1

Assembly Line Balancing

The assembly line is a production line where material moves continuously through a series of workstations where assembly work is performed.

2

Assembly Lines

• Principle of Interchangeability– individual components that make up a finished

product should be interchangeable between product units

• Division of Labor – complex activities divided into elemental tasks– work simplification– standardization– specialization

• Mass Production

3

Production Systems

Project ShopJob ShopFlow ShopAssembly Line

Continuous Flow

project networks

job shop scheduling

flow shop scheduling

assembly line balancing

– e.g. cyclic scheduling

single facility EOQ model

4

The Problem

1 2 3 4 5 6

flow of the line

station 1 station 2 station 3

Tasks

Assign work elements (tasks) to workstationsto minimize unit assembly costs (e.g. labor cost).

precedencerequirements

precedencerequirements

5

Cycle Time

The time between the completion of two successive products, assumed constant for all products for a given production line speed. The minimum value of the cycle time must be greater than or equal tothe longest station time.

A group of engineering management students discussing cycle times.

6

Problem Formulation

Assume production rate of P with m parallel lines.Then each line must produce a unit every m / P time units.

Set Cycle time = C <= m / P ; the time between completion oftwo successive units.

Example:Planned order release requires a production rate of 80 units per hours. Four (4) assembly lines are available.Therefore cycle time = C 4/80 = .05 hr. = 3 minutes

7

Station Time

Let ti = time to perform task i where i = 1,2,…,n Sj = station j time where

j

j ii I

S t

and Ij = {i | task i is assigned to station j}

Sj <= C

8

Performance Measureslet k = number of workstations; 1 <= k <= n

dj = C – Sj = delay (idle) time at station j

line efficiency: 1 100%

k

jj

S

LE xk C

line smoothness index: 2

max1

k

jj

SI S S

total idle time: 1 1 1

k k n

j j ij j i

IT d C S k C t

0 is perfectbalance

9

Minimizing Idle Timeminimizes assembly time per unit

1

1

max1

Min

. .

min ,

precedence requirements

n

ii

n

ii

n

ii

IT k C t

ts t k

Ck n

mt C t

P

j

j ii I

S t C

for IT = 0,must be integer

This looks likeit is NP-hard to me.

10

Precedence Relationships

• Precedence constraints– some tasks may have to be completed in a

particular sequence task i task j

• Zoning restrictions– some tasks cannot be performed at the same

workstation (divorces)– some tasks may be required to be performed at

the same workstation (marriages)

11

Our Very First Example

15 min

27 min

310 min

46 min

58 min

k = 5 C = 10 min. P = 6 per hr.

Performance MeasuresIT = 5(10) – 36 = 14 min.LE = 36/50 = 72%SI = 7.35

1

36n

ii

t

S1 S2 S3 S4S5

5 2

1

10 jj

SI S

12

Our Very First Example

15 min

27 min

310 min

46 min

58 min

k = 4 C = 12 min. P = 5 per hr.

Performance MeasuresIT = 4(12) – 36 = 12 min.LE = 36/48 = 75%SI = 7.48

1

36n

ii

t

S1 S2 S3 S4

13

Our Very First Example

15 min

27 min

310 min

46 min

58 min

k = 3 C = 14 min. P = 4.286 per hr.

Performance MeasuresIT = 3(14) – 36 = 6 min.LE = 36/42 = 85.7 %SI = 4.47

1

36n

ii

t

S1 S2S3

14

Our Very First Example

15 min

27 min

310 min

46 min

58 min

k = 2 C = 22 min. P = 2.72 per hr.

Performance MeasuresIT = 2(22) – 36 = 8 min.LE = 36/44 = 81.8 %SI = 8

1

36n

ii

t

S1 S2

15

Our Very First Example

15 min

27 min

310 min

46 min

58 min

k = 1 C = 36 min. P = 1.67 per hr.

Performance MeasuresIT = 1(36) – 36 = 0 min.LE = 36/36 = 100 %SI = 0

1

36n

ii

t

S1

Look, a perfectly

balanced line.

16

k C P IT LE SI5 10 min 6/hr 15 min 72 % 7.354 12 min 5/hr 12 min 75 % 7.483 14 min 4.28/hr 6 min 85.7 % 4.472 22 min 2.73/hr 8 min 81.8 % 81 36 min 1.67/hr 0 100 % 0

Chuck. Could you summarize all this for me? Just tell me what I need to

know!

17

Our Very Next Example Problem

Task 1 2 3 4 5 6 7 ti 5 3 4 3 6 5 2

Task 8 9 10 11 12 ti 6 1 4 4 7

2 3

1

54

6 9

10

7 8

11

12

precedence relationships

18

Trial and Error Approach

• Find minimum number of stations for a given cycle time

• Repeat for various cycle times• Select solution that minimizes idle time

12

1

50 2 5 5ii

t x x

cycle times feasible? min k 50 yes 1 25 yes 2 10 yes 5 5 no 10 tmax = 7 2 no 25

primefactors

1

1

Min

Then an integer

n

ii

n

ii

IT k C t

tk

C

19

2 3

1

54

6 9

10

7 8

11

12

I II III IV V VI VII

station task ti column sum cumulative sumI 1 5 5 5II 2 3

4 3 6 11III 3 4

5 6 10 21IV 6 5 5 26V 7 2

9 110 4 7 33

VI 8 611 4 10 43

VII 12 7 7 50

C = 10

IT = 7(10) – 50 = 20

LE = 50/70 = 71%

SI = 9.16

20

Heuristic

• place each task as far to the left as possible• no restriction of movement within a column• can move tasks further to the right• assign tasks to work stations such that the

sum of the times does not exceed C• always select task with longest time when

forming a workstation

21

task ti pred1 5

02 3

13 4

24 3

15 6

26 5

57 2

68 6

79 1

610 4

611 4

712 7

11

station task ti column sum cumulative sumI 1 5

2 3 8 8II 4 3

5 6 9 17III 3 4

6 5 9 26IV 7 2

9 110 4 7 33

V 8 611 4 10 43

VI 12 7 7 50

C = 10

IT = 6(10) – 50 = 10

LE = 50/60 = 83%

SI = 4.89

2 3

1

54

6 9

10

7 8

11

12

X

I II III IV V VI

22

2

31

5

4

6

9107

8 11 12

I II III IV V VI

C = 9

IT = 6(9) – 50 = 4

LE = 50/54 = 92.6%

SI = 2

station task ti column sum cumulative sumI 1 5

2 3 8 8II 4 3

5 6 9 17III 3 4

6 5 9 26IV 7 2

8 6 8 34V 10 4

11 4 8 42VI 9 1

12 7 8 50

Can movetasks to theright.

23

Positional Weight Method

• Find positional Weight (PW) for each task• Rank tasks based upon PW

– highest first

• Assign tasks to stations with highest rank first• Continue to assign tasks as long as time remains

– task does not violate precedence relationship– station time does not exceed cycle time

• Repeat until all tasks are assigned• Each task is assigned to the first feasible station

– greedy algorithm

24

Positional WeightPWi = time of the longest path from beginning of task i

through the remainder of network.

Task 1 2 3 4 5 6 7 PWi 34 27 24 29 26 20 15

Task 8 9 10 11 12 PWi 13 8 15 11 7

2 3

1

54

6 9

10

7 8

11

12

5

3

3

5

4

1

2 6

4

7

4

6

25

Task 1 2 3 4 5 6 7 PWi 34 27 24 29 26 20 15

Task 8 9 10 11 12 PWi 13 8 15 11 7

Rank Task PW1 1 342 4 293 2 274 5 265 3 246 6 207 7 158 10 159 8 1310 11 1111 9 812 12 7

Assume CT = 10

task ti pred1 5

02 3

13 4

24 3

15 6

26 5

57 2

68 6

79 1

610 4

611 4

712 7

11

26

Rank Task PW1 1 342 4 293 2 274 5 255 3 246 6 207 7 158 10 159 8 1310 11 1111 9 812 12 7

station task ti column sum cumulativeI 1 5

4 3 8 8II 2 3

5 6 9 17III 3 4

6 5 9 26IV 7 2

10 4 6 32V 8 6

11 4 10 42VI 9 1

12 7 8 50

C = 10

IT = 6(10) – 50 = 10

LE = 50/60 = 83.3%

SI = 5.09

I have long advocated the positional weight method

in order to achieve the best balance.

task ti pred1 5

02 3

13 4

24 3

15 6

26 5

57 2

68 6

79 1

610 4

611 4

712 7

11

27

An Integer Programming Approach

let xik = 1 if task i assigned to station k; 0 otherwise ci,k = cost coefficient of xi,k where cik < ci,k+1

1 1

1

1

1

. . 1,...,

1 1,...,

n K

ik iki k

n

i iki

K

ikk

h

ih jkk

Min c x

s t t x C k K

x i n

x x

precedence relationshipwhen task j precedes task i

28

Some Final Considerations• If significant idle time remains

– consider parallel lines with larger cycle times– use more than one worker per station (group stations)

• task variability– max station time = E[Si] + 2.33 STD[Si] <= C

– probability all stations complete on time .99k

• provide rework area• add buffers• use unpaced (asynchronous) lines • Max profit rather then minimize idle time

These are some very good final

considerations.