1. Lecture 15: Assembly Line Scheduling 3
© J. Christopher Beck 2005

2. Outline Flexible Flow Line Loading Algorithm
Example 6.4.1
Assembly Line Scheduling at Toyota
© J. Christopher Beck 2005

3. Flexible Flow Line with Bypass? ? ? ?
Stage 2
Stage 1
Maximize throughput
Minimize WIP
Figure 6.2
© J. Christopher Beck 2005

4. Flexible Flow Line Loading (FFLL)
Three phase (heuristic) algorithm for a flexible flow line with bypass environment
Allocate operations to machines
Load balancing
Sequence operations
Dynamic balancing
Assign release times
Backward/forward from bottleneck
© J. Christopher Beck 2005

5. Processing time at stage j
Example 6.4.1 1 4 3 2 5
Stage 1
Stage 2
Stage 3
Jobs 1 2 3 4 5
P(1) 6
P(2)
P(3)
Processing time at stage j
Jobs in the MPS
© J. Christopher Beck 2005

6. FFLL Step 1: Allocate Machines4 3 2 5
Stage 1
Stage 2
Stage 3
Jobs 1 2 3 4 5
P(1) 6
P(2)
P(3)
Processing time at stage j
Jobs in the MPS
© J. Christopher Beck 2005

7. FFLL Step 1: Allocate Machines4 3 2 5
Stage 1
Stage 2
Stage 3
Jobs 1 2 3 4 5
P(1) 6
P(2)
P(3)
Processing time at stage j
Jobs in the MPS
© J. Christopher Beck 2005

8. FFLL Step 1: Allocate Machines4 3 2 5
Stage 1
Stage 2
Stage 3
Jobs 1 2 3 4 5
P(1) 6
P(2)
P(3)
Processing time at stage j
Jobs in the MPS
© J. Christopher Beck 2005

9. FFLL Step 1: Allocate Machines4 3 2 5
Stage 1
Stage 2
Stage 3
Jobs 1 2 3 4 5
P(1) 6
P(2)
P(3)
Processing time at stage j
Jobs in the MPS
© J. Christopher Beck 2005

10. FFLL Step 1: Allocate Machines4 3 2 5
Stage 1
Stage 2
Stage 3
Jobs 1 2 3 4 5
P(1) 6
P(2)
P(3)
Processing time at stage j
Jobs in the MPS
© J. Christopher Beck 2005

11. FFLL Step 1: Allocate Machines4 3 2 5
Stage 1
Stage 2
Stage 3
Jobs 1 2 3 4 5
P(1) 6
P(2)
P(3)
Processing time at stage j
Jobs in the MPS
© J. Christopher Beck 2005

12. FFLL Step 1: Allocate Machines4 3 2 5
Stage 1
Stage 2
Stage 3
Jobs 1 2 3 4 5
P(1) 6
P(2)
P(3)
Processing time at stage j
Jobs in the MPS
© J. Christopher Beck 2005

13. Workloads Jobs 1 2 3 4 5 Wi p1j 6 9 p2j p3j 11 p4j 12 p5j 10 Total: 519
p2j
p3j 11
p4j 12
p5j 10
Total: 51
© J. Christopher Beck 2005

14. Calculate Target Workload
Fraction of the total workload of machine i that has entered system just after job k enters
Fraction of the total workload of all machines that has entered system just after job k enters
© J. Christopher Beck 2005

15. Calculate Target Workload
Jobs 1 2 3 4 5 Wi
p1j 6 9
p2j
p3j 11
p4j 12
p5j 10
0.25
0.20
0.16
0.17
0.22
W = 51
13/51
10/51
8/51
9/51
11/51
© J. Christopher Beck 2005

16. Calculate Overload Target workload on machine i
Overload/underload on machine i up to from job k
Actual workload on machine i
Sum overloads
Minimize
Greedy: Choose as the next job, the one that minimizes
© J. Christopher Beck 2005

17. Overload Jobs 1 2 3 4 5 Wi p1j 6 9 p2j p3j 11 p4j 12 p5j 10 0.25 0.209
p2j
p3j 11
p4j 12
p5j 10
0.25
0.20
0.16
0.17
0.22
W = 51
O11 = 6 – 0.25 * 9 = 3.75
© J. Christopher Beck 2005

18. Overload Jobs 1 2 3 4 5 Wi p1j 6 9 p2j p3j 11 p4j 12 p5j 10 0.25 0.209
p2j
p3j 11
p4j 12
p5j 10
0.25
0.20
0.16
0.17
0.22
W = 51
O11 = 6 – 0.25 * 9 = 3.75
O32 = 2 – 0.20 * 11 = -0.2
© J. Christopher Beck 2005

19. Overload Jobs 1 2 3 4 5 Wi O1j 3.75 9 O2j O3j -0.2 11 O4j 12 O5j 10
0.25
0.20
0.16
0.17
0.22
W = 51
© J. Christopher Beck 2005

20. Overload Jobs 1 2 3 4 5 Wi O1j 3.75 -1.8 -1.44 1.47 -1.98 9 O2j -2.25
1.2
-0.44
-1.53
3.02
O3j
0.25
-0.2
-0.76
1.13
-0.42 11
O4j
2.6
-1.92
0.96
-2.64 12
O5j -3 -2
4.4
-0.17
1.8 10
0.20
0.16
0.17
0.22
W = 51
© J. Christopher Beck 2005

21. Overload Job 4 goes first Jobs 1 2 3 4 5 Wi O1j 3.75 -1.8 -1.44 1.47
-1.98 9
O2j
-2.25
1.2
-0.44
-1.53
3.02
O3j
0.25
-0.2
-0.76
1.13
-0.42 11
O4j
2.6
-1.92
0.96
-2.64 12
O5j -3 -2
4.4
-0.17
1.8 10
0.20
0.16
0.17
0.22
3.8
3.73
4.82
Job 4 goes first
© J. Christopher Beck 2005

22. Now Find Next Job Careful – the pij’s need to be updated!
Fraction of the total workload of machine i that has entered system just after job k enters
Fraction of the total workload of all machines that has entered system just after job k enters
© J. Christopher Beck 2005

23. Calculate Target Workload
Jobs 1 2 3 4 5 Wi
p1j 6 9
p2j
p3j 11
p4j 12
p5j 10
W = 51
Since job 4 goes, the meaning of each cell changes: total load up to (and including) job j
What is cumulative load on m1 if we choose job 1 next?
© J. Christopher Beck 2005

24. Calculate Target Workload
Jobs 1 2 3 4 5 Wi
p1j 9
p2j
p3j 6 11
p4j 7 8 12
p5j 10
W = 51
© J. Christopher Beck 2005

25. Calculate Target Workload
Jobs 1 2 3 4 5 Wi
p1j 9
p2j
p3j 6 11
p4j 7 8 12
p5j 10
0.43
0.37
0.33
0.39
W = 51
22/51
19/51
17/51
20/51
© J. Christopher Beck 2005

26. Calculate Oij Jobs 1 2 3 4 5 Wi O1j 9 O2j O3j 11 O4j 12 O5j 10 0.43
0.37
0.33
0.39
W = 51
© J. Christopher Beck 2005

27. Calculate Oij Jobs 1 2 3 4 5 Wi O1j 5.13 -0.33 -0.51 9 O2j -3.87 1.49
-0.51 9
O2j
-3.87
1.49
O3j
1.27
0.93
0.37
0.71 11
O4j
1.84
3.56 -1
-1.68 12
O5j
-4.3
-3.7
2.67
0.1 10
0.43
0.33
0.39
W = 51
© J. Christopher Beck 2005

28. Calculate max Oij & Choose
Jobs 1 2 3 4 5 Wi
O1j
5.13
-0.33
-0.51 9
O2j
-3.87
1.49
O3j
1.27
0.93
0.37
0.71 11
O4j
1.84
3.56 -1
-1.68 12
O5j
-4.3
-3.7
2.67
0.1 10
0.43
0.33
0.39
8.67
4.86
5.37
2.69
W = 51
Job 5 goes next
© J. Christopher Beck 2005

29. And so on … Sequence chosen: 4, 5, 1, 3, 2
See P p. 132 for the rest (not much left)
You will need to read Sect 6.2 to understand MPS!
MPS = minimum part set
© J. Christopher Beck 2005

30. Does this sound familiar?
What Does Toyota Do?
Everything operates on Just-in-Time
Works to minimize WIP & tardiness
To do this, the most important objective is to keep the part consumption regular
The quantity of a given part consumed per hour should be constant
Solution: the next car to build is chosen to minimize error from the desired rate
Does this sound familiar?
© J. Christopher Beck 2005

31. Example Model 1 requires 1 grommit Model 2 requires 2 grommits
We will produce:
10 cars of model 1
15 cars of model 2
10 cars of model 3
© J. Christopher Beck 2005

32. Example
So we need
(10 * 1) + (15 * 2) + (10 * 3) = 70
If the consumption of grommits is to be constant, we need to consume 70/35 = 2 per car
Choose car j to minimize:
(Grommits use/cars built – 2)2
© J. Christopher Beck 2005

33. Back to Toyota Toyota tracks about 20 “parts” (subassemblies)
Minimizes the sum of squared error over all these parts
See p.134
© J. Christopher Beck 2005