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Makespan with Sequence Dependent Setup Time (MSDST)

Makespan with Sequence Dependent Setup Time (MSDST). 1 | s jk | C max. Introduction to MSDST. 1 machine N jobs All the jobs are released at time 0 Objective: minimize C max. Introduction to MSDST So what makes MSDST different?.

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Makespan with Sequence Dependent Setup Time (MSDST)

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  1. Makespan with Sequence Dependent Setup Time (MSDST) 1|sjk|Cmax

  2. Introduction to MSDST • 1 machine • N jobs • All the jobs are released at time 0 • Objective: minimize Cmax

  3. Introduction to MSDSTSo what makes MSDST different? • Each job has a required starting state and a completion state: • aj : required state of the machine in order to start job j • bj : state of the machine after the completion of job j

  4. Introduction to MSDSTSo what makes MSDST different? • Setup Cost! sjk = |bj – ak| • At time zero, the machine is at state b0 • After completing all the jobs, the machine must return to state a0

  5. Applications of MSDST • Metal processing, commercial printing, plastics production, chemical production, pharmaceutical and automobile manufacturing **Baker, 1974 K.R. Baker, Introduction to sequence and scheduling, Wiley, NY (1974). • Setup Operations Could Strongly Depend on the Immediately preceding process • cleaning up • changing tools • machine temperature

  6. Introduction to MSDSTSo what makes MSDST different? • If there 5 jobs and the optimal order is j3-j1-j2-j5-j4 Then the Cmax : s03 + p3+s31 + p1+s12 + p2+s25 + p5+s54 + p4+s40

  7. How to solve MSDST? Equivalent to Traveling Salesman Problem (TSP) N jobs => N+1 cities • Using Mix Integer Programming: TSP-Relaxation and Min-cut problem • Algorithm 4.4.5 given in the Textbook

  8. TSP Relaxation xij – 0 or 1, integer variable for all i<=n and j<=n (decision variable to decide whether to use the route (i, j)) costij = abs(bi-aj), for all i<=n-1 and j<=n Minimize SUMi{SUMj{costij*xij}} s.t. sumi x[i,j] = 1, for all j<=n sumj x[i,j] = 1, for all i<=n

  9. TSP Relaxation TSP Relaxation alone does not work!The Integer Program will cheat by finding subtours. However, we can force the IP to eliminate such subtours by adding these two constraints to the IP formulation: • sumij{x[i,j]} >=1 , where i is in subtour and j is not • sumij{x[j,i]} >=1 , where i is in subtour and j is not

  10. TSP Relaxation How do we find out subtours? We can solve another integer program “min-cut problem” to find the subtours

  11. TSP RelaxationMin-Cut Problem 0 <= fij <= 1, for all i<=n and j<=n yi – 0 or 1 integer variable for all i<=n xij – 0 or 1 variable for all i<=n and j<=n (answer to the TSP Relaxation problem) minimize SUMi{SUMj{xij*fij}} s.t. fij >= yi - yj fij >= yj - yi 1 <= sumi{yi} <= n-1

  12. TSP RelaxationMethod Strategy 1) Solve TSP Relaxation Integer Program 2) Using the output data, solve a min-cut problem to see if subtour exists. If no subtour found, exit. 3) If subtour is found - add subtour constraints to the original TSP Relaxation Integer Program and return to step 1).

  13. Algorithm 4.4.5 Define: Φ(j) = k as a salesman traversing directly from city j to city k Define: Φ’ = ΦI(j,k) as Φ’(k) = Φ(j), Φ’(j) = Φ(k), Φ’(l) = Φ(l) for all l not equal to j or k

  14. Algorithm 4.4.5 Lemma 4.4.1 If the swap I(j,k) causes two arrows that did not cross ealier to cross, then the cost of the tour increases and vice versa. The change in cost is: cΦI(j,k) = length of vertical overlap of intervals [bj,bk] and [aΦ(j), aΦ(k)]

  15. Algorithm 4.4.5 *Change in Cost Due to Swap I(j,k)

  16. Algorithm 4.4.5 Lemma 4.4.2 An optimal permutation mapping Φ’ is obtained if bj <= bk and aΦ(j) <= aΦ(k) • From the picture from the slide, we know that performing a swap which uncrosses the lines leads to a solution as good or beter than the previous.

  17. Algorithm 4.4.5 Does Lemma 4.4.1 and 4.4.2 gaurantee optimal tour? • It gives us the optimal permuatation mapping • Not necessarily feasible tour • Provide a lower bound

  18. Algorithm 4.4.5 How to obtain optimal sequence from optimal permutation mapping? • Identify the subtours • Select the cheapest arc (using lemma 4.4.1) that connects two subtours • Repeat above two steps until no more subtour exists

  19. Algorithm 4.4.5 *When performing swaps, the sequence of swaps can affect the final cost! Example : up, up, down We can prevent this by scheduling nodes with bj <= aΦ(j) in a decreasing order of bj bj > aΦ(j) in an increasing order of bj

  20. Algorithm 4.4.5Complete Algorithm • Given a set of data, aj and bj • Sort bj in increasing order, and move aj along with bj • Rank the moved aj’s from the smallest to the largest, ranking the smallest 0, next smallest 1, and so on, and call this permutation mapping, Φ*(j) • Now, sort aj’s in a similar fashion as bj’s • Formulate a graph by connecting each node j with Φ*(j) node • Formulate c Φ*(j) I(j, j+1), which is an interchange cost

  21. Algorithm 4.4.5 • Continued -Connect the nodes with undirected arcs -Divide newly inserted arcs into two groups -Sort the two groups, and run sequence interchanges according to the values in group 1 and group 2 -Obtain an optimal tour

  22. Running Time Test Average Running time Comparison

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