Designing an assembly line for modular products

Pergamon Computers ind. Engng Vol. 34, No. 1, pp. 37-52, 1998 © 1998 Elsevier Science Ltd. All rights reserved

PII: S0360-8352(97)00149-6 Printed in Great Britain 0360-8352/98 $19.00 + 0.00

DESIGNING AN ASSEMBLY LINE FOR MODULAR PRODUCTS

D A V I D W. H E and A N D R E W K U S I A K Intelligent System Laboratory, Department of Industrial Engineering, The University of Iowa, Iowa

City, Iowa 52242-1527, U.S.A.

Abstract--To respond to the challenge of agile manufacturing, companies are striving to provide a large variety of products at low cost. Product modularity has become an important issue. It allows to produce different products through combination of standard components. One of the characteristics of modular products is that they share the same assembly structure for many assembly operations. The special structure of modular products provides challenges and opportunities for operational design of assembly lines. In this paper, an approach for design of assembly lines for modular products is proposed. This approach divides the assembly line into two subassembly lines: a subassembly line for basic assembly operations and a subassembly line for variant assembly operations. The design of the subassembly line for basic operations can be viewed as a single product assembly line balancing problem and be solved by existing line balancing methods. The subassembly line for the variant operations is designed as a two-station fiowshop line and is balanced by a two-machine flowshop scheduling method. A three-station fiowshop line for a special structure of modular products is proposed and illustrated with an example. © 1998 Elsevier Science Ltd. All rights reserved.

INTRODUCTION

Standardization of components and the ability to produce a variety of products through combination of components are two of the benefits of product modulari ty (Ulrich and Tung [1]). Ulrich and Tung described five different ways that product modulari ty is used in current industrial practice. Component-swapping modulari ty can be achieved when two or more alternative types of components can be paired with the same basic product body to create different product variants. Examples of this type of modulari ty can be found in automobile industry where different audio cassette decks, windshield glass types, and wheel types for the same automobile model are produced. In computer industry, different hard disk types, monitor types, and keyboards are matched with the same basic CPU. Component-swapping modulari ty is often associated with the creation of product variety as perceived by the customer [1].

An example of product variants generated through component-swapping in toy industry is provided in Fig. 1.

The toy robots in Fig. 1 are assembled on the same assembly equipment. The body parts are uniform and are assembled automatically. The heads distinguish the products and are assembled manually.

To maximize manufacturing efficiency, assembly lines must be designed to fit the product structures. In this paper, a design approach of assembly lines for modular products is proposed.

The remainder of the paper is organized as follows. An assembly line design approach for modular products is developed in Section 2. Section 2.1 discusses the graph representation of the structure of modular products. The balancing and scheduling problems of assembly lines are discussed in Section 2.2. In Section 2.3, a two-station flowshop assembly line design approach is presented and illustrated with an example. Section 2.4 proposes a line design approach for modular products which is illustrated with an example. The performance evaluation of the assembly line is examined in Section 2.5. In Section 2.6, the limitations of the design approach presented in this paper are discussed. Section 3 develops a design approach for the three-station flowshop line for modular products with a special structure. A numerical example illustrates the design approach. Conclusions are presented in Section 4.

37

38 David W. He and Andrew Kusiak

"1

Fig. 1. Product variants accomplished with modular design.

DESIGN OF MODULAR PRODUCT ASSEMBLY LINES

Graph representation of product structures A product structure is represented by an acyclic digraph. Nodes represent operations and arcs

precedence relations between the operations. Each assembly operation corresponds to a physical component in a product. For example, the presence of a component is represented by one or more corresponding operations, e.g., loading the component, positioning the component on the base component, etc. In this paper it is assumed that the assembly structure of a modular product consists of two parts: basic structure and variant structure. Each product in a product family shares the same basic structure and differs from others by the variant structure. The operations in the basic structure are basic operations and the operations in the variant structure variant operations. Two examples of modular product structures are provided in Fig. 2 and 3.

In Fig. 2, two products share the same variant operations but different variant structures. In Fig. 3, two products share different variant operations and structures. Note that the order of the basic and variant structure in Figs 2 and 3 can be reversed. However, this type of product structure would violate the principle of design for assembly that variant features should appear as far up the assembly structure as possible [2].

Line balancing and scheduling problems

Line balancing and scheduling are two problems in operational design of assembly lines. Traditionally, operational design of assembly lines has been accomplished by the line balancing methods. The assembly line balancing problem is to assign operations to assembly stations to minimize: 1) the cycle time for a given number of stations, or 2) the number of stations for a given cycle time. Most assembly line balancing methods have been developed for a single pro-

PI:

P2:

Fig. 2. Modular products with the same variant operations but different variant structure.

Modular products 39

P3:

P4:

Fig. 3. Modular products with different variant operations and variant structures.

duct assembly line due to the fact that the assembly line was designed to produce a large volume of a single product.

A multi-product assembly line is more difficult to balance than a single product assembly line. Chow [3] discussed the difficulty of balancing a multi-product line. Since assembly operations assigned to stations vary from product to product, creating an uneven workload along the line, a single line is difficult to balance for all the products. There are two ways to balance a multi- product line: 1) balance the line for each product and product batches for inventory, 2) balance the line for all products by using average processing times. The first approach can create high volume of inventory and should be avoided in just-in-time manufacturing system. Since using the second approach, it is difficult to balance a line for all products, scheduling different products must be considered. The scheduling problem occurs when a number of different products are produced in an assembly line at the same time. The objective of the scheduling problem is to determine the sequence of products which maximizes the utilization of the assembly stations.

Assembly line balancing and scheduling have been considered as two separate but related problems [4, 5]. Most research on assembly line sequencing considers scheduling products after assembly operations have been assigned to the assembly stations [6-8, 4]. The difficulty associated with this procedure is that for a given assembly line configuration most resulting scheduling problems are difficult to solve (see [6] and [4]). Suboptimal solutions rather than optimal solutions for the scheduling problems are often obtained. Even in the case where the balancing and scheduling problems can be solved optimally, the optimality of the overall solution may not be obtained by solving the balancing and scheduling problem separately. Next, an assembly line design approach for modular products is developed. The design approach considers the assignment of assembly operations to the stations and scheduling products simultaneously.

Design of a two-station flowshop line

The special structure of modular products provides an opportunity to develop optimal procedure for design of assembly lines. For the product structure in Fig. 2, the assembly line can be treated as a single product assembly line since all the products in the product family share the same assembly operations. The existing line balancing methods for a single product assembly line can be used. For the product structure in Fig. 3, the line balancing and scheduling is difficult since each product has different assembly operations. However, the assembly line of this type of modular product structure has its own special characteristics, i.e., products share the same assembly structure for most of the assembly operations and each product differs from other products by a few variant assembly operations. This paper focuses on the line design problem for this type of modular product structure. Due to the special structure of modular products, the assembly line can be divided into two subassembly lines. The first subassembly line is designed to perform basic assembly operations and the second subassembly line is designed to perform variant assembly operations. Due to unequal production rates of the two subassembly


-- StOonll"" "l to°°M F Subassembly line for basic structure Subassembly line for variant structure

Fig. 4. Example of the assembly line for modular products.

lines, an inventory buffer between two subassembly lines may be needed. An example of the assembly line for modular product structure is shown in Fig. 4.

Consequently, each subassembly line is created by different design approaches. Since each product of a product family shares the same basic structure, the subassembly line for basic structures can be designed as a single product assembly line using the existing line balancing methods and no scheduling is necessary. The balancing problem is to minimize the number of stations for a given cycle time. The target cycle time is set by the production manager. The single product assembly line balancing problems have been extensively studied and a large number of solution approaches exists (see Ghosh and Gagnon [5]).

In the basic subassembly line, ordinary automated assembly equipment can be used. In contrast, due to the variety of assembly operations to be performed in the variant subassembly line, specialized assembly machines, assembly operators, and specialized auxiliary equipment such as feeders, jigs, and fixtures are required. The specialized equipment and skilled operators are ex- pensive. Therefore, reducing the total idle time in a variant subassembly line is important in reducing the production cost.

Due to the difficulty in balancing and scheduling a multi-product line, an approach is proposed to design the variant subassembly line. In designing this line, it is desired that the line be short and efficient. Due to the small number of variant operations, it is possible that the line can be often designed as a two-station assembly line. The line can be considered as a flowshop line and is balanced by a two-machine flowshop scheduling method. The scheduling problem that minimizes the makespan for a two-machine flowshop can be solved optimally by Johnson's algorithm [9].

It is assumed that the variant subassembly line is not paced and the products are dispatched into the line according to a minimum makespan flowshop schedule. The line operates as a flow-

(a)

PI:

(b)

Fig. 5. Variant product structures and the superimposed assembly graph: (a) variant structures of three products; (b) superimposed assembly graph.

Modular products 41

(a)

o @l--.- S1 $2

S1 Pl[ P2[ P3[ 1 2 3

$2 I el I P21P31 1 8 13 15

(b)

--.t@@ H@@I-"- S1 $2

Sl PIIP" 4 P3 I 3 4 7 ~¢'P3

I 3 8 13

(c)

S1 $2

P2

S1 I~l~2P3 I 5

$2 I e l I P31 4 8 10

(d)

S1 $2

S1

$2

el I P2 I P3 I 6 10 13

F-P;~ [ ~ ' P 3

6 8 10 1213

Fig. 6. Four line designs for Example 1: (a) line design 1; (b) line design 2; (c) line design 3; (d) line design 4.

shop. Once the operations of a product at the first station are completed, the product is immediately released to the buffer. If the second station is available, the product is immediately processed. If the second station is busy when the product enters the buffer, the product waits in the buffer until the station is available.

Before the design procedure is presented, a superimposed assembly graph is constructed from the variant assembly structures of products. Let the variant assembly structure of product Pi be represented by graph Gi with a set of Ni nodes and a set of Ai arcs. The nodes represent operations and the arcs precedence relations. Thus an arc from node (operation)f to node (operation) g indicates that operation f must be completed before starting operation g. It is assumed that Gi is acyclic. The variant assembly structures of N products can be combined into a superimposed assembly graph G with N nodes and A arcs, where

N=N1UN2U...UNN, A=AIUA2U...AN

An arc ~, g) in G is redundant if in addition to the arc itself there exists a chain from node f to node g. The redundant arcs are omitted.

The design procedure for the two-station flowshop line is presented next. Design Procedure 1 (DPI). Let P be the set of feasible partitions of operations among the two

stations, S1 and $2 be the first station and the second station of the two-station flowshop line, respectively. A feasible partition p = {p(S1), p(S2)}, where peP, and p(S1) and p(S2) are sets of operations assigned to S1 and $2, respectively. A partition is feasible if each station contains at least one operation and precedence relations among operations specified in the superimposed assembly graph are not violated. Let Cmax(P) be the minimum makespan schedule corresponding to the feasible partition p. For any p, Cmax(P) can be obtained by Johnson's algorithm. The


Table 1. The processing times (Example 1) Operation No. 1 2 3 4 ti 1 2 3 2

objective of the design procedure is to find p* = {p*(S1), p*(S2)} so that the makespan is minimal, i.e., Cmax(P*) = min{Cmax(P)[ for all peP}.

Due to the special structure of modular products, it is expected that the size of P is reason- ably small and the two-station line design procedure can be implemented by an exhaustive search. Next, an example is provided to illustrate the design procedure for the two-station flowshop line.

Example 1. Design a two-station flowshop line for three products with the variant structures in Fig. 5. The processing times are provided in Table 1.

For the variant assembly structure in Fig. 5, four two-station flowshop line designs are possible (see Fig. 6). For each line design, the makespan is computed by Johnson's algorithm. The optimal design with the minimum makespan of 10 time units is selected (see Fig. 6(c)).

Design an assembly line for modular products

It is assumed that the target throughput rate Rt is set for an assembly line by a production manager. Rt can serve as the minimum throughput rate required.

A modular product assembly line is designed in the following two consecutive steps.

1. The variant subassembly line is designed as a two-station flowshop line. Due to variation of the number of assembly operations and assembly structure, an ideal balance of a line can not be achieved. Products are scheduled to minimize the makespan. Let Qi be the production quantity of product Pi, N be total number of products, and Q = y]N=l Qi be the total quantity of products to be produced. For a serial two-station flowshop line, the throughput rate

P I :

P2:

P3:

P4:

Fig. 7. The assembly structure of four product variants.

Modular products

Table 2. The processing times of the assembly operations (Example 2)

43

Operation 1 2 3 4 5 6 7 8 9 10 No. t~ 3 1 2 2 1 2 I 3 3 3

Fig. 8. The superimposed assembly graph of variant operations.

can be computed as Rv= Q/Cmax, where Cma x is the makespan of the schedule. If Rt<Rv, a serial line is used. If Rv < Rt, M-parallel lines are used. The M is determined such that Rt<MRv. The value of M is set to 1 for the serial line.

2. Since each product shares the same basic operations and assembly structure, it is possible to design the basic subassembly line using the existing single product line balancing methods. The balancing problem is to minimize the number of stations for a given cycle time. The cycle time is computed as Cb = 1/(MRv).

Next, an example is provided to illustrate the assembly line design approach for modular products.

Example 2. Design an assembly line for four product variants, as shown in Fig. 7. The processing times of assembly operations are provided in Table 2.

The production quantity for each product is as follows: Q1=4, Q2=5, Q3=2, Q4=3. The superimposed assembly graph of variant operations is shown in Fig. 8.

Assume that the target throughput rate Rt =0.3. The two-station flowshop line obtained by DPI is shown in Fig. 9.

The corresponding minimum makespan schedule is: {P4(3), P2(5), P3(2), Pl(4)} with the minimum makespan of Cmax = 41. Note that Pi(Qi) denotes Qi units of product Pi, i = 1 . . . . . N.

The throughput rate is Rv = 14/41 = 0.341. Since Rt<_Rv, a serial two-station line design is selected for the variant subassembly line.

The cycle time for the basic subassembly line is Cb = 1/Rv = 1/0.341 = 3.0. Using the ranked positional weight algorithm (Chow [3]), the number of stations required for

basic subassembly line is determined as 3. The layout of the assembly line is shown in Fig. 10.

Performance evaluation of assembly lines for modular products It is important to quantify the balance efficiency of assembly lines for modular products. The

balance efficiency of an assembly line is impacted by the number of stations in the line, the idle time of the line, and the waiting of products in the line. Normally, the fewer the number of stations in a line and the less the idle time of the line, the more efficient is the line. In an

ieeeH=,-SeGk Sl 12

Fig. 9. A two-station line design.


@@H@a> @@@l-u-I@@ Basic subassembly line Variant subassembly line

I : Buffer

Fig. 10. The layout of the assembly line design in Example 2.

assembly line for modular products, buffers are used to absorb the unbalanced flow of the products. It is expected that the lesser the time the products wait in the buffer the better the balance efficiency of the line. Thus, the balance cost of a modular product assembly line is affected by three factors: the number of the stations, the idle time, and the waiting time.

The waiting time and idle time in a flowshop line

In contrast to a paced line, the flowshop line is not paced. To absorb the unbalanced flow of products, buffers between stations are used. Instead of remaining on each station in a fixed time interval (usually a cycle time), a product leaves the station to the succeeding buffer once its operations are completed at this station. The product waits in the buffer until the succeeding station is available. On the other hand, once a station completes a product it retrieves another product from the buffer. If the preceding buffer is empty, the station will be idle. It can be observed that, in a flowshop assembly line, some products may have to wait in the buffer before they are processed on the next station and some stations may be idle if no products are available in the preceding buffers. For a modular product assembly line, while the idle time of stations can be computed directly from a flowshop schedule, the waiting time of the products is computed from a no-delay flowshop schedule. To explain the above, consider the example of the three-station flowshop line in Fig. 11.

(a)

(b)

S1 P1 I P2l P3 I 3 5 8

S2 ~ ~ " i ] I 2 2 ~ 132 ~ 3 4 5 6 8 1O

$3 ~ P1 I P2 [ P3 [ - Time 4 10 14 19

S1

$2

$3

P2 I P3 [

7 9 p3

3 4 9 10 12 14 r DI3

PI I P2 I P3 1 - - 4 10 14 19 v Time

Fig. 11. Two schedules for the three products: (a) flowshop schedule; (b) no-delay flowshop schedule.

Modular products 45

Figure 11 (a) shows a flowshop schedule of the three-station flowshop assembly line. Note that 122 and 132 represent the idle time associated with products P2 and P3 on station $2 in the flowshop schedule, respectively. The idle time in a fiowshop schedule represents the time that a station remains idle after it completes the operations of a previous product and before it starts to perform the operations of a current product. For example, 122 represents the time that station $2 remains idle after it completes the operations of product P1 and before it starts to perform the operations of product P2. Figure 1 l(b) shows the corresponding no-delay flowshop schedule. D21, D22, and D32 represent the idle time of product P2 on stations SI and $2, and the idle time of product P3 on station $2 in a no-delay flowshop schedule. The idle time in the no-delay schedule represents the time that a product remains on the current station after its operations are completed and before it is moved to the next station. For example, D22 is the time that product P2 remains on station $2 after its operations are completed on $2 and before it is moved to station $3. Based on the definition of 122 and D22, one can see that the time that product P2 has to wait in the buffer of station S2 is D22-I22. Note that the times 112, I13, DI2, and D13 associated with the first product P1 are not considered as idle times.

Define: N = number of products;

M = number of stations; Qi = quantity of product i; Q = ~v=t Qi = total quantity of products; S = (1,2 . . . . . Q) = sequence of products;

rim processing time of ith product in S at station m. i = 1 . . . . . Q; m = 1 . . . . . M; Dj., = the idle time on station m between the ( i - 1)st product and the ith product of a no-delay flowshop

schedule; li,,, = the idle time of station m after it finishes processing the ( i - l)st product and before it starts to process

the ith product of a flowshop schedule; Wire = Di,.-lim, the waiting time of the ith product of S in the buffer of station m;

W= the total waiting time required to process products of S; 1= the total idle time required to process products of S.

The number of a station's buffer is the same as the corresponding station. Therefore, for example, the buffer number of the first station S1 is 1.

In general, Dim can be computed as follows [10]:

m - I m

Dim = Dil q- y~'Cis - - Z ' C i - I , s , s= I s=2

where

[zm I Dil = max2_<m<U Ti_l , s - %,0 for i = 2 . . . . ,Q ( s=2 s= l

(1)

The value of Iim can be computed as follows:

[ ~__~ ~ i-1 i-s~l } lira = m a x 0, I s ,m- I "-]- "Cs.m-1 -- Z T s m -- Is,m , s=l s= l s=l . =

where

m - I

Ilm=Z'Cls f o r m = 2 . . . . . M, I i l = O for i = l , 2 , . . . , Q s=l

(2)

Based on Equations (1) and (2), the values of W and I can be computed as follows:

Q M-I

W = E ~ ( O i m - l i m ) i=2 m=2

(3)

Q M i = (4)

i=2 m=2

46 D a v i d W. H e a n d A n d r e w K u s i a k

Table 3. The processing time rim, i = 1 . . . . . 14; m = 1,2,3 (Example 3)

Product ith product in S "~11 "c¢2 zi3

P4 1 3 1 3 174 2 3 1 3 P4 3 3 1 3 P2 4 3 4 3 P2 5 3 4 3 P2 6 3 4 3 P2 7 3 4 3 P2 8 3 4 3 P3 9 3 3 0 P3 10 3 3 0 P1 11 3 3 0 P1 12 3 3 0 PI 13 3 3 0 PI 14 3 3 0

For any two-station flowshop line and the corresponding product sequence obtained from DP1, the total waiting time and idle time can be computed from (3) and (4).

The balance cost of an assembly line for modular products

In an assembly line for modular products, the basic subassembly line is paced by a cycle time Cb and a buffer between the basic subassembly line and the variant subassembly line is used to absorb the unbalanced flow of products. It can be observed that some products coming from the basic subassembly line may have to wait in the buffer until the stations in the variant subassembly line become available and idle time may occur on the first station of the variant subassembly line due to the fact that the cycle time Cb differs from the processing time of products on the first station of the variant subassembly line. To compute the waiting time in the buffer and the idle time of the first station of the variant subassembly line, the basic subassembly line can be viewed as a station with all products having the same processing time Cb. Therefore, for a modular product assembly line with a two-station variant subassembly line, the idle and waiting time can be computed from (1)-(4) with M = 3.

Let a, b, and c be rate of the station cost, inventory holding cost, and idle time cost, respectively. Let E be the total number of stations in a modular product assembly line. The balance cost (BC) of an assembly line for modular products can be computed as follows:

BC = aE + b W + cI (5)

Next, an example is provided to illustrate the computation of the balance cost. The balance cost of the assembly line for modular products is compared with that of a paced assembly line.

Example 3. Consider the design problem in Example 2. For the assembly line in Fig. 10, the processing time rim for each product is provided in Table 3.

The total waiting and idle times are computed as follows:

Dsl = D61 = DTI = Dsl = Dgl = 1; D22 = D32 = D42 = 2; all other Dim = O. 122=132=142=2;143=3,153=163=173=183 = 1; all other lira=O; 1 = 13. W s I = W 6 1 = W 7 1 = W 8 1 = W 9 1 = ] - 0 = 1; W 2 2 = W 3 2 = W 4 2 = 2 - 2 = 0; W = 5.

Compared to a paced assembly with a fixed cycle time, the assembly line for modular products involves less idle time and requires fewer stations at the cost of WIP inventory. To see

Fig. 12. T h e s u p e r i m p o s e d g r a p h fo r t he p r o d u c t v a r i a n t s in Fig. 7.

Modular products 47

4 eeHe>eHec H- e®F Station 1 Station 2 Station 3 Station 4 Station 5 Station 6

Fig. 13. The paced assembly line with a fixed cycle time.

this, consider design of a paced assembly line for the four product variants in Fig. 7. The superimposed graph for the four product variants in Fig. 7 is shown in Fig. 12.

To achieve the same throughput rate as that of the assembly line for modular products in Fig. 10, the cycle time is set to Cb = 1/Rv = 1/0.341 = 3. The paced line with a fixed cycle time of 3 is shown in Fig. 13.

The total idle time for the paced line is computed as follows. Let li be the idle time associated with product Pi. For each unit of product P1, 6 units of idle time can occur. Therefore, Ii = 6 x 4 = 24 and similarly, •2=2 x 5 = 10, •3=6 × 2 = 12, •4=5 × 3 = 15. The total idle time is I~ +12 + I3 + I4 = 61.

In this example, one can see that to achieve the same throughput rate, a paced line results in greater idle time and more stations than the assembly line for modular products. However, the WIP inventory occurs in the assembly line for modular products. To determine the type of line to be selected, the balance cost BC is used. The line with the lowest balance cost is selected.

Let a = $100, b = $1 min -1, and c = $0.25 min -I. Then the balance cost for the modular product assembly line is BC(1) = $100 x 5 + $1 x 5 + $0.25 x 13 = $508.25. The balance cost for the paced line is BC(2) = $100 x 6 + $1 x 0 + $0.25 x 61 = $615.25. In this case, the modular product assembly line is selected.

The limitations of the design approach

Consider two cases that apply to the two-station variant assembly line design approach. The first case is concerned with modular products share a small number of basic assembly operations and differ in a large number of variant assembly operations. In the second case the modular products share a large number of basic assembly operations and differ in a small number of variant operations. In the first case, the cycle time of the basic subassembly line may be much lower than the processing time of products on the first station of the variant subassembly line. As a consequence, the balance cost of the line can be high. One solution to the problem is to

PI:

P3:

Fig. 14. Modular products with a linear variant structure.


design the variant subassembly line with more than two stations. The design of a variant subassembly line with more than two stations is beyond the scope of this paper. In the second case, since fewer variant assembly operations are assigned to the two stations of the variant subassembly line, the balance between the basic subassembly line and the variant subassembly line is expected to be better. In fact, from the principle of avoiding variants in assemblies (Andreasen et al. [2]), it is expected that the number of variant assembly operations would be low.

Scheduling an assembly line with more than two stations is difficult. Next, a design approach for a three-station line for a class of special modular products is developed.

D E S I G N O F A T H R E E - S T A T I O N L I N E F O R A CLASS O F SPECIAL M O D U L A R P R O D U C T S

In this section, modular products with a linear variant assembly structure are considered (see Fig. 14).

The three products in Fig. 14 share the same basic assembly structure and they differ in their variant assembly structures. The variant assembly structures are linear and variant operations are numbered in an increasing order from left to right. This type of product designs is evident in some industries, e.g., in the automotive industry, cars of a product family share the same basic body. In order to have a cruise control option a car must have an automatic transmission option installed. The stacked product structure discussed by Andreasen et al. [2] is a special case of the product structure in Fig. 14. A stacked product structure is a product structure where most of the components are laid in stacks and are finally secured by the internal cohesion of the components, e.g., plug contains and suction pumps [2]. When modular products have a linear variant structure, the variant subassembly line can be designed as a three-station flowshop line.

N o t a t i o n : J = number of operations; i= product index; j = operation index;

m = station index; 0 = processing time of operation j; t O- t/, if operation is required by product i; 0, otherwise;

tSim = processing time of product i at station m.

If one views each station of an assembly line as a machine, the scheduling problem of a three- station flowshop assembly line can be treated as a three-machine flowshop. Johnson's algorithm for the two-machine flowshop problem can be extended to a special case of the three-machine flowshop problem [11]. If

either min{tsil l i = 1 . . . . . N } >max{tsi2li = 1 . . . . . N }

or min{tsi3[i= 1 . . . . . N } > m a x { t s a i = 1 . . . . . N} (6)

i.e., the maximum processing time on the second machine is no greater than the minimum processing time on either the first or the third one, an optimal schedule can be found by letting

ai = ts i t + tsi2, bi = tsi2 + tsi3

and scheduling the products as if they were to be processed on two machines only but with the processing time of each product being ai and b i o n the first and second machine, respectively.

A three-station assembly line can be designed so that the processing times of the products satisfy condition (6). In this case, the scheduling problem can be easily solved.

Suppose that the condition min{tsi3: i = 1 . . . . . N} > max{tsa: i = 1,.. . ,N} is used to design the three-station assembly line. Let T~ = {ta, ta . . . . . t 0 . . . . , t~j} be the processing time vector of product Pi. The following heuristic procedure is developed for products with linear variant assembly structures.

Design Procedure 2 (DP2)

Step 0: In i t i a l i za t ion .

Let I be the set of products to be produced.

Modular products

Fig. 15. The variant structure of the three products.

Table 4. The processing times (Example 4)

49

Operation No. 1 2 3 4 5 ti 1 2 1 3 2

"* maxl Y~J=l to: i = l . . . . . N}. Set a = t = Let tma x be the maximum processing on station 2, and tmi n be the minimum processing

time on station 3. Set tma x = 0 a n d tmi n = 0.

Step 1: Assignment of operations of product a. 1 J Compute rave = ~Y~j=I taj, where tave is the average processing time of product a on a

station. Let tsik be the processing time of product i at station k. Assign sequentially operations of

product a to stations 1 and 2 such that tSal and tSa2~tave- Assign the remaining operations to station 3. Let Qk, k = 1,2,3, be the set of operations

assigned to station k. Set tma x = tSa2 a n d tmi n = tSa3.

Step 2:

if if if

For i¢/~{a), do:

ZjcQ2 tii > tmax then set tmax = Y~4cQ2 t~i; ~-~j~Q3 to-<tmi n then set tmin = Y~J'Q3 to'; tmin < tmax then remove operations in Q2 from left to right to Q3 until tmin~tmax-

Step 3: Compute ai = tSil + ts;2, bi = ts;2 + tsi3. The choice between a two-station line and three-station line is made based on two criteria.

First, the target throughput rate of the assembly line has to be met. For example, if a target throughput rate of 0.3 is required and the throughput rate of two-station line and three-station line is 0.25 and 0.35, respectively, then the three-station line design is selected. The second criteria to select the design with the smallest balance cost if both designs satisfy the target throughput rate.

Next, an example is provided to illustrate the application of the three-station flowshop line design approach.

Example 4. Design a three-station line for the three products with a linear variant structure as shown in Fig. 15. The processing times of variant assembly operations are provided in Table 4.

The production quantity for each product is as follows: Ql = 3, Q2 = 4, Q3 = 2. The processing time vector for each product is given as follows: T~ = { 1, 2, 1, 3, 2}, 7"2 = { 1, 0,

1, 3, 2}, 7"3={1, 2, 1, 0, 2}.

Step 0: Initialization.

Set I = {1, 2, 3}. a = i*=max{9, 7, 6} = 1.

tmax = train = 0.


- - ~ @ @ ~ - ~ B u f f e r ~ ~ - ~ Buffer ~ 4 ~ @ @ b

S1 $2 $3

Fig. 16. The three-station line.

SI

S2

$3

2 7 12 17 22 27 32 37 39 41 Time

product P2 ~ product P1 @ product P3

Fig. 17. The minimum makespan schedule of the three-station flowshop line.

Step 1: tave= 9/3 = 3. Assign operations of product 1 to stations 1, 2, and 3: QI = {1, 2}, tSll =3; Q2 = {3}, tSl2 = 1; Q3 = {4, 5}, tSl3 = 5. Set tmax = ts12 = I, train = tsl3 = 5.

Step 2: For i = 2 (product 2):

ts21 = 1, ts22 = 1, ts23 = 5; For i = 3 (product 3): ts31 = 3, ts32 = 1 and/min = ts33 = 2 since ts33 < tmin. Since train > tmax, the three-station line design is completed.

Step 3 :a1=3 + 1 = 4 , b1=1 + 5 = 6;

a2 = 1 + 1 = 2, b2 = 1 + 5 = 6; a3=3 + 1 = 4, b3=l + 2 = 3.

The three-station line design is shown in Fig. 16. The minimum makespan schedule obtained {P2(4), Pl(3), P3(2)} is shown in Fig. 17. The two-station line design obtained by DP1 is shown in Fig. 18. The corresponding minimum makespan schedule is shown in Fig. 19.

From Figs 18 and 19, one can see that the minimum makespans for the two-station flowshop line and three-station flowshop line are identical and thus the throughput rate is Rv=9/ 41 = 0.2. The cycle time for the basic subassembly line is Cb = 5.

4eee ,,,orHe e k S1 $2

Fig. 18. The two-station flowshop line.

M o d u l a r p r o d u c t s 51

S1

$2 2 7 1 2 1 7 2 2 2 7 3 2 3 7 3 9 4 1 T i m e

I p r o d u c t P 2 [ " 7 p r o d u c t P 1 ~ - - ~ p r o d u c t P 3

F i g . 19. T h e m i n i m u m m a k e s p a n s c h e d u l e o f t h e t w o - s t a t i o n f l o w s h o p l i ne .

The balance cost for both lines is computed next. For the three-station flowshop line:

D22 = D32 = D42 = D52 = 4 , D 6 2 = D 7 2 = D 8 2 = D92 = 2; D 2 3 = D 3 3 = D 4 3 = 4 . D53 = 6 . D 6 3 = D 7 3 = D 8 3 = D 9 3 = 4 ; all o the r Dim=O;

1,.2 = 132 = 142 = 152 = 4, 16z = 17z = 18:z = 192 = 2;

123 = 133 = 143 = 4 . 153 = 6 , 163 = 173 = / 8 3 = / 9 3 = 4; •54 = 2 , 194 = 3; all o the r 1.,, = 0; 1 - 67 ,

W~_2 - W32 = W42 = W52 = 4 - 4 = O. W62 = W72 = W82 = W92 = 2 - 2 = 0; W23 = W33 = W43 = 4 - 4 = 0, W53 = 6 - 6 = 0, W63 = W73 = W83 = W93 = 4 - 4 = 0; all o the r IV;,. = 0; W = 0 .

For the two-station flowshop line: D22 = D32 = D 4 2 = D 5 2 = 3 - 3 = 0, D 6 2 = D72 = D82 = 2 - 2 = 0, 0 9 2 = 1 - - 1 = 0; all o the r D;m = 0; 122 =132 ~142 = 1 5 2 - 3, 162=172=182=2, /92 = 1; 153 = 1, 1~3 = 1, 193 - 3: all o the r llm = 0; 1 - 24. W22 = W32 = W42 - W52 = 3 - 3 = 0, W62 = W72 = W82 = 2 - 2 = 0, Wgz = 1 - 1 = 0; all o the r IV;,,, = 0; W = 0 .

Assume that a = $100, b = $1 min -1, and c = $0.25 min -l. The difference between the balance cost of the three-station flowshop line and two-station flowshop line is computed as follows:BC(3)- BC(2) = $100 × 1 + $1 × ( 0 - 0 ) + $0.25 × (67- 24) = $110.75.

Since the two-station flowshop line has a lower balance cost than the three-station flowshop line, the two-station flowshop line is selected.

C O N C L U S I O N S

Product modularity has become an important issue for manufacturing companies that are striving to produce a large variety of products at low cost. It allows to produce different products through combination of standard components. One of the characteristics of modular product family is that its products share the same assembly structure for most of the assembly operations and a small number of variant operations may differentiate two different products. The special structure of modular products provides challenges and opportunities for operational design of assembly lines. In this paper, a new approach to design of assembly lines for modular products was proposed. This approach divides the assembly line into two subassembly lines: a subassembly line for basic assembly operations and a subassembly line for variant assembly operations. The design of the subassembly line for basic operations can be viewed as a single product assembly line balancing problem and solved by the existing line balancing methods. The subassembly line for variant operations is designed as a two-station flowshop line. The proposed design approach allows optimal scheduling of the assembly line and reduces the number of stations and the idle time and waiting times of the assembly line. For a special structure of modular products, a three-station flowshop line design approach was proposed and illustrated with an example. The three-station flowshop line design approach is an extension of the two-station flowshop line design approach.


Acknowledgements This research has been partially supported by funds from the National Science Foundation (Grant No. DDM-9215259) and Rockwell International Corporation

REFERENCES

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