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A study of the worst case ratio of a simple algorithm for simple assembly line balancing problem

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A study of the worst case ratio of a simple algorithm for simple assembly line balancing problem

1. 1. A study on a simple heuristic for Simple Assembly Line Balancing problem Monard VONG Optimization Lab. Seoul National University December 15th 2009 M. Vong (SNU) SALBP December 15th 2009 1 / 24
2. 2. Outline 1 Introduction to Assembly Line Problems 2 Solving SALBP 3 Worst case performance ratio and computational experiment M. Vong (SNU) SALBP December 15th 2009 2 / 24
3. 3. Concept of Assembly Line (AL) Figure: Tasks and precedence constraints Figure: Assembly Line Concept In assembly line balancing problem, the process (tasks The amount of time a and constraints) is known workpiece can be processed by a station, before the conveyor The goal is to ﬁnd a feasible belt moves the workpiece to assignment of tasks to the next station is called the stations. cycle time. M. Vong (SNU) SALBP December 15th 2009 3 / 24
4. 4. Simple Assembly Line Balancing Problem Now we consider the simplest case of an AL problem: Simple Assembly Line Balancing Problem The objective is to maximize the eﬃciency of the line. Characteristics 1 paced serial line; 2 deterministic (and integral) task time tj ; 3 assignment restrictions: precedence constraints ; From now on, we focus on SALBP-1: cycle time c is ﬁxed minimize the number of stations required This problem is NP-hard. M. Vong (SNU) SALBP December 15th 2009 4 / 24
5. 5. SALBP-1 as an optimization problem Input Set of task V = {1, . . . , n} with task time tj ∈ N, ∀j ∈ V Precedence digraph G = (V , A) (partial ordering of tasks) Cycle time c Goal Find number of station m to maximize the line eﬃciency. max tsum m·c with total task time tsum = n tj j=1 Output S1 , . . . , Sm , m subset of tasks such as ∪k∈1,...,m Sk = V and Sk is feasible for all station. M. Vong (SNU) SALBP December 15th 2009 5 / 24
6. 6. Notations n Number of tasks; V Set of task V = 1, . . . , n m Number of stations; index k = 1, . . . , m m∗ Optimal number of station m, m Lower, upper bound on m∗ tj Task time of task j = 1, . . . , n Pj Set of direct predecessors of task j Pj∗ Set of all predecessors of task j Fj∗ Set of all followers of task j Sk Set of task assigned to station k: Station load t(Sk ) t(Sk ) = j∈Sk tj , k = 1, . . . , m M. Vong (SNU) SALBP December 15th 2009 6 / 24
7. 7. State of the Art Lot of research done on SALBP since 50 years. Many heuristics and optimal procedures have been designed. Most research is focused on solving the problem with additional constraints. But there is few research on worst case ratio of heuristic algorithms. Queyranne(1985): no polynomial time algorithm achieves an absolute worst case performance ratio less than 3 2 M. Vong (SNU) SALBP December 15th 2009 7 / 24
8. 8. Outline 1 Introduction to Assembly Line Problems 2 Solving SALBP 3 Worst case performance ratio and computational experiment M. Vong (SNU) SALBP December 15th 2009 8 / 24
9. 9. Construction schemes for heuristic algorithm Most existing algorithm enumerates solutions by constructing them successively assigning tasks or subset of tasks to stations. Deﬁnition Availability A task j is available if all predecessor h ∈ Pj∗ have been assigned. Assignability An available task j is assignable to a station k if the current idle time of k is suﬃcient. Maximal Station Load A station load Sk is maximal if no available task is assignable to k. M. Vong (SNU) SALBP December 15th 2009 9 / 24
10. 10. Task-Oriented Greedy Heuristic for SALBP-1 Problem Adaptation of Next-Fit algorithm 1: topologically sort the tasks 2: k = 1 3: for i = 1 → n do 4: if t(Sk ) + ti ≤ c then 5: Sk ← i 6: else 7: k ←k +1 8: end if 9: end for M. Vong (SNU) SALBP December 15th 2009 10 / 24
11. 11. For instance c = 10 Figure: Precedence Graph Figure: Solution Obtained by Next-Fit M. Vong (SNU) SALBP December 15th 2009 11 / 24
12. 12. Adaptation of First Fit algorithm 1: topologically sort the tasks 2: m ← 1 3: for i = 1 → n do 4: if ∃k ∈ {1, . . . , m} such that tSk + tj ≤ c and the precedence constraints are respected then 5: Sk ← Sk + {i} 6: else 7: m ←m+1 8: end if 9: end for M. Vong (SNU) SALBP December 15th 2009 12 / 24
13. 13. For instance c = 10 Figure: Precedence Graph Figure: Solution Obtained by adaptation of First-Fit M. Vong (SNU) SALBP December 15th 2009 13 / 24
14. 14. c = 10 Figure: An optimal Solution with 6 stations M. Vong (SNU) SALBP December 15th 2009 14 / 24
15. 15. Outline 1 Introduction to Assembly Line Problems 2 Solving SALBP 3 Worst case performance ratio and computational experiment M. Vong (SNU) SALBP December 15th 2009 15 / 24
16. 16. Worst case performance ratio Notations I is an instance of our optimization problem. z ∗ (I ) is the optimal value solution. z H (I ) is the value of the solution obtained by heuristic H The worst case ratio of H is deﬁned as the smallest constant ρH such that z H (I ) ≤ ρH z ∗ (I ) for each instance of I ∈ I The asymptotic worst case ratio of H is deﬁned as the smallest constant ρH for which there exist another constant δ such that z H (I ) ≤ ρH z ∗ (I ) + δ for each instance of I ∈ I. M. Vong (SNU) SALBP December 15th 2009 16 / 24
17. 17. Upper Bound on the worst case ratio Proposition An upper bound of the worst case ratio of any heuristics which build maximal station load is 2. Proof. For any heuristic H which build maximal station load, for any station k built by this heuristic: t(Sk ) + t(Sk+1 ) > c mH n mH mH ·c t(Sk ) j=1 tj k=1t(Sk ) > 2 . However, m∗ ≥ k=1 c = c , then we obtain m < 2 · m∗ . H Corollary The worst case ratio of Next Fit is 2, and it is tight. M. Vong (SNU) SALBP December 15th 2009 17 / 24
18. 18. A 2 worst case instance for SALBP We will provide an instance for which, even the adaptation of First Fit algorithm to SALBP cannot beat the 2 worst case factor. 1C Let consider 0 < δ ≤ 6 k: 2k tasks of task time 3δ, 2k tasks of task time 1 C − 2δ, 2 2 ∗ (2k − 1) tasks of task time δ Figure: 2 Worst case instance M. Vong (SNU) SALBP December 15th 2009 18 / 24
19. 19. Figure: Optimal Solution Figure: Solution given by the algorithm A(I ) 2k lim = lim =2 k→∞ OPT (I ) k→∞ k + 1 M. Vong (SNU) SALBP December 15th 2009 19 / 24
20. 20. Ranked Positional Weight Using a topological sorting is not enough, the quality of the sorting may be improved. Ranking Positional Weight (RPW) of task i is the sum of the task time ti and the task time of all followers of i. RPWi = ti + j∈F ∗ tj i Tasks are sorted by non-increasing RPW. Example RPW1 = 38 RPW6 = 18 RPW2 = 23 RPW7 = 17 RPW3 = 32 RPW8 = 13 RPW4 = 27 RPW9 = 11 RPW5 = 22 RPW10 = 2 Sorting by non increasing RPW, the following order is considered. 1, 3, 4, 2, 5, 6, 7, 8, 9, 10 M. Vong (SNU) SALBP December 15th 2009 20 / 24
21. 21. Using RPW as topological sorting, the optimal solution is obtained for the previous worst case instance with the same ﬁrst ﬁt algorithm. Ranked Positional Weight algorithm 1: Sort the tasks by non-increasing ranked positional weight 2: Use First Fit rule What about the worst case of the improved algorithm then? M. Vong (SNU) SALBP December 15th 2009 21 / 24