# ch 7 assembly line balance by HC12110410446

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```									Assembly Line Balance

Balance         1
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.
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The Assembly Chart

The assembly chart
of a toolbox

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Time Standards Are Required for Every Task

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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.

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Plant Rate and Conveyor Speed

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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).
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Initial assembly line balancing of toolbox

Takt time (for 2,000 units per shift, considering 10%
downtime and 80% efficiency) = .173 minutes per unit.
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Assembly line balancing

1. 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
@ \$15.00 per hour
=    \$10,500.00 per year savings
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Assembly line balancing

Subassemblies that can be taken off the line must be:

For example, a 60 percent load on the 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.

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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.

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Assembly line balancing

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Assembly line balancing

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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 person’s time is in the total hours.
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Step-by-step procedure for completing the
assembly line balancing form

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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
cycle time
number of stations 
R
12. Average cycle time
cycle time
ave. cycle time 
# of stations
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Step-by-step procedure for completing the
assembly line balancing form

The percentage load 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.

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

Example: percent load 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.
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Step-by-step procedure for completing the
assembly line balancing form

14. Hours per unit:
100 % average cycle time
h.p.u. 
60 minutes per hour
Example: Hours per unit of the toolbox assembly line balance

.167
h.p.u.        .00278 hour per unit
60
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

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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. Balance                     20
Efficiency of the assembly line

Sum of hours per 1000
Line efficiency                                      100%
Sum of hours per 1000 line balance

or
Sum of hours per unit
Line efficiency                                      100%
Sum of hours per unit line balance

For our example:
0.05823
Line efficiency           100%  84%
0.06960
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Analysis of single model assembly lines

Production Rate is given by

Da
Rp 
50 S w H sh

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.

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Analysis of single model assembly lines

The cycle time can be determined as

60 E
Tc 
Rp

where Tc = cycle time of the line, min./cycle;
Rp = production rate, units/hr;
E = line efficiency;

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Analysis of single model assembly lines

The cycle rate can be determined as

60
Rc 
Tc

where Rc = cycle rate, cycles/hr;
Tc is in min./cycle;
Line efficiency E therefore defined as:
Rp
Tc
E   
Rc Tp
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Analysis of single model assembly lines

The number of workers on the line can be
determined as
WL
w
AT
where w = number of workers on the line;
WL = workload to be accomplished in a given time period.
AT = available time in the period.

WL  R pTwc      TWc = work content time, min/piece.

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Analysis of single model assembly lines

Using the previous equation, we also have

60 ETwc
WL 
Tc
The available time in the period, AT.   AT = 60E

Substitute these terms for WL and AT into w
equation, we can state:
Twc
w  minimun integer 
*

Tc
If we assume one worker per station, then this ratio also
gives the theoretical minimum number of workstations on
the line.
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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.
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Analysis of single model assembly lines

Example

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Analysis of single model assembly lines

Example

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Analysis of single model assembly lines

Solution:
(a) The total work content time is:
Twc = 4.0 min.

(b) The production rate is:
100 ,000
Rp                53 .33 units/hr
50 (5)( 7.5)
(c) The cycle time Tc with an uptime efficiency of 96% is:

60(0.96)
TC            1.08 min .
53.33
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Analysis of single model assembly lines

Solution:
(d) The theoretical minimum number of workers is given by:

Twc
w*  min int       3.7  4
Tc

(e) The average service time against which the line must
be balanced is:

Ts  Tc  TR  1.08  0.08  1.00 min .

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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
w
minimize ( wTs  Twc ) or minimize            (T  T
i 1
s   si   )
subject to:

(1)   T
k i
ek    Ts
and

(2) all precedence requirements are
obeyed.
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Analysis of single model assembly lines

The algorithms are:
1) Largest Candidate Rule
2) Kilbridge and Wester method
3) Ranked positional weights

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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.
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Largest Candidate Rule

Work elements sorted in descending order

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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:

Twc   4.0
E            0.8
wTs 5(1.0)

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Largest Candidate Rule
Work elements assigned to stations by LCR

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Analysis of single model assembly lines

Example

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Analysis of single model assembly lines

Kilbridge and Wester method

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Analysis of single model assembly lines

Ranked positional weights

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Analysis of single model assembly lines

Ranked positional
Largest Candidate   Kilbridge and
weights
Rule             Wester method

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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

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