HP 9g Manual

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HP 9g Statistics – Process Capability
Process Capability
Practice Solving Process Capability Problems

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HP 9g Statistics – Process Capability
Process capability
Any manufacturing process is subject to variability due to both assignable (e.g. a defective machine, the operator’s
mistakes) and non-assignable causes (e.g. the limited precision of machines, the operator’s skill). When the former are
eliminated, then the process is said to be in control. Variability is then predictable.
We can determine whether an in-control process meets the quality specifications by measuring the capability of the
process. A process is capable when it produces similar products for a sustained period of time, that is to say, when
almost all pieces are fabricated within a given tolerance. Assuming that the performance measure data is distributed
normally, then approximately 99.73% 1of all values will lie within three standard deviations on either side of the mean –-
this interval (6 sigma) is called the natural spread of the process. This is why the process capability index is defined as:
σ
−
=6LSLUSL Cp
where USL and LSL are the upper and lower specification limits, respectively. USL – LSL is called the tolerance spread.
If = 1, then 99.73% of pieces will be fabricated within the specification limits provided that the process mean is at the
midpoint of the specification range (
p
C
2)LSLUSL(m
+
=). The following index is used for assessing the capability of an
off-center process (i.e. when µ≠m):
)
LSL
,
USL
(minCpk σ
−
µ
σ
µ
−
=33
pk
Ccan be expressed in terms of by defining another index , , which is the scaled distance between the mean of
the process and the midpoint of the specification range:
p
Ca
C
2)LSLUSL( m
Ca−
µ−
=
Thus, =(1 – ). Note that , which implies that
pk
Cp Ca
C10 ≤≤ a Cppk CC
≤
. For a centered process (i.e. µ= m)
, and therefore
0=
a
Cppk CC
=
.
A process for which = 1 might produce up to defective pieces per million. While such a
process was considered to be right in the past, it might well deemed unacceptable nowadays. Today standards for both
and are that they must be greater than 1.33 for the process to be capable (that assures us of no more than 60
non-conforming parts per million) . But this value certainly depends on the kind of product being fabricated.
p
C270010997301 6=⋅− ).(
p
Cpk C
The Hp 9g provides functions to calculate the three statistics described above, as well as an estimation of defection per
million opportunities (ppm).
1On the HP 9g, this value can be calculated as 1 – 2·R(3), when x= 0 and σx = 1: `1(D-CL)=`1(1-VAR)=
E=1‡‡M1‡EE…=3=~e …M2+1=.
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HP 9g Statistics – Process Capability
Practice solving process capability problems
Example 1: As part of a process optimization effort, an industrial manager had to determine whether their
manufacturing process, for which µ= 19.35 and σ= 2.05, was capable. The population of data values is
assumed to be normally distributed. Theupper and lower specification limits of the process are 23.3 and
11, respectively. How could the HP 9g help her?
Solution: By calculating the Cpand Cpk statistics to measure the process capability. First of all, let’s clear any
previous statistical data in memory (press `1, then select D-CL and finally press =). Since it is
one single process, we’ll carry out all calculations in 1-VAR mode (i.e. `1, select 1-VAR and press
=). No data set is provided. Therefore, in order to enter µand σinto the calculator, we’ll have to use
the method described in the HP 9g learning module Statistics – Normal Distribution., that is to say: create a
data set consisting of two values µ+ σand µ– σ. To do this press:
E=19.35+2.05‡‡19.35-2.05‡
It is very important to bear in mind that, when using this trick, those variables shown in the STATVAR
menu that depend on the actual data or the number of data, n, will be meaningless. Only those which only
depend on the average and the population standard deviation (that is, in which n is not explicitly used in
their calculation) will be correct; namely: x, y, σx, σy, , , , , , ,P(t), Q(t), R(t)
and t . Any other variable involves the value of n or the data in their calculation. Particularly, note that ‘ppm’
is calculated by the HP 9g usingthe actual value of n, not the one ‘embedded’ in µ.
px
Cpkx Cax
Cpy
Cpky Cay C
The next step is to input the LSL and USL values. Press:
E, select LIMIT, =11‡23.3‡
Let’s determine if the process is located at the midpoint of the specification range. We can do it graphically
by plotting the process control chart: press ~e P2. The x-axis range is automatically set to
(LSL, USL). The resulting bell-shaped curve is clearly off-center. Alternatively, we can do it numerically:
35191517
211323 2..
.LSLUSL =µ≠=
+
=
+
Or simply by finding the value of , which is displayed by pressing:
a
C
~e ‡‡‡‡
=0.3577 ≠0. Therefore, the capability index that we must find is C
a
Cpk because Cpk, unlike Cp, accounts
for a shift in the mean of the process to either specification limit. All these statistics are in the STATVAR
menu, so press ~e ‡‡‡†† to display Cpk.
Answer: C
pk = 0.6423 (rounded to four decimal digits). Since Cpk is clearly smaller than 1.33 (even than 1, in fact) the
process is incapable: the process spread exceeds the tolerance spread, which assures that out-of-
tolerance products will be made.
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HP 9g Statistics – Process Capability
Example 2: The length of pieces made by certain machine is found to be normally distributed (µ= 3 and
σ= 0.001). The established upper and lower specification limits for this particular piece are 3.005 and
2.995, respectively. Assuming the manufacturing process is in control, determine whether the process is
capable or incapable.
Solution: Let’s input the process average and standard deviation, which will overwrite the ones used in the previous
example (remember to clear the statistical data first, if there are more than two values):
E=3+.001‡‡3-.001‡
And now let’s input USL and LSL
EE‡2.995‡3.005‡=
The process capability indices are all shown in the STATVAR menu: ~e :
= 2.842170943·10
a
C–12 , = 1.666666621 and = 1.666666621.
p
Cpk C
Answer: Since is almost zero, the process is centered. And since is greater than 1.33, the process is
capable.
a
Cp
C
Example 3: If the mean of the previous process shifts two standard deviations to the USL (µis now 3.002), will the
process still be capable? How many pieces will be defective?
Solution: Let’s input the new average by pressing:
E=3.002+.001‡‡3.002-.001‡
The process capability indices are now: ( ~e )
= 0.4
a
C, = 1.666666724 and = 1.000000035.
p
Cpk C
Defective pieces are those bigger than 3.005 and smaller than 2.995. The probability that a piece is
defective is: R(3.005) + P(2.995),2which can be calculated by pressing:
E…=3.005=~e …A9=, which stores R(3.005) in R)
E…=2.995=~e ……+W9=
R(3.005) + P(2.995) = 0.0013 + 0.0 = 0.0013, or expressed in pieces per million: 1300.
Answer: The process is now clearly off-center, and incapable because ≤1.33. The estimated number of
defective pieces is 1300 each million of pieces fabricated. pk
C
2Refer to the Hp 9g learning module Statistics – Normal Distribution.
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