approaches to line balancing optimal solutions active learning module 3 dr. césar o. malavé texas...

Approaches to Line balancingOptimal Solutions

Active Learning Module 3

Dr. César O. Malavé

Texas A&M University

Background Material

Modeling and Analysis of Manufacturing Systemsby Ronald G. Askin , Charles R. Standridge, John Wiley & Sons, 1993, Chapter 2.

Manufacturing Systems Engineering by Stanley B. Gershwin, Prentice – Hall,1994, Chapter 2.

Any good manufacturing systems textbook which has detailed explanation on reliable serial systems.

Lecture Objectives

At the end of this module, the students should be able to Explain the Optimal Solutions approach to line

balancing Find the optimal solutions to line balancing

problems using the above technique.

Time Management

3Assignment

10Team Exercise

4Practical Issues

50 MinsTotal Time

15Tree Exploration

10Tree Generation

5Readiness Assessment Test (RAT)

3 Introduction

Readiness Assessment Test (RAT)

Task Time (Seconds) Predecessor

A 45 -

B 11 A

C 9 B

D 50 A

E 15 D

F 12 C

G 12 C

H 12 E

I 12 E

J 8 F, G, H, I

K 9 J

Draw the precedence diagram for the tasks shown below

RAT – Solution

A

B

D

C

E

F

G

H

I

J K

Precedence Chart :

Optimal Solutions

Imagine a decision tree containing all possible sequences of tasks that obey the precedence constraints.

“0” node is called root of the tree, terminal nodes at the other end are called the leaves.

A unique path leads from root to each leaf.

Each leaf represents a complete sequence.

Each feasible sequence is represented by exactly one leaf.

Best solution is found by examining every path.

Tree Generation

Backtracking – Allows both symmetric exploration of the tree and efficient storage of our location and history during exploration.

General applicability is to an environment in which N sequential decisions are to be made.

In generating the tree of possible sequences, the order of performing the assembly tasks is important rather than the workstations.

Tree initially grows by selecting first alternative at each stage until there is a complete assembly sequence

Tree Generation – Cont…

Backtrack until we reach the first node of the unexplored branch.

Continue moving forward, decision by decision, as far as possible.

Process continues until all possible leaves are explored.

A task is fittable if it satisfies three conditions. Task must fit in the remaining idle time of the station. It must be currently unassigned. All its predecessors are assigned.

Tree Generation Algorithm0. Input Bound and task data

1. Setupk = 1, p = 0

ck = C, B = 0

i’ exist ?

2. Select new taskFind i’ = lowest i

i fittable, i>Bi > i* unless new

station .

3. Assign taskAi* = k, p = p+1

ck = ck – ti*

TAp = i*, B = 0

p = N ?

ck = C ?No

i* = i’

Yes

6. Sequence complete Save if best solution

Yes

p = 0 ?Yes

i* > 0 ?

No

Stop Enumeration

Complete

Yes

No

No

7. Backtrack to B.(Remove B from

Station K)If ck = C, k = k-1B = Tap, AB = 0

ck = ck+tB

p = p-1, i* = 0

No

4. Open New Stationk = k+1ck = C

Yes

Tree Exploration

Creating only those new nodes required to find or disprove the existence of a better solution.

In practice, we need not create each node of the tree.

At any partial sequence node, if we can establish that all completions of this partial sequence are non-optimal; then we will start backtracking immediately instead of completing a sequence.

Two Rules – Problem Structure Rules & Fathoming Rules.

Problem Structure Rules

Basic Optimality Principle – Never close a workstation while “fittable” tasks remain.

Open a new workstation once when necessitated by time.

Long task becomes a station only when no other task can feasibly fit in the same station.

A lower bound on the number of workstations needed is useful in determining if we have an optimal solution.

Fathoming Rules

These rules help us to prune the tree rapidly.

Allows us to implicitly consider all leaves without explicitly enumerating many of them.

Rules only be checked when a new workstation must be opened.

Rule 1 : Task Dominance

Suppose Task i can be feasibly replaced by a longer task j and all successors must also follow j.

If we substitute, we reduce the workload without losing any possible sequence completions.

Fathoming Rules – Cont…

Rule 2 : Station Dominance

Suppose we try to form a station that is identical to a “first” station that was explored earlier.

As no new sequences can be considered, new node is fathomed.

Rule 3 : Solution Dominance

We know that K0 gives a lower bound on optimal number of solutions.

We can stop once we have got a solution as that of lower bound

Fathoming Rules – Cont…

Rule 4 : Bound Violation

Suppose we have K workstations. This is optimal unless a K – 1 solution is found.

Let Ai be the station to which task i is assigned. If we require at most K – 1 stations, the upper bound is given by

Nodes containing at least one of Ai outside these bounds can be pruned from the tree.

C

ttKU iSj ji

i)(

Fathoming Rules – Cont…

Rule 5 : Excessive Idle Time

Whenever cumulative idle time exceeds

(K – 1)C – i ti ,we may not fathom the partial sequence

All the above mentioned fathoming rules are adapted from FABLE – Fast Algorithm for Balancing Lines Effectively described by Johnson

Optimal Solutions - ExampleTask Activity Assembly

TimeImmediate

Predecessor

a Insert Front Axle / Wheels

20 -

b Insert Fan Rod 6 a

c Insert Fan Rod Cover 5 b

d Insert Rear Axle / Wheels

21 -

e Insert Hood to Wheel Frame

8 -

f Glue Windows to top 35 -

g Insert Gear Assembly 15 c, d

h Insert Gear Spacers 10 g

i Secure Front Wheel Frame

15 e, h

j Insert Engine 5 c

k Attach Top 46 f, i, j

l Add Decals 16 k

Example Solution

Product Structure Rules :

1. Task Time AugmentationLongest Station = tk = 46

Shortest Station = tc = 5

tk + tc C. So task would require own station.

2. Solution Lower BoundLower Bound K0 = 3

Only one task exceeds C/2 and two tasks exceeding C/3. No better solution exists now..

Example Solution – Cont…

Fathoming Rules :

1. Task DominanceSeveral Tasks can be replaced.

Task Pairs satisfying the relationship are

(j, d), (j, e), (j, f), (j, g), (j, h), (j, i), (i, f), (e, a), (e, h), (e, g), (e, d)

2. Task Bound ViolationAssume current best solution has four stations.

The natural ordering being (a, b, c, d, e), (f, g, h), (i, j, k), (l)

Still need to search for three-station solution.

Bound violations will be useful.

Example Solution – Tree Generation

0

1

a

2

b

3

c

4

d

5

e

13

g

6

j

7f

8g

9h

10

g

11h

12i

14

e

15

f

16h

17i

18j

19k

20l

Team Exercise

Consider the five assembly tasks given below

Evaluate each sequence in the tree to determine the number of workstation required. List all optimal sequences

Task Time Immediate Predecessor

a 40 -

b 75 a

c 50 a

d 35 c

e 80 d

Exercise Solution

0 a

b c d e

c b d e

d b e

e b

Workstation Optimal

4

4

4

4

Yes

Yes

Yes

Yes

Practical Issues

C has been fixed so far but in reality we only forecast demand.

C should be set such that i ti / C is an integer or jus less than an integer.

Randomness in performance times exists.

Leti

2 Variance of performance times.

ij Correlation between tasks i and j.

sk Random variable for the time required by station k

for a cycle

Sk Set of tasks assigned to station k.

Practical Issues – Cont…

From basic statistics we know

When added to the station, the task must satisfy

k

k

Si ik

Si ik

sV

tsE

2)(

)(

CsVsE kk 2/1)(33.2)(

Assignment

Solve by implicit enumeration technique and find the optimal solution

g15h

e24g

d26f

c24e

c11d

a, b10c

-5b

-3a

PredecessorsTimeOperation