Ch-7 Assembly Line Balance

8/11/2019 Ch-7 Assembly Line Balance

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Balance 1

Assembly Line Balance


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Balance 2

Assembly analysis

Assembly Chart

It shows the sequence of operations in puttingthe product together. Using the exploded drawing

and the parts list, the layout designer will diagramthe assembly process.

The sequence of assembly may have severalalternatives.

Time standards are required to decide whichsequence is best. This process is known as assemblyline balancing.


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Balance 3

The Assembly Chart

The ass

embly chartof a toolbox


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Balance 4

Time Standards Are Required for Every Task


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Balance 5

Plant Rate and Conveyor Speed

Conveyor speed is dependent on the number andunits needed per minute, the size of the unit, thespace between units. Conveyor belt speed is

recorded in feet per minute.

Example:

Charcoal grill are in cartons 30X30X24 incheshigh. A total of 2,400 grills are required everyday.


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Balance 6

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 assemblers2. To identify the bottleneck operation3. To establish the speed of the assembly line4. To determine the number of workstations5. To determine the labor cost of assembly and packout6. To establish the percentage workload of each operator7. To assist in plant layout8. 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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1. Cost of balancingSubassemblies that cost too high can be taken off the

line.

SA3 could be taken off the assembly line and handledcompletely 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 year700 hours per year

@ $15.00 per hour

= $10,500.00 per year savings


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Balance 10


Subassemblies that can be taken off the line must be:

1. Poorly loaded. The less percent that is loaded.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 tiedto the other operators), we could save 40 percentof 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 betterlabor utilization, total cost must go down.


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2. Improvement of assembly lineImprove the busiest (100 percent) workstation first.

(a) The busiest workstation is P.O. It has .167 minute of work to doper packer. The next closest station is A1 with .155 minute ofwork. As soon as we identify the busiest workstation, we identify

it as the 100 percent station, and communicate that this timestandard is the only time standard used on this line from nowon. Every other workstation is limited to 360 pieces per hour.Even though other workstations could work faster, the 100percent 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 hourper 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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Balance 12



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Balance 13



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Balance 14


2. Improvement of assembly lineImprove 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 stationat 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 1.75people 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 thelowest total number of hours per unit. If we add an additional

person, that persons time is in the total hours.


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Balance 15

Step-by-step procedure for completing theassembly line balancing form


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9. R valueThe R value goes behind each operation. The plant rate is thegoal of each workstation, and by putting the R value on eachline (operation), one keeps that goal clearly in focus.

10. Cycle timeThe time standard.

11. Number of stations

12. Average cycle time

Rtimecyclestationsofnumber

stationsof#

timecycletimecycleave.


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Balance 17


13.Percentage load:

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 percentstation.

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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Balance 18


13.Percentage load:

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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Balance 19


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 peoplenumber, the hour per unit will be:

Two people = .00557 hour per unit

Three people = .00835 hour per unit

Four people = .01113 hour per unit

hourperminutes60

timecycleaverage%100h.p.u.

unitperhour.0027860

.167h.p.u.


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Balance 20


15. Piece per hour:Inversion of hours per unit.

16. Total hours per unitSum of the elements in column 14. For this example is .0696hour.

17. Average hourly wage rate, say $15 per hour

18. Labor cost per unit

Total hours X average hourly wage

19. Total cycle timeIt 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.


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Balance 21

Efficiency of the assembly line

%100balanceline1000perhoursofSum

1000perhoursofSumefficiencyLine

%100balancelineunitperhoursofSum

unitperhoursofSumefficiencyLine

or

For our example:

%84%100

0.06960

0.05823efficiencyLine


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Balance 22

Analysis of single model assembly lines

shw

ap

HS

DR

50

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.


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Balance 23


p

c

R

ET60

The cycle time can be determined as

where Tc= cycle time of the line, min./cycle;

Rp= production rate, units/hr;

E = line efficiency;


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Balance 24


The cycle rate can be determined as

where Rc= cycle rate, cycles/hr;

Tcis in min./cycle;

Line efficiency E therefore defined as:

cc T

R60

p

c

c

p

T

T

R

RE


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Balance 25


The number of workers on the line can bedetermined 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.

AT

WLw

wcpTRWL TWc = work content time, min/piece.


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Balance 26


Using the previous equation, we also have

The available time in the period,AT.

c

wc

T

ETWL60

AT = 60E

Substitute these terms for WL and AT into wequation, we can state:

c

wc

TTw integerminimun*

If we assume one worker per station, then this ratio alsogives the theoretical minimum number of workstations onthe line.


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Example

A small electrical appliance is to be produced on a singlemodel assembly line. The work content of assembling theproduct has been reduced to the work elements listed intable below along with other information. The line is to bebalanced for an annual demand of 100,000 units per year.The line will be operated 50 weeks/yr, 5 shifts/wk, and 7.5hrs/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 cyclewill be 0.08 min. Determine (a) total work content timeTwc, (b) required hourly production rate Rpto achieve theannual demand, (c) Cycle time, and (e) service time Tstowhich the line must be balanced.


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Example


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Example


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Solution:

units/hr33.53)5.7)(5(50

000,100pR

(b) The production rate is:

(c) The cycle time Tcwith an uptime efficiency of 96% is:

(a) The total work content time is:

Twc= 4.0 min.

.min08.133.53

)96.0(60CT


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Balance 31


Solution:

43.7intmin* c

wc

TTw

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

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

.min00.108.008.1 Rcs TTT


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Balance 32


The objective in line balancing is to distributethe total workload on the assembly line asevenly as possible among the workers

w

1i

s )minimizeor)(minimize siwcs T(TTwT

subject to:

ikek(1) sTT

and

(2)all precedence requirements areobeyed.


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Balance 33


The algorithms are:

1) Largest Candidate Rule

2) Kilbridge and Wester method

3) Ranked positional weights


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Balance 34

Largest Candidate Rule

Step 1:Rank the Teks in the descending order.

Step 2:Assign the elements to the worker at first stationby starting at the top of the list and selecting the firstelement that satisfies precedence requirements and doesnot cause the total sum of Tek at that station to exceed theallowable Ts; when an element is selected for assignmentto the station, start back at the top of the list forsubsequent assignments.

Step 3: when no more element can be assigned withoutexceeding Ts, then proceed to the next station.

Step 4:repeat steps 2 and 3 for as many additionalstations as necessary until all elements have beenassigned.


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Balance 35


Work elements sorted in descending order


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Balance 36


Solution:

The largest candidate algorithm is carried out as presentedin table below. 5 workers and stations are required in thesolution. Balance efficiency is computed as:

8.0)0.1(5

0.4

s

wc

wT

TE


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Balance 37


Work elements assigned to stations by LCR


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Balance 39


Kilbridge and Wester method


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Balance 40


Ranked positional weights


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Balance 41


Ranked positionalweights

Kilbridge andWester method

Largest CandidateRule


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Automation, Production Systems, andComputer-Integrated Manufacturing, ByMikell P. Groover, 3rdedition, c2008.

Manufacturing Facilities Design and MaterialHandling, By F. E. Meyers and M. P. Stephens,4th Edition, Prentice-Hall, Inc., 2010

Ch-7 Assembly Line Balance

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