ON-LINE SCHEDULING AND W.I.P. REGULATION Jean-Marie PROTH

Production background The production system works 24 hours a day. When an order appears in the production system, we have to provide the best delivery time in real time to the customer (less than 3 minutes). Previous schedule cannot be modified.

ON-LINE SCHEDULING AND W.I.P. REGULATION: THE JOB-SHOP CASE Since orders are scheduled as soon as they appear in the system, the job-shop case is a flow-shop for each one of the orders.

Basic problem m operations O 1, …, O m must be performed to complete a given order. Each operation is performed by one resource or several identical resources. Each resource is partially busy when the order appears in the system. We have to manufacture a product using the idle periods of the resources. We first arrange the idle windows in the increasing order of their lower limit.

Classifying the idle periods Three identical resources for one operation R1 R2 R O α1α1 β1β1 α2α2 β2β2 α3α3 β3β3 α4α4 β4β4 α5α5 β5β5 α6α6 β6β6 α7α7 β7β7 α8α8 β8β8 α9α9 β9β9

Problem setting The operations should be performed in the order O 1, O 2, …, O m. The time spent for performing O i, i=1, …, m belongs to [θ i, θ i +δ i ]. Product cannot be stored between two operations. Objective: Minimize the completion time.

EXAMPLE: O1O1 O2O2 O3O3 O4O4 Time BUSY PERIODS MANUFACTURING TIMES EXTENSIONS OF MAN. TIMES

THE PROBLEM FOR GIVEN IDLE WINDOWS Min x m+1 For i = 1, …, m

THE ALGORITHM

The k i value is the rank of the idle window in which we want to perform operation i, for i=1 to m.

A numerical example θ 1 =3 δ 1 =2 θ 2 =5 δ 2 =4 θ 3 =3 δ 3 =1 k 1 =1 k 2 =1 k 3 =1 t 1 =0 t 2 =Max ( 0, 0+3)=3 t 3 =Max (0, 3+5)=8 t 4 =8+3=11 x 4 =11 x 3 =Max (8, )=8 x 2 =Max (3, 8-5-4)=3 x 1 =Max(0, 3-2-3)=0 Since x 3 >β 2, k 2 =k 2 +1Since x 4 >β 3, k 3 =k 3 +1

θ 1 =3 δ 1 =2 θ 2 =5 δ 2 =4 θ 3 =3 δ 3 =1 k 1 =1 t 1 =0 t 2 =Max ( 15, 0+3)=15 t 3 =Max (10, 15+5)=20 t 4 =20+3=23 x 4 =23 x 3 =Max (20, )=20 x 2 =Max (15, )=15 x 1 =Max (0, )=10 Since x 4 >β 3, k 3 =k 3 +1=3Since x 2 >β 1, k 1 =k 1 +1=2 k 2 =2 k 3 =2

θ 1 =3 δ 1 =2 θ 2 =5 δ 2 =4 θ 3 =3 δ 3 =1 k 1 =2 k 2 =2 k 3 =3 t 1 =11 t 2 =Max ( 15, 11+3)=15 t 3 =Max (23, 15+5)=23 t 4 =23+3=26 x 4 =26 x 3 =Max (23, )=23 x 2 =Max (15, )=15 x 1 =Max(11, )=11 THIS SOLUTION IS OPTIMAL

θ 1 =3 δ 1 =2 θ 2 =5 δ 2 =4 θ 3 =3 δ 3 =1 THIS SOLUTION IS OPTIMAL

REMARKS 1.It is not always possible to extend the operation time. 2.A resource is busy until the end of the operation time (including its extension). ⇓ We transform the extension of the operation time into inventory time.

Two approaches are proposed: Approach 1: We are interested in managing only the inventory time between two operations. Approach 2: We are interested in managing both the inventory time and the number of parts in inventory between two operations.

APPROACH 1 We add a “storage resource” at the end of each operation. These storage resources are totally idle each time a new order appears in the production system.

R1 R2 R3 R4 Time BUSY PERIODS MAN. TIMES S1 S2 S3 STORAGE PERIODS

APPROACH 2 We add as many “storage resources” as the number of WIP units that are allowed at the exit of an operation. We keep the busy periods of these “storage operations”. The δ i and θ i are assigned as in approach 1.

EXAMPLE: R1 R2 R3 R4 Time

ON-LINE SCHEDULING AND W.I.P. REGULATION: THE ASSEMBLY SYSTEM CASE The algorithm for on-line scheduling and WIP management in assembly systems is based on the previous algorithm. The idea behind this algorithm is to adjust iteratively job-shop like systems.

EXAMPLE

S1S S5S S4S S3S S2S2 DECOMPOSITION

If we apply the job-shop algorithm to each one of the lines, there is little likelihood that a given assembly operation starts at the same time in the different schedules. We propose an iterative approach that adjust gradually the starting time of each assembly operation. This approach is based on the two following rules. The proposed algorithm converges to the optimal solution, i.e. to the minimal makespan.

RULE 1 In the resulting schedule: If a given assembly operation is scheduled in different windows, we restart the computation constraining this operation to be scheduled at the earliest in the last window. S1 S2 S4 kk+1k+2 We restart the algorithm with window k+2 in all the lines that contain this assembly operation. Assembly operation 7

RULE 2 In the resulting schedule: If a given assembly operation is scheduled in the same window whatever the line, then we restart the computation from this window after assigning to the lower bound of the window the greatest starting time of the operation. Configuration when restarting the scheduling New window