Selector efficiency

8/7/2019 Selector efficiency

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4.2.4Part Feeding Case Study (Asfahl, 1992)

4.2.4.1 Selector Efficiency

A general criterion for evaluating a mechanical part selector is “efficiency”

which is defined as follows:

O

I

F efficiency

F = (4.1)

where O F = output feed rate of correctly orientated parts

I F = total input feed rate

It is possible to calculate the efficiency from experimental data after a

selector system has been designed and fabricated. However, a growing body

of scientific data has made possible the synthesis of efficiencies from

available information on the individual selector mechanism employed by the

system. For each selector mechanism there is a “transition probability

matrix” that reveals the probability that a piece will change to orientation y

when it passes through a selector device when the part is in orientation x

prior to encountering the device.

The most popular mode of selector device is the rejector , which simply

discards an incorrectly oriented piece back into the center of the vibratory

bowl or other hopper. The pure rejector is a devise for which the transition

probability from a given entry state to the rejected state is 100 percent. A

device can be a pure rejector for some entry states and not others. If a

device is a pure rejector for all entry states, it can be call a “total rejector,”

meaning that it rejects every part that encounters it, regardless of

orientation. The concept of “total rejector” is introduced here merely to

clarify the term “pure rejector.” In real automation application, pure rejectors

would have utility; total rejectors would not.

The general case for a part orientation devise recognizes a probability of

transition from any feasible orientation to any feasible orientation, includingthe possibility of rejection. Therefore, an n by n + 1 matrix of probabilities is

generated, in which n represents the number of feasible orientations. For

example, a rectangular block has six feasible orientations as shown below.



Thus, there are 42 probabilities in a 6 x 7 matrix used to describe completely

the general orientation characteristics of a device that orients rectangular

blocks. With most real devices, most of the probabilities in the matrix are

either zero or one (0 or 100 percent). Now let us consider an example

combination of orientation devices and perform probability calculations to

determine the distribution of output orientations.

Example 1:



Solution :

Before proceeding with a quantitative analysis of the probabilities associated

with each selector device, the general characteristics of each selector can be

observed in Table 3.1. The phase one wiper can be seen to be a pure rejector

for the “erect lengthwise” orientation, an intuitively reasonable characteristic

when the table is compare with the diagram in Figure 3.17. The wiper is only

a 90 percent rejector for the “erect crosswise” orientation; the other 10percent are switched to “flat lengthwise,” which is still an incorrect

orientation. The four remaining input orientations are shown by the table to

be unaffected by the wiper. The phase two off-ramp is shown in Table 3.1 to

be pure rejector for orientation e and f, just as was indicated in the diagram,

and has no effect on the other four orientations. The final device is the phase

three inclined rail, which has the function of lifting the “flat lengthwise” block

to “on-edge lengthwise,” the correct orientation. From Table 3.1 it can be

seen that the rail is only 80 percent successful in this function, the other 20

percent being rejected from the track.



4.2.4.2 Efficiency vs. Effectiveness

A 29.8 percent efficiency might seem to be little disappointing, but

remember that the objective in the example was to orient piece parts

automatically into a single correct orientation when presented with six

possible input orientations, and that the input distribution of those six

orientations was very unfavorable, with only 5 percent of the input streamappearing in the correct orientation to begin with. The example of selector

system above had a challenging assignment, and most real industrial

applications of automatic assembly present similar challenges in part

orientation. It is fascinating to watch an industrial vibratory bowl at work,

and it is not uncommon at all to observe a large quantity of rejected

orientations being tossed back into the bowl for a retry.

Another way to view the selector example is in term of “effectiveness”

instead of efficiency.



4.2.4.3 Part Wear and Damage.A final consideration may be that the parts selector will overwork the parts,

oscillating and vibrating them and kicking many of them back for retry after

retry. If the effectiveness is virtually 100 percent, it is possible to use the

system efficiency to calculate the chances that a part will be tossed back k

times before reaching an acceptable orientation. The formula is:

1100 100

k

k

E E p

= −

(4.3)

where k p = = probability that the part will be kick back k times

E= efficiency

k = number of kickbacks

For previous example:

(4.2)



0

0

29.8 29.81 0.298

100 100 p

= − =

1

1

29.8 29.81 0.209

100 100

p

= − =

2

2

29.8 29.81 0.147

100 100 p

= − =

10

10

29.8 29.81 0.009

100 100 p

= − =

Thus, nearly out of hundred parts will be kicked back ten times before

achieving an acceptable orientation. The automation engineer must decide

whether or not this kind of treatment will damage the product.

4.2.5 Analysis on Production and Throughput (Asfahl,

1992)

The automatic assembly machine, whether it be rotary or in-line, produces a

completed assembly every time the machine indexes, regardless the number

of stations in the assembly process. To compute the (ideal) production rate

of an automatic assembly machine, one needs to know only the indexing

cycle time; the number of stations is immaterial. It is true that throughputtime, the time required to assemble a given assembly from start to finish, is

dependent upon the number of stations and the index time, but production

rate is not.

Example 2:

An automatic assembly machine is of the dial-indexing configuration, has 8

stations, and is driven by a geneva mechanism in which the driver has a

rotational speed of 30 rpm. What is the production and throughput time of this machine?

Solution:



In geneva mechanism, every revolution of the driver constitutes one

indexing of the assembly machine. Therefore, the production rate is 30

units/min.

Production time = 1/production rate = 1/30 units/min x 60sec/min

= 2s/unit

Throughput time = prod. Time x no. of stations

= 2s x 8 stations

= 16 s

4.2.5.1 Machine Jamming

It is easy to overlook the pitfalls of automation, and one of the most

notorious examples can result from the setting of automatic multistationassembly machines without due consideration to the potential effects of

station malfunction or jamming. Consider an eight-station rotary indexing

machine driven by a geneva mechanism in which the index time is three

seconds and the dwell time is five seconds. The reader may want to check

the validity of the ratio of index to dwell time. Under ideal operating

conditions (no malfunctions), this eight-station rotary indexing machine will

produce a completed assembly every eight seconds and will achieve a

corresponding production rate of 450 units/ hour.

In more realistic case, suppose that each station malfunctions on theaverage of once every 100 cycles, a seemingly tolerable rate of work

stoppage. One or more station malfunctions will immediately jam the

indexing machine, requiring an operator to make adjustments to restart the

machine. For our example, let us say that this adjustment and restart

process requires a mere ten minutes. The result is a drastic reduction in the

productivity of the automated assembly machine, as can be seen in the

following series of calculations.

If the chances of malfunction is one in 100, the chance that a given stations

will not malfunction in a given cycle is 99% or 0.99. But all eight stationsmust operate without malfunction to produce a completed assembly

successfully. So the probability of no malfunction in a given cycle is a

product of the chances of no malfunctions at each station during that cycle.

Multiplying the station success probabilities;

80.99 0.9227=



Therefore, out of 10,000 machine cycles, 9227 assemblies will be produced

without malfunction, with a cycle time of eight seconds per assembly. This

will consume

9227 8sec 73816sec 20.50hr × = =

In the other 773 cycles (10,000-9227), at least one station will malfunction

and require a 10-minute repair. This will consume

773 10 min/ 7730 min 128.83beakdown hr × = =

The total time to produce the 9227 assemblies is then

20.50 128.83 149.33hr hr hr + =

for the total of operating time plus malfunction downtime. But note which

figure is the larger ! The percent of downtime is

128.830.863 86.3%

149.33= =

of the total production time! The production rate has been reduce from the

ideal 450 units per hour calculated earlier to

922761.8 /

149.33

unitsunits hr

hr =

The efficiency of the assembly machine would be the ratio of the actualproduction rate to the ideal production rate, calculated as

61.8 /0.137 13.7%

450 /

units hr efficiency

units hr = = =

an amazing 86.3% drop in efficiency from introduction of only a slight(one

percent) station malfunction rate. Such is the world of automation, and the

automation engineer should be prepared to deal with the realities of system

malfunction and downtime when planning for a new automated assembly

machine.

4.2.5.2 Component Quality Control

The previous section discussed the disastrous effects of station malfunction

in an automatic assembly machine, but failed to explain why these

malfunctions might occur. The predominant cause of assembly station



malfunction is some random variation in the components being assembled—

variation of a magnitude that cannot be handled by the assembly machine. If

tighter specifications can be applied or closer quality control can be exacted

upon the components produced to existing specifications, the automation

engineer may be able to achieve astonishing improvements in assembly

machine production rates. Suppose in the example of the previous section

that 90 percent of the assembly malfunctions were due to faulty

components. Elimination of the component quality problem would then

reduce the station malfunction rate from one out of 100 cycles to one out of

10,000 cycles. Such station reliability corresponds roughly to “four sigma”

performance.

Probability of no station malfunction in a given typical cycle:

8[ ] 0.9999 0.9992 probability nostationmalfucntion = =

thus, in 10,000 machine cycles, 9992 assemblies will be produced with no

malfunction with a cycle time of eight seconds per completed assembly. This

will consume

9992 8 sec 79936 22.20hr × = =

In the other 8 cycles (10,000-9992), at least one station will malfunction and

require a 10-minute repair. This will consume

8 x 10 min = 80 min = 1.33 hr

From here try to calculate the total production time, downtime percentage,

production rate and the efficiency.

Example 3:

An in-line automatic transfer and assembly machine has 30 consecutive

assembly stations. The line is under control of a walking beam with an index

time of four seconds and a dwell of 20 seconds. Each station along the line

will operate without malfunction with a reliability of 0.999 when its hopper issupplied with quality components. Any defective component will cause a

station to jam, which in turn will precipitate a line jam because there is no

provision for in-process assembly storage along the line. A jam or

malfunction requires 10 minutes to correct.

(a) No station malfunction or jamming



Assuming no station malfunction or jamming at all, what is the ideal

production capability of this line? What is throughput time?

Solution:

Cycle time

4 sec 20 sec 24 secC

T index dwell = + =

Ideal production rate

3600 /150 /

24sec/

sc hr R unit hr

unit = =

Throughput time

30 24 sec/ 720sec 12 minT T station station= × = =

(b) Assuming station malfunction

Assuming ideal component quality, what is the production rate

considering station malfunction? What is the percent downtime? What is

the throughput time?

Solution:

Prob [a cycle will not malfunction] = 300.999 0.9704=

Prob [a cycle will malfunction] = 1 0.9704 0.0296− =

Time to produce 9704 assemblies

Total time = total successful cycle time + malfunction

correction time

=24 sec 1 10 min 1

9704 2963600 sec 60 min

hr hr units malfucntions

unit malfunction× × + × ×

=64.69hr+49.33hr = 114.02hr

Production rate

R = 9702 units/ 114.02hr = 85 unit/hr

Percent downtime

D = 49.33 hr/114.02 hr = 0.43 = 43%



Throughput time

1 60 min30

85 /T

T stationunits hr hr

= × ×

=21.18 min

(c) Assuming defective part jamming

Assuming component lot quality is at the level of 12 of one percent

defective, what is the effect of defective components upon production

rate?

Solution:

We now have two ways that the station can jam:

1. General station malfunction (reliability = 0.99)

2. Defective component (1

2of one percent =0.005)

For the situation in which neither of these possibilities occurs we must

compute the product of probabilities:

Prob[a given station will not jam at a given cycle] = (0.999) X Pob[no

defective component]

= (0.999)(1 – 0.005) = 0.999 X 0.995

= 0.994

Since the system has 30 stations, any one of which is capable of bringing

down the entire line, we must compute the product of probabilities that

individual stations will not jam to obtain the overall system success

probability for a given cycle:

Prob [a cycle will not jam] = 300.994 0.8349=

And the probability that the system will jam in a given cycle is the

complement:

Prob [a cycle will jam] = 1 0.8349 0.1651− =

Time to produce 8349 assemblies



Total time = total successfully cycles time + unjam time

=24sec 1 10 min 1

8349 16513600sec 60 min

hr hr units jams

unit jam× × + × ×

= 330.82hr

Production rate

R = 8349 units/ 330.82hr = 25.24 units/hr

Percent downtime

D = 275.16 hr/ 330.82 hr = 0.83 = 83%

Efficiency

E =25.24 / 0.17 17%150 /

actualproductionrate units hr

idealproductionrate units hr = = =

Throughput time

1 60 min30

25.24 /T

T stationsunits hr hr

= × ×

71.32 min 1.19hr = =

Thus, a component quality level of only ½ of one percent defectives in this

case reduces the production rate from 95 units/hr to 25.24 units/hr,

increases downtime from 43 percent to 83 percent, and throughput time

increases from a little over 20 minutes to over an hour.

So important is the quality and uniformity of components to the success of

assembly automation that some firms pay several times the standard price

for such items as screws, nuts, and bolts to purchase the finest quality

available for use in automated assembly.

Selector efficiency

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