approaches to line balancing optimal solutions active learning module 3 dr. césar o. malavé texas...
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Approaches to Line balancingOptimal Solutions
Active Learning Module 3
Dr. César O. Malavé
Texas A&M University
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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.
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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.
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Time Management
3Assignment
10Team Exercise
4Practical Issues
50 MinsTotal Time
15Tree Exploration
10Tree Generation
5Readiness Assessment Test (RAT)
3 Introduction
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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
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RAT – Solution
A
B
D
C
E
F
G
H
I
J K
Precedence Chart :
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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.
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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
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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.
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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
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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.
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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.
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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.
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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
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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)(
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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
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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
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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..
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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.
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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
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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
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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
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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.
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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)(
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Assignment
Solve by implicit enumeration technique and find the optimal solution
g15h
e24g
d26f
c24e
c11d
a, b10c
-5b
-3a
PredecessorsTimeOperation