Prof. Yuan-Shyi Peter Chiu

1

Material ManagementClass Note # 5

SCHEDULING

Prof. Yuan-Shyi Peter Chiu Feb. 2012

2

Schedulingis an important aspect of

operations control in both ..

Manufacturing Serviceindustries

§ S1§ S1 ：：IntroductionIntroduction ◇◇

3

Types of scheduling problems

¡]¢Ï¡^Job shop

scheduling

¡]¢Ð¡^Personnelscheduling

¡]¢Ñ¡^Facilities

scheduling

¡]¢Ò¡^Vehicle

scheduling

¡]¢Õ¡^Dynamicvs. static

scheduling

¡]¢Ô¡^Project

scheduling

¡]¢Ó¡^Vendor

scheduling

§ S1§ S1 ：：IntroductionIntroduction ◇◇

4

Important characteristics ofJob shop scheduling problems

The job arrival pattern - static ?

Number and variety of machines in the shop - identical ?

Number of workers in the shop - capacity ?

Particular flow patterns - a fixed order ?

Evaluation of alternative rules - no universal optimal ?

§ S5§ S5 ：： Characteristics of Job Shop Characteristics of Job Shop SchedulingScheduling

◇◇

5

Objectives of Job Shop Management

1.Meet duedates

2.MinimizeWIP

inventory

3.Minimizeaverage

flow time

4.Maximizemachine/

worker timeUtilization

7.Minimizeproductionand worker

costs

6.Reducesetup time

5.Provideaccurate jobstatus info

§ S5§ S5 ：： ObjectivesObjectives of Job Shop Managementof Job Shop Management

IT IS IMPOSSIBLE TO OPTIMIZE ALL 7 OBJECTIVES SIMULTANEOUSLY. CUSTOMER Service CUSTOMER Service Quality vs. Plant Efficiency / CostPlant Efficiency / Cost

◇◇

6

§ S6§ S6 ：： Flow Shop vs. Job ShopFlow Shop vs. Job Shop (1) FLOW SHOP(1) FLOW SHOP job M/C M/C M/C M/C M/C

1 2 3 4 m

. . .

...

n

Assembly Line different M/C = different operations.

(2) JOB SHOP(2) JOB SHOP job M/C M/C M/C M/C M/C

1 2 3 4 m... . . .n

{1, 3, 4, 1, m}{2, 4, 2, m-1}

PROBLEMS are extremely complex.All-purpose solution algorithms for solving general job shop problems do not exist.

◇◇

7

n

(3)(3) Parallel processingParallel processing vs. Sequential processingSequential processing

JOB M/C M/C M/C M/C1 2 3 m... Sequential . . .

JOB M/C M/C M/C M/C1

23

m

...1

12

3

3

3

mParallel processing

. . .n

§ S6§ S6 ：： Flow Shop vs. Job ShopFlow Shop vs. Job Shop ◇◇

8

§ S7§ S7 ：： Indicators of Performance EvaluationIndicators of Performance Evaluation

(1) FLOW TIME(1) FLOW TIME

i

1jji tF

job 1 t1 = 25 sec.job 2 t2 = 13 sec.job 3 t3 = 21 sec. . . .

3

1jj3 .sec59t3jobofcompletionF

(2) MEAN FLOW TIME(2) MEAN FLOW TIME

n

1i

i

1jj

n

1ii t

n1nFF

F

◇◇

9

(3) MAKESPAN(3) MAKESPAN ： F[n]

TIME REQUIRED TO COMPLETE ALL n jobs Minimizing to the Makespan is a common objective in multiple - m/c sequencing problem.

(4) TARDINESS(4) TARDINESS

max ( F[i] - D[i] , 0 ) where F[i] : Completion time of job i.

D[i] : Job i ’s Due Date.

Example ： Job F[i] D[i] Tardiness

1 3 10 0 2 7 12 0 3 14 13 1 4 16 13 3

§ S7§ S7 ：： Indicators of Performance Indicators of Performance EvaluationEvaluation ◇◇

10

(5) LATENESS(5) LATENESS

F[i] - D[i]

Job F[i] D[i] Lateness

1 3 10 -7 2 7 12 -5 3 14 13 1 4 16 13 3

(6) MINIMIZING(6) MINIMIZING AVG. TARDINESSAVG. TARDINESS

MAX. TARDINESSMAX. TARDINESS

ARE COMMON SCHEDULING OBJECTIVESARE COMMON SCHEDULING OBJECTIVES.

§ S7§ S7 ：： Indicators of Performance Indicators of Performance EvaluationEvaluation ◇◇

11

ti ; di

Wi = Fi － ti

Fi = Wi + ti

Li = Fi － di

Ti = max[Li , 0] Ei = max[-Li , 0] Tmax = max {T1 , T2 , … , Tn}

→ SPT minimizes

Mean flow Time. Mean waiting Time. Mean lateness. → EDD minimizes maximum lateness Lmax ～ Tmax. Lmax = max {L1 , L2 , L3 , … , Ln}

3 1 0 -3

-3 -2 -1 0

1

1 n

i

iF Fn

§ S8§ S8 ：： Notation Notation ◇◇

12

Common Scheduling Rules for single machine:Common Scheduling Rules for single machine:

(1)(1) FCFSFCFS

(2)(2) Shortest Processing Time (SPT)Shortest Processing Time (SPT)

(3) Earliest Due Date (EDD)(3) Earliest Due Date (EDD)

(4) CRITICAL RATIO (CR)(4) CRITICAL RATIO (CR)

◇◇

13

(1) Plane 1 2 3 4 5 t[i] 26 11 19 16 23 F[i] 95 11 46 27 69

makespan = 95

SPT = {2, 4, 3, 5, 1} = 49.6 F

(2) Plane 1 2 3 4 5 t[i] 26 11 19 16 23 # of Passengers 180 12 45 75 252 # of Passengers per minute 6.9 1.1 2.4 4.7 10.9 49 95 84 65 23 F[i]

432 564 552 507 252 # of Passengers

{5, 1, 4, 3, 2}

Example 8.2 – Example 8.2 – An example of priority rulesAn example of priority rules

14

20 40 60 80 100

100%

90%

76%

45%

23 49 65time

(3) ARRIVAL TIME = DUE TIME.

(4) PRIORITY e.g.continuing flights.

low fuel level.

carrying precious or perishable cargo.

% Passengers

15

§ S9§ S9 ：： EDD Scheduling EDD Scheduling minimizes the minimizes the maximum latenessmaximum lateness..

§ S10§ S10：： Moore’s (1968) algorithmMoore’s (1968) algorithm minimizes the number of Tardy jobsminimizes the number of Tardy jobs.

Step1 ： Sequence by earliest due date i.e. d[1] d≦ [2] d≦ [3] … d≦ ≦ [n]

Step2 ： Find the 1st tardy job in the current sequence, say job i. IF None exists go to Step 4

Step3 ： Consider jobs [1], [2], …, [i] Reject the job with largest tj .and Return to Step2.

Step4 ： Current sequence + rejected job(In any order.)

Applications ： chef, runway, preparing exam. garage, dock.

◇◇

16

Job# 2 3 1 5 4 6Di 6 9 15 20 23 30ti 3 4 10 10 8 6Fi 3 7 17 27 35 41

×× 1

Job# 2 3 5 4 6di 6 9 20 23 30ti 3 4 10 8 6Fi 3 7 17 25 31

×× 5

Example 8.3Example 8.3 Ti = { max ( F[i] - D[i] , 0 ) } > 0

17

2 3 4 6di 6 9 23 30ti 3 4 8 6Fi 3 7 15 21 Done!

2 - 3 - 4 - 6 - 5 - 1 1 - 5

2 3 4 6 5 1di 6 9 23 30 20 15ti 3 4 8 6 10 10Fi 3 7 15 21 31 41

Example 8.3Example 8.3

18

§.§. S10.1: S10.1: ClassClassProblems DiscussionProblems Discussion

Chapter 8 :Chapter 8 : # # 3,3, 4, 4, 5 5 p.413p.413 30, 30, 3232(a),(b)(a),(b), 34 , 34 p.449-450p.449-450

Preparation Time : 15 ~ 20 minutesPreparation Time : 15 ~ 20 minutesDiscussion : 15 minutesDiscussion : 15 minutes

19

§ S11§ S11 ：： Lawler’s Algorithm : PrecedenceLawler’s Algorithm : Precedence

min. max. gi(Fi) 1 i N≦ ≦

gi(Fi) = Fi – di = Li min. max. Lateness.

gi(Fi) = max (Fi – di , 0) Tardiness.

…

nextto

LAST

LAST

◇◇

20

JOBS:

1 2 3

45

6

JOB 1 2 3 4 5 6

ti 2 3 4 3 2 1

di 3 6 9 7 11 7

Example 8.4Example 8.4 ◇◇

21

JOBS 3

5

6 v = { 3, 5, 6 }Total Processing time of all jobs is 15 min iv {gi(Fi)}= {Fi-di}

miniv

= min {15-9 , 15-11 , 15-7}= min {6 , 4 , 8}= 4 ∴ JOB # 5 is scheduled last. (6th)

JOBS 3

6v = { 3, 6 }

Total Processing time is 13 min iv {gi(Fi)}= min {13-9 , 13-7}= min {4 , 6}= 4

∴ JOB # 3 is scheduled last. (5th)So far, { … , 3 , 5}


◇◇

22

JOBS 2

6v = { 2, 6 }

Total Processing time is 9 min iv {gi(Fi)}= min {9-6 , 9-7}= 2

∴ JOB # 6 is scheduled last. (4th)So far, { … , 6 , 3 , 5}

JOBS 2

4v = { 2, 4 }

Total Processing time is 8 min iv {gi(Fi)}= min {8-6 , 8-7}= 1

∴ JOB # 4 is scheduled 3rd

∴ So far, { … , 4 , 6 , 3 , 5}

JOBS → { , 2 , 4 , 6 , 3 , 5}∴

JOBS → { 1 , 2 , 4 , 6 , 3 , 5}∴

2

1


◇◇

23

Job# ti Fi di Ti (Tardiness) 1 2 2 3 0

2 3 5 6 04 3 8 7 16 1 9 7 23 4 13 9 45 2 15 11 4

∴ maximum tardiness is 4 days.


24

§.§. S12: S12: ClassClass Problems Discussion Problems Discussion

Chapter 8 : Chapter 8 : # # 6, 6, 7, 7, 8,8, 9 9 p.419p.419 1010

# # 3737,, p.451p.451

Preparation Time : 10 ~ 15 minutesPreparation Time : 10 ~ 15 minutesDiscussion : 10 minutesDiscussion : 10 minutes

◇◇

25

§ S13§ S13 ：： n job on m M/C’sn job on m M/C’s

(1) n JOBS Must BE PROCESSED ON 1 M/C’s

n …….. 3 2 1

1 ……..(n-2) (n-1) n

∴ there are n! possible ways.

(2) “n” JOBS Must BE PROCESSED ON “m” M/C’s

n …….. 3 2 1 n!

　　　 .……. n!

…….. n!

　　∴ there are (n!)m possible ways. #12 #12 p.428p.428

M/C

M/C１M/C 2

M/C m

…… ……

◇◇

26

(A) It provides better system performance in terms of both total flow time ( makespan ) and average flow time ( ).

mean IDLE time?

(B) SCHEDULING n Jobs on 2 M/C’s : if each Jobs must be processed in the order M/C-1 then M/C-2.

Results: The permutation schedule will minimize “makespan” and minimizes “ ”.

(C) Theorem 8.2: (p.422)

The optimal solution for scheduling n jobs on 2 M/C’s is always a permutation schedule.

F

§ S14§ S14 ：： Permutation schedules – characteristicsPermutation schedules – characteristics

F

27

§ S15§ S15 ：： 2 jobs on 2 M/C’s2 jobs on 2 M/C’s(A) All Possible Schedules for Two Jobs on Two Machines.

(B) Assumes that both jobs must be processed first on both jobs must be processed first on M/C#1 then on M/C#2M/C#1 then on M/C#2.

Machine 1 I JMachine 2 I J

Machine 1 J IMachine 2 J I

Machine 1 J IMachine 2 I J

Machine 1 I J Machine 2 J I

4 5 9

1 5 6

1 5 6 10

4 5 9 10

makespan idle flow time

9

6

10

10

42

44

72

95

12

11

5.52

65

52

55

82106

52

55

5.92109

28

§ S16§ S16 ：： Permutation Schedules - DefinitionPermutation Schedules - Definition

＊ Permutation SchedulesPermutation Schedules ＝ Same Sequence on

both (all) M/C’s

Total # of permutation schedules is exactly n!n!

◇◇

29

§ S17§ S17 ：： Johnson’s RuleJohnson’s Rule

(1) Definition:

M/C-A

M/C-B

Jobs must be processed first on M/C-A

then M/C-B

Ai : Processing Time of Job i on M/C-A

Bi : Processing Time of Job i on M/C-B

Job i procedes Job i+1Job i procedes Job i+1

if min (Aif min (Ai i , B, Bi+1i+1) < min (A) < min (Ai+1i+1,B,Bii))

2 M/C’s

◇◇TO MINIMIZE THE MAKESPAN

30

(2) Working Procedures for (2) Working Procedures for Johnson’s ruleJohnson’s rule::

1. List the values of Ai & Bi in 2 columns.

2. Find the smallest remaining element in the 2 columns.

If it appears in column A then schedule that job next.

If it appears in column B then schedule that job last.

3. Cross off the jobs as they are scheduled.

Stop when all jobs have been scheduled.

An easy way to implement Johnson’s Rule

◇◇

31

Five jobs are to be scheduled on two machines. The processing times are

Jobs Machine A Machine B

1 5 2 2 1 6 3 9 7 4 3 8 5 10 4


32

A #2 #4 #1 #3 #5

B #2 #4 #1 #3 #5

1 4 9 18 28

0 1 7 15 17 25 28 32

makespan=30

if using SPT in Ai then obtain the above 2-4-1-3-5 _____________________________________________________

IF BY JOHSON’S ROLE TO MINIMIZE THE MAKESPAN OR TOTAL FLOW TIME.

Rule: Job i precedes job i+1 if MIN(Ai,Bi+1)<(Ai+1,Bi)

in 2-4-3-5-1 order

(A)

(B) A #2 #4 #3 #5 #1

B #2 #4 #3 #5 #1

1 4 13 18 28

0 1 7 15 22 23 27 28 30


33

§ S18§ S18 ：： n jobs on 3 M/C’sn jobs on 3 M/C’s

(A)Objective : To minimize total flow time (“make span.”)

it is still true that a permutation schedule

is optimal.

(B) But it is not necessarily optimal for the case of F’ .

(C) 3 M/C’s can be reduced to 2 M/C’s (then using Johnson’s Rule to solve it)

if min Ai max B≧ i

or min Ci max B≧ i

“either one” of these conditions be satisfied.

then define iiiiii CBBBAA '' ;

◇◇

34

Consider the following job times for a three-machine problem. Assume that the jobs are processed in the sequence A-B-C

M/C’sJobs A B C 1 4 5 8 2 9 6 10 3 8 2 6 4 6 3 7 5 5 4 11

Min Ci=6

Max Bi=6

∴ Min Ci Max B≧ i

let

iii

iii

CBB

BAA

'

'


35

M/C’sJobs A’ B’ 1 9 13 2 15 16 3 10 8 4 9 10 5 9 15

Using Johnson’s Rule

5-1-4-2-3


36

5 10 15 20 25 30 35 40 45 500

B

C

A

5 4 1 2 3

5 4 1 2 3

5 4 1 2 3

51

5-4-1-2-3

5-1-4-2-3

5 10 15 20 25 30 35 40 45 500

B

C

A

5 1 4 2 3

5 1 4 2 3

51 4 2 3

51


37

§.§. S18.1: S18.1: ClassClass Problems Discussion Problems Discussion

Chapter 8 : Chapter 8 : # # 13, 13, 1414 p.428 p.428

38,38, 4040 p.451p.451

Preparation Time : 10 ~ 15 minutesPreparation Time : 10 ~ 15 minutesDiscussion : 10 minutes Discussion : 10 minutes

◇◇

◆◆

38

§ S19§ S19 ：： more on 3 M/C’s problemsmore on 3 M/C’s problems(A) If neither min A﹛ i max B≧ i Nor min Ci max B≧ i ﹜ Then Using

(B) TO MINIMIZE “MAKESPAN” OR “TOTAL FLOW TIME”

A PERMUTATION SCHEDULE IS OPTIMAL ON 3 M/C’s

iii BAA 'iii CBB '

and

Will usually give “REASONABLE” but

possibly “SUBOPTIMAL” result.

Simplifying 3 M/C’s problems 2 M/C’s.

◇◇

39

§ S20§ S20 ：： Akers’ procedures for solving Akers’ procedures for solving 2 Jobs on 2 Jobs on mm M/C’s problems M/C’s problems

(1) Draw a cartesian coordinate system.

Job 1 on X-axis

Job 2 on Y-axis

mark off the operation times

on X & Y axis

(2) BLOCK OUT AREAS for each M/C’s

(3)Determine a path from origin to the end of the final block.

move:

The path with minimum vertical distance is “Optimal”.

◇◇

40

JOB 1 JOB 2

Oper. Time Oper. Time

(A) Sanding 3 A 2

(B) Lacquering 4 B 5

(C) Polishing 5 C 3

A regional manufacturing firm produces a variety of household products. One is a wooden desk lamp. Prior to packing, the lamps must be sanded, lacquered, and polished. Each operation requires a different machine. There are currently shipments of two models awaiting processing. The times required for the three operations for each of the two shipments are


41

2B1A

2B

1A

2C 1B

1B1C

2C

1C2B

1B1B

2A2A

maximize the diagonal movement = min. horizontal vertical

Fig. 8-8 p.426Example 8.7Example 8.7 ◇◇

42


Fig. 8-9 p.426 ◇◇

43


44

2

2

10

8

6

4

12

14

16

4 6 8 10 12 14 16 18

(16,14)

(16,11)(7,11) Job 2

(BOB)

Job 1 (Reggie)

Time=16+4+3=23

A1 →B1 →C1 A2 A2 ↓ C1 D2 ↓ C2←D1←D1 ←D2 C2 B2Time=16+1+3=20

A2 →D2 →B2 →C2 →C2 →C2 →C1 →D1 A1 A1 A1 B1

A

B

C

D


45

§.§. S21: S21: ClassClass Problems Discussion Problems Discussion

Chapter 8 : Chapter 8 : # # 15,15,16 16 p.428p.428

3939 p.452p.452

Preparation Time : 25 ~ 30 minutesPreparation Time : 25 ~ 30 minutesDiscussion : 20 minutes Discussion : 20 minutes

◇◇

46

§ S22§ S22 ：： Parallel processing on identical M/C’s.Parallel processing on identical M/C’s.

SPT → minimize mean flow time.

LPT → minimize total flow time, or makespan.

( longest )

◇◇

The EndThe End

47

SchedulingPreview : Chap. 8 [ pp.413~442 ]Preview : Chap. 8 [ pp.413~442 ]

Problem:Problem: # 5, #32(a),(b), # 5, #32(a),(b),

#7, #8, #10,#7, #8, #10,

#37, #40, #39,#37, #40, #39,

#14, #15, #16#14, #15, #16

Prof. Yuan-Shyi Peter Chiu

Documents

Transcript of Prof. Yuan-Shyi Peter Chiu