Probability of demand satisfaction for hybrid production networks using a topological technique

Abstract

This article presents a topological technique for evaluating the performance of a hybrid flow-shop production system. Such a production system is modeled as a hybrid production network; each workstation of the hybrid production network has stochastic capacity levels. This article evaluates the probability of demand satisfaction as a performance indicator for the hybrid production network, in which the network has multiple lines and each workstation has a distinct defect rate. First, a topological transformation is utilized to transform a hybrid flow-shop production system into a network-structured hybrid production network; then, the hybrid production network is decomposed into several routes for flow analysis. Second, two procedures for two models (Model I considers the hybrid production network with parallel lines, Model II with intersectional lines) are designed to generate the minimal required capacity for workstations to satisfy the given demand. The probability of demand satisfaction is subsequently evaluated in terms of the minimal required capacities by applying the recursive sum of disjoint products algorithm.

Introduction

In a flow-shop production system, each workstation is made up of several identical machines. Such a production system is a typical hybrid flow-shop production system.^1,2 This implies that the workstation has stochastic capacity levels, with performance ranging from a perfectly working order to complete malfunction, due to the failure, partial failure, or maintenance of machines. Hence, the production system also involves stochastic capacity levels, which can be modeled by the capacitated-flow network.^3–5 A hybrid flow-shop production system with stochastic capacity levels is termed as a hybrid production network (HPN). In the HPN, arrows represent workstations and nodes represent inspection stations following the workstations. Some studies^6,7 have been devoted to constructing a hybrid flow-shop production system as an HPN. Those studies evaluated the demand satisfaction of an HPN on the basis of two assumptions: (1) no flow is increased or decreased during transmission, and (2) no cycle (return) is allowed. Such strong assumptions, however, are not appropriate for an HPN because of the possibility of damage, defect, and rework of products in real life.^8–10 In a practical HPN, defective products might be repaired (i.e. rework of defective products) or scrapped.^9–11 Therefore, how repair affects the amount of output products is an important issue to consider. More recently, Lin and Chang^12,13 proposed an HPN model with only one line to measure the performance of a production network taking single repair into consideration. However, their works lack applicability to multiple lines in most real-life cases.

In order to meet the practical needs of the HPN, this article considers three important characteristics: (1) multiple lines, (2) workstations’ distinct defect rates, and (3) multiple repairs. This article presents a topological technique to transform a production system into a network-structured HPN. The transformed HPN is decomposed into several routes: primary production routes and repair routes. The amount of raw materials and input flow of each workstation can be derived according to the decomposed routes. Subsequently, the minimal required capacity (MRC) is generated for workstations to meet a given demand. The probability of demand satisfaction (PDS) that the HPN could produce d units of product per unit time could be evaluated in terms of MRCs. Two models are considered: Model I considers the HPN with parallel lines, and Model II considers the HPN with intersectional lines, meaning that common workstations process work-in-process (WIP) from different lines. The rest of this article is structured as follows. Two models and assumptions are described in section “Problem statement and assumptions.” The model construction, which uses a topological technique to transform and decompose the HPN, is described in section “Topological model constructions.” Flow analysis and PDS evaluation are formulated in section “Input flow and PDS of HPN.” Two procedures are proposed in section “Procedures for two models” to generate the MRCs for demand in both models. A case-based example of a printed circuit board (PCB) is illustrated in section “Case study of a PCB production system.” The conclusion is summarized in section “Conclusion.”

Problem statement and assumptions

This article measures the performance of the HPN with multiple lines. This article analyzes the input flow of each workstation, where the input flow is the amount of raw materials or WIP each workstation processes per unit time. The PDS is evaluated for a specified demand d as a performance indicator. Two models are considered.

Model I for parallel lines

Two identical lines in parallel produce the same product type. That is, the ordered workstations and their functions in both lines are the same.

Model II for intersectional lines

Two different lines produce the same product type. Most workstations and their functions in both lines are the same. However, one of the two lines may produce a compact product type using fewer workstations. The other line produces more than one product type (regular product type using all workstations and compact product type using fewer workstations). Hence, a common workstation is utilized when producing a regular product type using both lines.

The models to evaluate the PDS of an HPN are developed based on the following assumptions:

1. All inspection stations (nodes) are perfectly reliable; no WIP or product is damaged by inspection nodes.

2. The capacity of each workstation (arrow) is a random variable according to a given probability distribution.

3. The capacities of different workstations (arrows) are statistically independent.

4. If the defective WIP after repair is still defective, then it is scrapped.

Topological model constructions

A common topological technique is proposed for both models to transform a production system into an HPN. The transformed HPN topology is decomposed into routes for flow analysis.

Activity-on-arrow-formed production system

An activity-on-arrow (AOA) diagram is utilized to represent a production system; each arrow denotes a workstation comprising several identical functional machines and each node denotes an inspection station following the workstation. Let (N, A) be a hybrid flow-shop production system with two lines, where N represents the set of nodes, and A = {a_i|i = 1, 2, …, n₁ + n₂} represents the set of arrows, where n₁ (n₂) is the number of workstations in the first line L₁ (the second line L₂). In Model I for instance, Figure 1 represents a production system with two lines in the AOA-formed diagram, in which L₁ = {a₁, a₂, …, a _{n 1} } and L₂ = {a_n1+1, a_n1+2, …, $a_{n_1+n_2}$ } produce the same type of product. The success rate of the ith workstation is p_i, where 0 ≤ p_i ≤ 1 (i.e. the defect rate is q_i = 1−p_i). In Figure 1, the two black nodes in each line represent the input inspection station (which checks raw materials) and the output inspection station (which checks final products). The white nodes between two workstations denote the WIP inspection stations to check whether it can enter the next process (success) or should be scrapped (failure). The meshed nodes indicate that defective WIP inspected by this inspection station can be repaired. Let a _{r j,u} denote the workstation whose defective WIP output can be repaired by the uth repair action in L_j, starting from a _{k j,u} . Thus, the first repair in L₁ (L₂) is that of the defective WIP output from a _{r 1,1} (a _{r 2,1} ), starting from a _{k 1,1} (a _{k 2,1} ). Similarly, the second repair in L₁ (L₂) is that of the defective WIP output from a _{r 1,2} (a _{r 2,2} ), starting from a _{k 1,2} (a _{k 2,2} ). For instance, in production line L₁ = {a₁, a₂, …, a₈} in which the first repair is that of the defective WIP from a₄ starting from a₂. Hence, a _{r 1,1}  = a₄ and a _{k 1,1}  = a₂ for L₁.

Figure 1.

AOA-formed production system in Model I.

Topological technique

The AOA-formed production system is transformed into an HPN to distinguish the input flows from the regular process (without repair) and the repair process. A topological technique is presented to transform the production system into a network-structured HPN. The transformed HPN is shown in Figure 2, in which all the possible routes are depicted in the network topology.

Figure 2.

Transformed HPN for Figure 1.

The HPN is further decomposed into routes. Considering that the jth line L_j can make two repairs, there are four combinations for these routes: (1) primary production route $L_j^{(\mathrm{P})}$ without repair; (2) repair route with only the first repair, which indicates that the defective WIP output from workstation a _{r j,1} is repaired starting from workstation a _{k j,1} ; (3) repair route with only the second repair, in which the defective WIP output from workstation a _{r j,2} is repaired starting from workstation a _{k j,2} ; and (4) repair route with both the first and the second repairs. Combinations (2) and (3) are termed as one-pass repair routes $L_j^{({\mathrm{R}}_1)}$ and $L_j^{({\mathrm{R}}_2)}$ , respectively, while combination (4) is termed as two-pass repair route $L_j^{({\mathrm{R}}_1{\mathrm{R}}_2)}$ . In the special case that a _{r j,1} is the same as a _{k j,1} , the defective WIP is repaired at the same workstation (Figure 3).

Figure 3.

Decomposition for L₁ in Figure 2: (a) primary production route without repair, (b) repair route with only the first repair, (c) repair route with only the second repair and (d) repair route with both the first and the second repairs.

Input flow and PDS of HPN

Common formulations are constructed for both models to derive the amount of raw materials and input flow of each workstation. The workload of each workstation and the MRCs for workstations to meet demand are determined according to the amount of raw materials. Thereafter, the PDS of an HPN is calculated in terms of such MRCs.

Amount of raw materials

The amount of raw materials should be pre-calculated according to the demand requirement. Because multiple lines are considered in this work, the demand d can be assigned to different lines. For each line L_j, suppose that I_j units of raw materials are able to produce O_j units of product; this article intends to obtain the relationship between I_j and O_j such that O_j satisfies d_j and $\sum _{j=1}^2{d}_j=d$ , where d_j is the assigned demand for L_j. Each I_j cannot exceed the maximum capacity of L_j, where the maximum capacity of L_j is M(L_j) =min{M_i|i: a_i∈L_j}. That is, the maximum capacity of L_j is determined by the bottleneck workstation at that line. The following constraint shows that the raw materials cannot exceed M(L_j)

$$I_j\le M(L_j)=\min \{M_i|i:a_i\in L_j\}$$

(1)

A set ${\Delta }_i^{-}$ is utilized to store the workstations prior to a_i. The function Φ(Δ) is the product of success rates of set Δ. That is

$$\mathrm{\Phi }(\Delta )=\underset{t:a_t\in \Delta }{\Pi }{p}_t$$

(2)

For instance, given ${\Delta }_6^{-}=\{a_1,a_2,a_3,a_4,a_5\}$ and $L_j^{(R_1)}=\{a_2,a_3,a_4,a_5,a_6,a_7,a_8\}$ . Then, $\mathrm{\Phi }({\Delta }_6^{-}\cap L_1^{(R_1)})=\underset{t:a_t\in {\Delta }_6^{-}\cap L_1^{(R1)}}{\Pi }{p}_t=p_2,p_3,p_4,p_5$ , in which the intersection set $\{{\Delta }_6^{-}\cap L_1^{(R_1)}\}$ is $\{a_2,a_3,a_4,a_5\}$ . In the special case, Δ is a null set Ø and Φ(Ø) is defined as 1. The amount of output products for L_j is

$$\begin{array}{c}\multicolumn{1}{c}{O_j=I_j\{\left[\mathrm{\Phi }\left(L_j^{(\mathrm{P})}\right)\right]+\left[\mathrm{\Phi }\left({\Delta }_{r_{j,1}}^{-}\right)(1-p_{r_{j,1}})\mathrm{\Phi }\left(L_j^{({\mathrm{R}}_1)}\right)\right]}\\ \multicolumn{1}{c}{+\left[\mathrm{\Phi }\left({\Delta }_{r_{j,2}}^{-}\right)(1-p_{r_{j,2}})\mathrm{\Phi }\left(L_j^{({\mathrm{R}}_2)}\right)\right]}\\ \multicolumn{1}{c}{+\left[\mathrm{\Phi }\left({\Delta }_{r_{j,1}}^{-}\right)(1-p_{r_{j,1}})\mathrm{\Phi }({\Delta }_{r_{j,2}}^{-}\cap L_j^{({\mathrm{R}}_1)})(1-p_{r_{j,2}})\mathrm{\Phi }\left(L_j^{({\mathrm{R}}_1{\mathrm{R}}_2)}\right)\right]\}}\end{array}$$

(3)

where the first term $[\mathrm{\Phi }(L_j^{(\mathrm{P})})]$ is the success rate of $L_j^{(\mathrm{P})}$ . The second term $[\mathrm{\Phi }({\Delta }_{r_{j,1}}^{-})(1-p_{r_{j,1}})\mathrm{\Phi }(L_j^{({\mathrm{R}}_1)})]$ and the third term $[\mathrm{\Phi }({\Delta }_{r_{j,2}}^{-})(1-p_{r_{j,2}})\mathrm{\Phi }(L_j^{({\mathrm{R}}_2)})]$ are the success rates of $L_j^{({\mathrm{R}}_1)}$ and $L_j^{({\mathrm{R}}_2)}$ , respectively. The last term $\mathrm{\Phi }({\Delta }_{r_{j,1}}^{-})(1-p_{r_{j,1}})\mathrm{\Phi }({\Delta }_{r_{j,2}}^{-}\cap L_j^{({\mathrm{R}}_1)})(1-p_{r_{j,2}})\mathrm{\Phi }(L_j^{({\mathrm{R}}_1{\mathrm{R}}_2)})$ is the success rate of $L_j^{({\mathrm{R}}_1{\mathrm{R}}_2)}$ . For convenience, let ${\Gamma }_{j,{\mathrm{R}}_1}=\mathrm{\Phi }({\Delta }_{r_{j,1}}^{-})(1-p_{r_{j,1}})$ , ${\Gamma }_{j,{\mathrm{R}}_2}=\mathrm{\Phi }({\Delta }_{r_{j,2}}^{-})(1-p_{r_{j,2}})$ , and ${\Gamma }_{j,{\mathrm{R}}_1{\mathrm{R}}_2}=\mathrm{\Phi }({\Delta }_{r_{j,1}}^{-})(1-p_{r_{j,1}})\mathrm{\Phi }({\Delta }_{r_{j,2}}^{-}\cap L_j^{({\mathrm{R}}_1)})(1-p_{r_{j,2}})$ for the following derivation. Given the input I_j, equation (3) can be simplified as

$$O_j=I_j\left\{\mathrm{\Phi }\left(L_j^{(\mathrm{P})}\right)+{\Gamma }_{j,{\mathrm{R}}_1}\mathrm{\Phi }\left(L_j^{({\mathrm{R}}_1)}\right)+{\Gamma }_{j,{\mathrm{R}}_2}\mathrm{\Phi }\left(L_j^{({\mathrm{R}}_2)}\right)+{\Gamma }_{j,{\mathrm{R}}_1{\mathrm{R}}_2}\mathrm{\Phi }\left(L_j^{({\mathrm{R}}_1{\mathrm{R}}_2)}\right)\right\}$$

(4)

Let ${\pi }_j=\mathrm{\Phi }(L_j^{(\mathrm{P})})+{\Gamma }_{j,{\mathrm{R}}_1}\mathrm{\Phi }(L_j^{({\mathrm{R}}_1)})+{\Gamma }_{j,{\mathrm{R}}_2}\mathrm{\Phi }(L_j^{({\mathrm{R}}_2)})+{\Gamma }_{j,{\mathrm{R}}_1{\mathrm{R}}_2}\mathrm{\Phi }(L_j^{({\mathrm{R}}_1{\mathrm{R}}_2)})$ . In other words, π_j is the overall success rate of L_j. Thus, equation (4) is simplified as

$$O_j=I_j{\pi }_j$$

(5)

It is necessary that O_j ≥ d_j for obtaining sufficient output that satisfies the demand of L_j. The following equation guarantees that the HPN can produce exact sufficient output O_j that satisfies demand d_j

$$I_j=\frac{d_j}{{\pi }_j}$$

(6)

Input flow of each workstation

Let $f_{j,i}^{(\mathrm{P})}$ , $f_{j,i}^{({\mathrm{R}}_1)}$ , $f_{j,i}^{({\mathrm{R}}_2)}$ , and $f_{j,i}^{({\mathrm{R}}_1{\mathrm{R}}_2)}$ represent the input flow of a_i in $L_j^{(\mathrm{P})}$ , $L_j^{({\mathrm{R}}_1)}$ , $L_j^{({\mathrm{R}}_2)}$ , and $L_j^{({\mathrm{R}}_1{\mathrm{R}}_2)}$ , respectively. Once the amount of raw materials I_j is determined, the input flow of each workstation can be formulated in terms of Φ(Δ) by

$$f_{j,i}^{(\mathrm{P})}=I_j\mathrm{\Phi }({\Delta }_i^{-})\mathrm{for}i\mathrm{such}\mathrm{that}{a}_i\in L_j^{(\mathrm{P})}$$

(7)

$$f_{j,i}^{({\mathrm{R}}_1)}=I_j{\Gamma }_{j,{\mathrm{R}}_1}\mathrm{\Phi }({\Delta }_i^{-}\cap L_j^{({\mathrm{R}}_1)})\mathrm{for}i\mathrm{such}\mathrm{that}{a}_i\in L_j^{({\mathrm{R}}_1)}$$

(8)

$$f_{j,i}^{({\mathrm{R}}_2)}=I_j{\Gamma }_{j,{\mathrm{R}}_2}\mathrm{\Phi }({\Delta }_i^{-}\cap L_j^{({\mathrm{R}}_2)})\mathrm{for}i\mathrm{such}\mathrm{that}{a}_i\in L_j^{({\mathrm{R}}_2)}$$

(9)

$$f_{j,i}^{({\mathrm{R}}_1{\mathrm{\&R}}_2)}=I_j{\Gamma }_{j,{\mathrm{R}}_1{\mathrm{\&R}}_2}\mathrm{\Phi }({\Delta }_i^{-}\cap L_j^{({\mathrm{R}}_1{\mathrm{\&R}}_2)})\mathrm{for}i\mathrm{such}\mathrm{that}{a}_i\in L_j^{({\mathrm{R}}_1{\mathrm{\&R}}_2)}$$

(10)

The summation of input flows is defined as the workload of each workstation, denoted by w_i. Hence, the workload of a_i is

$$w_i={\displaystyle \sum _{j=1}^2(f_{j,i}^{(\mathrm{P})}}+f_{j,i}^{({\mathrm{R}}_1)}+f_{j,i}^{({\mathrm{R}}_2)}+f_{j,i}^{({\mathrm{R}}_1{\mathrm{R}}_2)}),i=1,2,\mathrm{\&.},n_1+n_2$$

(11)

Note that the workload of a_i cannot exceed the maximal capacity M_i. If w_i > M_i, the PDS is evaluated to be 0.

Let x_i denote the capacity of each workstation a_i. According to assumption 2, the capacity x_i of a_i is a random variable, and thus, the HPN is stochastic. Assume that x_i takes possible values $0=x_{i1}<x_{i2}\dots <x_{i{\alpha }_i}=M_i$ with α_i denoting the number of possible capacities of a_i. Under the capacity state $X=(x_1,x_2,\dots ,x_{n_1+n_2})$ , constraint (12) is necessary to guarantee that a_i can process the input raw materials/WIP

$$x_i\ge w_i\mathrm{for}i=1,2,\dots ,n_1+n_2$$

(12)

PDS

Given the demand d, PDS(d) is the probability that the output product from the HPN is no less than d. Thus, PDS(d) is Pr{X|V(X) ≥ d}, where V(X) is defined as the maximum output under X. That is, all the workstations have to provide sufficient capacities to process the raw materials/WIP and eventually produce enough units of output products. However, it is time-consuming to list all X such that V(X) ≥ d and then sum up their probabilities to derive PDS(d). The minimal ones, say Y, in the set {X|V(X) ≥ d} are claimed to be an MRC for d. Hence, Y is an MRC for d if and only if (1) V(Y) ≥ d and (2) V(X) < d for any capacity state X such that X < Y. Given {Y₁, Y₂, …, Y_h}, the set of MRCs satisfying demand, with the corresponding sets B_v = {X|X ≥ Y_v} and v = 1, 2, …, h, the PDS(d) is

$$\mathrm{PDS}(d)=\mathrm{Pr}\left\{\bigcup _{v=1}^h{B}_v\right\}$$

(13)

Several methods, such as inclusion–exclusion method,^6,14,15 disjoint-event method,^14,16 state-space decomposition,^17–19 and recursive sum of disjoint products (RSDP) algorithm, ²⁰ may be applied to compute $\mathrm{Pr}\{\bigcup _{v=1}^h{B}_v\}$ . The RSDP algorithm has been proved^19,20 to be the most efficient among the above methods. Thus, this study applies the RSDP algorithm to calculate the PDS in terms of MRCs.

Procedures for two models

Two procedures are presented for both models to generate the MRCs for workstations to satisfy d. The PDS can be derived in terms of MRCs by the RSDP algorithm.

Procedure I: generate MRCs for d in Model I

In Model I, two parallel lines are given, say $L_1=\{a_1,a_2,\dots ,a_{n_1}\}$ and $L_2=\{a_{n_1+1},a_{n_1+2},\dots ,a_{n_1+n_2}\}$ (n₁ = n₂), to produce the same product type. The MRCs for d in Model I are derived with the following steps.

• Step 1. Find the maximum output for each route

  $$\begin{array}{c}\multicolumn{1}{c}{O_{1,\max }=\min \{M_i|i:a_i\in L_1\}{\pi }_1}\\ \multicolumn{1}{c}{O_{2,\max }=\min \{M_i|i:a_i\in L_2\}{\pi }_2}\end{array}$$

  (14)

• Step 2. Find the demand assignment (d₁, d₂) satisfying d₁ + d₂ = d under constraints d₁ ≤ O_1,max and d₂ ≤ O_2,max.

• Step 3. For each demand pair (d₁, d₂), do the following steps:

• Step 4. Those Y obtained in Step 3 are the MRCs for d. Calculate the PDS(d) in terms of MRCs by the RSDP algorithm.

Procedure II: generate MRCs for d in Model II

In Model II, two intersectional lines are given, say L₁ = {a₁, a₂, …, a _c , …, a _{n 1} } and L₂ = {a_n1+1, a_n1+2, …, a _c , …, $a_{n_1+n_2}$ } (n₁ > n₂), to produce the same product type, in which a _c is set as a common workstation (assumed a _c is installed in L₁) (Figure 4). In the following context, the common workstation is marked in bold to be easily identified.

Figure 4.

AOA-formed production system in Model II.

The MRCs for d in Model II are derived with the following steps:

• Step 1. Find the maximum output for each route

  $$\begin{array}{c}\multicolumn{1}{c}{O_{1,\max }=\min \{M_i|i:a_i\in L_1\}{\pi }_1}\\ \multicolumn{1}{c}{O_{2,\max }=\min \{M_i|i:a_i\in L_2\}{\pi }_2}\end{array}$$

  (18)

• Step 2. Find the maximum output of the common workstation a _c

  $$O_{\mathrm{cmax}}=M_{\mathrm{c}}\max \{{\pi }_1,{\pi }_2\}$$

  (19)

  If d ≤ O_cmax, go to Step 3; otherwise, stop (i.e. d > O_cmax indicates that the production network cannot satisfy such a demand level).

• Step 3. Find the demand assignment (d₁, d₂) satisfying d₁ + d₂ = d under constraints d₁ ≤ O_1,max and d₂ ≤ O_2,max.

• Step 4. For each demand pair (d₁, d₂), do the following steps:

• Step 5. Those Y obtained in Step 4 are the MRCs for d. Calculate the PDS(d) in terms of MRCs by the RSDP algorithm.

Case study of a PCB production system

A PCB production system is adopted to illustrate the PDS evaluation for both models. The function, success rate, and capacity data of each workstation are provided in Table 1. The capacity is measured in terms of pieces of boards processed per day (pcs/day). For instance, the specification of a deburring machine is estimated as 300,000 ft²/month. Hence, for the board size of 24″ × 24″, the capacity of the deburring machine is 75,000 pcs/month (i.e. 2500 pcs/day). The deburring workstation a₃ comprising four machines has five capacity levels, say {0, 2500, 5000, 7500, 10,000}. ¹³ The lowest level 0 corresponds to complete malfunction of all machines, while 10,000 is the highest level, at which all machines operate successfully. The capacity distribution is in the form of probability mass function, indicating that each workstation can provide a certain capacity level with a corresponding probability. For instance, the probability that a₃ can produce 5000 pcs/day (or more) is Pr{x₃≥5 000} = Pr{x₃=5000} + Pr{x₃ = 7500} + Pr{x₃ = 10,000} = 0.003 + 0.015  + 0.980 = 0.998.

Table 1.

Workstation data of PCB production system.

Workstation (ai)	Success rate (pi)	Capacity a	Probability	Workstation (ai)	Success rate (pi)	Capacity a	Probability
Shearing (a1/a11)	0.980/0.973	0	0.015/0.013	Chemical etching (a6/a16)	0.958/0.953	0	0.002/0.002
		4500	0.024/0.030			3000	0.003/0.002
		9000	0.961/0.957			6000	0.010/0.008
						9000	0.985/0.988
Automated drilling ( a 2 b /a12)	0.981/0.981	0	0.012/0.003	Resist stripping (a7/a17)	0.976/0.974	0	0.015/0.005
		5000	0.025/0.004			2500	0.020/0.005
		10,000	0.030/0.013			5000	0.020/0.020
		15,000	0.933/0.980			7500	0.030/0.080
						10,000	0.915/0.890
Deburring (a3/a13)	0.960/0.958	0	0.001/0.003	Scrubbing (a8/a18)	0.988/0.981	0	0.002/0.001
		2500	0.001/0.004			3500	0.007/0.008
		5000	0.003/0.009			7000	0.019/0.014
		7500	0.015/0.011			10,500	0.972/0.977
		10,000	0.980/0.973
Scrubbing (a4/a14)	0.985/0.980	0	0.002/0.005	Legend printing (a9/a19)	0.943/0.950	0	0.015/0.012
		3500	0.003/0.008			4500	0.015/0.017
		7000	0.011/0.008			9000	0.020/0.030
		10,500	0.984/0.979			13,500	0.950/0.941
Imaging (a5/a15)	0.956/0.962	0	0.001/0.003	Packaging (a10/a20)	0.990/0.987	0	0.001/0.001
		1500	0.001/0.003			4000	0.005/0.006
		3000	0.010/0.012			8000	0.015/0.015
		4500	0.012/0.015			12,000	0.979/0.978
		6000	0.012/0.018
		7500	0.030/0.021
		9000	0.934/0.928

a

The unit of capacity is processed per day.

b

a ₂ serves as the common workstation in Model II.

For both models, the PCB production system has to satisfy d = 7000 pcs/day, with output boards beingbatched by every 1000 boards. The PDSs for both models are derived in the following subsections.

Model I

In Model I, two identical lines, L₁ and L₂, are in parallel. The defective WIP outputs from a₈ and a₁₈ are repaired starting from a₆ and a₁₆, respectively (see Figure 5). The defective product outputs from a₁₀ and a₂₀ can be repaired by the same workstation. The first line L₁ is divided into one primary production route $L_1^{(P)}=\{a_1,a_2,a_3,a_4,a_5,a_6,a_7,a_8,a_9,a_{10}\}$ and three repair routes $L_1^{(R_1)}=\{a_6,a_7,a_8,a_9,a_{10}\}$ , $L_1^{(R_2)}=\{a_{10}\}$ , and $L_1^{(R_1{R}_2)}=\{a_{10}\}$ . Similarly, the second line L₂ is divided into one primary production route $L_2^{(P)}=\{a_{11},a_{12},a_{13},a_{14},a_{15},a_{16},a_{17},a_{18},a_{19},a_{20}\}$ and three repair routes $L_2^{(R_1)}=\{a_{16},a_{17},a_{18},a_{19},a_{20}\}$ , $L_2^{(R_2)}=\{a_{20}\}$ , and $L_1^{(R_1{R}_2)}=\{a_{20}\}$ . The MRCs for d = 7000 pcs/day in Model I is derived as follows.

• Step 1. Find the maximum output for each route

  $$\begin{array}{l}{O}_{1,\max }=\min \left\{M_i\right|i:a_i\in L_1\}{\pi }_1\\ O_{2,\max }=\min \left\{M_i\right|i:a_i\in L_2\}{\pi }_2\end{array}$$

• Step 2. Find the demand assignment (d₁, d₂) satisfying d₁ + d₂ = 7000 under constraints d₁ ≤ 6889.50 and d₂ ≤ 6828.93. Since the output boards are batched by every 1000 boards, the feasible demand pairs are D₁ = (6000, 1000), D₂ = (5000, 2000), D₃ = (4000, 3000), D₄ = (3000, 4000), D₅ = (2000, 5000), and D₆ = (1000, 6000).

• Step 3. For each demand pair (d₁, d₂), do the following steps:

  (a) For D₁ = (6000, 1000)

• Step 4. Six MRCs for d = 7000 are obtained from Step 3. The results are summarized in Table 3. PDS(7000) = 0.87931 by the RSDP algorithm.

Figure 5.

PCB production system in Model I.

Table 2.

Results of Step 3 for (d₁, d₂) = (6000, 1000) in Model I.

	a 1	a 2	a 3	a 4	a 5	a 6	a 7	a 8	a 9	a 10
$f_{1,i}^{(\mathrm{P})}$	7837.99	7681.23	7535.29	7233.88	7125.37	6811.85	6525.75	6369.14	6292.71	5934.02
$f_{1,i}^{({\mathrm{R}}_1)}$	0.00	0.00	0.00	0.00	0.00	76.43	73.22	71.46	70.60	66.58
$f_{1,i}^{({\mathrm{R}}_2)}$	0.00	0.00	0.00	0.00	0.00	0.00	0.00	0.00	0.00	59.34
$f_{1,i}^{({\mathrm{R}}_1{\mathrm{\&}\mathrm{R}}_2)}$	0.00	0.00	0.00	0.00	0.00	0.00	0.00	0.00	0.00	0.67
wi	7837.99	7681.23	7535.29	7233.88	7125.37	6888.28	6598.97	6440.60	6363.31	6060.61
yi	9000	10,000	10,000	10,500	7500	9000	7500	7000	9000	8000
	a 11	a 12	a 13	a 14	a 15	a 16	a 17	a 18	a 19	a 20
$f_{2,i}^{(\mathrm{P})}$	1317.92	1282.34	1257.97	1205.14	1181.03	1136.15	1082.76	1054.60	1034.57	982.84
$f_{2,i}^{({\mathrm{R}}_1)}$	0.00	0.00	0.00	0.00	0.00	20.04	19.10	18.60	18.25	17.33
$f_{2,i}^{({\mathrm{R}}_2)}$	0.00	0.00	0.00	0.00	0.00	0.00	0.00	0.00	0.00	12.78
$f_{2,i}^{({\mathrm{R}}_1{\mathrm{\&}\mathrm{R}}_2)}$	0.00	0.00	0.00	0.00	0.00	0.00	0.00	0.00	0.00	0.23
wi	1317.92	1282.34	1257.97	1205.14	1181.03	1156.19	1101.85	1073.20	1052.81	1013.17
yi	4500	5000	2500	3500	1500	3000	2500	3500	4500	4000

Table 3.

MRCs generated from Step 4 in Model I.

Demand pair	MRC
D1 = (6000, 1000)	Y1 = (9000, 10,000, 10,000, 10,500, 7500, 9000, 7500, 7000, 9000, 8000, 4500, 5000, 2500, 3500, 1500, 3000, 2500, 3500, 4500, 4000)
D2 = (5000, 2000)	Y2 = (9000, 10,000, 7500, 7000, 6000, 6000, 7500, 7000, 9000, 8000, 4500, 5000, 5000, 3500, 3000, 3000, 2500, 3500, 4500, 4000)
D3 = (4000, 3000)	Y3 = (9000, 10,000, 7500, 7000, 6000, 6000, 5000, 7000, 4500, 8000, 4500, 5000, 5000, 7000, 4500, 6000, 5000, 3500, 4500, 4000)
D4 = (3000, 4000)	Y4 = (4500, 5000, 5000, 7000, 4500, 6000, 5000, 3500, 4500, 4000, 9000, 10,000, 7500, 7000, 6000, 6000, 5000, 7000, 4500, 8000)
D5 = (2000, 5000)	Y5 = (4500, 5000, 5000, 3500, 3000, 3000, 2500, 3500, 4500, 4000, 9000, 10,000, 7500, 7000, 6000, 6000, 7500, 7000, 9000, 8000)
D6 = (1000, 6000)	Y6 = (4500, 5000, 2500, 3500, 1500, 3000, 2500, 3500, 4500, 4000, 9000, 10,000, 10,000, 10,500, 7500, 9000, 7500, 7000, 9000, 8000)

MRC: minimal required capacity.

Model II

In Model II, the drilling workstation a ₂ (the same function as a₁₂) is set as a common workstation to process WIP from both a₁ and a₁₁ (see Figure 6). The first line L₁ is divided into one primary production route $L_1^{(P)}=\{a_1,a_2,a_3,a_4,a_5,a_6,a_7,a_8,a_9,a_{10}\}$ and three repair routes $L_1^{(R_1)}=\{a_6,a_7,a_8,a_9,a_{10}\}$ , $L_1^{(R_2)}=\{a_{10}\}$ , and $L_1^{(R_1{R}_2)}=\{a_{10}\}$ . Similarly, the second line L₂ is divided into one primary production route $L_2^{(P)}=\{a_{11},{\mathit{a}}_2,a_{13},a_{14},a_{15},a_{16},a_{17},a_{18},a_{19},a_{20}\}$ and three repair routes $L_2^{(R_1)}=\{a_{16},a_{17},a_{18},a_{19},a_{20}\}$ , $L_2^{(R_2)}=\{a_{20}\}$ , and $L_1^{(R_1{R}_2)}=\{a_{20}\}$ . The MRCs for d = 7000 pcs/day in Model II is derived as follows.

• Step 1. Find the maximum output for each route

  $$\begin{array}{c}\multicolumn{1}{c}{O_{1,\max }=9000\times 0.76550=6889.50}\\ \multicolumn{1}{c}{O_{2,\max }=9000\times 0.75877=6828.93}\end{array}$$

• Step 2. Find the maximum output of the common workstation a _c ( a ₂ )

  $$\begin{array}{c}\hfill O_{\mathrm{cmax}}=15000\times \max \{0.76550,0.75877\}=11482.50,\hfill \\ \multicolumn{1}{c}{d=7000<O_{\mathrm{cmax}}=11482.50\mathrm{and}\mathrm{go}\mathrm{to}\mathrm{Step}3}\end{array}$$

• Step 3. Find the demand assignment (d₁, d₂) satisfying d₁+d₂ = 7000 under constraints d₁ ≤ 6889.50 and d₂ ≤ 6828.93. Since the output boards are batched by every 1000 boards, the feasible demand pairs are D₁ = (6000, 1000), D₂ = (5000, 2000), D₃ = (4000, 3000), D₄ = (3000, 4000), D₅ = (2000, 5000), and D₆ = (1000, 6000).

• Step 4. For each demand pair (d₁, d₂), do the following steps:

  (a) For D₁ = (6000, 1000)

• Step 5. Six MRCs for d = 7000 are obtained from Step 4. The results are summarized in Table 5. PDS(7000) = 0.86194 is derived by the RSDP algorithm.

Figure 6.

PCB production system in Model II.

Table 4.

Results of Step 4 for (d₁, d₂) = (6000, 1000) in Model II.

	a 1	a 2	a 3	a 4	a 5	a 6	a 7	a 8	a 9	a 10
$f_{1,i}^{(\mathrm{P})}$	7837.99	7681.23	7535.29	7233.88	7125.37	6811.85	6525.75	6369.14	6292.71	5934.02
$f_{1,i}^{({\mathrm{R}}_1)}$	0.00	0.00	0.00	0.00	0.00	76.43	73.22	71.46	70.60	66.58
$f_{1,i}^{({\mathrm{R}}_2)}$	0.00	0.00	0.00	0.00	0.00	0.00	0.00	0.00	0.00	59.34
$f_{1,i}^{({\mathrm{R}}_1{\mathrm{\&}\mathrm{R}}_2)}$	0.00	0.00	0.00	0.00	0.00	0.00	0.00	0.00	0.00	0.67
wi	7837.99	7681.23	7535.29	7233.88	7125.37	6888.28	6598.97	6440.60	6363.31	6060.61
yi	9000	−	10,000	10,500	7500	9000	7500	7000	9000	8000
	a 11	a 2	a 13	a 14	a 15	a 16	a 17	a 18	a 19	a 20
$f_{2,i}^{(\mathrm{P})}$	1317.92	1282.34	1257.97	1205.14	1181.03	1136.15	1082.76	1054.60	1034.57	982.84
$f_{2,i}^{({\mathrm{R}}_1)}$	0.00	0.00	0.00	0.00	0.00	20.04	19.10	18.60	18.25	17.33
$f_{2,i}^{({\mathrm{R}}_2)}$	0.00	0.00	0.00	0.00	0.00	0.00	0.00	0.00	0.00	12.78
$f_{2,i}^{({\mathrm{R}}_1{\mathrm{\&}\mathrm{R}}_2)}$	0.00	0.00	0.00	0.00	0.00	0.00	0.00	0.00	0.00	0.23
wi	1317.92	1282.34	1257.97	1205.14	1181.03	1156.19	1101.85	1073.20	1052.81	1013.17
yi	4500	-	2500	3500	1500	3000	2500	3500	4500	4000

Table 5.

MRCs generated from Step 5 in Model II.

Demand pair	MRC
D1 = (6000, 1000)	Y1 = (9000, 10,000, 10,000, 10,500, 7500, 9000, 7500, 7000, 9000, 8000, 4500, 2500, 3500, 1500, 3000, 2500, 3500, 4500, 4000)
D2 = (5000, 2000)	Y2 = (9000, 10,000, 7500, 7000, 6000, 6000, 7500, 7000, 9000, 8000, 4500, 5000, 3500, 3000, 3000, 2500, 3500, 4500, 4000)
D3 = (4000, 3000)	Y3 = (9000, 10,000, 7500, 7000, 6000, 6000, 5000, 7000, 4500, 8000, 4500, 5000, 7000, 4500, 6000, 5000, 3500, 4500, 4000)
D4 = (3000, 4000)	Y4 = (4500, 10,000, 5000, 7000, 4500, 6000, 5000, 3500, 4500, 4000, 9000, 7500, 7000, 6000, 6000, 5000, 7000, 4500, 8000)
D5 = (2000, 5000)	Y5 = (4500, 10,000, 5000, 3500, 3000, 3000, 2500, 3500, 4500, 4000, 9000, 7500, 7000, 6000, 6000, 7500, 7000, 9000, 8000)
D6 = (1000, 6000)	Y6 = (4500, 10,000, 2500, 3500, 1500, 3000, 2500, 3500, 4500, 4000, 9000, 10,000, 10,500, 7500, 9000, 7500, 7000, 9000, 8000)

MRC: minimal required capacity.

Conclusion

This article constructs a hybrid flow-shop production system as an HPN using the topological technique to evaluate the PDS. Three practical characteristics are considered: (1) multiple lines, (2) workstations’ distinct defect rates, and (3) multiple repairs. Two models are studied in this article. Model I considers parallel lines producing the same product type; Model II considers intersectional lines with common workstations producing the same product type. Two procedures are developed to generate all MRCs that satisfy demand d for both models. In terms of MRCs, the PDS is derived using the RSDP algorithm.

Future research can be devoted to conducting a sensitivity analysis to investigate which workstation in an HPN is most important for improving the PDS. One possible method of sensitivity analysis is to increase the capacity of one workstation at a time (the others retain the same conditions). Once the capacity of a workstation is increased, the PDS is also improved. Eventually, the production manager can investigate which workstation increases the PDS most, making it the most important one in the HPN.

One more possible issue for future research is to address competence management in an HPN. Competence management can involve a more multidimensional and comprehensive approach.^21,22 For instance, experience knowledge management, skills-gap analysis, succession, planning competency analysis, and profiling can be involved in the continuous improvement of competence management. These can be used with the PDS proposed in this article to provide a more comprehensive assessment.