Production System Performance Improvement by Assembly Line-seru Conversion

Abstract

Seru, as a new effective and flexible production system, has been successfully proved in practice and received more and more attention from academic communities. However, line-seru conversion is an interesting but difficult problem because the product demand is uncertain and worker’s skill levels are different. In this paper, we propose a multi-objective model to describe line-seru problem with one product type but many orders for minimizing the total flow time and the total labor cost. Taking line-seru conversion problem in the electronics industries as an example, how much time and cost can be reduced and how many efficiencies can be improved by line-seru conversion are analyzed. According to the solutions of the proposed multi-objective model, seru has a great performance and different seru combinations can bring unequal reduction of time and cost.

1 Introduction

In the volatile and uncertain market, such as electronics industry, the business is in a rapidly changing environment because of its shorter product life cycles, more unpredictable product types, and the variable production volumes with the fast innovation technology [9]. There are two main aspects of demand changes: product variety and product volume [8]. To cope with volatile demand, many electronic giants attempt to apply Toyota Production System (TPS) to satisfy fluctuations in customers demand but failed, and the economies of scale and shorter cycle time brought by mass production are disappearing. The efficiency and low cost of the conveyor assembly line will greatly decrease when met with volatile demands [1]. Hence, traditional TPS system applied in conveyor assembly line could not satisfy the volatile customers demand because of the fast updates of products. In this situation, to produce more variety, flexible volume and high-value-added products, Sony invited an expert, Hitoshi Yamada, to solve this problem after failing to test the new manufacturing approaches including TPS and one-person production organization [12].

In 1992, several short lines replaced the long assembly conveyor line to produce the entire product in Sony [12], and that made a big success. In 1994, Kon Tatsuyoshi, a former staff of Sony, first called the term “seru seisan” for such an innovation of the production management system [11]. In Japan, “seru Seisan” is really popular in some manufacturing industries as the most powerful and practical way under the volatile environment [7]. In fact, seru has many advantages: it can reduce labor cost, lead time, setup time, work-in-process (WIP) inventories, semi-finished and finished product inventories, equipment cost and space. In addition, seru also increases profits, and improves product quality and motivates workers in a great way [10].

Seru is an assembly organization, which is consisted of required equipment and workers, to produce one or more products (only one product type in most cases). A seru has three characteristics: firstly, Kanketsu, which means all tasks are completed in a seru; Secondly, Majime, which means all required resources are placed close to reduce unnecessary movement; thirdly, Jiritsu, which means self-management and learning organization [12]. What’s more, there are three types of seru: Divisional seru, Rotating seru and Yatai [5]. Many famous Japanese companies such as Sony, Canon, Panasonic, NEC, Fujtsu, Sharp and Sanyo have adopted seru and not only acquire economic benefits but also environment benefits.

As seru has continuously achieved big success in not only Japan but also Korea and China, more and more attention have paid to performance indicators of seru. Several papers analyse the performance of line-seru conversion affected by operational factors. Johnson [2] adopted a previous theory to illustrate why assembly serus have a better performance than traditional conveyor assemble line. He studied the simulation models to observe the marginal impact when the operational factors are changed in this conversion. Kaku et al. [3] constructed a theoretical model considering human factors in the conversion. They argued the cross-trained workers should be the key role in the conversion. By using human memory ability they analysed the cross-training of workers quantitatively and found that information support system is benefit for improving the cross-training effect. Yu et al. [13] proposed a 64-array experiment and used three non-dominated solutions to find operational factors or interactions between them which may improve the performance. They suggested several insights about the formation of assembly serus and load serus based on the experimental results.

For the methods to achieve better line-seru performance, many researchers use a multi-objective model to optimize two line-seru performances: the total throughput time and the total labor hour. Kaku et al. [4] researched 24 cases about line-seru conversion and concluded that this conversion can greatly adapt multi-item small-sized products. They proposed a linear weight method to solve the multi-objective problem and determine the most profitable layout of cells and workers in it. Yu et al. [15] constructed a multi-objective model for two goals and adopted an improved exact algorithm by transferring the multi-objective optimization into the single objective optimization. Liu et al. [6] proposed a three-stage heuristic algorithm with nine steps to solve this optimal problem and took several computational cases to validate the performances of model and algorithm by MATLAB programming. Yu et al. [14] proved that the line-seru conversion problem is NP-hard and the non-dominated or pareto-optimal front of the multi-objective problem is non-covex. They developed a non-dominated sorting genetic algorithm which can solve large size problems in a reasonable time to solve multi-objective problem and used several numerical examples to verify the reliability. However, the papers mentioned above just consider time as the optimal objective and suppose that orders can not be split in general. In this paper, the labor cost is introduced as optimal objective and will be calculated with two levels of worker skill, and we also plan to find whether the efficiency increase or not if split the orders. Additionally, the most appropriate quantity of each seru for the best balance will be pointed out.

This paper is aim to show seru is more suitable for the electronics industry by comparing the efficiency of seru and the conveyor line. The paper is organized as follows. In Sect. 95.2, we will present the problem description, including the brief explanation and introduction of variables. Then, the multi-objective model and assumptions as well as notation will be given in detail. In Sect. 95.3, the efficiency of seru and the conveyor line will be shown under a numerical example in electronics industry. Furthermore, in the Sect. 95.4, we will make a comparison between these two product systems in serval aspects to illustrate the advantages of line-seru conversion. The conclusion and future research will be put forward in Sect. 95.5.

2 Problem Statement

The problem considered in this paper is from a practical production problem of low efficiency and flexibility in some assembly areas. As conveyor assembly line has been widely accepted, it has been found many disadvantages increasingly. For example, it should continue working to keep high efficiency and low cost; workers may have much idle time and so on. Therefore, in electronic assembly area, seru has been paid more and more attention because of better performance in some aspects. Seru is a small production assembly organization, it can be constructed, dismantled, and reconstructed quickly. However, when we apply it in practical production, we may face how to design seru and assign workers in it for the best profit rate. In this paper, we introduce a multi-objective model to describe this line-seru conversion problem and propose a solution to solve the numerical example.

3 Modelling

As mentioned above, there are three types of seru: divisional seru, rotating seru and yatai, and Yin et al. [12] introduced three types in detail: Divisional seru is a short assembly line and composed by several partially cross-trained workers responsible for several tasks. Rotating seru is commonly a U-shaped assembly line which is equipped fixed stations for cross-trained workers walking from one to another to perform all required tasks of a product. Yatai is a special seru as there is only one worker in it to produce required products. In this paper, rotating seru and yatai are considered and divisional seru is a further issue to discuss. Assumption is that one product type is produced which has different known orders. Two types of assembly system are shown in the Fig. 95.1, including a pure seru system and a pure conveyor line system. In traditional conveyor assembly line, products are produced constantly in line. But when it comes to seru, we may face how to design serus and assign workers in it for the best profit rate. Hence, this paper considers two objectives: to minimize the total flow time and the total labor cost. And in this section, a multi-objective mathematical model of line-seru will be developed.

Fig. 95.1

Two types of assembly system

3.1 Problem Assumptions

Following assumptions are considered in line-seru to construct the multi-objective model:

• The orders of product are known and constant;

• The number of assembly tasks is the same to two types of assembly system;

• If the assembly system is a conveyor line, just one conveyor line is considered;

• The number of workers is same with the number of tasks on the conveyor line;

• A worker only does one assembly task in conveyor line;

• The number of workers in each seru may be different but limited;

• A worker assigned in a seru can operate all the tasks required;

• The setup time between two batches is zero.

3.2 Notation

Here, the assembly problem based on one product with different product orders is considered, in which workers have been assigned to the conveyor line or seru already. Following notation are used in the multi-objective model.

Indices

i::

Index of set workers ($i=1,2,\ldots ,W$).

j::

Index of set production units ($j=1,2,\ldots ,J$).

m::

Index of set product orders ($m=1,2,\ldots ,M$).

k::

Index of set the sequence of product orders in a seru ($k=1,2,\ldots ,M$).

Parameters

$B_{mn}$::

Size of batch n in product order m ($n=1,2,\ldots ,N$).

TP::

Total time of all processes.

$\eta _{i}$::

Upper bound on the number of tasks for worker i in a cell. If a worker is assigned to a number of tasks over than it, the task time will become longer than ever.

$C_{i}$::

Coefficient of variation of worker i’s increased task time after the line-seru conversion as he or she is from a specialist to a completely cross-trained worker.

$Y_{i}$::

The cost of worker i per second.

$\varepsilon _{i}$::

Worker i’s coefficient of influencing level of doing multiple assembly tasks.

$\beta _{i}$::

Skill level of worker i for each task of product.

Decision variables

$X_{ij}$ ::

If worker is assigned to cell, $X_{ij}=1$, otherwise, $X_{ij}=0$.

$Z_{mnjk}$::

If batch of product order is assigned to cell in sequence, $Z_{mnjk}=1$, otherwise, $Z_{mnjk}=1$. In addition, if $k=0$, $Z_{mnjk}$=0.

Variables

$FS_{mn}$::

 Flow time of product batch n of order m in a seru.

$FSB_{mn}$::

 Begin time of product batch n of order m in a seru.

$T_{bp}$::

 Process time of bottleneck.

3.3 Formulation

Here is an manufacturing assembly problem: there exists a traditional belt conveyor assembly line with various assembly stations, and workers are assigned at each station according to a traditional job design method but they have had abilities to finish all tasks which are easy to learn. Meanwhile, there are also some serus with the same tasks. The assembly plan with one type product has M orders, W workers have been assigned in two product systems already. In addition, the orders are produced according to an First come first service (FCFS) principle.

At first, each worker is specialized with original work in conveyor assembly line, and need training to become as a cross-trained worker. If tasks assigned to a worker are beyond $\eta _{i}$, the flow time will be longer as the worker is not familiar with too many tasks. The details are given:

$$\begin{aligned} C_{i}= \left\{ \begin{array}{ll} 1+\varepsilon _{i}\times (W-\eta _{i}),&{} W\ge \eta _{i},\\ 1,&{}\text{ otherwise }. \end{array} \right. \end{aligned}$$

(95.1)

Subsequently, the flow time of a product varies with workers’ skill levels. $TP\times C_{i}\times \beta _{i}$ means the time for worker i to finish one product. $B_{mn}$ means the size of batch n in order m. Hence, the flow time can be represented as follows:

$$\begin{aligned} FS_{mn}=\frac{B_{mn}\times \max (T_{P}\times C_{i}\times \beta _{i})}{\sum _{i=1}^W\sum _{j=1}^J\sum _{k=1}^M X_{ij}\times Z_{mnjk}}. \end{aligned}$$

(95.2)

Then, since each worker has different skill levels, so according to $\beta _{i}$, all workers are classified in two classes. The cost of different workers is following:

$$\begin{aligned} Y_{i}= \left\{ \begin{array}{ll} a,&{}\text{ if } \,\,\, \beta _{i}\ge 1,\\ b,&{}\text{ otherwise }. \end{array} \right. \end{aligned}$$

(95.3)

Finally, because there is no waiting time and setup time, the start time of each batch is the sum of flow time for all preview product batches assembled in the same seru.

$$\begin{aligned} \text {FSB}_{mn}=\sum _{s=1}^{M-1}\sum _{n=1}^N\sum _{j=1}^J\sum _{k=1}^M FS_{mn}\times Z_{mnjk} \times Z_{snj(k-1).} \end{aligned}$$

(95.4)

(1) Total Flow Time Minimization Objective

The total flow time is depend on the finish time of the last product batch, so we can calculate the finish time of all orders in the following function:

$$\begin{aligned} \text {TFT}=\min \{\max \limits _{mn}(FSB_{mn} + FS_{mn})\}, \end{aligned}$$

(95.5)

where $FS_{mn}+ FSB_{mn}$ is the finish time of batch n in order m.

Besides, when workers are assigned into different serus, we should make sure that all workers are assigned to assemble the required products. Thus,

$$\begin{aligned} \sum _{j=1}^J\sum _{i=1}^W X_{ij}=W, \forall (i,j). \end{aligned}$$

(95.6)

About the worker assignment rule, each worker should be assigned to one and only one seru. That means workers are not allow to help other serus when needed, so

$$\begin{aligned} \sum _{j=1}^J X_{ij}=1, \forall i. \end{aligned}$$

(95.7)

If there is no worker in this seru, we cannot assign assembly task to it, i.e.,

$$\begin{aligned} \sum _{m=1}^M\sum _{k=1}^M Z_{mnjk}=0, \left\{ \forall j\mid \sum _{i=1}^W X_{ij}=0\right\} . \end{aligned}$$

(95.8)

In addition, the product orders must be assigned sequentially, and

$$\begin{aligned} \sum _{j=1}^J\sum _{k=1}^M Z_{mnjk}\le \sum _{j'=1}^J\sum _{k'=1}^M Z_{(m-1)nj'k'}, m=2,\ldots ,M. \end{aligned}$$

(95.9)

(2) Total Labor Cost Minimization Objective

The total labor cost is the sum of the cost of each worker. As mentioned above, we can know how much time each worker spends on assembling, so the objective function is:

$$\begin{aligned} \text {TLC}=\min \sum _{m=1}^M\sum _{n=1}^N\sum _{i=1}^W\left( \sum _{j=1}^J\sum _{k=1}^M FS_{mn}\times X_{ij}\times Z_{mnjk}\times Y_{i}\right) . \end{aligned}$$

(95.10)

Therefore, combining the Eqs. (95.5)–(95.10), we have:

$$\begin{aligned} \begin{array}{l} \text {TFT}=\min \{\max \limits _{mn}(\text {FSB}_{mn} + FS_{mn})\}\\ \text {TLC}=\min \sum \limits _{m=1}^M\sum _{n=1}^N\sum _{i=1}^W(\sum _{j=1}^J\sum _{k=1}^M FS_{mn}\times X_{ij}\times Z_{mnjk}\times Y_{i})\\ \begin{array}{ll} s.t.&{}\left\{ \begin{array}{ll} \sum _{j=1}^J\sum _{i=1}^W X_{ij}=W, &{}\forall (i,j)\\ \sum _{j=1}^J X_{ij}= 1, &{}\forall i\\ \sum _{m=1}^M\sum _{k=1}^M Z_{mjk}=0,&{}\{\forall j\mid \sum _{i=1}^W X_{ij}=0\}\\ \sum _{j=1}^J\sum _{k=1}^M Z_{mjk}\le \sum _{j'=1}^J\sum _{k'=1}^M Z_{(m-1)j'k'}, &{}\,m=2,\ldots ,M.\\ \end{array} \right. \end{array} \end{array} \end{aligned}$$

(95.11)

4 Solution Method

From the descriptions of TLC and TFT, all the parameters are involved in minimization objective and there are many satisfied solutions which are hard to enumerated, as the proposed multi-objective model have been proved as NP-hard. Therefore, in this paper, we try to compare two production systems by the satisfying solutions in the same situation. The orders and workers are known, what we need to do is to find how many serus are formed and how to assign workers in it for the objectives. FCFS principle are applied in seru design, and supposed that orders can not be split at first. The procedure should be processed as follows:

Step 1. :

J serus should be set up and workers are allocated in them on average.

Step 2. :

Find the worker who has the longest flow time in each seru, thus these workers are bottlenecks of each seru, and bottlenecks are arranged in the order of smallest to largest to seru 1, seru $2,\ldots$, serus J.

Step 3. :

At first, order 1 are arranged to seru 1, order 2 are arranged to seru 2 and so on. Thus, calculating the finish time of each order.

Step 4. :

Then, according to FCFS principle, the next order will be arranged to the seru which has least flow time and the order after next are arranged to the seru which has the second least flow time. Repeat this process, until all orders are allocated.

Step 5. :

The slowest finish time is as the TFT, and TLC is the sum of each seru’s labor cost.

On the other hand, considering orders split, how to separate orders to get an optimal results is the most challenging problem. If all serus finish production tasks at the same time, the idle time will be less. Therefore, the most suitable task arrangement for each seru should be calculated. Weighted average is accepted for the best balance due to different workers’ skill levels. The procedure is:

Step 1. :

J serus should be set up and workers are allocated in them.

Step 2. :

Find the worker who has the longest flow time in each seru. These workers are bottleneck of each seru, and bottlenecks are arranged in the order of smallest to largest to seru 1, seru $2,\ldots$, serus J. Recording the workers as $w_{1}$, $w_{2}$, $\dots$, $w_{J}$.

Step 3. :

Calculating the best allocation for the best balance. $\frac{1}{\beta _{w_{j}}\times C_{w_{j}}}$ as the weight and the sum of orders is 401, correspondingly, we can get the appropriate tasks of workers. In a seru, $\frac{401}{\sum _{j=1}^J\frac{1}{\beta _{w_{j}}\times C_{w_{j}}}}\times \frac{1}{\beta _{j}\times C_{j}}$ is one best allocation of seru j.

Step 4. :

Using FCFS principle to arrange tasks and satisfy each seru’s best allocation until all orders are finished. If the quantity of order is less than the best allocation, the order can be allocated to this seru. But if in contrast, the order should be split to satisfy the best allocation for best balance.

Step 5. :

The slowest finish time is as the TFT, and TLC is the sum of each seru’s labor cost.

5 Application

5.1 Data

Here, considering line-seru problem in assembly process of mobile phone, we will analyze how much time and cost can be reduced and how many efficiencies can be improved by line-seru conversion. Assembly process of mobile phone includes 12 workers and 8 orders, and the related data is in Tables 95.1 and 95.2, respectively.

Table 95.1 Process time lists

Table 95.2 The parameters of the example [13]

5.2 Numerical Examples

Since the solution method have been given in the above section, so we can compare two production systems as follows.

(1) Conveyor Assembly Line Production System

There are 8 orders and the quantity of required products is 401 in total. In the conveyor assembly line, 12 processes are involved and the bottleneck is process k. We can calculate the finish time and labor cost as:

$$\text {TFT}=TP+\sum _{m=1}^8\times T_{bp}=10179; \text {TLC}=\text {TFT}\times 12\times b=122148b.$$

(2) Seru Production System

${\textcircled {\small {1}}}$ Orders cannot be split

Supposed that orders cannot be split so that there are 8 serus at most, thus we design four serus shown in the Table 95.3, where $\{1,2,3,4,5,6,7,8,9,10,11,12\}$ means a rotating seru with 12 workers in it.

Table 95.3 Several serus formations

According to FCFS principle, the orders are arranged to appropriate serus as shown in the Fig. 95.2.

Fig. 95.2

Arrangements of orders (1)

In Fig. 95.2 (1), there is only one seru, so all orders are arranged to it and calculation is simple. The bottleneck is worker 8 and the flow time is 205, $\text {TFT}$ is $401\times 205/12=6850$. About $\text {TLC}$, there 5 workers’ skill levels less than 1, so $\text {TLC}$ is $6850\times (5a + 7b)= 34250a+47950b$. In Fig. 95.2(3), there are 3 serus. seru 1 has the least flow time and seru 3 has the most flow time. Order 1 is finished by seru 1, order 2 is seru 2 and order 3 is seru 3. When arranged order 4, seru 2 finishes tasks first, so this order is responsible by seru 2. According to this principle, seru 1 is responsible for order 1, 5, 8; seru 2 is 2, 4, 7; seru 3 is 3, 6. The finish time of each order can be seen in the Fig. 95.2(3). Therefore, we can get:

$$\begin{aligned} TFT_{1}=6850 ; TLC_{1}=34250a+47950b,\\ TFT_{2}=6864; TLC_{2}=33609a+47337b,\\ TFT_{3}=7622; TLC_{3}=33408a+48652b,\\ TFT_{4}=7031; TLC_{4}=33066a+46734b.\\ \end{aligned}$$

${\textcircled {\small {2}}}$ Orders can be split

As four seru formations are listed, we can deduce that when all workers complete required tasks together the $\text {TFT}$ may be less because of little idle time. So we split the orders and arrange them into different serus for the best balance, the best balance as the Table 95.4 shown.

Table 95.4 Several serus formations when orders can be splitted

According to FCFS principle, appropriate orders arrangement are shown in the Fig. 95.3.

Fig. 95.3

Arrangements of orders (2)

In Fig. 95.3, seru formation $\{\{1,2,3,4,5,6\}$, $\{7,8,9,10,11,12\}\}$, and the bottlenecks are worker 2 and worker 8 respectively. $\frac{401}{\left( \frac{1}{\beta _{2}\times C_{2}} + \frac{1}{\beta _{8}\times C_{8}}\right) }\times \frac{1}{\beta _{2}\times C_{2}}$ is 203, which is the number of products arranged to seru 1, and the suitable task arranged to seru 2 is 198. Orders are allocated to seru 1 and 2 one by one and may be split in order to satisfy arrangement. Order 8 is 52 but it is split into 45 and 7 which are allocated to the different serus. In Fig. 95.3(2), we calculated the finish time after each order arrangement and the results are signed. In Fig. 95.3(6), there are 8 serus and orders are split into several parts for the best balance. Therefore, we can obtain the results as:

$$\begin{aligned} TFT_{1}=6850;~~ TLC_{1}=34250a+47950b,\\ TFT_{2}=6800;~~ TLC_{2}=33794a+47188b,\\ TFT_{3}=6717;~~ TLC_{3}=33548a+46916b,\\ TFT_{4}=6696;~~ TLC_{4}=33251a+46519b,\\ TFT_{5}=6561;~~ TLC_{4}=32447a+45235b,\\ TFT_{6}=6401;~~ TLC_{4}=31549a+44049b,\\ TFT_{7}=6369;~~ TLC_{4}=31520a+44070b.\\ \end{aligned}$$

Table 95.5 Optimal solutions in two assembly systems

Hence, we can get the optimal solutions as Table 95.5. From the above results, we can see that seru has higher efficiency and lower cost than the conveyor assembly line. What’s more, whether the orders can be separated or not has a great influence. If the order cannot be split, we will find each seru varies a lot in finish time because it may be responsible for the different orders. For example, in Fig. 95.2(3), seru 1 and seru 2 are responsible for 2 orders, but seru 3 for 3 orders. That means seru 1 and seru 2 have lots of idle time while seru 3 is still working. When separate the order, however, each seru has nearly same finish time. In this situation, $\text {TFT}$ and $\text {TLC}$ are reducing increasingly.

5.3 Comparison and Analysis

Two production systems in this paper represents the transform from the assembly line to the seru. The assembly line becomes shorter and shorter, and the workers’ skills and efficiency become higher and higher. Finally, in some systems, the yatai may be formed. Here are the detailed comparisons in several aspects:

(1) Efficiency

The ideal type of the assembly system is that every station has the same workload but it is not practical because every process is different and workers’ efficiency is affected by various factors. In fact, the assembly system is unbalance with a bottleneck. In the above example of the conveyor assembly line, the productivity is limited by the bottleneck process k. In the other processes, there exists idle time which is a waste of productivity. The total flow time of the conveyor assembly line is 10179. However, in the rotating serus, every worker is responsible for production from begin to end. Although workers are affected by the slowest worker, the idle time is much less than the conveyor assembly line. In the yatai, the bottleneck worker maybe still the lowest one but the others can work faster and better because everyone is in different serus. Without others’ effects, workers in their own serus can arrange their work plans based on the factory’s schedule. As only one worker in seru, there is no idle time because all processes are completed by that worker one by one. There are no waiting time and efficiency lost, the productivity rate is nearly 100 %. As the Table 95.5 shown, seru has higher efficiency. If the orders are not split, the total flow time is 6850, the efficiency has improved 32.7 %. When orders are split, the efficiency has improved 37.4 % compared with the conveyor assembly line, and 7 percent compared with not split orders. Therefore, better design of seru can get better economic benefit.

(2) Workers’ Activity

In the conveyor assembly line, every worker completes a simple process based on the theory of division of labor after a short and easy train. Task is monotonous and dull. In the idle time, workers can only wait. Therefore, the activities are low and the turnover rate of workers is high. However, seru is a self-learning organization and there are several workers or only one worker in each seru, workers have more freedom about how to complete required tasks. They are not only trained into cross-trained but also learn some management knowledge because they may need making some decisions under the authorization system. As a result, workers’ enthusiasm are inspired and have better and better performance.

(3) Labor Cost

In the multi-objective model, labor cost is one of the optimal objectives. From the above comparisons, we can conclude that when orders are split the labor cost may be the lowest. The labor cost is related to total flow time and workers’ salary level. We suppose that two levels of workers’ salary are a and b and $a=1.2b$. According to Table 95.5, the total labor cost of the conveyor assembly line is 122148b, and the cost of whether split orders are 89050b and 81894b respectively. The cost has decreased a lot by 32.7 % compared with the conveyor assembly line. Also, when we split orders, the cost decrease by 7 percent. Training cross-trained workers may be time-consuming but worthwhile because production efficiency will be very high.

6 Conclusion

In this paper, we discuss the problem of production system performance improvement based on the bi-objective model with time and cost minimization. Then, we design the solution method for comparison of assembly line-seru conversion when considering orders split or not. Finally, we test the model and solution by an numerical example, and the results validates that seru production system with order split has a better performance. Although we prove that seru has high efficiency and line-seru conversion is necessary, many practical factors are not considered in this paper such as the cost of worker training, production of multi-types product, random arrival of orders and so on. All areas are very important and worth our equal concern in the future.