Workflow in a flow shop

• Each job consists m operators in different machine.
• The flow of work is unidirectional.
• Machines in a flow shop = 1,2,…….,m
• The operations of job i , (i,1) (i,2) (i ,3)…..(i, m)
• Not processed by machine k , P( i , k) = 0

  ) ! n ( k ) ! n ( ^m k : constraint (

  Baker p.137

  

Type 2.

  Input Input Input Input

Type 1.

  Input output output output output output

  Input _Machine output _{1 Machine 2 Machine 3 Machine M-1 Machine M} ….

  or ₄ Workflow in a flow shop _{Machine 1 Machine 2 Machine 3 Machine M-1 Machine M} ….

  1 3 2 4 5

  ∵ routing problem)

   Flow shop  Job shop n! - flow shop permutation schedule n!.n! …….n!

  2 3 1 5 4 2 3 1 5 4

  M

  . . . .

  2 .

  1

  Baker p.136 The processing sequence on each machine are all the same.

  2 Definitions • Contains m different machines.

  Mata Kuliah: Penjadwalan Produksi Teknik Industri – Universitas Brawijaya Flow Shop Scheduling

• Job shop _m

3 Flow Shop Scheduling

  Ex. Johnson’s Rule

  Baker p.142 Step 1 : Find min {t , t } Stage U Min t jk Assignment Partial Schedule _{i i1 i2} Step

  2 a : If the min t requires machine 1 , place the job in the first available position 1 1,2,3,4,5 t 3=[1] 3 x x x x ₃₁ in sequence . go to step 3 .

  2 1,2,4,5 t 2=[5] 3 x x x 2 ₂₂ Step 2 b : If the min t requires machine 2 , place the job in the first available position 3 1,4,5 t ^{11 1=[2] 3 1 x x 2} in sequence . go to step 3 .

  4 4,5 t 5=[4] 3 1 x 5 2 ₅₂ Step 3 : Remove the assigned job from considerat ion and return to step 1 until all positions in sequence are filled .

  5 4 t 4=[3] 3 1 4 5 2 ₁₁ Note: Johnson’s rule can find an optimum with two machines ₅ Flow shop problem for makespan problem. ₆ Ex. The B&B for Makespan Problem

  Job Job1 Job2 Job3 Job4 Job5 The Ignall-Schrage Algorithm (Baker p.149)

• A lower bound on the makespan associated with any

  t

  3

  5

  1

  6

  7

  j1

  completion of the corresponding partial sequence σ is

  t

  6

  2

  2

  6

  5

  j2

  obtained by considering the work remaining on each machine. To illustrate the procedure for m=3.

  Job3 Job1 Job4 Job5 Job2

  M=24 For a given partial sequence σ, let

  q

  3

  1 = the latest completion time on machine 1 among jobs in σ.

  1

  4

  5 2 q

  2 = the latest completion time on machine 2 among jobs in σ. q 3 = the latest completion time on machine 3 among jobs in σ.

  3

  1

  4

  5

  2  t

  The amount of processing yet required of machine 1 is j ¹

  24 ^j _{7 8}  ' 

  9 The Ignall-Schrage Algorithm

        

  9

  13

  12

  2

  

   

     

  1 min

      

      

         

      

  1

  11

  5

  6 ₁₂ Ex.

  11

  45

                

  14 } t t { min t q b _{' j ' j}   

  9 ) min 10 7 (

  13

  12

  2

  22

  Partial Sequence ( q1 , q2 , q3 ) (b1,b2,b3) B 1xxx ( 3 , 7 , 17 ) ( 37 , 31 , 37 ) 37 2xxx ( 11 , 12 , 17 ) ( 45 , 39 , 42 ) 45 3xxx ( 7 , 16 , 29 ) ( 37 , 35 , 46 ) 46 4xxx ( 10 , 22 , 24 ) ( 37 , 41 , 52 ) 52 12xx ( 14 , 15 , 22 ) ( 45 , 38 , 37 ) 45 13xx ( 10 , 19 , 32 ) ( 37 , 34 , 39 ) 39 14xx ( 13 , 25 , 27 ) ( 37 , 40 , 45 ) 45 132x ( 21 , 22 , 37 ) ( 45 , 36 , 39 ) 45 134x ( 20 , 32 , 34 ) ( 37 , 38 , 39 )

  17

  15

  7

  14

  3

  2

  39 _{1 2 1 2 1}

  2

  12

  In the most favorable situation, the last job 1) Encounters no delay between the completion of one operation and the start of its direct successor, and 2) Has the minimal sum (t ^j2 +t ^j3 ) among jobs j belongs to

  …….. q ² q ^{3 b 2} Job in σ’

  4 t j1 3 11 7 10 t j2

  3

  2

  1

  j

  11 Ex. B&B

  …….. …….. q ¹ q2 q ^{3 b}

_M3

¹ _M2 t k2 t k3 ……..

  m=3 For the first node: σ =1

  

M3

t k1 t k2 t k3 ……..

    ' j ^{3 j 3 3} t q b } b , b , b max{ B _{3 2 1  10} The Ignall-Schrage Algorithm

_M1

_M2

        

  } t { min t q b _{3 j ' j ' j 2 j 2 2}    

  } t t { min t q b ^{3 j 2 j} _{' j ' j} ^{1 j 1 1     }    

  σ’ Hence one lower bound on the makespan is A second lower bound on machine 2 is A lower bound on machine 3 is The lower bound proposed by Ignall and Schrage is

  4 1 9 12 t j3 10 5 13 2

  37 ) 37 , 31 ,

  3

  7 t t q 3 t q

  11

  12

  13

  1

  2

  3

  28 3 b T he bound lower 17 t t t q

  B 37 max(

  6

  37

  22 7 b

  2

  31

  20 17 b

  37

     

  Traditional B&B:

• The computational requirements will be severe for large problems
• Even for relatively small problems, there is no guarantee that the solution can be obtained quickly,
• can obtain solutions to large problems with limited computational effort.
• Computational requirements are predictable for problem of a given size.
₁₆ CDS

  2 j

  ' t

  1 j

  2. It generally creates several schedules from which a “best” schedule can be chosen. The CDS algorithm corresponds to a multistage use if Johnson’s rule applied to a new problem, derived from the original, with processing times and . At stage 1,

  1. It use Johnson’s rule in a heuristic fashion

  Its strength lies in two properties:

  (Campbel, Dudek and Smith)

  Heuristic Approaches

  15 Heuristic Approaches

  13 Ex. B&B P 1xxx 2xxx 3xxx 4xxx _{B=37 B=45 B=46 B=52}

  1 Consider the following four-job three-machine problem

  6 t j3 12 16 7

  4 t j1 13 7 26 2 t j2 3 12 9

  3

  2

  1

  b. Find the min makespan using the modified algorithm. j

  a. Find the min makespan using the basic Ignall-Schrage algorithm. Count the nodes generated by the branching process.

  12xx 13xx 14xx _{B=45 B=45 B=39} 132x 134x _{B=45 B=39}

₁₄

Hw.

  ' t ' t t and t ' t  

  17 CDS

  Set K=2

  t t

• ^* _{1 1 , 2 , * 1 , 1 , i i}  t t
• ^* _{2 , 2 , i i}

   t t CDS

  19 ^{Job i Mesin 1 Mesin 3 Mesin 1 Mesin 2 1 4 5 7 8 2 3 4 6 7 3 2 6 3 7 4 5 2 8 5 5 6 7 10 11 6 1 3 9 Job i Mesin 1 Mesin 2 Mesin 3 11 K=1 K=2 1 4 3 5 2 3 3 4 3 2 1 6 4 5 3 2 5 6 4 7 6 1 8 3} Set K=1

  4 _{1 , 1} ¹ _{1 , 1} ^* _{1 , 1}   

    t t t _{k k}

  5 _{3 , 1} ¹ _{1 1 3 , 1} ^* _{2 , 1}   

      t t t _{k k}

  7

     

  3

  4 _{2 , 1 1 , 1} ² _{1 , 1} ^* _{1 , 1}      

    t t t t _{k k}

  8

  3

  5 ^{2 , 1 3 , 1 2 1 1 3 , 1 * 2 , 1}      

      t t t t _{k k}

  CDS CDS

   ^K _{k k m i i}

  t t

• ^* _{1 , 1 ,}

  In other words, Johnson’s rule is applied to the first and mth operations and intermediate operations are ignored. At stage 2,

   

  , 1 m j jm 2 j 2 j 1 j

  1 j

  ' t t t and t t ' t

  

     

  That is, Johnson’s rule is applied to the sums of the first two and last two operation processing times. In general at stage i,

   

      i

   ^K _{k k i i}

  1 k , 1 k m j 2 j i

  1 k jk 1 j

  ' t t and t ' t ₁₈ Step 1. Set K=1. Hitung (m=jumlah mesin):

  dan Step 2.

  Gunakan Algoritma Johnson untuk penentuan urutan pekerjaan dengan menyatakan dan Step 3.

  Hitung makespan untuk urutan tersebut. Catat jadwal dan makespan yang dihasilkan Step 4.

  Jika K=m-1 maka pilih jadwal dengan makespan terpendek sebagai jadwal yang digunakan, lalu stop. Jika K<m-1 maka K=K+1 dan kembali ke Step 1.

   

  

20 CDS

  21

²²

Palmer Palmer proposed the calculation of a slope index, s j

  t t if

  1 t t if 1 e

  t t if

  jm 1 j jm 1 j j

  

    

    

  Where

   

    

  , 1 k j jk 1 m k 1 j j 

  Generalizing from this structure, Gupta proposed that for m>3 , the job index to be calculated is } t t { min e s

  1 t t if 1 e ₂₄ Gupta

  3 j 1 j 3 j 1 j j

  , for each job. _{1 , j 2 , j , 2 m j , 1 m j m , j j} ) t ( 1 m t )

  

    

    

  Where

    

  3 j 2 j 2 j 1 j j j

  Gupta thought a transitive job ordering in the form of follows that would produce good schedules. Where } t t , t t min{ e s

   s s s   

  1 [

  Then a permutation schedule is constructed using the job ordering ] n [ ] ] 2 [

    

  ( 3 m t ) ( 5 m t ) ( 3 m t ) ( 1 m s           

23 Gupta

  P.C. Chang. IEM. YZU.

  10

  1

  j

  2. Use Palmer, Gupta, CDS to solve this problem.

  1. Use Ignall-Schrage & McMahon-Burton to solve

  Let

  26 HW.

  CDS: 3-5-4-1-2 M=35

  1 s _{5 4 3 2 1}              

  1 s

  3

  2

  1 s

  12

  1 s

  13

  1 s

  11

  5

  2

  4

  4

  1

  Kenneth R. Baker. Duke University. John Wiley & Sons. 1974.

  Referensi • Introduction to Sequencing and Scheduling.

  31 , , , , , xxx xxx b b b b b of P P

  13

  5

  4

  3

  2

  2

  5 t

  2

  } 3 ,  1 { 

  2 5 7 11 t j3 6 5 7 13 10

  3

  9 t j2

  6

  8 11 7

  j1

  3

  1

  25 Ex.

  9

  1

  2

  6 t j3

  5

  9

  1

  8

  5 t j2

  3

  8

  4

  6

  5 t j1

  4

  3

  2

  1

  Palmer: j

  5

  6

  2

  6

  36 M

  Gupta:

  M s s s s s t t t m t m s _{j j j j j}

  1 _{5 4 3 2 1 1 3 1 3}                    

  1

  2

  2

  8

  4

     

  2

  2

  3

  5

  4

  2

  1

  37

• Production Scheduling. PPT: Course Material.