Makespan with Sequence Dependent Setup Time (MSDST)

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Makespan with Sequence Dependent Setup Time
(MSDST)

• 1sjkCmax

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Introduction to MSDST

• 1 machine
• N jobs
• All the jobs are released at time 0
• Objective minimize Cmax

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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

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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

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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

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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

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How to solve MSDST?

• Equivalent to Traveling Salesman Problem (TSP)
• N jobs gt N1 cities
• Using Mix Integer Programming TSP-Relaxation and
  Min-cut problem
• Algorithm 4.4.5 given in the Textbook

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TSP Relaxation

• xij 0 or 1, integer variable for all iltn and
  jltn (decision variable to decide whether to use
  the route (i, j))
• costij abs(bi-aj), for all iltn-1 and jltn
• Minimize SUMiSUMjcostijxij
• s.t. sumi xi,j 1, for all jltn
• sumj xi,j 1, for all iltn

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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
• sumijxi,j gt1 , where i is in subtour and j
  is not
• sumijxj,i gt1 , where i is in subtour and j
  is not

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TSP Relaxation

• How do we find out subtours?
• We can solve another integer program min-cut
  problem to find the subtours

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TSP RelaxationMin-Cut Problem

• 0 lt fij lt 1, for all iltn and jltn
• yi 0 or 1 integer variable for all iltn
• xij 0 or 1 variable for all iltn and jltn
  (answer to the TSP Relaxation problem)
• minimize SUMiSUMjxijfij
• s.t. fij gt yi - yj
• fij gt yj - yi
• 1 lt sumiyi lt n-1

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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).

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Algorithm 4.4.5

• Define F(j) k as a salesman traversing
  directly from city j to city k
• Define F FI(j,k) as
• F(k) F(j),
• F(j) F(k),
• F(l) F(l) for all l not equal to j or k

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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
• cFI(j,k) length of vertical overlap of
  intervals bj,bk and aF(j), aF(k)

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Algorithm 4.4.5

• Change in Cost Due to Swap I(j,k)

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Algorithm 4.4.5

• Lemma 4.4.2
• An optimal permutation mapping F is obtained
• if bj lt bk and aF(j) lt aF(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.

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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

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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

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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 lt aF(j) in a decreasing order of bj
• bj gt aF(j) in an increasing order of bj

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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 ajs from the smallest to the
  largest, ranking the smallest 0, next smallest 1,
  and so on, and call this permutation mapping,
  F(j)
• Now, sort ajs in a similar fashion as bjs
• Formulate a graph by connecting each node j with
  F(j) node
• Formulate c F(j) I(j, j1), which is an
  interchange cost

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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

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Running Time Test

• Average Running time Comparison

Number of Jobs CPLEX TSP-Relaxation JAVA Algorithm 4.4.5
10 0secs 0secs
20 10secs 0 sec
100 5 minutues 0 sec
200 gt10mintues 0 sec
500 Crash 2-5 secs
1000 Crash 30-40 secs
1500 Crash 4-5 minutes
2000 Crash gt10mintues