat not all values will lie in this range.

If a products function depends only on the single parameterx, then

its reliability is the probability thatx lies in the rangexL toxU. That is

R = Pr (xLxxU)

and this is represented by the area under the probability density

function which can be seen through the operating window.

Conversely the unreliability is the area under the curve outside the

window.

OPERATING WINDOW

2.12.2 Example: Paper feeder operating window

Many photocopiers have paper feeders which use the frictional


32/52



driving force of an elastomeric covered roll which sits on top of the pa-

per stack.

It is clear that if the normal force exerted by the roll on the paper is

too low, the paper will not move. This failure mode is called a misfeed.

Conversely, if the normal force is too high, more than one sheet will be

driven forward. This failure mode is called a multifeed. The normal

force thus becomes a quality variable which must be kept within defined

upper and lower specification limits.

Considering the normal force as a random variable, the proportion

of its probability density function that we can see through the windowframe formed by the upper and lower specification limits is the reliabil-

ity of the feeder for the normal force failure modes.

2.12.3 Supply and demand

Another type of system depends for its correct performance on the

demandx being less than the supplyxs.

R = Pr (x


33/52

RELIABILITY 2 - 33


The margin of safety is defined byy =xs -x.

Hence the reliability may be written

R = Pr (y > 0)

The operating window for y is then 0 y for a margin of safetyproblem.

2.12.4 Estimation of the reliability

1. Calculate the mean and variance ofy.

2. If the type of distribution fory is known, use a formula or table forits cumulative distribution function to calculate the area 0 y.Otherwise use a table for the normal distribution as follows:

3. Calculate the distancez of the mean ofy from zero in units of thestandard deviation ofy.

4. Read off the required probability (reliability or unreliability) in thetable.

The parameter z (measured in standard deviations ofy) is often

called the reliability index or safety index. It can be seen from the table

below that the reliability is quite sensitive to small changes in z for zgreater than about 2. A doubling ofz from 2.4 to 4.8 decreases the prob-

ability of failure by a factor of approximately 10 000!

You can visualize this geometrically by imagining what happens to

the area of the distribution fory 0 as you shift the distribution to theright.

Table 2: Probability of failure versus reliability index for a Normal

Distribution

RELIABILITY

INDEX Z

UNRELIABILITY

Q PER MILLION

RELIABILITY

INDEX Z

UNRELIABILITY

Q PER MILLION

0.00 500 000 2.33 10 000

0.67 250 000 3.10 1 000

1.00 160 000 3.72 100

1.28 100 000 4.25 10

1.65 50 000 4.75 1


34/52



Problem 2.23 Bolt strength reliability

A production run of bolts has a normally distributed ultimate ten-

sile strength with mean 100 MN and standard deviation 2 MN.

The applied load is expected to be normally distributed with mean

90 MN and standard deviation 4 MN.

What proportion may be expected to fail?

Problem 2.24 Buoyancy force reliability

A design calculation predicts that the buoyancy force B acting ona sonar device is normally distributed with mean 800 N and standard de-

viation 24 N, and that the weight force W is normally distributed with

mean 800 N and standard deviation 8 N.

The sonar device fails to operate as intended if: a) it sinks, or b) its

buoyancy force exceeds its weight force by more than 32 N.

Calculate the probability of failure to operate as intended (to 3 dec-

imal places).

Problem 2.25 Bearing fit

A mass-produced bearing of a journal bearing has a normally dis-

tributed diameter with mean 50 mm and tolerance 0.03 mm.The journal has a normally distributed diameter with mean 49.9

mm and tolerance 0.03 mm.The assembly fails if (a) the journal will not fit in the bearing, or

(b) the diametral clearance is greater than 0.1 mm.

1. Estimate the proportion of assemblies that might be expected to

fail if the manufacturer is very inexperienced in this area of manufac-

ture.

2. Estimate the proportion of assemblies that might be expected to

fail if the manufacturer is highly experienced in this area of manufac-

ture.

Problem 2.26 Shaft failure

A production run of shafts has a normally distributed failure torque


35/52

PRODUCTS OF RANDOM VARIABLES 2 - 35


with mean 100 kNm and variance 9 (kNm)2.

The applied torque is expected to be normally distributed with

mean 80 kNm and variance 16 (kNm)2.

What proportion of product may be expected to fail?

(Express your answer as number of failures per million)

Problem 2.27 Fitting of car doors and windshields

Discuss the potential application of the margin of safety concept to

the fitting together of components in the automobile industry, for exam-

ple, doors and windshields.

2.13 PRODUCTS OF RANDOM VARIABLES

To develop formulae for the moments of products of independent

random variables we use the fact that if x, y, are any independentran-

dom variables, then

E[x y ] = E[x] E[y]

The formulae developed below will be exact independentof the

type of distribution to which the random variables belong.The formulae are used by calculating the mean, variance and skew

in succession.

Suppose z is a product of any number of independent random vari-

ables xi: z = x1 x2 x3

2.13.1 The mean of a product

The mean of a product is a direct application of the formula above.

z = x1 x2 x3

E[z] = E[x1] E[x2] E[x3]

z = x1x2x3

The mean of a product of independent random variables is simply

the product of their means.


36/52



2.13.2 The variance of a product

The variance of a product is obtained by taking the expectation of

the square of z

z2 = x12 x2

2 x32

E[z2] = E[x12] E[x2

2] E[x32]

(z2 + z) = (x1

2 + x1) (x22 + x2) (x3

2 + x3)

To calculate the variance z of a product of independent randomvariables, first compute the product on the right hand side of the equa-

tion above and then subtract the square of the mean z2 calculated pre-

viously.

2.13.3 The skew of a product

z3 = x13 x2

3 x33

E[z3] = E[x13] E[x2

3] E[x33]

(z3 + 3 zz + sz) = (x1

3 + 3 x1x1 + sx1)

(x23 + 3 x2x2 + sx2) (x3

3 + 3 x3x3 + sx1)

To calculate the skew sz of a product of independent random vari-

ables, first compute the product on the right hand side of the equation

above and then subtract the term z3 + 3 zz calculated from the pre-

vious steps.

Higher moments are calculated in a similar fashion.

Problem 2.28 Volume of a cube

Calculate the mean and variance of the volume of a cube of side Lwhere the sides are machined independently by the same machining

process and are therefore considered to be identically distributed inde-

pendent random variables each with mean and variance .

Comment on how this calculation differs from one based simply on

the formula V = L3 (see below).


37/52

POSITIVE INTEGER POWERS 2 - 37


2.14 POSITIVE INTEGER POWERS

2.14.1 The mean of a positive integer power

We have already derived the formula for the Expectation of a pos-

itive integer power of a random variable in terms of central moments.

Since the expectation gives the mean value, we have immediately that

for z = xn:

2.14.2 Tables for a Normally distributed random variable

Since the central moments of a normal distribution can all be ex-

pressed in terms of its mean and variance (see the listing in the section

on the Normal distribution), its powers can therefore be expressed via

the above formula in terms of them also.

The tables below thus give exact formulae for calculating the mean,

variance and skew of positive integer powers of a normally distributed

random variable x with mean and variance .The entries in the table are expressed in the form

(first order approximation) (1 + terms in the variance ratio u)

where the variance ratio u has been defined as the square of the co-

efficient of variation u = v/2.

Table 3: Moments of a square

z = x2

z 2 (1 + u)

z 4 2 (1 + u/2)

sz 24 22 (1 + u/3)

z

n

i

E x x

( )i

[ ] x

( )n i

i 0=

n

=


38/52



Problem 2.29 Inconsistency?

If = 0, does this imply that z, z, sz are all zero?

2.15 GENERAL FUNCTIONS

2.15.1 Introduction

We complete our introductory discussion of probabilistic design by

describing a method (called the Moment Analysis Method) by which

you can calculate the moments of any differentiable function of inde-

Table 4: Moments of a cube

z = x3

z 3 (1 + 3 u)

z 9 4 (1 + 4 u + (5/3) u2)

sz 162 52 (1 + (16/3) u + 5 u2)

Table 5: Moments of a fourth power

z = x4

z 4 (1 + 6 u + 3 u2)

z 16 6 (1 + (21/2)u + 24 u2 + 6 u3)

sz 576 82 (1 + 16 u + (149/2) u2 + 99 u3 + 33 u4)

Table 6: Moments of a fifth power

z = x5

z 5 (1 + 10 u + 15 u2)

z 25 8 (1 + 20 u + 114 u2 + 180 u3 + (189/5) u4)

sz 1500 112 (1 + (97/3)u + 366 u2 + 1710 u3 + 2997 u4 + 1323 u5)


39/52

GENERAL FUNCTIONS 2 - 39


pendent random variables, and hence get an estimate of the variability

inherent in a given design. Indeed, all the formulae we have introduced

so far may be derived by this method.

The basic principle of the Moment Analysis Method is the specifi-

cation of each probability distribution by its set of moments in the form

{mean, variance, skew, kurtosis,...}. Then, if we wish to calculate a

function of several random variables, the moments of that function will

be functions of the moments of those several random variables.

The two techniques that we will use are

1. Expansion of the function in a Taylor series2. Application of the Expectation operator to the series

2.15.2 The basic algorithm

Calculation of the mean

1. Expand the function z = g(x, y, ...) as a Taylor series about the

mean values (x, y, ) of the independent random variables.

2. Calculate the mean of the function (z) by calculating the expec-tation of the terms in the expansion.

Calculation of the nth central moment

1. Expand the function [z - z]n as a Taylor series about the mean

values (x, y, ) of the independent random variables.

2. Calculate the nth central moment of the function (z, sz, kz, )by calculating the expectation of the terms in the expansion.

3. Calculate z and substitute for it in the expression.

2.15.3 The theoretical foundation

Assumptions

The fundamental assumptions upon which the method is based are:

1. The random variables x, y, are independent. (Very important!)

2. The pertinent information content of each of the distributions is

sufficiently well represented by a finite (small) number of moments.


40/52



3. The function is sufficiently well represented by a finite (small)

number of terms of its Taylor series.

Formulae

The fundamental formulae upon which the method is based are:

1. The mean z of a function z = g(x, y, ...) is the expectation of thefunction.

2. The nth central moment (n)z of a function z = g(x, y, ...) is the

expectation of (z - z)n

3. The expectation of a linear sum is the sum of the expectations of

the terms.

4. The expectation of a product of independent random variables is

the product of their expectations.

2.15.4 How to write down a Taylor series

In this section we discuss a mnemonic method for easily writing

down a Taylor series expansion of a function of several variables.

Suppose you have a function z = g(x, y, ...) and you wish to writedown the Taylor series for z expanded about the point: x = x, y = y,. A simple mnemonic way of doing this is as follows:

1. Write down the power series for exp(X+Y+):

1 + (X+Y+) + (1/2!)(X+Y+)2 + (1/3!)(X+Y+)3 +

2. Expand the terms:

1 + (X+Y+) + (1/2!)(X2+2XY+Y2+)

+ (1/3!)(X3+3X2Y+3XY2+Y3+) +

3. Make the following replacements:1 z[ ] g x y,( )=( )

Xn

xn

n

z

x x( )n


41/52



and so on for more products of more than two variables.

Remember that the notation [] means that the bracketed func-

tion is evaluated at the point x = x, y = y, .

2.15.5 How to write down the expectation of a function

Again suppose you have a function z = g(x, y, ...) and you wish to

write down an expression for the expectation E[z] of z. The normal pro-

cedure for doing this is:

1. Write down the Taylor series with x0 = x, y0 = y,

2. Apply the expectation operator to the series, remembering its

properties when it acts on a constant, a linear sum, and a product of in-

dependent random variables.

3. Make the following replacements:

The resulting expression is a series expressing E[z] in terms of the

moments of x, y, .

If the series terminates the expression will be exact. A polynomial

function, for example, will terminate.

2.15.6 The shortest way to write down the expectation

It may be somewhat shorter to first simplify the terms in our origi-nal mnemonic expansion

1 + (X+Y+) + (1/2!)(X2+2XY+Y2+) +

before replacing them with their corresponding terms in the Taylor se-

ries expansion.

We list possible simplification rules below, and illustrate them with

the example of a function of two variables for which we know only their

means and variances:

XnY

m

xn

ym

n m+( )

z

x x( )n

y y( )m

E xx

[ ]

0

E x x( )n

[ ] n( )x


42/52



1 + (X+Y) + (1/2!)(X2+2XY+Y2) + (1/3!)(X3+3X2Y+3XY2+Y3)

+ (1/4!)(X4+4X3Y+6X2Y2+4XY3+Y4)

+ (1/5!)(X5+5X4Y+10X3Y2+10X2Y3+5XY4+Y5) +

1. Any term involving a variable to the first power is zero since the

expectation E[x - x] is zero.1 + (1/2!)(X2+Y2) + (1/3!)(X3+Y3)

+ (1/4!)(X4+6X2Y2+Y4)

+ (1/5!)(X5+10X3Y2+10X2Y3+Y5) +

2. Any term involving a higher power leading to a moment for

which you have no information must be omitted.

In this example we only know means and variances, hence the ex-

pression reduces to

1 + (1/2!)(X2+Y2) + (1/4!)(6X2Y2)

3. If the coefficients resulting from the higher derivatives in the ex-

pansion are small enough compared to those resulting from the lower

ones, the corresponding terms may be neglected. This is often the case

for functions which are not too far off linear in the region near the point

= (x, y).

In this example we would look at the comparative size of

Assuming the term can be neglected the expression reduces to

1 + (1/2!)(X2+Y2)

leading finally to a general second order approximation for z:

It is evident from this process that the same form is valid for any

number of variables.

x2

y2

4

z

z g x y,( )

1

2--- x2

2

z

x y2

2

z

y+

+=

z g x y , ,( )1

2---

x2

2

z

xy

2

2

z

y + +

+=


43/52



Note carefully that the mean of a general function is only equal to

the function of the means as a first order approximation.

2.15.7 Calculation of the variance of a function

As an example of the method described for calculating higher order

moments of a differentiable function of random variables, we will cal-

culate an expression for the second order approximation to the variance

of z = g(x, y, ...), that is, E[(z - z)2].

1. Since (z - z

)2 is still a function of x, y, ..., we can let Z = (z -

z)2. E[(z - z)

2] then becomes Z which we can write down directlyfrom the result derived in the section above:

2. Evaluate [Z]:

3. Evaluate a typical second derivative:

4. Collect the terms and simplify to get finally:

This is a formula for the second order approximation to the vari-

ance of a differentiable function of any number of random variables. It

is one of the most important formulae in the area of probabilistic and ro-

bust design.

z Z Z[ ]1

2---

x2

2

Z

xy

2

2

Z

y + +

+= =

Z[ ] z[ ] z( )2 1

4

---

x2

2

z

xy

2

2

z

y + +

2

= =

x2

2

Z2

xz 2

z z( )x

2

2

z+

=

z x

z

2

x y

z

2

y + +

1

4---

x2

2

z

xy

2

2

z

y + +

2

=


44/52



2.15.8 Computer implementation

The Moment Analysis Method is ideally suited to being pro-

grammed using a symbolic mathematical programming language.

The size and complexity of the problems that can be tackled will

depend on the memory and speed of the computational devices used.

The accuracy of the result will in addition depend on the degree of lin-

earity of the function g(x, y, ) near the mean values of the independent

random variables (x, y, ), together with the number of momentsused to specify each of their distributions. The more linear the function

and the more moments used, the more accurate the results will be. Themost significant problems will arise when the function has a singularity

within the support of the distribution.

Problem 2.30 Mean of a function of one variable

Derive an expression for the mean of a general differentiable func-

tion of a single random variable up to and including the term involving

the kurtosis.

Letx be a symmetrically distributed random variable with mean 0,

variance 1 and kurtosis 1. Using the results above, determine an approx-imation to the mean ofex.

Problem 2.31 Variance of a function of one variable

Derive an expression for the variance of a general differentiable

function of a single random variable up to and including the term in-

volving the skew. Neglect terms of order higher than 3.

Letx be a symmetrically distributed random variable with mean 0

and variance 1. Using the results above, determine an approximation to

the variance ofex

.

Problem 2.32 Mean of a function of two variables

By using a Taylor series expansion about the mean values of ran-

dom variables x and y, derive an approximate formula for the mean val-

ue of z = g(x, y) in terms of the means and variances of x and y.


45/52

SUMMARY OF APPROXIMATE FORMULAE 2 - 45


Problem 2.33 Probabilistic means

The mean of a function is not generally the function of the means.

1. Describe the types of functions for which the mean of the func-

tion is the function of the means.

2. Discuss the ramifications of these results for ordinary tolerance

analysis.

3. Discuss the ramifications of these results for quality design.

2.16 SUMMARY OF APPROXIMATE FORMULAE

This section summarizes the first and second order approximations

to the mean z and variance z of a differentiable function z = g(x, y, ...)of independent random variables x, y, ...

If g(x, y, ...) is approximately linear in the region within several

standard deviations of the mean values (x, y, ) of (x, y, ...), then thefirst order approximation will often give satisfactory estimates. In any

case, the value of the extra term in the second order approximation may

always be computed to assess its significance.For estimates more accurate than that provided by the second order

approximation, moments of order higher than the variance will need to

be known.

2.16.1 First order approximation

Mean

Variance

z g x y , ,( )=

z xz

2

x yz

2

y + +=


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2.16.2 Second order approximation

Mean

Variance

Problem 2.34 Approximate formulae

Write down the formulae for the first and second order approxima-

tion to the variance of a function z = g(x1, x2, x3) of independent random

variables xi each with mean i and variance i.

What is the main assumption upon which the validity of these for-

mulae resides?

Problem 2.35 Right-angled bracket

An angled bracket has legs of nominal length B and C. The nominal

angle between them is = 90. The distance A between the ends of thelegs of the bracket is an important quality variable.

Using the formulae above, derive first order approximations for the

mean and variance of A in terms of the means and variances of B, C, and

.

Problem 2.36 Applications for probabilistic design

From your own experience, describe a possible application for

probabilistic design.

z g x y , ,( )1

2---

x2

2

z

xy

2

2

z

y + +

+=

z xz

2

x yz

2

y+ 1

4---

x2

2

z

xy

2

2

z

y+

2

=


47/52

GENERAL POWER FUNCTIONS 2 - 47


2.17 GENERAL POWER FUNCTIONS

In engineering we often deal with general power functions of the

form

In this section we take the formulae above for the approximate

mean and variance of z and sketch their application to functions of this

form.

2.17.1 The mean of a general power function

1. Calculate the second derivatives:

2. Substitute into the formula for the mean z and rewrite terms ofthe form /2 as squares of coefficients of variation:

2.17.2 The variance of a general power function

In a similar fashion we can substitute into the formula for the vari-

ance:

2.17.3 The coefficient of variation of a power function

From the above two results, we can now write down an expression

for the second order approximation to the square of the coefficient of

variation of z in terms of the squares of the coefficients of variation of

x, y, .

z cxmy

n=

x2

2

z

m m 1( )cxm 2

yn[ ] cx

my

n( )m m 1( )

1

x2

--------= =

z cxm

yn( ) 1

1

2--- m m 1( )x

2n n 1( )y

2+ +{ }+=

z cxm

yn

( )

2

m

2

x

2

n

2

y

2

+ +{ }

1

4--- m m 1( )x

2

n n 1( )y

2

+ +{ }

2

=


48/52



The first order approximation in coefficient of variation terms is

then simply:

2.17.4 The quotient of two random variables

We apply these formulae to obtain second order approximation ex-

pressions for the quotient of two random variables z = x/y:

The mean of a quotient

The variance of a quotient

2.17.5 The inverse of a random variable

We again apply these formulae to obtain second order approxima-

tion expressions for the inverse of a random variables z = 1/y:

The mean of an inverse

z2

m2x

2n

2y

2+ +{ }

1

4--- m m 1( )x

2n n 1( )y

2+ +{ }

2

11

2--- m m 1( )x

2n n 1( )y

2+ +{ }+

2---------------------------------------------------------------------------------------------------------------------------------------------=

z2

m2x

2n

2y

2+ +=

zxy----- 1 y

2+( )=

zxy-----

2

x2

y2

y4

+( )=

z1

y----- 1 y

2+( )=


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The variance of an inverse

Problem 2.37 The mean of a square

Suppose that the standard deviation of a random variablex is 10%

of the mean, that is .

Show that for z = x2 the first order approximation to the mean of z

underestimates the second order approximation by about 1%.

Problem 2.38 Volume of a cylinder

A mass-produced cylindrical container has an internal diameter of

{D = 2 m, D = 0.01 m2} and an internal length of

{L = 10 m, L = 0.04 m2}.

Compare the first and second order approximations to the mean

and variance of its volume.

Problem 2.39 Second moment of area

A quality variable z is related to independent random variables x

and y by z = a xp yq (where a, p and q are constants).

1. Using the first order formulae for the mean and variance of a

function of random variables, derive an expression for the approximate

coefficient of variation of z in terms of the coefficients of variation of x

and y.

2. A triangular beam cross-section has second moment of area

, where B and H are independent random variables with

means B, H and variances B, H respectively. Use the formula de-rived in 1. to derive an expression for the variance I of the second mo-ment of area in terms ofB, H, B, H.

Problem 2.40 Coefficients of variation

The radius and length of a cylinder are independent random vari-ables with identical coefficients of variation C.

z1

y-----

2 y

2y

4( )=

x 2 0.01=

I1

36------BH

3=


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1. Determine the coefficient of variation of the volume of the cyl-

inder, using both the first and second order coefficient of variation ap-

proximation formulae.

2. Comment on the increased accuracy provided by the second or-

der approximation.

Problem 2.41 Experimental formula

An experimentally derived formula for a quality variable W interms of design parametersX, Y,Zis

W = 2.657X0.2Y0.7Z0.1

The coefficients of variation ofX, YandZare each equal to k.

Determine an estimate of the coefficient of variation ofWin terms

ofk.

Problem 2.42 First order approximation

1. State the formula for the first order approximation to the vari-ance of a function of random variables.

2. From the formula in 1. above derive the formula for the varianceof the functionZ=X2 sinY, whereXand Yare independent random vari-ables.

3. From your result in 2. determine the standard deviation ofZifthe means and standard deviations ofXand Yare 2 units each.

Problem 2.43 Springs in series

A mass produced product contains two springs with stiffnesses K1and K2 connected in series.

1. Calculate the first order approximation to the variance of thecombined stiffness.

2. Calculate the first order approximation to the variance of a) A =

K1 K2, b) B = K1 + K2 and c) C = A/B.

3. Explain why this does not give the same result as in 1. (Hint: The

difference is not due to the inexactitude of the approximations used.)


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Problem 2.44 The minimum diameter of a rod in tension

Determine the minimum mean diameter D of a circular rod withultimate tensile strength S which will sustain a load P in tension with a

reliability of R.

The rod is to be made of 4130 steel of ultimate tensile strength S,

considered normally distributed with mean S = 1075 MPa, and vari-ance S = 900 MPa

2.

The manufacturing process is known to be capable of turning out

99.7% of the product to a tolerance of 1.5% of the rod diameter.The load P is the resultant of a large number of randomly varyingloads and hence can be considered to be normally distributed. Its mean

is P = 13200 N, and variance is P = 40 000 N2.

The reliability R (equivalent to the proportion of product within

specification) is required to ensure a probability of failure of less than

one per million of shafts produced.


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4 probabilistic design

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