© J. Christopher Beck 20051 Lecture 7: Shifting Bottleneck.

© J. Christopher Beck 2005 1

Lecture 7: Shifting Bottleneck


Outline Shifting Bottleneck Heuristic

Simplified version using example 5.4.2

Full version using example 5.4.2 Try it yourself on the “cut out” JSP


Shifting Bottleneck

Pick most loaded resource Find optimal one-machine schedule Pick next most loaded resource Find optimal one-machine schedule

consistent with previous one-machine schedules

(This is a bit simplified)


Example 5.4.2, p. 89

JSP, min Cmax

J1

J2

J3

10 8 4

8 3 5 6

4 7 3

Jobs Machines

Processing times

1 1,2,3 p11=10, p21=8, p31=4

2 2,1,4,3 p22=8, p12=3, p42=5 , p32=6

3 1,2,4 p13=4, p23=7, p43=3M1

M2

M4

M3


Simplified Shifting Bottleneck Heuristic (SSBH)

M = set of all machines M0 = set of “already scheduled”

machines Initially M0 is empty


SSBH Step 1

Find release date and due date of each operation

Remove all sequence constraints among activities in M – M0, use CPM to find CP and min. start time, max. end time for each activity Since M0 is initially empty, we only

have “conjunctive arcs”


SSBH Step 1: Find Cmax(M0)

10 8 4

8 3 5 6

4 7 3

J1

J2

J3

10 8 4

8 3 5 6

4 7 3

Cmax(M0)0 22 NOTE: This is ajob-based Gantt chart


SSBH Step 1: Find Release & Due Dates (CPM)

J1

J2

J3

10 8 4

8 3 5 6

4 7 3

Cmax(M0)0 22

[16 22]

J1

J2

J3

10 8 4

8 3 5 6

4 7 3

[0 10] [10 18] [18 22]

[0 8] [8 11] [11 16]

[0 12] [4 19] [11 22]


SSBH Step 2: Find Optimal 1-Machine Schedules

Using release and due dates, min. Lmax

[16 22]

M1

M2

M3

10

8

4

8

3

5

6

4

7

3

[0 10]

[10 18]

[18 22]

[0 8]

[8 11]

[11 16]

[0 12]

[4 19]

[11 22]

M4

J2J1 J3

10 3 4 Lmax(1) = 5

0 10 13

88 7J2 J3 J1 Lmax(2) = 5

0 8 15

46 Lmax(3) = 4

16 22

5 311 16

Lmax(4) = 0

Lj = Cj – dj

Lmax = max(Lj)


SSH Step 3: Add Machine to M0

Pick machine with highest Lmax

Use sequence found in Step 2 Lmax(1) = Lmax(2) = 5 Arbitrarily choose to add machine 1


SSHB Step 4: (SKIPPED)

Step 4 from SHB is skipped in the Simplified SSHB


SSHB Step 5: Termination

M0 ≠ M so goto Step 1


SSHB Overview

Step 1: Find Cmax(M0), release, and due dates

Step 2: Find optimal 1-machine schedules

Step 3: Add machine to M0 Step 4: (SKIPPED) Step 5: If M = M0, done. Else goto 1


8J2

6

5

SSHB Step 1 (Iteration 2) M0 = {M1} Find Cmax(M0) = 27, find release & due dates

M1

M2

M4

M3

10 3 4

8J1

4

3

7J3

[18 27]

M3 4 6

[18 27]

5 3[13 21] [24 27]

M4

M2 8 8 7

[10 23] [0 10] [17 24]

J2J1 J3


SSBH Step 2 (Iteration 2): Find Optimal 1-M Schedules


46 Lmax(3) = 1

18 24

5 313 24

Lmax(4) = 0

[18 27]

M3 4 6

[18 27]

5 3[13 21] [24 27]

M4

M2 8 8 7

[10 23] [0 10] [17 24]

J2J1 J3 88 7J2 J3J1 Lmax(2) = 1

0 8 1810


SSH Step 3 (Iteration 2): Add Machine to M0

Pick machine with highest Lmax

Use sequence found in Step 2 Lmax(2) = Lmax(3) = 1 Arbitrarily choose to add machine 2


SSHB Step 4 (Iteration 2): (SKIPPED)




M0 ≠ M so goto Step 1


8J2

6

5

SSHB Step 1 (Iteration 3) M0 = {M1, M2} Find Cmax(M0) = 28, find release & due dates

M1

M2

M4

M3

10 3 4

8J1

4

3

7J3

[18 28]

M3 4 6

[18 28]

5 3[13 22] [25 28]

M4


SSBH Step 2 (Iteration 3): Find Optimal 1-M Schedules


46 Lmax(3) = 0

18 24

5 313 24

Lmax(4) = 0

[18 28]

M3 4 6

[18 28]

5 3[13 22] [25 27]

M4


SSH Step 3 (Iteration 2): Add Machine to M0

Lmax(3) = Lmax(4) = 0 So you actually have a final schedule

by adding sequences from Step 2

8J2

6

5

M1

M2

M4

M3

10 3 4

8J1

4

3

7J3


SSHB Step 4 (Iteration 2): (SKIPPED)




M0 ≠ M so STOP


SSHB Overview



Step 3: Add machine to M0 Step 4: (SKIPPED) Step 5: If M = M0, done. Else goto 1


SHB Overview



Step 3: Add machine to M0 Step 4: Resequence machines M0

(except for new one) Step 5: If M = M0, done. Else goto 1


SHB Step 4

Let Mi be the most recently added resource to set M0

For each resource Mj in M0 \ {Mi} Define single machine problem using

sequence in M0 \ {Mj} If resequencing results in better

solution, keep new sequence


SSHB & SHB on Cut Out JSP

Run SSHB & SHB on “cut out” JSP that we have been used in previous lectures

Activities

Jobs

1 2 3 4

1 M1, 9 M2, 8 M3, 4

M4, 4

2 M1, 5 M2, 6 M4, 3

M3, 6

3 M3, 10

M1, 4 M2, 9

M4, 2

Opt = 33

© J. Christopher Beck 20051 Lecture 7: Shifting Bottleneck.

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