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1POPULAR MANUFACTURING SYSTEMS

Assembly Lines Group technology / cellular manufacturing

systems (GT /CMS) MRP 1 / MRP 2 supported systems Flexible manufacturing system (FMS) Just in time production systems (JIT) ERP supported manufacturing systems Agile manufacturing systems (AMS) Quick response manufacturing systems

(QRMS) Lean production systems (LEAN)

Facility Layout 2

Machine shop process layoutReceiving Grinders

Mills

Raw material. Large numberof

storage Assembly low volumeproducts

DrillsPlaners

FinishedInspection goods

Lathes Automatics storage

Part APart B

Facility Layout 3

Product layout

Raw material. FabricationReceiving storage line-part B

Fabrication Planerline-part A

Finished Lathe goods Drillstorage Mill

MillDrill

GrinderMill

Assembly line Automatic

Small numberofhigh volumeproducts

2WHICH IS BETTER ?

WHEN ?

WHY ?

ASSEMBLY LINES IN MANUFACTURING SYSTEMS

ASSEMBLY LINES Form a major part of manufacturing

systems Popular Offer tremendous advantage in terms of

higher productivity. A system of assembly line consists of >tasks to be performed, >the work stations at which various sets of

these tasks are performed and >conveyor that moves the product from

station to station

USED FOR PRODUCING Any product Generally in Automobile IndustryFinal Assembly, Engine Assembly and other majorAssemblies like Gear Box, Axle, Steering System

etc., Consumer durables like TV, AC units,

Refrigerators, DVD players, Washing Machines,Electrical motors, Fans etc.

Garment Industry ready made garments PCB assembly

3THE BEGINNING OF ASSEMBLY LINES IN FORD

Around 1913Magneto assemblyOld output : One piece in 5.5 HrsImproved output : One piece in 13 minutes

Around 1918Engine assemblyOld output : One piece in 8 HrsImproved output : One piece in 18 minutes

MASS MANUFACTURING

Assembly line - Henry Ford 1920s Low skilled labor, simplistic jobs,

no pride in work Interchangeable parts Lower quality Affordably priced for the average family Billions produced - identical

FORD PRODUCTION CYCLE - 1926

MONDAY7:00 PM Ore boat docks at the River Rouge plantTUESDAY10:55 AM Ore reduced to foundry iron 16 hours later12:55 PM Cylinder block is cast5:05 PM 58 machining operations on casting in 55 minutes6:00 PM Motor assembly takes an average of 97 minutes7:45 PM Finished motor loaded on railcars for the assembly plant.WEDNESDAY8:00 AM 4-hour assembly time at standardized assembly plant12:00 PM Dealer takes delivery of car

41 HOURS !

SOME DEFINITIONS

Assembly line is a moving conveyor passing through a series of workstations in a uniform time interval called cycle time.

A workstation is a location on the line at which the specific tasks assigned to the station are performed.

A number of specific operations (tasks) are performed in a specific sequence as required by the precedence requirements of the tasks to assemble a product.

The cycle time is the maximum time available to any workstation to complete all the tasks assigned to it.

The cycle time is dependent upon the rate of production desired of the assembly line.

4BASIC ISSUES

Identify the n tasks to be performed to assemble a product.

Identify S Work stations Distribute the n tasks to S workstations Load each station with tasks taking almost equal

time (Balancing) Deliver at least one product out of the line in

the time C ,the cycle time. Decide the value of cycle time based on the

output rate P desired.

EXAMPLE

Assembly of a certain product consists of 11 tasks with specific time and precedence requirements as shown in Table .

The required rate of output necessitates that the cycle time to be 10 Minutes.

Design the assembly line such that sum of the station idle times is minimized.

Activity i 1 2 3 4 5 6 7 8 9 10 11Time ti 1 5 4 1 5 6 2 4 3 5 3Immediate followers

3428

7 7 5 6 7 11 9 10 11 -

1

10

4

3

2

7 1165

8 9

Activity i 1 2 3 4 5 6 7 8 9 10 11

Time ti 1 5 4 1 5 6 2 4 3 5 3

Immediate follower

3,4,

2,8

7 7 5 6 7 11 9 10 11 -

5SOME BASIC CONCEPTS ? P= production rate required in unit time i=activity or the task , i=1, n j=workstation , j=1.S ti=time for activity i tij= time for activity i assigned to station j nj = number of activities assigned to station j Tj= time consumed by all activities at station j T= Total work content time C= Cycle time

WHAT ARE THE RELATIONSHIPS ?

ti vs Tj ti vs T Tj vs T C vs ti, ? C vs Tj ? C vs T ? C vs P

WHAT ARE THE RELATIONSHIPS?

Tj= time consumed by all activities at station j

= tij, i = 1 to nj, T= Total work content time = ti, i=1 to n

= Tj, j=1 to S ;also = tij, i = 1 to njand j=1to S

C= Cycle time 1/ P

WHAT ARE THE RELATIONSHIPS?

C = Max j ( Tj ) Therefore C = Tj + Ij , Where Ij = Idle time at Station j Under perfect condition ,S C = ti = Tj = T SC is the total time a product spends in the

line while passing through all stations.

6Station 1

Minutes per Unit 6

Station 2

7

Station 3

3

Assembly Line Balancing Concepts

Question: Suppose you load work into the three work stations below such that each will take the corresponding number of minutes as shown. What is the cycle time of this line?

WHAT ARE THE BASIC ISSUES IN DESIGNING ASSEMBLY LINES ?

MAXIMISE LINE EFFICIENCY ? MINIMISE TOTAL IDLE TIME ? MINIMISE BALANCE DELAY ? BALANCE WORK LOAD ACROSS ALL

STATIONS

SOME COMMON MEASURES LINE EFFICIENCY = T / SC or ti / SC or tij,/ SC BALANCE DELAY = Ij / SC or (SC- ti ) / SC= 1 ( ti / SC) = 1- SMOOTHNESS INDEX SI= SQRT ( ( Tj Max Tj)2), j = 1 to S

Or some times, ( why ?)= SQRT ( ( C Tj)2), j = 1 to S

THE MANAGERIAL ISSUE MAXIMISE LINE EFFICIENCY = MAXIMISE ( ti / SC)

MINIMISE BALANCE DELAY = MINIMISE (1 ( ti / SC))

BOTH ARE SAME !

WHAT ARE THE OPTIONS ?

7THE OPTIONS

FOCUS ON NUMBER OF STATIONS CYCLE TIME BOTH

THE POPULAR CASE

Keep C Constant and Minimise S

The input: Activities, time , precedence and the desired output rate ( Cycle time)

The Problem Given n tasks and their time, assign these

tasks to S number of work stations taking care that the total time required to complete the tasks assigned to each station does not exceed the cycle time C

Is it EASY ? Is it like putting n apples or mangoes in to

minimum number of bags (S) where the weight of the each bag should not exceed the predetermined weight (C) ??

IT IS RATHER DIFFICULT

WHY ?

8THERE ARE MANY CONSTRAINTS The technological precedence constraints Cycle time related constraintsC= Cycle time 1/ PC t i max ; Station related constraintsS = Int [ ( ti) / C ] Location constraints Grouping constraints such as compatibility Operational constraints such as fatigue, safety,

mismatch of tools / skills etc.,

WHAT ARE THE APPROACHES ?

COMMON SENSE ( ARBITRARY) ENGINEERING SENSE HEURISTIC ALGORITHMS MATHEMATICAL PROGRAMMING SIMULATION EVOLUTIONARY ALGORITHMSGenetic Algorithm, Simulated Annealing,

ANC , etc.,

HEURISTICS Vs OPTIMIZATION

Heuristic procedures generally allow for a broader problem definition, but may not guarantee optimal solution.

Optimizing procedures generally have used more narrowly defined problems, but guarantee optimal solution for that problem

Examples of optimizing procedures Dynamic programming 0-1 Integer programming Branch and bound techniques.

BASIS FOR THE HEAURISTICS Decide the feasible number of work

stations Build the work stations one by one by

assigning the activities How activities are assigned ? a) Selection rule b) Eligibility check ( cycle time and

precedence)c) Assignment policy ( Tie breaking rules)

9SELECTION RULE Any idea ?

What information do we have about the activities ?

Primary information Time Number of predecessors Number of successors Technological Secondary information Can be derived using different logic.

COMMONLY USED SELECTION RULES Smallest time activity first Largest time activity first Least number of predecessors first Largest number of followers first Least number of followers first Task with greatest sum of task times of its

predecessors first Positional weight Pre-imposed priority And so on

THERE ARE MANY ALGORITHMS

We will see some of the popular algorithms through examples

10

EXAMPLE 1 Example of Line Balancing

Youve just been assigned the job of setting up an assembly line for an electric fan with the following tasks:

Task Time (Mins) Description PredecessorsA 2 Assemble frame NoneB 1 Mount switch AC 3.25 Assemble motor housing NoneD 1.2 Mount motor housing in frame A, CE 0.5 Attach blade DF 1 Assemble and attach safety grill EG 1 Attach cord BH 1.4 Test F, G

Example of Line Balancing: Structuring the Precedence Diagram

Task PredecessorsA None

A

B A

B

C None

C

D A, C

D

Task PredecessorsE D

E

F E

F

G B

G

H F, G

H

IF DESIRED OUTPUT IS NOT GIVEN

11

Example of Line Balancing: Precedence Diagram

A

C

B

D E F

GH

2

3.25

1

1.2 .5

11.4

1

Question: Which process step defines the maximum Question: Which process step defines the maximum rate of production?

Answer: Task C is the cycle time of the line and therefore, the maximum rate of production.

Example of Line Balancing: The Bottleneck

Task Time (Mins) Description PredecessorsA 2 Assemble frame NoneB 1 Mount switch AC 3.25 Assemble motor housing NoneD 1.2 Mount motor housing in frame A, CE 0.5 Attach blade DF 1 Assemble and attach safety grill EG 1 Attach cord BH 1.4 Test E, G

Max Production = Production time per day

Bottleneck time=

420 mins3.25 mins / unit

=129 unitsMax Production = Production time per day

Bottleneck time=

420 mins3.25 mins / unit

=129 units

IF DESIRED OUTPUT IS GIVEN

t

Example of Line Balancing: Determine Cycle Time

Required Cycle Time, C = Production time per periodRequired output per period

Required Cycle Time, C = Production time per periodRequired output per period

C = 420 mins / day100 units / day

= 4.2 mins / unitC = 420 mins / day100 units / day

= 4.2 mins / unit

Question: Suppose we want to assemble 100 fans per day. What would our cycle time have to be?

Answer:

12

Example of Line Balancing: Determine Theoretical Minimum Number of Workstations

Question: What is the theoretical minimum number of workstations for this problem?

Answer: Theoretical Min. Number of Workstations, N

N = Sum of task times (T)

Cycle time (C)

t

t

Theoretical Min. Number of Workstations, N

N = Sum of task times (T)

Cycle time (C)

t

t

N = 11.35 mins / unit

4.2 mins / unit= 2.702, or 3tN =

11.35 mins / unit4.2 mins / unit

= 2.702, or 3t

Example of Line Balancing: Rules To Follow for Loading Workstations

Assign tasks to station 1, then 2, etc. in sequence. Keep assigning to a workstation ensuring that precedence is maintained and total work is less than or equal to the cycle time. Use the following rules to select tasks for assignment.

Primary: Assign tasks in order of the largest number of following tasks

Secondary (tie-breaking): Assign tasks in order of the longest operating time

A

C

B

D E F

GH

2

3.25

1

1.2 .5

11.4

1

Station 1 Station 2 Station 3

Task Followers Time (Mins)A 6 2C 4 3.25D 3 1.2B 2 1E 2 0.5F 1 1G 1 1H 0 1.4

A

C

B

D E F

GH

2

3.25

1

1.2 .5

11.4

1


A (4.2-2=2.2)


13

A

C

B

D E F

GH

2

3.25

1

1.2 .5

11.4

1

A (4.2-2=2.2)B (2.2-1=1.2)



A

C

B

D E F

GH

2

3.25

1

1.2 .5

11.4

1

A (4.2-2=2.2)B (2.2-1=1.2)G (1.2-1= .2)

Idle= .2



A

C

B

D E F

GH

2

3.25

1

1.2 .5

11.4

1

(4.2-3.25=.95)C


(4.2-2=2.2) AA(2.2-1=1.2)A,BG(1.2-1= .2)A,B,G

Idle= .2


t

(4.2-3.25=.95) C

Idle = .95

A

C

B

D E F

GH

2

3.25

1

1.2 .5

11.4

1


A (4.2-2=2.2)AB (2.2-1=1.2)A,BG (1.2-1= .2)A,B,G

Idle= .2


14

t

C (4.2-3.25)=.95

Idle = .95

A

C

B

D E F

GH

2

3.25

1

1.2 .5

11.4

1

D (4.2-1.2=3) D


A (4.2-2=2.2)B (2.2-1=1.2)G (1.2-1= .2)

Idle= .2


A

C

B

D E F

GH

2

3.25

1

1.2 .5

11.4

1

C (4.2-3.25)=.95

Idle = .95

D (4.2-1.2=3) DE (3-.5=2.5) D,E


A (4.2-2=2.2)B (2.2-1=1.2)G (1.2-1= .2)

Idle= .2


A

C

B

D E F

GH

2

3.25

1

1.2 .5

11.4

1

C (4.2-3.25)=.95

Idle = .95

(4.2-1.2=3)D(3-.5=2.5)D,E(2.5-1=1.5)D,E,F


A (4.2-2=2.2)B (2.2-1=1.2)G (1.2-1= .2)

Idle= .2


A

C

B

D E F

GH

2

3.25

1

1.2 .5

11.4

1

C (4.2-3.25)=.95

Idle = .95

(4.2-1.2=3)D(3-.5=2.5)D,E(2.5-1=1.5)D,E,F(1.5-1.4=.1) D,E,F,H

Idle = .1


A (4.2-2=2.2)B (2.2-1=1.2)G (1.2-1= .2)

Idle= .2


15

Example of Line Balancing: Determine the Efficiency of the Assembly Line

Efficiency =Sum of task times (T)

Actual number of workstations (Na) x Cycle time (C)Efficiency =

Sum of task times (T)Actual number of workstations (Na) x Cycle time (C)

Efficiency =11.35 mins / unit(3)(4.2mins / unit)

=.901Efficiency =11.35 mins / unit(3)(4.2mins / unit)

=.901

SOLVE THE SAME PROBLEM USING Smallest time activity first Largest time activity first Least number of predecessors Least number of followers first ( successors) Positional weight Task with greatest sum of task times of its

predecessors first COMPARE YOUR SOLUTIONS

EXAMPLE 2

USE RANK POSITIONAL WEIGHT METHOD

RANK POSITIONAL WEIGHT METHOD (RPW)

Step 1: For each activity find the positional weight (PWi)

PWi = Time of the activity i plus the sum of the times of all activities following this activity.

Step 2: Rank the activities on the basis of PWi, starting with the activity with the maximum PWi.

In case of tie, give priority to the activity with larger processing time.

This rank is called Rank Positional Weight (RPW)

16

STEPS 1 & 2 OF RPW METHOD

Activity I 1 2 3 4 5 6 7 8 9 10 11

Time ti 1 5 4 1 5 6 2 4 3 5 3

Immediate

Follower3,4

,2,8

7 7 5 6 7 11 9 10 11 -

Positionalweights

39 10 9 17 16 11 5 15 11 8 3

RPW 1 7 8 2 3 5 10 4 6 9 11

HOW RPW IS USED ?

Step 3: Use RPW and find the order in which the

activities are to be assigned to the workstations

The activity with rank 1 will be the first in the sequence with others following accordingly.

The sequence in which the activities are to be assigned is:

Sequence in which activities are to be assigned to the workstation

Activity 1 4 5 8 6 9 2 3 10 7 11

Rank 1 2 3 4 5 6 7 8 9 10 11

17

Step 4: Take Workstation 1 and assign the activities from the sequence obtained in Step 3, by taking care that at each station, the sum of the times of activities assigned does not exceed the cycle time.

Step 5: Take the next Workstation and repeat Step 4.

Step 6: Repeat Step 5 until all activities have been assigned.

The Solution is Cycle time = 10

STATIONI II III IV V

TASK 1, 4, 5 8, 6 9,2 3,10 7, 11TIME 7 10 8 9 5

Results: There are five workstations, identified with the tasks as follows.

Station I II III IV VActivities 1,4,5 8,6 9,2 3,10 7,11Total time 7 10 8 9 5Idle time 3 0 2 1 5

Analysis

Measure Value

Total work content time 39

Total throughput time for a piece 5 x 10 = 50

Total idle time 11

Line Efficiency (39 / 50) x100 = 78 %

Balance delay (11/ 50) x 100= 22 %

Smoothness index (9+0+4+1+25) 1/2

EXAMPLE 3

Company operates one shift per dayAvailable time per shift is 450 minutesDemand is 75 units/day

cycle time = 450/75 = 6 minutes/partS= 20/6 = 3.33 = 4 stations

18

69

DATA

Task Time Required PrecedesA 2.2 min. B, C, DB 3.4 EC 1.7 ED 4.1 FE 2.7 FF 3.3 GG 2.6 --

70

Precedence Diagram

A

B

C

D

E

F G

TWO PHASE METHOD THE APPROACH

Uses two phases In phase1:>The sequence in which the activities areplaced is determined .> Using this sequence, the activity-time andthe cycle time , work stations are built.> The solution obtained is the feasiblesolution.>The efficiency of this solution is measured in

terms of smoothness index.>Now go to Phase 2

19

STEPS IN PHASE 1 Prepare immediate predecessor matrix PMATRIX Prepare immediate follower matrix FMATRIX Use these two matrices and obtain the sequence in which the tasks are to be assigned

THE STEPS1. Identify the element with zero predecessor in the

PMATRIX and place it in the first available place in the sequence ( in case of tie use LPT)

2. Go to the row corresponding to this activity in FMATRIX and identify all the follower activities.

3. Return to the PMATRIX and locate the row(s) of these activities.

4. Remove the activities just placed in the sequence, from this row ( these rows). Now you will have at least one activity with zero predecessor.

5. Place this activity ( use tie breaking rule if necessary) in the next available place in the sequence

6. Repeat steps 2 to 5 until all activities are placed in the sequence

Activity IMMEDIATE PREDECESSOR MATRIXPMATRIX

TIMEMIN.

Activity IMMEDIATE FOLLOWER MATRIXFMATRIX

ANIL 2.2 A B C D

B A 3.4 B EC A 1.7 C ED A 4.1 D FE B C 2.7 E FF D E 3.3 F GG F 2.6 G Nil

SEQUENCE 1 2 3 4 5 6 7

Activity

20


TIMEMIN.

Activity

IMMEDIATE FOLLOWER MATRIXFMATRIX

A NIL 2.2 A B C DB A 3.4 B EC A 1.7 C ED A 4.1 D FE B C 2.7 E FF D E 3.3 F GG F 2.6 G Nil

SEQUENCE 1 2 3 4 5 6 7

Activity A


TIMEMIN.



SEQUENCE 1 2 3 4 5 6 7

Activity A D


TIMEMIN.



SEQUENCE 1 2 3 4 5 6 7

Activity A D


TIMEMIN.


A NIL 2.2 A B C DB A 3.4 B E

C A 1.7 C ED A 4.1 D FE B C 2.7 E FF D E 3.3 F GG F 2.6 G Nil

SEQUENCE 1 2 3 4 5 6 7

Activity A D B

21


TIMEMIN.


ANIL 2.2 A B C D

B A 3.4 B EC A 1.7 C ED A 4.1 D FE B C 2.7 E FF D E 3.3 F GG F 2.6 G Nil

SEQUENCE 1 2 3 4 5 6 7

Activity A D B C


TIMEMIN.



SEQUENCE 1 2 3 4 5 6 7

Activity A D B C E


TIMEMIN.



SEQUENCE 1 2 3 4 5 6 7

Activity A D B C E F


TIMEMIN.



SEQUENCE 1 2 3 4 5 6 7

Activity A D B C E F

22


TIMEMIN.



SEQUENCE 1 2 3 4 5 6 7Activity A D B C E F GTime 2.2 4.1 3.4 1.7 2.7 3.3 2.6Station 1 2 3 3 4 4 5Station time 2.2 4.1 5.1 6.0 2.6

PHASE 2 In phase 2: The solution obtained is improved to minimise

smoothness indexHow : By reallocation tasks among the stations The re-allocation may involve Shifting activities Interchanging activities ( two way or three way) Merging stations But care must be taken to ensure that the cycle time and precedence constraints are not violated.

Assembly Line Balancing Steps1. Determine tasks (operations)2. Determine sequence3. Draw precedence diagram4. Estimate task times5. Calculate cycle time 6. Calculate number of work stations7. Assign tasks 8. Calculate efficiency

PRACTICE PROBLEMS

23

A B

E H

C

DF G I

10 Min.

511

12

3 7 3

4

11

Precedence Diagram Example TRY WITH C= 15 min Longest task time - choose task with

longest operation time Most following tasks - choose task with

largest number of following tasks Ranked positional weight - choose task

where the sum of the times for each following task is longest

Shortest task time - choose task with shortest operation time

Least number of following tasks - choose task with fewest subsequent tasks

ALB Example:

Assume we wish to assemble 125 units per hour. Inorder to maintain required volume, we must produceone completed unit every 28.8 seconds

Assume we wish to assemble 125 units per hour. In

TRY WITH DIFFERENT HEURISTICS LPT with random tie breaker Most followers with tie breakers as LPT and

random Least followers with LPT as tie breaker Shortest processing time with random tie

breaker Least number of predecessors with random tie

breaker Random selection rule RPW Two phase method

24

93

The work elements, precedence requirements and time requirements to assemble a picture frame are shown here.

1. Construct a precedence diagram of the process and label task times.2. Set up an assembly line capable of producing 1,600 frames per 40-hour week.3. Calculate the efficiency and balance delay of the line.4. Calculate the maximum number of frames that can be assembled each week.5. Rebalance the line for maximum production. Indicate the composition of each station.6. Assume the company can sell as many frames as can be produced. If workers are paid

$8 an hour and the profit per frame is $5, should the production quota be set to the maximum? Assume one worker per station.

Elements Description Precedence Time (min)

ABCDEFGHI

Attach left frame side to top of frameAttach right frame side to bottom of frameAttach left and right frame subassembliesCut 8-inch by 10-inch glassCut 8-inch by 10-inch cardboardPlace glass into framePlace cardboard into frameSecure cardboard and glassApply descriptive label to glass

--

A, B--

C, DE, FF, GD

0.350.350.700.500.500.200.200.500.10

94

Synopsis of assembly line balancing

1. Set up a precedence table.

2. Compute required cycle time:C = (production time) / (required output in units)

3. Compute minimum cycle time:CMIN = time for longest task

4. Compute minimum number of stations:SMIN = (sum of all times) / C

95

Synopsis of assembly line balancing (cont.)

5. Select rule by which tasks are assigned to work stations.Examples:

(1) select tasks with longest operation times first or(2) select tasks with largest number of following tasks first

6. Assign tasks to the first work station until the sum of the task times are equal to the cycle time, or no other tasks are feasible. Repeat for stations 2, 3, until tasks are assigned.

7. Evaluate the efficiency of the balance:E = (sum of all times) / (actual nbr. of stations x C)

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