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<P align=3Dcenter><FONT face=3DArial size=3D6><B>PROCESS =
TOLERANCING:<BR>A SOLUTION TO=20
THE DILEMMA OF <BR>WORST-CASE VERSUS STATISTICAL =
TOLERANCING</B></FONT></P>
<P align=3Dcenter><A href=3D"http://www.variation.com/wayne.html"><FONT=20
face=3DHelvetica size=3D4><B>Dr. Wayne A. Taylor</B></FONT></A></P>
<BLOCKQUOTE>
  <P>When solving tolerancing problems, one must choose between =
worst-case=20
  tolerancing and statistical tolerancing. Both of these methods have =
their pros=20
  and cons. If worst-case tolerancing is used, all tolerances must be =
specified=20
  as worst-case tolerances. If statistical tolerancing is used, all =
tolerances=20
  must be specified as statistical tolerances. In reality, the behavior =
of=20
  certain inputs is best described using worst-case tolerances while the =

  behavior of other inputs is best described by statistical tolerances. =
Still=20
  other inputs are not adequately represented by either type of =
tolerance. This=20
  article reviews the two current methods of tolerancing along with =
their pros=20
  and cons. It then introduces a new method of tolerancing, called =
process=20
  tolerancing, which solves many of the problems associated with the =
current=20
  approaches. Process tolerancing represents a unified approach to =
tolerancing=20
  that encompasses both of the previous approaches. Using process =
tolerancing,=20
  worst-case tolerances and statistical tolerances can be combined into =
the same=20
  analysis. The flexibility provided by process tolerancing results in =
solutions=20
  to such common problems as multiple components from the same lot and=20
  off-center processes.</P></BLOCKQUOTE>
<BLOCKQUOTE>
  <P>KEY WORDS: Statistical Tolerance, Worst-Case Tolerance, Process =
Tolerance,=20
  Variation Transmission Analysis, Propagation of Error, and Six Sigma=20
  Quality.</P></BLOCKQUOTE>
<P align=3Dleft><FONT face=3DArial =
size=3D4><B>INTRODUCTION</B></FONT></P>
<P>Tolerance analysis is used during product design. For example, =
consider the=20
design of a pump. Flow rate is a key characteristic that must be =
controlled. It=20
is called an output variable. Flow rate is affected by such factors as =
diameter=20
of the piston, stroke length, motor speed and viscosity of the solution. =
These=20
factors are called input variables. The goal of tolerance analysis is to =
predict=20
the behavior of the output variable based on an understanding of how the =
inputs=20
behave. The behavior of the inputs can then be adjusted to cause the =
output to=20
behave as desired.</P>
<P>Tolerance analysis is equally important to process development. For =
example,=20
consider the installation of a new heat seal machine. Seal strength is =
the=20
output variable. Input variables include die temperature, seal time, and =

pressure. Using tolerance analysis, a process window can be determined =
for the=20
inputs that ensures that seal strength stays within its specification=20
limits.</P>
<P>Two methods of tolerancing currently exist: worst-case and =
statistical=20
tolerancing. Both have their pros and cons. These two methods will be =
reviewed=20
and compared. A new method of tolerancing, called process tolerancing, =
will then=20
be introduced. Process tolerancing represent a unified approach to =
tolerancing=20
that encompasses both the previous approaches.</P>
<P>The greatest benefit of process tolerancing is that it allows both =
worst-case=20
and statistical tolerances to be combined in the same analysis. Some =
inputs are=20
best described by worst-case tolerances. For example, consider solution=20
viscosity for the pump. Different customers will use the pump with =
different=20
solutions. Each customer wants the correct flow rate. As a result, a =
worst-case=20
tolerance best describes the behavior of solution viscosity. Some inputs =
are=20
best described by statistical tolerances. For example, consider die =
temperature=20
on the heat seal machine. Die temperature is controlled automatically =
using=20
close-loop feedback. This ensures it remains at the specified target. =
Its=20
behavior is best described using a statistical tolerance. The purpose of =
a=20
tolerance analysis is to predict the behavior of the output based on =
knowledge=20
of how the inputs behave. Key to its success is the ability to =
accurately=20
describe the behavior of the inputs. Using process tolerancing, you can =
select=20
the type of tolerance that best describes the behavior of each =
input.</P>
<P>Some inputs are not well represented by either worst-case or =
statistical=20
tolerances. For example, inputs from processes that are off-center or =
lack=20
statistical control. Process tolerancing offers much greater flexibility =
than=20
the current two methods and can be used to better represent such =
situations.=20
Because of this added flexibility, process tolerancing provides =
solutions to=20
such common problems as multiple components from the same lot and =
off-center=20
processes.</P>
<P align=3Dleft><FONT face=3DArial size=3D4><B>WORST-CASE =
TOLERANCES</B></FONT></P>
<P>Worst-case tolerances are specified in the form 1.00 <FONT=20
face=3DSymbol>=B1</FONT> 0.03". The tolerance half-width of 0.03" =
represents the=20
maximum amount that a "good" unit can deviate from target. This =
worst-case=20
tolerance can also be represented in interval form as (0.97", 1.03"). A =
unit=20
falling outside the range of 0.97" and 1.03" fails to meet the tolerance =
and is=20
called a nonconformance or defect. With worst-case tolerances, units of =
product=20
can fall anywhere in the tolerance interval. All units could be 0.97", =
all units=20
could be 1.03", or the units could be uniformly spread across the =
interval.=20
Nonsymmetrical tolerances can also be specified, but are ignored here =
because=20
they can be easily translated into equivalent symmetrical =
tolerances.</P>
<P>If worst-case tolerances are specified for the inputs, a worst-case =
tolerance=20
can be calculated for the output. Let the output be represented by =
<I>y</I> and=20
the inputs by <I>x<SUB>1</SUB>, =85, x<SUB>n</SUB></I>. Suppose <I>y</I> =
=3D=20
f(<I>x<SUB>1</SUB>, =85, x<SUB>n</SUB></I>) and that the inputs have =
worst-case=20
tolerances <I>t<SUB>x<SUB>i</SUB></SUB></I> <FONT =
face=3DSymbol>=B1</FONT> <FONT=20
face=3DSymbol>D</FONT> <SUB><I>x<SUB>i</SUB></I></SUB>. Then the =
resulting=20
worst-case tolerance for <I>y</I>, denoted <I>t<SUB>y</SUB></I> <FONT=20
face=3DSymbol>=B1</FONT> <FONT face=3DSymbol>D</FONT> =
<SUB><I>y</I></SUB>, is:</P>
<P><IMG height=3D42 alt=3D"Equation 1" hspace=3D30=20
src=3D"http://www.variation.com/techlib/ta-2_files/ta-2-e1.gif" =
width=3D197=20
align=3DtextTop> (Eq. 1)</P>
<P><IMG height=3D42 alt=3D"Equation 2" hspace=3D30=20
src=3D"http://www.variation.com/techlib/ta-2_files/ta-2-e2.gif" =
width=3D202=20
align=3DtextTop> (Eq. 2)</P>
<P><IMG height=3D43 alt=3D"Equation 3" hspace=3D30=20
src=3D"http://www.variation.com/techlib/ta-2_files/ta-2-e3.gif" =
width=3D82=20
align=3Dmiddle> (Eq. 3)</P>
<P><IMG height=3D43 alt=3D"Equation 4" hspace=3D30=20
src=3D"http://www.variation.com/techlib/ta-2_files/ta-2-e4.gif" =
width=3D89=20
align=3Dmiddle> (Eq. 4)</P>
<P>The terms <I>l<SUB>y</SUB></I> and <I>u<SUB>y</SUB></I> represent the =
minimum=20
and maximum values of <I>y</I> over the worst-case tolerances for the =
inputs.=20
The worst-case tolerance for <I>y</I> is the interval =
(<I>l<SUB>y</SUB></I>,=20
<I>u<SUB>y</SUB></I>). The target for <I>y</I>, <I>t<SUB>y</SUB></I>, is =
the=20
midpoint of this interval while <FONT =
face=3DSymbol>D</FONT><SUB><I>y</I></SUB> is=20
half the width of the interval. The following example demonstrates the =
use of=20
these equations.</P>
<P><B>Example:</B> Suppose that four identical components are stacked on =
top of=20
each other and that we are interested in controlling the overall height. =
Overall=20
height is denoted by <I>h</I>. Specification limits for overall height =
are from=20
3.9" to 4.1". The height of the four components are denoted by=20
<I>d<SUB>1</SUB></I>, =85, <I>d<SUB>4</SUB></I>. Suppose these =
components have a=20
tolerance of 1.00 <FONT face=3DSymbol>=B1</FONT> 0.03". Then:</P>
<P><IMG height=3D24 alt=3D"Equation 5" hspace=3D30=20
src=3D"http://www.variation.com/techlib/ta-2_files/ta-2-e5.gif" =
width=3D133=20
align=3Dtop> (Eq. 5)</P>
<P><IMG height=3D60 alt=3D"Equation 6" hspace=3D30=20
src=3D"http://www.variation.com/techlib/ta-2_files/ta-2-e6.gif" =
width=3D427=20
align=3DtextTop> (Eq. 6)</P>
<P><IMG height=3D60 alt=3D"Equation 7" hspace=3D30=20
src=3D"http://www.variation.com/techlib/ta-2_files/ta-2-e7.gif" =
width=3D432=20
align=3DtextTop> (Eq. 7)</P>
<P><IMG height=3D41 alt=3D"Equation 8" hspace=3D30=20
src=3D"http://www.variation.com/techlib/ta-2_files/ta-2-e8.gif" =
width=3D222=20
align=3Dmiddle> (Eq. 8)</P>
<P><IMG height=3D41 alt=3D"Equation 9" hspace=3D30=20
src=3D"http://www.variation.com/techlib/ta-2_files/ta-2-e9.gif" =
width=3D228=20
align=3Dmiddle> (Eq. 9)</P>
<P>The resulting worst-case tolerance for <I>h</I> is 4.00 <FONT=20
face=3DSymbol>=B1</FONT> 0.12" or (3.88", 4.12"). This tolerance is =
shown in Figure=20
1 along with the specification limits. Worst-case tolerancing, predicts =
that the=20
specification limits will not be meet and that the tolerance for =
component=20
height must be tightened.</P>
<P align=3Dcenter><FONT face=3DHelvetica><B><IMG height=3D168 =
alt=3D"Figure 1"=20
src=3D"http://www.variation.com/techlib/ta-2_files/ta-2-1.gif"=20
width=3D304></B></FONT></P>
<P align=3Dcenter><FONT face=3DHelvetica><B>Figure 1: Worst-case =
tolerance for=20
overall height<BR>4-component stack example</B></FONT></P>
<P>When the output is the sum of the inputs as in the above example, the =

worst-case tolerance equations simplify to:</P>
<P><IMG height=3D25 alt=3D"Equation 10" hspace=3D30=20
src=3D"http://www.variation.com/techlib/ta-2_files/ta-2-e10.gif" =
width=3D129=20
align=3DtextTop> (Eq. 10)</P>
<P><IMG height=3D25 alt=3D"Equation 11" hspace=3D30=20
src=3D"http://www.variation.com/techlib/ta-2_files/ta-2-e11.gif" =
width=3D158=20
align=3Dmiddle> (Eq. 11)</P>
<P>Equation 11 is frequently cited in textbooks on tolerancing. However, =
when=20
more complex equations are involved, Equations 1-4 must be used =
instead.</P>
<P align=3Dleft><FONT face=3DArial size=3D4><B>STATISTICAL =
TOLERANCES</B></FONT></P>
<P>Statistical tolerances are determined by a target, =
<I>t<SUB>x</SUB></I>, and=20
maximum standard deviation, <I><FONT =
face=3DSymbol>s</FONT><SUB>x</SUB></I>. In=20
this article they are denoted by <I>t<SUB>x</SUB></I> <FONT =
face=3DSymbol>=A3</FONT>=20
<I><FONT face=3DSymbol>s</FONT><SUB>x</SUB></I>. They are more commonly =
denoted by=20
<I>t<SUB>x</SUB></I> <FONT face=3DSymbol>=B1</FONT> <FONT =
face=3DSymbol>D</FONT>=20
<SUB><I>x </I></SUB><IMG height=3D9=20
src=3D"http://www.variation.com/techlib/ta-2_files/ta-2-st.gif" =
width=3D17=20
align=3Dtop>. When using the <IMG height=3D12=20
src=3D"http://www.variation.com/techlib/ta-2_files/ta-2-st.gif" =
width=3D23> symbol,=20
it is most common to have <FONT face=3DSymbol>D</FONT> =
<SUB><I>x</I></SUB> =3D=20
3<I><FONT face=3DSymbol>s</FONT><SUB>x</SUB></I>. However, this =
relationship is=20
not formalized in the standards and the six sigma approach uses <FONT=20
face=3DSymbol>D</FONT> <SUB><I>x</I></SUB> =3D 6<I><FONT=20
face=3DSymbol>s</FONT><SUB>x</SUB></I>. Throughout this article we will =
assume=20
that <FONT face=3DSymbol>D</FONT> <SUB><I>x</I></SUB> =3D 3<I><FONT=20
face=3DSymbol>s</FONT><SUB>x</SUB></I>. Using this interpretation, both =
1.00 <FONT=20
face=3DSymbol>=A3</FONT> 0.01" and 1.00 <FONT face=3DSymbol>=B1</FONT> =
0.03" <IMG=20
height=3D9 =
src=3D"http://www.variation.com/techlib/ta-2_files/ta-2-st.gif" =
width=3D17=20
align=3Dtop> denote the statistical tolerance with target =
<I>t<SUB>x</SUB></I> =3D=20
1.00" and maximum standard deviation <I><FONT=20
face=3DSymbol>s</FONT><SUB>x</SUB></I> =3D 0.01".</P>
<P>The worst-case tolerance 1.00 <FONT face=3DSymbol>=B1</FONT> 0.03" =
and the=20
statistical tolerance 1.00 <FONT face=3DSymbol>=B1</FONT> 0.03" <IMG =
height=3D9=20
src=3D"http://www.variation.com/techlib/ta-2_files/ta-2-st.gif" =
width=3D17=20
align=3Dtop> both assume that units of product fall in the interval =
(0.97",=20
1.03"). However, the statistical tolerance is more restrictive. The =
statistical=20
tolerance further requires that the units of product have an average of =
1.00"=20
and a standard deviation no greater than 0.01". Statistical tolerancing =
assumes=20
centered processes. It further assumes that the inputs vary independent =
of each=20
other, i.e., are not correlated. Figure 2 shows a production batch that =
meets=20
the worst-case tolerance 1.00 <FONT face=3DSymbol>=B1</FONT> 0.03" but =
fails to meet=20
the assumptions of the corresponding statistical tolerance 1.00 <FONT=20
face=3DSymbol>=B1</FONT> 0.03" <IMG height=3D9=20
src=3D"http://www.variation.com/techlib/ta-2_files/ta-2-st.gif" =
width=3D17=20
align=3Dtop>.</P>
<P align=3Dcenter><FONT face=3DHelvetica><B><IMG height=3D168 =
alt=3D"Figure 2"=20
src=3D"http://www.variation.com/techlib/ta-2_files/ta-2-2.gif"=20
width=3D323></B></FONT></P>
<P align=3Dcenter><FONT face=3DHelvetica><B>Figure 2: Production batch =
satisfying=20
the requirements of a <BR>worst-case tolerance but not a statistical=20
tolerance</B></FONT></P>
<P>If statistical tolerances are specified for the inputs, a statistical =

tolerance can be calculated for the output. This amounts to determining =
the=20
average and standard deviation of the output. Simulations can always be =
used to=20
predict these two values. However, there also exist a variety of methods =
for=20
deriving approximate and sometimes exact equations for the average and =
standard=20
deviation. Details can be found in Taylor (1991), Cox (1986) and Evans =
(1975).=20
These approaches have a variety of names including statistical tolerance =

analysis, propagation of errors and variation transmission analysis.</P>
<P>Returning to the case where <I>y</I> =3D f(<I>x<SUB>1</SUB>, =85,=20
x<SUB>n</SUB></I>), equations can be derived for the average and =
standard=20
deviation of <I>y</I> in terms of the average and standard deviation of =
the=20
<I>x<SUB>i</SUB></I>=92s. These will be denoted <FONT =
face=3DSymbol>m</FONT>=20
<SUB><I>y</I></SUB> (<FONT face=3DSymbol>m</FONT> <SUB><I>x<FONT=20
size=3D2><SUB>1</SUB></FONT></I></SUB>, =85, <FONT =
face=3DSymbol>m</FONT>=20
<SUB><I>x<FONT size=3D2><SUB>n</SUB></FONT></I></SUB>, <FONT =
face=3DSymbol>s</FONT>=20
<SUB><I>x<FONT size=3D2><SUB>1</SUB></FONT></I></SUB>, =85, <FONT=20
face=3DSymbol>s</FONT> <SUB><I>x<FONT =
size=3D2><SUB>n</SUB></FONT></I></SUB>) and=20
<FONT face=3DSymbol>s</FONT> <SUB>y</SUB> (<FONT face=3DSymbol>m</FONT>=20
<SUB><I>x<FONT size=3D2><SUB>1</SUB></FONT></I></SUB>, =85, <FONT=20
face=3DSymbol>m</FONT> <SUB><I>x<FONT =
size=3D2><SUB>n</SUB></FONT></I></SUB>, <FONT=20
face=3DSymbol>s</FONT> <SUB><I>x<FONT =
size=3D2><SUB>1</SUB></FONT></I></SUB>, =85,=20
<FONT face=3DSymbol>s</FONT> <SUB><I>x<FONT =
size=3D2><SUB>n</SUB></FONT></I></SUB>)=20
respectively. The statistical tolerance for <I>y</I> can then be =
calculated as=20
follows:</P>
<P><IMG height=3D25 alt=3D"Equation 12" hspace=3D30=20
src=3D"http://www.variation.com/techlib/ta-2_files/ta-2-e12.gif" =
width=3D178=20
align=3Dmiddle> (Eq. 12)</P>
<P><IMG height=3D25 alt=3D"Equation 13" hspace=3D30=20
src=3D"http://www.variation.com/techlib/ta-2_files/ta-2-e13.gif" =
width=3D183=20
align=3Dmiddle> (Eq. 13)</P>
<P>When the output is the sum of the inputs, the statistical tolerancing =

equations simplify to:</P>
<P><IMG height=3D25 alt=3D"Equation 14" hspace=3D30=20
src=3D"http://www.variation.com/techlib/ta-2_files/ta-2-e14.gif" =
width=3D152=20
align=3DtextTop> (Eq. 14)</P>
<P><IMG height=3D32 alt=3D"Equation 15" hspace=3D30=20
src=3D"http://www.variation.com/techlib/ta-2_files/ta-2-e15.gif" =
width=3D169=20
align=3Dmiddle> (Eq. 15)</P>
<P>Multiplying both sides of Equation 15 by 3 gives:</P>
<P><IMG height=3D32 alt=3D"Equation 16" hspace=3D30=20
src=3D"http://www.variation.com/techlib/ta-2_files/ta-2-e16.gif" =
width=3D173=20
align=3Dmiddle> (Eq. 16)</P>
<P>Equation 16 is frequently cited in textbooks on tolerancing. It still =
holds=20
even if we assume <FONT face=3DSymbol>D</FONT> <SUB><I>x</I></SUB> =3D =
6<I><FONT=20
face=3DSymbol>s</FONT><SUB>x</SUB></I>. When more complex equations are =
involved,=20
Equations 12 and 13 should be used instead.</P>
<P><B>Example (continued):</B> Returning to the example of four =
components=20
stacked on top of each other, assume each component has a statistical =
tolerance=20
of 1.00 <FONT face=3DSymbol>=A3</FONT> 0.01". Then the statistical =
tolerance for=20
overall height is:</P>
<P><IMG height=3D25 alt=3D"Equation 17" hspace=3D30=20
src=3D"http://www.variation.com/techlib/ta-2_files/ta-2-e17.gif" =
width=3D353=20
align=3Dtop> (Eq. 17)</P>
<P><IMG height=3D32 alt=3D"Equation 18" hspace=3D30=20
src=3D"http://www.variation.com/techlib/ta-2_files/ta-2-e18.gif" =
width=3D430=20
align=3Dmiddle> (Eq. 18)</P>
<P><IMG height=3D24 alt=3D"Equation 19" hspace=3D30=20
src=3D"http://www.variation.com/techlib/ta-2_files/ta-2-e19.gif" =
width=3D102=20
align=3Dtop> (Eq. 19)</P>
<P>The resulting statistical tolerance for <I>h</I> is 4.00 <FONT=20
face=3DSymbol>=A3</FONT> 0.02" or (3.94", 4.06"). This tolerance is =
shown in Figure=20
3 along with the specification limits. Statistical tolerancing predicts =
that the=20
specification limits will be easily met. No change is required to the =
tolerance=20
for component height.</P>
<P align=3Dcenter><FONT face=3DHelvetica><B><IMG height=3D168 =
alt=3D"Figure 3"=20
src=3D"http://www.variation.com/techlib/ta-2_files/ta-2-3.gif"=20
width=3D304></B></FONT></P>
<P align=3Dcenter><FONT face=3DHelvetica><B>Figure 3: Statistical =
tolerance for=20
overall height<BR>4-component stack example</B></FONT></P>
<P align=3Dleft><FONT face=3DArial size=3D4><B>COMPARISON OF THE TWO =
APPROACHES TO=20
TOLERANCING</B></FONT></P>
<P>In the 4 component stack example, worst case tolerancing results in a =

tolerance for overall height of 4.00 <FONT face=3DSymbol>=B1</FONT> =
0.12" while=20
statistical tolerancing results in a tolerance of 4.00 <FONT=20
face=3DSymbol>=B1</FONT> 0.06"<SUB><I> </I></SUB><IMG height=3D9=20
src=3D"http://www.variation.com/techlib/ta-2_files/ta-2-st.gif" =
width=3D17=20
align=3Dtop>. The worst-case tolerance is double the width of the =
statistical=20
tolerance. Statistical tolerancing predicts that the specification =
limits will=20
be met and that the tolerance for component height does not need to be=20
tightened. Worst-case tolerancing predicts that the specification limits =
will=20
not be met and that the tolerance for component height must be =
tightened. The=20
two methods of tolerancing produce different results and that can lead =
to=20
opposite conclusions. Selecting which method of tolerancing to use is an =

important decision. This section examines the pros and cons of each=20
approach.</P>
<P>Worst-case tolerancing is the safer approach. If the inputs are =
within their=20
respective tolerances, the output is guaranteed to be within its =
worst-case=20
tolerance. This is especially important for products like heart values =
or=20
critical components on airplanes. However, this guarantee comes at a =
high cost.=20
Worst-case tolerancing guards against a worst-case scenario that is =
highly=20
unlikely if not impossible in most situations. Worst-case tolerancing =
guards=20
against all inputs simultaneously being at the same extreme. In the =
4-component=20
stack example, this would require that all four components be either =
0.97" or=20
1.03". For this to occur repeatedly, would require the entire batch of=20
components to be at one of the two extremes. This could only occur if =
the=20
process producing the components had zero variation.</P>
<P>Suppose instead that the process producing the components had a =
standard=20
deviation of 0.005". As shown in Figure 4, the process average must stay =
between=20
0.985" and 1.015". Otherwise some components would fall outside the =
tolerance=20
limits. The two normal curves in Figure 4 represent the worst-case =
conditions=20
for component height when the standard deviation is 0.005".</P>
<P align=3Dcenter><FONT face=3DHelvetica><B><IMG height=3D168 =
alt=3D"Figure 4"=20
src=3D"http://www.variation.com/techlib/ta-2_files/ta-2-4.gif"=20
width=3D323></B></FONT></P>
<P align=3Dcenter><B><FONT face=3DHelvetica>Figure 4: Worst-case =
conditions for=20
component height when </FONT><FONT face=3DSymbol>s</FONT><FONT =
face=3DHelvetica> =3D=20
0.005"</FONT></B></P>
<P>When component height has an average of 0.985" and a standard =
deviation is=20
0.005", overall height will have an average of 3.94" and a standard =
deviation of=20
0.0025". Likewise when component height has an average of 1.015" and a =
standard=20
deviation is 0.005", overall height will have an average of 4.06" and a =
standard=20
deviation of 0.0025". These two sets of conditions represent the =
worst-case=20
scenarios for overall height. They are shown in Figure 5. As a result of =
the=20
variation, so long as the components meet their tolerance, overall =
height will=20
easily meet the specification limits.</P>
<P align=3Dcenter><FONT face=3DHelvetica><B><IMG height=3D166 =
alt=3D"Figure 5"=20
src=3D"http://www.variation.com/techlib/ta-2_files/ta-2-5.gif"=20
width=3D304></B></FONT></P>
<P align=3Dcenter><B><FONT face=3DHelvetica>Figure 5: Resulting =
worst-case=20
conditions for overall height when </FONT><FONT =
face=3DSymbol>s</FONT><FONT=20
face=3DHelvetica> =3D 0.005"</FONT></B></P>
<P>Worst-case tolerancing generally overestimates the range of the =
outputs. This=20
can result in the unnecessary tightening of tolerances for the inputs, =
which=20
drives up costs. Worst-case tolerancing is the safer but more costly of =
the two=20
approaches.</P>
<P>Statistical tolerancing results in narrower tolerances for the =
outputs. In=20
the 4-component stack example, statistical tolerancing resulted in a =
tolerance=20
for overall height that was half as wide as that obtained using =
worst-case=20
tolerancing. As a result, the tolerances of the inputs do not have to be =

tightened as frequently or by as much. Statistical tolerancing is the =
less=20
costly of the two approaches. When applicable, we would like to use =
statistical=20
tolerancing. However, statistical tolerancing assumes centered =
processes.=20
Violation of this assumption can result in defective product.</P>
<P>To see how, let us return to the 4-component stack example. Figure 6 =
shows=20
how component height is suppose to behave when a statistical tolerance =
of 1.0=20
<FONT face=3DSymbol>=B1</FONT> 0.03"<SUB><I> </I></SUB><IMG height=3D9=20
src=3D"http://www.variation.com/techlib/ta-2_files/ta-2-st.gif" =
width=3D17=20
align=3Dtop> is specified (<FONT face=3DSymbol>m</FONT> =3D 1.0", <FONT=20
face=3DSymbol>s</FONT> =3D 0.01"). It also shows the corresponding =
behavior of=20
overall height (<FONT face=3DSymbol>m</FONT> =3D 4.0", <FONT =
face=3DSymbol>s</FONT> =3D=20
0.02").</P>
<P align=3Dcenter><FONT face=3DHelvetica><B><IMG height=3D168 =
alt=3D"Figure 6"=20
src=3D"http://www.variation.com/techlib/ta-2_files/ta-2-6.gif"=20
width=3D713></B></FONT></P>
<P align=3Dcenter><FONT face=3DHelvetica><B>Figure 6: How component =
height and=20
overall height are suppose to behave</B></FONT></P>
<P>Consider what would happen if the component height actually behaved =
as in=20
Figure 7 (<FONT face=3DSymbol>m</FONT> =3D 0.976", <FONT =
face=3DSymbol>s</FONT> =3D=20
0.002"). All the component heights are within the tolerance. However, =
the=20
process is off-center. Figure 7 also shows the resulting behavior of =
overall=20
height (<FONT face=3DSymbol>m</FONT> =3D 3.904", <FONT =
face=3DSymbol>s</FONT> =3D=20
0.004"). As a result of the components being off-center, the overall =
height of=20
some units exceeds the specification limits and all the units exceed the =

statistical tolerance of 3.94" to 4.06". Statistical tolerancing, while =
less=20
costly, can underestimate the range of the outputs.</P>
<P align=3Dcenter><FONT face=3DHelvetica><B><IMG height=3D168 =
alt=3D"Figure 7"=20
src=3D"http://www.variation.com/techlib/ta-2_files/ta-2-7.gif"=20
width=3D713></B></FONT></P>
<P align=3Dcenter><FONT face=3DHelvetica><B>Figure 7: How component =
height and=20
overall height actually behave</B></FONT></P>
<P>To counter the tendency of statistical tolerances to underestimate =
the range=20
of the outputs, Bender (1962) recommends multiplying statistical =
tolerances by a=20
factor of 1.5. For the 4-component stack example, this results in a =
tolerance of=20
4.0 <FONT face=3DSymbol>=B1</FONT> 0.09" or (3.91", 4.09"). The =
resulting tolerance=20
is half way between the statistical and worst-case tolerances. It is =
probably a=20
more accurate estimate of the actual behavior of overall height. =
However, the=20
factor 1.5 is somewhat arbitrary and certainly not appropriate for every =

problem.</P>
<P>Another way of adjusting for the tendency of statistical tolerances =
to=20
underestimate the range of the outputs is to estimate separately the =
amount that=20
the average varies around the target and then calculate the total =
variation of=20
the input as follows:</P>
<P><IMG height=3D32 alt=3D"Equation 20" hspace=3D30=20
src=3D"http://www.variation.com/techlib/ta-2_files/ta-2-e20.gif" =
width=3D172=20
align=3DabsMiddle> (Eq. 20)</P>
<P>Benderizing tolerances amounts to assuming that the process average =
has a=20
standard deviation equal to 112% of that of the within lot standard=20
deviation.</P>
<P>These two approaches are only appropriate when the process varies =
around the=20
target over longer periods of time. Neither approach properly accounts =
for a=20
process that is consistently run off-target. For example, sheeting =
thickness=20
might purposely be run closely to the lower specification limit in order =
to=20
reduce material usage.</P>
<P align=3Dleft><FONT face=3DArial size=3D4><B>THE =
DILEMMA</B></FONT></P>
<P>Statistical tolerancing is less costly. We would like to use it when=20
appropriate. If statistical tolerancing is selected, statistical =
tolerances must=20
be specified for all inputs. However, statistical tolerances are not =
appropriate=20
for all inputs. To be safe, if some inputs are best represented by =
worst-case=20
tolerances, worst-case tolerancing should be used instead. But if =
worst-case=20
tolerancing is selected, worst-case tolerances must be specified for all =
inputs,=20
even those meeting the assumptions of statistical tolerancing. This =
dives up=20
cost. What is needed is a method of combining the two types of =
tolerances into a=20
single analysis. Process tolerancing, the topic of the next section will =
allow=20
us to do just this.</P>
<P align=3Dleft><FONT face=3DArial size=3D4><B>PROCESS =
TOLERANCING</B></FONT></P>
<P>Process tolerancing is a new approach to tolerancing that is =
implemented in=20
Taylor (1997). A process tolerance consists of a minimum average, =
maximum=20
average and maximum standard deviation. For example, it might be =
specified that=20
the average for component height be between 0.99" and 1.01" and that the =

standard deviation be no more than 0.0067". This process tolerance is =
shown in=20
Figure 8. The two normal curves represent the minimum and maximum =
average with=20
the maximum standard deviation.</P>
<P align=3Dcenter><FONT face=3DHelvetica><B><IMG height=3D168 =
alt=3D"Figure 8"=20
src=3D"http://www.variation.com/techlib/ta-2_files/ta-2-8.gif"=20
width=3D323></B></FONT></P>
<P align=3Dcenter><FONT face=3DHelvetica><B>Figure 8: Process tolerance =
for=20
component height</B></FONT></P>
<P>Process tolerances will be denoted by t<SUB><I>x</I></SUB> <FONT=20
face=3DSymbol>=B1</FONT> <FONT face=3DSymbol>D</FONT> =
<SUB><I>x</I></SUB> <FONT=20
face=3DSymbol>=A3</FONT> <FONT face=3DSymbol>s</FONT> =
<SUB><I>x</I></SUB>. The term=20
t<SUB><I>x</I></SUB> <FONT face=3DSymbol>=B1</FONT> <FONT =
face=3DSymbol>D</FONT>=20
<SUB><I>x</I></SUB> denotes an interval containing the average. The =
minimum=20
average is t<SUB><I>x</I></SUB> - <FONT face=3DSymbol>D</FONT> =
<SUB><I>x</I></SUB>=20
and the maximum average is t<SUB><I>x</I></SUB> + <FONT =
face=3DSymbol>D</FONT>=20
<SUB><I>x</I></SUB>. The maximum standard deviation is <FONT=20
face=3DSymbol>s</FONT> <SUB><I>x</I></SUB>. For example, the process =
tolerance=20
shown in Figure 8 is denoted by 1.0 <FONT face=3DSymbol>=B1</FONT> 0.01 =
<FONT=20
face=3DSymbol>=A3</FONT> 0.0067". Process tolerances ensure that =
individual units=20
all fall in the range (t<SUB><I>x</I></SUB>-<FONT face=3DSymbol>D</FONT> =

<SUB><I>x</I></SUB>-3<FONT face=3DSymbol>s</FONT> <SUB><I>x</I></SUB>,=20
t<SUB><I>x</I></SUB>+<FONT face=3DSymbol>D</FONT> =
<SUB><I>x</I></SUB>+3<FONT=20
face=3DSymbol>s</FONT> <SUB><I>x</I></SUB>). As shown in Figure 8, the =
process=20
tolerance 1.0 <FONT face=3DSymbol>=B1</FONT> 0.01 <FONT =
face=3DSymbol>=A3</FONT> 0.0067"=20
ensures individual components fall in the interval (0.97", 1.03").</P>
<P>Process tolerances differ from worst-case tolerances in that =
worst-case=20
tolerances represent requirements for individual components while =
process=20
tolerances represent requirements for the process producing the =
components.=20
Hence its name. Process tolerances are more similar to statistical =
tolerances.=20
However, process tolerances do not require that the process remain =
centered.=20
Instead, process tolerances specify an operating window that the process =
average=20
must remain within.</P>
<P>If process tolerances are specified for the inputs, a process =
tolerance can=20
be calculated for the output. This requires equations for the average =
and=20
standard deviation. Returning to the case where <I>y</I> =3D =
f(<I>x<SUB>1</SUB>,=20
=85, x<SUB>n</SUB></I>), let these two equations be represented by <FONT =

face=3DSymbol>m</FONT> <SUB><I>y</I></SUB> ( <FONT =
face=3DSymbol>m</FONT>=20
<SUB><I>x<FONT size=3D2><SUB>1</SUB></FONT></I></SUB>, =85, <FONT=20
face=3DSymbol>m</FONT> <SUB><I>x<FONT =
size=3D2><SUB>n</SUB></FONT></I></SUB>, <FONT=20
face=3DSymbol>s</FONT> <SUB><I>x<FONT =
size=3D2><SUB>1</SUB></FONT></I></SUB>, =85,=20
<FONT face=3DSymbol>s</FONT> <SUB><I>x<FONT =
size=3D2><SUB>n</SUB></FONT></I></SUB>)=20
and <FONT face=3DSymbol>s</FONT> <SUB>y</SUB> (<FONT =
face=3DSymbol>m</FONT>=20
<SUB><I>x<FONT size=3D2><SUB>1</SUB></FONT></I></SUB>, =85, <FONT=20
face=3DSymbol>m</FONT> <SUB><I>x<FONT =
size=3D2><SUB>n</SUB></FONT></I></SUB>, <FONT=20
face=3DSymbol>s</FONT> <SUB><I>x<FONT =
size=3D2><SUB>1</SUB></FONT></I></SUB>, =85,=20
<FONT face=3DSymbol>s</FONT> <SUB><I>x<FONT =
size=3D2><SUB>n</SUB></FONT></I></SUB>).=20
These two equations can be obtained using the same methods used to =
calculate=20
statistical tolerances. Further let the process tolerance for =
x<SUB>i</SUB> be=20
denoted t<SUB><I>x<SUB>i</SUB></I></SUB> <FONT face=3DSymbol>=B1</FONT> =
<FONT=20
face=3DSymbol>D</FONT> <SUB><I>x<SUB>i</SUB></I></SUB> <FONT =
face=3DSymbol>=A3</FONT>=20
<FONT face=3DSymbol>s</FONT> <SUB><I>x<SUB>i</SUB></I></SUB>. Then the =
resulting=20
process tolerance for y, denoted t<SUB><I>y</I></SUB> <FONT =
face=3DSymbol>=B1</FONT>=20
<FONT face=3DSymbol>D</FONT> <SUB><I>y</I></SUB> <FONT =
face=3DSymbol>=A3</FONT> <FONT=20
face=3DSymbol>s</FONT> <SUB><I>y</I></SUB>, is:</P>
<P><IMG height=3D53 alt=3D"Equation 21" hspace=3D30=20
src=3D"http://www.variation.com/techlib/ta-2_files/ta-2-e21.gif" =
width=3D238=20
align=3Dtop> (Eq. 21)</P>
<P><IMG height=3D53 alt=3D"Equation 22" hspace=3D30=20
src=3D"http://www.variation.com/techlib/ta-2_files/ta-2-e22.gif" =
width=3D242=20
align=3Dtop> (Eq. 22)</P>
<P><IMG height=3D53 alt=3D"Equation 23" hspace=3D30=20
src=3D"http://www.variation.com/techlib/ta-2_files/ta-2-e23.gif" =
width=3D245=20
align=3Dtop> (Eq. 23)</P>
<P><IMG height=3D43 alt=3D"Equation 24" hspace=3D30=20
src=3D"http://www.variation.com/techlib/ta-2_files/ta-2-e3.gif" =
width=3D82=20
align=3Dmiddle> (Eq. 24)</P>
<P><IMG height=3D43 alt=3D"Equation 25" hspace=3D30=20
src=3D"http://www.variation.com/techlib/ta-2_files/ta-2-e4.gif" =
width=3D89=20
align=3Dmiddle> (Eq. 25)</P>
<P>If the inputs satisfy their respective process tolerances, then =
<I>y</I> is=20
guaranteed to satisfy the process tolerance given by Equations =
21-25.</P>
<P>When the output is the sum of the inputs, the process tolerancing =
equations=20
simplify to:</P>
<P><IMG height=3D25 alt=3D"Equation 26" hspace=3D30=20
src=3D"http://www.variation.com/techlib/ta-2_files/ta-2-e26.gif" =
width=3D98=20
align=3Dtop> (Eq. 26)</P>
<P><IMG height=3D25 alt=3D"Equation 27" hspace=3D30=20
src=3D"http://www.variation.com/techlib/ta-2_files/ta-2-e11.gif" =
width=3D158=20
align=3Dtop> (Eq. 27)</P>
<P><IMG height=3D32 alt=3D"Equation 28" hspace=3D30=20
src=3D"http://www.variation.com/techlib/ta-2_files/ta-2-e15.gif" =
width=3D169=20
align=3Dtop> (Eq. 28)</P>
<P>Equation 27 is the same as Equation 11 for worst-case tolerances. =
Equation 28=20
is the same as Equation 15 for statistical tolerances. A process =
tolerance=20
represents a worst-case tolerance for the average and a statistical =
tolerance=20
for the variation around the average. When more complex equations are =
involved,=20
Equations 21 through 25 should be used instead.</P>
<P><B>Example (continued):</B> Returning to the example of four =
components=20
stacked on top of each other, assume each component has a process =
tolerance of=20
1.00 <FONT face=3DSymbol>=B1</FONT> 0.01 <FONT face=3DSymbol>=A3</FONT> =
0.00667". This=20
tolerance was shown in Figure 8. It is roughly equivalent to the =
worst-case=20
tolerance 1.00 <FONT face=3DSymbol>=B1</FONT> 0.03" and the statistical =
tolerance=20
1.00 <FONT face=3DSymbol>=A3</FONT> 0.01". All three tolerances result =
in units in=20
the range (0.97", 1.03"). The resulting process tolerance for overall =
height is=20
calculated as follows:</P>
<P><IMG height=3D25 alt=3D"Equation 29" hspace=3D30=20
src=3D"http://www.variation.com/techlib/ta-2_files/ta-2-e17.gif" =
width=3D353=20
align=3Dtop> (Eq. 29)</P>
<P><IMG height=3D25 alt=3D"Equation 30" hspace=3D30=20
src=3D"http://www.variation.com/techlib/ta-2_files/ta-2-e30.gif" =
width=3D213=20
align=3Dtop> (Eq. 30)</P>
<P><IMG height=3D32 alt=3D"Equation 31" hspace=3D30=20
src=3D"http://www.variation.com/techlib/ta-2_files/ta-2-e31.gif" =
width=3D370=20
align=3Dmiddle> (Eq. 31)</P>
<P>The resulting process tolerance for <I>h</I> is 4.00 <FONT=20
face=3DSymbol>=B1</FONT> 0.04 <FONT face=3DSymbol>=A3</FONT> 0.01334". =
This tolerance is=20
shown in Figure 9 along with the specification limits. Individual units =
of=20
product are expected to fall in the interval (3.92", 4.08"). Process =
tolerancing=20
predicts that the specification limits will be met. No change is =
required to the=20
tolerance for component height.</P>
<P align=3Dcenter><FONT face=3DHelvetica><B><IMG height=3D168 =
alt=3D"Figure 9"=20
src=3D"http://www.variation.com/techlib/ta-2_files/ta-2-9.gif"=20
width=3D304></B></FONT></P>
<P align=3Dcenter><FONT face=3DHelvetica><B>Figure 9: Process tolerance =
for overall=20
height<BR>4-component stack example</B></FONT></P>
<P>The results of the three different approaches to tolerance analysis =
are=20
summarized in Figure 10. Process tolerancing predicts the overall =
heights will=20
be in the interval (3.92", 4.08"). This falls between the worst-case =
tolerance=20
of (3.88", 4.12") and the statistical tolerance of (3.94", 4.06").</P>
<P align=3Dcenter><FONT face=3DHelvetica><B><IMG height=3D265 =
alt=3D"Figure 10"=20
src=3D"http://www.variation.com/techlib/ta-2_files/ta-2-10.gif"=20
width=3D615></B></FONT></P>
<P align=3Dcenter><FONT face=3DHelvetica><B>Figure 10: Comparison of =
three=20
approaches to tolerance analysis<BR>4-component stack =
example</B></FONT></P>
<P align=3Dleft><FONT face=3DArial size=3D4><B>A UNIFIED APPROACH TO=20
TOLERANCING</B></FONT></P>
<P>Process tolerancing represent a unified approach to tolerancing that=20
encompasses both the previous approaches. The process tolerance=20
<I>t<SUB>x</SUB></I> <FONT face=3DSymbol>=B1</FONT> <FONT =
face=3DSymbol>D</FONT>=20
<SUB><I>x</I></SUB> <FONT face=3DSymbol>=A3</FONT> 0 is in fact a =
worst-case=20
tolerance. In this case, Equations 21-25 simplify to Equations 1-4. The =
process=20
tolerance <I>t<SUB>x</SUB></I> <FONT face=3DSymbol>=B1</FONT> 0 <FONT=20
face=3DSymbol>=A3</FONT> <FONT face=3DSymbol>s</FONT> =
<SUB><I>x</I></SUB> is in fact a=20
statistical tolerance. In this case, Equations 21-25 simplify to =
Equations=20
12-13. Worst-case tolerancing guards against the worst-case conditions =
which can=20
only occur in the unlikely event of zero variation (<FONT =
face=3DSymbol>s</FONT>=20
<SUB><I>x</I></SUB> =3D 0). Statistical tolerances make the restrictive =
assumption=20
of centered processes (<FONT face=3DSymbol>D</FONT> <SUB><I>x</I></SUB> =
=3D 0). In=20
one sense statistical tolerances represent the best-case conditions. =
Neither of=20
these two extreme cases adequately describes all inputs. Process =
tolerancing=20
provides a middle ground. The purpose of a tolerance analysis is to =
predict the=20
behavior of the output based on knowledge of how the inputs behave. Key =
to its=20
success is the ability to accurately describe the behavior of the =
inputs.=20
Process tolerancing, provides the needed flexibility to accurately =
describe the=20
behavior of the inputs and to therefore accurately predict the behavior =
of the=20
outputs.</P>
<P>Even if you never specify a process tolerance, you can still benefit =
from=20
process tolerancing. Process tolerancing allows both worst-case and =
statistical=20
tolerances to be combined in the same analysis. Some inputs are best =
described=20
by worst-case tolerances. Other inputs are better described by =
statistical=20
tolerances. However, if worst-case tolerancing is used, all tolerances =
must be=20
specified as worst-case tolerances and if statistical tolerancing is =
used, all=20
tolerances must be specified as statistical tolerances. Using process=20
tolerancing, you can select the type of tolerance for each input that =
best=20
describes its behavior. The decision you must make changes from "Which =
type of=20
tolerancing should I use?" to "Which type of tolerance best describes =
the=20
behavior of each input?" The following seven examples illustrate how =
tolerances=20
might be specified for different circumstances.</P>
<P><B>Example 1 =96 Usage Temperature (environmental input):</B> The =
flow rate of=20
a pump is affected by usage temperature. The pump is intended for use =
over a=20
temperature range of 40<FONT face=3DSymbol>=B0</FONT> F to 120<FONT=20
face=3DSymbol>=B0</FONT> F. A worst-case tolerance of 80<FONT =
face=3DSymbol>=B0</FONT> F=20
<FONT face=3DSymbol>=B1</FONT> 40<FONT face=3DSymbol>=B0</FONT> F best =
describes the=20
behavior of usage temperature since individual pumps could be run =
continuously=20
at each of the temperature extremes. In general, inputs representing =
usage and=20
environmental conditions as well as other inputs outside the control of =
the=20
designer should be specified as worst-case tolerances.</P>
<P><B>Example 2 =96 Die Temperature (automatically controlled =
input):</B> Die=20
temperature affects the seal strength of a heat seal machine. The heat =
sealer=20
continuously monitors die temperature and automatically adjusts it. This =
ensures=20
that die temperature is centered on the target. The behavior of die =
temperature=20
is best described using a statistical tolerance. Suppose the target is =
150<FONT=20
face=3DSymbol>=B0</FONT> F and that the standard deviation of die =
temperature is=20
5<FONT face=3DSymbol>=B0</FONT> F. Then the statistical tolerance =
150<FONT=20
face=3DSymbol>=B0</FONT> F <FONT face=3DSymbol>=A3</FONT> 5<FONT =
face=3DSymbol>=B0</FONT> F=20
could be specified. In general, inputs with automatic controls to keep =
them=20
centered should be specified as statistical tolerances.</P>
<P><B>Example 3 =96 Motor Speed (manually controlled input):</B> The =
flow rate of=20
a pump is affected by motor speed. The motors are purchased from a =
supplier. A=20
capability study performed on the supplier=92s process indicates that =
their=20
process is in control with a standard deviation of 0.1 rpm. The =
statistical=20
tolerance 18.0 rpm <FONT face=3DSymbol>=A3</FONT> 0.1 rpm might be =
considered.=20
However, there is a concern that the capability study, run during the =
production=20
of a single lot, is not representative of the full range of variation =
that will=20
be experience over more extended periods of time. One possibility is to=20
Benderize the tolerance by multiplying it by 1.5. This results in the=20
statistical tolerance 18.0 rpm <FONT face=3DSymbol>=A3</FONT> 0.15 rpm. =
Another=20
approach is to allow an operating window for the average. Suppose that =
an <IMG=20
height=3D21 =
src=3D"http://www.variation.com/techlib/ta-2_files/ta-2-xr.gif"=20
width=3D43> control chart is used to keep the process centered. If the =
process=20
shifts off-target by 1.5<FONT face=3DSymbol>s</FONT> , the first point =
following=20
the shift has a 70% chance of falling outside the control limits. This =
control=20
chart is capable of keeping the process within 1.5<FONT =
face=3DSymbol>s</FONT> of=20
target. A process tolerance of 18 <FONT face=3DSymbol>=B1</FONT> 0.15 =
<FONT=20
face=3DSymbol>=A3</FONT> 0.1 rpm could be specified. This allows a <FONT =

face=3DSymbol>=B1</FONT> 1.5<FONT face=3DSymbol>s</FONT> operating =
window around the=20
target. In general, manually controlled processes are best represented =
by=20
process tolerances specifying the operating window that can be =
maintained.=20
Stable processes and processes controlled using control charts typically =
hold a=20
<FONT face=3DSymbol>=B1</FONT> 1.5<FONT face=3DSymbol>s</FONT> operating =
window.</P>
<P><B>Example 4 =96 Line voltage (unstable process):</B> The line =
voltage effects=20
the seal strength of a sonic sealing machine. Voltage fluctuates between =
120=20
<FONT face=3DSymbol>=B1</FONT> 10 volts. It cycles up and down. It can =
run for long=20
periods of time in the upper end or the lower end of the range. This =
input is=20
best described using the worst-case tolerance 120 <FONT =
face=3DSymbol>=B1</FONT> 10=20
volt. In general, unstable processes should be handled using worst-case =
or=20
process tolerances. If process tolerances are used, the minimum and =
maximum=20
averages should represent the extremes seen over extended periods of =
time.</P>
<P><B>Example 5 =96 Sheeting Thickness (consistently off-target =
input):</B>=20
Sheeting thickness affects seal strength of a heat sealer. The =
requirements have=20
been specified as 15 <FONT face=3DSymbol>=B1</FONT> 1 mil (thousandth of =
an inch).=20
Suppose the sheeting supplier purposely runs below the target in order =
to reduce=20
material costs. They average 14.6 mils with a standard deviation of 0.2 =
mils.=20
Off-center processes violate the assumption of statistical tolerancing.=20
Statistical tolerances should not be used is this case. The process =
tolerance 15=20
<FONT face=3DSymbol>=B1</FONT> 0.4 <FONT face=3DSymbol>=A3</FONT> 0.2 =
mil accounts for=20
the off-center average. In general, only process or worst-case =
tolerances should=20
be used for consistently off-target processes.</P>
<P><B>Example 6 =96 4-Component Stack (multiple identical =
components):</B> In the=20
4-component stack problem, the 4-components are identical. As a result, =
finished=20
units consist of four components coming from the same supplier lot. A =
capability=20
study performed on the supplier=92s process indicates that their process =
is in=20
control with a standard deviation of 0.00667". Using historical data it =
is=20
estimated that the process average varies from lot to lot with a =
standard=20
deviation of 0.0075". The total standard deviation is then =
<SUB><SUB><IMG=20
height=3D30 =
src=3D"http://www.variation.com/techlib/ta-2_files/ta-2-sdt.gif"=20
width=3D206></SUB></SUB>". However, specifying a statistical tolerance =
of 1.0=20
<FONT face=3DSymbol>=A3</FONT> 0.01" and performing a tolerance analysis =
can produce=20
misleading results. When the lot average is low, all the components tend =
to be=20
low. When the lot average is high, all the components tend to be high. =
This=20
creates a correlation between the inputs violating the assumption of=20
independence. Statistical tolerances should not be used is this case. A =
process=20
tolerance of <BR>1.0 <FONT face=3DSymbol>=B1</FONT> 0.015 <FONT =
face=3DSymbol>=A3</FONT>=20
0.00667" could be used instead. It accounts for the effect of below and =
above=20
average lots. In general, only process and worst-case tolerances should =
be used=20
for products containing multiple identical components or correlated =
inputs.</P>
<P><B>Example 7 =96 Spring Force (deterioration over time):</B> The =
closure force=20
of a car door hinge is affected by spring force. The springs are =
purchased from=20
a supplier. The target is 10 lbs. A capability study performed on the =
supplier=92s=20
process indicates that their process is in control with a standard =
deviation of=20
0.2 lbs. A process tolerance of 10 <FONT face=3DSymbol>=B1</FONT> 0.3 =
<FONT=20
face=3DSymbol>=A3</FONT> 0.2 lb. could be used, allowing a <FONT=20
face=3DSymbol>=B1</FONT> 1.5<FONT face=3DSymbol>s</FONT> operating =
window around the=20
target. This represents the manufacturing variation. However, in this =
case the=20
springs are also prone to deterioration over time. Life testing of the =
springs=20
indicates that they deteriorate by 1 lb. over a 5 year period. =
Deterioration is=20
best represented by a worst-case tolerance since the closure force must =
be=20
within specifications at both the beginning and the end of the 5 year =
period.=20
Deterioration is best represented by the worst-case tolerance 10 +0 =961 =
lb.=20
Combining these two tolerances results in 10 + 0.3 =961.3 <FONT=20
face=3DSymbol>=A3</FONT> 0.2 lb. In general, deterioration of an input =
should be=20
represented as a worst-case tolerance. The deterioration effect should =
then be=20
combined with the manufacturing variation to obtain the final =
tolerance.</P>
<P align=3Dleft><FONT face=3DArial size=3D4><B>SIX SIGMA =
QUALITY</B></FONT></P>
<P>There is an important link between six sigma quality and process =
tolerancing.=20
Six sigma quality, a term introduced at Motorola, requires that a =
measurable=20
characteristic such as flow rate or seal strength have no more the 3.4 =
defects=20
per million units produced. This is in turn translated into the =
requirements=20
that the average and standard deviation meet the conditions specified by =
the=20
process tolerance t<SUB><I>y</I></SUB> <FONT face=3DSymbol>=B1</FONT> =
<FONT=20
face=3DSymbol>D</FONT> <SUB><I>y</I></SUB> <FONT =
face=3DSymbol>=A3</FONT> <FONT=20
face=3DSymbol>s</FONT> <SUB><I>y</I></SUB> where:</P>
<P><IMG height=3D41 alt=3D"Equation 32" hspace=3D30=20
src=3D"http://www.variation.com/techlib/ta-2_files/ta-2-e32.gif" =
width=3D111=20
align=3Dmiddle> (Eq. 32)</P>
<P><IMG height=3D41 alt=3D"Equation 33" hspace=3D30=20
src=3D"http://www.variation.com/techlib/ta-2_files/ta-2-e33.gif" =
width=3D117=20
align=3Dmiddle> (Eq. 33)</P>
<P><IMG height=3D25 alt=3D"Equation 34" hspace=3D30=20
src=3D"http://www.variation.com/techlib/ta-2_files/ta-2-e34.gif" =
width=3D74=20
align=3Dtop> (Eq. 34)</P>
<P>USL is the upper specification limit and LSL is the lower =
specification=20
limit. This process tolerance is shown in Figure 11.</P>
<P align=3Dcenter><FONT face=3DHelvetica><B><IMG height=3D156 =
alt=3D"Figure 11"=20
src=3D"http://www.variation.com/techlib/ta-2_files/ta-2-11.gif"=20
width=3D307></B></FONT></P>
<P align=3Dcenter><FONT face=3DHelvetica><B>Figure 11: Six Sigma Process =

Tolerance</B></FONT></P>
<P>For the process tolerance in Figure 11, the worst-case scenario is =
when the=20
average is 1.5<FONT face=3DSymbol>s</FONT> off-target resulting in an =
average that=20
is 4.5<FONT face=3DSymbol>s</FONT> from the nearest specification limit. =
Based on=20
the normal distribution, when the process average is 4.5<FONT=20
face=3DSymbol>s</FONT> from the specification limit, the defect rate is =
3.4=20
defects per million. This is the definition of six sigma quality. The =
name comes=20
from the fact that six sigma quality requires specifications that are =
<FONT=20
face=3DSymbol>=B1</FONT> 6<FONT face=3DSymbol>s</FONT> wide.</P>
<P>Since the requirements for six sigma quality are specified in the =
form of a=20
process tolerance, it is only natural that process tolerancing be used =
to ensure=20
that these tolerances are achieved. Process tolerancing provides an =
alternative=20
to the combined dynamic and static mean tolerance analysis procedure =
described=20
in Harry and Lawson (1992).</P>
<P align=3Dleft><FONT face=3DArial size=3D4><B>SUMMARY</B></FONT></P>
<P>Worst-case tolerancing tends to overestimate the variation of the =
output.=20
Cost suffers when the variation of the output is overestimated. =
Statistical=20
tolerancing, tends to underestimate the output variation. Quality =
suffers when=20
the variation of the output is underestimated. Using process =
tolerancing, it is=20
possible to accurately predict the behavior of the output, providing the =
desired=20
quality at a lower cost.</P>
<P>Process tolerancing provides a unified approach to tolerancing with=20
encompasses both worst-case and statistical tolerancing. It allows =
worst-case=20
tolerances to be used for some inputs and statistical tolerance are used =
for=20
others. No longer does one have to use the same type of tolerance for =
all=20
inputs. The purpose of a tolerance analysis is to predict the behavior =
of the=20
output based on knowledge of how the inputs behave. Key to its success =
is the=20
ability to accurately describe the behavior of the inputs. Using process =

tolerancing, you can describe behaviors for inputs not adequately =
described by=20
either worst-case or statistical tolerances. This provides solutions to =
commonly=20
occurring problems such as off-center processes and multiple identical=20
components.</P>
<P align=3Dleft><FONT face=3DArial size=3D4><B>REFERENCE =
LIST</B></FONT></P>
<P>Bender, A. (1962). Benderizing Tolerances - A Simple Practical =
Probability=20
Method of Handling Tolerances for Limit-Stack-Ups. <I>Graphic =
Science</I>=20
Dec.</P>
<P>Cox, N. D. (1986). <I>How to Perform Statistical Tolerance =
Analysis</I>.=20
Milwaukee: <A href=3D"http://www.asq.org/">ASQC Quality Press</A>.</P>
<P>Evans, D. H. (1975). Statistical Tolerancing: The State of the Art.=20
<I>Journal of Quality Technology</I> 7 (1): 1-12.</P>
<P>Harry, M. J. and Lawson J. R. (1992). <I>Six Sigma Producibility =
Analysis and=20
Process Characterization</I>. New York: Addison-Wesley.</P>
<P>Taylor, W. A. (1991). <I><A=20
href=3D"http://www.variation.com/ovr-book.html">Optimization and =
Variation=20
Reduction in Quality</A>.</I> Libertyville, IL: Taylor Enterprises, =
Inc.</P>
<P>Taylor, W. A. (1997). <A=20
href=3D"http://www.variation.com/var-soft.html"><I>VarTran User =
Manual</I></A>.=20
Libertyville, IL: Taylor Enterprises, Inc.</P>
<P align=3Dcenter>&nbsp;</P>
<P align=3Dcenter><STRONG>Presented at Fall Technical Conference,=20
1995</STRONG></P><!--webbot bot=3D"Include" u-include=3D"../bottom.html" =
tag=3D"BODY" startspan -->
<HR align=3Dleft color=3D#000080 SIZE=3D5>

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