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## Assembly Line Balance

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Title: Assembly Line Balance

1
Assembly Line Balance
2
Assembly analysis
• Assembly Chart
• It shows the sequence of operations in putting
the product together. Using the exploded drawing
and the parts list, the layout designer will
diagram the assembly process.

The sequence of assembly may have several
alternatives. Time standards are required to
decide which sequence is best. This process is
known as assembly line balancing.
3
The Assembly Chart
The assembly chart of a toolbox
4
Time Standards Are Required for Every Task
5
Plant Rate and Conveyor Speed
• Conveyor speed is dependent on the number and
units needed per minute, the size of the unit,
the space between units. Conveyor belt speed is
recorded in feet per minute.

Example Charcoal grill are in cartons
30X30X24 inches high. A total of 2,400 grills are
required every day.
6
Plant Rate and Conveyor Speed

7
Assembly line balancing
The purpose of the assembly line balancing
technique is   1. To equalize the work load
among the assemblers 2. To identify the
bottleneck operation 3. To establish the speed of
the assembly line 4. To determine the number of
workstations 5. To determine the labor cost of
assembly and packout 6. To establish the
percentage workload of each operator 7. To assist
in plant layout 8. To reduce production
cost   The assembly line balancing technique
builds on The assembly chart Time
standards Takt time (minutes/piece)
(Plant rate, R value, Pieces/minutes).

8
Initial assembly line balancing of toolbox
Takt time (for 2,000 units per shift, considering
10 downtime and 80 efficiency) .173 minutes
per unit.
9
Assembly line balancing
• Cost of balancing
• Subassemblies that cost too high can be taken off
the line.
• SA3 could be taken off the assembly line and
handled completely separate from the main line
and we can save money. SA3 .250 240 pieces per
hour and .00417 hour each. If balanced, the
standard would be 180 pieces per hour and .00557
hour each.
• .0057 balanced cost
• - .00417 by itself cost
• .00140 savings hour per unit
• X 500,000 units per year
• 700 hours per year
• _at_ 15.00 per hour
• 10,500.00 per year savings

10
Assembly line balancing

Subassemblies that can be taken off the line must
be 1. Poorly loaded. The less percent that is
assembly line balance would indicate 40 percent
lost time. If we take this job off the assembly
line (not tied to the other operators), we
could save 40 percent of the cost. 2. Small
parts that are easily stacked and stored.
3. Easily moved. The cost of transportation and
the inventory cost will go up, but because of
better labor utilization, total cost must go
down.
11
Assembly line balancing

2. Improvement of assembly line Improve the
busiest (100 percent) workstation first.
(a) The busiest workstation is P.O. It has .167
minute of work to do per packer. The next
closest station is A1 with .155 minute of
work. As soon as we identify the busiest
workstation, we identify it as the 100
percent station, and communicate that this time
standard is the only time standard used on
this line from now on. Every other
workstation is limited to 360 pieces per hour.
Even though other workstations could work
faster, the 100 percent station
limits the output of the whole assembly
line. (b) The total hours required to assemble
one finished toolbox is .06960 hour. The
average hourly wage rate times .06960 hour
per unit gives us the assembly and packout labor
cost. Again, the lower this cost is the
better the line balance is.
12
Assembly line balancing

13
Assembly line balancing

14
Assembly line balancing

2. Improvement of assembly line Improve the
busiest (100 percent) workstation first.
Look at the 100 percent station (P.O.). If we
add a fourth packer, we will eliminate the 100
percent station at P.O. Now the new 100 percent
(bottleneck station) is A1 (93 percent). By
adding this person, we will save 7 percent of 25
people or 1.75 people and increase the percent
load of everyone on the assembly line (except
P.O.). We might now combine A1 and A2, and
further reduce the 100 percent. The best answer
to an assembly line balance problem is the lowest
total number of hours per unit. If we add an
additional person, that persons time is in the
total hours.
15
Step-by-step procedure for completing the
assembly line balancing form

16
Step-by-step procedure for completing the
assembly line balancing form

9. R value The R value goes behind each
operation. The plant rate is the goal of
each workstation, and by putting the R value on
each line (operation), one keeps that goal
clearly in focus. 10. Cycle time The
time standard. 11. Number of
stations 12. Average cycle time
17
Step-by-step procedure for completing the
assembly line balancing form

tells how busy each workstation is
compared to the busiest workstation. The
highest number in the average cycle time column
12 is the busiest workstation and,
therefore, is called the 100 percent
station. Now every other station is compared
to this 100 percent station by
dividing the 100 percent average station time
into every other average station time. The
percent load is an indication of where
more work is needed or where cost
reduction efforts will be most fruitful. if the
100 percent station can be reduced by 1
percent, then we will save 1 percent for
every workstation on the line.
18
Step-by-step procedure for completing the
assembly line balancing form

of the toolbox assembly line balance In Figure
4-11, the average cycle times reveals that .167
is the largest number and is designated the 100
percent workstation. The percentage load of
every other workstation is determined by dividing
.167 into every other average cycle
time Operation SSSA1 .153 / .167 92
percent SSA1 .146 / .167 87
percent SSA2 .130 / .167 78
percent and so on.
19
Step-by-step procedure for completing the
assembly line balancing form

14. Hours per unit     Example Hours per
unit of the toolbox assembly line balance The
.167 time standard is for one person, if
considering the people number, the hour per unit
will be Two people .00557 hour per
unit Three people .00835 hour per unit Four
people .01113 hour per unit
20
Step-by-step procedure for completing the
assembly line balancing form

15. Piece per hour Inversion of hours
per unit. 16. Total hours per unit Sum
of the elements in column 14. For this example is
.0696 hour. 17. Average hourly wage
rate, say 15 per hour 18. Labor cost per
unit Total hours X average hourly wage 19.
Total cycle time It tells us what a
perfect line balance would be. Our
example 3.494 minutes divided by 60 minutes per
hour equals .05823 hour per unit.
21
Efficiency of the assembly line

or
For our example
22
Analysis of single model assembly lines
Production Rate is given by
where Rp average hourly production rate,
units/hr Da annual demand, units/year Sw
number of shifts/week Hsh hrs/shift. This
equation assume 50 weeks per year.
23
Analysis of single model assembly lines
The cycle time can be determined as
where Tc cycle time of the line, min./cycle Rp
production rate, units/hr E line efficiency
24
Analysis of single model assembly lines
The cycle rate can be determined as
where Rc cycle rate, cycles/hr Tc is in
min./cycle Line efficiency E therefore defined
as
25
Analysis of single model assembly lines
The number of workers on the line can be
determined as
where w number of workers on the line WL
workload to be accomplished in a given time
period. AT available time in the period.
TWc work content time, min/piece.
26
Analysis of single model assembly lines
Using the previous equation, we also have
The available time in the period, AT.
AT 60E
Substitute these terms for WL and AT into w
equation, we can state
If we assume one worker per station, then this
ratio also gives the theoretical minimum number
of workstations on the line.
27
Analysis of single model assembly lines
Example
A small electrical appliance is to be produced on
a single model assembly line. The work content of
assembling the product has been reduced to the
work elements listed in table below along with
other information. The line is to be balanced for
an annual demand of 100,000 units per year. The
line will be operated 50 weeks/yr, 5 shifts/wk,
and 7.5 hrs/shift. Manning level will be one
worker per station. Previous experience suggests
that the uptime efficiency for the line will be
96, and repositioning time lost per cycle will
be 0.08 min. Determine (a) total work content
time Twc, (b) required hourly production rate Rp
to achieve the annual demand, (c) Cycle time, and
(e) service time Ts to which the line must be
balanced.
28
Analysis of single model assembly lines
Example
29
Analysis of single model assembly lines
Example
30
Analysis of single model assembly lines
Solution
• The total work content time is
• Twc 4.0 min.

(b) The production rate is
(c) The cycle time Tc with an uptime efficiency
of 96 is
31
Analysis of single model assembly lines
Solution
(d) The theoretical minimum number of workers is
given by
(e) The average service time against which the
line must be balanced is
32
Analysis of single model assembly lines
The objective in line balancing is to distribute
the total workload on the assembly line as evenly
as possible among the workers
subject to
and (2) all precedence requirements are obeyed.
33
Analysis of single model assembly lines
• The algorithms are
• Largest Candidate Rule
• Kilbridge and Wester method
• Ranked positional weights

34
Largest Candidate Rule
Step 1 Rank the Teks in the descending
order. Step 2 Assign the elements to the worker
at first station by starting at the top of the
list and selecting the first element that
satisfies precedence requirements and does not
cause the total sum of Tek at that station to
exceed the allowable Ts when an element is
selected for assignment to the station, start
back at the top of the list for subsequent
assignments. Step 3 when no more element can be
assigned without exceeding Ts, then proceed to
the next station. Step 4 repeat steps 2 and 3
for as many additional stations as necessary
until all elements have been assigned.
35
Largest Candidate Rule
Work elements sorted in descending order
36
Largest Candidate Rule
Solution The largest candidate algorithm is
carried out as presented in table below. 5
workers and stations are required in the
solution. Balance efficiency is computed as
37
Largest Candidate Rule
Work elements assigned to stations by LCR
38
Analysis of single model assembly lines
Example
39
Analysis of single model assembly lines
Kilbridge and Wester method
40
Analysis of single model assembly lines
Ranked positional weights
41
Analysis of single model assembly lines
Ranked positional weights
Kilbridge and Wester method
Largest Candidate Rule
42
Analysis of single model assembly lines
• Automation, Production Systems, and
Computer-Integrated Manufacturing, By Mikell P.
Groover, 3rd edition, c2008.
• Manufacturing Facilities Design and Material
Handling, By F. E. Meyers and M. P. Stephens, 4th
Edition, Prentice-Hall, Inc., 2010