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Lecture 9: Simplified Shifting Bottleneck
© J. Christopher Beck 2008
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Outline Simplified Shifting Bottleneck Heuristic Try it out
Example 5.4.2 Try it out © J. Christopher Beck 2008
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Readings P Ch 5.4 © J. Christopher Beck 2008
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Shifting Bottleneck Heuristic Method for JSP
makespan minimization Bottleneck resource: the most used resource the one in which the activities are “latest” in some sort of relaxation of the problem Idea: solve a series of 1-machine problems from the most to least loaded resource © J. Christopher Beck 2008
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Simplified Shifting Bottleneck (SSBH)
Find optimal 1-machine schedule for each unscheduled machine subject to any already scheduled resources Identify bottleneck resource Keep optimal 1-machine sequence for bottleneck resource © J. Christopher Beck 2008
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Simplified Shifting Bottleneck Heuristic (SSBH)
M = set of all machines M0 = set of “already scheduled” machines Initially M0 is empty “already scheduled” means all activities on that resource have been sequenced © J. Christopher Beck 2008
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SSBH Overview Step 1: Find Cmax, release, and due dates
CPM Step 2: Find min Lmax 1-machine schedules Step 3: Add max Lmax machine to M0 Step 4: (SKIPPED – see Lecture 10) Step 5: If M = M0, done. Else goto 1 © J. Christopher Beck 2008
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Example 5.4.2, p. 89 JSP, min Cmax 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=3 M1 J1 10 8 4 M2 8 J2 3 5 6 M3 M4 © J. Christopher Beck 2008 J3 4 7 3
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SSBH Step 1 Find release date and due date of each operation
Use CPM to find CP and min. start time, max. end time for each activity © J. Christopher Beck 2008
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SSBH Step 1: Find Cmax 10 8 4 8 3 5 6 4 7 3 © J. Christopher Beck 2008
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SSBH Step 1: Find Cmax [0 [10 [10 [18 [18 [22 10 8 4 [0 [8 [8 [11 [11
[16 [16 [22 8 3 5 6 [0 [4 [4 [11 [11 [14 4 7 3 © J. Christopher Beck 2008
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SSBH Step 1: Find Cmax rj = min start dj = max end [0 0] [10 10]
[18 18] [18 18] [22 22] 10 8 4 [0 0] [8 8] [8 8] [11 11] [11 11] [16 16] [16 16] [22 22] 8 3 5 6 [0 8] [4 12] [4 12] [11 19] [11 19] [14 22] 4 7 3 rj = min start dj = max end © J. Christopher Beck 2008
![SSBH Step 1: Find Cmax rj = min start dj = max end [0 0] [10 10]](http://slideplayer.com/slide/8584123/26/images/12/SSBH+Step+1%3A+Find+Cmax+rj+%3D+min+start+dj+%3D+max+end+%5B0+0%5D+%5B10+10%5D.jpg "[18 18] [18 18] [22 22] [0 0] [8 8] [8 8] [11 11] [11 11] [16 16] [16 16] [22 22] [0 8] [4 12] [4 12] [11 19] [11 19] [14 22] rj = min start. dj = max end. © J. Christopher Beck")
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SSBH Step 1: Find Release & Due Dates (CPM)
Questions? J1 10 8 4 release date [0 10] [10 18] [18 22] due date 8 J2 3 5 6 [0 8] [8 11] [11 16] [16 22] J3 4 7 3 [0 12] [4 19] [11 22] © J. Christopher Beck 2008
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SSBH Step 2: Find Min Lmax 1-Machine Schedules
Using release and due dates, min. Lmax 10 3 4 10 3 4 Lmax(1) = 5 13 M1 [0 10] [8 11] [0 12] M2 J1 8 J2 8 J3 7 8 7 J2 J3 J1 Lmax(2) = 5 15 [10 18] [0 8] [4 19] 4 6 M3 4 6 Lmax(3) = 4 16 22 [18 22] [16 22] M4 5 3 5 3 11 16 Lmax(4) = 0 Lj = Cj – dj Lmax = max(Lj) [11 16] [11 22] © J. Christopher Beck 2008
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SSBH 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 © J. Christopher Beck 2008
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SSBH Step 4: (SKIPPED) Step 4 from SBH is skipped in the Simplified SBH © J. Christopher Beck 2008
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SSBH Step 5: Termination
M0 ≠ M so goto Step 1 © J. Christopher Beck 2008
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SSBH Overview Step 1: Find Cmax, release, and due dates
CPM Step 2: Find min Lmax 1-machine schedules Step 3: Add max Lmax machine to M0 Step 4: (SKIPPED) Step 5: If M = M0, done. Else goto 1 © J. Christopher Beck 2008
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SSBH Step 1 (Iteration 2): Find Cmax
10 8 4 8 3 5 6 4 7 3 © J. Christopher Beck 2008
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SSBH Step 1 (Iteration 2): Find Cmax
[0 [10 [10 [18 [18 [22 10 8 4 [0 [8 [10 [13 [13 [18 [18 [24 8 3 5 6 [13 [17 [17 [24 [24 [27 4 7 3 © J. Christopher Beck 2008
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SSBH Step 1 (Iteration 2): Find Cmax
[0 0] [10 10] [10 15] [18 23] [18 23] [22 27] 10 8 4 [0 2] [8 10] [10 10] [13 13] [13 16] [18 21] [18 21] [24 27] 8 3 5 6 [13 13] [17 17] [17 17] [24 24] [24 24] [27 27] 4 7 3 © J. Christopher Beck 2008
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SSBH Step 1 (Iteration 2) M0 = {M1} Find Cmax, release & due dates M2
8 7 [10 23] [0 10] [17 24] J2 J1 J3 [18 27] M3 4 6 5 3 [13 21] [24 27] M4 © J. Christopher Beck 2008
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SSBH Step 2 (Iteration 2): Find Min Lmax 1-M Schedules
Using release and due dates, min. Lmax M2 8 7 [10 23] [0 10] [17 24] J2 J1 J3 8 7 J2 J3 J1 Lmax(2) = 1 18 10 [18 27] M3 4 6 4 6 Lmax(3) = 1 18 24 either schedule OK 5 3 [13 21] [24 27] M4 5 3 13 24 Lmax(4) = 0 © J. Christopher Beck 2008
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SSBH 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 © J. Christopher Beck 2008
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SSBH Step 4 (Iteration 2): (SKIPPED)
Step 4 from SHB is skipped in the Simplified SSHB © J. Christopher Beck 2008
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SSBH Step 5: Termination
M0 ≠ M so goto Step 1 © J. Christopher Beck 2008
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SSBH Step 1 (Iteration 3): Find Cmax
10 8 4 8 3 5 6 4 7 3 © J. Christopher Beck 2008
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SSBH Step 1 (Iteration 3): Find Cmax
[0 [10 [10 [18 [18 [22 10 8 4 [0 [8 [10 [13 [13 [18 [18 [24 8 3 5 6 [13 [17 [18 [25 [25 [28 4 7 3 © J. Christopher Beck 2008
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SSBH Step 1 (Iteration 3): Find Cmax
[0 0] [10 10] [10 10] [18 18] [18 24] [22 28] 10 8 4 [0 2] [8 10] [10 11] [13 14] [13 17] [18 22] [18 22] [24 28] 8 3 5 6 [13 14] [17 18] [18 18] [25 25] [25 25] [28 28] 4 7 3 © J. Christopher Beck 2008
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SSHB Step 1 (Iteration 3) M0 = {M1, M2}
Find Cmax= 28, find release & due dates [18 28] M3 4 6 5 3 [13 22] [25 28] M4 © J. Christopher Beck 2008
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SSBH Step 2 (Iteration 3): Find Min Lmax 1-M Schedules
Using release and due dates, min. Lmax [18 28] M3 4 6 4 6 Lmax(3) = 0 18 24 5 3 [13 22] [25 27] M4 5 3 13 24 Lmax(4) = 0 © J. Christopher Beck 2008
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SSBH Step 3 (Iteration 3): Add Machine to M0
Lmax(3) = Lmax(4) = 0 So you actually have a final schedule by adding sequences from Step 2 M1 10 3 4 J2 8 8 J1 J3 7 M2 M3 6 5 4 M4 3 © J. Christopher Beck 2008
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SSHB Step 4 (Iteration 3): (SKIPPED)
Step 4 from SHB is skipped in the Simplified SHB © J. Christopher Beck 2008
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SSHB Step 5: Termination
M0 = M so STOP © J. Christopher Beck 2008
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SSBH Overview Step 1: Find Cmax, release, and due dates
CPM Step 2: Find min Lmax 1-machine schedules Step 3: Add max Lmax machine to M0 Step 4: (SKIPPED) Step 5: If M = M0, done. Else goto 1 © J. Christopher Beck 2008
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SSBH Example JSP Run SSBH on JSP from previous lectures Jobs
Processing times J0R0[15] J0R1[50] J0R2[60] 1 J1R1[50] J1R0[50] J1R2[15] 2 J2R0[30] J2R1[15] J2R2[20] © J. Christopher Beck 2008
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