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? ?= ? = ? .