Page 1

978- 1- 4673- 1023- 9/ 12/ $31. 00 ©2012 IEEE

U. S. Pasaribu1, a, H. Husniah 2 , B. P. Iskandar 3

1Department of Mathematics, Bandung Institute of Technology, Bandung, Indonesia 40132

2Graduate Program at Department of Industrial Engineering, Bandung Institute of Technology, Bandung, Indonesia 40132

3Department of Industrial Engineering, Bandung Institute of Technology, Bandung, Indonesia 40132

(audjianna@math.itb.ac.id )

Abstract— This paper deals with a stochastic system and

maintenance service contract. We consider a situation where an

agent offers more then one service contract option and a

company as the owner has to select the optimal option. This case

is typically found in a mining industry where the original

equipment manufacturer (OEM) is the only maintenance service

provider. As repair time is related to the revenue of the company,

service contract options need to consider the repair time target.

We consider the repair time target from both the owner and

OEM point of views. As an illustration, we give a numerical

example with weibull failure distribution

Keywords- Maintenance, service contract, non-cooperative

game theory

I.

INTRODUCTION

In mining industry, an availability of heavy equipments e.g.

dump-truck, excavator, tow truck are critical in achieving the

revenue of the company. The equipment deteriorates with

usage and age and finally fails to operate as intended. As the

result, no revenue is generated. High availability of the

equipment is needed for achieving the revenue of the company.

To get this achievement, preventive maintenance (PM) actions

are performed using age based or conditioned based

maintenance – to reduce the likelihood of failure and down

time. Beside that corrective maintenance (CM) actions are

taken after failure occurs, which restores the failed equipment

to the operational state. The maintenance program is aimed at

not only to sustain the performance (e.g. reliability) of the

equipment according to the intended function but also to obtain

optimum business profitability. In a mining industry,

availability of dump trucks is a key measure, which influences

significantly the revenue of a company.

Many companies do in-house maintenance service with

limited skilled labors and maintenance facilities, due to

expensive investment and this affects their capability in

performing a preventive maintenance. As a result, performing a

maintenance service in-house seems to be in-economical

solution, and as alternative way an out-sourcing maintenance

service, either preventive or corrective, could be the better

solution. The benefits of an out-sourcing maintenance service

are two folds: to assure the maximum availability and to reduce

the maintenance cost.

Maintenance service contract has received attention in the

literature such as in [1], [2], [3] and [4]. They formulated

decision problems using as a Stackelberg game theory model to

obtain an optimal cost strategy with the agent as a leader and

consumer as the follower. However, they did not consider any

PM action. Further, [5], [6], [7] and [8], involved a PM policy

in the contract. Another researcher such as [9] studied the

contract that considers the periodic inspection and CM. He also

considered three contract options and the optimal strategy for

selecting the options is obtained which maximizes the expected

profit for both the agent and the owner. All those contracts

considered a penalty based on down time for each failure – i.e.

a penalty cost incurs the agent (or OEM) when the actual down

time to fix the failed equipment is greater than the target value.

In this paper, we develop [8] from the manufacturer’s

perspective and the customer’s perspective, which considers

discrete PM action and the penalty. This purposed PM is more

realistic since the company usually does PM in periodic time.

Similar with [8], the penalty is based on the repair time, if it is

lower than the target repair time the OEM incurs the penalty

cost.

The paper is organized as follows. Section II gives the

methodology which includes model formulation and model

analysis. Section III deals with the result of the solution and

numerical examples for the case where the product has a

Weibull failure distribution. The discussion presents in Section

IV. Finally, a brief conclusion and a discussion for future work

are presented in Section V.

II.

METHODOLOGY

Let L denotes as a life time product and let the OEM and

the consumer have two options.

Option

0

O : In the interval[0, )

in-house PM to undertake a preventive maintenance. If the

product fails, OEM will charge the consumer for the cost of

repair

s

C whenever the failure is repaired by the OEM. There

is no penalty cost to the OEM if the repair time exceeds τ unit

time.

Option

1

O : For a fixed cost of service contract

OEM agrees to repair all the failures with PM and CM along

L , the consumer does an

G

P , the

Stochastic System in Discrete

Preventive Maintenance and Service Contract

Bandung, Indonesia, September 23- 26, 2012

2012 IEEE Conference on Control, Systems and Industrial Informatics (ICCSII)

250

Page 2

( )

r x

L

( )

L

o

mr

( )

L

h

mr

Failure rate

the interval [0, )

reach an operational condition by τ unit time after the time it

failed, the OEM should pay a penalty cost (see Table I below).

L , without any cost. If a fail product could not

TABLE I.

OEM OPTIONS

Option in [0,L)

Option O0 Option O1

PM

In House PM Service Contract

CM Service Contract Service Contract

The consumer must choose the option

}

01

,O O . And the OEM has to determine the optimal cost

structure (service contract cost

C for option

0

O . The values of service contract cost

repair cost

m

C , will be formulated through a non-cooperative

game theory using Nash equilibrium.

*

O taken from the

set{

G

P for option

1

O and repair cost

mG

P and

A. Model Formulation for Product Failures and PM Policy

We use a black-box approach to model product failure.

And it will be modelled by its distribution function. We

consider a dump truck as a repairable product and every

failure is rectified by a minimal repair. With a minimal repair,

the failure rate after repair is the same as that before it fails. It

is assumed that the rectification time is relatively small

compared to its mean time between failures, so that it can be

neglected. As a result, the failure occurs as a Non-

Homogenous Poisson Process (NHPP) with the intensity

function ( )

r x [10].

The PM policy follows [11] by assuming a PM occurs at

discrete time instants 12

,,...,

τ ττ

PM is that it results in a rejuvenation of the item so that it

effectively reduces the age of the item. The reduction in the

age depends on the maintenance effort ( )

maintenance effort is constrained so that 0

corresponds to no PM and M is the upper limit of

maintenance effort. Larger value of m corresponds to greater

maintenance effort. Let

1

−≥

, PM action. See [12] for the concept of

virtual age. If maintenance effort level m is used, the virtual

age is given by

1

,... and 0

j

τ = . The effect of

m used. The

mM

≤

.

≤

0m =

jv− denote the virtual age of the item

after ( 1) , 2

st

jj

1110

110

111

()before the PM with 0(1 )a

( )(m) after the PM for option (1 )b

( )(m) after the PM for option (1 )c

th

jjj

τ

j

τ

hhth

jjhjj

ooth

jjojj

vvjv

vvvjO

vvjO

ττ

δ

δττ

−−−

−−

−−

=+−=

==+−

=+−

⎧

⎪⎨

⎪⎩

with ( )

δ

function of m with (0)

as the PM effort is increased, the effect of aging is reduced.

With mM

=

at every PM action, then the item is restored

back to as good as new after each PM action. If

,1

jj

vj

τ=≥ .

[0,1],0( )

m

( )

m

δ

1

oh

m

δ

δ

δ∈≤<

1

≤ . ( )

δ

)

M

=

m

. This implies that

is a decreasing

= and (0

0

m = , then

We assume that the level of PM effort used is the same

throughout the life. As a result, with PMm, the item’s virtual

age at time x is given by

1110

1111

,,j=1,2,...for option (2 )

a

( )

v x

,,j=1,2,...for option (2 )

b

h

jjjj

o

jjjj

vxxO

vxxO

τττ

τττ

−−−

−−−

⎧

⎪

+ −≤<

=⎨

⎪

⎩

+ −≤<

Since failure are repaired minimally and repair times are

negligible, the rate of occurrence of failures (ROCOF) is given

by

(

=

⎨

⎪

⎩

)

()

1110

1111

,,j=1,2,...for option (3 )a

[ ( )] r v x

,,j=1,2,...for option (3 )b

h

jjjj

o

jjjj

r vxxO

r vxxO

τττ

τττ

−−−

−−−

+−≤<

+−≤<

⎧

⎪

In other words, failures over time occur according to a Non-

stationary Poisson Process with intensity function is given by

the failure rate for the virtual age.

Let

L for option

ROCOF functions with PM effort m given in (4) and (5).

Their ROCOF are defined as follow:

01

,

n n denote the number of PM actions over

O and

1

O respectively. We consider two

[0, )

0

000

0

r v

11

),

1

0

0

(),

( )x 1,..., (4)

(

h

jjjj

h

m

h

nnn

r vxx

rjn

xxL

τ

τ

τ

τ

τ

−−−

+

+

−

−

≤

≤

<

≤

==

⎧⎪⎨

⎪ ⎩

111

0

r v

11

),

1

1

0

( ),

( )x1,..., (5)n

(

o

jjjj

o

m

o

nnn

r vxx

rj

xxL

τ

τ

τ

τ

τ

−−−

+

+

−

−

≤

≤

<

≤

==

⎧⎪⎨

⎪ ⎩

with

,1

j

jj

τ = Δ≥ and

01

,

n n is the largest integer less than

L Δ .

Figure 1. ROCOF for options

10,OO

251

Page 3

B. Maintenance Cost

As in [8], to get high revenue both the OEM and the

customer have to construct maintenance cost for each option.

For option

0

O , we assume that

maintenance cost per unit of time and

repair cost. We denote Cτ as the expected penalty cost of the

OEM if the repair time exceedsτ , and

repair and corrective cost in the interval [0, )

consumer does not buy a service contract, and

pricing of product.

pm

C is the average of

C

is a preventive

m

s

C is the average of

L when the

C denotes the

b

The cost of CM for option

where

m

C is the average cost of each rectification. And for

option

1

O the cost of PM is

C

function of the repair cost Y (r.v with the distribution

function ( )

G y ) and τ . Let

rectify the i-th failure and for option

rectified by minimal repair and the repair time is negligible,

the expected number of failure in the interval[0, )

is given by

1

O can be expressed as

m

C ,

1 pm

n . The penalty cost is a

iY denotes the time needed to

O , since failures are

1

L ,

()

10,

NL

( )

x dx

0

L

o

m r

∫

failure in the interval [0, )

∫

()

max 0,

i

=

∑

. And for option

0

O , the expected number of

L

,

()

00,

NL

is given by

( )

x dx

0

L

h

m r

. The penalty cost in the interval [0, )

L is given

by

{}

10,N

1

L

i

CY

τ

τ−

.

1) Customer’s decision problem

We assume that

the operational condition after the i-th failure and the expected

number of failures in the interval [0, )

option

1

O is

1

N . Consumer profit upon choosing the option

O is (

,

G

O P Cτ

ω

, and the value is given by

∑

∑

iY denotes the time needed to return to

L whenever choose the

1

)

1;

()

{}

()

1

1

1

1

1

;,

max 0,

N

GibG

i

N

i

i

O P CR L

⎢

⎣

YCP

CY

τ

τ

ω

τ

=

=

⎡⎤

⎥

⎦

=−−−+

+−

(6)

Suppose that the number of failures in the interval [0, )

choosing the option

0

O is

choosing the option

0

O P C

ω

∑

L upon

0

;

N . The consumer profit upon

)

00

,

s

is given by O , (

()

0

0000

1

;,

N

sibs

i

O P C R L

⎢

⎣

YCPC N

ω

=

⎡⎤

⎥

⎦

=−−−−

(7)

2) OEM’s decision problem

OEM revenue for option

0

O is given by:

()()

00,

bsmsm

OCCCN CC

π=+−>

(8)

OEM revenue for option

1

O :

1

G

( )=

P - [Penalty cost] - [CM cost] - [PM cost]

O

π

(9)

In our model, by using Nash bargaining solution for both

options

0

O and

1

O , OEM and consumer will negotiate the

pricing of service contract

G

P and the cost of

m

C , respectively.

C. Model Analysis

We assume that OEM and consumer have the same

attitudes to risk, in order to make the solution reach equilibria.

1) Customer’s expected profit

From (6) then the expected profit of the consumer upon

choosing the

1

O option,

(

EO

ω

⎡

⎣

)

1

⎤

⎦, is given by

()()( )

() ( )

11

0

1

0,1

-0,1

bG

EOR L

⎢

⎣

NLG ydy

CPC N

τ

L G ydy

τ

∫

ω

⎡

⎣

∞

∞

⎡⎤

⎥

⎦

⎧

⎨

⎩

⎫

⎬

⎭

=−−−

⎤

⎦

⎡

⎣

⎤

⎦

⎧

⎨

⎩

⎫

⎬

⎭

−+−

⎡

⎣

⎤

⎦

∫

(10)

And from (7), the expected profit of the consumer upon

choosing the option

0

O ,

(

EO

ω

⎡

⎣

)

0

,

⎤

⎦is given by

()

()( )

()

00

0

00

0,1

0,

bs

EOR L

⎢

⎣

NLG ydy

CPC NL

ω

⎡

⎣

∞

⎡⎤

⎥

⎦

⎧

⎨

⎩

⎫

⎬

⎭

⎤ =

⎦

−−−

⎡

⎣

⎤

⎦

−−−

∫

(11)

2) OEM’s expected revenue

The revenue function OEM for option

Since the failure follows the NHPP, then as in [10]

0

O is given in (8).

()

()

()

()

00,

0

0

0,

0,

!

n

NL

NLe

P NLn

n

−

⎡

⎣

⎤

⎦

==

Hence the expected revenue of the OEM whenever the

consumer chose the option

0

O is

given by

()

0

EO

π

⎡

⎣

⎤

⎦, and its value is

()()

()

000,

bsm

EOCCCNL

π

⎡

⎣

⎤ =

⎦

+−

(12)

Where

()

00,NL is given by

()

0

1

00

0

0

∑∫

1

00

1

0,

()() (13)

j

jn

n

L

h

j

h

njn

j

NL

r vxdxr vx dx

τ

ττ

ττ

+

−

=

=

=+−++−

∫

252

Page 4

From (9) then the expected revenue of the OEM,

is given by

(

[

=Penalty cost

G

PEE

−−

()

1

EO

π

⎡

⎣

⎤

⎦,

)

][][]

1

=

CM costPM cost (14)

EO

E

π

⎡

⎣

⎤

⎦

−

The expected number of failures

()

10,NL is given by

()

1

1

11

1

1

100

1

0,()() (15)

j

jn

n

∑∫

L

o

j

h

njn

j

NLr vxdx r vxdx

τ

ττ

ττ

+

−

=

=+−++−

∫

Expected cost of rectification is expressed as

(

mm

ECC NL

=

)

10,

(16)

Expected cost of penalty

()() ( )

g y

τ

−

10,ECC N

τ

Lydy

τ

τ

∫

∞

⎧

⎨

⎩

⎫

⎬

⎭

=

(17)

Using integral by part, we have

() ( )

g y dy

τ−

( )

1yG ydy

τ

∫

τ

∫

∞∞

==−

⎡

⎣

⎤

⎦

(18)

Thus (17) becomes

()( )

10,1 ECC N

τ

LG ydy

τ

τ

∫

∞

=−

⎡

⎣

⎤

⎦

(19)

Expected PM cost is

1

pmpm

ECCn

=

(20)

Then, total expected revenue of the OEM in (14) becomes

()() ( )

)

L

(

11

11

0,1

0,

G

mpm

EOPC N

τ

LG ydy

C NCn

τ

∫

π

⎡

⎣

∞

=−−−

⎤

⎦

⎡

⎣

⎤

⎦

−−

(21)

III.

RESULT

We assume that the consumer uses the product up to L , so

that there would be a negotiation between the consumer and

the manufacturer to determine the value of the service contract

cost,

G

P and the repair cost,C

obtained using the method of Nash equilibrium. In principle

this method can be used whenever there is a negotiation

between the two parties. In the presence of negotiation, for

every option, the consumer and the manufacturer receipt the

same profit if [ ]

EE

ωπ=

. Firstly we will compare the

expected profit of the consumer and manufacturer with the

respect to option

1

O and secondly we obtain

expected revenue of the OEM on option

m

. The optimal solution is

[ ]

*

G

P . Hence the

O becomes

1

()

[]

()( )

()

1

1

0

11

1

2

1

2

[ [ R L 0,{1}]]

[ 0,]

b pmm

EO

NLG ydy

CCn C NL

π

∞

=

−

−−−

⎡

⎣

⎤

⎦

++

∫

(22)

The expected profit of the OEM on option

0

O after we get

*

m

C becomes

()

()( )

()

00

0

00

(0,{1 })

0,.

sb

EOR LNLG ydy

C NL e CP

π

⎡

⎣

∞

⎧

⎨

⎩

⎫

⎬

⎭

⎤ =

⎦

−−−

⎡

⎣

⎤

⎦

−−−

∫

(23)

We consider that the failure distribution follows the Weibull

distribution with 1

α = and

β =

repair time y has a Weibull distribution with

and

0.5

β =

,

1

β <

(decreasing time of repair). And

2

. We also consider that

0.5

α =

( )m

0.04

(1),0

m

m em

η

τ =

at discrete time

δ

−

==

(years), we assume that maintenance is carried out

τ , with

0.33

Δ =

. Table II gives the pricing of

+≥

and integer. For 5L =

(years) and

j

service contract

O respectively. The sensitivity analysis of the maximum

expected revenue is carried out by varying the gradient of

failure rate latter called as degradation PM level from

0.05

η =

to

1.00

η =

. Fig. 2 shows the resulting plot of the

expected revenue [profit] for the OEM [consumer] for

O options based on the results of Table III. Solution of the

numeric is solved using MAPLE 9.5 of the Waterloo.

*

G

P and

*

m

C the repair cost for option

1

O and

0

1

O and

0

TABLE II.

SERVICE CONTRACT AND REPAIR COST

(

G

⎥

8.05 7.17

8.03 7.18

8.01 7.19

7.99 7.21

7.96 7.23

7.94 7.24

7.92 7.26

7.89 7.27

7.87 7.29

7.85 7.30

7.84 7.31

η

*

G

P

)

*

1;,

m

EO P C

ω

⎢

⎣

⎡⎤

⎦

*

m

C

()

*

m

00

;,EO P C

ω

⎢

⎣

⎡⎤

⎥

⎦

0.05

0.15

0.25

0.35

0.45

0.55

0.65

0.75

0.85

0.95

1.00

22.88

22.88

22.88

22.88

22.88

22.88

22.88

22.88

22.88

22.88

22.88

7.25

7.23

7.21

7.18

7.16

7.14

7.15

7.09

7.07

7.05

7.03

The result in Tabel II is given using parameter value as

follow.

Parameter

b

C

R

m

C

s

C

pm

CCτ

0P

Value

(103 $)

5 4 0.1 1 0.05 0.5 2

253

Page 5

Figure 2. Expected OEM revenue for various degradation PM levels based

on Table II

IV. DISCUSSION

Table II presents the decreasing value of expected

revenue [profit] of OEM [consumer] as the degradation PM

level η increases (worsen). In Fig. 2, we show that option

is the optimal option for the consumer. It also shows in Fig. 2,

that option

1

O gives a higher expected value for the OEM

(consumer) compared to option

the degradation preventive maintenance level the greater the

discrepancy between the expected revenues of the different

option.

1

O

0

O . It is clear that the larger

V.

CONCLUSION

In this paper, we have studied a maintenance service

contract for a warranted product with discrite PM. Some

insights are discussed and one can extend to service contract

for product sold with two-dimensional warranty. This topic is

currently under investigation.

ACKNOWLEDGMENT

Part of this work is funded by the DGHE, Ministry of

National Education, the Republic of Indonesia through Hibah

Desentralisasi 2012 .

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194–203, 1988.

η η

Ε[π(Ε[π(Ο Οι ι)] )]

1

O

0

O

254