School of Engineering
Scheduled maintenance policy for
minimum cost
- A case study
Växjö, Spring 2011
Authors: Mohamad Tabikh and Ammar Khattab
Case company: DynaMate IntraLog AB
Scheduled maintenance policy for minimum cost – A case study
Linnaeus University – Spring 2011
Organisation/Organization
Författare/Authors
Linnaeus University
Mohamad Tabikh
School of Engineering
Ammar Khattab
Department of Terotechnology (Systemekonomi)
Dokumenttyp/Type of document
Examensarbete/Degree Project
Handledare/Tutor
Examinator/Examiner
Matias T. Hailemariam
Imad Alsyouf
Titel och undertitel/Title and subtitle
Scheduled maintenance policy for minimum cost – A case study
Sammanfattning/Abstract
This report evaluate the maintenance policies that been applied within specific industrial company,
Taken into considerations all corrective and preventive maintenance costs ,in addition to optimise best
preventive maintenance schedule for minimum cost.
Dynamate Intralog AB was the surveyed company that been encountered high maintenance cost
compatible with less productivity, therefore obtaining maintenance schedule policy for minimum cost
was the best solution for their problem, then by calculating their corrective and preventive
maintenance cost the optimum time was acquired. Finally, the maintenance schedule approve that
organized maintenance based on optimum time enhance the productivity and minimize the company
maintenance cost.
Nyckelord/Keywords
Corrective maintenance, Preventive maintenance, Maintenance Schedule, optimum time
Utgivningsår/Year of issue
2011
Språk/Language
English
Antal sidor/Number of pages
58
Internet/WWW
http://www.lnu.se
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Scheduled maintenance policy for minimum cost – A case study
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Acknowledgements
First of all, we would like to take the opportunity to express our gratitude to
everybody who has helped us with this thesis.
In this context, our special thanks refer to our tutor, Matias T. Hailemariam and our
examiners, Imad Alsyouf, Lars Erikson and Anders Ingwald as they have guided and
helped us with our work.
Moreover, we appreciated the cooperation with Dynamate Intralog AB
presented by Mr. Gaith that helps us during collecting the data
Finally, we are grateful for our families and friends who have supported us
during this period of time, firstly to Bea Cerezo Valera for her constructive
comments, then to Idriss, Nurdan, Akif, Nurdos, Hande and Zehra Aksoy.
Växjö, June 2011
Mohamad Tabikh and Ammar Khattab
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Scheduled maintenance policy for minimum cost – A case study
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Table of contents
1. INTRODUCTION ......................................................................................................................... 8
1.1 Background ......................................................................................................................... 8
1.2 Problem discussion ............................................................................................................. 9
1.3 Problem formulation .......................................................................................................... 9
1.4 Purpose ............................................................................................................................. 10
1.5 Relevance .......................................................................................................................... 10
1.6 Limitations ........................................................................................................................ 10
1.7 Time Frame ....................................................................................................................... 10
2. METHODOLOGY ...................................................................................................................... 12
2.1 Research strategy ............................................................................................................. 12
2.2 Research design ................................................................................................................ 12
2.3 Data collection .................................................................................................................. 13
2.3.1 Semi-structured interviews......................................................................................... 14
2.3.2 Secondary analysis ..................................................................................................... 14
2.4 Validity and reliability ...................................................................................................... 14
2.5 Generalizing ...................................................................................................................... 15
3. THEORETICAL FRAMEWORK.................................................................................................... 16
3.1 Impact of Maintenance .................................................................................................... 16
3.2 Corrective Maintenance ................................................................................................... 16
3.3 Preventive Maintenance .................................................................................................. 17
3.4 Block policy ....................................................................................................................... 18
3.5 Age policy .......................................................................................................................... 18
3.6 Weibull distribution.......................................................................................................... 19
3.7 Maintenance optimization ............................................................................................... 22
3.8 Schedule maintenance ..................................................................................................... 23
3.9 Life Cycle Cost (LCC) .......................................................................................................... 24
4. EMPIRICAL DATA ..................................................................................................................... 26
4.1 General description .......................................................................................................... 26
4.2 The logistics outsourcing .................................................................................................. 26
4.3 The empty boxes process ................................................................................................. 27
4.4 Robot performance and critical components .................................................................. 28
4.5 Maintenance strategies within Dynamate Intralog AB ................................................... 28
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Scheduled maintenance policy for minimum cost – A case study
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4.6 Corrective maintenance for essential components ........................................................ 29
4.7 Preventive maintenance for essential components........................................................ 30
4.7.1 Weekly preventive maintenance ................................................................................ 30
4.7.2 Monthly preventive maintenance .............................................................................. 32
4.8 Corrective and preventive maintenance related cost ..................................................... 33
4.9 Production list for palletizing robot ................................................................................. 34
5. ANALYSIS ................................................................................................................................. 35
5.1 Maintenance cost calculations ......................................................................................... 35
5.2 Distribution for the failure data ....................................................................................... 38
5.3 Optimum time to PM actions ........................................................................................... 39
5.4 Maintenance schedule based on optimum time ............................................................. 41
6. RESULTS ................................................................................................................................... 42
7. CONCLUSION ........................................................................................................................... 43
References................................................................................................................................... 44
Appendix 1 .................................................................................................................................. 46
Appendix 2 .................................................................................................................................. 47
Appendix 3 .................................................................................................................................. 48
Appendix 4 .................................................................................................................................. 50
Appendix 5 .................................................................................................................................. 51
Appendix 6 .................................................................................................................................. 52
Appendix 7 .................................................................................................................................. 53
Appendix 8 .................................................................................................................................. 54
Appendix 9 .................................................................................................................................. 55
Appendix 10 ................................................................................................................................ 56
Appendix 11 ................................................................................................................................ 57
Appendix 12 ................................................................................................................................ 58
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List of tables, figures and images
List of tables
Table 1. Time Frame .................................................................................................................... 11
Table 2. The value of m ............................................................................................................... 40
Table 3. The optimum time for each component to perform PM .............................................. 40
Table 4. The optimum time to perform PM ................................................................................ 41
List of figures
Figure 1. Maintenance-related time ............................................................................................. 9
Figure 2. Corrective maintenance cycle ...................................................................................... 17
Figure 3. Example of age policy ................................................................................................... 18
Figure 4. Curve shows the total cost with increasing PM actions to an optimal time ................ 22
Figure 5. TPL circulation .............................................................................................................. 27
Figure 6. No. of failures for each critical component during 5190 hours ................................... 38
Figure 7. The time at which the components have failed ........................................................... 39
Figure 8. Current PM Time and Optimum time .......................................................................... 41
List of images
Image 1. The wooden box ........................................................................................................... 27
Image 2. The photocell in palletizing robot................................................................................. 31
Image 3. The brake motor ........................................................................................................... 31
Image 4. The gearbox .................................................................................................................. 32
Image 5. The air bellows ............................................................................................................. 32
Image 6. The control list .............................................................................................................. 33
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Scheduled maintenance policy for minimum cost – A case study
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List of abbreviations
CBM
Condition based monitoring
CM
Corrective maintenance
LCC
Life Cycle Cost
MLE
Maximum Likelihood Estimation
MTTF
Mean time to failure
PM
Preventive maintenance
TPL
Third-party logistics
tp
Time interval between PM actions
Cf/Cp
Cost ration
Cf
Cost of failure
Cp
Cost of preventive maintenance
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Scheduled maintenance policy for minimum cost – A case study
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1. INTRODUCTION
This chapter will present an overview about the report, which includes: Background, problem
discussion, problem formulation, purpose, relevance, limitations and time table.
1.1 Background
The definition of maintenance may combine the technical and administrative actions
that intended to retain an item or restore it to a state in which it can perform a
required function. Maintenance provides crucial support for heavy and capitalintensive industry by keeping machinery and equipment in a safe operating condition.
Therefore, maintenance plays a main role in sustaining long-term profitability and
competitiveness for an organization (Aditya and Uday, 2006).
Thus, the market demands on products with higher quality, faster delivery and more
reliable characteristics led to the development of manufacturing systems combining
automation (robotic machines), integration and flexibility. Accordingly this complex
system becoming more vulnerable to various kinds of disturbances, such as
components failure and incidental faults, but maintenance strategies consistent with
high applicable performance may mitigate it. The assimilation of maintenance
management could increase the productive time (availability), reliability, and optimise
performance due to many policies that correlated with high maintainability
techniques. There are two common maintenance strategies dealing with industrial
machinery known as corrective and preventive maintenance, the examples of
corrective maintenance (CM) are configuration of mechanical part and control
equipment when the robot break down (Fix when it’s break). In other hand, preventive
maintenance (PM) can be classified into three categories: time driven, predictive, and
equipment driven. Scheduled maintenance derived from periodic, time intervals of the
system to replace the part. The concept behind it is that wear parts have a fixed
number of cycles to failure which is converted to operating time. By determining the
optimum time to maintain or replace the part with minimum total cost (Mobley,
2004).
Furthermore, maintenance strategy1 cannot be set up farther cost effective analysis
because the economical view tracking the entire process and organization take into
consideration all the direct and hidden cost behind their maintenance strategy. The
total cost of the maintenance (CM and PM) depends on the number of components
replaced during operating period of the system (Chitra, 2003).
1
Maintenance strategy: It is plans which direct the maintenance management towered a desired future
state (Levitt, 2003).
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Scheduled maintenance policy for minimum cost – A case study
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In order to establish an effective Preventive maintenance (time based), failure data are
required and it does not have to be based on Expert opinions anymore. Using
statistical methods, scheduled maintenance can be optimized, taking into
consideration the three types of maintenance: preventive maintenance, inspection
maintenance and predictive maintenance (Didson, 1994).
1.2 Problem discussion
Many organizations have been failed in implementing PM due to absent of clear
program that should be followed. In addition, many organizations which have program
failed also because of not having activity that ensure the program accomplishment.
(Mobley, 2004) In order to establish an effective programme, failure data are required
to be analysed to calculate the optimum time (Didson, 1994).
There is a survey within industrial sectors covered different production companies
1997 to highlight the maintenance time and in which trend; the present assessment is
shared between planning, prevention and correction. The percentages were
distributed according to how much of their maintenance time was spent on those
factors. Figure 1 show that about half of the maintenance time is spent on corrective
actions and two-fifths on preventive or condition based monitoring (CBM). The
optimum figure for CM is considered to not exceed 30-40 percent and most firms
probably knows this but are still not changing strategies or techniques (Jonsson, 1997).
Figure 1. Maintenance-related time
Source: Jonsson, 1997, pp.242
1.3 Problem formulation
The problem can be formulated in this thesis as following:
 Is it possible to determine the optimum period to perform PM actions by
analysing a set of failure data?
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1.4 Purpose
The goal of this report is to identify the optimum time that a certain PM actions should
take a place by analyzing the company maintenance policy towards their robot
availability.
1.5 Relevance
In today’s business, a valuable utilization of the machinery (robots) and incorporated
maintenance policies2 are a prerequisite for keep on challenging into the market and
gain profits. Thus, all the efforts regarding to decrease idle time and minimize delivery
delays should be made.
The relevance of this report includes the importance of implementing the right
maintenance policy at the right time in order to avoid unnecessary PM action which
could affect the machine availability.
1.6 Limitations
Maintenance management is wide subject and it’s difficult to consider all the elements
within it, therefore, our focusing will be only on the company applied method. In
addition to, visits limitation to company that narrowing our observations.
1.7 Time Frame
The preliminary work of this thesis started on week nine by deciding the thesis topic
and case company. Afterwards, the authors of this thesis started searching scientific
articles, books, etc. regarding the area of study.
In order to follow up with submission dates, a time plan was developed at the
beginning of week ten. The time frame for the accomplishment of this work is shown
in the following table.
2
Maintenance policy: An identified step used to implement the maintenance actions (Levitt, 2003).
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w.15
w.16
w.17
w.18
w.19
w.20
w.21
Introduction
Methodology
Theory
Empirical findings
Analysis
Results
Conclusions
Review and hand-in
Submission
of chapters
1-3
Submission
of chapters
1-5
Submission
of final
thesis
Table 1. Time Frame
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Scheduled maintenance policy for minimum cost – A case study
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2. METHODOLOGY
In this chapter different methods had been selected by the authors in order to reveal the
process of the research. The chapter contains the research strategy, the research design, data
collection, validity and reliability, as well as generalizing.
2.1 Research strategy
Ghauri and Grønhaug (2005) classify two different approaches of research by
distinguishing between inductive and deductive approach. These two ways are used by
the researchers in order to identify if their statements are either true or false. If the
statement is established as true, it can be used as the basis for theories. Once the
statement is identified as true or false, it is time to draw the conclusion.
The deductive approach is when the researcher deduces a hypothesis based on what is
known in a specific field, and by collecting data the researcher will reach his or her
findings. Some researchers prefer the inductive approach which is the opposite from
deductive approach. In the inductive approach “theory is the outcome of research”
(Bryman and Bell, 2007, p.14).
According to Saunders et al. (2007), there is another approach which is called
abductive. This approach is the combination of deductive and inductive approaches.
The abductive approach consists on elaborating the theoretical framework in order to
explain a specific case, and later testing this theory on other cases.
An abduction research approach which combines deduction and induction approaches
will be used in this thesis. The deduction approach will be used to collect different
theories, such as life based maintenance, related to the research question in order to
obtain conclusions. The induction approach will be used to verify the conclusions in the
case study of this thesis making new schedule maintenance after applying optimum
time selection methodology.
2.2 Research design
Research design may be classified as exploratory, descriptive and causal. The
exploratory design is used to clarify the research problem by gathering and providing
all the information. In the descriptive design the research problem is known, but focus
on the description of different characteristics of the problem. The causal design is used
to obtain evidence of cause and effect problems (Ghauri and Grønhaug, 2005).
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Scheduled maintenance policy for minimum cost – A case study
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As the purpose of this thesis is to determine the exact optimum period to perform PM
actions, the causal method will be used. The authors will confront the cause-and-effect
problems by determining which policy is the most suitable for the robot.
Moreover, Bryman & Bell (2007) assert that a research design provides a structure for
the collection and analysis of data and there are five different types: experimental
design, cross-sectional or social survey design, longitudinal design, case study design
and comparative design.
Case study design “is concerned with the complexity and particular nature of the case
in question”. A case study can be: a single organization, a single location, a person, or a
single event (Bryman & Bell, 2007, p.62).
In this thesis a case study design will be applied because the data will be collected from
an industrial company located in Sweden. For this reason this design is the most
suitable for the thesis in order to obtain appropriate data and develops analytic and
problem solving skills. Furthermore, this kind of research requires a tangible approach
whilst connected to failures and statistical distributions, although it allows for
exploration of solutions for complex issues and applying new knowledge. The case
study here enables us to observe the current situation and monitoring all the factors
surrounded by it.
2.3 Data collection
There are two methods for collecting data, qualitative and quantitative. Qualitative
method is the collection and analysis of data by words; whereas quantitative method is
through statistics and mathematics i.e. researchers employ measurement (Bryman and
Bell, 2007). In addition, Creswell (2009) states that the combination of quantitative and
qualitative approaches leads to mixed methods research. Employing both methods
provides an extensive understanding of research problems.
This thesis is a result of both qualitative and quantitative research, as the authors
investigate and evaluate maintenance policy in the industrial sector using data which is
verbally coded as well as statistics in order to provide cost effective solutions to the
case study. Therefore both methods are suitable in order to create a deeper
understanding about our topic.
For this paper, data was collected through interviews with different employees at the
company and by analyzing statistics and documents that have been collected by the
case study. The methods that will be applied in this thesis in order to get the most
significant data are detailed below.
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Scheduled maintenance policy for minimum cost – A case study
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2.3.1 Semi-structured interviews
Bryman and Bell (2007, p.213) state that a semi-structured interview “refers to a
context in which the interviewer has a series of questions that are in the general form
of an interview schedule but is able to vary the sequence of questions”. Interviewers
have the opportunity to ask further questions depending on the interviewees’
answers. When a researcher is collecting data by conducting semi-structure interviews,
he or she has to control the situation, ask the most important questions and be able to
adapt to the situation (Ghauri and Grønhaug, 2005).
The authors of this paper will interview several workers in the company with the
purpose of obtaining expert information about maintenance actions. The interviews
will consist of ten questions of different types such as open and closed questions. See
appendix (1). Thus, we will reduce the duration of the interview as well as making it
easier for interviewees to answer. The questions will be accurate in order to reach our
purpose. These questions mostly been formulated based on recent observations and
meeting with production and logistics manager, in addition to recent data collected
from last year of robot effectiveness such as availability, productivity and reliability
(failure data).
2.3.2 Secondary analysis
According to Bryman and Bell (2007), secondary analysis involves the analysis of data
by researches who have not participated in the compilation of the data. Other
researchers such as companies or other kinds of organizations have collected them for
their own purposes. Secondary analysis entails the analysis of quantitative as well as
qualitative data.
In this thesis will be performed a secondary analysis of a large amount of information
that our case study have collected for several years and thereby will reveal relevant
knowledge on our thesis.
2.4 Validity and reliability
There are different important criteria in order to evaluate a research. On the one hand
there is validity and on the other hand reliability. Validity is “concerned with the
integrity of the conclusions that are generated from a piece of research” (Bryman and
Bell, 2007, p.41). Moreover, reliability is “concerned with the question of whether the
results of a study are repeatable” (Bryman and Bell, 2007, p.40). Reliability is
connected with quantitative research since it is relevant to know whether if a measure
is stable or not (Bryman and Bell, 2007).
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This thesis will provide the validity of the data by collecting accurate and relevant
information from various articles and books in order to find the most appropriate
theories for our thesis. Furthermore, the questions for the interview will be formulated
in a careful way to get the best relevant answers. Therefore the ways that been
performed for designing the questions taken into considerations the internal, external
and content validity. To provide reliability, the authors will interview different
operators and group leaders in order to know how the maintenance affects the robot.
The questions are related to every part of the process with the aim of achieving a
wider perspective.
2.5 Generalizing
Generalization of results involves that the researcher create a representative sample in
order to generalize the results to other groups or cases beyond than the one of the
research (Bryman and Bell, 2007). The qualitative method as well as quantitative can
be generalized, some of the reasons may be the situation of the case study or the type
of the research (Saunders et al., 2007).
The Generalizability of the results can be guaranteed from the implemented
researching approaches and specific methods adapted only in company’s applied same
maintenance strategies and have almost similar cost ratios. Hence, these factors may
rely on optimum time method in order to perform preventive maintenance schedule.
By the way, we cannot generalize the following results statements because of various
systems adaptation and different element may involve the optimum time model
evaluation.
Consequently, may this method encountered an internal and external blocks through
implementation stage such as human errors or learning curve minimization, especially
this way based on high corrective cost over preventive cost, and in such companies a
certain corrective action still benefiting more than life based maintenance.
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Scheduled maintenance policy for minimum cost – A case study
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3. THEORETICAL FRAMEWORK
This chapter contains theories about maintenance management in order to provide a clear
picture of the theories which will be used in the research.
3.1 Impact of Maintenance
Maintenance has been always considered as extra unnecessary actions which mean
cost for managers nevertheless; it is an existence cost that could be hidden.
Maintenance forms a portion from the total operating plant. Recently Maintenance
has been highlighted due to it is impact in industrial field (Mobley, 2004).
Nowadays organizations are not only satisfied with keep machines in good conditions
further, maintenance actions could be planned and totally efficient. Maintenance
sometimes could be critical aspect in an organization it could turn unnecessary cost
into profits. In united state over than $600 billion has been spent on critical plant
systems although it has been increased to $800 billion by 10 years (Mobley, 2004).
3.2 Corrective Maintenance
Corrective maintenance or breakdown maintenance, which attempt to minimize the
severity of equipment failures once they occur, and this tactic focuses on maintenance
procedures that bring equipment back to production in the shortest time. Such
alternatives include standby machines, spare parts inventories, and worker
reassignments. Furthermore, corrective maintenance plays a vital role in maintenance
management whilst the failures occurrences could not be allocated and prevented.
Hence, there are many corrective maintenance techniques to eliminate the severity of
machine malfunction; those techniques include increases service crews, inventory
buffers, and machine redundancy (Sheu and Krajewski, 1994). The evaluation of
corrective maintenance requirements based on tangible data developed from previous
system faults and repair experience with an item, controlling the current corrective
maintenance activity, and prediction of future corrective maintenance requirements
through statistical analysis of repair history that categorized by quantitative
methodologies. The classic corrective maintenance cycle runs from the point of failure
detection through verification of restoration, as shown in Figure 2 (Langford, 2007).
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Scheduled maintenance policy for minimum cost – A case study
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Figure 2. Corrective maintenance cycle
Source: Langford, 2007, pp.60
3.3 Preventive Maintenance
Preventive maintenance is a very wide concept there is no specific accurate definition,
but it can be generalized as all actions that should be taken on the machine in order to
minimize the occurrence of unexpected downtime (Duffuaa et al., 1998).
Mean time to failure (MTTF) or the bathtub curve shows that there is a high probability
of occurrence of failure in the first phase which is the installation phase. After a period
of time the machine goes into the Normal phase where it supposes to work as it
designed. Then the cost of maintain the machine goes higher and higher with time
because of wear out. In preventive maintenance machine repaired according to
schedules based on MTTF statistically (Mobley, 2004).
Implementing the preventive maintenance could be varies, since it is a wide subject
which has more than one edge. Some of the preventive maintenance practices do not
exceed what the manual of the machine says. On the other hand it could be a
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comprehensive preventive maintenance starting with replacing a component and
ending with minor adjustment of a minor component. Depending on the type of the
machine and the type of the component, the production system, and many other
factors play a role in implementing preventive maintenance (Mobley, 2004).
3.4 Block policy
Maintenance policies are a set of protocols that followed in order to reduce the
number of unexpected stoppages. Performing PM actions at certain point of time
regardless of component’s condition defined as a Block policy. Block policy aims to
maintain the component so that the failure could be avoided as much as possible that
may lead to a catastrophic failure (Thomas.H.Savit, 1993).
A fixed time intervals is identified for PM actions to take a place, these actions are
scheduled and is being implemented periodically. This policy wins normally when the
cost of preventive maintenance is lower comparing with other PM cost polices.
Moving towered from fixed intervals that are scheduled to a dynamic scheduling for
those intervals based on operating hours, defined as age policy (Sherwin, 2010).
3.5 Age policy
This policy is defined as perform a preventive maintenance at t p (time interval between
PM actions) hours of continuous operating. t p could be finite or infinite, in case of
infinite tp no preventive maintenance scheduled. On the other hand, when failure
occurs before tp PM take a place and rescheduling for the next PM actions must be
done.
Figure 3. Example of age policy
Source: Duffuaa el at., 1998, pp.60
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Figure above shows two operating cycle for a component, first cycle component has
been operating till tp and scheduled PM took a place. Next cycle component has been
fail prior tp, that case PM actions has been rescheduled for the next cycle. The
objective of this model is that tp is being optimum and PM actions (replacement, break
down maintenance) are performed with minimum cost per unit time (Duffuaa et al.,
1998).
The cost of PM actions and break down maintenance are associated with the Total
expected cost per cycle which is:
Where R(tp) is the probability that the component will survives till t p (Duffuaa el at.,
1998).
3.6 Weibull distribution
In probability and statistics Weibull distribution is the most commonly used model in
modern reliability engineering. Weibull used in statistical analysis due its flexibility and
ability to deal with small sample size in order to evaluate lifetime of a system,
component. It was named after Waloddi Weibull. Weibull and lognormal are called the
lifetime distribution because they tends better in representing the measurement of
product life. Weibull distribution is mostly applicable in manufacturing industries and
can be applied in a variety of forms of parameters (Murthy et al., 2004).
Exponential function is a special case from Weibull distribution. It is widely used to
represent the lifetime for a set of data and also a modelling trend with decreasing or
increasing failure rate the probability density function f(t) for Weibull distribution is
give as:
f(t) = β/ηβ(t-ɣ)β-1exp[-(t-ɣ/η)β]
And
R(t) = exp[-(t-ɣ/η)β]
Where: γ = Location parameter
β = Shape parameter
η = Scale parameter
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The Weibull distribution model gives an insight of how the distribution will look like. By
using the three parameters mentioned above as an input. Obtaining a graph from
those inputs could be possible which will show the probability that a certain
component will fail at a certain point in time. The two parameters are obtained by
setting γ=0 which give:
f(t) = β/ηβ(t)β-1exp-(t/η)β
Weibull distribution one-parameter occurs when the value of shape parameter β is
being determined or assumed. The one-parameter Weibull p.d.f is obtained by setting
γ = 0 and assumed β = k = constant.
f(t) = k/ηβ(t)k – 1exp-(t/η)k
The first parameter that is used is called the Location parameter denoted by γ
(Gamma). This determines where the graph will start and in most cases it always
assumed to be zero showing that it is the minimum time to failure. When gamma
parameter is set to be zero, this is called the two-parameter Weibull distribution.
The shape parameter is the second parameter usually denoted by β. This parameter
determines the shape of the graph and known as the Weibull slope. The β parameter is
the function of the hazard rates (Murthy et al., 2004).
β=1, constant hazard rate and Fit to exponential function.
β<1, Decreasing hazard rate.
β>1, Increasing hazard rate.
β=3.5, fit to normal distribution.
With some values of (β), the equation will reduce to other distribution model. For
example if β=1, the p.d.f of the three-parameter, Weibull will reduce to the twoparameter exponential distribution with the probability density function f(t) given as:
The different slope will approach the value of η at different F(t) depending on time.
The last parameter in Weibull distribution is called the scale parameter denoted by η.
The scale parameter is the point at which 63.2% of the products have failed. The
higher the number of η value, the more the graph is stretched over period of time
(Murthy et al., 2004).
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Scheduled maintenance policy for minimum cost – A case study
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Parameters estimation methods
According to Dodson (1994), there are four most commonly used methods to estimate
Weibull parameters:
-
Maximum Likelihood estimation
MLE refers to Maximum Likelihood Estimation, one of the most widely statistics
method to estimate Weibulls’ parameters. Based on maximize the value that maximize
the probability of a data. Let X1,X2,…..Xn be independent random variables which are
the representations for the probability density function f(x,Ɵ).
The Likelihood function is being maximized by a natural logarithm in order to simplify
the calculations.
-
Moment estimation
This method is used in estimation parameters by matching the moment of the sample
to the moment defined by distribution. In case of Weibull two parameters, first and
second moment for a sample data would be variance and mean which equal to:
And
-
Probability and Hazard plotting
Both of them are a graphical method used in order to estimate the Weibull
parameters. The cumulative distributions are being linearized by a logarithmic
transformation. Median rank is being used in probability approach. Furthermore if it
would be a manual approach it would require special papers. But due to high
technique in computers, linearization can easily be done.
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3.7 Maintenance optimization
Figure 4. Curve shows the total cost with increasing PM actions to an optimal time
Source: Levitt , 2003, pp.12
Figure above shows the minimum overall cost with increasing PM costs to an optimum
time. Notice that cost becomes stabilized with time since PM actions going no more
effective.
L is the time interval which the total cost is the minimum.
Implementing a maintenance policy does not mean avoiding extra cost all the time.
Defining the right time interval in order to PM actions take a place is exactly what all
organizations looking for. With time PM costs increased and become ineffective then,
it would preferable to have breakdown instead of running ineffective PM actions. This
caused by age of the system wearing out (Levitt, 2003).
The area denoted by X is all organizations interest in which the total cost during this
time interval would be the minimum. As mentioned before, determining this time
interval is the biggest challenge. Dodson (1994) asserts that minimizing the total cost
per unit time in order to find the optimum time according to the following equation:
Where: Cp is the cost of preventive maintenance.
Cr is the cost of failure.
T is the time between preventive maintenance actions.
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Instead of minimizing the previous equation every time by usual numerical routine,
Dodson (1994) has developed a table that can easily used and applicable by following
assumptions:
-
Time to failure follows Weibull distribution.
Preventive maintenance is performed at time = T with cost equal to Cp.
If a failure occur before time = T cost of failure is incurred.
The last and the most important condition is when the component is being
maintained it is return into its’ initial state “As good as new”.
The optimum time between preventive maintenance actions can be easily calculated
by the following formula:
Where: m is the cost ration Cf/Cp.
is the shape parameter.
is the location parameter.
3.8 Schedule maintenance
When preventive maintenance is being mentioned, a number of fixed PM actions done
every month, quarter or season are coming to our minds. Actually PM actions are
based on two main aspects procedure and discipline.
Procedure is that the right actions are being carefully taken at the right time. Discipline
is that all actions are planned and under control. Discipline is the check aspect for PM
actions; hence it could not be overlooked. Failing of implementing a scheduled
maintenance for some organization mainly comes from discipline aspect, which is
being ignored. Scheduled maintenance is one step forward of improving PM actions to
be in accurate need of it (Mobley, 2004).
According to Mobley (2004), there are six main elements for accurate procedure:

Listing of components plus the intervals that they should received a preventive
maintenance.

A schedule for a year that breaks down by month.

Person responsibility to do the work.

Updating the records for the actions that had done, and when next action due
to.

Do any corrective action when it is needed.
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Some PM actions could take a place every interval weather it is necessary or not,
hence, unnecessary stoppage may occurs. Since scheduled maintenance aim to reduce
unplanned stoppages, scheduled maintenance should be based on real actual need of
PM actions.
3.9 Life Cycle Cost (LCC)
Life Cycle Cost is a method to calculate the total cost of a structure during its lifetime.
A system will not be considered as an economical when the maintenance costs are
high. From an economic perspective, the purpose of cost optimization is decreasing
the total life cycle cost (Sarma and Adeli, 2002). LCC analysis is applied with the aim of
selecting the most cost effective approach from different alternatives in order to reach
the lowest long-term cost (Negrea et al., 2007).
According to Negrea et al. (2007, p.217), the LCC includes the following steps:
1. Define the problem which requires LC
2. Alternatives and acquisition/sustaining costs
3. Prepare cost breakdown structure
4. Select analytical cost model
5. Collect cost estimates as well as cost models
6. Make cost profiles for each year of study
7. Make break, even charts for alternatives
8. Pareto charts of vital few cost contributors
9. Analysis of high costs
10. Study risks of high cost items
11. Select preferred course of action using LCC
Furthermore, Liu et al. (2010) state that the value of LCC can be obtained by
calculating the equation below.
CLCC = CI + CO + CM + CF + CD
Where CI is the investment costs, CO is the operation costs, CM is the maintenance
costs, CF is the failure costs and CD is the disposal costs. But only analyzing the LCC one
by one is not enough since the impact of the system will be ignored. It is necessary to
include the cost of externalities (Cexter), which results in the following equation.
CLCC = CI + CO + CM + CF + CD + Cexter
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LCC can be discounted using the present value method at a certain interest rate (i)
after n years, in the present time. The following equation shows the present value of
LCC:
PC LCC = F C LCC / (1 + i)n
Where F C LCC is the future value of LCC and is equal to CLCC calculated above, and PC LCC is
the present value of LCC discounted.
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4. EMPIRICAL DATA
In this chapter, a description of case company, the production process is presented. Moreover,
data gathered from the interviews and observations are included through this part.
4.1 General description
Dynamate Intralog AB is a Swedish company established in 2000, the company is
located in Oskarshamn city and it includes 60 employees in total that distributed into
managers, technicians and workshop workers. The company main roles are
categorized into three functions: providing warehousing, transportation and
production support services.
Dynamate Intralog AB considered Daughter Company of heavy industrial firm known
as Scania that placed in Södertalje; Scania objectives are to deliver optimised heavy
trucks, buses, and engines. Thus, obtaining logistical services from Dynamate Intralog
AB are critical issues among Scania’s production system because of intensive workload
within manufacturing and assembling lines.
4.2 The logistics outsourcing
Dynamate Intralog AB is a third-party logistics (TPL) provider that transports ondemand the containers that required in Scania’s production line, and those containers
contain boxes filled up with truck elements such as electronic items, steering wheels,
doors, and so on.
After supplier delivered containers to Dynamate Intralog AB, the unloaded process
start taken place by workers in which they sort the boxes according to their marked
labels and then shifted to warehouse area by forklifts for later supply. On other hand,
Scania send their empty boxes by trucks to Dynamate in order to store them after
robots processing them, and these procedures related to limited and overloaded
materials in Scania’s facility. The figure below shows TPL circulation between supplier,
Scania and Dynamate.
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Supplier
Dynamate
Intralog
Scania
AB
Figure 5. TPL circulation
Proceeding from the figure, Dynamate highlighted a core value through TPL circulation,
for that reason Dynamate was our interest of study and we are going to present the
empty boxes process in details after it arrives to company and then to robots.
4.3 The empty boxes process
The container arrived to Dynamate Intralog AB from Scania filled by empty boxes (30),
the workers unloading following container and transported through conveyor belt
towards the workstation whilst fork-lifts waiting to carry those empty boxes to
palletizing robot. Then the wooden boxes have been checked by taken lids away to be
ensured that boxes don’t contain any forgotten items by Scania inspectors. The next
step is to organize all boxes upon the production line and let the chain driver take
them into the main robot function which is splitting the box elements into pallet and
collar. This action done by robot arm (Kragplockaren) by folding the collar and place it
in collars area for packaging and transportation purposes, then pallet moved to pallets
region for same goal. This picture shows how the wooden box looks like:
Collar
Pallet
Image 1. The wooden box
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According to Mrs. Norberg (Maintenance Technician of Dynamate Intralog AB), she
defined empty boxes process as simple and complicated system; its simple while all
steps are clear and there is no delay between different operations, and its very
complicated in case of components failure occurred to palletizing robot that stop the
whole production line. The company refers their time and money loses to machine
wear-out, therefore they are depending quite much in maintenance strategies, and in
such a case the corrective and preventive policy have been applied. The problem still in
high rate and need more integrated and flexible maintenance approach in order to
minimize time and money loses.
4.4 Robot performance and critical components
During the interview with Mr. Weli Zubair (Production and Logistic Manager of
Dynamate Intralog AB), he states various kind of failures types encountered the
machine components and he mentioned frequent stoppages because of these
breakdowns. Furthermore, the observations show many disturbances blocking the
robot efficiency in terms of time (availability), consequently the overall equipment
effectiveness declined by limiting the operational time. Hence studying robot critical
components are important to optimize system reliability and conducting best
maintenance practices. According to Mr. Weli Zubair, components can be ranked and
listed as following:
Photocell, gearbox, brake motor, guide rails, chain, turntable, and robot arm; the
system performance proportionally affected by these components availability. Those
elements are going to be clarified in details in later parts.
4.5 Maintenance strategies within Dynamate Intralog AB
Proceeding from Mrs. Norberg statement about maintenance importance in their daily
work and how may reduce the declined production, they been conducted a preventive
maintenance schedule beside a corrective actions policy while failure suddenly occurs
and need to be replaced or fixed instantly. For that reason, weekly and monthly
preventive maintenance procedures have been accomplished through various ways of
problem solving by maintenance technicians and the instruction list have to be
followed by all robot workers even though they are not maintenance engineers. In
addition to all these preventive and corrective actions, the time and money loses still
gradually arising and affects production capacity negatively wherein the accuracy in
scheduling maintenance may enhance those factors positively.
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4.6 Corrective maintenance for essential components
During our visit to the company, we have tried to collect data as much as it’s possible
about robot critical components and its stoppage time for approximately two year
interval. As we mentioned above, the critical components selection based on interview
with production manager that listed by him, consistency with his daily experience. First
component was the photocell and its reflexes, this item exposed to internal and
external errors done by human through accidental crash, or by environmental issues
and not dismissed the electrical contact, in which all situations led to non processing
system and breakdowns to other mechanical parts. Thus, photocell needs a certain
type of corrective maintenance known as replacement by new, and this route
considering money and time consumption, for instance every hour spent in corrective
maintenance its costs 450 SEK and new photocell costs 1500 SEK.
Sequentially, detecting the photocell been broken or non-functioning takes around 30
minutes and 1 hour to be fixed or replaced although the spare parts reserved in
storage and buffer areas, but the consequences from broken photocell involve other
hindrance operations and components to be responding. For example, whilst photocell
not running perfectly the whole process flows through production line deactivated and
passed wooden box may damaged because of mistiming robotic movement, and in
such a case if this wooden box damage and stuck in the robot arm need around 1 hour
to remove it. The dataset about photocell divided into number of failures and their
times to failure been collected through interview and it will be presented in appendix
(3).
Second component is brake motor incurred mechanical failure such friction between
brake cylinder and vanes caused by unstable screws that being running for long term,
plus the erosion appeared on the surface of brake metal. Accordingly, this kind of
failure evaluated as real problem because of time needed to repair the fault and in
case of replacement by new one. In both cases it goes along 3 days to reactivate the
production engine back because of outsourcing demand for brake motor elements as
well as importing maintenance engineers who has capabilities dealing with complex
repair. Nevertheless, detecting the failure takes around 1 hour to be located. Dataset
been gathered and generated based on times to failure and number of failures. See
appendix (4).
Third component is gearbox, its very fundamental components and hard to allocate
the failure immediately might took 2 hours to detect, and in some cases the oil leakage
from the gearbox gives a deterministic sign to stop the robot and recurred the minor
failure before entering the major stoppage that long 2-4 days until the robot reactivate
again. This long period due to gearbox parts outsourcing or buying a new one, that
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type of failure affect other components indeed. Dataset about this component can be
found in appendix (5).
Fourth component known as air bellows, the failure here surrounded by rusty
conditions because of corrosion basis, and that can be known through air leakages.
This is the hardest failure to detect because the system still processing and long farther
in same state down to suddenly breakdown. That’s taking almost 5 hours to detect the
failure and need 4-6 hours to restore the functional circumstances. Data set about this
component shown in appendix (6).
Finally, control list component has over-flooded fat problem suit a sliding collars and
that’s actually a serious problem been observed tangibly through our visit, also the
wear-out and corrosion appeared clearly. When this problem occurred the robot arm
stop working however attempted to reinstall again, and required the worker to stop
machine then try to remove the collar manually and put it in collar area then restart
the machine. This mission takes 5-10 minute to reuse the robot arm, but main problem
was control list and here the maintenance technician has to come and repair or
replace the control list. The time needed split into two branches either item is
available in storages and this operation to repair required 2-3 hours or they have to
wait at least 2 days to receive such a component. Data set of this component shown in
appendix (7).
4.7 Preventive maintenance for essential components
Mrs. Norberg focused on preventive maintenance practices within Dynamate Intralog
AB; she splits the preventive maintenance schedule into two periods, weekly and
monthly. She has talked about other component entailed by this scheduled
maintenance but she gave us information about the critical component that been
ranked by Mr. Weli Zubair.
4.7.1 Weekly preventive maintenance
Photocells and reflexes have to be cleaned every Thursday by a cloth for safety reason;
otherwise any smash may prohibit signalling waves from access reflexes. The worker
has to follow the cleaning instruction step by step and not endanger photocell
positioning mode. This operation doesn’t require long time and high effort, perhaps
takes 20 minutes approximately and the machine does not need to stop. Moreover,
the worker has to replace the photocell almost every two and a half weeks, which may
take up to 1 hour of inactivated robot through replacement operation.
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Photocell
Image 2. The photocell in palletizing robot
Brake motor has to check every Thursday by measuring the distance between the vane
and brake cylinder, the measurement has to shown 0, 2 mm through using feeler
gauge otherwise the screws around the vane must adjust the difference between
them. This operation longs from 15 minutes – 1 hour but the whole machine being
stopped. The replacement screws take place every 2 weeks and the whole robot being
stopped for almost 1 hour, and the screws cost 100 SEK.
Vane
Image 3. The brake motor
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4.7.2 Monthly preventive maintenance
Gearbox undergoes among monthly check to notice the oil leakage and to check if
there is chaotic sound release from it, in such a case the mechanical technicians has to
monitor such a situation. The robot still working during this kind of observations and
inspections, but in extreme cases the machine has to stopped, for example the sound
test showed unusual resonance waves. The inspection takes 30 minutes, besides the
oiling for 20 minutes.
Image 4. The gearbox
Air bellows checked monthly by monitoring the air outlet paths if there is no air
leakage through those channels besides checking the cracks that would happen
throughout chemical reactions, for instance corrosion and erosion. The inspection
longs 30-50 minutes by stopping the machine.
Outlet path
Image 5. The air bellows
Control list criticality considered because of its role importance, the control list has to
be checked from surface cracked or lubricant troubles, for that reason the machine has
to be stopped form 30 minutes- 1 hour and dried from over-flood fat, then pour new
viscosity oil.
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Control list
Image 6. The control list
4.8 Corrective and preventive maintenance related cost
The data have been obtained about corrective maintenance cost covered the following
categories; labour and new item cost. (The stoppage time cost will be discussed in
analysis part).
According to appendix (3) the photocell shows 43 number of failures through 5190
hours, the labour cost as mentioned before was 450 SEK and replacement cost is 1500
SEK.
Then, Brake motor appendix (4) shows 3 number of failures during 5109 hours, the
labour cost approximately 500 SEK and parts cost about 2000-6000 SEK for instance
vane or cylinder damage, but in case of replace it with new item cost 8000 SEK. After
that, the gearbox repair may cost approximately 6000 SEK and can be less when its
minor fix such as small parts within the gearbox, labour cost at least 500 SEK, the
number of failures doesn’t occurs so often which is only 2 times see appendix (5). On
other hand, the air bellows take place quite much in failure number it reach 6 times,
and the labour cost 500 SEK beside the repair and replacement costs that ranging from
1000 SEK- 5000 SEK. At last, control list required more attention because of its
importance related to robot arm, the labour cost is 450 SEK and the replacement or
repair cost ranging from 1000 SEK- 4000 SEK.
The preventive cost classified into time spent in preventive maintenance in terms of
money and the stoppage time cost that are going to be considered in analysis part as
well.
Preventive maintenance cost for photocell within 1 week is 75 SEK and no need for
machine stoppage by cleaning the dirt upon photocell. In paradox with brake motor
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that need to stop the whole robot to make preventive maintenance and it costs up to
500 SEK. The gearbox required only vision inspection to be surely about functioning
conditions, and does not required any spending cost in this activity, because it might
be included through daily work instructions. The air bellows require 225 SEK even as
considered monthly and it’s very sensitive to be checked correctly. Finally the
preventive maintenance cost for control list component may range around 500 SEK
divided into labour and lubricant cost.
4.9 Production list for palletizing robot
Mr. Weli Zubair provided us by data about production list during week 6, this
information contains total number of the boxes that been produced among this week,
the stoppage time and operational time been given. He informed us about company
maximum production value and it was 89 boxes during 1 hour and the current
situation show opposite or bad production rate because of external disturbances and
internal failures. Appendix (2) represent 1067 boxes been produced in one shift during
(week 6) which is reflect negatively to company ultimate goal.
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5. ANALYSIS
This chapter comprising the data that have been collected about maintenance practices and
their correlations to costs, the assumptions been conducted in this part to obtain systematic
calculations.
5.1 Maintenance cost calculations
As we mentioned before the data been gathered within 5190 hours interval of time,
the assumption here related to production unit whereas each box equal to 1 SEK, and
the dataset record 43, 3, 2, 6 and 7 failures number of components respectively.
The total maintenance cost contains of two main factors which are corrective and
preventive maintenance cost, so the calculations will take each component one by
one, starting with photocell, the formulas consists of the following combinations:
(1) Photocell

Corrective cost: Labour cost /h + Stoppage time cost + Replacement item cost
Labour cost /h = 450 SEK
Stoppage time/ one failure number = 2.5 hours
New photocell cost 1500 SEK
Labour cost during 5190 hours = 450×2.5×43 = 48,375 SEK
New photocells cost through 5190 hours = 1500×43 = 64,500 SEK
Stoppage time = 2.5×43 = 107, 5 hours and in terms of money, as the assumption
induce the time variable by each box cost 1 SEK, then Stoppage time cost = 89×1×107,5
= 9,567.5 SEK
Therefore, Corrective cost = 48,375+64,500+9,567.5 = 122,442.5 SEK

Preventive cost: Labour cost + Stoppage time cost+ Replacement item cost
Labour cost during 5190 hours = 150×5190\60 = 12,975 SEK
Replacement item cost = 1500×35= 52,500 SEK
Stoppage time = 34.6 hours, which mean = 89×1×34.6= 3,079.4 SEK
Therefore, preventive cost = 12,975+ 52,500+3,079.4= 68,554.4 SEK
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(2) Brake motor

Corrective cost: Labour cost /h + Stoppage time cost + Replacement item cost
Labour cost /h = 450 SEK
Stoppage time/ one failure number = 30 hours
New parts cost 6000 SEK
Labour cost during 5190 hours = 450×30×3 = 40,500 SEK
New parts cost through 5190 hours= 6000×3 = 18,000 SEK
Stoppage time = 30×3= 90 hours, then Stoppage time cost = 89×1×90= 8,010 SEK
Therefore, Corrective cost = 40,500+18,000+8,010 = 66,510 SEK

Preventive cost: Labour cost + Stoppage time cost+ Replacement item cost
Labour cost during 5190 hours = 450×5190\60 = 38,925 SEK
Replacement item cost = 100×5190\120= 4,325 SEK
Stoppage time = 173 hours, which mean = 89×1×173= 15,397 SEK
Therefore, preventive cost =38,925 +4,325 +15,397 = 58,647 SEK
(3) Gearbox

Corrective cost: Labour cost /h + Stoppage time cost + Replacement item cost
Labour cost /h = 450 SEK
Stoppage time/ one failure number = 30 hours
New parts cost 6000 SEK
Labour cost during 5190 hours = 450×30×2 = 27,000 SEK
New parts cost through 5190 hours= 6000×2 = 12,000 SEK
Stoppage time = 30×2= 60 hours, then Stoppage time cost = 89×1×60= 4,800 SEK
Therefore, Corrective cost =27,000 +12,000 +4,800 = 43,800 SEK
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
Preventive cost: Labour cost + Stoppage time cost+ Replacement item cost
Labour cost during 5190 hours = 56.25×5190\60 = 4,865.625 SEK
Replacement item cost = 0 SEK
Stoppage time = 28.6 hours, then Stoppage time cost = 89×1×28.6= 2,551.3 SEK
Therefore, preventive cost = 4,865.625 + 2,551.3 = 7416.9 SEK
(4) Air bellows

Corrective cost: Labour cost /h + Stoppage time cost + Replacement item cost
Labour cost /h = 450 SEK
Stoppage time/ one failure number = 11 hours
New parts cost 2500 SEK
Labour cost during 5190 hours = 450×11×6 = 29,700 SEK
New parts cost through 5190 hours= 2500×6 = 15,000 SEK
Stoppage time = 11×6= 66 hours, then Stoppage time cost = 89×1×66= 5,874 SEK
Therefore, Corrective cost =29,700 +15,000 +5,874 = 50,574 SEK

Preventive cost: Labour cost + Stoppage time cost+ Replacement item cost
Labour cost during 5190 hours = 56.25×5190\60 = 4,865.625 SEK
Replacement item cost = 0 SEK
Stoppage time = 72 hours, then Stoppage time cost = 89×1×72= 6,408 SEK
Therefore, preventive cost = 4,865.625 + 6,408= 11,273.625 SEK
(5) Control list

Corrective cost: Labour cost /h + Stoppage time cost + Replacement item cost
Labour cost /h = 450 SEK
Stoppage time/ one failure number = 3 hours
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New parts cost 2000 SEK
Labour cost during 5190 hours = 450×3×7 = 9,450 SEK
New parts cost through 5190 hours= 2000×7 = 14,000 SEK
Stoppage time = 33×7= 231 hours, then Stoppage time cost = 89×1×231= 20,559 SEK
Therefore, Corrective cost =9,450 +14,000 +20,559 = 44,009 SEK

Preventive cost: Labour cost + Stoppage time cost+ Replacement item cost
Labour cost during 5190 hours = 112.5×86.5=9,731.25 SEK
Lubricant cost = 500 SEK
Stoppage time = 21.625 hours, then Stoppage time cost = 89×1×21.625= 1,924.6 SEK
Therefore, preventive cost = 9,731.25 +500+1,924.6 = 10,233.1746 SEK
5.2 Distribution for the failure data
Figure 6. No. of failures for each critical component during 5190 hours
A graph above gives a holistic view about the failures that occurred during 5190
operating hours. While X axis represent the components and Y axis represents the
numbers of failure. From the graph it is so obvious that photo cell has the most
frequent failure comparing with the rest of the components.
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Figure 7. The time at which the components have failed
A graph above shows failure time for critical components. X axis represents the
number of failures while Y axis represents the time at which the component has been
failed. Failures have been registered over a 5190 operating hours. It is obviously shown
that Guide ways for example has registered three failures during these operating
hours. Those failures registered in different hours.
Due to large Number of failures for the ‘photo cell’ component it has been taken out of
the graph in order to clarify the other components.
5.3 Optimum time to PM actions
In order to find the optimum time for each components, we need to use the following
equation 3.7. Under the following assumptions:
-
The component as good as new after performing the maintenance actions.
Cf is incurred if the component fail before time=T.
Preventive maintenance is performed with Cp on a component at time =T.
Time to fail follows a Weibull distribution.
T= (m.η) + γ
Gamma is zero here for all components, thus value of m can be easily found from the
table appendix (8).
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Table 2. The value of m
Source: Dodson, 1994
Finding the value of m, substitute it with equation 1 we have the optimum time to
replace, maintain the component. Table 4 bellow shows the optimum time for each
component. For the value of B see appendixes (9, 10, 11, and 12).
Cc/Cp
β
η
m
γ
Opti T
= (m.η) +
Ration cost
γ
Photocells and
reflexes
1.786063
9.3
121.8
2.229
0
97.8054
brake Motor
1.134073
1.36
4178
2.229
0
9312.762
Gear Box
5.905432
1.26
4667
0.574
0
2678.858
Air Bellows
4.486046
1.48
4048
0.746
0
3019.808
Control list
4.30062
2.75
4570
0.521
0
2380.97
Table 3. The optimum time for each component to perform PM
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5.4 Maintenance schedule based on optimum time
Maintenance scheduling could be done based on the calculating the optimum time. Graph
bellow shows if we could run PM actions for more than one component at the same time in
order to minimize the total PM costs. Notice Photo cell has been taken out with respect to
scale.
Figure 8. Current PM Time and Optimum time
It is obvious that there are some unnecessary PM actions which mean extra
unnecessary cost. PM could be rescheduled and implemented in groups for multiple
components.
Components
Time to perform PM
Photocells and reflexes
97.8054
Break Motor
9312.762
Gear Box
2678.858
Air Bellows
3019.808
Control List
2380.97
Table 4. The optimum time to perform PM
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6. RESULTS
In this chapter the final result of the thesis analysis presented. With respect to the data
collected from the company.
As results, critical components have been determined based on an interview with the
production manager. Those components are photocells and reflexes, Break motor,
Gear box, Air bellows, and Control list. Failure data for these components have been
followed a Weibull distribution. In addition estimating the value of beta and eata for
each component are essential to calculate the optimum time.
The optimum time for the critical components have been determined to let PM actions
take a place. Optimum hours are 98, 9313, 2679, 3020 and 2381 hours for photocells
and reflexes, Break motor, Gear box, Air bellows, and Control list respectively.
Based on the calculation, schedule for those critical components has been made,
aiming to reduce the total number of PM actions. A group of actions has been
scheduled together as table 4.
The number of PM actions could be reduced by knowing the exact time, therefore
unnecessary cost could be avoided. In the previous table, a PM could be performed
after 2200 operating hours for Control list and Gear Box. Instead of apply PM every
900 hours. It is obvious that two round of PM unnecessary. There for those costs could
be avoided, the total cost for PM actions for the whole components would be
Cost of PM for Component x1, 2,…n * Number of PM
Cost of PM After scheduling 87570.7 sek which is definitely less than the current one
since the total no. of planned stoppages has been reduced.
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7. CONCLUSION
In this chapter the conclusion of our thesis which answer the problem formulation, criticism to
our thesis and suggestions
Company experience is one method to identify the critical components. Using FMECA
for example could make the analysis more accurate. Recording the failures of the
machine is one step forward to enhance the availability of it. Analyse a historical data
is the second step to improve the maintenance policy.
In order to avoid extra unnecessary cost Optimum time has been scheduled; PM
actions should be implemented at the right time that could effective. By determining
the optimum time a group of PM actions could be implemented at once.
The main aim for scheduling the PM actions is to identify the right time interval that
should the PM actions performed. Based on analysis and knowing the behaviour of the
system or component. Readability and availability of the data play big role in the result
part and sometimes could affect a critical decision that the organization should make.
Subsequently, the reasons behind the failure types are the core of analyzing the data
by Weibull, and the way that trends through dealt with certain reasons of faults and
not the all causes. Thus, FMEA is an integrated reliability analysis tool in order to cover
all the critical failures and their consequences in the system. For instance the external
factors may enhance the failure occurrence and the environment indeed, therefore, to
manage better solutions these factors have to be considered.
Recommendations
The company needs to enhance their workers to have enough knowledge about those
components and have best maintenance practices manual to monitor all the machine
elements in case of stoppage and breakdowns.
Furthermore, the inspection is very fundamental for avoiding system idle and
malfunctions components later on, hence may condition based maintenance be
suitable to be investigate in future.
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Scheduled maintenance policy for minimum cost – A case study
Linnaeus University – Spring 2011
References
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Scheduled maintenance policy for minimum cost – A case study
Linnaeus University – Spring 2011
Appendix 1
QUESTIONNAIRE
1- What is the logistic connection between Dynamate and Scania along the supply
chain?
2- Is there any lead-time or barriers through production process flow?
3- Is the problem classified into following categorizations:
a- Logistic
b- Machinery and Maintenance
c- Human resources
4- What is the production process according to robot main functions?
5- Is the robot stoppage time related to:
a- Failure components
reasons
b- company policy such as just in time
c- other
6- Is the declined production whilst the robot deactivates involved critical
components Breakdowns and what are these components?
7- What are the maintenance strategies and policies towards these components,
and if there any analysis tools dealing with?
8- Is there any human factors and untrained worker may affect negatively the
palletizing robotic system?
9- How much money and time were spending in corrective maintenance respect
to failure components?
10- How much money and time were spending in preventive maintenance and if it
been scheduled?
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Scheduled maintenance policy for minimum cost – A case study
Linnaeus University – Spring 2011
Appendix 2
Number of boxes that been produced
Disturbance time or stoppages/minutes
First day of work is 22 from 26 boxes
70 minutes
Second day of work is 26 from 35 boxes
120 minutes
Third day of work is 26 from 39 boxes
60 minutes
Fourth day of work is 23 from 34 boxes
40 minutes
This table present the boxes number that been produced in week 6
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Scheduled maintenance policy for minimum cost – A case study
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Appendix 3
Component
No. of failures during 5190 hours
Photo cell
43
Failure No.
at T, hour
1
119
2
224
3
355
4
455
5
590
6
719
7
835
8
964
9
1085
10
1218
11
1332
12
1459
13
1581
14
1689
15
1818
16
1914
17
2031
18
2156
19
2291
20
2381
21
2521
22
2612
23
2701
24
2820
25
2917
26
3049
27
3161
28
3285
29
3418
30
3546
31
3654
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Scheduled maintenance policy for minimum cost – A case study
Linnaeus University – Spring 2011
32
3745
33
3885
34
3990
35
4085
36
4196
37
4303
38
4420
39
4519
40
4640
41
4734
42
4847
43
4976
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Scheduled maintenance policy for minimum cost – A case study
Linnaeus University – Spring 2011
Appendix 4
Component
Brake Motor
Failure No.
1
2
3
No. of failures during 5190 hours
3
at T
2407
3641
4753
Table of failures for Brake motor during 5190
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Scheduled maintenance policy for minimum cost – A case study
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Appendix 5
Component
Gear Box
Failure No.
1
2
No. of failures during 5190 hours
2
at T
3588
4799
Table shows the failure time for Gear box
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Scheduled maintenance policy for minimum cost – A case study
Linnaeus University – Spring 2011
Appendix 6
Component
Gear Box
Failure No.
1
2
No. of failures during 5190 hours
2
at T
3588
4799
Table shows the failure time for Gear box
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Scheduled maintenance policy for minimum cost – A case study
Linnaeus University – Spring 2011
Appendix 7
Component
Control List
Failure No.
No. of failures during 5190 hours
7
at T
1
2700
2
2923
3
4
3444
5
4133
4456
6
4878
7
5045
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Scheduled maintenance policy for minimum cost – A case study
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Appendix 8
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Scheduled maintenance policy for minimum cost – A case study
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Appendix 9
Break Motor
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Appendix 10
Gear Box
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Scheduled maintenance policy for minimum cost – A case study
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Appendix 11
Air bellows
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Appendix 12
Control List
58