DOI 10.1007/s00170-005-2568-2 O R I G I N A L A R T I C L E
Muh-Cherng Wu · Wen-Jung Chang · Chie-Wun Chiou
Product-mix decision in a mixed-yield wafer fabrication scenario
Received: 26 August 2004 / Accepted: 3 November 2004 / Published online: 24 March 2006 ©Springer-Verlag London Limited 2006
Abstract The product mix decision problem for semiconductor manufacturing has been extensively studied in literature. How-ever, most of them are based on a high-yield scenario. Yet, in a low-yield manufacturing environment, some research claims that scrap low-yield lots in an early stage may produce more profit. Considering the early scrapping characteristics, this paper aims to solve the product mix decision problem for a mixed-yield sce-nario, which involves the simultaneous production of high-yield and low-yield products. A nonlinear mathematical program is de-veloped to model the decision problem. Two methods for solving the nonlinear program are proposed. Method 1 converts the non-linear program into a non-linear program by setting some variables as parameters. The method provides an optimal solution by exhaus-tively searching these parameterized variables and solving the LP models iteratively. Method 2 aims to reduce the computation com-plexity while providing a near optimal solution. Experiment re-sults show that method 2 is better than method 1, when aggre-gately considering solution quality and computation efforts. Keywords Mixed-yield scenario·
Product mix planning· Scrapping
1 Introduction
Semiconductor manufacturing is a capital-intensive industry. Building a semiconductor fab (factory) usually costs over 1 bil-lion dollars. How to select market demands to effectively utilize the production resources is therefore very important. Such a deci-sion, often called product mix planning problem, is to determine the production quantity for each type of product, given that the future demand of each product has been known.
For a semiconductor fab, different product mix decisions would generate different profits [1]. A fab, which was optimally M.-C. Wu (u) · W.-J. Chang · C.-W. Chiou
Department of Industrial Engineering and Management, National Chiao Tung University,
Hsin-Chu, Taiwan, ROC E-mail: mcwu@cc.nctu.edu.tw Fax: +886-35-731913
designed for a particular product mix, produces a maximum throughput in quantity for that product mix. However, the par-ticular product mix may not be the most profitable due to the change of market. Another product mix might become more profitable, yet producing less in throughput. Therefore, the prod-uct mix planning problem is a trade-off decision between choos-ing the most profitable products and maximizchoos-ing throughput.
Much literature on product mix decision has been published. Kasilingam [2] proposed a nonlinear programming model for solving the product mix decision in the presence of stochastic demand and alternate process plans. Hsu and Chung [3] ap-plied the theory of constraints (TOC) to solve the product mix decision problem. Malik and Sullivan [4] use activity-based cost-ing (ABC) to make product-mix and costcost-ing decisions. Kee and Schmidt, in their studies [5, 6], compared the differences of using ABC and TOC in making product mix decisions. Morgan and Daniels proposed methods for integrating product mix and technology adoption decisions [7]. The previous studies, though having established significant milestones, rarely addressed the impact of low production yield in developing their methods.
Low production yield is not unusual for a semiconductor fab, in particular at the stage of developing a new process or manu-facturing a new product [8]. The production yield for some new products at their launching stages, from our interview with in-dustry, may be as low as 30%. Wu et al. claimed that scrapping a low-yield wafer lot earlier may increase the profit of a fab [9]. That is, a wafer lot has to be scrapped if its yield after a process is lower than a predefined threshold. They developed a method for identifying the threshold of each process in order to maximize the total profit of a fab.
A competitive semiconductor fab always has to face the prob-lem of developing new processes and new products. Therefore, a fab may be situated in a scenario, which involves the production of both low-yield new products and high-yield mature products. Such a scenario is called a mixed-yield scenario. Low-yield new products usually have a higher profit margin, due to their advances in technology or functions. High-yield mature products on the other hand usually have lower profit margin. The product mix de-cision for such a mixed-yield scenario is very important; however,
it has been rarely concerned in literature. Chou and Hong [10] considered the factor of low production yield in solving the prod-uct mix decision for a semicondprod-uctor fab; yet their study did not consider the strategy of scrapping low-yield wafer lots.
This paper presents a method for making the product mix de-cision for a mixed-yield semiconductor fab, which adopts a low-yield-scrapping strategy. The product mix decision is formulated by a nonlinear programming model. This research proposes two methods to solve the nonlinear program. The first method obtains the optimal solution at the expense of requiring extensive com-putation. The second method provides a good or near-optimal solution and requires much less computation.
The remainder of the paper is organized as follows. Sect. 2 reviews the concept of scrapping low-yield lots. Sect. 3 discusses the proposed nonlinear programming model. Sect. 4 presents the two solution methods. Sect. 5 compares the solutions of two methods in various numerical experiments. Some concluding re-marks are given in Sect. 6.
2 Scrapping low-yield lots
Scrapping low-yield wafer lots have three main impacts [9]. First, the unit cost of processing a low-yield lot is lower than that of a high-yield lot. Second, the bottleneck of a fab might change when the fab yield greatly decreases. Third, the through-put of a fab might increase or decrease due to the scrapping of low-yield lots. Each impact is explained below.
2.1 Cost behavior of processing low-yield lots
In semiconductor manufacturing, wafers are transported in a cas-sette (called a lot), which normally carry 25 wafers. Due to yield problems in processing, some wafers in a lot may become out-of-specification and are removed from the lot. Such a lot, less than 25 wafers in number, is called a small lot. A lot carrying 25 wafers is herein called a full lot. To ensure the production qual-ity, small lots usually cannot be merged into a full lot in their remaining processing.
Workstations in a semiconductor fab can be classified into two types: batch and series. A series machine processes one wafer at a time. The series-processing cost of each wafer, whether it is carried in a small lot or in a full lot, is equal. A batch machine processes several lots at a time; for example, a furnace machine may process up to six lots at a time. Each lot equally shares the batch-processing cost. Therefore, the processing cost
per wafer for a small lot is higher than that of a full lot.
2.2 Change of fab bottleneck
In a semiconductor fab, the stepper is usually the most expen-sive machine and thus relatively few in number. The stepper workstation, being of series-type, is generally assumed to be the bottleneck of the fab in previous literature [3, 5, 6]. Such a bottle-neck hypothesis is valid in a high-yield environment. However, the bottleneck may change to a batch workstation in a low-yield environment [9].
The following example illustrates the change of the fab bottleneck due to yield variations. Let a fab be equipped as follows. Among all the series workstations, the bottleneck series-workstation is 10 000 wafer/month in capacity. Among all the batch workstations, the bottleneck batch-workstation is 800 lots/month in capacity, which is equivalent to 20 000 wafers/month if all lots are full lots (100% yield environment). The bottleneck series-workstation, in a high-yield environment, is therefore the bottleneck of the fab according to the theory of constraints (TOC) [11].
Now suppose the fab is in a low-yield situation, and the average wafer number per lot is ten wafers; namely, with 40% average yield. The bottleneck batch-workstation at most can produce 800 lots/month, which is now equivalent to produces only 8000 wafers/month. Remember that the bottleneck series-workstation can at most produce 10 000 wafers/month. The batch-workstation therefore becomes the bottleneck of the fab in the low-yield environment. That is, in the low-yield environment (40% yield), the fab bottleneck is a batch workstation, while in a high-yield environment (100% yield), the fab bottleneck is a se-ries workstation.
2.3 Impacts on throughput due to scrapping small lots
Setting an appropriate threshold for scrapping small lots could increase the profit of a fab [9]. In a typical low-yield fab, only a few critical operations accounts for the low yield character-istic. After each critical operation, a threshold is defined for scrapping small lots. For example, a threshold value of five im-plies that a small lot, less than five wafers in number, should be scrapped and cannot be processed for its remaining operations.
Scrapping small lots may impose positive or negative im-pacts on the fab throughput. First, the throughput of the fab might increase because scrapping small lots increases the aver-age wafer number per lot. This subsequently increases the output wafers of the bottleneck batch-workstation. If the fab bottleneck is now of batch type, then the throughput of the fab will increase by scrapping small lots. Conversely, the throughput of the fab might decrease because scrapping a small lot make its utilized capacity wasted. That is, the capacity, which has been utilized to produce the small lot, cannot produce any fab throughput.
3 Mathematical model
This section formulates a nonlinear programming model for the product mix decision in a mixed-yield scenario adopting the scrapping-small-lot policy. We first model the input/output rela-tionship for a fab of interest. The input denotes the number of wafers released to the fab, and the output denotes the number of finally produced wafers. The input/output relationship model is subsequently used to formulate the nonlinear program.
3.1 Input/output relationship
The process route of manufacturing a wafer is quite long, typic-ally involving hundreds of operations. These operations are
usu-ally segmented into several tens of groups (also called layers). In each layer of operations, an inspection has to be performed on each wafer to determine whether the wafer can be processed further. Therefore, each layer can be modeled as a small produc-tion system with a particular yield. A full wafer lot after passing a layer of operations, due to a yield problem, might reduce the number of its good wafers and become a small lot.
Consider a fab adopting the strategy of scrapping small lots. Letλi represent the number lots of product i released to the fab. Suppose a full lot carry m wafers, then m· λiwafers are released to the fab. After passing each layer, due to the yield problem, the number of good wafers in a lot might reduce. The number of small lots at the output of each layer is thus a distribution. Let
Wij= [wijk] represent such a distribution, in which w ij
k represents the number of small lots carrying k (0≤ k ≤ m) good wafers when product i passes layer j. Thus, Wi,0= [0, 0, · · · , λi] rep-resents the distribution of small lots at the input of layer 1, which denotes thatλilots or m·λiwafers have been released to the fab. Let Ai, j= [aijs,t] represent the yield distribution of product
i passing layer j. Aij is an(m + 1) × (m + 1) matrix, in which
aijs,t (0≤ s, t ≤ m) represent the probability of a lot with s good wafers becoming a lot with t good wafers. As stated, scrapping small lots based on appropriate thresholds might produce more profit. Let hijrepresent the scrapping threshold of product i pass-ing layer j. That is, a small lot with less than hij good wafers shall be scrapped. Such a scrapping strategy can be modeled by an(m + 1) × (m + 1) matrix, R(hi, j) = [rsij,t]. Here, rsij,t is a
bi-nary value (0 or 1), rsij,t= 1 denotes that a lot with s good wafers will become a lot with t good wafers after applying the scrapping strategy; conversely, rsij,t= 0 denotes that there is no such change. The values of rsij,tcan be determined as below.
if s> hi, j and s= t, then rsij,t= 1 if s> hi, j and s= t, then rijs,t= 0
if s≤ hi, j and t= 0, then rsij,t= 1 if s≤ hi, j and t= 0, then rijs,t= 0
Figure 1 shows the input/output relationship of a layer, which involves three modules: processing, inspection, and scrapping. The input/output relationship for each layer can therefore be modeled as follows.
Wi, j = Wi,( j−1)× Ai, j× R(hi, j)
Fig. 1. The input/output relationship of wafer lots distribution between layer
j− 1 and layer j
Suppose Ai, j, hi, j at each layer are known, for a given wafer re-lease Wi,0, we can determine the small lot distribution at each layer. Let S(Wi, j) =
m k=1
k· wijkrepresent the total number of good wafers and L(Wi, j) =
m k=1w
ij
k represent the total number of lots, for product i at layer j.
As stated, a semiconductor fab typically involves two types of workstations: batch and series. Let BNsrepresent the bottle-neck among the series workstations, and BNb represent the bot-tleneck among the batch workstations. According to TOC [11],
BNs and BNbare the two critical factors in constraining the fab throughput.
Let tsi, j represent the processing time of a wafer by BNsfor product i at layer j, 1≤ j ≤ Li, where Li denotes the number of layers for product i. Let ¯hi= [hi, j] represent the threshold at each layer j for product i. The utilized capacity of BNs can be computed as follows. Cap_S(λi, ¯hi) = n i=1 Li j=1 tsi, j· S(Wi, j−1)
Suppose a machine in the batch workstation BNb can process r lot per operation run. Let tbi,, j represent the operation time per run on BNb for product i at layer j (1 ≤ j ≤ Li). The utilized capacity of BNbcan be computed as follows.
Ca p_B(λi, ¯hi) = n i=1 Li j=1 tbi, j· L(Wi, j−1) r
To generalize the input/output relationship model, we assume only one wafer lot is released to the fab and define a unit vector
Ui, j =λ1i· Wi, j. That is, Ui, j = [uijk] represents the distribution of small lots at each layer for each product, when only one lot is released to the fab. Since Wi, j= Wi,( j−1)× Ai, j× R(hi, j) and Ui, j=λ1i· Wi, j, the formula Ui, j= Ui,( j−1)× Ai, j× R(hi, j) therefore holds.
3.2 Product mix decision
The above input/output relationship is used to formulate a nonlin-ear program for the product mix decision under a mixed-yield sce-nario, which applies a strategy of scrapping low-yield lots. Nota-tions of the model are given below. To facilitate referencing, some notations, which have been explained above, are also listed here.
A. Notations.
n Total number of product type m The number of wafers in a full lot Li Total number of layers for product i Pi Price of product i
FC Fixed cost of the fab in the concerned time horizon Ci, j= [cijk] a (m + 1) × 1 matrix, in which c
ij
k represents the processing cost of a lot carrying k good wafers for product i at layer j; Ci, j denotes the raw wafer cost per lot for k= 0.
di Minimum demand of product i Di Maximum demand of product i ATs The available capacity of BNs ATb The available capacity of BNb
tsi, j The operation time per wafer on BNsfor product i at layer j
tbi, j The operation time per run on BNbfor product i at layer j
r The number of lots per run on BNb
Ui, j = [uijk] The distribution of small lot when only one lot is released to the fab
Ai, j= [aijs,t] The yield matrix for product i at layer j
¯hi= [hi, j] The scrapping threshold for product i at each layer λi The number of lots released to the fab
B. Model Max n i=1 ⎡ ⎣Pi· S(Ui,Li) · λi − ⎛ ⎝Ui,0× Ci,0+ m j=1 Ui, j−1× Ci, j ⎞ ⎠ ·λi ⎤ ⎦− FC Subject to Ui, j = Ui,0 j k=1 (Ai,k× R(hi,k)) 1 ≤ i ≤ n; 1 ≤ j ≤ Li (1) S(Ui, j) = m k=1 kuijk 1≤ i ≤ n; 1 ≤ j ≤ Li (2) L(Ui, j) = m k=1 uijk 1≤ i ≤ n; 1 ≤ j ≤ Li (3) n i=1 ⎡ ⎣Li j=1 tsi, j· S(Ui, j−1) ⎤ ⎦ ·λi≤ ATs (4) n i=1 ⎡ ⎣Li j=1 tbi, j·L(Uir, j−1) ⎤ ⎦ ·λi≤ ATb (5) S(Ui,Li) · λi≥ di 1≤ i ≤ n (6) S(Ui,Li) · λi≤ Di 1≤ i ≤ n (7) hi, j≥ hi,kif j< k 1 ≤ i ≤ n (8) In the above formulation, the objective function models the profit of the fab, in which the term
Pi· S(Ui,Li) · λi
models the revenue, the term ⎛ ⎝Ui,0× Ci,0+ m j=1 Ui, j−1× Ci, j ⎞ ⎠ ·λi
describes the variable costs, and FC is the fixed cost.
Constraint 1 describes the input/output relationship of each layer. Constraint 2 is used to compute the number of good wafer outputted at each layer. Constraint Eq. 3 intends to compute the number of lots outputted at each layer. Constraints 4–5 request that the utilized capacity of BNsand BNbcannot be greater than its available capacity. Constraint 6 denotes the minimum output quantity for each product, requested by the management. Con-straint 7 describes the maximum market demand of each product. Constraint 8 prohibits an incompatible setting of threshold values. That is, a downstream layer cannot define a threshold value higher than its upperstream layer. For example, suppose the threshold for layer 1 is hi,1= 6 and that for layer 2 is hi,2= 8. Based on such a scrapping strategy, a lot with seven good wafers will pass layer 1 and be processed at layer 2. Yet, whatever the yield of layer 2 is, the lot will be surely scrapped after layer 2 because hi,2= 8. Such an incompatible threshold setting is un-doubtedly prohibited.
4 Solution methods
This section first analyzes the complicated characteristics in solving the mathematical program formulated in the above sec-tion. Two proposed methods for solving the mathematical pro-grams are then presented.
(A) Analysis of the mathematical program. The formulation in the above section is a nonlinear program, in which hi, j and λi are independent decision variables. Constraint 1, the multipli-cation of several matrices including R(hi, j), essentially gives the nonlinear relationship. Solving such a nonlinear program ap-pears to be quite complicated, and needs a method for making it simplified.
The nonlinear program can be simplified as a linear pro-gram if hi, jare given parameters. Namely, suppose hi, jare given parameters which meet Constraint 8, we can determine the vec-tor of Ui, j and subsequently the values of S(Ui, j) and L(Ui, j). Constraints 1–3 can therefore be removed. The above nonlin-ear program formulation now becomes a linnonlin-ear program, which involves only four constraints, 4–7. In the linear program, hi, j, Ui, j, S(Ui, j) and L(Ui, j) are treated as parameters, and λi are the decision variables. The two proposed methods for solving the nonlinear program are both based on such a simplification, and become solving linear programs.
(B) Method 1. The first method starts with determining a solution space of ¯hiand then solves the linear program for each element in the solution space. Let S represent such a solution space. An element in S, { ¯hi= [hi, j]|1 ≤ i ≤ n} is a set of vectors, which denotes a combination of thresholds for each product i at each critical layer j. Let C represent the number of critical layers and N(S) represent the number of elements in S. Then N(S) = mnC, if Constraint 8 is not considered, where m denotes the number of wafers in a full lot. Namely, assume that m= 25, C = 2, n = 2, then N(S) = 254= 390, 625. This implies that we have to solve the linear program 390 625 times to obtain the solution of the
product mix decision problem. Such a computation requirement seems acceptable.
However, the method is in nature computationally extensive. For a case with m= 25, C = 2, n = 10, a huge number of com-putation, N(S) = 2520, which appears to be infeasible to solve the problem in an acceptable time. Therefore, another approach to solve such a problem is by adopting some meta-heuristics
algorithms such as genetic algorithms [12], Tabu search
algo-rithms [13], or simulated annealing algoalgo-rithms [14]. The ba-sic ideas of these meta-heuristics algorithms are randomly and
wisely selecting some good elements in S in an non-exhaustive
search manner. Such approaches will greatly reduce the number of computations; however, the obtained solution is not an optimal solution. This research adopts the exhaustive search approach in the numerical experiments.
(C) Method 2. The other method solves the formulated nonlinear
program by a two-stage approach. The first stage determines the threshold ¯hi for each product i, and the second stage solves the linear program by taking the obtained ¯hias parameters.
In determining ¯hi, we assume that the fab produces only one product i. The bottleneck of the fab may be either of series-type or of batch-series-type. If the fab bottleneck is of series-series-type, then Constraint 5 in the nonlinear program model can be removed. Likewise, if the fab bottleneck is of batch-type, then Constraint 4 in the nonlinear program model can be removed. Following the discussion about method 1, we need only solve a linear program
mC times to obtain ¯hi for a particular product i based on a par-ticular assumption of fab bottleneck.
For a particular assumption of the fab bottleneck, we need to solve the linear program n· mC times, if Constraint 8 is not considered, to obtain the thresholds of the n products. There are two assumptions for the fab bottleneck. Therefore, we need to solve the linear program 2· n · mCtimes. For the above case with
m= 25, C = 2, n = 10, method 2 only need to solve the
lin-ear program 12 500 times, greatly less than 2520times required
by method 1. The basic idea of method 2 essentially adopts a problem-decomposition approach, and cannot guarantee an op-timal solution.
The required computation for method 2 surely can also be computationally extensive if C is a large number. However, in a typical fab, only a few layers are critical layers. That is, the number of critical layers (C) may only range from 1 to 3. There-fore, method 2 may not be so computationally extensive in real application.
5 Numerical examples
Without loss of generality, a hypothetic fab is first presented to illustrate the product mix decision problem. We then analyze whether the strategy of scrapping small lots is better than the no-scrapping strategy in a mixed–yield scenario. Subsequently, the scrapping thresholds determined in the first stage of method 2 are discussed. Finally the solution results of the two proposed methods for solving a product mix decision problem are compared.
(A) Fab data. The fab produces two products (n= 2), and each
product has 20 layers of which layer 2 and 3 are the two critical layers. The yield at each non-critical layer is 100%. The yields of the two critical layers are equal. Let y represent the aggre-gated fab yield and p represent the yield of a critical layer. If
y= 40%, then p = 63% (p = √y). The yield matrix Ai, j= [aijs,t]
is so designed based on the binomial distribution as follows:
aijs,t= Ctspt(1 − p)s−t for s≥ t, and aijs,t= 0 fors < t.
A full lot has 25 wafers (m= 25). The raw wafer cost is cij0 = $2000. The variable cost for processing at each layer,[cijk](1 ≤
k≤ 20), is shown in Table 1. The fixed cost is FC = $1.1 × 107. The parameters tsij and tbij are assumed to be constant, namely tsij= ts and tbij = tb. The capacity of bottleneck series-workstation is so designed that ATsts = 900 000 wafer-layers. Likewise, the capacity of the bottleneck batch-workstation is such designed, ATb·r
tb = 42, 000 lot-layers, which is equivalent to 1 050 000= 42 000 × 25 wafer-layers if all lots are full lots.
(B) Comparison between scrapping and no-scrapping. Suppose
that the fab is in a mixed-yield scenario. The first product is with high yield, y1= 90%; the second product is with low yield,
y2= 40%. Two scrapping alternatives are to be compared.
Alter-native 1 uses the strategy of scrapping small lots, and alterAlter-native 2 does not scrap any small lots. Let P1 and P2respectively
de-note the unit price of the first and the second product; d1, d2
respectively represent the lower bound of the production quantity of product 1 and product 2. Table 2 shows the profit difference of the two alternatives for d1= 0 and d2= 0. From the table, the
strategy of scrapping small lots will yield more profit than the no-scrapping strategy. The profit differences, in the example, range from 5.2% to 7.8%. This finding implies that scrapping small lots is also a good strategy in a mixed-yield scenario, just as that in a low-yield scenario [9].
Table 1. Variable processing cost at each layer
cij1= $283 c ij 6= $320 c ij 11= $358 c ij 16= $395 c ij 21= $433 cij2= $290 cij7= $328 cij12= $365 cij17= $403 cij22= $440 cij3= $298 cij8= $335 cij13= $373 cij18= $410 cij23= $448 cij4= $305 cij9= $343 cij14= $380 cij19= $418 cij24= $455 cij5= $313 cij10= $350 c ij 15= $388 c ij 20= $425 c ij 25= $463
Table 2. Profit comparison between scrapping and no-scrapping strategies
Profit of using scrapping Profit diff. Profit diff
strategy ($M) ($M) (%)
P1= $500 32.2 1.6 5.2%
P2= $2, 900
P1= $500 8.3 0.6 7.8%
Table 3. Scrapping thresholds for the low yield product determined in
method 2
Yield Bottleneck series Bottleneck batch
h2 h3 h2 h3 40% 7 4 12 7 50% 8 5 12 9 60% 9 6 14 11 70% 10 7 16 13 80% 0 0 18 15 85% 0 0 18 16 90% 0 0 19 17
(C) Determining thresholds of method 2. In method 2, the
scrap-ping threshold for each product, based on each assumption of fab bottleneck, should be first determined. Table 3 shows such results in various fab yields. Let h2, h3respectively represent the
thresh-old of the two critical layers, layer 2 and layer 3, of a particular product. Notice that the scrapping thresholds (h2 and h3) may
change when the fab bottleneck changes. Refer to the first row in Table 3, where the fab yield of a product is y= 40%. The thresh-olds are h2= 7, h3= 4 when the fab bottleneck is of series-type,
and h2= 12 and h3= 7 when the fab bottleneck is of batch-type.
Moreover, when the fab yield changes, the thresholds (h2
and h3) may also change. Refer to the first two rows in Table 3,
which address two fab yields y= 40% and y = 50%. When the fab yield increases, the threshold values in general will in-crease. However, when the fab yield is high enough (y> 80%), the threshold value becomes zero; namely, it requires no scrap-ping. This phenomenon can be interpreted as follows. When the fab yield increases, the number of small lots, less than a cer-tain threshold in the number of good wafers, may decrease. The scrapping threshold therefore may need to be raised to increase profit. Yet, when the fab yield is high (y= 81%, or p = 90%), the probability of producing small lots becomes quite low. There-fore, we may not need taking any scrapping strategy (referring to the last three rows of Table 3).
(D) Comparison of the two proposed methods. The two proposed
methods for solving the nonlinear program are compared. Let
hi2, hi3 represent the scrapping threshold of layer 2 and 3 of
Yield Method h12 h13 h22 h23 Profit ($) Profit diff ($)
p1= 90% Method 1 19 17 11 7 34 216 809 – p2= 40% Method_2_batch 19 17 12 7 34 214 340 −2469 Method_2_ series 0 0 7 4 34 067 098 −149 711 p1= 85% Method 1 18 16 12 7 34 598 533 – p2= 40% Method_2_batch 18 16 12 7 31 598 533 0 Method_2_ series 0 0 7 4 30 649 497 −949 036 p1= 70% Method 1 16 13 12 7 31 318 551 – p2= 40% Method_2_batch 16 13 12 7 31 318 551 0 Method_2_ Series 10 7 7 4 29 972 939 −1 345 612 p1= 60% Method 1 15 11 12 7 31 059 049 – p2= 40% Method_2_batch 14 11 12 7 31 058 649 −400 Method_2_ series 9 6 7 4 29 710 377 −1 348 672
Table 4. Solution comparisons between method
1 and method 2 (P1= $1440, P2= $2880)
Table 5. Comparison of computation time (* denotes by estimation)
Problem size Method Computation Number of times time in solving LP program
Fab 1 Method 2 2 min 1302
n= 2 Method 1 240 min 105 625
m= 25, C = 2
Fab 2 Method 2 5 min 1952
n= 3 Method 1 54 days* 34 328 125
m= 25, C = 2
product i (i= 1, 2). Table 4 shows the solution results of adopt-ing method 1 and method 2 in four mixed-yield scenarios. Here, method 1 uses an exhaustive search for determining hi2and hi3 and solve the linear program for each combination of (h12, h13,
h22, h23). The solution obtained by such an exhaustive search is
an optimal solution.
Table 4 shows the solution results of the two proposed methods. In the table, method_2_batch denotes that the fab bot-tleneck is assumed to be of batch workstation, and
method_2_series denotes that the fab bottleneck is assumed to
be of series workstation. The table shows that the fab bottle-neck is of batch-type when method 2 is used to solve each of the four mixed-yield scenarios. For the four mixed-yield sce-narios, the scrapping thresholds hij suggested by method 2 and method 1 only slightly differ. The solution quality of method 2 is very good, though not optimal, only about 0.007% less than the optimal solution provide by method 1.
Table 5 shows the computation efforts required by the two methods for two hypothetical fabs. Fab 1 produces two products (i.e., n= 2, m = 25, C = 2), and fab 2 produces three products (i.e., n= 3, m = 25, C = 2). In solving the product mix deci-sion problem of fab 1, method 1 needs about 240 min, (4 hours), which solves 105 625 LP programs and method 2 needs only 2 min, which solves 1302 LP programs. In solving the problem of fab 2, method 2 takes about 5 min, which solves 1952 LP pro-grams; method 1 needs a formidable computation time, about 54 days by estimation (34 328 125/105 625 *4/24). The computer program for the above experiments is coded in Matlab software,
running on a personal computer equipped with CPU 3.0 GHZ and DRAM 512 MB.
From the numerical experiments, method 2 seems better than method 1 due to its much less computation requirements and its good solution quality. The computation efforts would be greatly saved in Method 2 when the problem size increases.
6 Concluding remarks
This paper presents a product mix decision problem in a mixed-yield semiconductor fab, which simultaneously produces high yield and low yield products. Scrapping low-yield wafer lots dur-ing the process may increase profit. Yet, this strategy has not been included in the previous literature on product mix decision. In the semiconductor industry, new products and new processes have been continuously developed and have low yields in its early stage. This research therefore includes the scrapping strat-egy in addressing the product mix decision problem.
A nonlinear program is developed to model such a product mix decision problem. The decision problem involves two sets of decision variables: scrapping thresholds (hij) and product release quantity (λi). Two solution methods are proposed to solve the nonlinear program, by setting the scrapping threshold variables as parameters and convert the nonlinear program into a linear program. Method 1 uses an exhaustive search to identify hij for all products and solve the associated linear program extensively. Method 2 individually determines hij for each product i, and then solve the associated linear program. Method 1 provides an optimal solution, but is computationally extensive. Though not providing an optimal solution, method 2 is very good in solu-tion quality, only less than the optimal solusolu-tion by 0–0.007% and requires much less computation than method 1. This research therefore suggests the use of method 2 in solving the product mix decision problem.
This research in modeling the capacity of bottleneck work-stations adopts a static capacity model. That is, the factor of
cycle time is not addressed. Future extension of this research in-cludes the consideration of cycle time in dealing with the mixed-yield product mix decision problem. Another extension to this research is developing some meta-heuristics algorithms such as genetic algorithms, Tabu search algorithms, and simulated an-nealing algorithms to enhance method 1.
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