Optimisation of the preventive maintenance plan of a series components system with Weibull hazard function Текст научной статьи по специальности «Компьютерные и информационные науки»

came into service on the 1st of May 2003. Used radars were produced by Danish corporation Terma Electronic AS (frequency 9735 MHz). Mean time between failure and mean time to repair given by HITT have been used to construct VTS Zatoka radars' subsystem reliability model (Table 1). Times to repair given in Table 1, concern the situation in which service team is near damaged radar. In fact, sustaining service

Table 1. MTBFi - mean time between failure and MTTRi - mean time to repair [5]

Device/component MTBFi [h] I, MTTR, [h]

Antenna 50 000 0,00002 3

Single radar transmitter 13 000 0,000077 0,5

Receiver 17 390 0,000058 0,34

Video processor 40 000 0,000025 0,25

Radar processor 20 000 0,00005 0,25

Data transmitter 87 500 0,000011 0,5

team near each radar is very expensive. For further consideration we assume that service team is in Tri-city i.e. in Gdynia, Sopot or Gdansk. Taking into account access time, time needed to fix what device is damaged and mean time of device' exchange, mean time to repair of the single radar (MTTRS) is about 3 and a half hours. After considering frequency of failures of particular parts of radars, the mean time of exchanging damaged part is about 40 minutes.

To find single radar reliability we assume that the radar is a series system. This means that the failure of one component of radar causes the failure of the whole radar. Two radars placed at Harbourmaster office buildings have five elements (antenna, single radar transmitter, receiver, video processor, radar processor), three others additionally have data transmitter. We assume that component reliability functions are exponential and given by the equation

R (t) = exp (-1 • t), t > 0.

(11)

When we put data from Table 1 to (11), and then to equation (1), as a result we obtain the single radar reliability function for radars at harbourmaster office buildings

RH (t) = exp(-0,000229 • t), t > 0,

and for three others radars

RO (t) = exp(-0,000235 • t), t > 0 .

(12)

MTBF = J R(t) dt.

(14)

According to equations (12)-(14) and to Table 1, we obtain mean time between failures of single radar at harbourmaster office buildings

MTBFH =

1

0,000229

= 4366h » 182 days,

and mean time between failures of three other radars

MTBFO =

1

0,000235

= 4255h » 177 days.

3.2. Reliability of radar system

In order to evaluate radars system reliability, we can use different approaches. First, we can assume that subsystem is series. VTS Zatoka radars' subsystem is working when all five radars are working. According to equation (3) with parameters

< 2 3

n = 5 q = = ^

and equations (12) and (13), we obtain the system reliability function

RS (t) = [exp[-0,000229t]]2[exp[-0,000235t]]3

= exp[-0,001163t], t > 0. (15)

Mean time between failures of that system is given by equation

¥ 1

MTBF, = f RS (t) dt =-

* 0 0,001163

= 860^ » 36 days. (16)

System availability is given by equation [1]

MTBF„

G = -

MTBFS + MTTRS

(17)

(13)

and after substituting MTBFS = 860h, MTTRs = 3,5h, amounts G = 0,9959.

Mean time between failures of single radar is given by equation

¥

1

0,9 -0,8 -0,7 -0,6 -R 0,5 0,4 0,3 0,2 0,1 0

G

MTBF„

SMN

0

500

1000 t [h]

1500

2000

Figure 6. Radars' subsystem's reliability functions

Another way of describing reliability of radars' subsystem is assumption that system is „m out of n". We can assume that system is working if at least four out of five radars are working. If we take into account that particular radars are series systems we obtain a non-homogenous series-„4 out of 5" system.

The above assumption is acceptable because ranges of any four radars covered fairways to ports in Gdansk and Gdynia and most traffic (nearly entire) is concentrated in those fairways.

According to equations (10) and Table 1, reliability function of such system is given by equation

Rsmn = R51,5,6,6,6 (t) = 2 exp(-0,000934t)

+ 3 exp(-0,000928t) - 4exp(-0,001163t). (18)

RSMN(t) function is showed on Figure 4.

Mean time to failure of series-„m out of n" system

MTBFsmn is given by equation

MTBFsmn = J Rsmn (t) dt.

(19)

The mean time to failure of above described system according to equations (18) and (19) equals

MTBFsmn + MTTRs

(21)

and hence

1935

Gsmn = q19 __ = 0,9982. 1935 + 3,5

As we can see system defined as series-„m out of n" has both higher reliability and availability than a series system.

4. Conclusion

As we can see from the performed analysis evaluation of the system reliability depends on taken assumptions. Reliability functions are significantly different one from the other (Figure 6), so choosing proper method of describing of system reliability structure is very important.

Whatever method was chosen, thanks to reliability of components of radars, VTS Zatoka radars system is highly reliable. Access to spare parts and organization of service has significant matter for availability of the subsystem. In order to sustain acceptable availability it is necessary to provide the service support located in Tri-City. It allows for quick reaction in case of failures and for repairing damaged parts of radars.

References

[1] Budny, T. (2006). Radar subsystem reliability in

the VTS Zatoka. Navigational Faculty Works 18, Gdynia.

[2] Kolowrocki, K., Blokus, A. Milczek, B., Cichocki,

A., Kwiatuszewska-Sarnecka, B. & Soszynska, J., (2005). Asymptotic approach to complex systems reliability analysis, Two-state non-renewal systems. Gdynia Maritime University Publishing House, Gdynia.

[3] Stupak, T., Wawrach, R., Dziewicki, M. & Mrzyglod, J. (2004). Research on availability of VTS Zatoka radars. Maritime Office, Gdynia.

MTBFsmn = J R515,5,6,6,6(t) dt = 2 •

0

1

0,000934

+ 3 •-

1

0,000928

- 4 •

1

0,001163

= 1935h @ 81days.

(20)

The availability of the series-„m out of n" system is given by equation

¥

Duarte José Caldeira

Mathematical Department, Instituto Politécnico de Setúbal/Escola Superior de Tecnologia de Setúbal, Portugal 1ST - Unit of Marine Technology and Engineering, Lisbon, Portugal

Soares Carlos Guedes

1ST - Unit of Marine Technology and Engineering, Lisbon, Portugal

Optimisation of the preventive maintenance plan of a series components system with Weibull hazard function

Keywords

availability, hazard function, Weibull distribution, preventive maintenance, series components Abstract

In this paper we propose an algorithm to calculate the optimum frequency to perform preventive maintenance in equipment that exhibits Weibull hazard function and constant repair rate in order to ensure its availability. Based on this algorithm we have developed another one to solve the problem of maintenance management of a series system based on preventive maintenance over the different system components. We assume that all components of the system still exhibit Weibull hazard function and constant repair rate and that preventive maintenance would bring the system to the as good as new condition. The algorithm calculates the interval of time between preventive maintenance actions for each component, minimizing the costs, and in such a way that the total downtime, in a certain period of time, does not exceed a predetermined value

1. Introduction

Throughout the years, there has been tremendous pressure on manufacturing and service organizations to be competitive and provide timely delivery of quality products. In many industries, heavily automated and capital intensive, any loss of production due to equipment unavailability strongly impairs the company profit. This new environment has forced managers and engineers to optimise all sectors involved in their organizations.

Maintenance, as a system, plays a key role in achieving organizational goals and objectives. It contributes to reducing costs, minimizing equipment downtime, improving quality, increasing productivity, and providing reliable equipment that are safe and well configured to achieve timely delivery of orders to costumers. In addition, a maintenance system plays an important role in minimizing equipment life cycle cost. To achieve the target rate of return on investment, plant availability and equipment effectiveness have to be maximized.

Grag and Deshmukh [38] had recently review the literature on maintenance management and points out that, next to the energy costs, maintenance costs can be the largest part of any operational budget. A brief bibliographic review, (Andrews & Moss [1], Elsayed [5], McCormick [9] and Modarres et al [10]), is enough to conclude that the discipline known as reliability was developed to provide methods that can guarantee that any product or service will function efficiently when its user needs it. From this point of view, reliability theory incorporates techniques to determine what can go wrong, what should be done in order to prevent that something goes wrong, and, if something goes wrong, what should be done so that there is a quick recovery and consequences are minimal.

So, reliability has several meanings. However it is usually associated to the ability of a system to perform successfully a certain function. To measure quantitatively the reliability of a system it is used a probabilistic metric, which we state next.

Reliability of a system is the probability that a system will operate without failure for a stated period of time under specified conditions.

Another measure of the performance of a system is its availability that reflects the proportion of time that we expect it to be operational. Availability of a system is the probability to guarantee the intended function, that is, the probability that the system is normal at time t. The availability of a system is a decreasing function of the failure rate and it is an increasing function of the repair rate.

According to Elsayed [5], reliability of a system depends mainly in the quality and reliability of its components and in the implementation and accomplishment of a suitable preventive maintenance and inspection program. If failures, degradation and aging are characteristics of any system, however, it is possible to prolong its useful lifetime and, consequently, to delay the wear-out period carrying out maintenance and monitoring programs. This type of programs leads necessarily to expenses and so we are taken to a maintenance optimisation problem.

Basic maintenance approaches can be classified as:

• Unplanned (corrective): this amounts to the replacement or repair of failed units;

• Planned (preventive):

• Scheduled: this amounts to performing inspections, and possibly repair, following a predefined schedule;

• Conditioned: this amounts to monitor the health of the system and to decide on repair actions based on the degradation level assessed.

In the unplanned, corrective strategy, no maintenance action is carried out until the component or structure breaks down. Upon failure, the associated repair time is typically relatively large, thus leading to large downtimes and high costs. In this approach, efforts are undertaken to achieve small mean times to repair (MTTRs).

To avoid failures at occasions that have high cost consequences preventive maintenance is normally chosen. This allows that inspections and upgrading can be planned for periods, which have the lowest impact on production or availability of the systems. The main function of planned maintenance is to restore equipment to the "as good as new" condition; periodical inspections must control equipment condition and both actions will ensure equipment availability. In order to do so it is necessary to determine:

• Frequency of the maintenance, substitutions and inspections

• Rules of the components replacements

• Effect of the technological changes on the replacement decisions

• The size of the maintenance staff

• The optimum inventory levels of spare parts

There are several strategies for maintenance; the one we have just described and that naturally frames in what has been stated is known as Reliability Centered Maintenance - RCM. Gertsbakh [7] reviews some of the most popular models of preventive maintenance. In theory, maintenance management, facing the problems stated above, could have benefited from the advent of a large area in operations research, called maintenance optimisation. This area was founded in the early sixties by researchers like Barlow and Proschan. Basically, a maintenance optimisation model is a mathematical model in which both costs and benefits of maintenance are quantified and in which an optimal balance between both is obtained. Well-known models originating from this period are the so-called age and the block replacement models. Valdez-Flores & Feldman [12] presents a comprehensive review of these approaches. Dekker [2] gives an overview of applications of maintenance optimisation models published so far and Duffuaa [4] describes various advanced mathematical models in this area that have "high potential of being applied to improve maintenance operations". More recently, Nakagawa [11] summarizes the results of maintenance policies and Garg and Deshmukh [6] describe the existence of 24 papers on the quantitative approach to maintenance optimisation As we have already mentioned, one of the most critical problems in preventive maintenance is the determination of the optimum frequency to perform preventive maintenance in equipment, in order to ensure its availability.

The Preventive Maintenance policies are adapted to slow the degradation process of the system while the system is operating and to extend the system life. A number of Preventive Maintenance policies have been proposed in the literature. These policies are typically to determine the optimum interval between preventive maintenance tasks to minimize the average cost over a finite time span. But in many areas one complains about the gap between theory and practice. Practitioners say maintenance optimisation models are difficult to understand and to interpret [2]. Vatn et al [13] claim there exists a vast number of academic papers describing narrow maintenance models, which are rarely, if ever, used in practice. Most of these papers are too abstract, and the models that could be useful are difficult to identify among this large number of suggested models.

In this paper we propose an algorithm to solve the previous problem for equipment that exhibits Weibull hazard function and constant repair rate. Based on this algorithm we have developed another one to optimise maintenance management of a series system based on preventive maintenance over the different system components.

This is a problem with many applications in real systems and there are not many practical solutions for it. The main objective of this paper is to present an optimisation model understandable by practitioners, simple and useful for practical applications. Duarte and Craveiro [3] have already outlined a solution for this problem for equipment that exhibit linearly increasing hazard rate and constant repair rate. We assume that all components of the system still exhibit Weibull hazard function and constant repair rate and that preventive maintenance would bring the system to the as good as new condition. We define a cost function for maintenance tasks (preventive and corrective) for the system. The algorithm calculates the interval of time between preventive maintenance actions for each component, minimizing the costs, and in such a way that the total downtime, in a certain period of time, does not exceed a predetermined value.

2. Previous concepts and results

In this section we present the classical concept of availability, while describing how to calculate it. Point-wise availability of a system at time t, A(t), is the probability of the system being in a working state (operating properly) at time t. The unavailability of the system, Q(t), is Q(t) = 1 - A(t).

The pointwise availability of a system that has constant failure rate l and constant repair rate m is

A(t ) =

m

■+-

i

m +1 m +1

exp[-(m +l)t ]

(1)

Example. A system is found to exhibit a constant failure rate of 0,000816 failures per hour and a constant repair rate of 0,02 repairs per hour.

Using formula 1, the availability of such a system (see Figure 1) is obtained as

A(t) = 0,9608 + 3.9201 x 10-2 exp( - 2,0816 x 10-21).

and the limiting availability is

lim A(t) = 0,9608.

0.99

0.97

0.95 0

Figure 1. The availability function

It should be notice that, in this in case, we do not have almost any variation in the value of component's availability for t > 200.

We can therefore conclude that, to guarantee a value of availability A, known the constant repair rate, m, the value of the constant failure rate of the system it will have to satisfy the relationship

m

m +1

û i

m(i - a)

A

(3)

and the limiting availability is

3. Model and assumptions

Suppose a system is found to exhibit an increasing hazard rate,

A = lim A(t) = —L.

t ®+¥ m +1

(2)

The second parcel in formula (1) decreases rapidly to zero as time t increases; so, we can state

A(t)

m

m +1

and this means that the availability of such a system is almost constant.

ai \P-i

h(t) = b[t I '9> 0, b> 0, t > 0,

and a constant repair rate m-

Our goal is to determine the interval time between preventive maintenance tasks (we assume that the system is restored to the "as good as new" condition after each maintenance operation) in such a way that the availability of the system is no lesser than A. The main idea for the solution of this problem consists of determining the time interval during which the

increasing hazard rate can be substituted by a constant failure rate in order to guarantee the pre-determinate availability level.

Applying the mean value theorem of integral calculus to the function

R i \ b —

W) = q(t) '0>0'R>0't-

and a constant repair rate m. To guarantee an availability for the system equal or greater than A the interval of time between two consecutive preventive maintenance tasks must be equal or lesser than

R-j m(i - a)qr .

we obtain

cr-i^r-1P-1 dt = 1x

0 0P

^ —= 1x 0P

= 1x

^ x = 0 v x = .

(4)

Substituting 1 in (4) by its approximate value given in formula 3, we have

x=MÎmiL^ q P.

We can therefore conclude that in the time interval

Example. A system is found to exhibit an increasing hazard rate, h(t) = 5 x 10-8 x t1,25, and a constant repair rate m(t) = 4 x 10-2. What should be the greatest time interval between preventive maintenance tasks (we assume that the system is restored to the "as good as new" condition after each maintenance operation) in such a way that the availability of the system is at least 98%? If the system had a constant failure rate, to guarantee such availability it should be

1 = 4 x 10 -2 (1 - P.98) = 0 163.

0,98

We want to calculate the instant x in order to satisfy the following condition

J0x5 x 10-8 x tU5dt = 0,0008163x

0, b

1 mi1 - A)0b

5 x 10-8 x x2'25

2,25

= 0,0008163x

the hazard functions,

R i \ P-1

h(t) = I ,0 > 0,R > 0,t > 0

and

h(t) =

m(1 - a)

guarantee approximately the same value of availability. What we have just demonstrated can formally be stated on the following form:

Proposition: Let S be a system exhibiting an increasing hazard rate,

P-1

h(t) = pft | ,0 > 0, R > 0,t > 0

^ x = 0 v x = 4488.

We can therefore conclude that the system must be restored to the "as good as new" condition after each maintenance task every 4488 hours in order to achieve the availability target of 98%. Figure 2 illustrates this example.

Figure 2. Hazard functions h(t)=5x10~8t125 and h(t) =0,0008163 over the interval [0, 4488]

x

P

t

P

0

0

P

4. Optimisation of the preventive maintenance plan of a series components system

In this section we will present a model for the preventive maintenance management of a series system.

The system is composed by a set of n components in series as Figure 3 shows.

Figure 3. A series system of n components

The cost of each preventive maintenance task is cmpf and the cost of each corrective maintenance task is

cmcj.

Since the availability of the system consisting of n components in series requires that all units must be available (assuming that components" failures are independent), system availability a is

n

a = ua,

7=1

where Al is the availability of component i. Applying proposition presented in section 3 we can write that the availability of each component / is Al over the interval

Let Tj, x2.....x„ be the time units between preventive

maintenance tasks on components 7, 2..... n,

respectively (Figure 4); assuming that these actions will restore periodically the components to the "as good as new" condition, they will have, therefore, consequences at the reliability and availability levels of the system.

°| 1^1 I 1*7. I I I I I*« I 1^7 I I t¡j I t<j pm I fn I |t17l

ZhtHIHIKHHHHHHHH]

Component 1

Component 2 -

—

( Component 3

T

C emportent n

Figure 4. A preventive maintenance plan. Our goal is to calculate the vector

|j 1 1 2 T3 * 1 n ]

in such a way that the total down time in a certain period of time does not exceed a predetermined value, that is to say, that it guarantees the specified service level and simultaneously minimizes the maintenance costs.

We assume that each component has a Weibull hazard function,

A,.(0 = — e,

V07/

e, >o,p, >o,f>o

and a constant repair rate mi(t) = \ii.

0. P;

-MzMj*

and its hazard function can be approximated by the constant function

(1-4)

h,{t) =

Then, the expected number of failures in that time interval is

P/-1

\

The objective function (defined as a cost function per unit time) is

el. ! ,. ! .

subject to

>4,) = 2

7 = 1

cmp¡

>,(!-A,) P, EE5

A, ' •!.

EM, >A,

7=1

0 <4 <1,7 = 1,2,» ,n.

5. Numerical example

The model described on section 4 was implemented to a three components series system.