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Integrated product and process design: a multi-objective modeling framework
Robotics and Computer Integrated Manufacturing 18 (2002) 157–168
Integrated product and process design: a multi-objective modeling framework Nanua Singh* Department of Industrial and Manufacturing Engineering, Integrated Product Development Lab, Wayne State University, Detroit, MI 48202, USA Received 1 August 2000; received in revised form 1 January 2001; accepted 1 August 2001
Abstract In this paper we present an analytical multi-objective framework for the concurrent design of product and processes. The objective is to simultaneously consider the tolerance specification on the product or the component dimensions along with the selection of the manufacturing processes. For this purpose we consider three objectives: to minimize unit cost, to minimize quality loss and to minimize manufacturing lead time. We characterize the properties of the non-dominated solutions. These solutions provide flexibility needed in an agile manufacturing environment. The min–max approach is used to obtain trade-off solutions. r 2002 Published by Elsevier Science Ltd. Keywords: Design for manufacturing system; Tolerance design; Concurrent engineering; Quality engineering; Concurrent product and process design
1. Introduction The twenty-first century business environment can be characterized by expanding global competition, and increasing variety/low in demand for products. The globalization of economic activity has brought about a major change in the attitudes of the customers. Customer individualism is certain to become the central theme of business. What we are going to witness is an era of mass customization. It means manufacturing products for the mass market in such a way that products are customized for each individual in that market. In the seventies, the cost of products used to be the main lever for obtaining competitive advantage. Later in eighties, the quality superseded cost and became an important competitive dimension. Now low unit cost and high quality of products no longer solely define the competitive advantage of manufacturing enterprises. Today, the customer takes both minimum cost and high quality for granted. Factors such as delivery performance, customization and environmental issues such as waste generation are assuming a predominant role in defining the success of organizations in terms of *Tel.: +1-313-577-7586; fax: +1-313-577-8833. E-mail address: [email protected] (N. Singh). 0736-5845/02/$ - see front matter r 2002 Published by Elsevier Science Ltd. PII: S 0 7 3 6 - 5 8 4 5 ( 0 1 ) 0 0 0 3 0 - 8
increased market share and profitability [1]. The question is: what can we do under these changing circumstances to stay in business and retain competitive advantages? As a first step what is needed is the development of the right business strategy to meet the challenges of the present and future market. In doing so, a manufacturing organization not only has to understand what customers want, but also has to develop internal mechanisms to instantly respond to the changes in the products demanded by customer wants. This requires a paradigm shift in everything our factories do. That means, not only do they have to make use of the state-of-the-art technologies and concepts in building products, but also think in the reverse direction [2]. Reverse direction means building products that realize customer expectations. That is, when an organization is deciding about business plans, it has to address such questions as the following: Will the customer find any change in what one does as a result of using this? Will the customer be able to define any benefit? From a customer point of view, a company has to respond to smaller and smaller market niches very quickly with standardized products that will get built in lower and lower volume. In other words, we can say that a future successful manufacturing organization would be a virtual corporation that is
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instantaneously responsive to customer needs. This view is shared by industry-led consortium on twenty-first century manufacturing enterprise strategy [3]. The next step is to determine the right kind of resources to support the business strategy. This requires the right choice of people, technology and business processes. The products are designed to satisfy certain functional requirements. The primary purpose of tolerancing is to help achieve the stated functional requirements. The tolerance design is the process of setting up acceptance regions for the design variables meeting certain criteria. In a competitive environment, these criteria include minimum cost, high quality and minimum lead time. The tolerances are the link between design and manufacturing. It is therefore important that we exploit the interaction between design and manufacturing by jointly considering the selection of processes as well as product tolerances. The tolerance design problem has a strong relationship with product cost, quality and lead time. Inappropriate tolerance may either reduce the quality or generate more scrap thereby increasing the product cost and lead time. A number of papers have appeared in the area of tolerancing. An exhaustive literature review on tolerancing is given in Zhang and Huq [4]. Most of the papers have considered cost minimization as the criterion for the optimal specification of the tolerances. What is however required is the consideration of other equally important issues such as quality and manufacturing lead time. Since the tolerances are the link between the design and manufacturing, it is also equally important to select the manufacturing processes along with the tolerance allocation. Singh [1] introduced the concepts of unit cost, quality and lead time to analyze the integrated product and process design problem in a concurrent engineering environment. DeVour et al. [5] provide a methodology for the robust design using models for the mean, variance and loss. Kapur [6] provides a general understanding of the taguchi loss function approach. However, no formal treatment to simultaneously determine the optimal tolerances as well as the manufacturing processes has been considered. In view of the requirements stated above, the primary objective of our work is to provide a paradigm shift by concurrently attempting the product and process design and jointly considering the criteria of minimum cost, maximum quality and minimum manufacturing lead time. In this paper we therefore provide a generalized multi-objective framework for the integrated product and process design problem. Generalized models for the following cases are presented: * * *
Single component single design variable case. Single component multiple design variables case. The assembled product Case.
This multiobjective framework for product and process design by considering unit cost, quality and lead time simultaneously can be used as a flexible decision support system tool to support a mass customization business strategy.
2. The mathematical model for integrated product and process design In sequential engineering approach, the product design group is responsible for designing the product and consequently determining the tolerances. The process design group then selects the machine(s) which produce(s) the product. The manufacturing engineering department will manufacture the product. The product(s) not meeting the tolerances are scrapped. The good quality products are shipped to the customers. The information on unit cost, scrap rates generated and the lead time are fed back to the product design group which may possibly modify the design and adjust the tolerances. This process of feedback normally takes a long period to modify the product design and adjust the tolerances to satisfy the needs of the customers. This delay may have a number of repercussions: For example, the process may either produce low quality products with very large tolerances or the manufacturing cost may be too high due to the wrong selection of very expensive machines or may result in the scrapping of too many products with very small tolerances. Small adjustments may be difficult to make at this stage since manufacturing processes have already been selected. Radical changes such as changing the product design or the manufacturing processes may turn out to be a very expensive proposition. The solution to these problems lies in adopting what is known as concurrent engineering approach. The basic idea in concurrent engineering approach is to integrate the manufacturability, cost, quality, lead time and marketing information with the design process. In the present paper we develop a modeling methodology to obtain the optimal tolerances as well as manufacturing process considering unit cost, quality and lead time. Since this modeling approach considers various design and manufacturing factors at the beginning stage of the design cycle, it eliminates the need for design changes and adjustments. Further, the trade-off among quality, cost and lead time can be evaluated according to the marketing information representing the needs of the customers. There is a major paradigm shift in decision making in the area of design and manufacturing of products and processes due to global competition. This has led to a shift from decisions based solely on the cost to decisions
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jointly based on the cost, quality and lead time considerations. Therefore, there is a need to develop such models to jointly design the products and processes considering cost, quality and lead time information. Accordingly, in this section we present three models: the unit cost of production model, the quality model and the lead time model. These models are then used for selecting the optimal tolerance and machine-tools using a multi-objective decision making approach.
Therefore, SCðd1 ; d2 ; jÞ ¼ FðZ l Þ þ 1 FðZu Þ ¼ F½ðT mðjÞ d1 Þ=sð jÞ þ 1 F½ðT mð jÞ þ d2 Þ=sð jÞ:
&
Note: In most of the cases, it is desired to have m ¼ T and d1 ¼ d2 ¼ d: Then, the fraction of scrap is given by
2.1. The cost model
SCðd; jÞ ¼ F½d=sð jÞ þ 1 F½d=sð jÞ:
In the development of the models, we provide the following preliminaries and notation. The manufacturing process requires the use of Yi units of raw material to produce Yo units of finished products. Ys units are scrapped in the process which are beyond the tolerance limits.
For simplicity, we assume m ¼ T and d1 ¼ d2 ¼ d in the rest of this paper. From the relationships among input, output and scrap we have
Definition 1. The tolerance limit of the components/ product is (nominal value Ttolerance d1 ; nominal value T+tolerance d2 ). The components/products having dimensions beyond these limits are scrapped.
Ys ¼ Yi SCðd; jÞ ¼ Yo SCðd; jÞ=½1 SCðd; jÞ:
Definition 2. The quality/design characteristic of the product produced by jth manufacturing process (for example, jth machine) is characterized by a probability distribution with mean mð jÞ and standard deviation sð jÞ:
Yo ¼ Yi Ys ¼ Yi SCðd; jÞYi ¼ Yi ½1 SCðd; jÞ; and We assume that the unit cost consists of the cost of raw material, the cost of manufacturing (include setup cost and labor cost), and the salvage value (income of selling scraped product). Accordingly, the unit cost model can be shown to be Xo ðd; jÞ ¼ Ki ðd; jÞ½Xi þ f ð jÞ Ks ðd; jÞXs ;
ð2Þ
Lemma 1. The fraction of scrap (SC) of product is the function of d1 , d2 and j as follows.
where j is the jth manufacturing process selected for producing the product, Xo ðd; jÞ the unit cost when using jth process with tolerance d; Xi the unit raw material cost, f ð jÞ the unit processing cost for jth process, Xs the unit salvage value. Ki ¼ Yi =Yo ¼ 1=½1 SCðd; jÞ; Ks ¼ Ys =Yo ¼ SCðd; jÞ=½1 SCðd; jÞ; SCðd; jÞ is the scrap rate given in formula (1), Yo the amount of output (required amount), Yi the amount of raw material required to produce the desired, Yo ; Ys the amount of scrapped product. From the cost equation we observe that the total unit cost is a function of d and j: The objective is to have minimal possible unit cost.
SCðd1 ; d2 ; jÞ ¼ F½ðT mð jÞ d1 Þ=sð jÞ
2.2. The quality model
The manufacturing process variables are process mean and standard deviation whereas tolerance d are design variables. In a cost minimization model, the objective is to select the process and design variables such that the cost function is minimized.
þ 1 F½ðT mð jÞ þ d2 Þ=sð jÞ;
ð1Þ
where FðxÞ is cumulative distribution function (c.d.f.) of probability distribution with mean equal to 0 and standard deviation equal to 1.
Proof. Let Z u ¼ ðT m þ d2 Þ=s; Z l ¼ ðT m d1 Þ=s: The fraction of scrap is given by definition as SCðd; jÞ ¼ Ys =Yi :
In developing a quality model we consider two issues. One is the fact that the more the number of defects, the worse is the perception of customer regarding the quality production capability of the company. Accordingly, there is a perceived loss because of the shear volume of defects. Another quality factor is the loss to the society which is described by the well-known Taguchi loss function. In developing the quality model, we consider both the issues together as follows: Qloss ðd; jÞ ¼ Qs ðd; jÞ þ Qt ðdÞ ¼ A SCðd; jÞ þ B d 2 :
ð3Þ
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Using Taguchi’s Loss function concept, Qt ðdÞ ¼ Bd 2 ; where B is the parameter and d is the tolerance. The loss Qs ðd; jÞ is due to the scrap rate. The high scrap rate will damage the credibility of the company on quality. Therefore, we consider Qs ðd; jÞ ¼ A SCðd; jÞ; where A is the parameter and SC is the scrap rate for given tolerance d and jth machine. SCðd; jÞ; j and d have the same meanings in cost model and A and B are constant parameters. The objective is to minimize the value of Qloss ðd; jÞ: 2.3. The lead time model The response time to deliver the products to the customers has added an important dimension to the competitiveness. The manufacturing lead time represents a significant portion of the total response time. In this section we develop a manufacturing lead time model as a function of manufacturing and design decision variables as follows: Assuming different manufacturing processes having different setup time Tsetup ð jÞ and different unit processing time Tp ð jÞ the manufacturing time on jth machine equals Tsetup ð jÞ plus Tp ð jÞ times the amounts of raw material Yi processed to manufacture the desired output Yo : Accordingly, the total manufacturing lead time to manufacture a lot is Tðd; jÞ ¼ Tsetup ð jÞ þ Tp ð jÞYi ¼ Tsetup ð jÞ þ Tp ð jÞYo =½1 SCðd; jÞ;
the functions of design and manufacturing decision variables. Usually, from the marketing information as well as from the customer requirements we are aware of the cost and tolerance of the competitors’ product and customers’ requirement for lead time being less than To ; tolerance less than do and cost less than Co : The company may want the Quality loss to be less than Qo : These requirements can be considered as the constraints. The feasible ranges within which To ; Co and Qo should be selected are discussed in Section 4.2. We consider Xo ðd; jÞ; Qloss ðd; jÞ and Tðd; jÞ as three objectives to be minimized. The customer requirements are considered as the constraints. The three objectives are conflicting in nature. Therefore, there is a need to obtain trade-off solutions. The multi-objective integrated product and process design problem can be stated as follows: Minimum fXo ðd; jÞ; Qloss ðd; jÞ; Tðd; jÞg d;j
ð5Þ
subject to Xo ðd; jÞpCo ; Qloss ðd; jÞpQo ; Tðd; jÞpTo ; 0pdpdo j ¼ 1; 2; :::; n;
ð4Þ
where Yi ; Yo ; SCðd; jÞ; j and d have the same meanings as defined in the cost model. Usually Yo is known and Yi is a function of Yo ; j and d. The objective is to have minimum possible lead time Tðd; jÞ:
3. Multi-objective model for the integrated product and process design problem The concept of cost, quality and lead time have become the key determinants of market share and profitability for any organization. As we have seen in our models the design and manufacturing decisions influence cost, quality and lead time which in turn influence market share and profitability. Therefore, there is a need to study all these issues in an integrated manner. Our integrated product and process design model makes it possible to integrate the cost, quality and lead time into the product and process design. The tolerance of the product and the selection of the manufacturing process are determined simultaneously. What we have here is a systems concept on integrated product and process design by considering cost, quality and lead time models in a multi-objective environment. All the relations of unit cost, quality and lead time are
where Co ; Qo ; To ; do are requirements from customers. n is the number of alternative manufacturing processes (or machines). The expressions of Xo ðd; jÞ; Qloss ðd; jÞ and Tðd; jÞ are given by Eqs. (1)–(3), respectively. Multi-objective mathematical programming methods can be used to solve this model and generate a set of non-denominated solutions. The preferred solution can be obtained by using min–max method or weighted min–max methods. Although it is possible to use the loss function in the cost model, we prefer to consider the loss function as a separate objective. The primary argument is that this loss function provides a good measure of the quality loss, but it is difficult to convert it into cost.
4. The characterization of non-dominated solutions for the multi-objective integrated product and process design model In this section we develop an algorithm to obtain the non-dominated solutions. We study the properties of the models developed and prove a number of results. Such a characterization is useful in efficiently developing the algorithm. Further, we use the min–max method to obtain trade-off solutions. The weighted min–max
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approach is used for the sensitivity analyses methods to get the optimal solution. 4.1. The characterization of non-dominated solutions In this section we provide the following results: Lemma 2. For given j, the fraction of scrap, SCðd; jÞ, is a decreasing function of tolerance d. Proof. Since FðxÞ is the cumulative distribution function of probability distribution with mean equal to 0 and standard deviation equal to 1, FðxÞ is an increasing function of x: For given j: sð jÞ is constant and Fðd=sð jÞÞ is an increasing function of d and Fðd=sð jÞÞ is a decreasing function of d. By Lemma 1, SCðd; jÞ ¼ Fðd=sð jÞÞ þ 1 Fðd=sð jÞÞ: So SCðd; jÞ is a decreasing function of d too. & Proposition 1. The sum of raw material cost (Xi ) and processing cost ( f ) is usually greater than the salvage value (Xs ), that is Xi þ f Xs > 0.
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The first derivative of T with respect to SC is qT=q SC ¼ðTsetup ð jÞ þ Tp ð jÞYo Þ=ð1 SCÞ2 > 0: So, Tðd; jÞ is the increasing function of SCðd; jÞ: By Lemma 2, SCðd; jÞ is the decreasing function of tolerance d. So, Tðd; jÞ is the decreasing function of tolerance d and attains its minimal value T ð jÞ at do because do is the upper boundary of d. & Lemma 5. For given j, the quality loss Qloss ðd; jÞ is a quadratic function of tolerance d and has unique minimal value Q ð jÞ at point d where d n ¼ A½fðd n =sð jÞÞ þ fðd n =sð jÞÞ=sð jÞ2B
ð6Þ
and the fðxÞ ¼ qFðxÞ=qx is the probability density function of the manufacturing process. sð jÞ is the standard deviation of jth manufacturing process. Proof. From the quality model, we have Qloss ðd; jÞ ¼ Qs ðd; jÞ þ Qt ðdÞ ¼ A SCðd; jÞ þ B d 2 :
Proof. From cost model (2), we have,
For given j; the Qs ðd; jÞ ¼ A SCðd; jÞ is an asymptotically decreasing function and Qt ðdÞ ¼ Bd 2 is an asymptotically increasing function of d: Therefore, the Qloss ðd; jÞ has unique minimal value. The first derivative of Qloss to d is
Xo ðd; jÞ ¼ Ki ðd; jÞ½Xi þ f ð jÞ Ks ðd; jÞXs
qQloss ðd; jÞ=qd ¼ qA SCðd; jÞ=qd þ 2Bd
Lemma 3. For given j and upon imposing Proposition 1, the cost function Xo ðd; jÞ is a decreasing function of tolerance d and attains its minimal value C ð jÞ at do .
¼ ½Xi þ f ð jÞ SCðd; jÞXs =½1 SCðd; jÞ: The first derivative of Xo with respect to SC is qXo =qSC ¼ðXs þ Xi þ f ð jÞÞ=ð1 SCðd; jÞÞ2 : By Proposition 1, qXo =qSC > 0: So, Xo ðd; jÞ is the increasing function of SCðd; jÞ: By Lemma 2, SCðd; jÞ is a decreasing function of tolerance d. So, Xo ðd; jÞ is the decreasing function of tolerance and attains its minimal value C ð jÞ at do because do is the upper boundary of d. & Remark. In some cases, we may have to pay to dispose off the scrap. Accordingly, Xs will be negative. In that case, Theorem 1 is correct even without imposing Proposition 1. Lemma 4. For given j and Yo > 0, the manufacturing lead time Tðd; jÞ is a decreasing function of tolerance d and reaches its minimal value T ð jÞ at do .
¼ qAðFðd=dð jÞÞ þ 1 Fðd=dð jÞÞÞ=qd þ 2Bd ¼ A½fðd=dð jÞÞ þ fðd=dð jÞÞ=dð jÞ þ 2Bd; where fðxÞ ¼ qFðxÞ=qx is the probability density function. fðxÞ > 0 for any x: Let qQloss ðd; jÞ=qd ¼ 0; we have d A½fðd =sð jÞÞ þ fðd =sð jÞÞ=sð jÞ2B: When dod ; we have qQloss ðd; jÞ=qdo0: When d ¼ d ; we have qQloss ðd; jÞ=qd ¼ 0: When d > d ; we have qQloss ðd; jÞ=qd > 0: Therefore, we have the unique minimal point at d : & Definition 3. The point (d 0 ; j 0 ) is a non-dominated solution if and only if, there is no d and j (dA½0; do and jAf1; 2; y; ng), which satisfies the following relations: Xo ðd 0 ; jÞ > Xo ðd; jÞ;
Proof. From the manufacturing lead time model (4), we have,
Qloss ðd 0 ; jÞ > Qloss ðd; jÞ;
Tðd; jÞ ¼ Tsetup ð jÞ þ Tp ð jÞYo =½1 SCðd; jÞ:
Tðd 0 ; jÞ > Tðd; jÞ:
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Theorem 1. For given j, the (d; j) is a non-dominated solution if and only if dA½d ; do , where d can be obtained by formula (6) and do is the upper boundary of d. Proof. For any dA½0; d Þ; by Lemmas 3–5, we have Xo ðd; jÞ > Xo ðd ; jÞ;
Table 1 Determining the feasible ranges of the Co ; Qo ; and To values j
Unit cost
Lead time
Quality loss
jc
jt
jq
C ð jc Þ Cð jt Þ Cð jq Þ
Tð jc Þ T ð jt Þ Tð jq Þ
Qð jc Þ Qð jt Þ Q ð jq Þ
Qloss ðd; jÞ > Qloss ðd ; jÞ; Tðd; jÞ > Tðd ; jÞ: By Definition 3, dA½0; d Þ is the dominated solution region for minimum fXo ðd; jÞ; Qloss ðd; jÞ; Tðd; jÞg: For any dAðdo ; NÞ; d is an infeasible solution. For any dA½d ; do ; by Lemmas 3–5 we have Xo ðd; jÞoXo ðd ; jÞ; Qloss ðd; jÞ > Qloss ðd ; jÞ;
½T ð jt Þ; maxfTð jc Þ; Tð jq Þg and that of Qo is ½Q ð jq Þ; maxfQð jt Þ; Qð jc Þg: Proof. If Co oC ð jc Þ; then there is no feasible solution because C ð jc Þ is a minimal value. However, if Co > maxfCð jt Þ; Cð jq Þg; then the feasible range will include the dominated solution. Therefore, the selection range for Co is [C ð jc Þ; maxfCð jt Þ; Cðjq Þg]. Similar results are valid for both To and Qo : &
Tðd; jÞoTðd ; jÞ: Therefore, dA½d ; do is a non-dominated solution region for minimum {Xo ðd; jÞ; Qloss ðd; jÞ; Tðd; jÞ}. & Definition 4. Let d ð jÞ ¼ fA½fðd=dð jÞÞ þ fðd=dð jÞÞ= fdð jÞ2Bgg be the lower boundary of non-dominated solution region for given j. Theorem 2. When j is a variable, all the non-dominated solutions of ðd; jÞ lie in the region S ¼f½d ð jÞ; do ; j ¼ 1; 2; y; ng.
4.3. The min–max optimal solution We can use the min–max method to obtain the min– max optimal solution of multi-objective optimization. Definition 6. We define the relative increments of Xo ; Qloss and T as Zx ¼ ðXo C ð jc ÞÞ=C ð jc Þ; Zq ¼ ðQloss Q ð jq ÞÞ=Q ð jq Þ
Proof. By Theorem 1, Sð jÞ ¼ ½d ð jÞ; do is the nondominated solution region for j: Therefore, S ¼ Sð1Þ,Sð2Þ,?,SðnÞ includes all the non-dominated solutions. &
respectively.
These results are useful in developing an efficient algorithm to obtain non-dominated solutions.
The Xo ; Qloss and T represent any non-dominated solution.
4.2. On selecting the feasible ranges of Co, Qo, and To
4.3.1. The min–max method A point (d ; j ) is min–max optimal, if for every ðd; jÞADJ; the following formula is satisfied:
We provide an approach to obtain the feasible ranges within which the values of Co ; Qo ; and To should be explored. Definition 5. Let minimal unit cost C ð jc Þ ¼ minfC
ð jÞg; the minimal manufacturing lead time T ð jt Þ ¼ minfT ð jÞg and minimal quality loss Q ð jq Þ ¼ minfQ ð jÞg; where C ð jÞ; T ð jÞ; Q ð jÞ are the minimal values when given j and the jc is the j which has the minimal unit cost C ðjc Þ; jt is the j which has the minimal lead time T ðjt Þ and jq is the j which has the minimal Quality loss Q ðjq Þ: These results are represented in Table 1. Remark. From Table 1, we can see that the range of Co is ½C ð jc Þ; maxfCð jt Þ; Cð jq Þg; the range of To is
and Zt ¼ ðT T ð jt ÞÞ=T ð jt Þ;
Zðd n ; j n Þ ¼ min fmaxfZx ; Zq ; Zt gg; ðd; jÞADJ
ð7Þ
where DJ is a set of non-dominated solutions obtained from the multi-objective model (4). 4.3.2. The weighted min–max method A point (d ; j ) is weighted min–max optimal, if for every ðd; jÞADJ; the following formula is satisfied: Zðd n ; j n Þ ¼ min fmaxfWx Zx ; Wq Zq ; Wt Zt gg; ðd; jÞADJ
ð8Þ
where DJ is a set of the non-dominated solutions obtained from model (4 ) and Wx ; Wq and Wt are the weights attached to the unit cost, quality loss and lead time, respectively. These weights reflect the aspirations
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of the decision makers which essentially represent the competitive environment that may exist at the time. 4.4. Algorithms Using the results from the previous section, we develop the two algorithms: Algorithm I to obtain the non-dominated solutions and Algorithm II to obtain the trade-off solutions from the non-dominated solutions generated by Algorithm I using the min–max as well as the weighted min–max approaches. 4.4.1. Algorithm I to obtain the non-dominated solutions From Theorem 2, we know that all the nondominated solutions lie in the region S ¼ f½d ð jÞ; do ; j ¼ 1; 2; y; ng: However, when considering all the j simultaneously, there may be some dominated solution. We now present the algorithm for obtaining the nondominated solutions as follows. We search the nondominated solutions only in region S ¼ f½d ð jÞ; do ; 0 j ¼ 1; 2; y; ng by a certain increment in d: We let S denote the set of non-dominated solutions. We place first solution ( j ¼ 1; d ¼ d ð1Þ) into S 0 ; then compare the other solutions in S one by one. If the new solution is dominated, then it is not added to S0 : If the new solution is non-dominated in S 0 ; then it is added to S0 and it eliminates all solutions in S 0 which are dominated by the new solution. Finally, we get the non-dominated solution set S 0 : Step 1: Determine d ð jÞ; 8jð j ¼ 1; 2; y; nÞ using Eq. (6). Step 2: Specify the step size Dd ¼ ðdo min fd ð jÞgÞ=m; m is the number of steps desired. The maximum possible number of non-dominated solutions are m n: Step 3: Set j ¼ 1; d ¼ d ð jÞ; ko ¼ 1; Xo ð1Þ ¼ Qloss ð1Þ ¼ Tð1Þ ¼ M; M is a very big number. Step 4: Evaluate the objective function vector (Xo ðd; jÞ; Qloss ðd; jÞ; Tðd; jÞ); and let k ¼ 1: Step 5: If Xo ðd; jÞ > Xo ðkÞ and Qloss ðd; jÞ > Qloss ðkÞ and Tðd; jÞ > TðkÞ; goto step 8. If Xo ðd; jÞoXo ðkÞ and Qloss ðd; jÞoQloss ðkÞ and Tðd; jÞoTðkÞ; goto step 7 otherwise k ¼ k þ 1. If kpko ; goto step 5. Step 6: Place the new non-dominated solution in the set S 0 : ko ¼ ko þ 1; Xo ðko Þ’Xo ðd; jÞ; Qloss ðko Þ’Qloss ðd; jÞ; Tðko Þ’Tðd; jÞ; Pd ðko Þ’d; Pj ðko Þ’j; goto step 8. Step 7: Substitute the dominated solution with the new point. Let Xo ðkÞ’Xo ðd; jÞ; Qloss ðkÞ’Qloss ðd; jÞ; TðkÞ’ Tðd; jÞ; Pd ðkÞ’d; Pj ðkÞ’j. Step 8: Let k0 ’ko . Step 9: If Xo ðd; jÞoXo ðk0 Þ and Qloss ðd; jÞoQloss ðk0 Þ and Tðd; jÞoTðk0 Þ; goto step 10 otherwise k0 ¼ k0 1. If k0 > k; goto step 9; otherwise goto step 11.
163
Step 10: Let Xo ðk0 Þ’Xo ðko Þ; Qloss ðk0 Þ’Qloss ðko Þ; Tðk0 Þ’Tðko Þ; Pd ðk0 Þ’Pd ðko Þ; Pj ðk0 Þ’Pj ðko Þ; ko ¼ ko 1. Step 11: If dpdo ; then d ¼ d þ Dd; goto step 4; Otherwise j ¼ j þ 1; d ¼ d ð jÞ: If jpn; goto step 4; otherwise end. The set S0 now contains only non-dominated solutions. This algorithm only needs the computer’s memory for S 0 which may be much less than S: 4.4.2. Algorithm II for obtaining the min–max solutions Most often there is a need to know trade-off solutions that reflect the aspirations of the decision makers. In this paper we adopt the min–max and weighted min–max methods to get trade-off solutions. The following algorithm developed in this section obtains the min– max solutions. Step 1: Calculate the relative increment Zx ; Zq ; and Zt ; 8ðd; jÞAS 0 : Step 2: If using weighted min–max method, then multiply the weight to the relative increments: Wx Zx ðd; jÞ; Wq Zq ðd; jÞ; Wt Zt ðd; jÞ; where Wxþ Wq þ Wt ¼ 1 Step 3: For each ðd; jÞAS 0 ; calculate Zðd; jÞ ¼ maxfWx Zx ðd; jÞ; Wq Zq ðd; jÞ; Wt Zt ðd; jÞg: Step 4: 8ðd; jÞAS 0 ; calculate Zðd ; j Þ ¼ minfZðd; jÞg: ðd; jÞAS0 : Then (d ; j ) is min–max or weighted min–max optimal solution. Step 5: If the sensitive analysis of weight is needed, then change the weights Wx ; Wq and Wt and goto step 2; otherwise stop.
5. Integrated product and process design with multiple design variables When a part has more than one design variable, we need to select a vector variable D ¼ ðd1 ; d2 ; y; dm Þ to let {Xo ð j; DÞ; Qloss ð j; DÞ; Tð j; DÞ} to be minimum. In this case, the scrap rate SCð j; DÞ will be different from that of a tolerance variable case. The scrap rate is the probability of a product being scrapped. When the design variables are independent from each other, SCð j; DÞ will be equal to the product of SCð j; dl Þ: Therefore, we have m Y SCð j; DÞ ¼ 1 ½1 SCð j; dlÞ: l¼1
Then Ki ¼ Yi =Yo ¼ 1=½1 SCð j; DÞ; Ks ¼ Ys =Yo ¼ SCð j; DÞ=½1 SCð j; DÞ; Xo ð j; DÞ ¼ Ki ð j; DÞ½Xi þ f ð jÞ Ks ð j; DÞXs ;
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Qloss ð j; DÞ ¼ A SCð j; DÞ þ
m X
Bl dl2 ;
Xo ð j; Dn Þ ¼
l¼1
¼
Tð j; DÞ ¼ Tsetup ð jÞ þ Tp ð jÞYo =½1 SCð j; DÞ:
i;D
Qloss ð j; Dn Þ ¼
ð9Þ ¼
subject to
Xo ð j; Dk Þ fKi ð j; Dk Þ½Xi þ f ð jÞ Ks ð j; Dk ÞXs g;
p X k¼1 p X
Qloss ð j; Dk Þ fA SCð j; Dk Þ þ
k¼1
Xo ð j; DÞpCo ; Tð j; DÞpTo ;
Tð j; Dn Þ ¼
Qloss ð j; DÞpQo ; j ¼ 1; 2; :::; n; D ¼ ðd1 ; d2 ; :::; dm Þ; 0pdl pdlo ;
k¼1 p X k¼1
Accordingly, the multi-objective programming model will be as follows: Minimum fXo ð j; DÞ; Qloss ð j; DÞ; Tð j; DÞg
p X
l ¼ 1; 2; :::; m;
where Co ; To ; Qo ; dlo are customer requirements. n is the number of alternative manufacturing processes. m is the number of tolerance variables.
¼
p X k¼1 p X
m X
Bl dl2 g;
l¼1
Tð j; Dk Þ fTsetup ð jÞ þ Tp ðjÞYo =½1 SCð j; Dk Þg:
k¼1
Accordingly, the multi-objective programming model for the assembly will be as follows: Minimum fXo ð j; Dn Þ; Qloss ð j; Dn Þ; Tð j; Dn Þg; n j;D
ð10Þ
subject to FðDn ÞoF; 6. Integrated product and process design for product assemblies
Xo ð j; Dn ÞpCo ; Tð j; Dn ÞpTo ;
When a product consists of p parts and the kth part has mk tolerances (k ¼ 1; 2; y; p), we have to decide m1 þ m2 þ ? þ mp tolerances. Therefore, we need to select all dkl (kth part and lth tolerance, l ¼ 1; 2; y; mk ; k ¼ 1; 2; y; p) to let fXo ð j; D Þ; Qloss ð j; D Þ; Tð j; D Þg be minimum. We have the assembly function F ¼ FðD Þ: The function F is the tolerance vector of product and D is the tolerance metrices of parts. Therefore, F represents a group of assembly function formulas. For example,
Qloss ð j; Dn ÞpQo ; j ¼ 1; 2; :::; n; Dn ¼ ðD1 ; D2 ; :::Dl Þ;
d1l þ d2l þ ? þ dpl ¼ Fl ;
mðkÞ Y
oodkj odkjo ;
j ¼ 1; 2; y; mk ; k ¼ 1; 2; y; l
where Fo ; Co ; To ; SCo ; and dkjo are requirements. n is the number of alternative manufacturing processes. mðkÞ or mk is the number of tolerance variables of kth part. l is the number of parts.
l ¼ 1; 2; y; m:
The assembly function is the requirement that has to be satisfied. We, therefore, add the assembly function to the constraints. When the design variables are independent from each other, SCð j; Dk Þ will be equal to the product of SCð j; dkl Þ: Therefore, we have
SCð j; Dk Þ ¼ 1
Dk ¼ ðdk1 ; dk2 ; :::; dk mðkÞÞ;
½1 SCð j; dkl Þ:
l¼1
Then Ki ð j; Dk Þ ¼ Yi =Yo ¼ 1=½1 SCð j; Dk Þ; Ks ð j; DkÞ ¼ Ys =Yo ¼ SCð j; DÞ=½1 SCð j; DkÞ;
7. The illustrative examples 7.1. Example 1 ABC company plans to produce 1000 units of a cylindrical part. The company wants to know which tolerance is the optimal considering simultaneously the objecives of minimum unit cost, quality loss and lead time. And what machine should be selected to produce these parts? The company has the following information: unit cost of raw material Xi ¼ $10:00; unit salvage valueðsell scrapped productÞXs ¼ $2:00: There are three kinds of machine-tools that can be used to produce the parts. The distribution of dimensions is normally distributed Nð0; sð jÞÞ (Table 2).
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N. Singh / Robotics and Computer Integrated Manufacturing 18 (2002) 157–168 Table 2 Data for example 1 j
Machine
Standard deviation sð jÞ (in)
Unit process cost f ð jÞ
Setup time Tsetup ð jÞ (h)
Unit process time Tp ð jÞ (h)
1 2 3
Turret Lathe Engine Lathe NC Machine
0.003 0.001 0.0005
$3 $10 $15
2 3 4
0.2 0.17 0.15
The loss function parameters A is 100 and B is 100/ 0.0032 (0.003 is maximal acceptable do ). Hundred is a scale value. From marketing information, the company knows the unit cost Xo should be less than $28, the lead time T should be less than 400 h, the tolerance d should be less than 0.003 in and the quality loss Qloss should be less than 140 to obtain non-dominated solution and min–max optimal solution. Solution. To obtain non-dominated solution and min– max optimal solution, we follow the steps of Algorithms I and II. Further, we know that do ¼ 0:003; and FðxÞ is standard normal distribution. Algorithm I: Step 1: Determine d ð jÞ; 8jð j ¼ 1; 2; 3Þ using Eq. (6). d ¼ A ½fðd =ð jÞÞ þ fðd =sð jÞÞ=sð jÞ 2B ¼ 100 ½fðd =sð jÞÞ þ fðd =sð jÞÞ=ðsð jÞ 200=0:0032 Þ when j ¼ 1; d ¼ fðd =0:003Þ0:003 ¼> d ð1Þ ¼ 0:0011: By the same approach, get d ð2Þ ¼ 0:0014: d ð3Þ ¼ 0:001: Step 2: Specify the step size Dd ¼ ðdo min fd ð jÞgÞ=m; m is the number of steps desired. Let m ¼ 100; Dd ¼ ð0:003 minf0:0011; 0:0014; 0:001gÞ=100 ¼ 2 105 : Step 3: Set j ¼ 1; d ¼ d ðjÞ; ko ¼ 1; Xo ð1Þ ¼ Qloss ð1Þ ¼ Tð1Þ ¼ M; M is a very big number. Let M ¼ 109 Step 4: Evaluate the objective function vector (Xo ðd; j), Qloss ðd; jÞ; Tðd; jÞÞ; and let k ¼ 1: SCð0:0011; 1Þ ¼ 0:714; Xo ð0:0011; 1Þ ¼ ðXi þ f ð1Þ SCð0:0011; 1Þ Xs Þ=ð1 SCð0:0011; 1ÞÞ ¼ ð2 þ 10 2 0:714Þ=ð1 0:714Þ ¼ 36:96; Tð0:0011; 1Þ ¼ Tsetup ð1Þ þ Tp ð1Þ Yo =½1 SCð0:0011; 1Þ ¼ 2 þ 0:2 1000=ð1 0:714Þ ¼ 701; Qð0:0011; 1Þ ¼ 100 SCð0:0011; 1Þ þ ð100=0:0032 Þ
0:00112 : Step 5: Since Xo ðd; jÞ oXo ðkÞ and Qloss ðd; jÞoQloss ðkÞ and Tðd; jÞoTðkÞ; goto step 7. Step 7: Substitute the dominated solution with the new point.
Let Xo ð1Þ’Xo ð0:0011; 1Þ; Qloss ð1Þ’Qloss ð0:0011; 1Þ; Tð1Þ’Tð0:0011; 1Þ; Pd ð1Þ’d ¼ 0:0011; Pj ð1Þ’j ¼ 1. Step 8: Let k0 ¼ ko ¼ 1. Step 9: Since k0 not > k; goto step 11. Step 11: Since dpdo ; then d ¼ d þ Dd; goto step 4. y y when d > do and j ¼ 3; end. Set S 0 now contains only non-dominated solutions. Then, we have the non-dominated solution ðd; jÞAS 0 ¼ fdA½0:0026; 0:003; j ¼ 1g,fdA½0:0014; 0:003; j ¼ 2g, fdA½0:001; 0:003; j ¼ 3g: Algorithm II for getting the min–max optimal solution: Step 1: Calculate the relative increments Zx ; Zq ; and Zt ; 8ðd; jÞAS 0 as Table 3. Step 2: If using weighted min–max method, then multiply the weight to the relative increments: Wx Zx ðd; jÞ; Wq Zq ðd; jÞ; Wt Zt ðd; jÞ; where Wxþ Wq þ Wt ¼ 1: The different weight are shown in Table 4 and the result is shown in Table 5. Step 3: For each ðd; jÞAS 0 ; calculate Zðd; jÞ ¼ maxfWx Zx ðd; jÞ; Wq Zq ðd; jÞ; Wt Zt ðd; jÞg: The result is shown in Table 3. Step 4: 8ðd; jÞAS0 ; calculate the Zðd ; j Þ ¼ min fZðd; jÞg ¼ 0:42 and d ¼ 0:0014; j ¼ 3: ðd; jÞAS0 : The ðd ; j Þ is min–max or weighted min–max optimal solution. Step 5: If the sensitive analysis of weight is needed, then change the weights Wx ; Wq and Wt and goto step 2; otherwise stop. From Table 3, the min–max optimal value Zðd ; j Þ ¼ minfmaxfZx ; Zq ; Zt gg ¼ 0:42; and the min–max optimal solution is ð j ; d Þ ¼ ð3; 0:0014Þ: A sensitivity analysis can be carried out attaching different weights to the objectives. Then the weighted min–max method can be used. Three cases are discussed in Table 4. Then we have the weighted min–max solution in Table 5. From Table 5, we can see the different weighted min– max solutions when the weights of cost, lead time and quality loss are different. These solutions will help
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Table 3 Eleven non-dominated solutions j; d
Cost
Time
Quality
Zx
Zt
Zq
MaxZx ; Zq ; Zt
1, 1, 2, 2, 2, 2, 2, 3, 3, 3, 3,
19.9 18.1* 23.5 21.4 20.5 20.2 20.0 26.1 25.1 25.0 25.0
348 315 226 206 199 195 194 181 175 174* 174*
113.7 131.7 37.9 43.2 56.6 76.0 100.3 15.7* 22.3 36.0 44.5
0.10 0* 0.30 0.18 0.13 0.12 0.10 0.44 0.39 0.38 0.38
1.00 0.81 0.3 0.18 0.14 0.12 0.11 0.04 0.01 0* 0*
6.24 7.39 1.41 1.75 2.61 3.84 5.38 0* 0.42 1.29 2.43
6.24 7.39 1.41 1.75 2.61 3.84 5.38 0.44 0.42* 1.29 2.43
0.0026 0.0030 0.0014 0.0018 0.0022 0.0026 0.0030 0.0010 0.0014 0.0018 0.0022
Table 4 Different weights for the min–max solution
Table 5 The weighted min–max solutions
Case
Wx (cost)
Wt (time)
Wq (quality)
1 2 3
0.7 0.2 0.1
0.2 0.1 0.7
0.1 0.7 0.2
select the design and process decision variables reflecting the aspirations of the decision makers. Essentially, the weighted min–max approach provides a flexible scheme to obtain the solutions that represent the different kinds of decision makers. 7.2. Example 2 In this example, we assume that there are tolerances for two-dimensions. There are three kinds of machinetools that can be used. The variates of two-dimensions are Normally distributed Nð0; sð1; jÞÞ and Nð0; sð2; jÞÞ: The data are same as in example 1 except the following Table 6. Solution. We know that do ¼ 0:003; and FðxÞ is a standard normal distribution. Obtain the minimal unit cost C ð jÞ; lead time T ð jÞ and quality loss Q ð jÞ for three objectives separately for j ¼ 1; 2, 3. SCðDo ; jÞ ¼ 1 ð1 SCðd1o ; jÞÞð1 SCðd2o ; jÞÞ;
Case
Weighted min–max value
j ; d
Unit cost
Lead time
Quality loss
1 2 3
0.175 0.088 0.044
2, 0.0018 3, 0.0010 3, 0.0010
21.4* 26.1 26.1
206 181* 181*
43.2 15.7* 15.7*
Calculate Q ð jÞ ¼ 100 SCðd ð jÞ; jÞ þ ð100=:052Þd ð jÞ2 then get Q ð1Þ ¼ 101; d ð1Þ ¼ 0:0005; Q ð2Þ ¼ 60:9; d ð2Þ ¼ 0:0018; Q ð3Þ ¼ 30:7; d ð3Þ ¼ 0:0014: (2) Obtain the minimal unit cost C ð jc Þ; lead time T ð jt Þ and Quality loss Q ð jq Þ for three objectives seperately. Accordingly C ð jc Þ ¼ minfC ð jÞg ¼ 22:51 and jc ¼ 2; dc ¼ do ¼ 0:003: T ð jt Þ ¼ minfT ð jÞg ¼ 174
SCðDo ; 1Þ ¼ 0:534; SCðdo ; 2Þ ¼ 0:025; SCðdo ; 3Þ ¼ 0:000036: Calculate C ð jÞ¼ðXi þf ð jÞSCðDo ; jÞXs Þ=ð1SCðDo ; jÞÞ then get C ð1Þ ¼ 29:89; C ð2Þ ¼ 22:51; C ð3Þ ¼ 28:00: Calculate T ð jÞ ¼ Tsetup ð jÞ þ Tp ð jÞYo =½1 SCðdo ; jÞ then get T ð1Þ ¼ 665; T ð2Þ ¼ 228; T ð3Þ ¼ 174:
and jt ¼ 3; dt ¼ do ¼ 0:003: Q ð jq Þ ¼ minfQ ð jÞg ¼ 30:71 and jq ¼ 3; dq ¼ d ð3Þ ¼ 0:0014: (3) Get the non-dominated solution region DJ: By Theorems 1 and 2, we can search the nondominated solution in region f½d ðjÞ; do ; j ¼ 1; 2; 3g ¼ fdA½0:0005; 0:003; j ¼ 1g,fdA½0:0018; 0:003; j ¼ 2g
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N. Singh / Robotics and Computer Integrated Manufacturing 18 (2002) 157–168 Table 6 Data for example 2 j
Machine
Standard deviation sð1; jÞ (in)
Standard deviation sð2; jÞ (in)
Unit process cost f ð jÞ
Unit process time Tpð jÞ (h)
1 2 3
Turret Lathe Engine Lathe NC Machine
0.003 0.0012 0.0007
0.0015 0.0002 0.0002
$5 $12 $18
0.3 0.2 0.15
Table 7 Min–max optimal solution for different component tolerances
d1 (part 1) j (machine) d2 (part 2) j (machine) Unit cost Lead time Quality loss a
d1 p0:003; d2 p0:003:a
d1 þ d2 p0:003
d1 þ d2 p0:002
d1 þ d2 p0:005
0.0016 3 0.0012 2 52.4 379 55
0.0015 3 0.0013 3 57 310 48
0.0011 3 0.0009 3 61.4 335 41
0.0015 3 0.0015 3 56.9 309 53
d1 and d2 are independent of each other.
,fdA½0:0014; 0:003; j ¼ 3g: For any fdA½0:0005; 0:003; j ¼ 1g we find that the solution is dominated. And in region fdA½0:0018; 0:003; j ¼ 2g,fdA ½0:0014; 0:003; j ¼ 3g every solution is nondominated solution. Then we have the non-dominated solution ðd; jÞADJ ¼ fdA½0:0018; 0:003; j ¼ 2g,fdA ½0:0014; 0:003; j ¼ 3g: (4) Get the min–max optimal solution Using the same approach as in example 1, we can get the min–max value Zðd ; j Þ ¼ minfmaxfZx ; Zq ; Zt gg ¼ 0:27; and the min–max optimal solution is ð j ; d Þ ¼ ð3; 0:0014Þ: The unit cost is 28.5. The lead time is 177. The quality loss is 38. 7.3. Example 3 In this case we assume that two tolerances d1 and d2 are for two different parts which are assembled together into a product. Further, we assume that the product tolerance is fixed at 0.003 in. The relation of product tolerance and parts tolerance is d1 þ d2 p0:003: We need to select the parts tolerances and the machine-tools for the three objectives (minimal unit cost, lead time and quality loss). Solution. (1) Obtain the minimal unit cost C ð jc Þ; lead time T ð jt Þ and Quality loss Q ð jq Þ for three objectives seperately. Accordingly C ð jc Þ ¼ minfC ð jÞg ¼ 49:06 and jc ¼ 2; d1 c ¼ 0:0017; d2 c ¼ 0:0013: T ð jt Þ ¼ minfT ð jÞg ¼ 307:72
and jt ¼ 3; d1 t ¼ 0:0011; d2 t ¼ 0:0009: Q ð jq Þ ¼ minfQ ð jÞg ¼ 41:23 and jq ¼ 3; d1 q ¼ 0:0011; d2 q ¼ 0:0009: (2) Get the min–max optimal solution. Using the same approach as in example 1, we can get the min–max value Zðd1 ; d2 ; j Þ ¼ minfmaxfZx ; Zq ; Zt gg ¼ 0:16393; and the min–max optimal solution is ð j ; d1 ; d2 Þ ¼ ð3; 0:0015; 0:0013Þ: The unit cost is 57.11. The lead time is 310. The quality loss is 47.92. (3) Discussion: The following Table 7 gives the min–max optimal solutions at different relationships between the tolerance d1 of part 1 and tolerance d2 of part 2. From Table 7, we can see that the assembly relations affect the min–max optimal solution. The tight assembly tolerance (d1 þ d2 p0:002) gives a worse solution. The loose assembly tolerance (d1 p0:003; d2 p0:003:) gives a better solution. Therefore, the assembly relations should be carefully considered in practical situations.
8. Conclusions In this paper we presented a multi-objective framework for the concurrent tolerance design. The model integrates the cost, quality, and lead time into the design process. The multi-objective programming methodology presented in this paper can be applied to help make decisions on selecting the product tolerances as well as
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the manufacturing processes. The modeling approach is quite comprehensive and realistic since competitive marketing information on cost, quality and lead time is built into the models.
References [1] Singh N. Concurrent engineering in systems approach to computer-integrated design and manufacturing. New York: Wiley Inc., 1996 (Chapter 4). [2] Davido WH, Malone MS. The virtual corporation: structuring and revitalizing the corporation of the 21st century. Harper Business 1992.
[3] Goldman S, Preiss K, editors. 21st Century Manufacturing Enterprise Strategy: An Industry-Led View, Iacocca Institute, Lehigh University, Harold S. Mohler Laboratory # 200, Bethlehem, PA, 1991. [4] Zhang HC, Huq ME. Tolerancing techniques: the state of the art. Int. J. Prod. Res. 1992;30(9):2111–35. [5] DeVour RE, Hunter RG, Sutherland JW. A methodology for robust design using models for the mean, variance, and loss. Proceedings of the Symposium on Quality and Performance, ASME Bound Volume, December 1989. [6] Kapur K. Quality engineering, tolerance design. In: A. Kusiak (Ed.) John Wiley & Sons Inc. 1992. pp 287–306. Concurrent Engineering: Automation, Tools & Techniques.
Integrated product and process design: a multi-objective modeling framework Nanua Singh* Department of Industrial and Manufacturing Engineering, Integrated Product Development Lab, Wayne State University, Detroit, MI 48202, USA Received 1 August 2000; received in revised form 1 January 2001; accepted 1 August 2001
Abstract In this paper we present an analytical multi-objective framework for the concurrent design of product and processes. The objective is to simultaneously consider the tolerance specification on the product or the component dimensions along with the selection of the manufacturing processes. For this purpose we consider three objectives: to minimize unit cost, to minimize quality loss and to minimize manufacturing lead time. We characterize the properties of the non-dominated solutions. These solutions provide flexibility needed in an agile manufacturing environment. The min–max approach is used to obtain trade-off solutions. r 2002 Published by Elsevier Science Ltd. Keywords: Design for manufacturing system; Tolerance design; Concurrent engineering; Quality engineering; Concurrent product and process design
1. Introduction The twenty-first century business environment can be characterized by expanding global competition, and increasing variety/low in demand for products. The globalization of economic activity has brought about a major change in the attitudes of the customers. Customer individualism is certain to become the central theme of business. What we are going to witness is an era of mass customization. It means manufacturing products for the mass market in such a way that products are customized for each individual in that market. In the seventies, the cost of products used to be the main lever for obtaining competitive advantage. Later in eighties, the quality superseded cost and became an important competitive dimension. Now low unit cost and high quality of products no longer solely define the competitive advantage of manufacturing enterprises. Today, the customer takes both minimum cost and high quality for granted. Factors such as delivery performance, customization and environmental issues such as waste generation are assuming a predominant role in defining the success of organizations in terms of *Tel.: +1-313-577-7586; fax: +1-313-577-8833. E-mail address: [email protected] (N. Singh). 0736-5845/02/$ - see front matter r 2002 Published by Elsevier Science Ltd. PII: S 0 7 3 6 - 5 8 4 5 ( 0 1 ) 0 0 0 3 0 - 8
increased market share and profitability [1]. The question is: what can we do under these changing circumstances to stay in business and retain competitive advantages? As a first step what is needed is the development of the right business strategy to meet the challenges of the present and future market. In doing so, a manufacturing organization not only has to understand what customers want, but also has to develop internal mechanisms to instantly respond to the changes in the products demanded by customer wants. This requires a paradigm shift in everything our factories do. That means, not only do they have to make use of the state-of-the-art technologies and concepts in building products, but also think in the reverse direction [2]. Reverse direction means building products that realize customer expectations. That is, when an organization is deciding about business plans, it has to address such questions as the following: Will the customer find any change in what one does as a result of using this? Will the customer be able to define any benefit? From a customer point of view, a company has to respond to smaller and smaller market niches very quickly with standardized products that will get built in lower and lower volume. In other words, we can say that a future successful manufacturing organization would be a virtual corporation that is
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instantaneously responsive to customer needs. This view is shared by industry-led consortium on twenty-first century manufacturing enterprise strategy [3]. The next step is to determine the right kind of resources to support the business strategy. This requires the right choice of people, technology and business processes. The products are designed to satisfy certain functional requirements. The primary purpose of tolerancing is to help achieve the stated functional requirements. The tolerance design is the process of setting up acceptance regions for the design variables meeting certain criteria. In a competitive environment, these criteria include minimum cost, high quality and minimum lead time. The tolerances are the link between design and manufacturing. It is therefore important that we exploit the interaction between design and manufacturing by jointly considering the selection of processes as well as product tolerances. The tolerance design problem has a strong relationship with product cost, quality and lead time. Inappropriate tolerance may either reduce the quality or generate more scrap thereby increasing the product cost and lead time. A number of papers have appeared in the area of tolerancing. An exhaustive literature review on tolerancing is given in Zhang and Huq [4]. Most of the papers have considered cost minimization as the criterion for the optimal specification of the tolerances. What is however required is the consideration of other equally important issues such as quality and manufacturing lead time. Since the tolerances are the link between the design and manufacturing, it is also equally important to select the manufacturing processes along with the tolerance allocation. Singh [1] introduced the concepts of unit cost, quality and lead time to analyze the integrated product and process design problem in a concurrent engineering environment. DeVour et al. [5] provide a methodology for the robust design using models for the mean, variance and loss. Kapur [6] provides a general understanding of the taguchi loss function approach. However, no formal treatment to simultaneously determine the optimal tolerances as well as the manufacturing processes has been considered. In view of the requirements stated above, the primary objective of our work is to provide a paradigm shift by concurrently attempting the product and process design and jointly considering the criteria of minimum cost, maximum quality and minimum manufacturing lead time. In this paper we therefore provide a generalized multi-objective framework for the integrated product and process design problem. Generalized models for the following cases are presented: * * *
Single component single design variable case. Single component multiple design variables case. The assembled product Case.
This multiobjective framework for product and process design by considering unit cost, quality and lead time simultaneously can be used as a flexible decision support system tool to support a mass customization business strategy.
2. The mathematical model for integrated product and process design In sequential engineering approach, the product design group is responsible for designing the product and consequently determining the tolerances. The process design group then selects the machine(s) which produce(s) the product. The manufacturing engineering department will manufacture the product. The product(s) not meeting the tolerances are scrapped. The good quality products are shipped to the customers. The information on unit cost, scrap rates generated and the lead time are fed back to the product design group which may possibly modify the design and adjust the tolerances. This process of feedback normally takes a long period to modify the product design and adjust the tolerances to satisfy the needs of the customers. This delay may have a number of repercussions: For example, the process may either produce low quality products with very large tolerances or the manufacturing cost may be too high due to the wrong selection of very expensive machines or may result in the scrapping of too many products with very small tolerances. Small adjustments may be difficult to make at this stage since manufacturing processes have already been selected. Radical changes such as changing the product design or the manufacturing processes may turn out to be a very expensive proposition. The solution to these problems lies in adopting what is known as concurrent engineering approach. The basic idea in concurrent engineering approach is to integrate the manufacturability, cost, quality, lead time and marketing information with the design process. In the present paper we develop a modeling methodology to obtain the optimal tolerances as well as manufacturing process considering unit cost, quality and lead time. Since this modeling approach considers various design and manufacturing factors at the beginning stage of the design cycle, it eliminates the need for design changes and adjustments. Further, the trade-off among quality, cost and lead time can be evaluated according to the marketing information representing the needs of the customers. There is a major paradigm shift in decision making in the area of design and manufacturing of products and processes due to global competition. This has led to a shift from decisions based solely on the cost to decisions
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jointly based on the cost, quality and lead time considerations. Therefore, there is a need to develop such models to jointly design the products and processes considering cost, quality and lead time information. Accordingly, in this section we present three models: the unit cost of production model, the quality model and the lead time model. These models are then used for selecting the optimal tolerance and machine-tools using a multi-objective decision making approach.
Therefore, SCðd1 ; d2 ; jÞ ¼ FðZ l Þ þ 1 FðZu Þ ¼ F½ðT mðjÞ d1 Þ=sð jÞ þ 1 F½ðT mð jÞ þ d2 Þ=sð jÞ:
&
Note: In most of the cases, it is desired to have m ¼ T and d1 ¼ d2 ¼ d: Then, the fraction of scrap is given by
2.1. The cost model
SCðd; jÞ ¼ F½d=sð jÞ þ 1 F½d=sð jÞ:
In the development of the models, we provide the following preliminaries and notation. The manufacturing process requires the use of Yi units of raw material to produce Yo units of finished products. Ys units are scrapped in the process which are beyond the tolerance limits.
For simplicity, we assume m ¼ T and d1 ¼ d2 ¼ d in the rest of this paper. From the relationships among input, output and scrap we have
Definition 1. The tolerance limit of the components/ product is (nominal value Ttolerance d1 ; nominal value T+tolerance d2 ). The components/products having dimensions beyond these limits are scrapped.
Ys ¼ Yi SCðd; jÞ ¼ Yo SCðd; jÞ=½1 SCðd; jÞ:
Definition 2. The quality/design characteristic of the product produced by jth manufacturing process (for example, jth machine) is characterized by a probability distribution with mean mð jÞ and standard deviation sð jÞ:
Yo ¼ Yi Ys ¼ Yi SCðd; jÞYi ¼ Yi ½1 SCðd; jÞ; and We assume that the unit cost consists of the cost of raw material, the cost of manufacturing (include setup cost and labor cost), and the salvage value (income of selling scraped product). Accordingly, the unit cost model can be shown to be Xo ðd; jÞ ¼ Ki ðd; jÞ½Xi þ f ð jÞ Ks ðd; jÞXs ;
ð2Þ
Lemma 1. The fraction of scrap (SC) of product is the function of d1 , d2 and j as follows.
where j is the jth manufacturing process selected for producing the product, Xo ðd; jÞ the unit cost when using jth process with tolerance d; Xi the unit raw material cost, f ð jÞ the unit processing cost for jth process, Xs the unit salvage value. Ki ¼ Yi =Yo ¼ 1=½1 SCðd; jÞ; Ks ¼ Ys =Yo ¼ SCðd; jÞ=½1 SCðd; jÞ; SCðd; jÞ is the scrap rate given in formula (1), Yo the amount of output (required amount), Yi the amount of raw material required to produce the desired, Yo ; Ys the amount of scrapped product. From the cost equation we observe that the total unit cost is a function of d and j: The objective is to have minimal possible unit cost.
SCðd1 ; d2 ; jÞ ¼ F½ðT mð jÞ d1 Þ=sð jÞ
2.2. The quality model
The manufacturing process variables are process mean and standard deviation whereas tolerance d are design variables. In a cost minimization model, the objective is to select the process and design variables such that the cost function is minimized.
þ 1 F½ðT mð jÞ þ d2 Þ=sð jÞ;
ð1Þ
where FðxÞ is cumulative distribution function (c.d.f.) of probability distribution with mean equal to 0 and standard deviation equal to 1.
Proof. Let Z u ¼ ðT m þ d2 Þ=s; Z l ¼ ðT m d1 Þ=s: The fraction of scrap is given by definition as SCðd; jÞ ¼ Ys =Yi :
In developing a quality model we consider two issues. One is the fact that the more the number of defects, the worse is the perception of customer regarding the quality production capability of the company. Accordingly, there is a perceived loss because of the shear volume of defects. Another quality factor is the loss to the society which is described by the well-known Taguchi loss function. In developing the quality model, we consider both the issues together as follows: Qloss ðd; jÞ ¼ Qs ðd; jÞ þ Qt ðdÞ ¼ A SCðd; jÞ þ B d 2 :
ð3Þ
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Using Taguchi’s Loss function concept, Qt ðdÞ ¼ Bd 2 ; where B is the parameter and d is the tolerance. The loss Qs ðd; jÞ is due to the scrap rate. The high scrap rate will damage the credibility of the company on quality. Therefore, we consider Qs ðd; jÞ ¼ A SCðd; jÞ; where A is the parameter and SC is the scrap rate for given tolerance d and jth machine. SCðd; jÞ; j and d have the same meanings in cost model and A and B are constant parameters. The objective is to minimize the value of Qloss ðd; jÞ: 2.3. The lead time model The response time to deliver the products to the customers has added an important dimension to the competitiveness. The manufacturing lead time represents a significant portion of the total response time. In this section we develop a manufacturing lead time model as a function of manufacturing and design decision variables as follows: Assuming different manufacturing processes having different setup time Tsetup ð jÞ and different unit processing time Tp ð jÞ the manufacturing time on jth machine equals Tsetup ð jÞ plus Tp ð jÞ times the amounts of raw material Yi processed to manufacture the desired output Yo : Accordingly, the total manufacturing lead time to manufacture a lot is Tðd; jÞ ¼ Tsetup ð jÞ þ Tp ð jÞYi ¼ Tsetup ð jÞ þ Tp ð jÞYo =½1 SCðd; jÞ;
the functions of design and manufacturing decision variables. Usually, from the marketing information as well as from the customer requirements we are aware of the cost and tolerance of the competitors’ product and customers’ requirement for lead time being less than To ; tolerance less than do and cost less than Co : The company may want the Quality loss to be less than Qo : These requirements can be considered as the constraints. The feasible ranges within which To ; Co and Qo should be selected are discussed in Section 4.2. We consider Xo ðd; jÞ; Qloss ðd; jÞ and Tðd; jÞ as three objectives to be minimized. The customer requirements are considered as the constraints. The three objectives are conflicting in nature. Therefore, there is a need to obtain trade-off solutions. The multi-objective integrated product and process design problem can be stated as follows: Minimum fXo ðd; jÞ; Qloss ðd; jÞ; Tðd; jÞg d;j
ð5Þ
subject to Xo ðd; jÞpCo ; Qloss ðd; jÞpQo ; Tðd; jÞpTo ; 0pdpdo j ¼ 1; 2; :::; n;
ð4Þ
where Yi ; Yo ; SCðd; jÞ; j and d have the same meanings as defined in the cost model. Usually Yo is known and Yi is a function of Yo ; j and d. The objective is to have minimum possible lead time Tðd; jÞ:
3. Multi-objective model for the integrated product and process design problem The concept of cost, quality and lead time have become the key determinants of market share and profitability for any organization. As we have seen in our models the design and manufacturing decisions influence cost, quality and lead time which in turn influence market share and profitability. Therefore, there is a need to study all these issues in an integrated manner. Our integrated product and process design model makes it possible to integrate the cost, quality and lead time into the product and process design. The tolerance of the product and the selection of the manufacturing process are determined simultaneously. What we have here is a systems concept on integrated product and process design by considering cost, quality and lead time models in a multi-objective environment. All the relations of unit cost, quality and lead time are
where Co ; Qo ; To ; do are requirements from customers. n is the number of alternative manufacturing processes (or machines). The expressions of Xo ðd; jÞ; Qloss ðd; jÞ and Tðd; jÞ are given by Eqs. (1)–(3), respectively. Multi-objective mathematical programming methods can be used to solve this model and generate a set of non-denominated solutions. The preferred solution can be obtained by using min–max method or weighted min–max methods. Although it is possible to use the loss function in the cost model, we prefer to consider the loss function as a separate objective. The primary argument is that this loss function provides a good measure of the quality loss, but it is difficult to convert it into cost.
4. The characterization of non-dominated solutions for the multi-objective integrated product and process design model In this section we develop an algorithm to obtain the non-dominated solutions. We study the properties of the models developed and prove a number of results. Such a characterization is useful in efficiently developing the algorithm. Further, we use the min–max method to obtain trade-off solutions. The weighted min–max
N. Singh / Robotics and Computer Integrated Manufacturing 18 (2002) 157–168
approach is used for the sensitivity analyses methods to get the optimal solution. 4.1. The characterization of non-dominated solutions In this section we provide the following results: Lemma 2. For given j, the fraction of scrap, SCðd; jÞ, is a decreasing function of tolerance d. Proof. Since FðxÞ is the cumulative distribution function of probability distribution with mean equal to 0 and standard deviation equal to 1, FðxÞ is an increasing function of x: For given j: sð jÞ is constant and Fðd=sð jÞÞ is an increasing function of d and Fðd=sð jÞÞ is a decreasing function of d. By Lemma 1, SCðd; jÞ ¼ Fðd=sð jÞÞ þ 1 Fðd=sð jÞÞ: So SCðd; jÞ is a decreasing function of d too. & Proposition 1. The sum of raw material cost (Xi ) and processing cost ( f ) is usually greater than the salvage value (Xs ), that is Xi þ f Xs > 0.
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The first derivative of T with respect to SC is qT=q SC ¼ðTsetup ð jÞ þ Tp ð jÞYo Þ=ð1 SCÞ2 > 0: So, Tðd; jÞ is the increasing function of SCðd; jÞ: By Lemma 2, SCðd; jÞ is the decreasing function of tolerance d. So, Tðd; jÞ is the decreasing function of tolerance d and attains its minimal value T ð jÞ at do because do is the upper boundary of d. & Lemma 5. For given j, the quality loss Qloss ðd; jÞ is a quadratic function of tolerance d and has unique minimal value Q ð jÞ at point d where d n ¼ A½fðd n =sð jÞÞ þ fðd n =sð jÞÞ=sð jÞ2B
ð6Þ
and the fðxÞ ¼ qFðxÞ=qx is the probability density function of the manufacturing process. sð jÞ is the standard deviation of jth manufacturing process. Proof. From the quality model, we have Qloss ðd; jÞ ¼ Qs ðd; jÞ þ Qt ðdÞ ¼ A SCðd; jÞ þ B d 2 :
Proof. From cost model (2), we have,
For given j; the Qs ðd; jÞ ¼ A SCðd; jÞ is an asymptotically decreasing function and Qt ðdÞ ¼ Bd 2 is an asymptotically increasing function of d: Therefore, the Qloss ðd; jÞ has unique minimal value. The first derivative of Qloss to d is
Xo ðd; jÞ ¼ Ki ðd; jÞ½Xi þ f ð jÞ Ks ðd; jÞXs
qQloss ðd; jÞ=qd ¼ qA SCðd; jÞ=qd þ 2Bd
Lemma 3. For given j and upon imposing Proposition 1, the cost function Xo ðd; jÞ is a decreasing function of tolerance d and attains its minimal value C ð jÞ at do .
¼ ½Xi þ f ð jÞ SCðd; jÞXs =½1 SCðd; jÞ: The first derivative of Xo with respect to SC is qXo =qSC ¼ðXs þ Xi þ f ð jÞÞ=ð1 SCðd; jÞÞ2 : By Proposition 1, qXo =qSC > 0: So, Xo ðd; jÞ is the increasing function of SCðd; jÞ: By Lemma 2, SCðd; jÞ is a decreasing function of tolerance d. So, Xo ðd; jÞ is the decreasing function of tolerance and attains its minimal value C ð jÞ at do because do is the upper boundary of d. & Remark. In some cases, we may have to pay to dispose off the scrap. Accordingly, Xs will be negative. In that case, Theorem 1 is correct even without imposing Proposition 1. Lemma 4. For given j and Yo > 0, the manufacturing lead time Tðd; jÞ is a decreasing function of tolerance d and reaches its minimal value T ð jÞ at do .
¼ qAðFðd=dð jÞÞ þ 1 Fðd=dð jÞÞÞ=qd þ 2Bd ¼ A½fðd=dð jÞÞ þ fðd=dð jÞÞ=dð jÞ þ 2Bd; where fðxÞ ¼ qFðxÞ=qx is the probability density function. fðxÞ > 0 for any x: Let qQloss ðd; jÞ=qd ¼ 0; we have d A½fðd =sð jÞÞ þ fðd =sð jÞÞ=sð jÞ2B: When dod ; we have qQloss ðd; jÞ=qdo0: When d ¼ d ; we have qQloss ðd; jÞ=qd ¼ 0: When d > d ; we have qQloss ðd; jÞ=qd > 0: Therefore, we have the unique minimal point at d : & Definition 3. The point (d 0 ; j 0 ) is a non-dominated solution if and only if, there is no d and j (dA½0; do and jAf1; 2; y; ng), which satisfies the following relations: Xo ðd 0 ; jÞ > Xo ðd; jÞ;
Proof. From the manufacturing lead time model (4), we have,
Qloss ðd 0 ; jÞ > Qloss ðd; jÞ;
Tðd; jÞ ¼ Tsetup ð jÞ þ Tp ð jÞYo =½1 SCðd; jÞ:
Tðd 0 ; jÞ > Tðd; jÞ:
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Theorem 1. For given j, the (d; j) is a non-dominated solution if and only if dA½d ; do , where d can be obtained by formula (6) and do is the upper boundary of d. Proof. For any dA½0; d Þ; by Lemmas 3–5, we have Xo ðd; jÞ > Xo ðd ; jÞ;
Table 1 Determining the feasible ranges of the Co ; Qo ; and To values j
Unit cost
Lead time
Quality loss
jc
jt
jq
C ð jc Þ Cð jt Þ Cð jq Þ
Tð jc Þ T ð jt Þ Tð jq Þ
Qð jc Þ Qð jt Þ Q ð jq Þ
Qloss ðd; jÞ > Qloss ðd ; jÞ; Tðd; jÞ > Tðd ; jÞ: By Definition 3, dA½0; d Þ is the dominated solution region for minimum fXo ðd; jÞ; Qloss ðd; jÞ; Tðd; jÞg: For any dAðdo ; NÞ; d is an infeasible solution. For any dA½d ; do ; by Lemmas 3–5 we have Xo ðd; jÞoXo ðd ; jÞ; Qloss ðd; jÞ > Qloss ðd ; jÞ;
½T ð jt Þ; maxfTð jc Þ; Tð jq Þg and that of Qo is ½Q ð jq Þ; maxfQð jt Þ; Qð jc Þg: Proof. If Co oC ð jc Þ; then there is no feasible solution because C ð jc Þ is a minimal value. However, if Co > maxfCð jt Þ; Cð jq Þg; then the feasible range will include the dominated solution. Therefore, the selection range for Co is [C ð jc Þ; maxfCð jt Þ; Cðjq Þg]. Similar results are valid for both To and Qo : &
Tðd; jÞoTðd ; jÞ: Therefore, dA½d ; do is a non-dominated solution region for minimum {Xo ðd; jÞ; Qloss ðd; jÞ; Tðd; jÞ}. & Definition 4. Let d ð jÞ ¼ fA½fðd=dð jÞÞ þ fðd=dð jÞÞ= fdð jÞ2Bgg be the lower boundary of non-dominated solution region for given j. Theorem 2. When j is a variable, all the non-dominated solutions of ðd; jÞ lie in the region S ¼f½d ð jÞ; do ; j ¼ 1; 2; y; ng.
4.3. The min–max optimal solution We can use the min–max method to obtain the min– max optimal solution of multi-objective optimization. Definition 6. We define the relative increments of Xo ; Qloss and T as Zx ¼ ðXo C ð jc ÞÞ=C ð jc Þ; Zq ¼ ðQloss Q ð jq ÞÞ=Q ð jq Þ
Proof. By Theorem 1, Sð jÞ ¼ ½d ð jÞ; do is the nondominated solution region for j: Therefore, S ¼ Sð1Þ,Sð2Þ,?,SðnÞ includes all the non-dominated solutions. &
respectively.
These results are useful in developing an efficient algorithm to obtain non-dominated solutions.
The Xo ; Qloss and T represent any non-dominated solution.
4.2. On selecting the feasible ranges of Co, Qo, and To
4.3.1. The min–max method A point (d ; j ) is min–max optimal, if for every ðd; jÞADJ; the following formula is satisfied:
We provide an approach to obtain the feasible ranges within which the values of Co ; Qo ; and To should be explored. Definition 5. Let minimal unit cost C ð jc Þ ¼ minfC
ð jÞg; the minimal manufacturing lead time T ð jt Þ ¼ minfT ð jÞg and minimal quality loss Q ð jq Þ ¼ minfQ ð jÞg; where C ð jÞ; T ð jÞ; Q ð jÞ are the minimal values when given j and the jc is the j which has the minimal unit cost C ðjc Þ; jt is the j which has the minimal lead time T ðjt Þ and jq is the j which has the minimal Quality loss Q ðjq Þ: These results are represented in Table 1. Remark. From Table 1, we can see that the range of Co is ½C ð jc Þ; maxfCð jt Þ; Cð jq Þg; the range of To is
and Zt ¼ ðT T ð jt ÞÞ=T ð jt Þ;
Zðd n ; j n Þ ¼ min fmaxfZx ; Zq ; Zt gg; ðd; jÞADJ
ð7Þ
where DJ is a set of non-dominated solutions obtained from the multi-objective model (4). 4.3.2. The weighted min–max method A point (d ; j ) is weighted min–max optimal, if for every ðd; jÞADJ; the following formula is satisfied: Zðd n ; j n Þ ¼ min fmaxfWx Zx ; Wq Zq ; Wt Zt gg; ðd; jÞADJ
ð8Þ
where DJ is a set of the non-dominated solutions obtained from model (4 ) and Wx ; Wq and Wt are the weights attached to the unit cost, quality loss and lead time, respectively. These weights reflect the aspirations
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of the decision makers which essentially represent the competitive environment that may exist at the time. 4.4. Algorithms Using the results from the previous section, we develop the two algorithms: Algorithm I to obtain the non-dominated solutions and Algorithm II to obtain the trade-off solutions from the non-dominated solutions generated by Algorithm I using the min–max as well as the weighted min–max approaches. 4.4.1. Algorithm I to obtain the non-dominated solutions From Theorem 2, we know that all the nondominated solutions lie in the region S ¼ f½d ð jÞ; do ; j ¼ 1; 2; y; ng: However, when considering all the j simultaneously, there may be some dominated solution. We now present the algorithm for obtaining the nondominated solutions as follows. We search the nondominated solutions only in region S ¼ f½d ð jÞ; do ; 0 j ¼ 1; 2; y; ng by a certain increment in d: We let S denote the set of non-dominated solutions. We place first solution ( j ¼ 1; d ¼ d ð1Þ) into S 0 ; then compare the other solutions in S one by one. If the new solution is dominated, then it is not added to S0 : If the new solution is non-dominated in S 0 ; then it is added to S0 and it eliminates all solutions in S 0 which are dominated by the new solution. Finally, we get the non-dominated solution set S 0 : Step 1: Determine d ð jÞ; 8jð j ¼ 1; 2; y; nÞ using Eq. (6). Step 2: Specify the step size Dd ¼ ðdo min fd ð jÞgÞ=m; m is the number of steps desired. The maximum possible number of non-dominated solutions are m n: Step 3: Set j ¼ 1; d ¼ d ð jÞ; ko ¼ 1; Xo ð1Þ ¼ Qloss ð1Þ ¼ Tð1Þ ¼ M; M is a very big number. Step 4: Evaluate the objective function vector (Xo ðd; jÞ; Qloss ðd; jÞ; Tðd; jÞ); and let k ¼ 1: Step 5: If Xo ðd; jÞ > Xo ðkÞ and Qloss ðd; jÞ > Qloss ðkÞ and Tðd; jÞ > TðkÞ; goto step 8. If Xo ðd; jÞoXo ðkÞ and Qloss ðd; jÞoQloss ðkÞ and Tðd; jÞoTðkÞ; goto step 7 otherwise k ¼ k þ 1. If kpko ; goto step 5. Step 6: Place the new non-dominated solution in the set S 0 : ko ¼ ko þ 1; Xo ðko Þ’Xo ðd; jÞ; Qloss ðko Þ’Qloss ðd; jÞ; Tðko Þ’Tðd; jÞ; Pd ðko Þ’d; Pj ðko Þ’j; goto step 8. Step 7: Substitute the dominated solution with the new point. Let Xo ðkÞ’Xo ðd; jÞ; Qloss ðkÞ’Qloss ðd; jÞ; TðkÞ’ Tðd; jÞ; Pd ðkÞ’d; Pj ðkÞ’j. Step 8: Let k0 ’ko . Step 9: If Xo ðd; jÞoXo ðk0 Þ and Qloss ðd; jÞoQloss ðk0 Þ and Tðd; jÞoTðk0 Þ; goto step 10 otherwise k0 ¼ k0 1. If k0 > k; goto step 9; otherwise goto step 11.
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Step 10: Let Xo ðk0 Þ’Xo ðko Þ; Qloss ðk0 Þ’Qloss ðko Þ; Tðk0 Þ’Tðko Þ; Pd ðk0 Þ’Pd ðko Þ; Pj ðk0 Þ’Pj ðko Þ; ko ¼ ko 1. Step 11: If dpdo ; then d ¼ d þ Dd; goto step 4; Otherwise j ¼ j þ 1; d ¼ d ð jÞ: If jpn; goto step 4; otherwise end. The set S0 now contains only non-dominated solutions. This algorithm only needs the computer’s memory for S 0 which may be much less than S: 4.4.2. Algorithm II for obtaining the min–max solutions Most often there is a need to know trade-off solutions that reflect the aspirations of the decision makers. In this paper we adopt the min–max and weighted min–max methods to get trade-off solutions. The following algorithm developed in this section obtains the min– max solutions. Step 1: Calculate the relative increment Zx ; Zq ; and Zt ; 8ðd; jÞAS 0 : Step 2: If using weighted min–max method, then multiply the weight to the relative increments: Wx Zx ðd; jÞ; Wq Zq ðd; jÞ; Wt Zt ðd; jÞ; where Wxþ Wq þ Wt ¼ 1 Step 3: For each ðd; jÞAS 0 ; calculate Zðd; jÞ ¼ maxfWx Zx ðd; jÞ; Wq Zq ðd; jÞ; Wt Zt ðd; jÞg: Step 4: 8ðd; jÞAS 0 ; calculate Zðd ; j Þ ¼ minfZðd; jÞg: ðd; jÞAS0 : Then (d ; j ) is min–max or weighted min–max optimal solution. Step 5: If the sensitive analysis of weight is needed, then change the weights Wx ; Wq and Wt and goto step 2; otherwise stop.
5. Integrated product and process design with multiple design variables When a part has more than one design variable, we need to select a vector variable D ¼ ðd1 ; d2 ; y; dm Þ to let {Xo ð j; DÞ; Qloss ð j; DÞ; Tð j; DÞ} to be minimum. In this case, the scrap rate SCð j; DÞ will be different from that of a tolerance variable case. The scrap rate is the probability of a product being scrapped. When the design variables are independent from each other, SCð j; DÞ will be equal to the product of SCð j; dl Þ: Therefore, we have m Y SCð j; DÞ ¼ 1 ½1 SCð j; dlÞ: l¼1
Then Ki ¼ Yi =Yo ¼ 1=½1 SCð j; DÞ; Ks ¼ Ys =Yo ¼ SCð j; DÞ=½1 SCð j; DÞ; Xo ð j; DÞ ¼ Ki ð j; DÞ½Xi þ f ð jÞ Ks ð j; DÞXs ;
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Qloss ð j; DÞ ¼ A SCð j; DÞ þ
m X
Bl dl2 ;
Xo ð j; Dn Þ ¼
l¼1
¼
Tð j; DÞ ¼ Tsetup ð jÞ þ Tp ð jÞYo =½1 SCð j; DÞ:
i;D
Qloss ð j; Dn Þ ¼
ð9Þ ¼
subject to
Xo ð j; Dk Þ fKi ð j; Dk Þ½Xi þ f ð jÞ Ks ð j; Dk ÞXs g;
p X k¼1 p X
Qloss ð j; Dk Þ fA SCð j; Dk Þ þ
k¼1
Xo ð j; DÞpCo ; Tð j; DÞpTo ;
Tð j; Dn Þ ¼
Qloss ð j; DÞpQo ; j ¼ 1; 2; :::; n; D ¼ ðd1 ; d2 ; :::; dm Þ; 0pdl pdlo ;
k¼1 p X k¼1
Accordingly, the multi-objective programming model will be as follows: Minimum fXo ð j; DÞ; Qloss ð j; DÞ; Tð j; DÞg
p X
l ¼ 1; 2; :::; m;
where Co ; To ; Qo ; dlo are customer requirements. n is the number of alternative manufacturing processes. m is the number of tolerance variables.
¼
p X k¼1 p X
m X
Bl dl2 g;
l¼1
Tð j; Dk Þ fTsetup ð jÞ þ Tp ðjÞYo =½1 SCð j; Dk Þg:
k¼1
Accordingly, the multi-objective programming model for the assembly will be as follows: Minimum fXo ð j; Dn Þ; Qloss ð j; Dn Þ; Tð j; Dn Þg; n j;D
ð10Þ
subject to FðDn ÞoF; 6. Integrated product and process design for product assemblies
Xo ð j; Dn ÞpCo ; Tð j; Dn ÞpTo ;
When a product consists of p parts and the kth part has mk tolerances (k ¼ 1; 2; y; p), we have to decide m1 þ m2 þ ? þ mp tolerances. Therefore, we need to select all dkl (kth part and lth tolerance, l ¼ 1; 2; y; mk ; k ¼ 1; 2; y; p) to let fXo ð j; D Þ; Qloss ð j; D Þ; Tð j; D Þg be minimum. We have the assembly function F ¼ FðD Þ: The function F is the tolerance vector of product and D is the tolerance metrices of parts. Therefore, F represents a group of assembly function formulas. For example,
Qloss ð j; Dn ÞpQo ; j ¼ 1; 2; :::; n; Dn ¼ ðD1 ; D2 ; :::Dl Þ;
d1l þ d2l þ ? þ dpl ¼ Fl ;
mðkÞ Y
oodkj odkjo ;
j ¼ 1; 2; y; mk ; k ¼ 1; 2; y; l
where Fo ; Co ; To ; SCo ; and dkjo are requirements. n is the number of alternative manufacturing processes. mðkÞ or mk is the number of tolerance variables of kth part. l is the number of parts.
l ¼ 1; 2; y; m:
The assembly function is the requirement that has to be satisfied. We, therefore, add the assembly function to the constraints. When the design variables are independent from each other, SCð j; Dk Þ will be equal to the product of SCð j; dkl Þ: Therefore, we have
SCð j; Dk Þ ¼ 1
Dk ¼ ðdk1 ; dk2 ; :::; dk mðkÞÞ;
½1 SCð j; dkl Þ:
l¼1
Then Ki ð j; Dk Þ ¼ Yi =Yo ¼ 1=½1 SCð j; Dk Þ; Ks ð j; DkÞ ¼ Ys =Yo ¼ SCð j; DÞ=½1 SCð j; DkÞ;
7. The illustrative examples 7.1. Example 1 ABC company plans to produce 1000 units of a cylindrical part. The company wants to know which tolerance is the optimal considering simultaneously the objecives of minimum unit cost, quality loss and lead time. And what machine should be selected to produce these parts? The company has the following information: unit cost of raw material Xi ¼ $10:00; unit salvage valueðsell scrapped productÞXs ¼ $2:00: There are three kinds of machine-tools that can be used to produce the parts. The distribution of dimensions is normally distributed Nð0; sð jÞÞ (Table 2).
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N. Singh / Robotics and Computer Integrated Manufacturing 18 (2002) 157–168 Table 2 Data for example 1 j
Machine
Standard deviation sð jÞ (in)
Unit process cost f ð jÞ
Setup time Tsetup ð jÞ (h)
Unit process time Tp ð jÞ (h)
1 2 3
Turret Lathe Engine Lathe NC Machine
0.003 0.001 0.0005
$3 $10 $15
2 3 4
0.2 0.17 0.15
The loss function parameters A is 100 and B is 100/ 0.0032 (0.003 is maximal acceptable do ). Hundred is a scale value. From marketing information, the company knows the unit cost Xo should be less than $28, the lead time T should be less than 400 h, the tolerance d should be less than 0.003 in and the quality loss Qloss should be less than 140 to obtain non-dominated solution and min–max optimal solution. Solution. To obtain non-dominated solution and min– max optimal solution, we follow the steps of Algorithms I and II. Further, we know that do ¼ 0:003; and FðxÞ is standard normal distribution. Algorithm I: Step 1: Determine d ð jÞ; 8jð j ¼ 1; 2; 3Þ using Eq. (6). d ¼ A ½fðd =ð jÞÞ þ fðd =sð jÞÞ=sð jÞ 2B ¼ 100 ½fðd =sð jÞÞ þ fðd =sð jÞÞ=ðsð jÞ 200=0:0032 Þ when j ¼ 1; d ¼ fðd =0:003Þ0:003 ¼> d ð1Þ ¼ 0:0011: By the same approach, get d ð2Þ ¼ 0:0014: d ð3Þ ¼ 0:001: Step 2: Specify the step size Dd ¼ ðdo min fd ð jÞgÞ=m; m is the number of steps desired. Let m ¼ 100; Dd ¼ ð0:003 minf0:0011; 0:0014; 0:001gÞ=100 ¼ 2 105 : Step 3: Set j ¼ 1; d ¼ d ðjÞ; ko ¼ 1; Xo ð1Þ ¼ Qloss ð1Þ ¼ Tð1Þ ¼ M; M is a very big number. Let M ¼ 109 Step 4: Evaluate the objective function vector (Xo ðd; j), Qloss ðd; jÞ; Tðd; jÞÞ; and let k ¼ 1: SCð0:0011; 1Þ ¼ 0:714; Xo ð0:0011; 1Þ ¼ ðXi þ f ð1Þ SCð0:0011; 1Þ Xs Þ=ð1 SCð0:0011; 1ÞÞ ¼ ð2 þ 10 2 0:714Þ=ð1 0:714Þ ¼ 36:96; Tð0:0011; 1Þ ¼ Tsetup ð1Þ þ Tp ð1Þ Yo =½1 SCð0:0011; 1Þ ¼ 2 þ 0:2 1000=ð1 0:714Þ ¼ 701; Qð0:0011; 1Þ ¼ 100 SCð0:0011; 1Þ þ ð100=0:0032 Þ
0:00112 : Step 5: Since Xo ðd; jÞ oXo ðkÞ and Qloss ðd; jÞoQloss ðkÞ and Tðd; jÞoTðkÞ; goto step 7. Step 7: Substitute the dominated solution with the new point.
Let Xo ð1Þ’Xo ð0:0011; 1Þ; Qloss ð1Þ’Qloss ð0:0011; 1Þ; Tð1Þ’Tð0:0011; 1Þ; Pd ð1Þ’d ¼ 0:0011; Pj ð1Þ’j ¼ 1. Step 8: Let k0 ¼ ko ¼ 1. Step 9: Since k0 not > k; goto step 11. Step 11: Since dpdo ; then d ¼ d þ Dd; goto step 4. y y when d > do and j ¼ 3; end. Set S 0 now contains only non-dominated solutions. Then, we have the non-dominated solution ðd; jÞAS 0 ¼ fdA½0:0026; 0:003; j ¼ 1g,fdA½0:0014; 0:003; j ¼ 2g, fdA½0:001; 0:003; j ¼ 3g: Algorithm II for getting the min–max optimal solution: Step 1: Calculate the relative increments Zx ; Zq ; and Zt ; 8ðd; jÞAS 0 as Table 3. Step 2: If using weighted min–max method, then multiply the weight to the relative increments: Wx Zx ðd; jÞ; Wq Zq ðd; jÞ; Wt Zt ðd; jÞ; where Wxþ Wq þ Wt ¼ 1: The different weight are shown in Table 4 and the result is shown in Table 5. Step 3: For each ðd; jÞAS 0 ; calculate Zðd; jÞ ¼ maxfWx Zx ðd; jÞ; Wq Zq ðd; jÞ; Wt Zt ðd; jÞg: The result is shown in Table 3. Step 4: 8ðd; jÞAS0 ; calculate the Zðd ; j Þ ¼ min fZðd; jÞg ¼ 0:42 and d ¼ 0:0014; j ¼ 3: ðd; jÞAS0 : The ðd ; j Þ is min–max or weighted min–max optimal solution. Step 5: If the sensitive analysis of weight is needed, then change the weights Wx ; Wq and Wt and goto step 2; otherwise stop. From Table 3, the min–max optimal value Zðd ; j Þ ¼ minfmaxfZx ; Zq ; Zt gg ¼ 0:42; and the min–max optimal solution is ð j ; d Þ ¼ ð3; 0:0014Þ: A sensitivity analysis can be carried out attaching different weights to the objectives. Then the weighted min–max method can be used. Three cases are discussed in Table 4. Then we have the weighted min–max solution in Table 5. From Table 5, we can see the different weighted min– max solutions when the weights of cost, lead time and quality loss are different. These solutions will help
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Table 3 Eleven non-dominated solutions j; d
Cost
Time
Quality
Zx
Zt
Zq
MaxZx ; Zq ; Zt
1, 1, 2, 2, 2, 2, 2, 3, 3, 3, 3,
19.9 18.1* 23.5 21.4 20.5 20.2 20.0 26.1 25.1 25.0 25.0
348 315 226 206 199 195 194 181 175 174* 174*
113.7 131.7 37.9 43.2 56.6 76.0 100.3 15.7* 22.3 36.0 44.5
0.10 0* 0.30 0.18 0.13 0.12 0.10 0.44 0.39 0.38 0.38
1.00 0.81 0.3 0.18 0.14 0.12 0.11 0.04 0.01 0* 0*
6.24 7.39 1.41 1.75 2.61 3.84 5.38 0* 0.42 1.29 2.43
6.24 7.39 1.41 1.75 2.61 3.84 5.38 0.44 0.42* 1.29 2.43
0.0026 0.0030 0.0014 0.0018 0.0022 0.0026 0.0030 0.0010 0.0014 0.0018 0.0022
Table 4 Different weights for the min–max solution
Table 5 The weighted min–max solutions
Case
Wx (cost)
Wt (time)
Wq (quality)
1 2 3
0.7 0.2 0.1
0.2 0.1 0.7
0.1 0.7 0.2
select the design and process decision variables reflecting the aspirations of the decision makers. Essentially, the weighted min–max approach provides a flexible scheme to obtain the solutions that represent the different kinds of decision makers. 7.2. Example 2 In this example, we assume that there are tolerances for two-dimensions. There are three kinds of machinetools that can be used. The variates of two-dimensions are Normally distributed Nð0; sð1; jÞÞ and Nð0; sð2; jÞÞ: The data are same as in example 1 except the following Table 6. Solution. We know that do ¼ 0:003; and FðxÞ is a standard normal distribution. Obtain the minimal unit cost C ð jÞ; lead time T ð jÞ and quality loss Q ð jÞ for three objectives separately for j ¼ 1; 2, 3. SCðDo ; jÞ ¼ 1 ð1 SCðd1o ; jÞÞð1 SCðd2o ; jÞÞ;
Case
Weighted min–max value
j ; d
Unit cost
Lead time
Quality loss
1 2 3
0.175 0.088 0.044
2, 0.0018 3, 0.0010 3, 0.0010
21.4* 26.1 26.1
206 181* 181*
43.2 15.7* 15.7*
Calculate Q ð jÞ ¼ 100 SCðd ð jÞ; jÞ þ ð100=:052Þd ð jÞ2 then get Q ð1Þ ¼ 101; d ð1Þ ¼ 0:0005; Q ð2Þ ¼ 60:9; d ð2Þ ¼ 0:0018; Q ð3Þ ¼ 30:7; d ð3Þ ¼ 0:0014: (2) Obtain the minimal unit cost C ð jc Þ; lead time T ð jt Þ and Quality loss Q ð jq Þ for three objectives seperately. Accordingly C ð jc Þ ¼ minfC ð jÞg ¼ 22:51 and jc ¼ 2; dc ¼ do ¼ 0:003: T ð jt Þ ¼ minfT ð jÞg ¼ 174
SCðDo ; 1Þ ¼ 0:534; SCðdo ; 2Þ ¼ 0:025; SCðdo ; 3Þ ¼ 0:000036: Calculate C ð jÞ¼ðXi þf ð jÞSCðDo ; jÞXs Þ=ð1SCðDo ; jÞÞ then get C ð1Þ ¼ 29:89; C ð2Þ ¼ 22:51; C ð3Þ ¼ 28:00: Calculate T ð jÞ ¼ Tsetup ð jÞ þ Tp ð jÞYo =½1 SCðdo ; jÞ then get T ð1Þ ¼ 665; T ð2Þ ¼ 228; T ð3Þ ¼ 174:
and jt ¼ 3; dt ¼ do ¼ 0:003: Q ð jq Þ ¼ minfQ ð jÞg ¼ 30:71 and jq ¼ 3; dq ¼ d ð3Þ ¼ 0:0014: (3) Get the non-dominated solution region DJ: By Theorems 1 and 2, we can search the nondominated solution in region f½d ðjÞ; do ; j ¼ 1; 2; 3g ¼ fdA½0:0005; 0:003; j ¼ 1g,fdA½0:0018; 0:003; j ¼ 2g
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N. Singh / Robotics and Computer Integrated Manufacturing 18 (2002) 157–168 Table 6 Data for example 2 j
Machine
Standard deviation sð1; jÞ (in)
Standard deviation sð2; jÞ (in)
Unit process cost f ð jÞ
Unit process time Tpð jÞ (h)
1 2 3
Turret Lathe Engine Lathe NC Machine
0.003 0.0012 0.0007
0.0015 0.0002 0.0002
$5 $12 $18
0.3 0.2 0.15
Table 7 Min–max optimal solution for different component tolerances
d1 (part 1) j (machine) d2 (part 2) j (machine) Unit cost Lead time Quality loss a
d1 p0:003; d2 p0:003:a
d1 þ d2 p0:003
d1 þ d2 p0:002
d1 þ d2 p0:005
0.0016 3 0.0012 2 52.4 379 55
0.0015 3 0.0013 3 57 310 48
0.0011 3 0.0009 3 61.4 335 41
0.0015 3 0.0015 3 56.9 309 53
d1 and d2 are independent of each other.
,fdA½0:0014; 0:003; j ¼ 3g: For any fdA½0:0005; 0:003; j ¼ 1g we find that the solution is dominated. And in region fdA½0:0018; 0:003; j ¼ 2g,fdA ½0:0014; 0:003; j ¼ 3g every solution is nondominated solution. Then we have the non-dominated solution ðd; jÞADJ ¼ fdA½0:0018; 0:003; j ¼ 2g,fdA ½0:0014; 0:003; j ¼ 3g: (4) Get the min–max optimal solution Using the same approach as in example 1, we can get the min–max value Zðd ; j Þ ¼ minfmaxfZx ; Zq ; Zt gg ¼ 0:27; and the min–max optimal solution is ð j ; d Þ ¼ ð3; 0:0014Þ: The unit cost is 28.5. The lead time is 177. The quality loss is 38. 7.3. Example 3 In this case we assume that two tolerances d1 and d2 are for two different parts which are assembled together into a product. Further, we assume that the product tolerance is fixed at 0.003 in. The relation of product tolerance and parts tolerance is d1 þ d2 p0:003: We need to select the parts tolerances and the machine-tools for the three objectives (minimal unit cost, lead time and quality loss). Solution. (1) Obtain the minimal unit cost C ð jc Þ; lead time T ð jt Þ and Quality loss Q ð jq Þ for three objectives seperately. Accordingly C ð jc Þ ¼ minfC ð jÞg ¼ 49:06 and jc ¼ 2; d1 c ¼ 0:0017; d2 c ¼ 0:0013: T ð jt Þ ¼ minfT ð jÞg ¼ 307:72
and jt ¼ 3; d1 t ¼ 0:0011; d2 t ¼ 0:0009: Q ð jq Þ ¼ minfQ ð jÞg ¼ 41:23 and jq ¼ 3; d1 q ¼ 0:0011; d2 q ¼ 0:0009: (2) Get the min–max optimal solution. Using the same approach as in example 1, we can get the min–max value Zðd1 ; d2 ; j Þ ¼ minfmaxfZx ; Zq ; Zt gg ¼ 0:16393; and the min–max optimal solution is ð j ; d1 ; d2 Þ ¼ ð3; 0:0015; 0:0013Þ: The unit cost is 57.11. The lead time is 310. The quality loss is 47.92. (3) Discussion: The following Table 7 gives the min–max optimal solutions at different relationships between the tolerance d1 of part 1 and tolerance d2 of part 2. From Table 7, we can see that the assembly relations affect the min–max optimal solution. The tight assembly tolerance (d1 þ d2 p0:002) gives a worse solution. The loose assembly tolerance (d1 p0:003; d2 p0:003:) gives a better solution. Therefore, the assembly relations should be carefully considered in practical situations.
8. Conclusions In this paper we presented a multi-objective framework for the concurrent tolerance design. The model integrates the cost, quality, and lead time into the design process. The multi-objective programming methodology presented in this paper can be applied to help make decisions on selecting the product tolerances as well as
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the manufacturing processes. The modeling approach is quite comprehensive and realistic since competitive marketing information on cost, quality and lead time is built into the models.
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[3] Goldman S, Preiss K, editors. 21st Century Manufacturing Enterprise Strategy: An Industry-Led View, Iacocca Institute, Lehigh University, Harold S. Mohler Laboratory # 200, Bethlehem, PA, 1991. [4] Zhang HC, Huq ME. Tolerancing techniques: the state of the art. Int. J. Prod. Res. 1992;30(9):2111–35. [5] DeVour RE, Hunter RG, Sutherland JW. A methodology for robust design using models for the mean, variance, and loss. Proceedings of the Symposium on Quality and Performance, ASME Bound Volume, December 1989. [6] Kapur K. Quality engineering, tolerance design. In: A. Kusiak (Ed.) John Wiley & Sons Inc. 1992. pp 287–306. Concurrent Engineering: Automation, Tools & Techniques.