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

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

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

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

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

FFLL Step 1: Allocate Machines 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

FFLL Step 1: Allocate Machines 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

FFLL Step 1: Allocate Machines 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

FFLL Step 1: Allocate Machines 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

FFLL Step 1: Allocate Machines 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

FFLL Step 1: Allocate Machines 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

FFLL Step 1: Allocate Machines 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

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

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

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

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

Overload Jobs 1 2 3 4 5 Wi p1j 6 9 p2j p3j 11 p4j 12 p5j 10 0.25 0.20 9 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

Overload Jobs 1 2 3 4 5 Wi p1j 6 9 p2j p3j 11 p4j 12 p5j 10 0.25 0.20 9 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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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