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8/4/2019 Modeling of Automatic Lathe Management System
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.
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;
8/4/2019 Modeling of Automatic Lathe Management System
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
8/4/2019 Modeling of Automatic Lathe Management System
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.
8/4/2019 Modeling of Automatic Lathe Management System
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
8/4/2019 Modeling of Automatic Lathe Management System