 Open Access
 Total Downloads : 164
 Authors : Sakon Wongmongkolrit, Bordin Rassameethes, Wisut Supitak
 Paper ID : IJERTV3IS120933
 Volume & Issue : Volume 03, Issue 12 (December 2014)
 Published (First Online): 29122014
 ISSN (Online) : 22780181
 Publisher Name : IJERT
 License: This work is licensed under a Creative Commons Attribution 4.0 International License
Spare Parts Management for Corrective Maintenance of the Complex System under Uncertainty of Failures
Sakon Wongmongkolrit
International Graduate Program Industrial Engineering Faculty of Engineering, Kasetsart University Bangkhen, 10900, Bangkok, Thailand
Bordin Rassameethes
Department of Operations Management
Faculty of Business Administration, Kasetsart University, Bangkhen 10900, Bangkok, Thailand
Wisut Supitak Department of Industrial Engineering
Faculty of Engineering, Kasetsart University Bangkhen, 10900, Bangkok, Thailand
Abstract Spare part is one of the main factors in the maintainability. Therefore, it is very importance to appropriately supply the spares. If spares are acquired more than necessary then it would make the inventory cost, which is a sunk cost incurred. If insufficient supply of spares, it will cause a lack of spare parts for maintenance actions. And this will result in cost of lost opportunity to make profit during machine stopped while waiting for spares. Importantly, the appropriated purchase and storage of parts in inventory will be helpful to save costs. This study is mainly focused and applied the principle of payoff matrix using for the spare parts procurement for a number of appropriations. The main idea is to think of the supply for spare parts on the basis of multiple machines with applying these with payoff matrix principle. This study also takes the opportunity cost when machine stopped production in the absence period of spares to be added into the equation. The principle and concept of this study is based on a balance and tradeoff between the inventory cost of spare parts and the cost of losing in opportunity. In addition, the raised sample of this research was basically supposed under conceptual of general form which is able to apply this principle to all types of machines.
Keywordsspare part, maintenance, inventory, opportunity cost

INTRODUCTION
Spare part is one of the main components for maintenance actions. It can probably be grouped by type of using or unit of measurement, these are 2 types: (1) the use of a piece or discrete unit as such pulley, switch, motor, so on; and, (2) the use of not in a piece or unit of measurement is continuous such as conveyor belt, chain, electrical cable, etc.
This researchmainly focuses on spares of the discrete unit or as a piece. The consideration for the inventory of these partsis applied the payoff matrix principle that proposes of findingthe reorder point and stocking level. And, machine failures are applied the Binomial distribution. In our point of view, we see that distribution is appropriated for this research.
Furthermore, the equation of total cost is also included the opportunity cost which is effectedfrom machine downtime due to the absence spares in inventory into the equation. The total cost is the combination of the cost of lost in opportunity to make the profit during downtime and inventory cost. The tradeoff between two portions is issued in term of effective total cost. This idea is to satisfy the total cost equation of a staph even more than in past research.

RETERATURES REVIEW AND RELEVANT RESEARCHES
There are many researches and articles which are focused on spare parts' inventory management for the corrective maintenance actions. Almost of them mainly apply the EOQ approach such as [1], [2], [3], [4], etc. However, many articles as well have adopted the payoff matrix which are applied to find the reorder point and number of stocking spares such as [5], [6], [7] but there are articles not many like [8], [9] have considered by the cost of losing chance of stopping production in case of the lack of spare parts included in the equation of the total cost structure. And most of the researches in this approach are rather based on the principle of EOQ more than apply the payoff matrix for answers but mostly of them did not consider any of the tradeoff between the cost of losing in opportunity and cost of storage. The overall of cost structure did not fulfill these into the equation as well. As a result, the previous researchesare mainly focused on three main costsas the cost of materials, cost of purchase, and storage costs, primarily in finding out about the inventory of spare parts for maintenance. The result is a lack of completeness in the equation for the total cost structure.
Thus, this research study is conducted on this concept,and principles to fulfill this niche of the previous researches. The equation of total cost for spare parts inventory management for maintenance is more complete and can be used as a study guide for research in the future.

DEFINITION OF VARIABLES
All variables appearing in this paper should be clarified before using in order to easy understand. So, the variables are:
TC is the grand total cost or total cost for all periods. E[TC(n)] is expected cost of sequence n.
TCt is total cost at period t.
ICt is total spare parts cost at period t. UC is spare part unit cost.
Qt is spare parts purchasing at period t. SCt is total ordering cost at period t.
OC is ordering costper one time.
St is policy of ordering at period t; St [0,1] , Integer; by: St= 0not buy, St= 1 buy at period t
HC is holding cost per unit per period.
HCt is total holding cost a period t.
h is holding cost fraction (percentage of unit cost).
is remaining of spares at period t.
LC is the lost cost during machine stoppage per machine per period.
LCt is total lost cost at period t.
PC is production lost cost during spares waiting time.
EC is expedited cost in case of express purchase when there is lack of spares in inventory.
MDT is maintenance downtime.
PR is production rate of each machine per period.
PU is unit cost to produce the product.
e is the multiplication time of express purchase compare with ordinary purchase.
SPt is the number of spare parts shortage at period t.
j is joint probability at sequence j.
j is running number of types of failuresfrom period 1 to period T (or considered period), and j = 1, 2, 3, , k
n is running number of possible sets of solutions, and n
= 1, 2, , total numbers of possible sets of solution.
k is total numbers of types of failures or states of nature, and k = (M+1)T
M is amount of machines in the system.
T is machine life time or considered period.
=1
X is number of failures.

COST STRUCTURE
For this study, already includes the effect of losing cost in opportunity during machine stoppage (in case of spares shortage) into the equation. So the structure of total cost is composed of four subcost facilities as:

Spare parts cost, this is the total cost of materials or unit price multiplied by the number of pieces of spare parts used. So, the cost equation is:
= (1)

Ordering cost,this is the total cost to purchase spare parts. This includes relevant costs such as the cost of transportation, packaging, etc., So the cost equation is:
= (2)

Holding cost,this is the total cost of storing the parts in stock. This includes relevant costs such as the cost of storage facilities, interest, depreciation, etc. So the cost equation is:
= (3)
And: =
= (4)

The lost cost of machine stoppage, this cost is composed of two portions. The opportunity cost from stopping prodction (in case of parts shortage) and cost to purchase spare parts fast. So, the cost equation is:
= +
(5)
By:
=
(6)
And:
=
(7)
In case of shortage of spares,this cost will occur. Suppose parts shortages at periodt of SPtpieces.From equation (6),the lost cost from the lack of spare parts on hand at period t is:
= (8)

Total cost, The total cost includes the cost of sub section 4 of the foregoing. The equation as mentioned, including at periodt is:
= + + + (9) Or: = + + + (10)
And the equation for the total cost of all times from period 1 to considered periodT is:
Xt is number of failure at period t.
=
(11)
For is the number of remaining parts at periodt, this is equal to the number of parts ordered at periodt plus the amount of parts left from the period t – 1 subtract by the number of machine failures at periodt.
Or: =
+
(12)
1

NUMERICAL RESULTS
By: Max[ ,0]when > 0 ; Holding cost will occur.
Assume that the number of machines in the system is 2
Min[ ,0] when < 0 ;Lost cost will occur.
For: <0 ; – =
From equation (11), if machine has probability of failures that is not continuous. The total cost equation needs to be changed to be the total expected cost. And rewriting as a form of the expected value is:
machines and there are 2 considered period, with the following conditions:

Prob. of failures = 0.2

Spare part unit price is 10,000 Baht.

Ordering cost is 1,000 Baht per time.

Holding cost is 1,200 baht per unit per period.

Opportunity cost 500,000 Baht per machine per one period).
Note: 1. The results and numerical calculation are shown in
=1
E =
=1
()
(13)
table 2.
2. Thai currency is Baht and 32.5 Baht is equal to 1
V. JOINT PROBABILITY
From equation (13), joint probabilityj is the probability that the machines will fail according to each type sequence j. These are the format of all types of machines (in the system) fail. Regarding to this study, machines failures are discrete probabilistic which the Binomial distribution, if probability that one machine is broken (probability of failure) which is equal to P. So, the joint probability is used in the case of M machines will fail equal to X machines that is:
= () (14)
In case of each machine is independent. And failures of each period are also independent. Therefore, the joint probability of failures from period 1 to period T is:
1, 2, , = 1 2
US$ (on December 2014).
From tables 2, the lowest total expected cost is 37,361.9 Baht when purchases spares on period 1 equal to 2 units and purchases spares on period 2 equal to 1 unit.
The examples shown in point 1 of Table 2, this is the joint probability of the machine failures on period 1 equal to 2 machines and on period 2 equal to 1 machine, that is equal to
0.013. In point 2 in Table 2, the total expected cost is equal to 6,825 Baht. For the case of buying spares on period 1 equal to 1 unit and period 2 equal to 2 units when machine fail on period 1 equal to 2 machines and on period 2 equal to 1 machines, this total cost is equal to 55,456 Baht.


CONCLUSION AND FUTURE RESEARCH
Application of payoff matrix advantage is to guarantee that the value is the optimal and dependability of 100%, but the payoff matrix for multiple machines and for many considered periods; these will incur the problem of magnitude for the matrix size. By increasing the number of machines
=
=1
( )
(15)
and the considered period will cause of incurring of rapid growth of matrix size. As a result, the payoff matrix is used
Then, the joint probability of machine failures for each type is:
to calculate the answer cannot be done directly. Due to the size of the problem is much more than that calculated by this method directly.
=1
=
( )
(16)
So, the simulation will be alternative in order to reduce the total number of failures statuses or decrease in the
Comparing equations (13) and (16), the sequence n is run
from number 1 to total number of possible sets of solution. The sequence j is run from 1 to (M+1)T or equal to total number for all types of machine failures.
VI. PAYOFF MATRIX
The payoff matrix is also known as decision table. This is thought on the basis of the calculation and find answers to all cases (or complete search), then find the best one for the answer.The purchasing strategy or format of ordered spares which gives the lowest total expected cost will be selected. The equation of the expected cost with multiple machines is:
numbers of states of nature. And to calculate by using the payoff matrix, then it can increase the number of machines and the number of considered periods.
However, this approach also has limitations. Since it cannot reduce the number of possible sets of solution to be so when increasing the number of machines and the number of periods considered up to a certain level, the simulation and calculation with the payoff matrix, it will cause of problems such as calculating the payoff matrix directly. If required, the calculated to supply spares for a system with multiple machines and have a lot of periods to consider. The appropriate method is to create the heuristics used to
=1
1, , =
=1
( , ) (17)
calculate the results. Although, the heuristics can to tackle even the size of the matrix size and can be calculated easily
Where, 1, , is expected cost for all purchasing
patterns or possible sets of solution. And F( , ) is the total cost function (this is already mentioned in section IV).
Note: see table 1.
and quickly for supply spares than the payoff matrix, but the heuristics method has drawbacks. As the results, it does not guarantee that the results are the best or optimal. But the
results from this method, it is close to the best that is good enough to use it in practice.
ACKNOWLEDGMENT
Thank you to Prof. Dr. WisutSupitak who gives the idea for this research. And the Department of Industrial Engineering International Program, Kasetsart University to support and facilitate for this research as well.
REFERENCES

S. Wongmongkolrit and B. Rassameethes, The Modification of EOQ Model under the Spare Parts Discrete Demand: A Case Study of Slow Moving Items,Proceeding of World Congress of Engineering and Computer Science (WCECS 2011),vol.2, pp. 11791184, 2011.

A.T. De Almeida,Multicriteria decision making on maintenance: Spares and contracts planning, European Journal of Operational Research,vol.129, pp. 23524, 2001.

W. Supitak, Determination of Safety Stock Inventory for Different Demand Patterns, Journal of Eastern Asia University, vol. 5 (1), pp. 2430, 2011.

A.Eaves, B.Kingsman,Forecasting for the ordering and stockholding of spare parts, Journal of the Operational Research Society, vol.55, pp.431437, 2004.

R.J. Tersine, Principle of Inventory and Materials Management 4th ed, PTR Prentice Hall, Englewood Cliff, New Jersey, USA, 1994.

P.R. Sharma and S. Shukla,Using Matrix to Solving the Probabilistic Inventory Models (Demand Model), International Archives of Applied Sciences and Technoogy,vol.4(3),pp. 0915, 2013.

M. Vaziri and M. Sodhi,Application of game theory on Inventory level decision making, International Journal of Business and Economics Research,vol.3(6), pp. 211219, 2014.

M. Schroter and T. Spengler,Designing control management systems for parts recovery and spare parts management in the final phase within closedloop supply chains, International Journal of Integrated Supply Management, vol.1, pp. 158179, 2004.

G.Harry L.Pintelon and A.Seidmann,Product lot sizing with machine breakdowns, Management Science,vol.38(1), pp. 104123, 1992.
TABLE I. PAYOFF MATRIX
Prob. of failure T1 T2 No. of failures (Xt1, , Xtj ) Tt (State of nature) TT 
1 2 
j 
K 

Parts Stocking (Qi1,., QiT) 
0 1 . M 
0 
1 . 
M 
.. 
0 
1 .M 

or Purchasing Strategy 
0 0 . 0 : 0 0 
1 0 
1 . : 0 
1 0 
.. .. 
M : M 
M .M : M .M 
Expected Cost E(Qi1,., QiT) 

: 
: 
: 
: 

T1 T2 . TT 
0 0 
0 
0 
0 
.. 
M 
M .M 

Q11 Q12 Q1T 
a11 
a12 
a13 
. 
a1j 
. 
a1K 
E(Q11,, Q1T) 

Q21 Q22 Q2T 
a21 
a22 
a23 
. 
a2j 
. 
a2K 
E(Q21,, Q2T) 

: : : 
: 
: 
: 
: 
: 
: 

Qi1 Qi2 QiT 
ai1 
ai2 
ai3 
. 
aij 
. 
aiK 
E(Qi1,, QiT) 

: : : 
: 
: 
: 
: 
: 
: 

QB1 QB2 QBT 
aB1 
aB2 
aB3 
. 
aBj 
. 
aBK 
E(QB1,, QBT) 
TABLE II. NUMERICAL RESULT
Prob. of failure 
0.41 
0.205 
0.026 
0.205 
0.1024 
0.013 
0.026 
0.013 
0.002 
1.00 

Possible sets of solution Period 
Period 
States of nature 1 
Total expected cost 

1 2 
0 0 
1 0 
2 0 
0 1 
1 1 
2 1 
0 2 
1 2 
2 2 

1 
2 
Row No. 
Expected cost 2 

0 
0 
1 
0 
1E+05 
25600 
1E+05 
102400 
19200 
25600 
19200 
3200 
400,000.0 

0 
1 
2 
4997 
1E+05 
25912 
2253 
52326 
12941 
13082 
12941 
2418 
231,768.0 

0 
2 
3 
9585 
1E+05 
26199 
4547 
53473 
13084 
537.6 
6669 
1634 
222,920.0 

1 
0 
4 
5489 
2253 
13082 
2499 
52326 
12941 
13112 
12941 
2418 
117,059.5 

1 
1 
5 
10486 
4751 
13394 
4997 
2252.8 
6682 
593.9 
6682 
1635 
51,473.3 

1 
2 
6 
15073 
7045 
13681 
7291 
3399.7 
6825 
880.6 
409.6 
851.2 
55,456.0 

2 
0 
7 
10568 
4792 
537.6 
5038 
2273.3 
6669 
599 
6684 
1634 
38,794.6 

2 
1 
8 
15565 
7291 
849.9 
7537 
3522.6 
409.6 
911.4 
425 
851.2 
37,361.9 

2 
2 
9 
20152 
9585 
1137 
9830 
4669.4 
553 
1198 
568.3 
67.2 
47,760.0 

3 
0 
10 
15647 
7332 
855 
7578 
3543 
412.2 
916.5 
427.5 
851.5 
37,561.9 

3 
1 
11 
20644 
9830 
1167 
10076 
4792.3 
568.3 
1229 
583.7 
69.12 
48,960.0 

4 
0 
12 
20726 
9871 
1172 
10117 
4812.8 
570.9 
1234 
586.2 
69.44 
49,160.0 