assembly line balance ok.ppt

8/10/2019 assembly line balance OK.ppt

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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 putting theproduct together. Using the eploded drawing and

the parts list! the layout designer will diagram theassembly process.

"he sequence of assembly may ha#e se#eralalternati#es.

"ime standards are required to decide whichsequence is best. "his process is $nown as assemblyline balancing.


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Balance %

"he Assembly Chart

"he

assembly chartof a toolbo


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Balance &

"ime 'tandards Are (equired for )#ery "as$


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Balance *

+lant (ate and Con#eyor 'peed

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

recorded in feet per minute.

Example:

Charcoal grill are in cartons %-%-2& incheshigh. A total of 2!&-- grills are required e#ery day.


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Balance /

+lant (ate and Con#eyor 'peed


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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 workstations. To determine the labor cost of assembly and packout!. To establish the percentage workload of each operator". To assist in plant layout#. To reduce production cost

The assembly line balancing technique builds on:The assembly chart$Time standards$Takt time %minutes&piece' %(lant rate) R *alue)

(ieces&minutes'.


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Initial assembly line balancing of toolbo

"a$t time for 2!--- units per shift! considering 1-3

downtime and -3 efficiency4 5 .10% minutes per unit.


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1.+ost of balancing,ubassemblies that cost too high can be taken off the

line.

,-3 could be taken off the assembly line and handledcompletely separate from the main line and we can sa*e money.,-3 .2 / 24 pieces per hour and .41" hour each. 0f balanced)the standard would be 1# pieces per hour and ." hour each.

." balanced cost

.41" by itself cost.14 sa*ings hour per unit

X ) units per year" hours per year

$15.00 per hour

/ 1). per year sa*ings


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Balance 1-


,ubassemblies that can be taken off the line must be:

1. (oorly loaded. The less percent that is loaded.or e5ample) a ! percent load on the assembly

line balance would indicate 4 percent lost time.0f we take this 6ob off the assembly line %not tiedto the other operators') we could sa*e 4 percentof the cost.

2. ,mall parts that are easily stacked and stored.3. 7asily mo*ed. The cost of transportation and the

in*entory cost will go up) but because of betterlabor utilization) total cost must go down.


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Balance 11


2. 0mpro*ement of assembly line 0mpro*e the busiest %1 percent' workstation first.

%a' The busiest workstation is (.8. 0t has .1!" minute of work to doper packer. The ne5t closest station is -1 with .1 minute of

work. -s soon as we identify the busiest workstation) we identify

it as the 1 percent station) and communicate that this timestandard is the only time standard used on this line from nowon. 7*ery other workstation is limited to 3! pieces per hour.7*en though other workstations could work faster) the 1

percent station limits the output of the whole assembly line.

%b' The total hours required to assemble one finished toolbo5 is.!9! hour. The a*erage hourly wage rate times .!9! hourper unit gi*es us the assembly and packout labor cost. -gain)

the lower this cost is the better the line balance is.


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Balance 12



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Balance 1%



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Balance 1&


2. 0mpro*ement of assembly line 0mpro*e the busiest %1 percent' workstation first.

ook at the 1 percent station %(.8.'.

0f we add a fourth packer) we will eliminate the 1 percent stationat (.8.

;ow the new 1 percent %bottleneck station' is -1 %93 percent'.


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Balance 1*

'tep7by7step procedure for completing theassembly line balancing form


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Balance 1/


9. R *alueThe R *alue goes behind each operation. The plant rate is thegoal of each workstation) and by putting the R *alue on each

line %operation') one keeps that goal clearly in focus.

1. +ycle time The time standard.

11. ;umber of stations

12. -*erage cycle time

Rtimecyclestationsofnumber =

stationsof#

timecycletimecycleave. =


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Balance 10


13.(ercentage load:

The percentage load tells how busy each workstation is

compared to the busiest workstation.

The highest number in the a*erage cycle time column 12 is

the busiest workstation and) therefore) is called the 1 percentstation.

;ow e*ery other station is compared to this 1 percent

station by di*iding the 1 percent a*erage station time into

e*ery other a*erage 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 1 percent station

can be reduced by 1 percent) then we will sa*e 1 percent for

e*ery workstation on the line.


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Balance 1


13.(ercentage load:

75ample: percent load of the toolbo5 assembly line balance

0n igure 411) the a*erage cycle times re*eals that .1!" is the

largest number and is designated the 1 percent workstation.

The percentage load of e*ery other workstation is determined by

di*iding .1!" into e*ery other a*erage cycle time:

8peration ,,,-1 / .13 & .1!" / 92 percent

,,-1 / .14! & .1!" / #" percent

,,-2 / .13 & .1!" / "# percent

and so on.


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Balance 16


14.?ours per unit:

75ample: ?ours per unit of the toolbo5 assembly line balance

The .1!" time standard is for one person) if considering the people

number) the hour per unit will be:

Two people / ." hour per unit

Three people / .#3 hour per unit

our people / .1113 hour per unit

hourperminutes60

timecycleaverage%100h.p.u.=

unitperhour.0027860

.167h.p.u. ==


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Balance 2-


1. (iece per hour:0n*ersion of hours per unit.

1!. Total hours per unit

,um of the elements in column 14. or this e5ample is .!9!hour.

1". -*erage hourly wage rate) say 1 per hour

1#. abor cost per unit

Total hours @ a*erage hourly wage

19. Total cycle time 0t tells us what a perfect line balance would be. 8ur e5ample 3.494 minutes di*ided by ! minutes per hour

equals .#23 hour per unit.


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Balance 21

)fficiency of the assembly line

%100balanceline1000perhoursofSum

1000perhoursofSumefficiencyine =

%100balancelineunitperhoursofSum

unitperhoursofSumefficiencyine =

or

8or our eample9

%8!%100

0.06"60

0.082$efficiencyine ==


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Balance 22

Analysis of single model assembly lines

shw

ap

HS

DR

#0

=

+roduction (ate is gi#en by

where Rp5 a#erage hourly production rate! units:hr;

Da

5 annual demand! units:year;

Sw5 number of shifts:wee$;

Hsh5 hrs:shift.

"his equation assume *- wee$s per year.


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Balance 2%


p

c

R

ET60

=

"he cycle time can be determined as

where Tc5 cycle time of the line! min.:cycle;

Rp

5 production rate! units:hr;

E 5 line efficiency;


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Balance 2&


"he cycle rate can be determined as

where Rc5 cycle rate! cycles:hr;

Tc

is 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 2*


"he number of wor$ers on the line can bedetermined as

where w5 number of wor$ers on the line;

WL = wor$load to be accomplished in a gi#en time period.

T5 a#ailable time in the period.

AT

WLw =

wcpTRWL= TWc = wor$ content time! min:piece.


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Balance 2/


Using the pre#ious equation! we also ha#e

"he a#ailable time in the period!T.

c

wc

T

ETWL60

=

T = !"E

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

c

wc

TTw = integerminimun

If we assume one wor$er per station! then this ratio alsogi#es the theoretical minimum number of wor$stations onthe line.


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Balance 20


)ample

A small electrical appliance is to be produced on a singlemodel assembly line. "he wor$ content of assembling theproduct has been reduced to the wor$ elements listed intable below along with other information. "he line is to bebalanced for an annual demand of 1--!--- units per year."he line will be operated *- wee$s:yr! * shifts:w$! and 0.*hrs:shift.


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Balance 2


)ample


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Balance 26


)ample


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Balance %-


'olution9

units&hr$$.#$'#.7'(#(#0

000)100==pR

b4 "he production rate is9

c4 "he cycle time Tcwith an uptime efficiency of 6/3 is9

a4 "he total wor$ content time is9

Twc5 &.- min.

.min08.1$$.#$

'"6.0(60==CT


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Balance %1


'olution9

!$.7intmin ==c

wc

TTw

e4 "he a#erage ser#ice time against which the line mustbe balanced is9

d4 "he theoretical minimum number of wor$ers is gi#en by9

.min00.108.008.1 === Rcs TTT


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Balance %2


"he ob>ecti#e in line balancing is to distributethe total wor$load on the assembly line ase#enly as possible among the wor$ers

=

*

1i

s 'minimi+eor'(minimi+e siwcs T(TTwT

sub>ect to9

i,e,(1' sTT

and

24all precedence requirements areobeyed.


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Balance %%


"he algorithms are9

14Largest Candidate (ule

24?ilbridge and @ester method

%4(an$ed positional weights


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Balance %&

Largest Candidate (ule

Step 1:(an$ the "e$s in the descending order.

Step :Assign the elements to the wor$er at first stationby starting at the top of the list and selecting the firstelement that satisfies precedence requirements and doesnot cause the total sum of "e$ at that station to eceed theallowable "s; when an element is selected for assignmentto the station! start bac$ at the top of the list forsubsequent assignments.

Step !: when no more element can be assigned withouteceeding "s! then proceed to the net station.

Step ":repeat steps 2 and % for as many additionalstations as necessary until all elements ha#e beenassigned.


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Balance %*


@or$ elements sorted in descending order


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Balance %/


'olution9

"he largest candidate algorithm is carried out as presentedin table below. * wor$ers and stations are required in thesolution. Balance efficiency is computed as9

8.0'0.1(#

0.!===

s

wc

wT

TE


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Balance %0


@or$ elements assigned to stations by LC(


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Balance %


)ample


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Balance %6


?ilbridge and @ester method


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Balance &-


(an$ed positional weights


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Balance &1


(an$ed positionalweights

?ilbridge and@ester method

Largest Candidate(ule


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#tomation$ %rod#ction Systems$ andComp#ter&'ntegrated (an#)act#ring! By

assembly line balance ok.ppt

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