modeling of automatic lathe management system

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American Journal of Scientific Research

ISSN 1450-223X Issue 12 (2010), pp.16-22© EuroJournals Publishing, Inc. 2010

http://www.eurojournals.com/ajsr.htm

Modeling of Automatic Lathe Management System

Lim Eng Aik Institut Matematik Kejuruteraan, Universiti Malaysia Perlis

Kangar, Perlis, Malaysia

E-mail: [email protected]

Abstract

In this paper, an automatic lathe management system was discussed, and an

effective algorithm was proposed aided with easy to execute the scheme, to reduceeffectively the cost of inspection intervals, cutting tools replacement policy and the loss

comes from replacing individual parts. For Model I , we obtained an inspection time-gap of,τ 0 = 18 and the time-gap for cutting tools replacement of, τ 1 = 342, which leads to thecorresponding optimal solution of cost of individual parts, C = 2.42 Ringgit. The model for

Model I is then tested for its stability with Monte Carlo simulation. While for Model II , the

inspection time-gap, τ 0 = 11, and the time-gap for cutting tools replacement, τ 1 = 242 wereobtained. The optimal solution for cost for individual parts in Model II is, C = 3.68 Ringgit.

Lastly, for Model III , we applied our proposed improvement to the model, and an expected

cost for individual part is reduced to 2.72 Ringgit was observed. Discussion on changes of

inspection intervals, parameter sensitivity and error analysis was provided in this paper.

Keywords: Monte-Carlo Simulation, Optimization, Modeling, Sensitivity Analysis, ErrorAnalysis

IntroductionIn industry, lathe management is an essential task to perform. Due to the failure of the cutting tools will

cause breakdown such as process failure, which is completely random during operation [1, 2]. The

process is mainly monitored by staff. They need to perform checking whether the parts fail duringoperation process failure, and replacing cutting tools after a certain period of time performing the

operations [3, 4]. Figure 1 and 2 shows an example of the appearance and cutting process of lathe.

Therefore, when performing inspection for parts of tools within the inspection time-gap, if theyfound a non-compliant component that causes fault occurred, then the inspection will immediately stop

to locate the fault is repaired, if there is no fault found, then production line will continues theoperation. When the cutting tool reaches the end of its time-gap for tools replacement, the tool needs to

replace with new one even though there is no fault occurs from the equipment. The illustration of lathecutting tools is showed in Figure 3.

Obviously, for the case of periodic replacement, if the inspection time-gap is too large, it may

cause failure of equipment extend in time, which resulting in increased losses in production cost; if theinspection time-gap is too small, this will cause an increment in inspection fees. The main problem

now is to find the optimal inspection time-gap, and the time-gap for cutting tools replacement to

achieve better process efficiency and reduce cost. The best process efficiency can be expressed as thesmallest expected loss from individual parts, and we defined a cycle of replacement as the period

between tool replacements, then Cost of expected loss from individual parts, C is defined as;



Modeling of Automatic Lathe Management System 17

C =Expected loss within 1 cycle

Total expected cycle(1)

Figure 1: Appearance of lathe

Figure 2: Cutting process of lathe

Figure 3: Illustration for lathe cutting tool

Model DescriptionTo set-up our model, we placed a few assumptions as follows:

1. The time of producing an individual part is set to 1.



18 Lim Eng Aik

2. We are not considering the fault detection time and time regulator to replace the cutting tool.

3. The process is restored to initial state, after replacement of cutting tools or adjustment is being

made to fault location.4. Staff immediately specified the substandard parts during an inspection, and confirm the

production process is a failure.

5. Each automatic lathe machine only consists of one cutting tool.

The symbol conventions involved are:

ƒ The cost of losses on producing a nonconforming product output with ƒ = 102.04 Ringgit/ piece

T Cost of inspection t = 5.10 Ringgit/inspection

D Average cost of fault adjustment d = 153.06 Ringgit/times

K Cost of replacing a new cutting tool in zero fault condition k = 510.20 Ringgit /unit replace

X Zero failure production process time

F ( x) Distribution function of X

p( x) Probability density function of X

0τ Inspection time-gap

1τ Time-gap for cutting tools replacement

E ( L) Expected total cost for one cycle

E (T ) Expected total cycleC Cost of expected loss cost for individual parts

Preparation of the ModelTo set-up the model, we recorded the statistical analysis of a 100 time of cutting tool failure. From the

frequency distribution histogram, we are able to examining the level of significance when α = 0.10, thetime of cutting tool working properly are comparable to normal distribution N ( µ, σ

2), where µ = 306.12,

σ 2

= 50.922.

To compute the probability distribution for zero failure process time, we set the decisive role of cutting tool failure as 95% loss. We believe that, a long working hour along with whole trouble-free

process distribution is equivalent to a zero-failure cutting tool distribution, i.e., X ~ N (0.95 µ, (0.95σ )2

).Lastly, for the cutting tool replacement policy, an inspection should be performed before the

constant cutting tool replacement is proceeded. If failure was found during an inspection, repairing

work will immediately carry on, otherwise, they will directly progress to constant cutting tool

replacement. For practicable convenience, constant cutting tool replacement cycle can be set at the first

m-examination, that if 0τ is constant, 1τ = m 0τ (m = 1,2, ...).

Set-Up of ModelsModel I

If 1 X τ > Loss: L1 = mt + k ;

If nτ0 < X ≤ (n + 1) 0τ (n = 0,1,2, ..., m -1)

Loss: Ln = (n + 1) t + d + [(n + 1) τ 0 - X ] ƒ

Expected total cost for one cycle E ( L) = ∑∫∫−

=

+∞

+

1

0

)1(

1

0

0

0

)()(m

n

n

nn

m

dx x p Ldx x p Lτ

τ

τ

Expected total cycle E (T ) = ∑ ∫∫−

=

+∞

++

)1(

0

)1(

00

0

00

)()1()(m

n

n

nmdx x pndx x pm

τ

τ τ

τ τ




For best efficiency, we need to find an equivalent to 0τ and 1τ , so that C =)(

)(

T E

L E minimum.

From the test, we computed that 0τ =18 (i.e., producing 18 parts per inspection), 1τ = 342 (i.e., produce

342 parts per cutting tool replacement), and the expected cost for individual parts C = 2.42 Ringgit.

Model II

Similar steps as in Model I , but for this model, we set the values for 0τ =11, 1τ = 242 and C = 1.88

Ringgit.

Model III

Taking into account that the equipment used within the period of replacement can be separated intostable and unstable equipment. Here, stability refers to a minor fault, but the so-called failure is major

fault, which refers as unstable. We improved the inspection time-gap to extend the inspection time

within the stable period, while reducing the inspection time during the unstable period, resulting in

higher efficiency. Here, the inspection interval 0τ and the time, x , is denoted by 0τ ( x ). If the failure

rate is large, then the number of inspection per unit time )( xn is also increasing. Note

that,)(

1)(0

xn x =τ , we define

)(1

)()(

xF

x p

t

f xn

−⋅= (2)

where dx x p )( is the probability of equipment failure bounded in ( x , x + dx ) and )(1 xF − is the

probability of zero failure. Therefore, ( t , t + dt ) is the boundary of conditional probability for

equipment failure, where)(1

)(

xF

dx x p

−is represent the zero failure condition. Here,

2

2

( 290.81)

2 47.611

( ) 2 47.61

x

e p x π

−

×

= (3)

Based on the above analysis, the inspection method can be expressed as follow:

1st time of inspection time-gap

=

)0(

11

nd ; (

)0(

1

nexpressed as

1

(0)nround value to the

nearest integer, d is the i -th times of inspections time-gap)

2nd time of inspection time-gap

=

)(

1

1

2d n

d ;

3rd time of inspection time-gap

+=

)(

1

12

3d d n

d ;

...

n-th time of inspection time-gap1 2 1

1

( ... )n

n n

d n d d d

− −

=

+ + + .

The results are listed in Table 1.

By substituting id into eq. (1), C =)(

)(

T E

L E , the cost of expected loss for individual parts, C =

2.72 Ringgit, this cost is less than the expected loss cost in eq. (1). From Model II , for constant

replacement of cutting tool 1τ = 242 cases, we found that, the frequency of inspection was greatly

reduced at the 72nd

,114th

,146th

,172nd

,194th

,214th

,232nd

and 242nd

parts of inspection.



20 Lim Eng Aik

Table 1: Model I Simulation Results

i 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

id 72 42 32 26 22 20 18 16 15 14 13 12 12 11 10

i 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30

id 10 10 9 9 9 8 8 8 8 7 7 7 7 7 7

i 31 32 33 34 35 36 37 38 39 40 41 42 43 …

id 7 6 6 6 6 6 6 6 6 6 5 5 5 …

Model Simulation and ResultsMonte-Carlo Simulation Test

For Model I , we adopt the Monte-Carlo method to simulate the inspection. Concrete steps are as

follows:

1) Zero failure process time distribution for 2~ (290.81 47.61 X N ，）, are generated using Monte-

Carlo simulation of 1000 times with a pseudo-random number,

)1000...,21(，，

=i X i .2) Given a constant inspection time-gap 0τ and cutting tool replacement time-gap 1τ , the i X value

can be calculated using the corresponding cost for i L

[ ]( )

−+

+

⋅+

+

=

f X t X

d

t d

t k

L

i

i

i

00

0

0

1

0

1

%τ τ τ

τ

τ

τ

τ

[ ]

[ ]

[ ] 1

1

1

τ

τ

τ

<

=

>

i

i

i

X

X

X

where [ ]i X express as round value of i X to the nearest integer, while % express as remainder of i X .3) Calculate the cost of expected loss for individual parts, C is defined as follow:

∑

∑

=

==

1000

1

1000

1

i

i

i

i

T

L

C with

+

=0

0

1

1 τ τ

τ

ii X T

[ ]

[ ]

1

1

i

i

X

X

τ

τ

≥

<

4) To find the optimal values for 0τ , 1τ , let ），2000(0 ∈τ and ），（ 100001 ∈τ , both values can be

obtained from the cost of expected loss for individual parts, C .

Based on the calculation, we obtained the optimal inspection time-gap, 180 =τ , and time-gap

for constant replacement of cutting tool, 1τ = 378, which yield the corresponding cost of expected loss

for individual parts of 2.12 Ringgit. Model I result using Monte-Carlo simulation results are compared(see Table 2). From Table 2, we observed that the simulation results using the Model I is comparable

with Monte-Carlo simulation. So, we conclude that Model I result is more stable.

Table 2: Monte-Carlo Simulation Results

0τ 1τ C

Model I 18 342 2.42

Monte Carlo simulation 18 378 2.12




Sensitivity Analysis

For normal distribution of 2~ (290.81 47.61 X N ，）, we change P (P is a proportion of substandard

products produce from the normal production process) and q (q is a proportion unqualified products

produce when production process failure occurs) and obtained the results are shown in Table 3.

Figure 4 shows the scatter plot of P-C (Table 3) and q-C (Table 4). We observed that the cost

the loss of individual components impacts, C , are related to the impact from the changes of P and q.

From Figure 4(a), One can sees, there is an essential linear relationship between P and C , and in

addition, the impact of P on C was significantly higher than q (see Figure 4(b)). From the patterns

C q − , when q value reached about 0.4, the increment of q to C had a minor effect or almost constant.

In short, C response to the changes in P, while the response is very slow in q. Therefore, in practical

management, if one hold a good control on the value P, then one can have a better control of the cost of

expected loss for individual parts. Thus, to maximize the efficiency, we have to minimize P so that C is

minimized.

Table 3: Results for P-C

P 0.01 0.012 0.014 0.016 0.018 0.02 0.022 0.024 0.026 0.028 0.03 0.035

0τ 13 13 12 12 12 11 11 11 10 10 9 7

m 22 22 22 22 22 22 22 22 22 22 22 22

1τ 286 286 264 264 264 242 242 242 220 220 198 154

C 3.21 3.32 3.41 3.51 3.59 3.68 3.76 3.84 3.90 3.96 4.00 4.04

Table 4: Results for q-C

q 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9

0τ 12 12 12 11 11 11 11 11 11

m 22 22 22 22 22 22 22 22 22

1τ 264 264 242 242 242 242 242 242 242

C 3.60 3.64 3.66 3.68 3.68 3.68 3.68 3.68 3.67

Similarly, by altering the misconception on the process failure which leads to discontinue of the

production process and resulting to the loss of 765.30 Ringgit for each discontinue made, and we found

that changes made by in this factor on C is not obvious.

Figure 4: Cutting process of lathe



22 Lim Eng Aik

Error AnalysisOur analysis on the trouble-free process time is to approximate the distribution of working hours when

there is no fault on the cutting tools. This is due to the actual process of working hours without failures

are unknown. If we try to use normal distribution to describe this situation, with uncertainty on its

mean value µ , on σ , ultimately, will cause the error for the results. Using the results from Model I to

analyze the errors, as in Table 5, where ∆ represent the relative error, with %1000

0×

−

C

C C (taking C 0

= 4.75). Table 5 shows the mean µ , standard deviation σ of smaller changes that have a minor affect

on the results, so it is reasonable to apply the method that has given in this paper, to approximate the

zero failure process time.

Table 5: Statistical Analysis Results

µ 290.81 290.81 295.91 301.02 306.12 306.12

2σ 48.38 50.92 48.38 48.38 48.38 50.92

C 2.42 2.50 2.36 2.30 2.24 2.32

∆ 0 3.2% 2.5% 5% 7.4% 4.2%

ConclusionBased on the models and replacement cutting tool time-gap through the inspection time-gap traversal

search, we are able to obtain an optimal solution for constant inspection time-gap. With a limited

number of Monte-Carlo simulation, shows the simulation results contain a larger finite precision error.

For Model III , we can only provide a single effective solution, but it did not give the optimal solution.

Thus, we propose an easy to execute scheme to facilitate staff during practices of the inspection

program.

References[1] Thomas, A., and Artiba, A., 2009. “Modeling and control of productive system: concepts and

application”, International Journal of Production Economics Vol. 212 (1), pp. 1-3.

[2] Yeo, S. H., 1995. “A multipass optimization strategy for CNC lathe operations”, International

Journal of Production Economics Vol. 40 (2), pp. 209-218.

[3] Pan, G., Xu, H., Kwan, C. M., Liang, C., Haynes, L., and Geng, Z., 1996. “Modeling and

intelligent chatter control strategies for lathe machine”, Control Engineering Practice 4(12), pp.

1647-1658.

[4] Jan, T. K., Krzysztof, M., 1995. “Modeling cutting process in dynamic stability analysis of

machine tools”, International Journal of Machine Tool and Manufacture 35(4), pp. 535-545.

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