3. Manufacturing 3 - Assembly Line Balancing Techniques

Manufacturing 3

Course Structure

  • Introduction
  • Plant Layout
  • Design and Scheduling of Flow Process Systems
    • Line Balancing
    • The Kilbridge and Wester Method
    • The Ranked Positional Weight Method
    • More Complex Situations – Mixed Model and Multi Model
  • Layout for Batch Production – Singleton’s Method
  • Scheduling
  • Computer Assisted Production Planning
  • Just-in-Time Systems
  • Synchronous Production Systems and Theory of Constraints
  • Digital Manufacturing and Work Measurement

Learning Objectives

  • Define performance measures.
  • Use the Kilbridge and Wester method to balance a line.
  • Use the Ranked positional weight method to balance a line.

Measuring Performance

  • Every organization has to measure performance.
  • If managers do not take measures, they have no idea how good the operations are.
  • "If you can’t measure, you can’t manage."

Performance Measures

  • Capacity: The maximum amount of a product that can be made in a given time.
  • Utilization: Measures the proportion of available capacity that is actually used.
  • Production: The total amount of a product that is made.
  • Productivity: The amount produced for each unit of resource used.
  • Efficiency: The ratio of actual output to possible output.
  • Effectiveness: Shows how well an organization sets and achieves its goals.
Class Exercise
  • A workshop has two machines designed to make 100 units each in a nine-hour shift.
  • During one shift, the machines worked for eight hours and made a total of 140 units.
  • What measures of performance can you give?
    • Production is the total amount of a product that is made: 140 units.
    • Efficiency=actual outputpossible output=1402×100=0.777=77.7%Efficiency = \frac{actual \ output}{possible \ output} = \frac{140}{2 \times 100} = 0.777 = 77.7 \%
    • Productivity=number of units mademachine time used=1402×8=8.75 units per machine hourProductivity = \frac{number \ of \ units \ made}{machine \ time \ used} = \frac{140}{2 \times 8} = 8.75 \ units \ per \ machine \ hour

Assembly Line Balancing

  • Line balancing seeks to reduce the waiting time caused by unbalanced production times between individual work centers and ensures production is the same at all processes, both in quantities and timing

Kilbridge and Wester Line Balancing - Procedure

  • Step 1: Construct the precedence diagram so that the nodes with identical precedence are arranged vertically in columns.
  • Step 2: List the elements in order of their columns. If an element can be located in more than one column, list all the columns by the element to show the transferability of the element to get a K&W starting table.
  • Step 3: To assign elements to workstations, start with the column I elements. Continue to the assignment procedure in order of column number until the cycle time is reached. Go on until all elements are allocated.
Kilbridge and Wester line balancing

Balance the following assembly process for a workstation cycle time of 36 seconds using the K&W line balancing procedure.

What is Transferability?

  • Step 1 and 2: K&W starting Table.

    • Column Number, Element No, Transferability, Element Duration (sec), Column Duration (sec), Cumulative Duration (sec)
    • I: 0, 1, 2, (6+5+8 = 19), 19
    • II: 3, 4, 5, 6, 7, (9+5+4+5+6=29), (19+29 = 48)
      • w/ element 8: 3, 7
    • III: 8, 9, 10, (10+5+6 = 21), (21+48 = 69)
      • IV - VI: 8, 10
    • IV: 11, (2), (2+69 = 71)
    • V: 12, (5), (5+71 = 76)
    • VI: 13, (4), (4+76 = 80)
    • VII: 14, (12), (12+80 = 92)
    • VIII: 15, (10), (10+92=102)
    • IX: 16, 17, 18, (5+15+10 = 30), (30+102 = 132)
      • X : 17, 18
    • X: 19, (5), (5+132 = 137)
    • XI: 20, (6), (6+137 = 143)

  • Step 3: To assign elements to workstations, start with the column I elements. Continue to the assignment procedure in order of column number until the cycle time is reached. Go on until all elements are allocated.

Step 3 K&W line balancing (C = 36sec)

SubStep 1: Establish work elements for Workstation 1

  • Is there a duration in the column “Cumulative duration” of the table equal to the cycle time of 36sec? Answer NO
  • Select the largest duration in the column “Cumulative duration” less than 36sec Answer 19 for column I
  • Subtract this value from the target cycle time Answer 36–19=17 sec36 – 19 = 17 \ sec
  • Do any elements in column II either individually or collectively, have a duration of 17? Answer NO, nearest is 16sec for elements 4, 6 and 7, which will give a possible time of 35sec for workstation 1.
  • Workstation 1 consists of work elements 0, 1, 2, 4, 6, 7
  • Work elements 3, 5 and 8 both shift to next columns
  • Now re-compute K&W Table and reset cumulative time after WS 1
    • Column Number, Element No, Transferability, Element Duration (sec), Column Duration (sec), Cumulative Duration (sec)
    • III: 3, 5, 9, 10, (9+4+5+6=24), 24
      • III – V (with 8): 3
      • Transferability with 8: 10
    • IV: 8, 11, (10+2 = 12), 36
      • V - VI
    • V: 12, (5), 41
    • VI: 13, (4), 45
    • VII: 14, (12), 57
    • VIII: 15, (10), 67
    • IX: 16, 17, 18, (5+15+10 = 30), 97
      • X : 17, 18
    • X: 19, (5), 102
    • XI: 20, (6), 108

Kilbridge and Wester line balancing (WS 36sec)

Step 2: Establish work elements for Workstation 2

  • Is there a duration in the column “Cumulative duration” of the table equal to the cycle time of 36sec? Answer YES, for columns III and IV
  • Allocate the elements in these columns to Workstation 2
    • Reset cumulative duration column -> Now re-compute K&W Table
    • Column Number, Element No, Transferability, Element Duration (sec), Column Duration (sec), Cumulative Duration (sec)
    • V: 12, (5), 5
    • VI: 13, (4), 9
    • VII: 14, (12), 21
    • VIII: 15, (10), 31
    • IX: 16, 17, 18, (5+15+10 = 30), 61
      • X : 17, 18
    • X: 19, (5), 66
    • XI: 20, (6), 72

Kilbridge and Wester line balancing (C = 36sec)

Step 3: Establish work elements for Workstation 3

  • Is there a duration in the column “Cumulative duration” of the table equal to the cycle time of 36sec? Answer NO
  • Select the largest duration in the column “Cumulative duration” of the table less than the cycle time of 36sec? Answer 31sec for columns V, VI, VII and VIII
  • Subtract this value from the target cycle time (36–31)=5 sec(36 – 31) = 5 \ sec
  • Does one or more of the work elements in column (IX) equal 5sec? Answer YES, work element 16
    • Now re-compute K&W Table
    • Column Number, Element No, Transferability, Element Duration (sec), Column Duration (sec), Cumulative Duration (sec)
    • IX: 17, 18, (15+10 = 25), 25
      • X : 17, 18
    • X: 19, (5), 30
    • XI: 20, (6), 36

Kilbridge and Wester line balancing (C = 36sec)

Step 4: Establish work elements for Workstation 4

  • Is there a duration in the column “Cumulative duration” of the table equal to the cycle time of 36sec? Answer YES, 36sec for columns IX, X and XI
  • Allocate these work elements to workstation 4
  • Now re-compute K&W Table

Allocation of Work Elements to Workstations

  • Workstation 1: 0, 1, 2
  • Workstation 2: 3, 4, 5, 6, 7
  • Workstation 3: 8, 9, 10, 11, 12, 13, 14, 15, 16
  • Workstation 4: 17, 18, 19, 20

Class Exercise, Tutorial 1 Q1

The following figure shows the necessary precedence relationships of 12 work elements which together constitute the total work content for the assembly of a product. Using the balancing technique devised by Kilbridge and Wester, design a line to produce as near as possible to and no less than three products per hour.

The Ranked Positional Weight Method

  • Logical extension to the Kilbridge and Wester Method
  • More formalized method for allocating work elements to work stations
  • Essentially, a weighting is calculated for each element
  • The weighting represents the tendency for the work element to be placed towards the beginning of the assembly line

Ranked Positional Weights Method - Procedure

A single sequence is constructed. A task is prioritized by cumulative assembly time associated with itself and its successors. Tasks are then assigned to the lowest numbered feasible workstation.

  • Step 1: Calculate the ranked positional weight value (RPW) for each element by summing the element’s time together with the time values for all the elements that follow it in the arrow chain of the precedence diagram.
  • Step 2: List the elements in descending order of their RPW.
  • Step 3: Assign elements to stations according to RPW, avoiding precedence constraint and time-cycle violations.

The Ranked Positional Weight Method

Design an assembly line with the minimum number of workstations to provide a cycle time of 0.55 hours for the given work element precedence diagram.

Tasks to complete:

  1. Develop a precedence and positional weights table
  2. Create a table of elements in order of positional weights
  3. Assign elements to work stations
Precedence and Positional Weights Table
  • Element No, Total Element Time, Positional Weight, Predecessors
  • Element: 0, time: 0.32, successors: 1,2,5,6,3,4,8,7,9,10; positional weight 1.72
  • Element: 1, time: 0.10, successors: 3,4,6,5,7,8,9,10; positional weight 1.65
  • Element: 2, time: 0.20, successors: 5,6,3,4,8,7,9,10; positional weight 1.40
  • Element: 3, time: 0.05, successors: 4,8,7,9,10; positional weight 0.87
  • Element: 4, time: 0.10, successors: 8,7,9,10; positional weight 0.82
  • Element: 5, time: 0.23, successors: 6,3,4,8,7,9,10; positional weight 1.20
  • Element: 6, time: 0.20, successors: 3,4,8,7,9,10; positional weight 0.92
  • Element: 7, time: 0.05, successors: 9,10; positional weight 0.45
  • Element: 8, time: 0.32, successors: 7,9,10; positional weight 0.72
  • Element: 9, time: 0.10, successor: 10; positional weight 0.40
  • Element: 10, time: 0.30, no successors; positional weight 0.30

Elements in Order of Positional Weights

  • Element No, Total Element Time, Positional Weight, Predecessors
  • 0, 0.32, 1.72, -
  • 1, 0.1, 1.65, -
  • 2, 0.2, 1.4, 0, 1
  • 5, 0.23, 1.2, 2
  • 6, 0.2, 0.92, 5
  • 3, 0.05, 0.87, 1
  • 4, 0.1, 0.82, 3
  • 8, 0.32, 0.72, 4, 6
  • 7, 0.05, 0.45, 5
  • 9, 0.1, 0.4, 7, 8
  • 10, 0.3, 0.3, 9

Assign Elements to Work Stations

Assigning Work Elements to WS 1
  • Elements are allocated to workstations in order of POSITIONAL WEIGHTS without violating precedence constraints
  • Element 0 has the highest positional weight of 1.72 and is allocated to WS 1
  • Allocation is made taking into account that it has no immediate predecessors yet unassigned and that the remaining workstation time is greater than its element time
  • Element 1 is the next to be considered for allocation due to it having the next highest positional weight
  • It can be allocated because it has no immediate predecessors and its element time of 0.10 is less than the remaining workstation time of 0.23
  • Element 2 is the next to be considered for allocation due to it having the next highest positional weight
  • It cannot be allocated because its element time of 0.2 is greater than the remaining time for WS 1 of 0.13
  • Similarly elements 5 and 6 are next on the PW but their element times are also greater than what is left for WS 1
  • Element 3 can be allocated as its time of 0.05 is less than the time remaining.
  • Remaining WS 1 time cannot be used due to precedence constraints

Class Exercise

Repeat the process to assign further work elements to further workstations

Remember C=0.55

Calculate balancing loss and efficiency

  • t=total element timeC=Mt = \frac{total \ element \ time }{C} = M
  • Balancing loss=100%×C−tC=10.5%Balancing \ loss = 100 \% \times \frac{C - t}{C} = 10.5 \%
  • Efficiency=100%−Balancing loss=89.5%Efficiency = 100 \% - Balancing \ loss = 89.5 \%
  • Actual Efficiency:Balancing loss=100%×C−tC=7.1%Actual \ Efficiency: Balancing \ loss = 100 \% \times \frac{C - t}{C} = 7.1 \%
  • Normal Efficiency=100%−Balancing loss=92.9%Normal \ Efficiency = 100 \% - Balancing \ loss = 92.9 \%

Summary

  • Define performance measures
  • Use the Kilbridge and Wester method to balance a line
  • Use the Ranked positional weight method to balance a line