Scheduling for fabrication and assembly in a two-machine flowshop with a fixed job sequence

Abstract

This paper studies a problem of scheduling fabrication and assembly operations in a two-machine flowshop, subject to the same predetermined job sequence on each machine. In the manufacturing setting, there are n products, each of which consists of two components: a common component and a unique component which are fabricated on machine 1 and then assembled on machine 2. Common components of all products are processed in batches preceded by a constant setup time. The manufacturing process related to each single product is called a job. We address four regular performance measures: the total job completion time, the maximum job lateness, the total job tardiness, and the number of tardy jobs. Several optimality properties are presented. Based upon the concept of critical path and block schedule, a generic dynamic programming algorithm is developed to find an optimal schedule in O(n ⁷) time.

Appendices

Appendix A: Generalization of Algorithm BA-1 for problem $F2|(c_{j},u_{j},a_{j}),s\mbox{-}\mathit{batch,fix\_seq}|C_{\max}$

There are five steps (a)–(e) in Algorithm BA-1 proposed by Cheng and Wang (1999). In Step (a), the jobs are sequenced according to Johnson’s rule. Step (b) denotes C ⁽¹⁾(j,k) and C ⁽²⁾(j,k) as the minimum machine-1 and machine-2 completion times, respectively, among the partial schedules which consist of jobs 1,…,j and have exactly k batches of common components. The dynamic programming recursive formula and initial values are given in Step (c) and (d), respectively. The optimal schedule whose objective value is min_1≤k≤n C ⁽²⁾(n,k) can be obtained in Step (e).

To solve problem $F2|(c_{j},u_{j},a_{j}),s\mbox{-}\mathit{batch,fix\_seq}|C_{\max}$, we directly use the preassigned job sequence in Step (a) and generalize the recursive formula in Step (c) as

For j=1,…,n and k=1,…,j:

$$C^{(1)}(j,k)=ks+p_{c,[1:j]}+p_{u,[1:j]}; $$

$$\begin{aligned} C^{(2)}(j,k) =&\min_{1\leq i\leq j-k+1} \begin{cases} \max \begin{cases} C^{(1)}(j-i,k-1)+s+p_{c,[j-i+1:j]}\\ \quad{}+\displaystyle\max_{j-i+1\leq h\leq j} \{p_{u,[j-i+1:h]}+p_{a,[h:j]}\},\\ C^{(2)}(j-i,k-1)+p_{a,[j-i+1:j]}. \end{cases} \end{cases} \end{aligned}$$

(3)

Note that i is the number of common components in the last batch of the considered partial schedule.

Appendix B: Reduction of the time complexity of Algorithm BA-1

The recursive equation (Eq. (3)) in Step (c) of Algorithm BA-1 retains O(n ²) states, each of which needs O(n ²) time, and the time complexity of Algorithm BA-1 is O(n ⁴). Since the fixed job sequence is determined in Step (a), we can have an O(n ²) preprocessing procedure for function L(⋅,⋅) inserted between Step (a) and (b). In doing so, Eq. (3) can be modified by replacing max _{j−i+1≤h≤j} {p _{u,[j−i+1:h]}+p _a,[h:j]} with p _u,j−i+1+L(j−i+1,j) and the run time for Step (c) can be reduced to O(n ³). This modification can be justified by the concept of critical path. If the last critical path of a considered partial schedule occurs on one of the jobs {j−i+1,…,j}, which can be determined by $\operatorname{arg\,max}_{j-i+1\leq h\leq j} \{p_{u,[j-i+1:h]}+p_{a,[h:j]}\}$, the time span from the completion of u _j−i+1 to that of a _j can be obtained by L(j−i+1,j).

Appendix C: Example of algorithm DP

Example 2

Consider the following instance with n=2: (p _c,1,p _u,1,p _a,1)=(1,2,2), (p _c,2,p _u,2,p _a,2)=(2,3,4) and s=1. Algorithm DP is demonstrated for the objective function of total completion time as follows:

Stage 1.:

(Preprocessing for L(j ₁,v), δ(j ₁,j ₂,h ₁,h ₂,k ₁) and L′(j ₁,j ₂,v,h ₁,h ₂,k ₁))

• For 1≤j ₁≤v≤2,

  L(1,1)=p _a,1=2;

  L(1,2)=max{L(1,1),p _u,[2:2]}+p _a,2=max{2,3}+4=7;

  L(2,2)=p _a,2=4.

• For 1≤h ₁≤h ₂<j ₁≤j ₂≤2 and $\lceil\frac{j_{1}-h_{2}-1}{j_{1}-1}\rceil\le k_{1}\le j_{1}-h_{2}-1$,

  δ(2,2,1,1,0)=max{0,p _a,[1:1]−s−p _u,[2:2]−p _c,[2:2]}=max{0,2−1−3−2}=0.

• For 1≤h ₁≤h ₂<j ₁≤v≤j ₂≤2 and $\lceil\frac{j_{1}-h_{2}-1}{j_{1}-1}\rceil\le k_{1}\le j_{1}-h_{2}-1$,

  L′(2,2,2,1,1,0)=max{L(2,2),δ(2,2,1,1,0)+p _a,[2:2]}=max{4,0+4}=4;

• For 1≤j ₁≤v≤j ₂≤2,

  L′(1,1,1,0,0,0)=L(1,1)=2;

  L′(1,2,1,0,0,0)=L(1,1)=2;

  L′(1,2,2,0,0,0)=L(1,2)=7;

  L′(2,2,2,0,0,0)=L(2,2)=4.

• For other values of j ₁,j ₂,v,h ₁,h ₂,k ₁, we set L′(j ₁,j ₂,v,h ₁,h ₂,k ₁)=∞.

Stage 2.:

(Algorithm DP)

• Initialization:

• g(0,0,0,0,0)=0;

• For other values of j,i,i ₁,k _L ,k _R , we set g(j,i,i ₁,k _L ,k _R )=∞.

• Recursion: For 1≤k _L ≤i≤i ₁≤j≤2 and $\lceil\frac{j-i_{1}}{j}\rceil\leq k_{R}\leq j-i_{1}$,

• (j,i,i ₁,k _L ,k _R )=(1,1,1,1,0)

• $\quad(j,j',i',i'_{1},k_{L},k'_{L})=(1,0,0,0,1,0)$

•   (L(1,1)=2=p _a,[1:1])∧

•   ((j′=0=i−1)∧(p _a,[0:0]=0<(1−0)s+p _u,[1:1]+p _c,[1:1]=1+2+1=4))

  $$\begin{aligned} z_1=f_1(1,0,0,0,1,0) =&g(0,0,0,0,0)+(1-0)(s+p_{c,[1:1]}+p_{u,[1:1]}) \\ &{}+L'(1,1,1,0,0,0) \\ =&0+(1+1+2)+2=6. \end{aligned}$$

•   g(1,1,1,1,0)=min{z ₁}=6.

• (j,i,i ₁,k _L ,k _R )=(2,1,1,1,1)

•  (j,j′,i,i ₁,k _L ,k _R )=(2,1,1,1,1,1)

•    s+p _u,[2:2]+p _c,[2:2]+L′(2,2,2,1,1,0)=1+3+2+4=10>p _a,[1:2]=6

•     z ₁=∞.

•   g(2,1,1,1,1)=min{z ₁}=∞.

• (j,i,i ₁,k _L ,k _R )=(2,1,2,1,0)

• $\quad(j,j',i',i'_{1},k_{L},k'_{L})=(2,0,0,0,1,0)$

•    L(1,2)=7>p _a,[1:2]=6

•     z ₁=∞.

•   g(2,1,2,1,0)=min{z ₁}=∞.

• (j,i,i ₁,k _L ,k _R )=(2,2,2,1,0)

• $\quad(j,j',i',i'_{1},k_{L},k'_{L})=(2,0,0,0,1,0)$

•   (L(2,2)=4=p _a,[2:2])∧

•   ((j′=0<i−1=1)∧(L′(1,2,1,0,0,0)=2<p _u,[2:2]=3))

  $$\begin{aligned} z_1=f_1(2,0,0,0,1,0) =&g(0,0,0,0,0)+(2-0)(s+p_{c,[1:2]}+p_{u,[1:1]}) \\ &{}+L'(1,2,1,0,0,0)+L'(1,2,2,0,0,0) \\ =&0+2(1+3+2)+2+7=21. \end{aligned}$$

• $\quad(j,j',i',i'_{1},k_{L},k'_{L})=(2,1,0,0,1,0)$

•  (L(2,2)=4=p _a,[2:2])∧

•  ((j′=1=i−1)∧(p _a,[0:1]=2<(1−0)s+p _u,[1:2]+p _c,[1:2]=1+5+3=9))

  $$\begin{aligned} z_2=f_1(2,1,0,0,1,0) =&g(1,0,0,0,0)+(2-1)(s+p_{c,[1:2]}+p_{u,[1:2]}) \\ &{}+L'(2,2,2,0,0,0) \\ =&\infty. \end{aligned}$$

• $\quad(j,j',i',i'_{1},k_{L},k'_{L})=(2,1,0,1,1,0)$

•  (L(2,2)=4=p _a,[2:2])∧

•  ((j′=1=i−1)∧(p _a,[0:1]=2<(1−0)s+p _u,[1:2]+p _c,[2:2]=1+5+2=8))

  $$\begin{aligned} z_3=f_1(2,1,0,1,1,0) =&g(1,0,1,0,0)+(2-1)(s+p_{c,[1:2]}+p_{u,[1:2]}) \\ &+L'(2,2,2,0,1,0) \\ =&\infty. \end{aligned}$$

• $\quad(j,j',i',i'_{1},k_{L},k'_{L})=(2,1,1,1,1,0)$

•  (L(2,2)=4=p _a,[2:2])∧

•  ((j′=1=i−1)∧(p _a,[1:1]=2<(1−0)s+p _u,[2:2]+p _c,[2:2]=1+3+2=6))

  $$\begin{aligned} z_4=f_1(2,1,1,1,1,0) =&g(1,1,1,0,0)+(2-1)(s+p_{c,[1:2]}+p_{u,[1:2]}) \\ &+L'(2,2,2,1,1,0) \\ =&\infty. \end{aligned}$$

•   g(2,2,2,1,0)=min{z ₁,z ₂,z ₃,z ₄}=21.

• (j,i,i ₁,k _L ,k _R )=(2,2,2,2,0)

• $\quad(j,j',i',i'_{1},k_{L},k'_{L})=(2,1,1,1,2,1)$

•   (L(2,2)=4=p _a,[2:2])∧

•   ((j′=1=i−1)∧(p _a,[1:1]=2<(2−1)s+p _u,[2:2]+p _c,[2:2]=1+3+2=6))

  $$\begin{aligned} z_1=f_2(2,1,1,1,2,1) =&g(1,1,1,1,0)+(2-1)(2s+p_{c,[1:2]}+p_{u,[1:2]}) \\ &{}+L'(2,2,2,1,1,0) \\ =&6+(2+3+5)+4=20. \end{aligned}$$

•   g(2,2,2,2,0)=min{z ₁}=20.

• Goal: $\min_{1\leq k_{L} \leq i\leq i_{1}\leq2, \lceil\frac{2-i_{1}}{2}\rceil\leq k_{R}\leq2-i_{1}}\{g(2,i,i_{1},k_{L},k_{R})\}=g(2,2,2,2,0)=20$.

• The corresponding optimal schedule can be constructed by backtracking the recursion, as illustrated in Fig. 7.

  Fig. 7

  

  Optimal schedule with g(2,2,2,2,0)=20