THE OPTIMUM UPPER SCREENING LIMIT AND OPTIMUM MEAN FILL
LEVEL TO MAXIMIZE EXPECTED NET PROFIT IN THE CANNING
PROBLEM FOR FINITE CONTINUOUS DISTRIBUTIONS
Except where reference is made to the work of others, the work described in this
dissertation is my own or was done in collaboration with my advisory
committee. This dissertation does not include proprietary or
classified information.
___________________________________________
Victoria Spooner Jordan
Certificate of Approval:
_________________________ ______________________
Alice E. Smith Saeed Maghsoodloo, Chair
Professor Professor
Industrial Engineering Industrial Engineering
_________________________ ______________________
Jorge Valenzuela Stephen L. McFarland
Associate Professor Dean
Industrial Engineering Graduate School
THE OPTIMUM UPPER SCREENING LIMIT AND OPTIMUM MEAN FILL
LEVEL TO MAXIMIZE EXPECTED NET PROFIT IN THE CANNING
PROBLEM FOR FINITE CONTINUOUS DISTRIBUTIONS
Victoria Spooner Jordan
A Dissertation
Submitted to
the Graduate Faculty of
Auburn University
in Partial Fulfillment of the
Requirements for the
Degree of
Doctor of Philosophy
Auburn, Alabama
May 11, 2006
iii
THE OPTIMUM UPPER SCREENING LIMIT AND OPTIMUM MEAN FILL
LEVEL TO MAXIMIZE EXPECTED NET PROFIT IN THE CANNING
PROBLEM FOR FINITE CONTINUOUS DISTRIBUTIONS
Victoria Spooner Jordan
Permission is granted to Auburn University to make copies of this dissertation at its
discretion, upon request of individuals or institutions and at their expense. The
author reserves all publication rights.
______________________________
Signature of Author
______________________________
Date of Graduation
iv
VITA
Victoria Spooner Jordan, daughter of Joe and June Spooner, was born January 26,
1963, in Athens, Georgia. She graduated from Auburn High School in 1980. She
received a Bachelor of Science degree in Statistics from the University of Kentucky in
May, 1983, a Master of Science degree in Industrial Engineering from Auburn University
in 1987, and a Master of Business Administration from the Ohio State University in
1989. After working for Ampex Corporation and General Electric Company in Quality
Assurance, she was a Senior Staff Associate for Luftig & Warren, a management
consulting firm. She entered Graduate School, Auburn University, in 2000. She married
T. Frank Jordan, son of Terry and Fayron Jordan, on January 1, 1995. Together they
have three children: Taylor Frank Jordan, born April 2, 1997; Eben Glenn Jordan, born
October 13, 1998; and Audrey Elise Jordan, born July 15, 2001.
v
DISSERTATION ABSTRACT
THE OPTIMUM UPPER SCREENING LIMIT AND OPTIMUM MEAN FILL
LEVEL TO MAXIMIZE EXPECTED NET PROFIT IN THE CANNING
PROBLEM FOR FINITE CONTINUOUS DISTRIBUTIONS
Victoria Spooner Jordan
Doctor of Philosophy, May 11, 2006
(M.B.A., Ohio State University, 1989)
(M.S., Auburn University, 1987)
(B.S., University of Kentucky, 1983)
125 Typed Pages
Directed by Saeed Maghsoodloo
The ?canning problem? occurs when a process has a minimum specification such
that any product produced below that minimum incurs a scrap/rework cost and any
product over the minimum incurs a ?give-away? cost. The objective of the canning
problem is to determine the target mean for production that minimizes both of these
costs. An upper screening limit can also be determined; above which give-away cost is
so high that reworking the product maximizes net profit.
Examples of the canning problem are found in the food industry (filling jars or
cans) and in the metal industry (thickness.)
vi
In this dissertation, continuous, finite range space distributions are considered,
specifically the Uniform and Triangular distributions. For the Uniform distribution, an
optimum upper screening limit and an optimum value for the mean fill level is found
using three net profit models. Each model assumes a fixed selling price and a linear cost
to produce, but costs differ as follows:
square4 Model 1 uses fixed rework/scrap and reprocessing costs
square4 Model 2 has linear rework/scrap and reprocessing costs, and
square4 Model 3 has fixed rework/scrap and reprocessing costs but adds an
additional, higher cost associated with a limited capacity of the container.
A discussion is included relating the selection of an optimum set point for the mean to
process capability.
For the Triangular distribution, an optimum upper screening limit and an optimum
value for the mean fill level is found for both the symmetrical and skewed cases using a
net profit model that has fixed rework/scrap and reprocessing costs.
vii
ACKNOWLEDGEMENTS
The author would like to thank Frank Jordan for his confidence, patience, and
positive attitude, and Taylor, Eben, and Audrey Jordan for their loving encouragement.
Thanks also to Joe and June Spooner for instilling self-confidence and a love of learning,
and to Terry and Fayron Jordan for their support. The author is grateful to Dr. James
Hool for suggesting the topic and to Dr. Saeed Maghsoodloo for his encouragement and
attention to details. This work is dedicated to the memory of Cynthia Spooner Hankes
who provided life lessons in grace, strength, and perspective.
viii
Style manual or journal used: Journal of Quality Technology
Computer software used: MS Excel
MS Word
MathType
ix
TABLE OF CONTENTS
List of Tables ??????????????????????......................... xii
List of Figures ???????????????????????????... xiii
List of Notation ..??????????????????????????.... xv
1.0 Introduction ??????????????????.?????????.. 1
2.0 Literature review ?????????????????????????... 3
3.0 Uniform distribution ...??????????????...???????..... 14
3.1 Uniform underlying distribution - constant scrap cost ?.......................... 14
3.1.1 Optimum upper screening limit ????????????. 17
3.1.2 Optimum target set point for mean ??????????? 22
3.2 Uniform underlying distribution - linear scrap cost ???????...? 39
3.2.1 Optimum upper screening limit ????????????. 40
3.2.2 Optimum target set point for mean ??????????? 42
3.3 Fixed scrap cost with capacity constraint ????????????... 55
4.0 Relationship between canning problem and process capability ?..?????... 59
4.1 Relationship between process capability and the selection
of oU and o? ??????????????????????. 60
10
TABLE OF CONTENTS (CONTINUED)
4.2 Relationship between PC and the selection of o? ?????????. 60
4.2.1 Process Capability Measure, PC .?????????..?...... 60
4.2.2 Process Capability Measure, PLC ?.??????????... 62
4.2.3 Process Capability Measure, PMC .???????????.. 62
4.3 Relationship between PLC and the selection of o? ?????????. 63
5.0 Triangular distribution ??????????????????????? 64
5.1 Symmetric Triangular underlying distribution ??????????? 66
5.1.1 Optimum upper screening limit ????????????... 67
5.1.2 Optimum target set point for mean ???????????.. 70
5.2 Skewed Triangular underlying distribution ?.??????????... 80
5.2.1 Optimum upper screening limit ???????????...? 82
5.2.2 Optimum target set point for m ??.?????????.?.. 84
5.2.3 Sensitivity analysis of skewness ???????????..?. 87
5.3 Process capability with underlying Triangular distribution ?...?. 88
6.0 Conclusion ??????????????????????????...... 91
6.1 Summary of Results ?????????????????????. 91
6.2 Practical Applications ???????????????...????.. 93
6.3 Recommendations for Future Work ????????????..??.. 94
7.0 References ?????????????????????..?????..... 95
11
TABLE OF CONTENTS (CONTINUED)
Appendices ????????????????????????????? 98
Appendix A: Formulas for Excel spreadsheet for calculating expected net profit
for various levels of ? with a Uniform distribution and fixed scrap cost ?... 99
Appendix B: Formulas for Excel spreadsheet for calculating expected net profit
for various levels of ? with a Uniform distribution and linear scrap cost ? 100
Appendix C: Formulas for Excel spreadsheet for calculating expected net profit
for various levels of m with a symmetric Triangular distribution. ????.. 101
Appendix D: Derivation of the first four moments of the skewed Triangular
Distribution ?????????????????????????. 102
xii
LIST OF TABLES
TABLE 1. Net Profit Model Comparison from Literature Review ??????.?. 11
TABLE 2. Case 1: Expected Net Profit for Ranges of ? when 2 ok U L? ? ??... 26
TABLE 3. Case 2: Expected Net Profit for Ranges of ? when 2 ok U L> ? ??.... 32
TABLE 4. Case 1: Expected Net Profit for Ranges of ? when 2k U L? ? ???.. 42
TABLE 5. Case 2: Expected Net Profit for Ranges of ? when 2k U L> ? ???.. 49
TABLE 6. Example of expected net profit calculations for symmetric triangular
distribution with 2k U L> ? with a fixed value of U ?????????..???.. 76
TABLE 7. Expected net profit calculations for symmetric triangular distribution with
2k U L> ? with a calculated value of U ?????????????????? 78
TABLE 8. om and the expected net profit for various levels of skewness ??..??. 88
xiii
LIST OF FIGURES
FIGURE 1. References that Address the Canning Problem With 100 % Inspection ?. 8
FIGURE 2. Other References that Address the Canning Problem ???????.? 9
FIGURE 3. Net Profit Model When U LR R> ???????????????.. 21
FIGURE 4. Distribution of fill level when 2 ok U L? ? and k L? + ? ?????.. 23
FIGURE 5. Distribution of fill level when 2 ok U L? ? , k L? ? ? and oL k U?? + ?
???????????????????????????.?. 23
FIGURE 6. Distribution of fill level when 2 ok U L? ? , k L? ? ? and ok U? + ? ... 24
FIGURE 7. Distribution of fill level when 2 ok U L? ? , oL k U?? ? ? , and ok U? + ?
????????????????????????????.. 24
FIGURE 8. Distribution of fill level when 2 ok U L? ? and ok U? ? ? ?????. 25
FIGURE 9. Example of Expected Net Profit for Different Values of ? when 2 ok U L? ?
?????????????????????????..??? 30
FIGURE 10. Distribution of fill level when 2 ok U L> ? , k L? ? ? and ok U? + ? . 31
FIGURE 11. Example of Expected Net Profit for Different Values of ? when
2 ok U L> ? and U LR R< ???????????????????..???.... 36
FIGURE 12. Example of Expected Net Profit for Different Values of ? when
2 ok U L> ? and U LR R> ???????????????????????.. 37
FIGURE 13. Example of Expected Net Profit for Different Values of ? when
2 ok U L> ? and L UR R R= = ....................................................................................... 38
xiv
LIST OF FIGURES (CONTINUED)
FIGURE 14. Example of Expected Net Profit for Different Values of ? when
2k U L? ? ?????????????????????????????. 47
FIGURE 15. Distribution of Fill Level when 2k > U ? L ??????????.... 47
FIGURE 16. Net Profit Equation when U LR R> ??????????????. 48
FIGURE 17. Example of Expected Net Profit for Different Values of ? when 2k > U ? L,
o L k? = + ?????????????????????????????.. 53
FIGURE 18. Example of Expected Net Profit for Different Values of ? when 2k > U ? L
and o U k? = ? ??????????????????????????.?. 54
FIGURE 19. Example of Expected Net Profit for Different Values of ? when oCAP U<
and 2k > CAP ? L ??????????????????????????... 58
FIGURE 20. The triangular probability density function with lower specification, L, and
upper screening limit, U ????????????????????????. 65
FIGURE 21. The triangular probability density function with lower specification, L, and
upper screening limit, U when 1m k L? < and 2L m k U< + ? ?????????. 70
FIGURE 22. The triangular probability density function with lower specification, L, and
upper screening limit, U when 1m k L? ? and 2m k U+ ? ??????????... 71
FIGURE 23. The triangular probability density function with lower specification, L, and
upper screening limit, U when 1L m k U? ? < and 2m k U+ > ????????.... 71
FIGURE 24. Expected net profit model for symmetric triangular distribution with
2k U L> ? and a fixed value of U ????????????????????. 77
FIGURE 25. Expected net profit model for symmetric triangular distribution with
2k U L> ? and a calculated value for U ?????????????????... 79
xv
LIST OF NOTATIONS
X = the random variable, fill level
x = a specific value of X
A = selling price of an acceptable product
C = variable cost per unit to produce
Co = fixed cost per unit to produce
D = selling price of a non-conforming product
( )E P x? ?? ? = expected net profit
L = lower specification limit
m = mode of distribution
P(x) = net profit function
LR = cost to scrap or rework product if x < L
UR = cost to scrap or rework product if x > U
R = cost to scrap or rework product when LR = UR
U = upper specification limit (if given)
oU = optimum upper screening limit
o? = optimum target set point for mean
1
1.0 Introduction
An interesting problem in process optimization is the ?canning problem? which
attempts to define the optimum set point for the mean of a manufacturing process to
minimize scrap and cost (or maximize net profit). The examples used typically refer to
filling jars or cans such as in the food industry. In each case, there is a minimum
requirement or specification set by the consumer. Any product produced below that
minimum is either scrapped or reworked and therefore incurs an associated cost. On the
other hand, any product over the minimum is extra product that is given to the customer
beyond the minimum requirement (and has thus been labeled ?give-away? cost). The
give-away cost is proportional to the distance between the existing fill level and the
minimum specification. If a target is set too low, product will be rejected as not meeting
the customer requirement (lower specification limit) and reworking or scrap costs will be
incurred. If a target is set too high, product will meet customer requirements but at the
added expense of ?giving away? more material than necessary. The objective of the
canning problem is to determine the target mean for production that minimizes both of
these costs, given that process variability is known and in a state of statistical control.
In some cases, an upper screening limit, U, can also be determined. This limit
identifies a level above which give-away cost is so high that net profit is maximized by
reworking the product.
2
Examples of the canning problem can be found in ?fill processes? such as the
amount of coffee in a jar, paint in a can, etc. Other examples come from the metal
industry such as steel, aluminum, and copper where metal thickness (gauge) has a
required minimum, but additional gauge just adds to the cost.
In this paper, continuous, finite range distributions, specifically the Uniform and
Triangular distributions will be considered. The Uniform distribution is presented as a
common, symmetric finite range distribution and the Triangular distribution is presented
with varying levels of skewness. An optimum upper screening limit (Uo) and an
optimum value for the mean fill level (?o) will be determined for two net profit models ?
one with fixed rework costs and one with linear rework costs. In the case of the net profit
model with fixed rework costs, the results for the Uniform distribution will be compared
with a generalized optimum mean and upper screening limit developed by Liu and
Raghavachari (1997) for any continuous fill distribution.
3
2.0 Literature Review
Springer (1951) introduced the concept of determining an optimum target for a
process given fixed costs associated with product falling outside of specifications (both
lower specification limit (LSL) and upper specification limit (USL.)) He identified a
general approach for calculating a process mean target that minimizes the total cost of
rejection for both a Normal and Pearson Type III (or Gamma) distribution. Bettes (1962)
wrote a similar article that identified a method for determining the process target when a
set lower specification but an arbitrary upper specification is given. He specifically used
a ?foodstuff? example in which items below the lower specification and above the upper
specification were reprocessed at a fixed cost. Both Springer?s and Bette?s solutions
involved a tabulation of factor W . W is a factor which varies depending on the values of
( )U L
?
? and L
U
C
C , the cost ratio between rejecting a part with fill level < L and rejecting a
part with fill level > U. The optimum mean is o L W? ?= + when L UC C? and
o U W? ?= ? when L UC C? where
1 log
2
L
U
CU LW
U L C
?
?
? ??? ? ? ?= +
? ?? ? ? ??? ? ? ? ? ?. (Note:
Springer uses the notation LC and UC to define the costs associated with rejected
material. In subsequent articles and in this research, the notation LR and UR is used.)
To avoid tabulation, Nelson (1979) presented a simplified approach to
Springer?s solution using a nomograph.
4
Hunter and Kartha (1977) identified a similar problem with a lower specification
limit, where items produced below the specification were sold at a reduced price and
give-away cost was linear. Since their approach does not provide a closed-form solution,
Nelson (1978) provides an approximating function for use in a calculator or computer
allowing an error of about three-decimal accuracy.
The ?canning problem? is not specific to cans. Any process that has a minimum
specification and costs associated with ?under-fill? (and ?over-fill?) can be classified as a
?canning problem.? Another application of the canning problem is the production of
steel beams. The beams have minimum web and flange widths. Beams produced below
the width specification cannot be sent to customers. Those with width levels above the
minimum use more steel (and therefore cost significantly more) than those produced at
minimum. What makes the steel beam example different from the traditional canning
problem is that beams produced below the minimum cannot be reworked as in the typical
canning problem (more steel cannot be added). The rejected beams are either sold at a
reduced price or scrapped and reprocessed. Beams with thickness levels greater than U
are most likely melted down and reprocessed. Carlsson presented this example (1984)
with the two classes of rejects previously mentioned and give-away cost is measured on a
cost per unit basis.
Bisgaard, Hunter, and Pallesen (1984) pointed out that Hunter and Kartha?s
assumption that under-filled items can be sold for a fixed price implied that even empty
cans could be sold. They expanded Hunter and Kartha?s model to determine the optimum
mean, but instead of using a constant selling price for under-filled items, they used a
proportional price. They also addressed the possibility and associated costs of reworking
5
under-filled items. Dodson (1993) provided a similar model with his specific example of
rolled aluminum sheet metal. Using the characteristic of footage on a finished coil, he
identified two costs: when footage is below the lower specification, the entire coil is
scrapped (at a cost proportional to footage) and when product is produced above the
upper specification, the extra footage is scrapped. He provided a method using graphs or
a spreadsheet to identify the target process mean.
Gohlar (1987) first coined the term ?canning problem?. He addressed the specific
case where over-filled cans are sold for a fixed price (and thus incur a ?give-away? cost)
and under-filled cans are emptied and refilled at a fixed reprocessing cost. Gohlar and
Pollock (1988) added the determination of an upper specification limit for cases where a
manufacturer may choose to empty and refill expensive product when the fill level is too
high. Gohlar (1988) provided a Fortran program to calculate these values.
Many variations of the original models have been written to include other
constraints. Schmidt and Pfeifer (1989) determined the cost benefit associated with
reducing variability based on a percentage reduction in standard deviation. Their model
extends Gohlar (1987) in which under-filled product is emptied and reprocessed at a
fixed cost and over-filled product is sold at the regular price. In 1991, Schmidt and
Pfeifer also extended the analysis to include limited capacity as a constraint. Usher,
Alexander, and Duggins (1996) recognized that handling rejects reduces efficiency, so
they extended Gohlar and Pollock (1988) to include the effect that the target mean and
upper limit have on the efficiency of a production line. Cain and Janssen (1997)
identified a target value when there is asymmetry in the cost function in the cases of
asymmetric linear, asymmetric quadratic, and combined linear and quadratic costs. Pulak
6
and Al-Sultan (1997) developed Fortran programs for solving nine selected targeting
models: Hunter and Kartha ?77, Bisgaard, Hunter, and Pallesen ?84, Carlsson ?84, Gohlar
?87, Schmidt and Pfeifer ?91, Boucher and Jafari ?91 (sampling), Gohlar and Pollock ?88,
Arcellus and Rahim ?90 (sampling), and Al-Sultan ?94 (two machines in series with
sampling.)
Many articles have been written that extend the target selection problem to
processes that are subjected to acceptance sampling rather than 100% inspection.
Carlsson (1989) identified a method for determining the target mean when using variable
data, a one-sided specification, a known variance, and a sampling plan such as MIL-STD-
414B. Lee and Elsayed (2002) calculated optimum process mean and screening limits by
maximizing profit when a 2-stage screening process is used with a surrogate variable.
Lee, Hong, and Elsayed (2001) calculated the optimum process mean and screening
limits for a correlated variable under single- and 2-stage screening. In their article, the
single screening was based on the quality characteristic being measured and the 2-stage
screening was being done on a correlated variable first, then on the quality characteristic
of interest.
Recent articles have explored the objective functions and use of a fixed variance.
Pfeifer (1999) identified two competing objectives: expected profit per fill attempt and
expected profit per can to be filled. Rather than setting the first derivative of expected
profit equal to zero and solving for the optimum target, he evaluated expected profit over
a range of values and found one that maximized expected profit using spreadsheets and
search routines. Misiorek and Barnett (2000) examined the effect of a change in variance
on the solution of optimum mean and expected profit. They also explored the
7
implications to ?Weights and Measures? requirements. In this article, over-fill was either
recaptured or lost and under-filled containers were emptied and material re-used or they
were ?topped-up?. Kim, Cho, and Phillips (2000) calculated the optimum mean while
keeping the process capability at a predetermined level. They used a cost function that
increased as variance decreased which seems counter to Taguchi?s loss function and
would be difficult to quantify in practical application.
Liu and Raghavachari (1997) generalized the determination of an optimal process
mean for the canning problem and an upper screening limit for any continuous
distribution. They used a simple profit model given by Schmidt and Pfeifer (1991) and
determined an optimal value of U (upper screening limit) and ? which maximized the
expected net profit for any continuous fill distribution. Their work addressed infinite-
range distributions that they truncated on the low end.
A flow chart of articles reviewed for this research follows:
8
FIGURE 1. References that Address the Canning Problem With 100 % Inspection
LSL and USL
given
Springer, 1951
Nelson, 1979
Dodson, 1993
Bettes, 1962
Gohlar & Pollack, 1988
Gohlar, 1988
Liu & Raghavachari, 1997
Reason for
Reject
Set Mean
Set Mean &
USL
Under-fill
Over-fill
Both Selling Price
Constant
Proportional
Hunter & Kartha, 1977
Nelson, 1978
Bisgaard, Hunter, &
Pallesen, 1984
Misiorek & Barnett, 2000
Carlsson, 1984
Gohlar, 1987
Cain & Janssen, 1997
LSL only
Optimization
Goal?
9
FIGURE 2. Other References that Address the Canning Problem
Other
Variations
Lim Capacity?Schmidt & Pfeifer, 1991
Line Efficiency?Usher, Alexander, &
Duggins, 1996
Fortran Programs?Pulak & Al-Sultan, 1997
Competing Obj Functions ? Pfeifer, 1999
With
Sampling Bai & Lee, 1993
Tang & Lo, 1993
Lee & Kim, 1994
Lee & Jang, 1997
Lee, Hong, & Elsayed, 2001
Lee & Elsayed, 2002
Correlated
Variables
Multi-Class
Inspection
Carlsson, 1989
Boucher & Jafari, 1991
Tango & Lo, 1993
Al-Sultan, 1994
Impact of Change in
Variance
Schmidt & Pfeifer, 1989
Misiorek & Barnett, 2000
Kim, Cho, & Phillips, 2000
10
Since many different models of net profit have been used in literature (with many
different forms of notation), they are summarized in Table 1 using the following notation:
X = the random variable, fill level
x = a specific value of X
A = selling price of an acceptable product
D = selling price of a non-conforming product
C = variable cost per unit to produce
Co = fixed cost per unit to produce
LR = cost to scrap or rework product if x < L
UR = cost to scrap or rework product if x > U
R = cost to scrap or rework product if LR = UR
L = lower specification limit
U = upper specification limit (if given)
oU = optimum upper screening limit
o? = optimum target mean
m = the modal point
11
TABLE 1. Net Profit Model Comparison from Literature Review
Author (Date) of
Reference
Net Profit Model Calculate oU in
addition to o?
Springer (1951),
Nelson (1979)
Springer and Nelson
do not use a net
profit model, just the
cost of rejection:
- LR if x < L
- UR if x > U
No
Bettes (1962) First application of
give-away cost, no
net profit model,
costs are as follows:
- LR if x < L
- Cx if x > oU
Yes
Hunter & Kartha (1977),
Nelson (1978)
D-Cx if x < L
A-Cx if x ? L
No
Carlsson (1984) (Differs from Hunter
& Kartha in that cost
to produce is both
fixed and variable.)
D-(Co+Cx) if x < L
A-( Co+Cx) if x ? L
No
Bisgaard, Hunter, &
Pallesen (1984)
(D-C)x-Co if x < L
A-Cx- Co if x ? L
No
Gohlar (1987) A-R-Cx if x < L
A-Cx if x ? L
No
12
TABLE 1 (continued). Net Profit Model Comparison from Literature Review
Author (Date) of
Reference
Net Profit Model Calculate oU in
addition to o?
Gohlar & Pollock (1988) ( ),P U? -R if x < L,
where ( ),P U? is
the expected profit at
the new level when
the can is refilled.
A-Cx if L ? x ? oU
( ),P U? -R if x >
oU
Yes
Schmidt & Pfeifer
(1991), Liu &
Raghavachari (1997)
A-Cx if L ? x ? U
-R otherwise
No
Yes
Dodson (1993) (A-C)x if L ? x ? U
- LR x if x < L
- UR x if x > U
No
The differences in Table 1 come from several sources. One variation is the
selling price. In some models, selling price is fixed, in others it is proportional, and in
still other models, there may be more than one selling price if rejected material can be
sold in a secondary market. Another variation in the models addresses how material is
handled if it is rejected. Some models assume a fixed rejection cost (whether due to
scrap or rework) and some models treat rejection cost as proportional to x (again this cost
may be due to scrap or rework.) In the following chapter, the two net profit models used
13
assume a fixed selling price, no secondary market for selling rejected product, and fixed
rejection costs for the first model and proportional rejection costs for the second model.
14
3.0 Uniform distribution
Most of the work in the literature regarding the canning problem has focused on
the Normal distribution with many different net profit functions. All assume a
continuous distribution and an infinite range. In actual practice, however, finite
distributions may sometimes be more plausible models for a canning process. Few
authors use a truncated distribution and those who do (Liu and Raghavachari, 1997) only
truncate on the low end of the distribution. This research is preliminarily focused on the
application of the canning problem to two standard statistical distributions that have a
finite range: the Uniform distribution (with two different net profit models) and the
Triangular distribution (with fixed rework/reprocessing costs.)
3.1 Uniform underlying distribution ? constant scrap cost
In every case, analysis should begin with a distribution fit ( 2? Goodness of Fit
Test for example) of the data?s dimension (fill level, metal thickness, etc.), to determine
the appropriate distribution to use in the analysis. In this chapter, it is assumed that such
an analysis has been completed and data are found to be most closely approximated by a
Uniform distribution.
In this first application, the basic profit function used in the Liu and Raghavachari
(1997) article is used, so results for the continuous Uniform distribution can be compared
to their?s for any continuous distribution for that particular profit model. In the net profit
function, first introduced by Schmidt and Pfeifer (1991), the cost of reworking product
15
below L or above U is fixed and the product is only sold if fill level, x, falls within the
limits of L and U. A slight modification of Schmidt and Pfeifer?s (1991) net profit model
is defined below. This net profit model is generalized such that RL need not be equal to
RU.
The Net Profit function:
,
,( )
,
0,
L
U
R a x L
A Cx L x UP x
R U x b
otherwise
? ? , A = revenue received for an acceptable container, and C = the production cost per
unit of ingredient. The constants A, LR , UR , C, and L are known and > 0. In the Liu and
Raghavachari (1997) article, L UR R R= = , but since the cost of rejecting a unit with fill
level less than L may be different from the cost of rejecting a unit with fill level greater
than U, a slight generalization is made for this model. For example, in the steel industry,
if sheets of steel are being produced to a minimum thickness specification, L, then sheets
with thickness less than L may be scrapped or sold in a secondary market, while sheets
16
with thickness greater than U, may be reprocessed, melted down and used as raw
material. The machine?s ?fill level? per attempt will be used throughout this research to
describe the random variable, X. The unit of measure will depend on the application. For
example, ?per unit? may be per ounce (in the case of filling a container) or per fraction of
an inch (in the case of steel thickness.)
In order to determine the optimum value of U, Uo, the equation for maximizing
expected profit must be determined and then the first derivative with respect to U is set
equal to zero to solve for Uo.
The Uniform distribution has the following probability density function:
1 ,
( )
0,
a x bf x b a
otherwise
? ? ??
= ??
??
1 ,
2
0,
a x bk
otherwise
? ? ??
= ?
??
where 2b a k? = .
Graphically, the density function appears as:
( )f x
x
a ? b
17
where a = ? - k, b = ? + k, and 2k is a constant value that describes the spread of the
Uniform distribution such that b ? a = 2k, and variance of X is ( )
2
3
kV x = (Appendix D.)
3.1.1 Optimum upper screening limit
With respect to the canning problem, there is a cost ( LR ) when the quantity falls
below the lower specification, L (L ? a). There is also an arbitrary upper screening limit,
U (U < b), such that the profit ( A Cx? ) above U is actually less than the cost to rework
or reprocess when fill level is too high, UR .
( )f x
x
a=?-k L U b=?+k
Assume that the filling machine variability is such that the process range is k? ? where
a k?= ? and b k?= + . The objective is to maximize the expected net profit, which is
obtained first by combining the equation for the net profit, ( )P x , and the probability
density function of x given by ( ) 12f x k= , a x b? ? .
Assuming that L U?? ? , a k L?= ? ? , and b k U?= + ? , the expected net
profit is given by
18
( ) ( )
kL U
UL
k L U
RR A CxE P X dx dx dx
b a b a b a
?
?
+
?
?= ? + + ?? ?
? ? ? ? ?? ? ?
= - |LL kR xb a ??? +
21
2 |
U
L
CxAx
b a
? ??
? ?? ? ? - |
kU
U
R x
b a
?+
?
= ( ) ( ) ( ) ( )
2 21
2 2L U
C U LR k L A U L R U k
k ? ?
? ??
? ?? ? + ? ? + ? ?
? ?? ?
= ( ) ( )12 2 2U L U LCU CLU R A R k R k L A Rk ? ?? ?? ? ? ?+ ? + ? ? + + ? ?? ? ? ?? ?? ? ? ?
? ?
(1)
In order to solve for the optimum value of the upper limit, UU , the first derivative
of the expected profit equation with respect to U is set equal to zero:
[ ] ( )( ) 12 UE P x R A CUU k? = + ?? (2)
The second derivative with respect to U is 2Ck? which is < 0. Since the second
derivative is < 0, the function is strictly concave and setting the first derivative = 0 will
yield a maximum expected net profit.
Setting Equation (2) equal to zero,
( )12 U UR A CUk + ? = 0 ? ( )U UR A CU+ ? = 0
or,
19
UU R AU C+= (3)
In the case where R = RU = RL, the optimum point obtained in Equation (3) matches the
solution found by Liu and Raghavachari (1997) for any distribution.
Substituting UU into the expected net profit equation, Equation (1), yields the
following optimum expected net profit:
( )oE P X? ?? ? =
( ) ( )12 2 2U UU L U LR A R A CLR A R k R k L A Rk C ? ?? ?+ +? ?? ? ? ?+ ? + ? ? + + ? ?? ?? ?? ?? ? ? ?? ?? ?
? ?
= ( ) ( ) ( )
2
1
2 2 2
U
L U L
R A CLR k R k L A R
k C ? ?
? ?+ ? ?
? ?+ ? ? + + ? ?? ?
? ?? ?? ?
= ( ) ( ) ( )
2
1
2 2 2
U
L U L U L
R A CLR R k R R L A R
k C ?
? ?+ ? ?
? ?+ ? ? + + ? ?? ?
? ?? ?? ? .
Because the variance of the Uniform density is given by ( ) ( )
2 2
12 3
b a kV x ?= = ,
( ) ( ) ( ) ( )
2
1 3
2 22 3
U
o L U L U L
R A CLE P X R R R R L A R
C ? ??
? ?+ ? ?
? ?? ? = + ? ? + + ? ?? ?? ?
? ?? ?? ? (4)
20
In the case where L UR R R= = , Equation (4) simplifies further to
( ) ( )
21
2 22 3o
R A CLE P x L A R R
C?
? ?+ ? ?
? ? = + ? ? ?? ?? ?? ?
? ?? ?? ? . (5)
Equations (4) and (5) show that as process variability is reduced, the expected net profit
increases. Further, ( )oE P x? ?? ? decreases as costs (R and C) increase.
If U LR R> , it is clear based on the net profit model in Figure 3, that when
2 Uk U L> ? , fill level should be positioned below L rather than above UU , so in
determining the optimum value for ? the probability of Ux U> will be zero. However, if
( )L UR P U? > , a higher profit can be realized if ? is set lower than UU k? . UU was
determined to be the ?break-even? point for costs related to UR , but in the case where
U LR R> , even within the range between L and UU , the cost for a given fill attempt, Cx ,
may be higher than the selling price, A, such that net profit becomes negative, and if A-Cx
< -RL, then the cost to rework when fill level is less than L is actually less than the cost to
produce and sell at the given fill level (below UU .)
21
FIGURE 3. Net Profit Model When U LR R> .
In the case where U LR R> , an alternative upper limit exists at UL = LR AC+ for the
purpose of determining o? . UU , in Equation (3), was determined to be the ?break-even?
point for costs related to UR , but when U LR R> , another ?break-even? point occurs
when ( )LR P x? = . This occurs when LA Cx R? = ? and leads to an alternative upper
screening limit for determining o? :
L
L
R AU
C
+= (6)
L
L
R AU
C
+= is the ?break-even point? for net profit with the cost of reprocessing
material with fill level below L. If it costs less to scrap and reprocess under-fills than to
fill and sell product with fill level > LU , then the production process should be centered
such that reject occur below L rather than above LU .
UR?
LR?
A Cx?
L
LU UU
( )P x
x
22
So, the optimum value for the upper screening limit is
( )min , min , ULo L U R AR AU U U C C++? ?= = ? ?? ? (7)
3.1.2 Optimum target set point for the mean
The next step is to determine an optimum machine set point for the mean. Of
course, for a Uniform distribution, 2a b? += , so the actual mean is defined. In this
chapter, the optimum set point for the mean will be determined that maximizes expected
net profit per unit. Rather than fixing a or b, in this analysis, the assumption is that the
filling machine has a fixed amount of variability, but the set point can be adjusted such
that b ? a = 2k is fixed, and the optimum mean, o? , is accordingly determined.
The ideal (though technically impossible) optimum situation would occur if the
process mean was centered at L and there was no variability. The fact that variability
exists in all processes, however, makes the canning problem a practical issue.
For a Uniform distribution with L and k given, the optimum set point for ? , o? , is
based on the equations for expected net profit. There are two cases to be considered:
Case 1: 2 ok U L? ? and Case 2: 2 ok U L> ? . Within each case, the equations for
expected net profit are different, for different ranges of ? .
Case 1: 2 ok U L? ? :
When 2 ok U L? ? , there are five different scenarios:
23
a) k L? + ?
FIGURE 4. Distribution of fill level when 2 ok U L? ? and k L? + ? .
b) k L? ? ? and oL k U?? + ?
FIGURE 5. Distribution of fill level when 2 ok U L? ? , k L? ? ? and oL k U?? + ? .
L
k? ? k? +
oU
( )f x
x
L
k? ? k? +
oU
( )f x
x
24
c) k L? ? ? and ok U? + ?
FIGURE 6. Distribution of fill level when 2 ok U L? ? , k L? ? ? and ok U? + ? .
d) oL k U?? ? ? and ok U? + ?
FIGURE 7. Distribution of fill level when 2 ok U L? ? , oL k U?? ? ? , and ok U? + ? .
L
k? ? k? +
oU
( )f x
x
L
k? ? k? +
oU
( )f x
x
25
e) ok U? ? ?
FIGURE 8. Distribution of fill level when 2 ok U L? ? and ok U? ? ? .
These scenarios and the formulas for expected net profit are summarized in Table 2:
L
k? ? k? +
oU
( )f x
x
26
TABLE 2. Case 1: Expected Net Profit for Ranges of ? when 2 ok U L? ? .
Scenario Range of ? Expected Net Profit, ( )E P x? ?? ?
a k L k?? ? ? LR?
b L k L k?? ? < + ( ) ( )
( ) ( )2 2
1
2
2
L
L
R k A k
Ck R A L k L
? ?
?
? ?? + + ?? ?
? ?? ?+ ? + ?
? ?? ?? ?
c oL k U k?+ ? < ? A C??
d o oU k U k?? ? < + ( )( ) ( )
( )22
1
2
2
U o U
o
A R U k A R
Ck U k
?
?
? ?+ ? + ? ?? ?
? ?? ?? ?
? ?? ?? ?
e oU k ?+ ? UR?
Expected Net Profit equations are calculated below and an attempt is made to find
through differentiation the optimum set point for ? , o? , to maximize expected net profit:
Case 1 (a) k L k?? ? ? :
( ) ( )k LkE P x R f x dx?? +?? ? = ?? ? ? ( ) ( ){ }12 L LR k k Rk ? ?? ?= ? + ? ? = ?? ?
So, there are no critical points in( ),k L k? .
27
Case 1 (b) L k L k?? ? < + :
( ) ( ) ( )( )L kLk LE P x R f x dx A Cx f x dx?? +?? ? = ? + ?? ? ? ?
( ) ( ) ( )2 212 2L CR L k A k L k Lk ? ? ?? ?? ?? ? ? ?= ? ? ? + + ? ? + ?? ?? ? ? ? ? ?? ?
( ) ( ) ( ) ( )2 212 2L L CR k A k R A L k Lk ? ? ?? ?? ?= ? + + ? + ? + ?? ?? ?? ?
Now, in order to find the optimum value of ?, the equation for expected net profit is
differentiated with respect to ?:
( ) ( ) ( )12 LE P x R A C kk ??? ? ? ? ?= + ? +? ? ? ??
The second derivative with respect to ?, ( )
2
2 2
CE P x
k?
? ? ? = ?
? ?? , is less than zero, so
setting the first derivative equal to zero will result in a maximum:
( ) ( )1 02 LR A C kk ?? ?+ ? + = ?? ? Lo LR A k U kC? += ? = ?
But, L k L k?? ? ? + , and 2 ok U L? ? is assumed in Case 1. Since
( )min ,o U LU U U= , L oU U? , so L oU k U k L k? ? ? > + , and o LU k? ? ? in this
range.
Case 1 (c) oL k U k?+ ? < ? :
( ) ( )( )kkE P x A Cx f x dx?? +?? ? = ?? ? ?
( ) ( ) ( ) ( )2 212 2CA k k k kk ? ? ? ?? ?? ?? ?= + ? ? ? + ? ?? ?? ? ? ?? ?
28
A C?= ?
So, there are no critical points in( ), oL k U k+ ? .
Case 1 (d) o oU k U k?? ? < + :
( ) ( ) ( )( )o
o
U k
Uk UE P x A Cx f x dx R f x dx
?
?
+
?
? ? = ? + ?? ? ? ?
( ) ( ) ( )2212 2o o U oCA U k U k R k Uk ? ? ?? ?? ?? ? ? ?= ? ? ? ? ? ? + ?? ?? ? ? ?? ?? ?
( )( ) ( ) ( )2212 2U o U oCA R U k A R U kk ? ?? ?? ?= + ? + ? ? ? ?? ?? ?? ?
Now, differentiating with respect to ?,
( ) ( ) ( )12 UE P x C k R Ak ??? ? ? ? ?= ? ? +? ? ? ??
Here, the second derivative with respect to ? is 02Ck > , so setting the first derivative
equal to zero will result in a minimum, not a maximum net profit.
Case 1 (e) oU k ?+ ? :
( ) ( )k UkE P x R f x dx?? +?? ? = ?? ? ?
( ) ( ){ }12 U UR k k Rk ? ?? ?= ? + ? ? = ?? ?
So, there are no critical points when oU k? > ? .
29
In each case, differentiation does not lead to a solution for o? , so the extreme
points for each interval are calculated to determine the optimum value of , o? ? , to
maximize ( )E P x? ?? ? .
Case 1(a): At L k? = ? , ( ) LE P x R? ? = ?? ?
Case 1(b): At L k? = ? , ( ) LE P x R? ? = ?? ? ,
At L k? = + , ( ) ( )E P x A C L k? ? = ? +? ?
Case 1(c): At L k? = + , ( ) ( )E P x A C L k? ? = ? +? ? ,
At oU k? = ? , ( ) ( )oE P x A C U k? ? = ? ?? ?
Case 1(d): At oU k? = ? , ( ) ( )oE P x A C U k? ? = ? ?? ? ,
At oU k? = + , ( ) UE P x R? ? = ?? ?
Case 1(e): At oU k? = + , ( ) UE P x R? ? = ?? ?
Since 2 ,o ok U L L k U k? ? + ? ? , so ( ) ( )oA C L k A C U k? + > ? ? . Because
oU is calculated using Equation (7), LL k U+ < , so ( ) LA C L k R? + > ? and
UL k U+ < , so ( ) UA C L k R? + > ? . Therefore, the optimum value of ? when
2 ok U L? ? , is:
o L k? = + (8)
and
( ) ( )oE P x A C L k? ? = ? +? ? (9)
30
Example 1 ( 2 ok U L? ? .)
Let A = $40, C = $0.10, RL =$5, RU = $6, k = 50, and L = 200. oU can be calculated using
Equation (7) to be oU = 450. Using Equation (8), ?o = 250. Various values for ? are
presented in Figure 9 with the corresponding expected net profit, ( )E P x? ?? ? , to show that
?o =250 does, in fact, give the highest expected net profit (at $15):
FIGURE 9. Example of Expected Net Profit for Different Values of ? when
2 ok U L? ? .
Uniform Distribution - Fixed Cost Profit Model
2k<=U-L, RL=5, RU=6
A=40, C=0.10, k=50, L=100, Uo=450
-10
-5
0
5
10
15
20
50 100 150 200 250 300 350 400 450 500 550 600
Value of Mu
E[
P(
x)]
31
Case 2: 2 ok U L> ?
As in Case 1, for Case 2 where 2 ok U L> ? , there are five different scenarios. The
ranges of ? change since 2 ok U L> ? , but the expected net profit equations remain the
same as in Case 1 with the exception of Case 2(c). Case 2(c) differs from Case 1(c) in
that with 2 ok U L> ? , fill level will fall below L and/or above oU as illustrated below:
c) k L? ? ? and ok U? + ?
FIGURE 10. Distribution of fill level when 2 ok U L> ? , k L? ? ? and ok U? + ? .
These scenarios and the formulas for expected net profit are summarized in Table 3:
L
k? ? k? +
oU
( )f x
x
32
TABLE 3. Case 2: Expected Net Profit for Ranges of ? when 2 ok U L> ? .
Scenario Range of ? Expected Net Profit, ( )E P x? ?? ?
a k L k?? ? ? LR?
b oL k U k?? ? < ? ( ) ( )
( ) ( )2 2
1
2
2
L
L
R k A k
Ck R A L k L
? ?
?
? ?? + + ?? ?
? ?? ?+ ? + ?
? ?? ?? ?
c oU k L k?? ? < + ( ) ( )
( ) ( )2 2
1
2
2
L o
o U o
R k L A U L
Ck U L R k U
?
?
? ?? ? + ? ?
? ?
? ?? ? + ?
? ?? ?
d oL k U k?+ ? < + ( )( ) ( )
( )22
1
2
2
U o U
o
A R U k A R
Ck U k
?
?
? ?+ ? + ? ?? ?
? ?? ?? ?
? ?? ?? ?
e oU k ?+ ? UR?
Expected Net Profit equations were calculated the same as with Case 1 with the exception
of 2(b) and 2(c).
Case 2 (b) oL k U k?? ? < ? :
As in Case 1, taking the first derivative with respect to ? of the equation for expected net
profit in the range L k L k?? ? < + , leads to L LR A k U kC? += ? = ? . However, when
2 ok U L> ? , L LR A k U kC? += ? = ? is a feasible value for o? .
33
Case 2 (c) oU k L k?? ? < + :
( ) ( ) ( ) ( )( )o
o
L U k
L Uk L UE P x R f x dx A Cx f x dx R f x dx
?
?
+
?
? ? = ? + ? + ?? ? ? ? ?
( ) [ ] ( )2 212 2L o o U oCR L k A U L U L R k Uk ? ?? ?? ?? ? ? ?= ? ? ? + ? ? ? ? + ?? ?? ? ? ?? ?? ?
( ) ( ) ( ) ( )2 212 2L o o U oCR k L A U L U L R k Uk ? ?? ?= ? ? + ? ? ? ? + ?? ?? ?
Now, in order to find the optimum value of ?, the equation for expected net profit is
differentiated with respect to ?:
( ) ( )12 L UE P x R Rk?? ? ? = ?? ??
which leads to no closed form solution for ? .
With the exception of Case 2(b), differentiation does not lead to a solution for
o? , so the extreme points for each interval are calculated to determine the optimum set
point for , o? ? .
Case 2(a): At L k? = ? , ( ) LE P x R? ? = ?? ?
Case 2(b): At L k? = ? , ( ) LE P x R? ? = ?? ? ,
At oU k? = ? ,
( ) ( ) ( ) ( )2 21 22 2L o o oCE P x R U L k A U L U Lk ? ?? ? = ? ? + ? ? ?? ? ? ?? ?
Case 2(c): At oU k? = ? ,
( ) ( ) ( ) ( )2 21 22 2L o o oCE P x R U L k A U L U Lk ? ?? ? = ? ? + ? ? ?? ? ? ?? ? ,
34
At L k? = + ,
( ) ( ) ( ) ( )2 21 22 2U o o oCE P x R U L k A U L U Lk ? ?? ? = ? ? + ? ? ?? ? ? ?? ?
Case 2(d): At L k? = + ,
( ) ( ) ( ) ( )2 21 22 2U o o oCE P x R U L k A U L U Lk ? ?? ? = ? ? + ? ? ?? ? ? ?? ? ,
At oU k? = + , ( ) UE P x R? ? = ?? ?
Case 2(e): At oU k? = + , ( ) UE P x R? ? = ?? ?
Clearly, the maximum expected net profit occurs either at LU k? = ? , from the
differentiation in Case 2(b), at L k? = + (from Case 2(c)) where
( ) ( ) ( )2 21[ ( )] 22 2L o o oCE P x R U L k A U L U Lk ? ?= ? ? + ? ? ?? ?? ? , or at oU k? = ? (from
Case 2(c)) where ( ) ( ) ( ) ( )2 21 22 2U o o oCE P x R U L k A U L U Lk ? ?? ? = ? ? + ? ? ?? ? ? ?? ? . Which
value of ? provides a higher expected net profit depends on the relationship between LR
and UR . If L UR R> , then the expected net profit when L k? = + is
( ) ( ) ( )2 21 22 2L o o oCR U L k A U L U Lk ? ?? ? + ? ? ?? ?? ? which is greater than the expected net
profit when oU k? = ? , or ( ) ( ) ( )2 21 22 2U o o oCR U L k A U L U Lk ? ?? ? + ? ? ?? ?
? ?
. This is
consistent with Case 2(b), because when U LR R> , o LU U= , so o LU k U k? = ? = ? .
Therefore, if L UR R> ,
o L k? = + (10)
35
and
( ) ( ) ( ) ( )2 21 22 2o L o o oCE P x R U L k A U L U Lk ? ?? ? = ? ? + ? ? ?? ? ? ?? ? (11)
If U LR R> , then
o oU k? = ? (12)
and
( ) ( ) ( ) ( )2 21 22 2o U o o oCE P x R U L k A U L U Lk ? ?? ? = ? ? + ? ? ?? ? ? ?? ? (13)
In the case where U LR R R= = , Equations (11) and (13) are equal and the optimum set
point for ? is any value in the range
[ ],o oU k L k? ? ? + (14)
and
( ) ( ) ( ) ( )2 21 22 2o o o oCE P x R U L k A U L U Lk ? ?? ? = ? ? + ? ? ?? ? ? ?? ? (15)
If o oU k? = ? , then 2oa U k L= ? < and ( )o ob U k k U= ? + = where there are no rejects
on the high side, but the proportion rejected on the low side is ( )22oL L U kp k? += .
Further, as oU L? approaches 2k , 0Lp ? . On the other hand, if o L k? = + , then
a L= and 2 ob L k U= + > where there are no rejects on the low side, but the proportion
rejected on the high side is ( )22 oU L k Up k+ ?= . Again, as oU L? approaches 2k ,
0Up ? . Note that, as expected, L Up p= .
36
Example 2 ( 2 ok U L> ? and L UR R> .)
Let A = $40, C = $0.10, RL = $6, RU = $5, k = 150, and L = 200. oU is calculated using
Equation (7) to be oU = 450. Using Equation (10), ?o = 350. Various values for ? are
presented in Figure 11 with the corresponding expected net profit, ( )E P x? ?? ? , to show that
calculating ?o by Equation (10) does, in fact, give the maximum expected net profit (at
$5.42):
FIGURE 11. Example of Expected Net Profit for Different Values of ? when
2 ok U L> ? and U LR R< .
Uniform Distribution - Fixed Cost Profit Model
2k>U-L, RU=5 ? and U LR R> .)
Let A = $40, C = $0.10, RL = $5, RU = $6, k = 150, and L = 200. oU can be calculated
using Equation (7) to be oU = 450. Using Equation (12), ?o = 300. Various values for ?
are presented in Figure 11 with the corresponding expected net profit, ( )E P x? ?? ? , to show
that ?o =300 does, in fact, give the highest expected net profit (at $5.42.)
FIGURE 12. Example of Expected Net Profit for Different Values of ? when
2 ok U L> ? and U LR R> .
Example 4 ( 2 ok U L> ? and L UR R R= = .)
Let A = $40, C = $0.10, RL = RU = $5, k = 150, and L = 200. Uo can be calculated using
Equation (7) to be Uo = 450. Using Equation (14), [ ]300,350o? ? . Various values for ?
Uniform Distribution - Fixed Cost Profit Model
2k>U-L, RU=6>RL=5
A=40, C=0.10, k=150, L=200, Uo=450
-10
-5
0
5
10
50 100 150 200 250 300 350 400 450 500 550 600
Value of Mu
E[
P(
x)]
38
are presented in Figure 13 with the corresponding expected net profit, ( )E P x? ?? ? , to show
that [ ]300,350o? ? does, in fact, give the highest expected net profit (at $5.42.)
FIGURE 13. Example of Expected Net Profit for Different Values of ? when
2 ok U L> ? and L UR R R= = .
To summarize, for fill level that follows a Uniform distribution when there is a
constant scrap cost, the optimum value for the upper screening limit was determined to be
( )min , min , ULo L U R AR AU U U C C++? ?= = ? ?? ?
The optimum target set point for the process mean was obtained for the various scenarios:
Case 1: 2 o ok U L L k?? ? ? = +
Case 2: 2 ok U L> ? :
Uniform Distribution - Fixed Cost Profit Model
2k>U-L, RU=RL=5
A=40, C=0.10, k=150, L=200, Uo=450
-6
-4
-2
0
2
4
6
50 100 150 200 250 300 350 400 450 500 550 600
Value of Mu
E[
P(
x)]
39
If U L oR R L k?< ? = +
If U L o oR R U k?> ? = ?
If [ ],U L o oR R U k L k?= ? ? ? +
3.2 Uniform underlying distribution ? linear scrap cost
In this section, the optimum upper screening limit and optimum target mean are
obtained for a different net profit model: one with a linear scrap cost. This net profit
model is appropriate when the majority of the scrap/rework/reprocessing cost is from the
cost of the material, for example in the gauge of steel beams. Using a model similar to
that used by Dodson (1993), the net profit model is:
,( )
,
Rx x LP x
A Cx x L
? , which means the per unit cost to
produce must be greater than the incremental cost to reprocess in order to calculate an
optimum upper screening limit. If UC R< , no upper screening limit is necessary,
because net profit will be higher to just produce all product with fill level greater than L,
sell at A and absorb the ?give-away? cost. However, UC R> implies that it costs more
to manufacture the part than to manufacture and scrap or rework it. Therefore, an upper
screening limit is not appropriate for this model. The remainder of this section will
assume a fixed value of U that is set by customer requirements or defined by limitations
of the container or the production equipment and that U is determined such that
( ) 0A C U? ? .
42
3.2.2 Optimum target set point for mean
To determine an optimum target mean, o? , the two cases (Case 1: 2k U L? ? and
Case 2: 2k U L> ? ) and the same five ranges of ? that were presented in Section 3.1.2
also apply for the linear model as presented in Tables 4 and 5.
Table 4 lists the formulas for expected net profit for different ranges of ? for
Case 1: 2k U L? ? :
TABLE 4. Case 1: Expected Net Profit for Ranges of ? when 2k U L? ? .
Scenario Range of ? Expected Net Profit, ( )E P x? ?? ?
a k L k?? ? ? LR ??
b L k L k?? ? < + ( ) ( )
( )
22
2 2
1 2
2
2
LR L k A k L
k C k L
? ?
?
?? ?? ?? ? + + ? ?
? ?? ?? ?
? ?
? ?? ?+ ?? ?
? ?? ?
c L k U k?+ ? < ? A C??
d U k U k?? ? < + ( ) ( )
( )
22
2 2
1 2
2
2
U
CA U k U k
Rk k U
? ?
?
? ?? ?? ?? ? ? ? ? ?
? ?? ?? ?? ?
? ?
? ?? ?+ ?? ?
? ?? ?
e U k ?+ ? UR ??
Expected Net Profit equations are calculated below and an attempt is made to find
through differentiation the optimum set point for ? , o? , to maximize expected net profit:
43
Case 1 (a) k L k?? ? ? :
( ) ( )k LkE P x R xf x dx?? +?? ? = ?? ? ? ( ) ( )2 212 2 L LR k k Rk ? ? ??? ?? ?= + ? ? = ?? ?? ?? ?
So, there are no critical points in( ),k L k+ .
Case 1 (b) L k L k?? ? < + :
( ) ( ) ( )( )L kLk LE P x R xf x dx A Cx f x dx?? +?? ? = ? + ?? ? ? ?
( ) ( ) ( )2 22 212 2 2LR CL k A k L k Lk ? ? ??? ?? ? ? ?? ?= ? ? + + ? ? + ?? ?? ?? ? ? ?? ?
Now, in order to find the optimum value of ?, the equation for expected net profit is
differentiated with respect to ?:
( ) ( ) ( )12 LE P x R k A C kk ? ??? ? ? ? ?= ? + ? +? ? ? ??
The second derivative with respect to ?, ( ) ( )
2
2
1 0
2 LE P x R Ck?
? ? ? = ? <
? ?? iff LC R> .
So, setting the first derivative equal to zero will result in a maximum when LC R> :
( ) ( )1 02 L
L
AR k A C k k
k R C? ? ?? ?? + ? + = ? = +? ? ? . But, when LC R> ,
L
A k k
R C? = + , so setting
the first derivative equal to zero will result in a maximum expected net profit if UR C> :
( ) ( ) ( ) ( )1 02 UU
U
A C R kE P x A C k R k
k C R? ? ??
+ +? ? ? ? ?= ? + ? ? + = ? =
? ? ? ?? ? . But this
value of ? leads to a maximum expected net profit in this range only if UR C> . And, if
UR C> , then
( ) 0U
U
A C R k
C R?
+ += <
? which is not feasible.
Case 1 (e) U k ?+ ? :
45
( ) ( )k UkE P x R xf x dx?? +?? ? = ?? ? ? ( ) ( )2 212 2U UR k k Rk ? ? ?? ?? ?= ? + ? ? = ?? ?? ?? ?
So, there are no critical points when U k? > ? .
In each case, differentiation does not lead to a solution for o? , so the extreme
points for each interval are calculated to determine the optimum value of , o? ? .
Case 1(a): At L k? = ? , ( ) ( ) 0LE P x R L k? ? = ? ? ? ? and clearly,
( ) ( )LA C L k R L k? + > ? ? and ( ) ( )UA C L k R U k? + > ? + , so the maximum expected
46
net profit occurs at L k? = + with ( ) ( )E P x A C L k? ? = ? +? ? . Therefore, the optimum
value of ? when 2 ok U L? ? , is:
o L k? = + (18)
and
( ) ( )E P x A C L k? ? = ? +? ? (19)
This conclusion is illustrated in Figure 14 below.
Example 5 ( 2k U L? ? .)
Let A = $40, C = $0.10, RL = $0.20, RU = $0.05, k = 50, L = 200, and / 400U A C= = .
Using Equation (18), o L k? = + = 250. Various values for ? are presented in Figure 14
with the corresponding expected net profit, ( )E P x? ?? ? to show that o L k? = + = 250 does,
in fact, give the highest expected net profit (at ( )oE P x? ?? ? =$15.)
47
FIGURE 14. Example of Expected Net Profit for Different Values of ? when
2k U L? ? .
Case 2: 2k U L> ?
If 2k U L> ? , the fill level distribution is illustrated by Figure 15:
( )f x
x
k? ? L ? k? + U
Fill Level
FIGURE 15. Distribution of Fill Level when 2k > U - L .
Uniform Distribution - Linear Cost Profit Model
2k<=U-L, RL=0.2, RU=0.05
A=40, C=0.10, k=50, L=200, U=400
-40
-30
-20
-10
0
10
20
50 100 150 200 250 300 350 400 450 500 550 600
Value of Mu
E[
P(
x)]
48
The optimum set point for the mean, o? , to maximize net profit will depend on the
relationship between the values of , , ,andL UR R C A as illustrated in Figure 16.
FIGURE 16. Net Profit Equation when U LR R>
If UR C> , it is clear that there is no optimum value for oU , as discussed in
Section 3.2.1. Even if UR C< , such that the optimum upper screening limit is defined by
Equation (17), the fact that the cost is a linear function of fill level, means that it would
be unlikely that a business would choose to allow fill level to run that high, since the cost
is so great. More likely, a fixed upper limit is given.
Table 5 lists the formulas for expected net profit for different ranges of ? when
2k U L> ? . Note that (as in Case 2 of the fixed cost model in Section 3.1.2) the ranges
of ? are different from Case 1 because 2k U L> ? .
LR x?
A Cx?
L
( )P x
x
UR x?
U
49
TABLE 5. Case 2: Expected Net Profit for Ranges of ? when 2k U L> ? .
Scenario Range of ? Expected Net Profit, ( )E P x? ?? ?
a L k? ? ? LR ??
b L k U k?? ? < ? ( ) ( )
( )
22
2 2
1 2
2
2
LR L k A k L
k C k L
? ?
?
?? ?? ?? ? + + ? ?
? ?? ?? ?
? ?
? ?? ?+ ?? ?
? ?? ?
c U k L k?? ? < + ( ) ( )
( ) ( )
2 2
22 2 2
1 2
2
2 2
L
U
R k L A U L
Rk C L U U k
?
?
? ?? ?? ? + ? +
? ?? ?? ?
? ?
? ?? ?? + ? +? ?
? ?? ?
d L k U k?+ ? < + ( ) ( )
( )
22
2 2
1 2
2
2
U
CA U k U k
Rk k U
? ?
?
? ?? ?? ?? ? ? ? ? ?
? ?? ?? ?? ?
? ?
? ?? ?+ ?? ?
? ?? ?
e U k ?+ ? UR ??
Expected Net Profit equations are calculated as in Case 1 with the exception of Case 2(c)
which is presented below:
Case 2 (c):
( ) ( ) ( ) ( )( )L U kL Uk L UE P x R xf x dx A Cx f x dx R xf x dx?? +?? ? = ? + ? + ?? ? ? ? ?
( ) ( ) ( ) ( )2 22 2 2 212 2 2 2UL RR Ck L A U L L U U kk ? ?? ?? ? ? ?= ? ? + ? + ? + ? +? ?? ? ? ?? ?
Now, differentiating with respect to ?,
50
( ) ( ) ( )12 L UE P x R k R kk ? ??? ? ? ? ?= ? ? +? ? ? ??
Here, the second derivative with respect to ? is ( )1 02 L UR Rk ? < iff U LR R> , so setting
the first derivative equal to zero will result in a maximum expected net profit if U LR R> :
( ) ( ) ( ) ( )1 02 L UL U
L U
k R RE P x R k R k
k R R? ? ??
+? ? ? ? ?= ? ? + = ? =
? ? ? ?? ? . But this value of
? leads to a maximum expected net profit in this range only if U LR R> . And, if
U LR R> , then
( ) 0L U
L U
k R R
R R?
+= <
? which is not feasible.
In each range of ? , differentiation does not lead to a solution for o? , so the
extreme points for each interval are calculated to determine the optimum value of , o? ? .
Case 2(a): At L k? = ? , ( ) ( ) 0LE P x R L k? ? = ? ? + ? ?
? ? ? ?
? ? ? ?
Otherwise, o U k? = ? .
Simplifying,
( ) ( )( ) ( ) ( ) ( )( ) ( )
2 2 2 2
U L
U L
U L R C U L U L R C U LA R L k A R U k
k k
? ? ? ?? ? + ? ? ++ ? + > + ? ? ?
? ? ? ?
? ? ? ?
( ) ( ) ( ) ( )2 2 2 2
4 4U L
U L U LR L k R U k
k k
? ? ? ?? ?
? ? ? ?? + > ? ?
? ? ? ?? ? ? ?
Therefore, the optimum value of ? when 2k U L> ? is:
o L k? = + if
( ) ( ) ( ) ( )2 2 2 2
4 4U L
U L U LR L k R U k
k k
? ? ? ?? ?
? ? ? ?? + > ? ?
? ? ? ?? ? ? ?
(20)
with ( ) ( ) ( )( ) ( )2 2Uo UU L R C U LE P x A R L kk ? ?? ? +? ? = + ? +? ?? ?
? ?
(21)
and
o U k? = ? if
( ) ( ) ( ) ( )2 2 2 2
4 4U L
U L U LR L k R U k
k k
? ? ? ?? ?
? ? ? ?? + ? ? ?
? ? ? ?? ? ? ?
(22)
with ( ) ( ) ( )( ) ( )2 2Lo LU L R C U LE P x A R U kk ? ?? ? +? ? = + ? ?? ?? ?
? ?
(23)
53
The following examples illustrate these conclusions:
Example 6: 2k U L> ? and ( ) ( ) ( ) ( )
2 2 2 2
4 4U L
U L U LR L k R U k
k k
? ? ? ?? ?
? ? ? ?? + > ? ?
? ? ? ?? ? ? ?
Let A = $40, C = $0.10, RL = $0.40, RU = $0.10, k = 75, and L = 100, U=200. In this case,
( ) ( ) ( ) ( )2 2 2 27.5 10.5
4 4U L
U L U LR L k R U k
k k
? ? ? ?? ?
? ? ? ?? + = ? > ? ? = ?
? ? ? ?? ? ? ?
, so based on
Equations (20 and 21), 175o L k? = + = with an expected net profit of $9.17. Various
values for ? are presented in Figure 17 with the corresponding expected net profit to
show that o L k? = + does, in fact, provide the maximum expected net profit.
FIGURE 17. Example of Expected Net Profit for Different Values of ? when 2k > U ? L,
o L k? = + .
Uniform Distribution - Fixed Cost Profit Model
2k>U-L, RU=0.1 ? and ( ) ( ) ( ) ( )
2 2 2 2
4 4U L
U L U LR L k R U k
k k
? ? ? ?? ?
? ? ? ?? + ? ? ?
? ? ? ?? ? ? ?
Let A = $40, C = $0.10, RL = $0.10, RU = $0.30, k = 75, L = 100, and U=200. In this
case, ( ) ( ) ( ) ( )
2 2 2 2
22.5 2.54 4U LU L U LR L k R U kk k
? ? ? ?? ?
? ? ? ?? + = ? ? ? ? = ?
? ? ? ?? ? ? ?
, so based on
Equations (22 and 23), 125o U k? = ? = with an expected net profit of $14.67. Various
values for ? are presented in Figure 18 with the corresponding expected net profit to
show that o U k? = ? does, in fact, provide the maximum expected net profit.
FIGURE 18. Example of Expected Net Profit for Different Values of ? when 2k > U ? L
and o U k? = ? .
Uniform Distribution - Fixed Cost Profit Model
2k>U-L, RU=0.3>RL=0.1
A=40, C=0.10, k=75, L=100, U=200
-60
-40
-20
0
20
10 30 50 70 90 110 130 150 170 190 210 230
Value of Mu
E[
P(
x)]
55
To summarize, for fill level that follows a Uniform distribution when there is a linear
scrap cost, the optimum target set point for the process mean was obtained for the various
scenarios:
Case 1: 2k U L? ?
o L k? = +
Case 2: 2k U L> ?
o L k? = + if
( ) ( ) ( ) ( )2 2 2 2
4 4U L
U L U LR L k R U k
k k
? ? ? ?? ?
? ? ? ?? + > ? ?
? ? ? ?? ? ? ?
o U k? = ? if
( ) ( ) ( ) ( )2 2 2 2
4 4U L
U L U LR L k R U k
k k
? ? ? ?? ?
? ? ? ?? + ? ? ?
? ? ? ?? ? ? ?
3.3 Fixed scrap cost with capacity constraint
In practice, there may be some maximum capacity, CAP, such that if fill level
exceeds CAP, an additional (usually very high) cost is incurred. This cost may be due to
spillage over the maximum capacity of the container, for example, which results in clean
up, downtime, restarting the equipment, etc.
The additional cost results in a new net profit equation:
( ) ( )min ,
L
U
R x L
A Cx L x U CAPP x
R U x CAP
Q x CAP
? ?
56
When the fill level is described by a Uniform distribution, the calculation of oU
as shown below is the same as that found in Section 3.1.1 where Equation (7) is
min ,U Lo R A R AU C C+ +? ?= ? ?
? ?
.
If oU CAP? , 2 ok U L> ? , and L U?? ? the expected net profit can be calculated as
follows:
( ) ( ) ( ) ( ) ( ) ( )
kL U CAP
L U
k L U CAP
E P x R f x dx A Cx f x dx R f x dx Q f x dx
?
?
+
?
? ? = ? + ? + ? + ?? ? ? ? ? ?
( ) ( ) ( ) ( ) ( )
2 21
2 2L U
C U LR k L A U L R U CAP Q CAP k
k ? ?
? ??
? ?= ? ? + ? ? + ? + ? ?
? ?? ?
( ) ( )
( )
21
2
2
U L U
L
CUU R A R k R CAP
k CLL A R Q CAP k
?
?
? ?? ?+ ? + ? ? +
? ?? ?? ?
? ?=
? ?? ?? ? + ? ?
? ?? ?? ?? ?
(24)
Since the second derivative with respect to U is 2Ck? which is < 0, setting the
first derivative of Equation (24) with respect to U equal to zero will provide an optimum
value of the upper limit, Uo:
( )1 02 UU o R AR A CU Uk C++ ? = ? =
Which is the same as Equation (3).
57
Differentiating Equation (24) with respect to ?, ( ){ } [ ]12 LE P x R Qk?? ? ? = ?? ??
which results in no closed form solution. The impact on the optimum set point for the
mean, o? , of a capacity constraint is to add to all of the scenarios in Section 3.1.2 the
condition that if CAP k? is less than the optimum value calculated for that scenario, then
o? = CAP k? . Specifically, the target set point for the process mean defined for the
various scenarios:
Case 1: ( )2 min , o ok CAP L U L L k?< ? ? ? = +
Case 2: ( )2 min , ok CAP L U L? ? ?
( )min ,U L oR R L k CAP k?< ? = + ?
( )min ,U L o oR R U k CAP k?> ? = ? ?
( ),min ,U L o oR R U k L k CAP k? ? ?= ? ? ? + ?? ?
When oCAP U< , the effect is that CAP replaces oU in equations for expected net profit
with the higher cost, Q, replacing UR . An example follows:
Example 8: oCAP U< and 2k CAP L> ?
Let A = $40, C = $0.10, $6LR = , 5UR = , CAP=400, Q=$500, k = 175, and L =
100. In this case, min , 450U Lo R A R AU C C+ +? ?= =? ?
? ?
, so oCAP U< , 2k CAP L> ? , and
L UR R> , so ( ) ( )min , min 275, 225 225o L k CAP k? = + ? = = . Various values for ?
58
are presented in Figure 19 with the corresponding expected net profit to show that
225o? = , does in fact lead to the maximum expected net profit at $12.
Uniform Distribution - Fixed Cost Profit Model with Capacity
2k >= CAP-L, CAP=400 < Uo=450
A=40, C=.1, RL=6, RU=5,k=175, L=100, CAP=400,Q=500
-120
-100
-80
-60
-40
-20
0
20
18
0
18
5
19
0
19
5
20
0
20
5
21
0
21
5
22
0
22
5
23
0
23
5
24
0
24
5
25
0
25
5
26
0
26
5
27
0
27
5
28
0
28
5
29
0
29
5
30
0
Value of Mu
E[
P(
x)]
FIGURE 19. Example of Expected Net Profit for Different Values of ? when oCAP U<
and 2k > CAP ? L.
59
4.0 Relationship between canning problem and process capability
The optimum target mean and upper screening limit for the canning problem is
determined to maximize expected net profit, and net profit is increased as the rejection
rate is decreased. The selection of o? to maximize expected net profit is consistent with
the goal of improving process capability. In this chapter, the relationship between the
canning problem and process capability is explored.
The term ?process capability? refers to the likelihood that a process meets
customer requirements or specifications as measured by a ratio of those requirements to
process variation. Before process capability can be determined for a process, the process
must be shown (e.g. using a control chart) to be stable over time. This means that only
common causes of variation are present and the process parameters can accurately be
estimated from empirical (or sample) data.
4.1 Relationship between process capability and the selection of oU :
There is no relationship between process capability and the selection of oU , since
oU depends only on the values of A, LR , UR , and C and not on process variation. There
is, however, a relationship between process capability and the selection of o? which is
presented in Section 4.2.
60
4.2 Relationship between process capability and the selection of o? :
Specific measures of process capability include ,PC ,PKC ,PLC ,PUC and PMC . Each
measure will be defined and discussed in this section as it relates to the canning problem.
4.2.1 Process Capability Measure, PC
PC is the most general capability index which compares the width of the
specifications to the spread of the distribution (i.e. the natural tolerance of the quality
characteristic, fill level.) It is generally calculated as the ratio between the blueprint (BT)
and natural tolerances (NT) of the process: P BT USL LSLC NT UNL LNL?= = ? where USL = upper
specification limit, LSL = lower specification limit, UNL = upper natural tolerance limit
of the distribution and LNL = lower natural tolerance limit of the distribution. The
natural tolerance is defined as the middle 99.73% of the distribution, so for a normal
distribution PC is calculated as 6P USL LSLC ??= . For a Uniform distribution,
( ).99865 .00135 .9973 2P
USL LSL USL LSLC
x x k
? ?= =
? . Since, for a Uniform distribution,
2
12 12
b a k? ?= = ,
2 2 3k ?= , and thus ( )
.9973 2 3P
USL LSLC
?
?= . The typically used definition of capability is
that a process (that is in-control) is ?potentially capable? if the width of the natural
tolerance is smaller than the width of the specifications. This corresponds to 1PC ? .
61
Since PC is independent of the position of ?, it is only an effective measure of
process capability if the target and the process mean are both centered inside the
specification, which is not the case in the canning problem. In cases where the mean is
not centered (as in the canning problem), PKC may be a better estimate. PKC will be
addressed in Section 4.2.2
Clearly, for any distribution, as variance decreases, the width of the natural
tolerances of the distribution decreases, and the value of PC increases. It has been shown
that in the canning problem, as variance decreases, net profit increases. This means that
as process capability is improved (and the value of PC increases), variability decreases,
and the mean can be set closer to L, so that ( )oE P x? ?? ? increases.
With the canning problem, a lower specification limit, L, is defined by the
customer, but (in the case of the fixed cost model) no upper specification is given, so PC
is not an appropriate measure of process capability. PC can only be calculated when an
upper specification exists (as in the case of the linear cost or capacity constrained
models), or when PC is calculated using oU as the upper specification limit. But even
when an upper specification limit is used, the target for the canning process is not the
center of the specifications. In cases of a one-sided specification, PKC is a better measure
and is discussed in the next section.
62
4.2.2 Process Capability Measure, PLC
In the canning problem, only a lower specification is given, so the PLC index can
be calculated. For a process that can be approximated with a normal distribution,
( )
3PL
LC ?
?
?= . In the case of the Uniform distribution, ( ) ( )
( ).00135 .49865 2PL
L LC
x k
? ?
?
? ?= =
? .
1PLC ? means that ? 0.135% of the distribution of the specific quality characteristic is
expected to fall below the lower specification.
Since process capability is inversely related to process standard deviation, it
relates directly to the canning problem. As the standard deviation decreases, PLC
increases, and the process mean can be moved closer to L. As the mean is moved closer
to L, cost decreases and net profit increases.
When a process has an upper specification, PUC can be calculated in a similar
fashion. In the case of a process with both an upper and lower specification limit,
( )min ,PK PL PUC C C= . Since the canning problem has only a lower specification, clearly
these measures of process capability do not apply.
4.2.3 Process Capability Measure, PMC
PKC , PLC , and PUC all assume the process target is centered inside the
specifications. PMC is a measure of process capability that considers deviation from
target rather than the center of the specifications. Since the canning problem has a target
not centered inside specification, it may be a better measure of process capability. In the
special case of the canning problem, the target is actually the lower specification limit if
63
there is no variation. Fill level below L is rejected, while fill level above L is accepted,
but net profit is highest when fill level is equal to L. Given that fill level has variation,
the target is actually some value above L, specifically o? . Once o? is determined, PMC
could be calculated as a measure of process capability. However,
( )2
21
P
PM
CC
T?
?
=
?+
where T = the target. So, to calculate PMC , one must first calculate PC which requires
both lower and upper specification limits.
4.3 Summary of Relationship between Canning Problem and Process Capability
Any measure of process capability compares the process variation to
specifications or requirements. Higher levels of process capability result when process
variation decreases. In the canning problem, this means that as process capability is
improved, variability decreases, and the mean can be set closer to L, so that ( )oE P x? ?? ?
increases.
At least one article has addressed the relationship between process capability and
the canning problem. Kim, Cho, and Phillips (2000) show an economic model to
determine the optimum process mean while maintaining a specific PC value when fill
level can be approximated by a Normal distribution. They assume that achieving a
smaller variance leads to an increase in manufacturing costs that they incorporate into the
net profit model. They provide a case study and sensitivity analysis.
64
5.0 Triangular distribution
Another finite distribution that can be a plausible model for a canning process is
the Triangular distribution. In this chapter, the Triangular distribution is studied to
determine an optimal set point for the mean of the production operation. An optimum
upper limit is first calculated which provides a ?cut-off? to maximize expected net profit
by minimizing ?give-away? cost.
Again, the basic profit function of Liu and Raghavachari (1997) article is used, so
that Triangular results can also be compared to their?s which pertained to continuous
distributions.
The Net Profit function:
,
( ) ,
,
L
U
R x L
P x A Cx L x U
R x U
? ??
?= ? ? ??
?? ??
where L = lower specification limit, below which the customer will not accept the
product. For example, if a jar is to be filled with 8 oz. of an ingredient, anything less than
L=8 oz. is not allowed. U = an upper limit (U is either set by the producer as an upper
specification limit, USL, or determined as a value that minimizes giveaway cost such that
any fill level above U costs more in giveaway cost than would be received in income.
65
LR = the rejection cost per container when fill level is less than L, UR = the rejection cost
per container when fill level is greater than U, A = revenue received for an acceptable
container, and C = the production cost per unit of ingredient. A, LR , UR , C, and L are
known and > 0.
A Triangular distribution has the following probability density function:
( )
( )( )
( )
( )( )
2 ,
2( ) ,
0,
x a a x m
b a m a
b xf x m x b
b a b m
otherwise
? ? ? ?
? ? ?
?
?
?
??= ? ?
? ? ?
?
?
?
?
??
where m is the mode and graphically the distribution appears below in Figure 20.
( )f x
a L m U b
FIGURE 20. The triangular probability density function with lower specification, L, and
upper screening limit, U.
66
With respect to the canning problem, there is a cost ( LR ) when the quantity falls
below the lower specification, L (L ? a), and a cost ( UR ) when the fill level falls above
an arbitrary upper specification, U (U < b). Giveaway cost above U is greater than the
cost of simply scrapping the unit ( UR .)
5.1 Symmetric Triangular underlying distribution
First, a symmetric Triangular distribution will be analyzed using the fixed rework
cost model from the previous chapter. Then the impact of skewness will be examined.
Assuming a symmetric Triangular distribution, the probability density function is given
below:
2
2
,
( ) ,
0,
x m k m k x m
k
m k xf x m x m k
k
otherwise
? +? ? ? ?
?
?
?
? + ??
= ? ? +?
?
?
?
?
??
where m is the modal point and 2k represents the spread of the distribution and it can be
verified (in the Appendix) that the ( ) 2
6
kV x = . In this chapter,
om , the optimum set point
for m is computed for the case where 1m k L? < , 2m k U+ > , and L m U? < .
67
5.1.1. Optimum upper screening limit
If there is no upper specification limit given, an optimum upper limit can be
calculated as in previous chapters. The equation for expected net profit for the
symmetrical triangular distribution is found below:
( )
( )
( )
2 2
2 2
L m
L
m k L
U m k
U
m U
x m k x m kR dx A Cx dx
k kE P x
m k x m k xA Cx dx R dx
k k
?
+
? ? + ? +? ? ? ?? + ? +
? ? ? ? ?? ? ? ??
=? ? ?? ?
+ ? + ?? ? ? ?? ? + ? =
? ? ? ?? ? ? ? ??
? ?
? ?
( ) ( )
( ) ( )
2 2 3 2 2
2 2 2
2 2 3
2 2 3 2 21
2 2 2 3 2
L m m
L
Lm k L
U m kU
U
mm U
x m k x m k x mx kxR A C
k m k x m k xmx kx x
A C R
?
+
? ?? ? ? ?? ? ? ?? ? ? ? ? ?? ? ? ?
? ?? ? ? ?? + ? ? + ?? ?
? ?? ? ? ? ? ?? ? ? ?? ?
=? ?
? ? ? ?? ?+ ? + ?? ? ? ?? ?? ? ? ?
? ? ? ?? ?? + ? +? ?
? ? ? ?? ?? ?? ? ? ?? ?
( ) ( ) ( ) ( )
( )
2 2 2 2
2 3 3 3 2 2 2 2 3 3 3 2
1
2 21
3 2 3 2 2 2 2 3 2 3 2
L
U
R L m k A m m L U k U L U L
k Rm m L mL kL mU kU U m mC m k U
? ?? ?? ? ? + ? + + + ? ? + ?? ?
? ?? ? ? ?? ?? ?
=? ?? ?
? ?? ? + ? + + ? ? + ? + ?? ?? ? ? ?
? ?? ?? ?
( )E P x? ?? ? =
( ) ( ) ( ) ( )
( ) ( ) ( ) ( )
2 2
2
2 3 3 3 2 2
2
2 2 21
3 2 2 2
L
U
m L m UR L m k A k L U
k m L U RL U
C m k m k m k U
? ?? ?? ?
? ?? ? ? ? + + ? ?? ? ? ?? ?
? ?? ?? ?? ?
? ?? ?+ +
? ?? ?? + ? + + ? + ?? ?
? ?? ?? ?
? ?? ?? ?
(25)
68
Equation (25) assumes that the distribution is such that m k L? < and m k U+ > and
shows that the expected net profit is inversely proportional to the ( )
2
6
kV x = .
Differentiating Equation (25) with respect to U, results in the following:
( ) ( ) ( ) ( ){ }221 UE P x A m k U CU m k CU R m k UU k? = + ? ? + + + + ? =? ? ? ? ? ?? ? ? ? ? ??
( )( ) ( ){ }221 U UA R m k A R C m k U CUk + + ? + + + +? ?? ?
To determine the optimum value of U, oU , the first derivative with respect to U is set
equal to zero. (This will be an optimum value for U if ( )2 2Uo Um kA RU UC ++< + = .)
( ){ } 0E P xU? = ?? ?? ??
( ) ( )( )2 0o U o UCU A R C m k U A R m k? + + + + + + =? ?? ?
Let ( )1 Uc A R C m k= + + +? ?? ? and ( )( )2 Uc A R m k= + + , then, 2 1 2 0o oCU c U c? + = , which means
that
2
1 1 24
2o
c c CcU
C
? ?=
( ) ( ) ( ) ( ) ( )( )2 4
2
U U U
o
A R C m k A R C m k C A R m kU
C
? ? ? ?+ + + ? + + + ? + +? ? ? ?
= =
69
( ) ( ) ( ) ( )( ) ( ) 22 2
2
U U UA R C m k A R C A R m k C m k
C
? ? ? ?+ + + ? + ? + + + +? ? ? ?
=
( ) ( ) ( ) ( ) 2
2
U UA R C m k A R C m k
C
? ? ? ?+ + + ? + ? +? ? ? ?
=
( ) ( ) ( ) ( )
2
U UA R C m k A R C m k
C
? ? ? ?+ + + ? + ? +? ? ? ?
Case 1: ( ) ( ) ( ) ( )2U UA R C m k A R C m kC? ? ? ?+ + + + + ? +? ? ? ? =
[ ]2 2 U UoA R A RUC C+ +? = (26)
This is consistent with the value found for oU in previous sections.
Case 2: ( ) ( ) ( ) ( )2U UA R C m k A R C m k m kC
? ?? ? ? ?+ + + ? + ? +? ? ? ?? ?
= + which is not < U
(since 2k U L> ? .)
So, the optimum value for oU is given in Equation (26).
70
5.1.2 Optimum target set point for m
The objective is to maximize the expected net profit. Since the profit function is
defined as:
,
( ) ,
,
L
o
U o
R x L
P x A Cx L x U
R x U
? ??
?= ? < .
Case 3:
( )f x
L m-k1 m U m+k2
FIGURE 23. The triangular probability density function with lower specification, L, and
upper screening limit, U when 1L m k U? ? ? and 2m k U+ > .
Due to the complexity of the distribution, the only case considered in this work is Case 2,
Figure 22, when L m U? ? , 1m k L? < , and 2m k U+ > .
72
To determine the optimum value of m to maximize the average net profit, the first
derivative of Equation (25) with respect to m is calculated below:
( )E P xm? =? ?? ??
( ) [ ]
( ) ( )2 2 2 2
2 ( )1
1
2
L o
o U o
R L m k A m L U
k C m L U R m k U
? ?? ? ? ? + +? ?? ?
? ?
? ? ? ?? + ? + ?? ?? ? ? ?
? ?? ?? ?
(27)
If the second derivative with respect to m is less than zero, then Equation (27) can be set
to zero to solve for the value of m that maximizes the expected net profit:
( )22 E P xm? =? ?? ?? [ ]21 2 2L UR A Cm Rk ? ? + ?
The above second derivative is less than zero only if 22L UR R Am C+ +< (28)
Setting Equation (27) equal to zero results in a quadratic equation:
( ) ( ) ( ) ( )2 2 221 12 02L o o U oR L m k A m L U C m L U R m k Uk ? ?? ?? ? ? ? + + ? + ? + ? = ?? ? ? ? ? ?? ?? ?? ? ? ? ? ?? ?
? ?
( ) ( ) ( ) ( ) ( )2 2 22 02L U L o U o oCCm R R A m R L k A L U R U k L U? ?? + + + + + + + ? ? + =? ?? ?
Let 3 2L Uc R R A= + +
and ( ) ( ) ( ) ( )2 24 2L o U o oCc R L k A L U R U k L U= + + + + ? ? + ,
then, 2 3 4 0o oCm c m c? + =
73
Thus, the optimum set point for m is given by,
2
3 3 44
2o
c c Ccm
C
? ?=
( )
( )
( ) ( ) ( )
2
22
2
2
4 2 2
2
L U
L U o
L o U o
o
R R A
R R A UL
C R L k A L U R U k C
m C
+ + ?
+ + ? ? ?? ?
+ + + + ? ? +? ?? ?
? ?? ?=
In the case where L UR R R= = ,
( ) ( ) ( ) ( ) 2222 2 2 4 2 2
2
o
o o
o
ULR A R A C R L U A L U C
m C
? ?? ?+ ? + ? + + + ? +
? ?? ?? ?
? ?=
( ) ( ) ( )( ) 2222 4 4 2 2
2
o
o
ULR A R A C R A L U C
C
? ?? ?+ ? + ? + + ? +
? ?? ?? ?
? ?=
( ) ( )( ) 222
2
2 2
o
o
ULR A C R A L U C
R A
C C
? ?? ?+ ? + + ? +
? ?? ?+ ? ?
? ?= ?
( )( )2 2 2
2
o oR A L U L UR A R A
C C C
+ + ? ?++ +? ?= ? ? +
? ?? ?? ? ? ? (29)
This requires ( )( )
2 2 2
2
o oR A L U L UR A
C C
+ + ? ?++? ? > ?
? ?? ?? ? ? ?
74
( ) ( )( ) ( )
2 2 2
2
2
o
o
C L UR A C R A L U ++ > + + ?
( ) ( )( ) ( )
2 2
2
2
o
o
C L UR A C R A L U? ?+? ?+ > + + ?
? ?? ?
In the case where L UR R R= = , the expected net profit Equation (25), can be simplified
as follows:
( )E P x? ?? ? =
( ) ( ){ } ( ) ( ) ( )
( ) ( )
2 2 2 2 2
2 33 3
2 2 2 2
1
2 21
3 2 2 3 3
o o o o
o
o o
R L m k m k U A m m L U k U L U L
k Um m k LC L U U L
? ?? ?? ? ? + + ? ? ? + ? ? + + ?? ? ? ?
? ?? ? ? ? ? ?? ?? ?
? ?? ?? ?
? ?? + + + ? ? +? ?? ?
? ?? ?? ?? ?
(30)
There are two possible values of m from Equation (29),
Case 1: ( )( )
2 2 2
1 2
o oR A L U L UR A R Am
C C C
+ + ? ?++ +? ?= + ? +
? ?? ?? ? ? ?
Case 2: ( )( )
2 2 2
2 2
o oR A L U L UR A R Am
C C C
+ + ? ?++ +? ?= ? ? +
? ?? ?? ? ? ?
To determine which value of m maximizes net profit, substitute each value of m into
Equation (28) which is the requirement for maximizing net profit.
75
When L UR R R= = ,
2
2
L UR R Am
C
+ +< ? R Am
C
+<
Clearly, this is only true for Case 2, so the optimum value of m must be:
( )( )2 2 2
2 2
o o
o
R A L U L UR A R Am m
C C C
+ + ? ?++ +? ?= = ? ? +
? ?? ?? ? ? ? (31)
The optimum ( )E P x? ?? ? is obtained by inserting om from Equation (31) into Equation
(25.)
Example 11.
An example similar to that used in the previous chapter follows. Assume that L =
the lower specification for fill level (L=100.) If an upper specification limit, USL, is
given for fill level, let U=USL (USL=200.) The revenue if fill level is between L and U
is A = $20. C is the unit cost to produce and C = $0.10. The scrap / reprocessing cost if
fill level is less than L or greater than U is R = $6. The spread of the distribution is
characterized by k such that the process limits are m - k and m + k and in this example, k
= 100.
76
TABLE 6. Example of expected net profit calculations for symmetric triangular
distribution with 2k U L> ? with a fixed value of U.
A C R k U L m E[P(x)]
20 0.1 6 100 200 100 139.1695 2.387498
20 0.1 6 100 200 100 10 -5.92167
20 0.1 6 100 200 100 20 -5.69333
20 0.1 6 100 200 100 30 -5.325
20 0.1 6 100 200 100 40 -4.82667
20 0.1 6 100 200 100 50 -4.20833
20 0.1 6 100 200 100 60 -3.48
20 0.1 6 100 200 100 70 -2.65167
20 0.1 6 100 200 100 80 -1.73333
20 0.1 6 100 200 100 90 -0.735
20 0.1 6 100 200 100 100 0.333333
20 0.1 6 100 200 100 110 1.276667
20 0.1 6 100 200 100 120 1.92
20 0.1 6 100 200 100 130 2.283333
20 0.1 6 100 200 100 140 2.386667
20 0.1 6 100 200 100 150 2.25
20 0.1 6 100 200 100 160 1.893333
20 0.1 6 100 200 100 170 1.336667
20 0.1 6 100 200 100 180 0.6
20 0.1 6 100 200 100 190 -0.29667
20 0.1 6 100 200 100 200 -1.33333
20 0.1 6 100 200 100 210 -2.355
20 0.1 6 100 200 100 230 -3.95833
20 0.1 6 100 200 100 240 -4.56
20 0.1 6 100 200 100 250 -5.04167
20 0.1 6 100 200 100 260 -5.41333
20 0.1 6 100 200 100 270 -5.685
20 0.1 6 100 200 100 280 -5.86667
20 0.1 6 100 200 100 290 -5.96833
20 0.1 6 100 200 100 300 -6
20 0.1 6 100 200 100 310 -6
77
According to Equation (29), the optimum set point for the mean is m = 139.17. At that
point, the expected net profit is $2.39.
A graph of expected net profit for the various values of m is shown in Figure 24.
With U set at 200, there is a decrease in net profit as the distribution approaches that
point. Clearly, profit would be higher if U was set higher. In the next example, an
optimum value of U, oU , is determined.
Expected Net Profit
Symmetric Triangular Distribution
-10
-5
0
5
13
9 20 50 80
11
0
14
0
17
0
20
0
23
0
26
0
29
0
32
0
Value of m
E[
P(
x)]
FIGURE 24. Expected net profit model for symmetric triangular distribution with
2k U L> ? and a fixed value of U.
Example 12.
Now the information from Example 11 is presented, but using the optimum upper
screening limit, oU , instead of a given upper limit,
20 6 260
.10
U
o o
A RU U
C
+ += ? = =
78
TABLE 7. Expected net profit calculations for symmetric triangular distribution with
2k U L> ? with a calculated value of U.
A C R k U L m E[P(x)]
20 0.1 6 100 260 100 146.8629 2.80481
20 0.1 6 100 260 100 0 -6
20 0.1 6 100 260 100 10 -5.92167
20 0.1 6 100 260 100 20 -5.69333
20 0.1 6 100 260 100 30 -5.325
20 0.1 6 100 260 100 40 -4.82667
20 0.1 6 100 260 100 50 -4.20833
20 0.1 6 100 260 100 60 -3.48
20 0.1 6 100 260 100 70 -2.65167
20 0.1 6 100 260 100 80 -1.73333
20 0.1 6 100 260 100 90 -0.735
20 0.1 6 100 260 100 100 0.333333
20 0.1 6 100 260 100 110 1.305
20 0.1 6 100 260 100 120 2.026667
20 0.1 6 100 260 100 130 2.508333
20 0.1 6 100 260 100 140 2.76
20 0.1 6 100 260 100 150 2.791667
20 0.1 6 100 260 100 160 2.613333
20 0.1 6 100 260 100 170 2.236667
20 0.1 6 100 260 100 180 1.68
20 0.1 6 100 260 100 190 0.963333
20 0.1 6 100 260 100 200 0.106667
20 0.1 6 100 260 100 210 -0.79167
20 0.1 6 100 260 100 220 -1.64
20 0.1 6 100 260 100 230 -2.42833
20 0.1 6 100 260 100 240 -3.14667
20 0.1 6 100 260 100 250 -3.785
20 0.1 6 100 260 100 260 -4.33333
20 0.1 6 100 260 100 270 -4.785
20 0.1 6 100 260 100 280 -5.14667
20 0.1 6 100 260 100 290 -5.42833
20 0.1 6 100 260 100 300 -5.64
79
TABLE 7 (continued). Expected net profit calculations for symmetric triangular
distribution with 2k U L> ? with a calculated value of U.
20 0.1 6 100 260 100 310 -5.79167
20 0.1 6 100 260 100 320 -5.89333
20 0.1 6 100 260 100 330 -5.955
20 0.1 6 100 260 100 340 -5.98667
20 0.1 6 100 260 100 350 -5.99833
20 0.1 6 100 260 100 360 -6
20 0.1 6 100 260 100 370 -6
20 0.1 6 100 260 100 380 -6
20 0.1 6 100 260 100 390 -6
20 0.1 6 100 260 100 400 -6
And, graphically, the expected net profit for various values of m (when an optimum value
of U, oU , is determined) is shown in Figure 25.
Expected Net Profit
Symmetric Triangular distribution
-8
-6
-4
-2
0
2
4
50 70 90 110 130 150 170 190 210 230 250 270 290 310 330 350 370 390
Value of m
E[
P(
x)]
FIGURE 25. Expected net profit model for symmetric triangular distribution with
2k U L> ? and a calculated value for U.
80
The optimum set point for m is om =146.86. At that point, the expected net profit
is $2.80 which is higher than it was in the previous example when U was set lower (at
200) than oU .
In general, the value of U should be set as the minimum of an upper specification
limit or the optimum value of U, Uo. min( , )oU USL U=
5.2 Skewed Triangular underlying distribution
In the case of a skewed triangular distribution, the lower limit of the distribution,
a, is set equal to 1m k? and the upper limit of the distribution, b, is set equal to 2m k+
such that the probability density function is:
( )
( )
( )
( )
1
1
1 1 2
2
2
2 1 2
2 ,
2( ) ,
0,
x m k m k x m
k k k
m k xf x m x m k
k k k
otherwise
? ? + ? ? ?
? +
?
?
?
+ ??= ? ? +
? +
?
?
?
?
??
81
The expected net profit is given by:
( ) ( )( ) ( ) ( )( )
( ) ( )( ) ( )( )
1
2
1 1
1 1 2 1 1 2
2 2
2 1 2 2 1 2
2 2
2 2
L m
L
m k L
m kU
U
m U
x m k x m kR dx A Cx dx
k k k k k k
m k x m k xA Cx dx R dx
k k k k k k
E P x
?
+
? + ? +? + ?
+ +
+ ? + ?+ ? + ? =
+ +
? ? ? ?=? ?
? ? ? ?? ? ? ? ? ?
? ? ? ?
? ? ? ?? ? ? ?
? ?
? ?
( )
( ) ( ) ( )
( ) ( )
( ) ( )
1
2
22 2 3
1
1 1
1 1 2
22 3 2
2
2 2
2 1 2
2 2 2 3 2
2
2 2 3 2
mL m
L
m k L L
UU m k
U
m Um
m k xx x xR m k x A m k x C
k k k
m k xx x xA m k x C R m k x
k k k
?
+
? ?? ??? ? ? ?
? ?? ? ? + ? ? ? ?? ?? ?? ? ? ? +
? ?? ? ? ? ? ?+
? ?? ?
? ?
? ?? ?+? ? ? ?? ?
+ ? ? ? ? + ?? ?? ?? ? ? ?+
? ? ? ?? ?? ?? ?
=
( )
[ ] ( ) ( )
( )( )
( )
( ) ( ) ( )( )
( )
2
2
1 1
2 23 31 1 2
1
2 2 23 3
2
2
2 1 2 2
2
( )2 2
2
3 2
2 2 3 2
2
L
U
m LR L m k A k m L
k k k L m m km L
C
U m m km U m UA k m U C
k k k R
U m k
? ?? ??
? ?? ? ? ? ? ? ?? ?
? ?? ?? ?? ?
+? ?+ ? ?
? ?? ??? ?+
? ?? ?
? ?? ?? ?
? ?? ?? ? ? +? ?
? ?? ?? + ? ? +? ?
? ?? ?? ?? ? ? ?? ?
+ ? ?
? ?? ? +? ?? ?? ?
(32)
Note that when the distribution is symmetrical, 1 2k k k= = and Equation (32) reduces to
Equation (25). Further, the ( )E P x? ?? ? is inversely proportional to the spread of the
distribution, 1 2k k+ .
82
5.2.1 Optimum upper screening limit
In order to determine the optimum value of U, the first derivative of the Equation
(32) with respect to U is calculated below:
( ) ( ) ( ) ( ) ( ){ }22 2 2
2 1 2
2
UE P x A m U k C U U m k R U m kU k k k
? ? ?= ? + + ? + ? ? +? ? ? ? ? ?
? ? ? ? ? ?? ?? + (33)
The second derivative with respect to U is less than zero (hence expected net profit is at a
maximum) only if
( ) ( )2 2
2 2 2
U UA R C m k m kA RU
C C
+ + + ++< = + (34)
Setting Equation (33) equal to zero to solve for the value of U, oU , that maximizes
expected net profit results in the following:
( ) ( ) ( )( ){ }
2
2 2
2 1 2
2 0
U UCU A R C m k U A R m kk k k ? + + + + + + =? ?? ?+
( ) ( )( )2 2 2 0U UCU A R C m k U A R m k? + + + + + + =? ?? ?
Let ( )5 2Uc A R C m k= + + +? ?? ? and ( )( )6 2Uc A R m k= + + , then 2 5 6 0o oCU c U c? + = ,
which means that
2
5 5 64
2o
c c CcU
C
? ?=
( ) ( ) ( )( )22 2 24
2
U U U
o
A R C m k A R C m k C A R m kU
C
+ + + ? + + + ? + +? ? ? ? ? ?? ? ? ? ? ?= =
83
( ) ( ) ( )( ) ( ) ( )( )222 2 2 22 4
2
U U U UA R C m k A R C A R m k C m k C A R m k
C
+ + + ? + + + + + + ? + +? ? ? ?? ? ? ? =
( ) ( ) ( )( ) ( )2 222 2 22
2
U U UA R C m k A R C A R m k C m k
C
+ + + ? + ? + + + +? ?? ? =
( ) ( ) ( ) 22 2
2
U UA R C m k A R C m k
C
+ + + ? + ? +? ? ? ?? ? ? ? =
( ) ( ) ( )2 2
2
U UA R C m k A R C m k
C
+ + + ? + ? +? ? ? ?? ? ? ? =
Case 1:
( ) ( ) ( )2 2
1 2
U UA R C m k A R C m kU
C
+ + + + + ? +? ? ? ?? ? ? ?= = ( )2
2
U UA R A R
C C
+ +=
Which is consistent with previous results.
Case 2:
( ) ( ) ( )2 2
2 2
U UA R C m k A R C m kU
C
+ + + ? + ? +? ? ? ?? ? ? ?= = ( )2
2
2
2
C m k m k
C
+ = +
However, Equation (34) requires that setting Equation (33) equal to zero will lead to an
optimum value of U iff ( )22 2U m kA RU C ++< + , i.e. oU = UA R
C
+ iff
( )2
22 2
U U Um kA R A R A R m k
C C C
++ + +< + ? < + . Therefore
2m k+ cannot be the optimum
value for U , and thus
84
oU = U
A R
C
+ (35)
5.2.2 Optimum target set point for m
Similarly, the optimum value of m, om , can be found by setting the first
derivative of Equation (32) with respect to m equal to zero:
( )
[ ] ( ) ( )
( )( )
( )
( ) ( ) ( )( )
( )
2
2
1 1
2 23 31 1 2
1
2 2 23 3
2
2
2 1 2 2
2
( )2 2
2
3 2
2 2 3 2
2
L
oo o
o
U
o
m LR L m k A k L m
k k k L m m km L
C
m U m m k
m U m UA k m U C
k k k R
U m k
? ?? ?? ??
? ?? ?? ? ? ? + ?? ?
? ?? ?? ?? ?? ?
? +? ?
+? ? ?? ?? ??
? ? ?? +? ?
?? ? ?? ?? ?? ?
??
? ? ?? ?? ? ? +? ?
? ? ?? ?? + ? ? +? ?
? ? ?? ?? ?? ? ? ?? ?
? + ? ?
? ? ?? ? +? ?
? ? ?? ?? ?
?
?
?
?
?
?
?
?
?
?
?
( ) [ ] ( )
( ) ( ) ( )
2 2
1 1 1
1 1 2
2 2
2 2 2
2 1 2
2 ( )
2
2
2
L
o
o U o
L mR L m k A m k L C mk
k k k
U mA m k U C mk R U m k
k k k
? ?? ?? ??? ?? ?? ? ? ? ? ? + +
? ?? ?? ?? ?+ ? ?
? ?? ?? ?
? ?? ?? ??? ?
? ?? ? ? ?? + ? ? ? + ? +? ?? ?? ? ? ?
? ?+ ? ?? ?? ?? ?
(36)
Note that in the case of the skewed triangular distribution, m is not equal to ( )E x ?=
(See Appendix D for the first four moments.)
The second derivative with respect to m is negative if
( )
( ) ( ) ( ) ( )1 2
1 2 1 2
2 0L UR A C m k A R C m k
k k k k
? ?? ? ? ?? + + ? ? + + +? ?+ <
? ?? ? ? ?+ ? ?
? ? ? ?? ?
85
( ) ( )
( )
2 1
1 2
L UR A k A R km
C k k
+ + +<
+ (37)
(Note that when 1 2k k k= = and L UR R R= = , Equation (37) simplifies to
o
R Am U
C
+< = as before.)
Setting Equation (36) equal to zero and solving for om results in
[ ] ( )
( ) ( )
2 2
1 1 1
1
2 2
2 2 2
2
1 ( )
2
1 0
2
L
o
o U o
L mR L m k A m k L C mk
k
U mA m k U C mk R U m k
k
? ?? ??? ?? ?? ? ? ? ? ? + +
? ?? ?? ?? ?? ?
? ?
? ?? ??? ?? ? ? ?? + ? ? ? + ? + =
? ?? ?? ? ? ?? ?? ?
? ?
( ) ( )
( ) ( ) ( ) ( )
2
1 2
1 2 1 2
22
1 1 2 2
1 2
1 1 1 1
2
1 1 0
2 2
L U
o
L o U o
C m R A Ck A Ck R m
k k k k
CUCLR L k A L k A U k R U k
k k
? ? ? ?+ + ? ? ? + ? + ? +
? ? ? ?? ? ? ?
? ?? ?+ + + ? + ? ? + ? =
? ?? ?? ? ? ?
( ) ( ) ( )( )
( )( )
2
2
1
1 2 1 2 1
2
2
2
1 1 1
2 2
1 0
2
L U
L
o
U o
R A R AC CLm m R A L k
k k k k k
CUR A U k
k
? ?+ +? ? ? ?+ ? + + + + ?
? ?? ? ? ?? ?? ? ? ?
? ?+ + ? ? =
? ?? ?
( ) ( ) ( ) ( )( )
( )( )
2
2 1
1 2 2 12
21 2 1 2 1 2
1 2
21 0
2
2
L
L U
o
U o
CLk R A L k
C k k k R A k R Am m
k k k k k k CUk R A U k
? ?? ?+ + ?
? ?? ?? ?+ + + + ? ?? ?
? + =? ?? ?
? ?? ? ? ?+ + ? ?
? ?? ?? ?? ?
86
( ) ( ) ( ) ( )( )
( )( )
2
1 2 2
2 1 2 1
2
1 2
2 2
02
L U L
o
U o
C k k CLm k R A k R A m k R A L k
CUk R A U k
+ ? ?? ?? + + + + + + ?
? ?? ? ? ?
? ?+ + ? ? =
? ?? ?
Let ( )1 27 2C k kc += , ( ) ( )8 2 1L Uc k R A k R A= + + + , and
( )( ) ( )( ) 229 2 1 1 22 2oL U o CUCLc k R A L k k R A U k? ?? ?= + + ? + + ? ?? ?? ?
? ? ? ?
, then 27 8 9 0o oc m c m c? + = ,
which means that
2
8 8 7 9
7
4
2o
c c c cm
c
? ?=
( ) ( )
( )
( ) ( ) ( )
( )( )
( )( )
( )
2 1
1 2
2
2 1
2
2 1 1 2 2
1 2
1 2
22
2
L U
o
L
L U
o
U o
k R A k R Am
C k k
CLk R A L k
k R A k R A C k k
CUk R A U k
C k k
+ + += ?
+
? ?? ?+ + ? +
? ?? ?? ?? ?
? ?+ + + ? + ? ?? ?
? ?? ?+ ? ?
? ?? ?? ?? ?
+
Given the requirement from Equation (37) that ( ) ( )( )2 1
1 2
L U
o
R A k A R km
C k k
+ + +<
+ , it follows
that
87
( ) ( )
( )
( ) ( )
( ) ( )( ) ( )( )
( )
2 1
1 2
2
2 1
22
1 2 2 1 1 2
1 2
2 2 2
L U
o
L U
o
L U o
k R A k R Am
C k k
k R A k R A
CUCLC k k k R A L k k R A U k
C k k
+ + += ?
+
? ?+ + + ?? ?
? ?? ?? ?? ?+ + + ? + + ? ?
? ?? ?? ?? ?? ? ? ?
? ?
+
(38)
Which reduces to Equation (31) when 1 2k k= and L UR R= .
Example 13.
For example, when A=20, C=.1, LR = 5, UR = 6, 1k = 20, 2k = 180, and L=100, oU =260
and om = 107.47 from Equation (38).
It can be shown (in the Appendix), that the mean of a skewed triangular distribution is
[ ] 2 13k kE X m? ?= = + , so the optimum value of the mean is given below:
2 1
3o o
k km? ?= + (39)
5.2.3 Sensitivity analysis of skewness
Note that the triangular distribution is negatively skewed when 1 2k k> and
positively skewed when 2 1k k> . Variations of Example 13 are presented below to
illustrate the impact of skewness on the calculation of om .
88
Example 14.
When A=20, C=0.10, LR = 5, UR = 6, and L=100, oU =260, values of om are shown in
Table 8 for various levels of skewness and the expected net profit is calculated by
substituting om for the value of m in Equation (32):
TABLE 8. om and the expected net profit for various levels of skewness
1k 2k om ( )E P x? ?? ?
20 180 107.47 3.334
80 120 136.06 2.953
100 100 144.43 2.948
120 80 156.32 2.781
180 20 208.00 1.755
As illustrated in this example, as the distribution becomes negatively skewed, the
optimum set point must be higher. In the canning problem, where only a lower
specification is given, a positively skewed distribution allows for a lower optimum set
point and a higher expected net profit.
5.3 Process capability with underlying Triangular distribution
As presented in Chapter 4, a process is generally deemed capable if 1.0PC ?
which means, if the process is centered inside the specifications, then the middle 99.73%
of the distribution will fall inside the specification limits. For a normal distribution, this
translates to / / 6PC BT NT BT ?= = . However, for a Triangular distribution, the limits
defined by 3? ?? do not define the middle 99.73% of the distribution. In fact, the limits
89
defined by 3? ?? actually fall outside the complete range of the distribution defined by
( )1 2,m k m k? + :
2 2
2 1 1 1 2 23
3 2
k k k k k km? ? ? + +? = + ?
Thus, the 6-sigma spread for a triangular distribution is given by
2 2
1 1 2 22
2
k k k k+ + . As a
result, as is the case with the Uniform distribution, the 6-sigma spread
( ) ( ) ( )22 2 2 21 1 2 2 1 2 1 2 1 22 k k k k k k k k k k+ + = + + + > + so that the 99.73% natural
tolerances of the Triangular distribution must be obtained from .99865x - .00135x .
It can be shown that the cdf of any Triangular distribution is given by
( )
( )
( )
( )
( )
1
2
1
1
1 1 2
2
2
2
1 2 2
2
0,
,
1 ,
1,
x m k
x m k m k x m
k k kF x
m k x m x m k
k k k
x m k
? ??
? ? +
? ? ? ?
? +?
= ?
+ ?? ? ? ? +
? +
?
? ? +?
Let px be the pth quantile (or fractile) of x, then upon inverting the above cdf, the
following percentile function is obtained:
( ) ( )
( ) ( )( )
1
1 1 2 1
1 2
1
2 2 1 2
1 2
, 0
1 , 1
p
km k k k k p p
k kx
km k k k k p p
k k
? ? + + ? ?
? +?
= ?
? + ? + ? ? ?
? +?
90
The PC index by definition is
.99865 .00135
P
USL LSLC
x x
?=
? , where the middle 99.73% of the
spread of the distribution is given by
( ).99865 .00135 2 1 2 1 2 10.036741x x k k k k k k? ?? = + + ? +? ?
91
6.0 Conclusion
The work presented in this dissertation extends the previous research on the canning
problem (which focused on infinite range distributions, specifically the normal
distribution, for fill level) to finite distributions. Three finite distributions were
analyzed: Uniform, Symmetric Triangular, and Skewed Triangular. In each case, an
optimum set point for the mean fill level was determined to maximize expected net
profit. When appropriate, an upper screening limit for fill level was also determined.
In the case of the Uniform distribution, three net profit models were studied: fixed
rework/reprocessing costs, linear rework/reprocessing costs, and capacity constrained.
Few closed form solutions were obtained by differentiation for determining an optimum
set point for the mean to maximize the expected net profit. However, the optimum set
point was determined to maximize expected net profit by evaluating expected net profits
at the extreme points for each range of ? .
6.1 Summary of Results
For fill level that follows a Uniform distribution when there is a constant scrap
cost, the optimum value for the upper screening limit was determined to be
( )min , min , ULo L U R AR AU U U C C++? ?= = ? ?? ?
92
The optimum target set point for the process mean was obtained for the various scenarios:
Case 1: 2 o ok U L L k?? ? ? = +
Case 2: 2 ok U L> ? :
If U L oR R L k?< ? = +
If U L o oR R U k?> ? = ?
If [ ],U L o oR R U k L k?= ? ? ? +
For fill level that follows a Uniform distribution when there is a linear scrap cost,
the optimum target set point for the process mean was obtained for the various scenarios:
Case 1: 2k U L? ?
o L k? = +
Case 2: 2k U L> ?
o L k? = + if
( ) ( ) ( ) ( )2 2 2 2
4 4U L
U L U LR L k R U k
k k
? ? ? ?? ?
? ? ? ?? + > ? ?
? ? ? ?? ? ? ?
o U k? = ? if
( ) ( ) ( ) ( )2 2 2 2
4 4U L
U L U LR L k R U k
k k
? ? ? ?? ?
? ? ? ?? + ? ? ?
? ? ? ?? ? ? ?
The target set point for the process mean, o? , of a profit model with a capacity
constraint was defined for the various scenarios:
Case 1: ( )2 min , o ok CAP L U L L k?< ? ? ? = +
Case 2: ( )2 min , ok CAP L U L? ? ?
( )min ,U L oR R L k CAP k?< ? = + ?
( )min ,U L o oR R U k CAP k?> ? = ? ?
93
( ), min ,U L o oR R U k L k CAP k? ? ?= ? ? ? + ?? ?
For both the symmetric and skewed Triangular distributions, a net profit model
with fixed rework/reprocessing costs was assumed. Assuming that L m U? ? ,
1m k L? < , and 2m k U+ > , the optimum set point for m was determined to be:
( ) ( )
( )
( ) ( )
( ) ( )( ) ( )( )
( )
2 1
1 2
2
2 1
22
1 2 2 1 1 2
1 2
2 2 2
L U
o
L U
o
L U o
k R A k R Am
C k k
k R A k R A
CUCLC k k k R A L k k R A U k
C k k
+ + += ?
+
? ?+ + + ?? ?
? ?? ?? ?? ?+ + + ? + + ? ?
? ?? ?? ?? ?? ? ? ?
? ?
+
which reduces to ( )( )
2 2 2
2
o o
o
R A L U L UR A R Am
C C C
+ + ? ?++ +? ?= ? ? +
? ?? ?? ? ? ? when 1 2k k= and
L UR R= .
Throughout the research, examples were provided and proofs, where necessary,
were outlined.
6.2 Practical Applications
This research shows that when fill level is not normally distributed, the optimum set
points for a canning problem can still be determined if the distribution can be modeled,
even if the distribution range is not infinite. In at least one case (steel thickness), the
Uniform distribution has semmed to be an appropriate fit. However, in most cases,
practical application of this work may be somewhat limited, since fill level is best
estimated by an infinite range distribution that is bounded rather than an actual finite
distribution.
94
6.3 Recommendations for Future Work
Given the similarities between the Uniform distribution, symmetric Triangular
distribution, and the Normal distribution (symmetric, continuous within the range, and
mean centered,) it may be interesting to compare the results presented here with results
when a Normal distribution is incorrectly assumed for fill level which is best modeled by
a Uniform distribution.. Fill level data generated from a Uniform distribution could be
used to calculate the optimum upper screening limit and the optimum mean using the
formulas presented here. Those results could then be compared with results from the
formulas from research using the same net profit model, but assuming a Normal
distribution of fill level. It would be interesting to see how the results differ in terms of
magnitude of o? and in terms of expected net profit.
Another extension of this work would be to extend the analysis of the Triangular
distribution to the other cases presented in Figures 21 and 23. and to more thoroughly
complete a sensitivity analysis of skewness using the formula for skewness in Appendix
D.
95
7.0 References
BETTES, D. C. (1962). ?Finding an Optimum Target Value in Relation to a Fixed
Lower Limit and an Arbitrary Upper Limit?. Applied Statistics 11, pp. 202-210.
BISGAARD, S., HUNTER, G. H., and PALLESEN, L. (1984). ?Economic Selection of
Quality of Manufactured Product?. Technometrics 26, pp. 9-18.
CAIN, M. and JANSSEN, C. (1997). ?Target Selection in Process Control Under
Asymmetric Costs?. Journal of Quality Technology 29, pp. 464-468.
CARLSSON, O. (1984). ?Determining the Most Profitable Process Level for a
Production Process Under Different Sales Conditions?. Journal of Quality Technology
16, pp. 44-49.
CARLSSON, O. (1989). ?Economic Selection of a Process Level Under Acceptance
Sampling by Variables?. Engineering Costs and Production Economics 16, pp. 69-78.
DODSON, B. L. (1993). ?Determining the Optimal Target Value for a Process with
Upper and Lower Specifications?. Quality Engineering 5, pp. 393-402.
GOHLAR, D. Y. (1987). ?Determination of the Best Mean Contents for a Canning
Problem?. Journal of Quality Technology 19, pp. 82-84.
GOHLAR, D.Y. and POLLOCK, S. M. (1988). ?Determination of the Optimal Process
Mean and the Upper Limit for a Canning Problem?. Journal of Quality Technology 20,
pp. 188-192.
96
GOHLAR, D.Y. (1988). ?Computation of the Optimal Process Mean and the Upper
Limit for a Canning Problem?. Journal of Quality Technology 20, pp. 193-195.
HUNTER, W. G. and KARTHA, C. P. (1977). ?Determining the Most Profitable Target
Value for a Production Process?. Journal of Quality Technology 9, pp. 176-181.
KIM, Y. J., CHO, B. R., and PHILLIPS, M. D. (2000). ?Determination of the Optimal
Process Mean with the Consideration of Variance Reduction and Process Capability?.
Quality Engineering 13, pp. 251-260.
LEE, M. K., HONG, S. H., and ELSAYED, A. E. (2001). ?The Optimum Target Value
Under Single and Two-Stage Screenings?. Journal of Quality Technology 33, pp. 506-
514.
LEE, M. K. and ELSAYED, A. E. (2002). ?Process Mean and Screening Limits for
Filling Processes Under Two-Stage Screening Procedure?. European Journal of
Operational Research 138, pp. 118-126.
LIU, W. and RAGHAVACHARI, M. (1997). ?The Target Mean Problem for an
Arbitrary Quality Characteristic Distribution?. International Journal of Production
Research 35, pp. 1713-1727.
MISIOREK, V. I. and BARNETT, N. S. (2000). ?Mean Selection for Filling Processes
Under Weights and Measures Requirements?. Journal of Quality Technology 32, pp.
111-121.
NELSON, L. S. (1978). ?Best Target Value for a Production Process?. Journal of
Quality Technology 10, pp. 88-89.
NELSON, L. S. (1979). ?Nomograph for Setting Process to Minimize Scrap Cost?.
Journal of Quality Technology 11, pp. 48-49.
97
PFEIFER, P. E. (1999). ?A General Piecewise Linear Canning Problem Model?.
Journal of Quality Technology 31, pp. 326-337.
PULAK, M. F. and AL-SULTAN, K. S. (1997). ?A Computer Program for Process
Mean Targeting?. Journal of Quality Technology 29, pp. 477-484.
SCHMIDT, R. L. and PFEIFER, P. E. (1989). ?Economic Evaluation of Improvements
in Process Capability for a Single-Level Canning Problem?. Journal of Quality
Technology 21, pp. 16-19.
SCHMIDT, R. L. and PFEIFER, P. E. (1991). ?Economic Selection of the Mean and
Upper Limit for a Canning Problem with Limited Capacity?. Journal of Quality
Technology 23, pp. 312-317.
SPRINGER, C. H. (1951). ?A Method for Determining the Most Economic Position of a
Process Mean?. Industrial Quality Control 8, pp. 36-39.
USHER, J. S., ALEXANDER, S. M., and DUGGINS, D. C. (1996). ?The Filling
Problem Revisited?. Quality Engineering 9, pp. 35-44.
98
APPENDICES
99
Appendix A: Formulas for Excel spreadsheet for calculating expected net profit for
various levels of ? with a Uniform distribution and fixed scrap cost.
H2=IF(2*E2<=F2-G2,G2+E2,IF(C2>D2,G2+E2,F2-E2))
I2==IF(H2<=G2-E2,-C2,IF(AND(H2=G2-E2),L2,IF(AND(H2=G2+E2),M2,IF(AND(H2>=F2-E2,H2=G2-
E2),L2,IF(AND(H2=F2-E2),O2,IF(AND(H2>=G2+E2,H2U-L
Case E[P(x)]
11.25 11.25 11.25 11.25 30 78 59.25
100
Appendix B: Formulas for Excel spreadsheet for calculating expected net profit for
various levels of ? with a Uniform distribution and linear scrap cost.
H2 =IF(2*E2<=F2-G2,G2+E2,IF(D2*((F2^2-G2^2)/(4*E2)-(G2+E2))>C2*((F2^2-
G2^2)/(4*E2)-(F2-E2)),G2+E2,F2-E2))
I2 =IF(H2<=G2-E2,-C2*H2,IF(AND(H2=G2-E2),L2,IF(AND(H2=G2+E2),M2,IF(AND(H2>=F2-E2,H2=G2-
E2),L2,IF(AND(H2=F2-E2),O2,IF(AND(H2>=G2+E2,H2U-L
Case E[P(x)]
15 15 15 15 15 77.5 77.5
101
Appendix C: Formulas for Excel spreadsheet for calculating expected net profit for
various levels of m with a symmetric Triangular distribution.
I2=IF(H2<=G2-E2,-C2,IF(AND(H2<=G2+E2,H2<=F2-E2,H2>G2-
E2),L2,IF(AND(H2<=G2+E2,H2>F2-E2),M2,IF(AND(H2>F2-
E2,H2>G2+E2,H2G2,(1/(E2^2))*(-(C2+A2)/2*(G2-H2+E2)^2-B2*(-
H2^3/3+(H2+E2)^3/6+G2^2/2*(H2-E2)-G2^3/3))+A2,(1/(E2^2))*((A2+C2)/2*(H2+E2-
G2)^2-B2*((H2+E2)^3/6-G2^2/2*(H2+E2)+G2^3/3))-C2)
M2= =(1/E2^2)*(-C2/2*(G2-H2+E2)^2-A2*(H2^2-H2*(G2+F2)-E2*(F2-
G2)+0.5*(F2^2+G2^2))-B2*(-H2^3/3+H2/2*(G2^2+F2^2)+E2/2*(F2^2-G2^2)-
(F2^3/3+G2^3/3))-C2/2*(H2+E2-F2)^2)
N2= =A2-B2/(E2^2)*(-H2^3/3+(H2-E2)^3/6+(H2+E2)^3/6)
O2= =IF(H2 .
108
If 2 1k k> , then 1 2k rk= , where 0 1r< < and 1
2
kr
k= . Substituting for 1 2k rk=
in Equaion (41) results in
( )( )
( )
2 2 2 2
2 2 2 2 2
3 1.52 2 2 2
2 2 2
2 2 50.08 k rk k r k rk
k rk r k
? ? + +=
+ +
( )( )
( )
2
1.52
1 2 2 50.08
1
r r r
r r
? + +=
+ +
Taking the first derivative of skewness,
( )( ) 1.52 3 23 0.08 2 3 3 2 1d d r r r r rdr dr? ?? ?= + ? ? + +? ?? ?
( )( ) ( ) ( )( )1.5 2.52 2 2 2 30.08 3 6 6 1 1.5 1 1 2 2 3 3 2r r r r r r r r r r? ?? ?= ? ? + + ? + + + + ? ?? ?? ?
( ) ( )( ) ( )( )2.52 2 2 2 30.08 1 3 6 6 1 1.5 1 2 2 3 3 2r r r r r r r r r r? ? ?= + + ? ? + + ? + + ? ?? ?
( ) ( )2.52 20.08 1 13.5 13.5r r r r?= + + ? ?
109
( )
( )2.52
114.58 0
1
r r
r r
+= ? <
+ +
for all r within 0 1r< < . Hence, the maximum of 3? occurs
at r = 0 and all triangular distributions have skewness values in the interval
30.32 0.32?? ? ? . Further, all right-triangular distributions have 1 0k = and 0r = ,
so that 3 0.32? = ? , and all left-triangular distributions have 1r = , so that 3 0.32? = .
When 1 2 0k k= = , 3 0? = .
Similarly, it can be shown that the kurtosis of all triangular distributions is given
by
4
4 4 3 3 2.4 3 0.60
xE ?? ?
?
? ??? ?= ? = ? = ? = ?
? ?? ?? ?? ?
? ?
.
So, to summarize, for the Triangular distribution
[ ] 2 13k kE X m ?= +
[ ] 2 21 2 1 218k k k kV X + +=
( )( )
( )
2 2
2 1 2 1 1 2
3 1.52 2
2 1 2 1
2 2 50.08 k k k k k k
k k k k
? ? + +=
+ +
and 4 4 3 0.60? ?= ? = ?
110
Note that for a symmetric Triangular distribution, the above equations simplify to
[ ]E X m= , [ ] 26kV X = , 3 0? = , and 4 4 3 0.60? ?= ? = ? .