# Operations Scheduling and Production – Activity Control

## Presentation on theme: "Operations Scheduling and Production – Activity Control"— Presentation transcript:

Operations Scheduling and Production – Activity Control
dr hab. inż., prof. nadzw. PWR Dorota Kuchta Operations Scheduling and Production – Activity Control

PROCESSING TIME (HOURS)
Job Shop Scheduling JOB PROCESSING TIME (HOURS) 1 8 2 4 3 7 5 6 JOB PROCESSING TIME (HOURS) 4 3 2 3 + 4 = 7 6 7 + 5 = 12 5 = 18 = 25 1 = 33

Common Scheduling Criteria
DEFINITION OBJECTIVES 1. Makespan Time to process a set of jobs Minimize makespan 2. Flowtime Time a job spends in the shop Minimize average flowtime 3. Tardiness The amount by which completion Minimize number of tardy jobs time exceeds the due date of a job Minimize the maximum tardiness

Scheduling with Due Dates
The optimal sequence is The optimal sequence is Scheduling with Due Dates JOB PROCESSING TIME DUE DATE 1 4 15 2 7 16 3 8 6 21 5 9 JOB FLOWTIME TARDINESS 1 4 2 4 + 7 = 11 3 = 13 5 = 19 = 22 13 average 13.8 3.6 JOB FLOWTIME DUE DATE TARDINESS 3 2 8 5 2 + 3 = 5 9 1 5 + 4 = 9 15 9 + 7 = 16 16 4 = 22 21 JOB DUE DATE LATEST START 1 15 11 2 16 9 3 8 6 4 21 5

Minimalizacja liczby opóźnionych elementów
Wstawić 1. zadanie do ciągu S Jeśli koniec wykonania ciągu przypada po terminie wykonania jego ostatniego elementu, wyrzucić najdłuższy element ciągu poza ciąg 3. Jeśli jeszcze są zadania nie ustawione, wstawić kolejne zadanie do ciągu, krok 2. W przeciwnym przypadku stop Proc. 2 4 1 3 due 5 6 7 8

Two – Machine Flowshop Problem
JOB SHEAR PUNCH 1 4 5 2 3 10 6

Two – Machine Flowshop Problem
JOB FLOWTIME 1 9 2 10 3 22 4 34 5 37

Two – Machine Flowshop Problem (Johnson’s Rule)
Since the minimum time is on the second machine, job 2 is scheduled last: __ __ __ __ 2 Next, we pick the second – smallest processing time. This is 2, which corresponds to job 5 on machine 1. Therefore, job 5 is scheduled first: 5 __ __ ___ 2 In the next step, we have a tie between job 1 on the shear and job 3 on the punch press. When ties occur, either job can be chosen. If we pick job 1, we then have 5 1 __ __ 2 Continuing with Johnson’s rule, the last two steps yield: 5 1 __ 3 2 JOB SHEAR PUNCH 1 4 5 2 3 10 6

Two – Machine Flowshop Problem (Johnson’s Rule)
Since the minimum time is on the second machine, job 2 is scheduled last: __ __ __ __ 2 Next, we pick the second – smallest processing time. This is 2, which corresponds to job 5 on machine 1. Therefore, job 5 is scheduled first: 5 __ __ ___ 2 In the next step, we have a tie between job 1 on the shear and job 3 on the punch press. When ties occur, either job can be chosen. If we pick job 1, we then have 5 1 __ __ 2 Continuing with Johnson’s rule, the last two steps yield: 5 1 __ 3 2

Job Data for Lynwood’s Job Shop
ARRIVAL TIME PROCESSING SEQUENCE (PROCESSING TIME) 1 L(10) - D(20) - G(35) 2 D(25) - L(20) - G(30) - M(15) 3 20 D(10) - M(10) 4 30 L(15) - G(10) - M(20)

Shop Status at Time T

Open shop dla dwóch maszyn
Wyznaczyć najkrótszy czas i odpowiedni element umieścić na tej maszynie, gdzie ten czas jest dłuższy (na tej drugiej) Uzupełnić szeregi na każdej maszynie w tej samej kolejności, co poprzednie

Priority Dispatching Rules for Job Shops
LP. RULE TYPE DESCRIPTION 1 Earliest release date Static Time job is released to the shop 2 Shortest processing time Processing time of operation for which job is waiting 3 Total work Sum of all processing times 4 Earliest due date Due date of job 5 Least work remaining Sum of all processing times for oparations not yet performed 6 Fewest operations remaining Number of operations yet to be performed 7 Work in next queue Dynamic Amount of work awaiting the next machine in a job's processing time 8 Slack time Time remaining until due date minus remaining processing time 9 Slack/ remaining operations Slack time divided by the number of operations remaining 10 Critical ratio Time remaining until due date divided by days required to complete job

Simulation of Dispatching Rules (lwr)

Simulation of Dispatching Rules

Simulation of Lynwood Manufacturing Problem

Simulation of Lynwood Manufacturing Problem

Simulation of Lynwood Manufacturing Problem

Simulation of Lynwood Manufacturing Problem

Simulation of Lynwood Manufacturing Problem

Bar Chart for Lynwood’s Job Shop

Simulation Results Using Least Work Remaining for Lynwood’s Job Shop
WAITING TIME COMPLETION TIME MACHINE IDLE TIME (MAKESPAN - PROCESSING TIME) 1 55 120 Lathe 75 2 90 Drill 65 3 25 45 Grind 4 110 Mill

Scheduling Consecutive Days Off
M T W F S 8 6 9 5 3 EMPLOYEE NO. M T W F S 1 8 6 9 5 3 M T W F S 7 5 8 3 EMPLOYEE NO. M T W F S 2 7 5 8 3

Scheduling Consecutive Days Off
M T W F S 6 4 7 5 3 EMPLOYEE NO. M T W F S 3 6 4 7 5 M T W F S 5 4 3 6 2

Scheduling Consecutive Days Off
EMPLOYEE NO. M T W F S 4 5 3 6 2 1 7 8 9 10

Scheduling Consecutive Days Off
EMPLOYEE M T W F S 1 x 2 3 4 5 6 7 8 9 10 Total

Vehicle Scheduling Customer 1 2 3 4 5 6 7 - 20 57 51 10 50 55 25 30 11
1 2 3 4 5 6 7 - 20 57 51 10 50 55 25 30 11 70 15 80 90 60 53 47 38 12

Vehicle Scheduling 1 2 3 4 5 6 7 - 26 61 58 15 87 51 37 50 -10 -45 62
1 2 3 4 5 6 7 - 26 61 58 15 87 51 37 50 -10 -45 62 -24 -25 57 100 103 130 10 93

Vehicle Scheduling ROUTE TIME 0 - 4 - 7 - 0 150 0 - 3 - 1 - 0 81
97 30 The total time required is reduced to 358 minutes, or about 6 hours, a savings of about 3.8 hours over the original schedule.