Download selective assembly to minimize surplus components by shifting the

SELECTIVE ASSEMBLY TO MINIMIZE SURPLUS COMPONENTS BY
SHIFTING THE PROCESS MEAN
Shun Matsuura
1
and Nobuo Shinozaki
2
1
Research Fellow of the Japan Society for the Promotion of Science,
Keio University, 3-14-1, Hiyoshi, Kohoku, Yokohama 223-8522, Japan
[email protected]
2
Keio University, 3-14-1, Hiyoshi, Kohoku, Yokohama 223-8522, Japan
[email protected]
Abstract
We consider a product assembled from two components when the quality characteristic is the clearance between the
matching components (or the sum of the dimensions of the matching components). Some variation is inevitable in
any production process, and the quality of the assembled product is dependent on the dimensional variability of the
component parts. A product is rejected when its clearance is outside a given specification range. Random assembly of
the matching components may lead to an unacceptably large number of rejected products. In such a situation,
selective assembly should be effective for reducing it. In this approach, the components are sorted into several classes
according to their dimensions, and the product is assembled by randomly selecting matching components from the
corresponding classes. Many previous works are focused on equal width and equal probability partitioning schemes.
When there is large difference between the variances of the two component dimensions, equal width partitioning will
result in large number of surplus components and equal probability partitioning will result in some rejected products.
Some authors have proposed the method of manufacturing the component with smaller variance at two (or more)
shifted means to cope with this difficulty. However, determination of optimal mean shift has not been addressed. This
paper treats with the problem of determining optimal mean shift in equal width and equal probability partitioning
schemes. Some numerical results are given which show that using optimal mean shift considerably reduces the
number of surplus components in equal width partitioning and may enable us to manufacture all products within the
specification range in equal probability partitioning. We also show some merits and demerits of equal width and
equal probability partitioning schemes.
Keywords: Clearance specification, Match gauging, Optimization, Statistical process control, Tolerance limit
INTRODUCTION
We consider a product assembled from two components when the quality characteristic is the difference of the
relevant dimensions of the matching components (i.e., clearance). Note that although we use the clearance as the
assembly dimension of interest, our discussion is equally valid in the case in which it is the sum of the dimensions of
the matching components. Some variation is inevitable in any production process, and the quality of the assembled
product is dependent on the dimensional variability of the component parts. The clearance of any product should be
within a given specification range. A product is rejected when its clearance is outside the specification. Random
assembly of the matching components may lead to an unacceptably large number of rejected products. In such a
situation, selective assembly should be effective for reducing it.
In this approach, any component with its dimension outside specified limits of the dimensional distribution is rejected,
and the remaining components are sorted into several classes according to their dimensions. The product is then
assembled by randomly selecting matching components from the corresponding classes. This approach enables the
assembly of high-precision products from relatively low-precision components, which may lead to a cost reduction
compared to the alternative of manufacturing the respective components at a higher level of precision.
A piston and cylinder assembly (Figure 1) is an example from an automobile industry in Japan. A matching pair of
piston and cylinder is chosen to satisfy a given clearance specification. If the clearance is too small, the oil film
between the cylinder wall and piston becomes too thin and piston scuffing occurs. If the clearance is too large, the
piston inclines in the cylinder and abnormal noise occurs. Random assembly of pistons and cylinders lead to an
unacceptably large number of products which do not satisfy the clearance specification. Thus, the automobile
industry has used selective assembly. Pistons and cylinders are sorted into several classes according to their outer and
inner diameters, respectively. The smaller pistons are matched with the smaller cylinders and the bigger pistons with
the bigger cylinders. Other applications of selective assembly include a pin and bushing assembly, a hole and shaft
matching, and a camshaft, valve, and valve lifter assembly.
1
Outer diameter
of piston
Inner diameter of
cylinder
Figure 1. Piston and Cylinder Assembly
A great deal of research and development effort has been devoted to the subject of selective assembly. Kwon et al.
(1999) studied optimal partitioning of the distributions of the two component dimensions to minimize expected
squared error loss, assuming that the two component dimensions had the same normal distribution. Mease et al.
(2004) extensively studied some optimal partitioning strategies for several loss functions and distributions. In
particular, they developed optimal partitioning for minimizing the expected squared error loss for the case in which
the two component dimensions are identically distributed. They gave equations for optimal partition limits, and
showed that the solution to them is unique when the density function of the dimensional distribution is strongly
unimodal. Matsuura and Shinozaki (2007) extended the results to the case in which measurement error is present.
Matsuura and Shinozaki (2009) studied optimal partitioning to maximize profit, including optimal choice of
truncation points of the dimensional distributions, in the presence and absence of measurement error. These papers
are focused on optimal partitioning of the dimensional distributions, and did not take account of the tolerance
constraint on the clearance.
In selective assembly when the dimensional variability of the two components are approximately equal and a
tolerance constraint on the clearance is given, equal width partitioning scheme, studied by Pugh (1986) for instance,
enables us to manufacture all products to satisfy the clearance specification. However, it is not unusual that the
variances of the two component dimensions are different. If it is the case and we use equal width partitioning scheme,
large number of surplus components are expected due to the difference between the numbers of components in the
corresponding classes. On the other hand, Fang and Zhang (1995) studied equal probability partitioning scheme in
which the corresponding classes have equal probability, that is, we have no surplus component. They proposed an
algorithm for equal probability partitioning to manufacture all products within the specification range. However,
equal probability partition to manufacture all products within the specification range does not exist when the
difference between the variances of the two component dimensions is large or the clearance specification is tight.
In the unequal variances case, some authors proposed the method of manufacturing the component with smaller
variance at two (or more) shifted means and mixing them. Although Matsuura and Shinozaki (2008) studied optimal
mean shift which minimizes clearance variation, they did not take account of the tolerance constraint on the clearance.
On the other hand, although Kannan et al. (1997) and Kannan and Jayabalan (2002) took account of the clearance
specification, they did not discuss optimal mean shift which minimizes the number of surplus components.
This paper studies optimal mean shift in equal width and equal probability partitioning schemes when the component
with smaller variance is manufactured at two shifted means and a tolerance constraint on the clearance is given. We
give models, notation, and assumptions in Section 2. In Sections 3 and 4, we describe equal width and equal
probability partitioning schemes and discuss optimal mean shift in both the schemes. Section 5 gives some numerical
results which show that using optimal mean shift considerably reduces the number of surplus components in equal
width partitioning and may enable us to manufacture all products within the specification range for tighter clearance
specification in equal probability partitioning. We also show some merits and demerits of equal width and equal
probability partitioning schemes. The final section gives concluding remarks.
2
MODELS, NOTATION, AND ASSUMPTIONS
X
U denote the dimensions of the components with larger and smaller variance, respectively. The
clearance X  U must be within the specification range [C   C  ] , where C is a given target
clearance and  is a given tolerance limit. Suppose that the process mean of U can be adjusted. Then we let
C  0 without loss of generality. We also suppose that X and U are normally distributed and we let
  SD[U ]  SD[ X ] ( 0    1 ).
Throughout this paper, we assume E[ X ]  0 and SD[ X ]  1 without loss of generality. Then, X is
distributed as N (01) . Let  ( x) denote the cumulative distribution function of N (01) . The component
with smaller variance is manufactured at two shifted means, and we let b be the two means. Then, the dimension
of the component with smaller variance, denoted by Y , follow the mixture distribution of U s with means b .
Let
and
Its cumulative distribution function is expressed as
1   y b 
 y  b 
FY ( y  b)   
  
 
2   
  
We also let  D be the common specification limits for the two component dimensions X and Y . A
component with its dimension in the intervals (   D ] and ( D ) is rejected before to the assembly
process. Let F and G denote the cumulative distribution functions of
respectively. They are expressed as
X
and
Y
after truncation at
D ,
 ( x)   ( D)
  D  x  D
 ( D)   ( D)
F ( y  b)  FY ( D  b)
G ( y  b)  Y
FY ( D  b)  FY ( D  b)
F ( x) 
{( yb )  ( yb )}  12 {(  D b )  (  D b )}

  D  y  D
{ ( Db )   ( Db )}  12 { (  D b )   (  D b )}
Let n denote the number of classes. The respective partition limits for X and Y are denoted
( x1 x2 … xn1 ) and ( y1  y2 … yn1 ) . In selective assembly, the components X  ( xi 1 xi ]
1
2
1
2
matched with the components
Y  ( yi 1 yi ]
as shown in Figure 2.
Component X
reject
reject
-D
x1
xn-1
D
・・・
x2
matching matching
xn-1
D
matching
Component Y
reject
reject
-D
y1
yn-1
D
y2
・・・
yn-1
Figure 2. Selective Assembly
3
D
by
are
We let
Xi
Yi
and
denote the truncated random variables of
X
and
Y
defined on
( xi 1 xi ] and
i th class of the
( yi 1  yi ] , respectively, where x0  y0   D and xn  yn  D . The probabilities of the
components X and Y , F ( xi )  F ( xi 1 ) and G( yi  b)  G( yi 1  b) , are denoted by pi
and
qi .
In Sections 3 and 4, we describe equal width and equal probability partitioning schemes and discuss optimal mean
shift in both the schemes.
EQUAL WIDTH PARTITIONING SCHEME
Equal width partitioning scheme partitions the distributions so that all classes have equal widths. The number of
classes needed to satisfy the clearance specification is
 2D 
n

  
 x 
We note that
Y
denotes the minimum integer which is not smaller than
x . The partition limits for X
and
are given as
( x1 x2 … xn1 )  ( y1 y2 … yn1 )  ( D  w  D  2w … D  w)
where
w
2D
n
. We easily see that
 X i  Yi  w   i  1 2… n
Thus, all products satisfy the clearance specification. However, some surplus components result from the difference
between
pi
and
qi . The probability that a component is used for assembly is
n
n
i 1
i 1
R(b)   min( pi  qi )   min F ( xi )  F ( xi 1 ) G ( yi  b)  G ( yi 1  b) 
We easily see that R (0)  1 holds when   1 . Thus, we see that we do not need to shift the process mean and
have no surplus component when the two component dimensions have the same variance. However, in the unequal
variances case (   1 ), b  0 does not necessarily maximize R (b ) and we should choose optimal mean shift
b
which maximizes
R (b ) , that is,
b  arg max R(b)
b 0
We give some properties of
b
and
Proposition 3.1. (The lower limit of
R(b ) .
R (b ) .) For any 
b,
and
D
where
f ()
and
g (  b)
R(b)   min( f ( x) g ( x  b))dx
D
denote the derivative functions of
Proposition 3.2. (The lower limit of
F ()
and
G (  b) , respectively.
R(b ) .) Let
D
b†  arg max  min( f ( x) g ( x  b))dx
b 0
Then, for any
D
,
D
R(b )  R(b† )  max  min( f ( x) g ( x  b))dx
b0
From
Proposition
3.2,
we
see
that
it
D
max b0  min( f ( x) g ( x  b))dx
D
for any
is
D
guaranteed
that
 . We notice that b†
4
R(b )
and
is
not
smaller
than
D
max b0  min( f ( x) g ( x  b))dx
do not depend on
max b0  min( f ( x) g ( x  b))dx
is sufficiently large, it may be an effective method that we use
D
D
D
without regard to the value of
 . Therefore, if
 , compared to the alternative of computing b
which depends on
In Section 5, we will give some numerical results which show that using optimal mean shift
increases
R (b ) , especially for the case in which 
b†
.
b considerably
is small.
EQUAL PROBABILITY PARTITIONING SCHEME
Equal probability partitioning is the partitioning in which the corresponding classes have equal probability, that is, no
surplus component exists. The goal is to find the number of classes n and the sets of partition limits
( x1 x2 … xn1 )
and
( y1  y2 … yn1 )
which satisfy the following conditions:
xi  yi 1   i  1 2… n
xi 1  yi   i  1 2… n
pi  qi  i  1 2… n
(1)
(2)
(3)
We note that if the conditions (1) and (2) are satisfied, then all products satisfy the clearance specification.
Fang and Zhang (1995) gave the following algorithm for finding equal probability partitioning which satisfies the
conditions (1)-(3). We note that
respectively.
1.
2.
F 1 ()
and
G 1 (  b)
denote the inverse functions of
x1   D  
If
F (  D   )  G (  D    b) ,
If
F ( D  )  G ( D    b) , then put y1   D  
We repeat the following for
then put
i  2 3 …
F ( yi  )  G( xi    b) ,
sequentially unless
xi 1  yi  
then put
and
and
and
G (  b) ,
y1  G 1 ( F ( D  )  b) .
x1  F 1 (G ( D    b)) .
D  xi  
and
F ()
and
D  yi  
1
yi 1  G ( F ( yi  )  b) ,
hold. If
and if
1
3.
F ( yi  )  G( xi    b) , then put yi 1  xi   and xi 1  F (G ( xi    b)) .
If D  xi   and D  yi   holds for some i , then put xi 1  yi 1  D and n  i  1
and finish the algorithm.
However equal probability partitioning which satisfies the conditions (1)-(3) does not exist under a condition as
discussed in the following proposition.
Proposition 4.1. The number of classes
n
and the partition limits
( x1 x2 … xn1 )
and
( y1  y2 … yn1 )
which satisfy the conditions (1)-(3) exist if and only if
  sup  F 1 (u )  G 1 (u  b)  
0 u 1
We see from this proposition that, letting
 (b)  sup  F 1 (u )  G 1 (u  b) 
0u 1
equal probability partitioning to manufacture all products within the specification range exists if and only if
   (b) . Thus, if we choose optimal mean shift b

which minimizes
 (b) , that is,

b  arg min sup  F 1 (u)  G 1 (u  b)  
b0
0u 1
then equal probability partitioning which satisfies the conditions (1)-(3) exists for tighter clearance specification.
We note that if   1 , then  (0)  0 holds, and therefore, we do not need to shift the process mean and can find
5
equal probability partitioning which satisfies the conditions (1)-(3) no matter how small
We also give the following proposition.

is.
Proposition 4.2. Suppose that
  sup  F 1 (u )  G 1 (u  b)  
0 u 1
Let
n

denote the number of classes determined by the algorithm for equal probability partitioning. Then, the
number of classes
n  n
and the partition limits which satisfy the conditions (1)-(3) do not exist.
From this proposition, we see that the algorithm minimizes the number of classes in the set of equal probability
partitions which satisfy the conditions (1)-(3).
NUMERICAL RESULTS
In this section, we give some numerical results on equal width and equal probability partitioning schemes for
D  3 and some values of  .
Tables 1-3 compare
R (0)
and
R(b )
for
  21 06
in equal width partitioning scheme. From these
tables, we see that using optimal mean shift results in considerable improvement on R (b ) . In other words, using
optimal mean shift considerably reduces the number of surplus components compared with not shifting. Especially
for the case in which  is small, the improvement is quite substantial. For example, although about 30% of
components are surplus without shift for
  03 , we have no surplus component by manufacturing
08559 .
  1 08 05 03 in equal probability partitioning scheme. We see
2
and
the component with smaller variance at two means
Table 4 compares  (0) and  (b ) for
that using optimal mean shift enables us to manufacture all products within the specification range for tighter
tolerance limit  in equal probability partitioning compared to not shifting. For example, when   05 ,

  12832 without
  03741 by manufacturing the component with smaller variance at two means 10194 .
although equal probability partitioning which satisfies the conditions (1)-(3) exists only for
shift, it exists for
Next we give a numerical example to compare equal width and equal probability partitioning schemes. We let
  05
  06 . From Table 2, if we use equal width partitioning scheme and b  06516 , then we
need 10 classes (the partition limits for X and Y are (24 18 12 06 0 061218 24) )
and about 10% of components are surplus. On the other hand, although equal probability partitioning which
satisfies the conditions (1)-(3) does not exist since  (0)  12832  06   if we use b  0 , it exists since
  06  03741   (b ) if we use b  10194 . Table 5 is the result obtained using the algorithm for

equal probability partitioning when we use b  10194 . From this table, we see that equal probability
partitioning reduces surplus components from 10% to 0% by increasing number of classes from 10 to 17
and
compared with equal width partitioning.
R (0) , b
Table 1.
and
R(b )
for
  08
R(b )

n
2
3
0.8957
0.5586
1
6
0.8957
0.5586
0.9925
0.6
10
0.9054
0.5418
0.9893
R (0)
b
in equal width partitioning
1
6
R (0) , b
Table 2.
R(b )
and
for
  05

n
2
3
0.7300
0.7595
1
6
0.7300
0.7595
0.9636
0.6
10
0.6829
0.6516
0.9084
R (0)
b
R (0) , b
Table 3.
R(b )
1
R(b )
and
for
  03
n
2
3
0.6854
0.8559
1
1
6
0.6854
0.8559
0.9572
0.6
10
0.4982
0.6356
0.8019
Table 4.

b
 (0) , b
 (0)
b
and

R(b )
 (b )

 (b )
in equal probability partitioning

1
0
0.8
0.4371
0.6269
0.0503
0.5
1.2832
1.0194
0.3741
0.3
1.9157
1.2378
0.7382
Table
5.
Equal
in equal width partitioning


R (0)
in equal width partitioning


probability
partitioning
under
  06
when
X
N (01)
N (10194 (05) )
2
Y
Class
Partition limits for
X
Partition limits for
Y
Clearance
Probability
min
max
min
max
min
max
1
-3.000
-2.774
-3.000
-2.400
-0.600
0.226
0.14
2
-2.774
-2.481
-2.400
-2.174
-0.600
-0.081
0.38
3
-2.481
-2.005
-2.174
-1.881
-0.600
0.169
1.60
4
-2.005
-1.281
-1.881
-1.444
-0.561
0.600
7.78
5
-1.281
-0.844
-1.444
-1.150
-0.131
0.600
9.96
6
-0.844
-0.550
-1.150
-0.917
0.073
0.600
9.21
7
-0.550
-0.317
-0.917
-0.682
0.132
0.600
8.47
8
-0.317
-0.082
-0.682
-0.281
-0.036
0.600
9.21
%
9
-0.082
0.198
-0.281
0.518
-0.600
0.478
11.12
10
0.198
0.423
0.518
0.798
-0.600
-0.095
8.57
11
0.423
0.677
0.798
1.023
-0.600
-0.121
8.71
12
0.677
1.026
1.023
1.277
-0.600
0.002
9.69
13
1.026
1.576
1.277
1.626
-0.600
0.299
9.53
14
1.576
2.226
1.626
2.013
-0.437
0.600
4.46
15
2.226
2.613
2.013
2.266
-0.041
0.600
0.86
16
2.613
2.866
2.266
2.504
0.110
0.600
0.24
17
2.866
3.000
2.504
3.000
-0.134
0.496
0.07
7
and
CONCLUSION
In selective assembly when the component with smaller variance is manufactured at two shifted means, this paper has
studied determining optimal mean shift in equal width and equal probability partitioning schemes. Some numerical
results have shown that using optimal mean shift considerably reduces the number of surplus components in equal
width partitioning and enables us to manufacture all products within the specification range for tighter clearance
specification in equal probability partitioning compared to not shifting. It has also been shown that by using equal
probability partitioning we have no surplus component but need larger number of classes compared with equal width
partitioning.
ACKNOWLEDGEMENTS
This research was supported by Grant-in-Aid for JSPS Fellows, 20  381.
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