mohd zaidan bin abdul aziz b050810281

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Mohd Zaidan bin Abdul Aziz B050810281Grace Hii Leh Ung B050810211Low Mei Ching B050810215Sawita bt Amir B050810051

SCHEDULING (Part II)

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1. Shortest Processing Time (SPT) Minimizes the mean flow time Minimizes waiting time Minimizes lateness for single-machine sequencing.2. Earliest Due Date Scheduling Minimizes the maximum lateness

4.4 Sequencing Theory for Single Machine

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3. Moore’s algorithm Minimizes number of tardy jobsStep 1:Sequence the jobs according to the earliest due date to obtain the initial solution

Step 2:Find the first tardy job in the current sequence, say job [i]. If none exists go to step 4.


d[1] d[2],…, d[n]

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Step 3:Consider jobs [1], [2], …, [i]. Reject the job with the largest processing time. Return to step2.

Step 4:Form an optimal sequence by taking the current sequence and appending to it the rejected jobs. The job appended to the current sequence scheduled in any order.

Reason: Consider number of tardiness jobs rather than tardiness

Reason: Largest effect on the tardiness of the Job[i]


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Job 1 2 3 4 5 6

Due date 15 6 9 23 20 30

Processing time 10 3 4 8 10 6

Job 2 3 1 5 4 6

Due date 6 9 15 20 23 30


Completion time 3 7 17 27 35 41

EXAMPLE 1:

Solution:

Longest processing time


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Job 2 3 5 4 6

Due date 6 9 20 23 30

Processing time 3 4 10 8 6

Completion time 3 7 17 25 31

EXAMPLE 1: Solution (cont.) Longest processing time

Job 2 3 4 6

Due date 6 9 23 30

Processing time 3 4 8 6

Completion time 3 7 15 21

Optimal sequence: 2, 3, 4, 6, 5, 1 or 2, 3, 4, 6, 1, 5Number of tardy jobs is two in either case.

No lateness


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Exercise 4.4.1:Seven jobs are to be processed through a single machine. The processing times and due dates are given below.

Job 1 2 3 4 5 6 7

Processing time 3 6 8 4 2 1 7

Due date 4 8 12 15 11 25 21

Determine sequence of the jobs in order to minimize a) Mean flow timeb) Number of tardy jobsc) Maximum lateness

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4. Lawler’s Algorithm Minimize maximum lateness Minimize maximum tardinessSubject to precedence constraint ( certain job must be completed before other jobs can begin)

)(maxmin1

iini

Fg

iiiii LdFFg )(

)0,max()( iiii dFFg


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Rule:Schedule job in reverse order. Step 1:At each stage, determine the set of jobs (named V) not require to precedes any other. Among set V, choose job k that satisfies:

n

i i

ivik

t

gg

1

))((min)(

Example: Job among V that has smallest tardiness, if arranged on position [n].

Processing time of current sequence.


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Step 2:Now, Job k scheduled last. Consider remaining jobs and again determine set of jobs that not require to precede any other remaining job.Step 3:After scheduling Job k, τ reduced by tk and job scheduled next to last is now determined.Step 4:Process is continued until all jobs are scheduled.


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1 2 3

4 5

6

Job 1 2 3 4 5 6


Due date 3 6 9 7 11 7

EXAMPLE 2:


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Solution:Step 1: Find the job scheduled last(sixth)

Job 1 2 3 4 5 6


Due date 3 6 9 7 11 7

τ = Total processing time = 2+3+4+3+2+1 = 15

Job 3 5 6

Tardiness 15-9=6 15-11=4 15-7=8

Min value

Hence, Job 5 scheduled last


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Solution: Cont.Step 2: Find the job scheduled fifth.

1 2 3

4 5

6

Job 1 2 3 4 6


Due date 3 6 9 7 7

τ = New total processing time = 15 – 2 = 13

Job 3 6Tardiness 13-9=4 13-7=6

Min valueHence, Job 6 scheduled fifth.


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Solution: Cont.Step 3: Find the job scheduled fourth.

1 2 3

4 5

6

Job 1 2 4 6

Processing time 2 3 3 1

Due date 3 6 7 7


Hence, Job 6 scheduled fourth.

Job 2 6

Tardiness 9-6=3 9-7=2

Min value


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Solution: Cont.Step 4: Find the job scheduled third.

1 2 3

4 5

6

Job 1 2 4

Processing time 2 3 3

Due date 3 6 7

2 4

Tardiness 8-6=2 8-7=1


Hence, Job 4 scheduled third.Min value


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Solution: Cont.Step 5: Find the job scheduled second.

1 2 3

4 5

6

Job 1 2

Processing time 2 3

Due date 3 6

Job Processing time

Flow time Due date Tardiness

124635

233142

2589

1315

36779

11

001244

The optimal sequence: 1-2-4-6-3-5

Maximum tardiness


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Exercise 4.4.2:Eight jobs are to be processed through a single machine. The processing times and due dates are given below.

Furthermore, assume the following precedence relationships must be satisfied:

Job 1 2 3 4 5 6 7 8

Processing Time 2 3 2 1 4 3 2 2

Due date 5 4 13 6 12 10 15 19

2 6 3

1 4 7 8

Determine the sequence in which jobs should be done in order to minimize maximum lateness subject to precedence restriction.

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• Analysis of the previous section which several jobs must be proceed on more than one machine.

•The optimal solution for scheduling n jobs on two machines is always a permutation schedule (that is, jobs are done in the same order on both machines).

•Permutation schedules provide better performances in term of both total and average flow time

• Minimization of the mean idle time in the system

4.5 Sequencing Theory for Multiple Machines

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• How to implement this rule:1. List value of A and B in to columns2. Find the smallest element in the 2 columns. If it appears in

column A, then schedule that at front of the sequence. If appears in column B, schedule that at the back of sequence

3. Find the remaining element in the two columns. If it appears on column A, then schedule that next job. If appears in column B, then schedule that job last.

4. Cross off the jobs as they are scheduled.

1. Johnson’s algorithm : Scheduling n jobs on two machines


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EXAMPLE:• Five jobs are to be scheduled on two machines. The processing

time are:

Job Machine A Machine B1 5 22 1 63 9 74 3 85 10 4

Machine A A2 A4 A3 A5 A1

Machine B B2 B4 B3 B5 B1

0 1 4 7 13 15 22 23 27 28 30

2-4-3-5-1


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Exercise 4.5.1: Six job are to be schedule on two machine. The processing time are

Job Machine A Machine B1 20 272 16 303 43 514 60 125 35 286 42 24

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2. Extension to three machine

• The 3 machine problem can be reduced to a 2 machine problem if the satisfied the following condition:min A ≥ max B or min C ≥ max B

• only either one of these conditions be satisfied. • Then reduced to 2 machine problem in the

following way:A’ = A + B and B’ = B + C


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EXAMPLE: Machine

Job A' B'1 9 132 15 163 10 84 9 105 9 15

MachineJob A B C1 4 5 82 9 6 103 8 2 64 6 3 75 5 4 11

Min A = 4 Max B = 6 Min C = 6 1-4-5-2-3

Machine A A1 A4 A5 A2 A3

Machine B B1 B4 B5 B2 B3

180 9 63 7122 27 32 42 47 52


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Exercise 4.5.2:The following four job must be processed through a three-machine flow shop

MachineJob A B C1 4 2 62 2 3 73 6 5 64 3 4 8

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3. Two job flow shop problem• Two jobs are processed through m machines.• Present graphical procedure for solving this problem. The step are:1. Draw a cartesian coordinate system with the processing times

corresponding the first job on the horizontal axis and the processing time corresponding to the second job on the vertical axis.

2. Block out areas corresponding to each machine at the intersection of the intervals marked for the machine on the two axes.

3. Determine the part from the origin to the end of the final block that does not intersect any of the blocks and that minimize the vertical movement.


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EXAMPLE:

• A regional manufacturing firm produces a variety of household products. One is a wooden desk lamp. Prior to packing, the lamp 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:

Job 1 Job 2Operation Time Operation Time

Sanding (A) 3 A 2Lacquering (B) 4 B 5Polishing ( C) 5 C 3


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A1 A2

B1 B2

C1 C2

3

531 42 1196 7 8 10 12

12

45

7

9

6

8

10

B

A

C

2 4 6 8 10 12 14

CB

Gantt ChartSolution

A

Total time = 12 + 3 = 15

Total time = 12 + (2 + 2) = 16


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3

531 42 1196 7 8 10 12

12

45

7

9

6

8

10

B

A

C Total time = 10 + 6 = 16

Total time = 10 + (3 + 2) = 15

2 4 6 8 10 12 14

CB

Gantt ChartSolution

A A1 A2

B1 B2

C1 C2


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Exercise 4.5.3:Two jobs must be proceeds through four machines in same order. The processing time in the required sequence are

Job 1 Job 2Machine Time Machine Time

A 5 A 2B 4 B 4C 6 C 3D 3 D 5

Determine how the two job should be schedule in order to minimize the total makespan and draw the gantt chart indicating the optimal schedule.

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Line balancing: the process of assigning tasks to workstations in such a way that the workstations have approximately equal time requirements.

Line Balancing

Cycle Time

Cycle time: maximum time allowed at each workstation tocomplete its set of tasks on a unit. amount of time allotted to each workstation determined in advance, based on the desired rate of production of assembly line.

Maximum cycle time = summation of the task times.Minimum cycle time = the longest task times.

4.6 ASSEMBLY LINE BALANCING

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Introduction (cont.)Factors that contribute to the

difficulty of the problem:- There are precedence

constraints- Some tasks cannot be

performed at the same workstation

DOT = timecycle = CT

CTOT = D rate,Output

rateoutput desired = Ddayper timeoperating OT


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• Assign tasks in order of most following tasks.

– Count the number of tasks that follow

• Assign tasks in order of greatest positional weight.

– Positional weight is the sum of each task’s time and the times of all following tasks.

Some Heuristic (intuitive) Rules:Line Balancing Rules


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Precedence diagram: Tool used in line balancing to display elemental tasks and sequence requirements

A Simple Precedence Diagrama b

c d e

0.1 min.

0.7 min.

1.0 min.

0.5 min. 0.2 min.

Figure 4.6.1

Precedence Diagram


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task timeof sum = tCT

t)( = N

Determine the Minimum Number of Workstations Required

What is the minimum number of workstations for the previous precedence diagram? (assume minimum cycle time)

mins 2.5 = t

35.2min 1

mins 2.5 = N


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WorkstationTimeRemaining Eligible

AssignTask

RevisedTime Remaining

StationIdle Time

1 1.00.90.2

a, ccnone

ac–

0.90.2

0.2

2 1.0 b b 0.0 0.0

3 1.00.50.3

de–

de–

0.50.3 0.3

0.5

EXAMPLE:Arrange tasks shown in Figure 4.6.1 into 3 workstations.

Use a cycle time of 1.0 minute Assign tasks in order of the most number of followers

Solution

Idle time per cycle


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%100*(N)(CT)cycleper timeIdle

= timeidle %

Efficiency = 100% – Percent idle time

Calculate Percent Idle Time

What’s the % idle time and efficiency for the above example?

%7.16%100*(3)(1.0)

0.5 = timeidlePercent

Efficiency = 100% - 16.7% = 83.3%


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A manager wants to assign workstations in such a manner that hourly output is 4 units. Working time is 56 minutes per hour. What is the cycle time?

Cycle time = operating time/output rate= 14 min

EXAMPLE:


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Assign the tasks above to workstations in the order of greatest positional weight.

Steps:1) Arrange the task in the decreasing order of positional weights.2) Find out the number of workstations

Number of workstations = = = 3.2 =4

timecycle s task timeof sum

1445

Task Task time

F 5

D 7

G 6

A 3

B 2

C 4

E 4

H 9

I 5

Arrange tasks in decreasing order of positional weight

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%64.19

%100*(4)(14)

11 = timeidlepercent

D=7

Cycle Time=14 min

F=5 G=6 A=3 B=2

C=4 E=4I=5H=9

14

Stati

on

I

II

III

IV

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A shop wants an hourly output of 33.5 units per hour. The working time is 60 minutes per hour. Assign the tasks using the rules:

a) In the order of most following task.b) In the order of greatest positional weight.

EXERCISE 4.6.1


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EXAMPLE:


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The job times & precedence relationship for this problem:

Task ImmediatePredecessor

Time

1 - 122 1 63 2 64 2 25 2 26 2 127 3,4 78 7 59 5 1

10 9,6 411 8,10 612 11 7

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Suppose that the company is willing to hire enough workers to produce one assembled machine every 15 minutes.

Sum of task time = 70 minutesMinimum no. Workstation = 70/15 = 4.67 5

Positional weight of task i: the time required to perform task i plus the times required to perform all task having task i ask a predecessor.

TASK POSITIONAL WEIGHT

1 70

2 58

3 31

4 27

5 20

6 29

7 25

8 18

9 18

10 17

11 13

12 7

TASK POSITIONAL WEIGHT

1 70

2 58

3 31

6 29

4 27

7 25

5 20

8 18

9 18

10 17

11 13

12 7


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Rank the step in the order of decreasing positional weight:Assume C = 15 min

But, minimum possible No. of workstation = 5!!!This method is heuristic possible that there is a solution with 5 stationsUse C = 16 min

No. of stations decreases 16 %, cycle time increases 7%Assume production day = 7 hrsC = 15 daily production level = 28 units/assembly operation

Station 1 2 3 4 5 6 Total idle timeTasks 1 2,3,4 5,6,9 7,8 10,11 12

Idle time 3 1 0 3 5 8 20

Station 1 2 3 4 5 Total idle timeTasks 1 2,3,4,5 6,9 7,8,10 11,12

Idle time 4 0 3 0 3 10

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C = 16 daily production level = 26.25 units/assembly operationManagement have to determine whether the decline in the production rate of 1.75

units/day/operation is justified by the savings realised with 5 rather than 6 stations.

Alternative choice:Stay with 6 stations, but use C = 13 min

13 min: minimum cycle time with 6 stations. Why not 12 min???!!Production rate = 32.3 units/day/operationIncreasing No. of stations from 5 to 6 substantial improvement in the throughput rate.

Station 1 2 3 4 5 6 Total idle timeTasks 1 2,3 6 4,5,7,9 8,10 11,12

Idle time 1 1 1 1 4 0 8

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EXERCISE 4.6.2

Consider the assembly line balancing problem represented by the figure above. Determine a balance for a. C = 20


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1. Single machine Uncertainty of processing times Exact completion time of one or more jobs may not be

predictableObjective: Minimize expected average weighted flow

time. Job i precedes job i+1 if

4.7 Stochastic Scheduling 4.7.1 Static Analysis

where job times are t1, t2, …, tn and weights are u1, u2, …, ui.

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Jobi 1 2 3 4 5 6 7 8 9 10

Processing Time, ti 1 10 5 2 8 7 8 4 3 6

Importance weight, ui 3 2 1 1 4 2 3 3 2 4

Due date, di 10 20 15 10 10 25 15 25 10 20

4.7.1 Stochastic Scheduling: Static Analysis

EXAMPLE:

Find the optimal sequence that minimize expected average weighted flow time.

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Jobi ti ui ti/ui di Ci Ti

1 1 3 0.3 10 1 08 4 3 1.3 25 5 09 3 2 1.5 10 8 0

10 6 4 1.5 20 14 04 2 1 2 10 16 65 8 4 2 10 24 147 8 3 2.7 15 32 176 7 2 3.5 25 39 143 5 1 5 15 44 292 10 2 5 20 54 34

∑Ci = 237

∑Ti= 114


Solution:

Expected average weighted flow time = 237/10= 23.7

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2. Multiple Machine• n jobs are to processed through two identical parallel

machines. Each jobs needs to be processed only once on either machine.

• The objective is to minimize the expected makespan.• Parallel is different from flow shop problem.• In flow shop, jobs are processed first one machine 1

then on machine 2.• Then the optimal sequence is to schedule the jobs

according to LEPT (longest expected processing time first) opposite with the SPT rule


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Dynamic: Jobs arrive randomly over time, and decisions must be made on an ongoing basis as how to schedule those jobs.When jobs arrive shop dynamically over time, queuing theory provides a means of analyzing the results. Standard M/M/1 queue applies to case of purely random arrivals to single machine with random processing times.

4.7.2 Stochastic Scheduling: Dynamic Analysis

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Selection independent of job processing times Same mean flow times, but differ variance

flow timesSelection dependent of job processing times Job times realized when job joins the queue

rather than when job enters service SPT results lowest expected flow time.

4.7.2 Stochastic Scheduling: Dynamic Analysis

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4.8.1 FINITE CAPACITY PLANNING (FCS)FCS - jobs are being scheduled through a number of work centers, each with one or more machines.• jobs can pass each other or change their order as they are processed, depending on their priority.• a job can be split into two or more parts if this will facilitate scheduling

Advantage:-The addition of capacity improves completion dates of jobs & reduces waiting times.

Bottleneck: a work centre whose capacity is less than the demand placed on it & less than the capacities of all other resources.

4.8 ADVANCED TOPIC FOR OPERATIONS SCHEDULING

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Batch operation:Job shop “shop”, “job”, “work center”Job “customer”, “patient”, “client”, “paperwork”Work centre “room”, “office”, “facility”, “skill specialty”

In a job shop, batch corresponds to what customer orders – can include one or several parts or items.

Each part or job is scheduled through the various machines & work centres according to the equipment & labour needed to process the job.

4.8.2 BATCH SCHEDULING

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Batch scheduling:• each batch flowing through a batch process typically moves along with many starts & stops, not smoothly – due to layout of batch process jobs or customers wait in line as each batch is transferred from one work centre to the next.

• batch scheduling problem: network or queues.

•Jobs or customers spend most of their time waiting in line – amount of time waiting varies with the load of the process

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Thank You ^^

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Jobi ti di Ci Ti

6 1 25 1 05 2 11 3 01 3 4 6 24 4 15 10 02 6 8 16 87 7 21 23 23 8 12 31 19

∑Ci = 90 ∑Ti= 31

4.4.1a) SPT sequence rule, Mean flow time

= 90/7 = 12.86

Solution:

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Job 1 2 5 3 4 7 6

Due date 4 8 11 12 15 21 25

Processing time 3 6 2 8 4 7 1

Completion time 3 9 11 19 23 30 31

Job 1 5 3 4 7 6

Due date 4 11 12 15 21 25


Completion time 3 5 13 17 24 25

Job 1 5 4 7 6

Due date 4 11 15 21 25


Completion time 3 5 9 16 17

4.4.1b) Moore ‘s algorithm.Optimal sequence: 1, 5, 4, 7, 6, 3, 2 or 1, 5, 4, 7, 6, 2, 3Number of tardy jobs is two in either case.

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Jobi ti di Ci Ti

1 3 4 3 02 6 8 9 15 2 11 11 03 8 12 19 74 4 15 23 87 7 21 30 96 1 25 31 6

∑Ci = 126

∑Ti= 31

4.4.1c) EDD sequence rule, Maximum lateness

= 9

Rule Sequence Mean flow time Number of tardy job

Maximum lateness

SPT 6-5-1-4-2-7-3 12.86 4 19

Moore’s 1-5-4-7-6-3-2 14.86 2 19

EDD 1-2-5-3-4-7-6 18 5 9

Summary:

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τ = 2+3+2+1+4+3+2+2 = 19

Job 3 8Tardiness 19-13=6 19- 19=0

Job 3 7Tardiness 17-13=4 17- 15=2

Job 1 2 3 4 5 6 7 8

Processing Time 2 3 2 1 4 3 2 2

Due date 5 4 13 6 12 10 15 19

2 6 3

1 4 7 8

2 6 3

1 4 7 8

τ = 19- 2= 17

Job 1 2 3 4 5 6 7

Processing Time 2 3 2 1 4 3 2

Due date 5 4 13 6 12 10 15

Solution:

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Job 3 4Tardiness 15-13=3 15- 6=9

2 6 3

1 4 7 8

τ = 17- 2 = 15

Job 1 2 3 4 5 6

Processing Time 2 3 2 1 4 3

Due date 5 4 13 6 12 10

Job 4 6Tardiness 13-6=7 13- 10=3

2 6 3

1 4 7 8

τ = 15-2 = 13

Job 1 2 4 5 6

Processing Time 2 3 1 4 3

Due date 5 4 6 12 10

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2 6 3

1 4 7 8

τ = 13-3 = 10

Job 1 2 4 5

Processing Time 2 3 1 4

Due date 5 4 6 12


2 6 3

1 4 7 8

τ = 10- 1= 9

Job 1 2 5

Processing Time 2 3 4

Due date 5 4 12

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Job Processing time

Flow time Due date Tardiness

2 3 3 4 0

1 2 5 5 0

4 1 6 6 0

6 3 9 10 0

3 2 11 13 0

7 2 13 15 0

8 2 15 19 0

5 4 19 12 7

4.4.2) Maximum tardiness: 7Optimal sequence: 2-1-4-6-3-7-8-5

2-1-3-5-6-4

4.5.1) Job Machine A Machine B1 20 272 16 303 43 514 60 125 35 286 42 24

A2 A1 A3 A5 A6 A4

B2 B1 B3 B5 B6 B4

73160 36 7946 114 130 156 158 182 216 228

MachineJob A B C1 4 2 62 2 3 73 6 5 64 3 4 8

4.5.2)

min A = 2 max B = 5 min C = 6 min C > max B = 6 > 5

Job A’ B’1 6 82 5 103 11 114 7 12

2-1-4-3

A2 A1 A4 A3

B2 B1 B4 B3

0 5 3315 26 291811 44

4.5.3) Job 1 Job 2Machine Time Machine Time

A 5 A 2B 4 B 4C 6 C 3D 3 D 5

14

12

10

8

6

4

2

12 2 4 6 8 10 14 16 18

A

B

C

DTotal time = 18 + 5 = 23

Total time = 18 + 2 = 20

A1 B1 C1 D2

A2 B2 C2 D2

0 2 6 7 11 17 2010 16

Sequence for Multiple Machines

14

12

10

8

6

4

2

2 4 6 8 10 12 14 16

A

D

B

C Total time = 16 + 4 = 20

Total time = 14 + (3 + 4) = 23

A1 A2

B1 B2

C1 C2D2 D1

ABCD

5 10 15 20

Different sequence

EXERCISE 4.6.1unit

hrunitshr

DOTCT min/80.1

/5.33min/60

65.00.15.08.07.06.05.04.1 t

433.3min8.1

min6

CTt

N

Workstation Time remaining

Eligible task Assigned Task

Revised Time Remaining

Station Idle Time

1 1.8 A A 0.4 0.4

2 1.8 BE

BE

1.30.5 0.5

3 1.8 DG

DG

1.10.1 0.1

4 1.8 CFG

CFG

1.20.70.2 0.2

Idle time per cycle = 1.2

(a)

Percent idle time = x 100%

%3.83%7.16%100

%%100

%7.16

%100)8.1(4

2.1

timeidleEfficiency

(b) Task Positional Weight

A 6

B 4.6

C 1.6

D 2.2

E 2.3

F 1.0

G 1.5

H 0.5

Task Positional Weight

A 1.4

B 0.5

E 0.8

D 0.7

C 0.6

G 1.0

F 0.5

H 0.5


B=0.5

433.38.1

6

CTt

N

B=0.5

Cycle Time=1.8 min

A=1.4 E=0.8 D=0.7

C=0.6 G=1.0H=0.5F=0.5

1.8St

atio

n

I

II

III

IV

0.4

0.3

0.2

0.3

Idle time

Idle time per cycle = 0.4 + 0.3 + 0.2 + 0.3 = 1.2

20CTask Immediate

predecessorTime Positional

Weight1 - 12 682 1 5 323 1 7 314 2 8 275 2,3 5 246 3 6 257 4,5,6 4 198 7 3 159 8 4 12

10 8 6 1411 9,10 8 8

68t

Arrange task in decreasing order of positional weight:

Task PositionalWeight

1 68

2 32

3 31

4 27

6 25

5 24

7 19

8 15

10 14

9 12

11 8

EXERCISE 4.6.2

Task 8=3 Task 10=6Task 6=6

Task 2=5

Cycle Time=20 min

Task 1=12 Task 3=7 Task 4=8

Task 5=5 Task 7=4Task 11=8Task 9=4

20

Stati

on

I

II

III

IV

3

-1

2

8

Idle time

Task 8=3 Task 10=6Task 5=5

Task 2=5

Cycle Time=20 min

Task 1=12 Task 3=7 Task 4=8

Task 7=4Task 11=8Task 9=4

20

Stati

on

I

II

III

IV

3

0

1

8

Idle time

Swap Task 6 & Task 5:

Task 6=6

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