Stochastic Modeling of a Concrete Mixture Plant with ... · PDF fileof preventive maintenance...

13

Available at

http://pvamu.edu/aam Appl. Appl. Math.

ISSN: 1932-9466

Vol. 9, Issue 1 (June 2014), pp. 13-27

Applications and Applied

Mathematics:

An International Journal

(AAM)

Stochastic Modeling of a Concrete Mixture Plant with

Preventive Maintenance

Ashish Kumar and Monika Saini Department of Mathematics

Manipal University Jaipur

Jaipur-303007, Raj (India)

S.C. Malik Department of Statistics

M.D. University

Rohtak-124001, Haryana (India)

[email protected]

Received: November 26, 2013; Accepted: February 6, 2014

Abstract

In this paper, a stochastic model for concrete mixture plant with Preventive Maintenance (PM) is

analyzed in detail by using a supplementary variable technique. In a concrete mixture plant eight

subsystems are arranged in a series. The system goes under PM after a maximum operation time

and work as new after PM. The time to failure of each subsystem follows a negative exponential

distribution while PM and repair time distributions are taken as arbitrary. A sufficient repair

facility is provided to the system for conducting PM and repair of the system. Repair,

maintenance and switch devices are perfect. All random variables are statistically independent.

Various measures of system effectiveness such as reliability, mean time to system failure

(MTSF), are derived using a supplementary variable technique. The numerical results for

reliability and availability are obtained for particular values of various parameters and costs.

Keywords: Concrete Mixture Plant; Reliability; Availability; Preventive Maintenance and

Supplementary Variable Technique

MSC 2010 No.: 90B25, 60K10

http://pvamu.edu/aam

mailto:[email protected]

14 Gurju Awgichew et. al

1. Introduction

Concrete mixture plants are widely used to produce various kinds of concrete including quaking

concrete and hard concrete, suitable for large or medium scale building works, road and bridge

works and precast concrete plants, etc. Basically, such plants are designed for the production of

all types of concrete, mixed cements, cold regenerations and inertisations of materials mixed

with resin additives. Due to the complexity of modern concrete mixture systems, which involve

high risks, the concept of reliability has become a very important factor in the overall system

design. While dealing with reliability-based design of machines and structures, we can study the

relative importance of mechanical and structural failures from the point of view of loss of human

lives. Reliability analysis of such a system helps us to obtain the necessary information about the

control of various parameters. Arekar et al. (2012), Kharoufeh et al. (2010), Proctor and Singh

(1976), Shakuntla et al. (2011), Malik (2008) and Uematsu and Nishida (1987) have analyzed

single-unit systems under a common assumption that the unit works continuously till failure

without undergoing PM.

The continued operation of the systems may reduce performance and reliability of the system.

Therefore, PM of the unit is necessary after a specific period of time at any stage of operation to

improve the reliability and availability of the system because the cost to repair the system after

its failure is greater than the cost of maintaining the system before its failure. Thus, the method

of preventive maintenance can be adopted to improve the reliability and profit of system. The

concept of preventive maintenance has been used by many researchers such as Malik and Nandal

(2010), Kumar et al. (2012) and Kumar and Malik (2012) while analyzing the redundant systems

with maximum operation time. It is also interesting to note that not much work related to the

reliability modeling of the concrete mixture plant subject to preventive maintenance has been

reported so far in the literature of reliability.

Most of the authors discussed the system possessing Markovian properties. The system having

non-Markovian property can be converted into a system having Markovian nature by introducing

a new variable called a supplementary variable. Initially, Cox (1955) used the supplementary

variable in analyzing a non-Markovian system and presented a systematic solution of reliability

and availability of that system using the supplementary variable technique. Gaver (1963) studied

a parallel redundant system with constant failure and arbitrary repair rates. Since then several

authors have studied the reliability of the various systems using supplementary variable

technique. Singh and Dayal (1991) used supplementary variable technique for problem

formulation. Alfa and Rao (2000) discussed the supplementary variable technique in stochastic

models. The concrete mixture plant with preventive maintenance has not been discussed so far

even though it plays an important role in our daily life and development of the infrastructure.

The assumption of constant failure, maintenance and repair rates may not be practical in any

industry. Keeping this in view, in the present study, we have considered eight-subsystems of the

concrete mixture plant with constant failure and arbitrary repair rates of the subsystems and

discussed the reliability modeling of concrete mixture plant with preventive maintenance using

supplementary variable technique. An attempt has also been made to discuss the availability of

this plant with respect to different failure and repair rates.

AAM: Intern. J., Vol. 9, Issue 1 (June 2014) 15

The paper has been organized as follows: Section 1 is introductory in nature. In Section 2, a

summary of the system and various notations of the subsystems are presented. The basic

assumptions, on which the present analysis is based, are also discussed in Section 2. The

mathematical formulation and solution of the differential-difference equation of Concrete

mixture plant developed using the supplementary variable technique, (assuming constant failure

and variable repair rates) presented in Section 3. Certain conclusions drawn from this analysis

are also discussed in the Section 4.

2. System Description, Notations and Assumptions

Concrete mixing plants are widely used to produce various kinds of concrete including quaking

concrete and hard concrete suitable for large or medium scale building works, road and bridge

works and precast concrete plants, etc. A concrete plant, also known as a batch plant, is a device

that combines various ingredients to form concrete. Some of these inputs include sand,

water, aggregate, fly ash, potash and cement. A concrete plant can have a variety of parts such as

a dosing system, mixer feeding belt conveyor, main chassis superstructure, mixing system,

cement silo, screw conveyor, electrical control system and insulated control cabinet.

In this paper, we consider concrete mixing plant consisting of eight sub-systems namely, a

dosing system, mixer feeding belt conveyor, main chassis superstructure, mixing system, cement

silo, screw conveyor, electrical control system and insulated control cabinet. The complete

description of the systems and their notations required in the mathematical formulation are as

follows:

2.1. System Description

2.1.1. Sub-system A (Dosing system)

It is a storehouse of the aggregates which are controlled by cylinders of two material discharging

hoppers. The size of the two doors is different. The size of the material doors can be adjusted.

2.1.2. Sub-system B (Mixer feeding belt conveyor)

It is a double surface ladder fence. It is jointing with main tower in such a way that it avoids the

shaking of the main tower making the exact weighing.

2.1.3. Sub-system C (Main chassis superstructure)

It is built from color steel sandwich board. Its main functioning is heat preservation and heat

insulation. It is equipped with dust catcher to avoid pollution.

2.1.4. Sub-system D (Mixing system)

In a mixing system various materials such as cement, sand or gravel, and water are combines in a

homogenous manner to form concrete. A typical concrete mixer uses a revolving drum to mix

the components.

2.1.5. Sub-system E (Cement silo)

It is used to store dry, bulk cement.

http://en.wikipedia.org/wiki/Concrete

http://en.wikipedia.org/wiki/Sand

http://en.wikipedia.org/wiki/Water

http://en.wikipedia.org/wiki/Construction_aggregate

http://en.wikipedia.org/wiki/Fly_ash

http://en.wikipedia.org/wiki/Potash

http://en.wikipedia.org/wiki/Cement

http://en.wikipedia.org/wiki/Cement

http://en.wikipedia.org/wiki/Water

http://en.wikipedia.org/wiki/Concrete


2.1.6. Sub-system F (Screw conveyor)

It is a duct along which material is conveyed by the rotational action of a spiral vane which lies

along the length of the duct.

2.1.7. Sub-system G (Electrical control system)

It is a collection of electronic devices that manage commands, directs or regulates the behavior

of the whole concrete plant.

2.1.8. Sub-system H (Insulated control cabinet)

In this cabin, electronic devices, fittings, sockets and switches exist.

2.2. Notations

A, B, C, D, E, F, G and H, indicate that the sub-system is working in full

capacity.

a, b, c, d, e, f, g and h, indicate the failed state of the sub-system.

αi, denotes the constant failure rate of the units, where

i = 1,2, …, 8.

αm, denotes the constant transition rate of the system.

( ) ( ) denote the repair rate of the unit and probability density

function, respectively, for the elapsed repair time ‘x’, where

i = 1, 2, …, 8.

( ) denotes the probability that at time t the system is

in good state.

( ) denotes the probability that at time t the system is in failed

state the elapsed repair time lies in the interval ( , ),x x

where 1,2, ,8.i

( ) denotes the probability that at time t the system is under PM,

the elapsed PM time is ‘y’.

( ) ( ) denote the preventive maintenance rate of the unit and

probability density function, for the elapsed maintenance

time ‘y’, respectively.

( ) Laplace transform of ( )


( ) ( ) [ ∫

( ) ]

( ) ( ) [ ∫

( ) ]

,

denotes the definite integral from 0 to x.

Figure 1: System Formulation

2.3. Assumptions

(i) Repair and failure rates are independent of each other and their unit is taken as per day.

(ii) Failure and repair rates of the subsystems are taken respectively as constant and variable.

(iii) Performance wise, a repaired unit is as good as new one for a specified duration.

(iv) Sufficient repair facilities are provided.

(v) Service of the subsystem includes repair and/or replacement.

(vi) Switch devices, repairs and preventive maintenances are perfect.

(vii) The distribution of preventive maintenance is considered as arbitrary.

Dosing system

Mixer feeding belt conveyor

Main chassis superstructure

Mixing system

Cement silo

Screw conveyor

Electrical control system

Insulated control cabinet


3. Formulation and Solution of Mathematical Model

By probability considerations and continuity arguments, we obtain the following difference-

differential equations governing the behavior of the system:

8 8

0 91 1 0 0

( ) [ ( , ) ( ) ] ( , ) ( ) .i m i i mi i

P t P x t x dx P y t y dyt

(1)

( ) ( , ) 0, where 1,2,...,8.i ix P x t i

t x

(2)

9( ) ( , ) 0.mx P y tt y

(3)

The boundary and initial conditions to be satisfied are given below

Boundary Conditions:

0(0, ) ( ), where 1,2,...,8,i iP t P t i (4)

0(0, ) ( ).m mP t P t (5)

Initial Conditions:

(0) 1, when 0,iP i (6)

(0) 0, when 0.iP i

By taking LT of equations (1)-(5) and using in (6), we get

8 8

0 9 91 1 0 0

( ) 1 ( , ) ( ) ( , ) ( ) .i m i ii i

s P s P x s x dx P y s y dy

(7)

( ) ( , ) 0, where 1,2,...,8i is x P x s i

x

(8)

9( ) ( , ) 0.ms y P y s

y

(9)

0(0, ) ( ), where 1,2,...,8.i iP s P s i

(10)

0(0, ) ( ).m mP s P s

(11)

Now, integrating equation (8) and further using in equation (10), we get


0

[ ( ) ]

( , ) (0, ) , where 1,2,...,8.

x

isx x dx

i iP x s P s e i

(12)

Integrating equation (9) and further using equation (11), we get

9

0

[ ( ) ]

( , ) (0, ) .

y

sy x dy

m mP y s P s e

(13)

By using equations (12-13) in equation (7), we get

8 8

0 0 0 91 1

( ) 1 [ ( ) ( )] ( ) ( ).i m i i mi i

s P s P s S s P s S s

(14)

8

01

(1 ( )) ( ) 0 .i ii

s S s P s

(15)

9 0[ (1 ( ))] ( ) 0.ms S s P s

(16)

0

1( ) ,

( )P s

T s

(17)

where

8

91

( ) (1 ( )) (1 ( )) .i i mi

T s s S s S s

(18)

Now, the Laplace Transformation of the probability that the system is in the failed state is given

by

1 11 1 1 0 1

0

1 ( ) ( )( ) ( , ) ( ) ,

( )

S s A sP s P x s dx P s

s T s

(19)

where

11

1 ( )( ) .

S sA s

s

Similarly

0 10

1 ( ) ( )( ) ( , ) ( ) .

( )

i ii i i


s T s

(20)

where

1 ( )

( ) , 2,3,4,5,6,7,8.ii

S sA s i

s

90

0

1 ( ) ( )( ) ( , ) ( ) .

( )

mm m m m


s T s

(21)

where

1 ( )

( ) .mm

S sA s

s


It is worth noting that

8

0

1( ) ( ) .i m

iP s P s

s

(22)

Evaluation of Laplace transforms of up and down state probabilities

The Laplace transforms of the probabilities that the system is in up (i.e., good) and down (i.e.,

failed) state at time ‘‘t’’ are as follows

0

8

81

1

1( ) ( ) ,

( )

( ) ( )

( ) ( ) ( ) .( )

up

i mi

down i mi

P s P sT s

A s A s

P s P s P sT s

(23)

Steady-State Probabilities

Using Abel’s Lemma in Laplace transforms, viz.

lim lim0

( ) ( ) ( )s t

sZ s Z t Z say

Provided the limit on the right hand side exists, the following time independent probabilities

have been obtained.

8 ' '9

1

8 '

18 ' '

91

1,

[1 (0) (0)]

(0)

.

[1 (0) (0)]

up

i i mi

i ii

down

i i mi

P

S S

S

P

S S

(24)

Reliability Indices

In order to obtain system reliability, consider repair rates (i.e., ( ) ) equal to zero. Using the

method similar to that in section 2, the differential–difference equations are:

8

01

( ) 0.i mi

P tt

(25)

Theorem 1.

The reliability of the system is given by


8

1( )

( ) .i m

it

R t e

(26)

Proof:

The proof of the Theorem 1 is given in the appendix.

Corollary 1.

The mean time to system failure (MTSF) is:

8

1

1.

i mi

MTSF

(27)

Proof:

Calculating ∫ ( )

implies the result ‘*’ given in the appendix.

Special Case (Availability)

When repair rates follows exponential time distribution. Setting 9 ( ) m

m

S ss

and

( ) ,ii

i

S ss

where i , i=1, 2, 3, ..., 8, are constant repair rates. Putting these values in equation

(17), we get

8

1

8 8 8 8

1 1 , 1 , 1

( )( )

.

( ) ( ) ( ) ( ) ( )

i mi

m i j m i m ii j i j i i m i

s s

s s s s s s s s

(28)

4. Numerical Analysis

Table-1: Effect of failure rate ( 1 ) on Reliability (R(t))

Time α2 α3 α4 α5 α6 α7 α8 αm R(t) for

α1=.001

R(t) for

α1=.005

R(t) for

α1=.05

1 0.002 0.003 0.004 0.05 0.03 0.0001 0.0002 .04 0.877832 0.874328 0.835855

2 0.002 0.003 0.004 0.05 0.03 0.0001 0.0002 .04 0.770589 0.764449 0.698654

3 0.002 0.003 0.004 0.05 0.03 0.0001 0.0002 .04 0.676448 0.668379 0.583973

4 0.002 0.003 0.004 0.05 0.03 0.0001 0.0002 .04 0.593808 0.584382 0.488117

5 0.002 0.003 0.004 0.05 0.03 0.0001 0.0002 .04 0.521263 0.510942 0.407995

6 0.002 0.003 0.004 0.05 0.03 0.0001 0.0002 .04 0.457582 0.44673 0.341025

7 0.002 0.003 0.004 0.05 0.03 0.0001 0.0002 .04 0.40168 0.390589 0.285047

8 0.002 0.003 0.004 0.05 0.03 0.0001 0.0002 .04 0.352607 0.341503 0.238258

9 0.002 0.003 0.004 0.05 0.03 0.0001 0.0002 .04 0.30953 0.298585 0.199149

10 0.002 0.003 0.004 0.05 0.03 0.0001 0.0002 .04 0.271715 0.261061 0.16646


Table 4. Effect of failure rate ( 4 ) on reliability (R(t))

Table 2. Effect of failure rate ( 2 ) on Reliability (R(t))


α2=.002

R(t) for

α2=.008

R(t) for

α2=.08

1 0.001 0.003 0.004 0.05 0.03 0.0001 0.0002 .04 0.877832 0.872581 0.811963

2 0.001 0.003 0.004 0.05 0.03 0.0001 0.0002 .04 0.770589 0.761397 0.659285

3 0.001 0.003 0.004 0.05 0.03 0.0001 0.0002 .04 0.676448 0.664381 0.535315

4 0.001 0.003 0.004 0.05 0.03 0.0001 0.0002 .04 0.593808 0.579726 0.434656

5 0.001 0.003 0.004 0.05 0.03 0.0001 0.0002 .04 0.521263 0.505858 0.352925

6 0.001 0.003 0.004 0.05 0.03 0.0001 0.0002 .04 0.457582 0.441402 0.286562

7 0.001 0.003 0.004 0.05 0.03 0.0001 0.0002 .04 0.40168 0.385159 0.232678

8 0.001 0.003 0.004 0.05 0.03 0.0001 0.0002 .04 0.352607 0.336082 0.188926

9 0.001 0.003 0.004 0.05 0.03 0.0001 0.0002 .04 0.30953 0.293259 0.153401

10 0.001 0.003 0.004 0.05 0.03 0.0001 0.0002 .04 0.271715 0.255892 0.124556

Table 3. Effect of failure rate ( 3 ) on Reliability (R(t))


α3=.003

R(t) for

α3=.009

R(t) for

α3=.06

1 0.001 0.002 0.004 0.05 0.03 0.0001 0.0002 .04 0.877832 0.872581 0.829195

2 0.001 0.002 0.004 0.05 0.03 0.0001 0.0002 .04 0.770589 0.761397 0.687564

3 0.001 0.002 0.004 0.05 0.03 0.0001 0.0002 .04 0.676448 0.664381 0.570125

4 0.001 0.002 0.004 0.05 0.03 0.0001 0.0002 .04 0.593808 0.579726 0.472745

5 0.001 0.002 0.004 0.05 0.03 0.0001 0.0002 .04 0.521263 0.505858 0.391997

6 0.001 0.002 0.004 0.05 0.03 0.0001 0.0002 .04 0.457582 0.441402 0.325042

7 0.001 0.002 0.004 0.05 0.03 0.0001 0.0002 .04 0.40168 0.385159 0.269523

8 0.001 0.002 0.004 0.05 0.03 0.0001 0.0002 .04 0.352607 0.336082 0.223487

9 0.001 0.002 0.004 0.05 0.03 0.0001 0.0002 .04 0.30953 0.293259 0.185315

10 0.001 0.002 0.004 0.05 0.03 0.0001 0.0002 .04 0.271715 0.255892 0.153662

Time α1 α2 α3 α5 α6 α7 α8 α m R(t) for

α4=.004

R(t) for

α4=.007

R(t) for

α4=.07

1 0.001 0.002 0.003 0.05 0.03 0.0001 0.0002 .04 0.877832 0.875202 0.821766

2 0.001 0.002 0.003 0.05 0.03 0.0001 0.0002 .04 0.770589 0.765979 0.675299

3 0.001 0.002 0.003 0.05 0.03 0.0001 0.0002 .04 0.676448 0.670387 0.554937

4 0.001 0.002 0.003 0.05 0.03 0.0001 0.0002 .04 0.593808 0.586724 0.456028

5 0.001 0.002 0.003 0.05 0.03 0.0001 0.0002 .04 0.521263 0.513503 0.374749

6 0.001 0.002 0.003 0.05 0.03 0.0001 0.0002 .04 0.457582 0.449419 0.307955

7 0.001 0.002 0.003 0.05 0.03 0.0001 0.0002 .04 0.40168 0.393332 0.253067

8 0.001 0.002 0.003 0.05 0.03 0.0001 0.0002 .04 0.352607 0.344246 0.207962

9 0.001 0.002 0.003 0.05 0.03 0.0001 0.0002 .04 0.30953 0.301285 0.170896

10 0.001 0.002 0.003 0.05 0.03 0.0001 0.0002 .04 0.271715 0.263685 0.140436


T Table 5. Effect of failure rate ( 5 ) on reliability (R(t))


α5=.05

R(t) for

α5=.09

R(t) for

α5=.15

1 0.001 0.002 0.003 0.004 0.03 0.0001 0.0002 .04 0.877832 0.843412 0.794295

2 0.001 0.002 0.003 0.004 0.03 0.0001 0.0002 .04 0.770589 0.711343 0.630905

3 0.001 0.002 0.003 0.004 0.03 0.0001 0.0002 .04 0.676448 0.599955 0.501125

4 0.001 0.002 0.003 0.004 0.03 0.0001 0.0002 .04 0.593808 0.506009 0.398041

5 0.001 0.002 0.003 0.004 0.03 0.0001 0.0002 .04 0.521263 0.426774 0.316162

6 0.001 0.002 0.003 0.004 0.03 0.0001 0.0002 .04 0.457582 0.359946 0.251126

7 0.001 0.002 0.003 0.004 0.03 0.0001 0.0002 .04 0.40168 0.303583 0.199468

8 0.001 0.002 0.003 0.004 0.03 0.0001 0.0002 .04 0.352607 0.256046 0.158437

9 0.001 0.002 0.003 0.004 0.03 0.0001 0.0002 .04 0.30953 0.215952 0.125846

10 0.001 0.002 0.003 0.004 0.03 0.0001 0.0002 .04 0.271715 0.182136 0.099959



α6=.03

R(t) for

α6=.093

R(t) for

α6=.12

1 0.001 0.002 0.003 0.004 0.05 0.0001 0.0002 .04 0.877832 0.824235 0.802278

2 0.001 0.002 0.003 0.004 0.05 0.0001 0.0002 .04 0.770589 0.679363 0.64365

3 0.001 0.002 0.003 0.004 0.05 0.0001 0.0002 .04 0.676448 0.559954 0.516386

4 0.001 0.002 0.003 0.004 0.05 0.0001 0.0002 .04 0.593808 0.461534 0.414285

5 0.001 0.002 0.003 0.004 0.05 0.0001 0.0002 .04 0.521263 0.380412 0.332372

6 0.001 0.002 0.003 0.004 0.05 0.0001 0.0002 .04 0.457582 0.313549 0.266655

7 0.001 0.002 0.003 0.004 0.05 0.0001 0.0002 .04 0.40168 0.258438 0.213931

8 0.001 0.002 0.003 0.004 0.05 0.0001 0.0002 .04 0.352607 0.213013 0.171632

9 0.001 0.002 0.003 0.004 0.05 0.0001 0.0002 .04 0.30953 0.175573 0.137697

10 0.001 0.002 0.003 0.004 0.05 0.0001 0.0002 .04 0.271715 0.144713 0.110471



α7=.0001

R(t) for

α7=.003

R(t) for

α7=.012

1 0.001 0.002 0.003 0.004 0.05 0.03 0.0002 .04 0.877832 0.87529 0.867448

2 0.001 0.002 0.003 0.004 0.05 0.03 0.0002 .04 0.770589 0.766133 0.752466

3 0.001 0.002 0.003 0.004 0.05 0.03 0.0002 .04 0.676448 0.670588 0.652725

4 0.001 0.002 0.003 0.004 0.05 0.03 0.0002 .04 0.593808 0.586959 0.566204

5 0.001 0.002 0.003 0.004 0.05 0.03 0.0002 .04 0.521263 0.51376 0.491153

6 0.001 0.002 0.003 0.004 0.05 0.03 0.0002 .04 0.457582 0.449689 0.426049

7 0.001 0.002 0.003 0.004 0.05 0.03 0.0002 .04 0.40168 0.393608 0.369576

8 0.001 0.002 0.003 0.004 0.05 0.03 0.0002 .04 0.352607 0.344521 0.320588

9 0.001 0.002 0.003 0.004 0.05 0.03 0.0002 .04 0.30953 0.301556 0.278093

10 0.001 0.002 0.003 0.004 0.05 0.03 0.0002 .04 0.271715 0.263949 0.241231




α8=.0002

R(t) for

α8=.004

R(t) for

α8=.122

1 0.001 0.002 0.003 0.004 0.05 0.03 0.0001 .04 0.877832 0.874503 0.777167

2 0.001 0.002 0.003 0.004 0.05 0.03 0.0001 .04 0.770589 0.764755 0.603989

3 0.001 0.002 0.003 0.004 0.05 0.03 0.0001 .04 0.676448 0.66878 0.4694

4 0.001 0.002 0.003 0.004 0.05 0.03 0.0001 .04 0.593808 0.58485 0.364802

5 0.001 0.002 0.003 0.004 0.05 0.03 0.0001 .04 0.521263 0.511453 0.283512

6 0.001 0.002 0.003 0.004 0.05 0.03 0.0001 .04 0.457582 0.447267 0.220336

7 0.001 0.002 0.003 0.004 0.05 0.03 0.0001 .04 0.40168 0.391136 0.171238

8 0.001 0.002 0.003 0.004 0.05 0.03 0.0001 .04 0.352607 0.342049 0.133081

9 0.001 0.002 0.003 0.004 0.05 0.03 0.0001 .04 0.30953 0.299123 0.103426

10 0.001 0.002 0.003 0.004 0.05 0.03 0.0001 .04 0.271715 0.261584 0.080379

T Table 9. Effect of Transition rate ( m ) on reliability (R(t))

Time α1 α2 α3 α4 α5 α6 α7 α8 R(t) for

αm=.04

R(t) for

αm=.12

R(t) for

αm=.32

1 0.001 0.002 0.003 0.004 0.05 0.03 0.0001 .0002 0.877832 0.810341 0.663451

2 0.001 0.002 0.003 0.004 0.05 0.03 0.0001 .0002 0.770589 0.656653 0.440167

3 0.001 0.002 0.003 0.004 0.05 0.03 0.0001 .0002 0.676448 0.532113 0.29203

4 0.001 0.002 0.003 0.004 0.05 0.03 0.0001 .0002 0.593808 0.431193 0.193747

5 0.001 0.002 0.003 0.004 0.05 0.03 0.0001 .0002 0.521263 0.349413 0.128542

6 0.001 0.002 0.003 0.004 0.05 0.03 0.0001 .0002 0.457582 0.283144 0.085281

7 0.001 0.002 0.003 0.004 0.05 0.03 0.0001 .0002 0.40168 0.229443 0.05658

8 0.001 0.002 0.003 0.004 0.05 0.03 0.0001 .0002 0.352607 0.185927 0.037538

9 0.001 0.002 0.003 0.004 0.05 0.03 0.0001 .0002 0.30953 0.150664 0.024905

10 0.001 0.002 0.003 0.004 0.05 0.03 0.0001 .0002 0.271715 0.12209 0.016523


Table-10: Availability of Concrete Mixture Plant w.r.t. failure rate (α1).

Set 1: α2=.002, α3=.003, α4=.004, α5=.05, α6=.03, α7=.0001, α8=.0002, αm=.04, β1=.05, β2=.9, β3=.3, β4=.59,

β5=.9, β6=.98, β7=.7, β8=.9, βm=1.2

α1

Set 1

In set 1 Replace α2=.002

by α2=.07


by α3=.05


by α2=.04

In set 1 Replace

α5=.05 by α5=.5

In set 1 Replace

α6=.03 by α6=.3

In set 1 Replace

α7=.0001 by

α7=.005

In set 1 Replace

α8=.0002 by α8=.02

In set 1 Replace

αm=.04 by αm=.19

0.001 0.876526 0.822082 0.770692 0.832026 0.609434 0.706026 0.87118 0.859943 0.789972 0.002 0.874992 0.820733 0.769506 0.830644 0.608692 0.705031 0.869665 0.858466 0.788726 0.003 0.873463 0.819388 0.768324 0.829266 0.607951 0.704038 0.868155 0.856995 0.787483 0.004 0.87194 0.818047 0.767145 0.827893 0.607213 0.703048 0.86665 0.855529 0.786245 0.005 0.870422 0.816711 0.76597 0.826525 0.606477 0.702061 0.865151 0.854067 0.785011 0.006 0.868909 0.815379 0.764798 0.825161 0.605742 0.701076 0.863656 0.852611 0.78378 0.007 0.867402 0.814051 0.76363 0.823801 0.605009 0.700095 0.862167 0.851159 0.782553 0.008 0.8659 0.812728 0.762465 0.822446 0.604278 0.699116 0.860683 0.849713 0.781331 0.009 0.864403 0.811409 0.761305 0.821096 0.603548 0.69814 0.859204 0.848271 0.780111

0.01 0.862911 0.810095 0.760147 0.819749 0.602821 0.697166 0.85773 0.846835 0.778896

Table-11: Availability of Concrete Mixture Plant w.r.t. repair rate (β1).

Set 2: α1=.001,α2=.002, α3=.003, α4=.004, α5=.05, α6=.03, α7=.0001, α8=.0002, αm=.04, β2=.9, β3=.3, β4=.59,

β5=.9, β6=.98, β7=.7, β8=.9, βm=1.2

β1

Set 2

In set 2 Replace

β2=.9 by β2=1.9

In set 2 Replace

β3=.3 by β3=1.3

In set 2 Replace

β4=.59 by β4=1.42

In set 2 Replace

β5=.9 by β5=2.5

In set 2 Replace

β6=.98 by β6=2.3

In set 2 Replace

β7=.7 by β7=1.7

In set 2 Replace

β8=.9 by β8=1.9

In set 2 Replace

βm=1.2 by βm=2.2

0.01 0.870422 0.871309 0.876289 0.873435 0.89822 0.883939 0.870486 0.870511 0.882055 0.02 0.874227 0.875121 0.880145 0.877266 0.902273 0.887863 0.874291 0.874316 0.885962 0.03 0.875502 0.8764 0.881438 0.87855 0.903631 0.889179 0.875567 0.875592 0.887272 0.04 0.876142 0.87704 0.882086 0.879194 0.904312 0.889839 0.876206 0.876231 0.887929 0.05 0.876526 0.877425 0.882476 0.879581 0.904721 0.890235 0.87659 0.876615 0.888323 0.06 0.876782 0.877682 0.882735 0.879839 0.904994 0.890499 0.876846 0.876872 0.888586 0.07 0.876965 0.877865 0.882921 0.880023 0.905189 0.890688 0.877029 0.877055 0.888774 0.08 0.877102 0.878003 0.88306 0.880161 0.905336 0.890829 0.877167 0.877192 0.888915 0.09 0.877209 0.87811 0.883168 0.880269 0.90545 0.89094 0.877274 0.877299 0.889025

0.1 0.877294 0.878196 0.883255 0.880355 0.905541 0.891028 0.877359 0.877385 0.889113

5. Conclusion

The results and system reliability are shown in Tables (1-10) which indicates that the reliability

of the system decreases with the increase of failure rates ( i ) and transition rate m w.r.t. time

and for fixed values of other parameters. Also, it is analyzed that there are sudden jumps in the

values of reliability function and over a long period of time the system becomes less and less

reliable. Table 10, shows that are availability of the system decreases with the increase of the

failure rate ( 1 ). Table 11, shows the behavior of steady state availability with respect to repair

rate ( 1 ) and observed that availability of the system increase with the increases of the repair rate

and preventive maintenance rate.


REFERENCES

Cox, D.R. (1955). Analysis of Non Markovian stochastic processes by the inclusion of

Supplementary variables, Proc. Comb. Phill. Soc., Vol. 51, pp. 433–441.

Gaver, D.P. (1963). Time to failure and availability of parallel system with repair, IEEE T.

Reliab. R-12, pp.30–38.

Proctor, C.L., Singh, B. (1976). A repairable 3- state device, IEEE Trans. Reliab., R-25, pp.

210- 211.

Uematsu, K, Nishida, T. (1987). One-unit system with a failure rate depending upon the

degree of repair, Mathematica Japonica,Vol. 32, No.1, pp.139-147.

Singh,J., Dayal,B. (1991). A 1-out of-N: G system with common cause failure and critical

human errors, Microelectron Reliab. Vol.31, pp.101–104.

Alfa, A.S., Rao, T.S.S. (2000). Supplementary variable technique in stochastic models,

Probab. Eng. Inform. Sci. Vol.14, pp. 203–218.

Malik, S.C. (2008): Reliability modeling and profit analysis of a single-unit system with

inspection by a server who appears and disappears randomly, Journal of Pure and Applied

Mathematika Sciences, Vol. LXVII, No. 1-2, pp. 135-146.

Kharoufeh, J.P., Solo, C.J. and Ulukus, M.Y. (2010). Semi-Markov models for degradation-

based reliability, IIE Transactions. Vol. 42, pp. 599–612.

Malik, S. C. and Nandal, P. (2010). Cost- Analysis of Stochastic Models with Priority to

Repair Over Preventive Maintenance Subject to Maximum Operation Time, Edited Book,

Learning Manual on Modeling, Optimization and Their Applications, Excel India

Publishers, pp.165-178.

Shakuntla S., Lal, A.K., Bhatia, S.S. and Singh, J. (2011). Reliability analysis of polytube

industry using supplementary variable technique, Applied Mathematics and Computation,

Vol.218, pp. 3981-3992.

Arekar, K., Ailawadi, S. and Jain, R. (2012). Reliability modeling for wear out failure period

of a single unit system, Journal of Statistical and Econometric Methods, Vol.1 (1), pp. 33-

41.

Kumar, A., Malik, S. C. (2012). Stochastic Modeling of a Computer System with Priority to

PM over S/W Replacement Subject to Maximum Operation and Repair Times.

International Journal of Computer Applications, Vol.43 (3), pp. 27-34.

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with Independent H/W and S/W Failures Subject to Maximum Operation and Repair

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APPENDIX

Derivation of Equations (1)-(3)

Assuming failure rates of the system are constant and repair rates are variable. By applying

supplementary variable technique, we develop the following differential difference equations

associated with the state transition diagram (fig. 1) of the system at time( ) and ( ) 8 8

0 01 1 0 0

( ) ( )[1 ] [ ( ) ( ) ] ( ) ( ) ( ).i m i i m mi i

P t t P t P t x dx t P t x dx t o t


( , ) ( , )[1 ( ) ] ( , ) 1,2,...,8.i i iP t t x x P t x x x o t x where i

( , ) ( , )[1 ( ) ] ( , ).m m mP t t y y P t y y y o t y

Proof of Theorem 1:

Taking Laplace transform of (25) and using (6) we get

8

01

( ) ( ) 1i mi

s P s

Using the initial conditions, the solution can be written as

0 8

1

1( ) ,

i mi

P s

s

8

1

1( ) ,

i mi

R s

s

Taking inverse Laplace transform, we get

8

1( )

( ) .i m

it

R t e

(*)

Operative State Failed State

Figure 2. State transition Diagram

ABCDEFGH

aBCDEFGH

ABcDEFGH

ABCdEFGH

ABCDeFGH

ABCDEfGH

ABCDEFgH

ABCDEFGh AbCDEFGH P1(x,t)

P2(x,t)

P3(x,t)

P4(x,t)

P5(x,t)

P6(x,t)

P7(x,t)

P8(x,t)

P0(t)

α1

α2

α3

α4

α5

α6

α7

α8 β1(x)

β2(x)

β3(x)

β4(x) β5(x)

β6(x)

β7(x)

β8(x)

Pm(x,t)

abcdefgh

αm

βm(y

)

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