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Sequence-Dependent Setup Problems

Sequence-Dependent Setup Problems. Contents 1. An algorithm which gives an optimal schedule with the minimum makespan with sequence-dependent setup times 1 | S jk | C max. Literature :

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Sequence-Dependent Setup Problems

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  1. Sequence-Dependent Setup Problems Contents 1. An algorithm which gives an optimal schedule with the minimum makespan with sequence-dependent setup times 1 | Sjk | Cmax • Literature: • Scheduling, Theory, Algorithms, and Systems, Michael Pinedo, Prentice Hall, 1995, or new: Second Addition, 2002, Chapter 3. • “Sequencing a One State-Variable Machine: A Solvable Case of the Travelling Salesman Problem” Gilmore, Gomory, Operations Research 12, 1964, pages 655-679

  2. Single machine: rj=0, no sequence dependent setup times  • 1 | Sjk | Cmax NP hard • a single real variable which describes each job • job j: to start the job the machine must be in state aj , • at the completion of the job the machine state is bj • sjk = | ak - bj | job k follows job j • Travelling Salesman Problem • with n+1 cities j0, j1, …,jn, j0 • k = (j) travelling salesman leaving city j for city k

  3. {0, 1, 2, 3}  {2, 3, 1, 0} (0) = 2 (1) = 3 (2) = 1 (3) = 0 0 2 0 2 3 1 3 1 {0, 1, 2, 3}  {2, 1, 3, 0} bk a(j) bj a(k) b2 a(2) a(1) cost of going from 1 to (1) is | a(1) - b1 | b1

  4. Swap I(j,k) applied to a permutation  produces another permutation ’: ’(k) = (j) ’(j) = (k) ’(l) = (l), l  j , k 5 5 1 1 4 4 2 3 2 3 6 7 6 7 11 11 8 8 10 9 10 9

  5. a(k) bk change in cost due to swap I(j, k) a(j) bj Lemma. If the swap causes two arrows that did nor cross earlier to cross, then the cost of the tour CI(j,k) increases and vice versa. CI(j, k) = || [bj, bk]  [a(j), a(k)] || Lemma. Let  * be a permutation that ranks a, that is j > k implies a(j) > a(k), then  * is optimal with respect to cost.

  6. Gundirected graph where arcs link the jth and (j)th nodes, j=1,…,n. Spanning tree - a minimal set of additional arcs that connect anunconnected graph. 2 2 2 3 3 3 1 1 1 4 5 4 5 4 5 8 8 8 6 6 6 14 14 14 7 7 7 13 13 13 9 9 9 10 16 10 16 10 16 15 15 15 11 12 12 12 11 11

  7. Theorem. Assume that Ghas p components. Let I(j1, k1), …, I(jp-1, kp-1), be the swaps corresponding to the arcs of a spanning tree G.The arcs may be taken in any order. Then the permutation ’ given by ’=  I(j1, k1), …, I(jp-1, kp-1) is a tour. Lemma. There is a minimal cost spanning tree for Gthat contains onlyarcs Rj, j+1.

  8. a(3)=11 b3=10 CI(1, 2) = || [1,6]  [2,4] || = 2 (4-2) = 4 CI(2, 3) = || [6,10]  [4,11] || = 2 (10-6) = 8 b2=6 a(2 )=4 a(1) =2 b1=1 a(3)=11 a(3)=11 b3=10 b3=10 b2=6 b2=6 a(2 )=4 a(2 )=4 a(1) =2 a(1) =2 b1=1 b1=1 I(1, 2) then I(2, 3) CI(2, 3) = || [6,10]  [2,11] || = 2 (10-6) = 8 I(2, 3) then I(1, 2) CI(1, 2) = || [1,6]  [2,11] || = 2 (6-2) = 8 CHANGED!

  9. A node is of Type 1 if bja(j) (arrow points up) A node is of Type 2 if bj>a(j) (arrow points down) A swap is of Type 1 if lower node is of Type 1 A swap is of Type 2 if lower node is of Type 2 If swaps I(j, j+1) of Type 1 are performed in decreasing order of the node indices, followed by swaps of Type 2 in increasing order of the node indices then a single tour is obtained without changing anyC* I(j, j+1) involved in the swaps

  10. Algorithm + Example 7 jobs Step 1. Arrange the bj in order of size and renumber the jobs so that b1b2 ... bn Arrange the aj in order of size. The permutation mapping *is defined by * (j) = k, k being such that ak is the jth smallest of the a.

  11. a4=45 b7=40 a6=34 b6=31 b5=26 a5=22 b4=19 a7=18 a3=16 b3=15 a1=7 b2=3 a2=4 b1=1

  12. Step 2. Form an undirected graph with n nodes and undirected arcs connecting the jth and *(j) nodes, j=1,…n . If the current graph has only one component then STOP ; otherwise go to Step 3. 2 1 3 7 4 6 5

  13. Step 3. Compute the swap costs C*I(j, j+1) for j=1,…,n C* I(j, j+1) = 2 max ( min (bj+1, a*(j+1)) - max (bj, a*(j)) ), 0 ) C* I(1, 2) = 2 max ( (3-4), 0 ) = 0 C* I(2, 3) = 2 max ( (15-7), 0 ) = 16 C* I(3, 4) = 2 max ( (18-16), 0 ) = 4 C* I(4, 5) = 2 max ( (22-19), 0 ) = 6 C* I(5, 6) = 2 max ( (31-26), 0 ) = 10 C* I(6, 7) = 2 max ( (40-34), 0 ) = 12

  14. 2 16 1 3 4 7 4 6 6 10 5 Step 4. Select the smallest value C* I(j, j+1) such that j is in one componentand j+1 in another. In case of a tie for smallest, choose any. Insert the undirected arc Rj, j+1 into the graph. Repeat this step untilall the components in the undirected graph are connected. 2 2 2 1 1 1 3 3 3 4 4 4 7 7 7 4 4 4 6 6 6 6 6 5 5 10 5

  15. Step 5. Divide the arcs added in Step 4 into two groups. Those Rj, j+1 for which bja(j) go in group 1, those for which bj>a(j) go in group 2. Step 6. Find the largest index j1 such that Rj1, j1+1 is in group 1. Find the second largest index, and so on, up to jl assuming there arel elements in the group. Find the smallest index k1 such that Rk1, k1+1 is in group 2. Find the second smallest index, and so on, up to km assuming there arem elements in the group. j1 = 3, j2 = 2, k1 = 4, k2 = 5

  16. Step 7. The optimal tour ** is constructed by applying the followingsequence of swaps on the permutation *: ** = * I(j1,j1+1) I(j2,j2+1) … I(jl,jl+1) I(k1,k1+1) I(k2, k2+1) … I(km,km+1) ** = * I(3,4) I(2,3) I(4,5) I(5,6) Type 1 Type 2

  17. a4=45 a4=45 b7=40 b7=40 a6=34 a6=34 b6=31 b6=31 b5=26 b5=26 a5=22 a5=22 b4=19 a7=18 b4=19 a7=18 a3=16 a3=16 b3=15 b3=15 a1=7 a1=7 b2=3 a2=4 b2=3 a2=4 b1=1 b1=1 * I(3,4) I(2,3) * I(3,4)

  18. a4=45 a4=45 b7=40 b7=40 a6=34 a6=34 b6=31 b6=31 b5=26 b5=26 a5=22 a5=22 b4=19 a7=18 b4=19 a7=18 a3=16 a3=16 b3=15 b3=15 a1=7 a1=7 b2=3 a2=4 b2=3 a2=4 b1=1 b1=1 * I(3,4) I(2,3) I(4,5) ** = * I(3,4) I(2,3) I(4,5) I(5,6)

  19. a4=45 b7=40 a6=34 b6=31 b5=26 a5=22 b4=19 a7=18 a3=16 b3=15 a1=7 b2=3 a2=4 b1=1 ** = * I(3,4) I(2,3) I(4,5) I(5,6) The optimal tour is: 1  2  7  4  5  6  3  1 The cost of the tour is: 3 + 15 + 5 + 3 + 8 + 15 + 8 = 57

  20. Summary • 1 | Sjk | Cmax , sjk = | ak - bj | a polynomial time algorithm is given • 1 | Sjk | Cmax is NP hard

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