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White Paper Impala Competence Center Asset Management
Investment Analysis – Life Cycle Cost Analysis:
Cost Driven Risk Management
Benny Lauwers Consultancy Manager Impala
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Content
1. Introduction ............................................................................................................. 3
2. Risk matrices........................................................................................................... 3
3. Converting the risks in a cost model .................................................................... 4
4. The notion “risk cost” ............................................................................................ 7 4.1. Costs that are charged ....................................................................................... 7 4.2. Determination of the year where the costs are charged..................................... 8
5. Decision ................................................................................................................. 15
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1. Introduction
Risk management involves not only the evasion of risks, but also the acceptance of certain risks.
The trend of accepting and controlling risks instead of merely reducing them has been going for some
time.
In the world of asset management risk analyses are often used to determine the critical failure modes of an asset, and how these failure modes can be made less critical. In this case critical is defined as a
situation where the failure of an asset and the consequences for the company are viewed as hazardous.
Failure modes can be made less critical by reducing the frequency and consequences of the failure. This can be done by changing the asset and making use of more reliable components. Another method is
setting up a predictive maintenance programme that shows the erosion or wear, and can enable action before the asset fails. Throughout the white paper the word ‘action’ will be used. Please keep in mind that this concept can be interpreted in different manners in practise.
The methodology that will be described in this paper can provide support in the decision whether an
action should be taken based on life cycle cost. The life cycle cost is based on the determination of the cost of risk, which is the product of the probability of failure and the costs of the failure consequences.
2. Risk matrices
When executing a risk analysis, often use is made of risk matrices. A risk matrix shows the probability of
failure with reference to the consequences of the failure. The combination fo this probability and consequence provides an analysed failuremode in a specific place in the risk matrix. The risk matrix often shows a critical zone, where the frequency of the failure and/or the consequences are very large;
and a non-‐critical zone where they are small. The critical and non-‐critical zones are seperated by a criticalityborder.
In some risk matrices one can find several degrees or levels in the transition of the critical to non-‐critical
zone. In this paper we will confine ourselves to a simple matrix, with a critical and non-‐critical zone, but please be aware that the reasoning is identical to the risk matrices that have several degrees or levels.
Consequence: Unavailability / Failure of the asset Probability of failure
(1/x) 16 hours 8 hours 4 hours 1 hour
6 months (1/2) Non-‐critical zone
3 months (1/4) 1 month (1/12) 1 week (1/52) Critical zone
Figure 1: Example of a risk matrix with a critical zone and a non-critical zone
The classical method of working when a failuremode enters the critical zone is that several actions will
have to be undertaken to ensure that the failuremode is moved from this zone. This can be done by
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taking preventive actions, that lower the probability of failure, but it is also possible to take damage reducing actions, that lower the consequences of the failure. The preference is usually given to the
preventie actions, before taking damage reducing measures.
Currently, however, does the undertaking of these actions involve an evaluation of the costs of these actions? And to what extent are they compared to the added value that the actions will bring? Usually
neither of these questions are asked.
In many cases a convulsive search for solutions is done that will keep the failuremode outside of the critical zone, regardless the costs of these solutions. Experience teaches us that it can be more cost
efficitent to accept the failuremode in the critical zone, as the costs of the actions do not outweigh their actual financial added value. This is known as ‘cost driven risk management’.
The method that uses life cycle costs to decide whether an action can be taken will be described
further, and is the subject matter of this paper. This methodology can also be used to compare several different possible actions, and to enable the choice for an action with the highest added value.
3. Converting the risks in a cost model
As was described in the introduction, a risk matrix is a summary of the possible probabilities and the probable consequences.
An assumption is made that every consequence, such as a production stagnation of a specific time, a
quality problem, an environmental or safety incident, etc. can be converted to a specific cost.
When we set up a matrix, where the consequence is shown as a function of the cost, and we combine
the probability of failure with these consequence costs, it is known as the ‘risk costs’.
The matrix can then be set up as follows:
RC = Risk costs= Probability x Costs
(in this example we use costs on a yearly basis to simplify the matrix \)
Consequence: Unavailability / Failure of the asset Probability of failure
(1/x) 16 hrs = 3.200 € 8 hrs = 1.600 € 4 hrs = 800 € 1 hrs = 200 €
6 months (1/2) RK = 2 x 3.200€ RK = 2 x 1.600€ RK = 2 x 800€ RK = 2 x 200€ 3 months (1/4) RK = 4 x 3.200€ RK = 4 x 1.600€ RK = 4 x 800€ RK = 4 x 200€ 1 month (1/12) RK = 12 x 3.200€ RK = 12 x 1.600€ RK = 12 x 800€ RK = 12 x 200€ 1 week (1/52) RK = 52 x 3.200€ RK = 52 x 1.600€ RK = 52 x 800€ RK = 52 x 200€
Figure 2: Example of a risk matrix where probability is set out against consequencecost
The risk costs in the red area will not be accepted, and is defined as critical from here on. The risk costs
in the green area are seen as acceptable and are not seen as critical
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According to the commonly used methodologies we would like to emphasise that when analysis shows that the critical zone has been reached, an action will be necessary to ensure that the critical zone will
be exited.
Through taking actions we will attempt to lower the risk costs in such a manner that it becomes acceptable. An important aspect to mark is that any action that will be taken will also involve action
costs.
Consequence: Unavailability / Failure of the asset Probability of failure
(1/x) 16 hrs = 3.200 € 8 hrs = 1.600 € 4 hrs = 800 € 1 hrs = 200 €
6 months (1/2) RC = 2 x 1.600€ 3 months (1/4) Cost of action 1 month (1/12) 1 week (1/52) RC = 52 x 1.600€
Figure 3: Lowering the riskcosts + costs of action
A cost efficient action thus is an action where the costs, added to the new acceptable risk costs, are
lower than the original, unacceptable risk costs.
This can also be expressed as follows:
Cost of action + Non-‐critical risk cost < Critical risk cost
Or
CA + CNC < CC (1)
Ca = Cost of the action CNC = Non-‐critical risk cost CC = Critical risk cost
When we look at the example (figure 3) this becomes:
Cost of the Action + RC: 2 x 1.600€ < RC 52 x 1.600€
If the critical risk costs turn out to be lower than the non-‐critical risk costs added to the cost of action, one can question the added value of the action. It may be better to (temporarily) accept the risk costs. It is evident that a strategic choice can be made to undertake the action, regardless of the cost
efficiency of the action.
In the reasoning so far we have assumed a risk cost that can be lowered on one occasion by taking a specific action.
In structural asset management, however, it is important that the costs of an action are specified for
the total life cycle of the asset. The action costs will have to be written off over the remaining life time
that the asset will be in use.
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This is why the notion ‘Net Present Value’ will be introduced. The Net Present Value (NPV) takes into account all the different cash flows, coupled to the asset, that flow in and out in a number of years. The
NPV discounts these future cash flows to their current or present value. The sum of all these discounted yearly cash flows is the Net Present Value.
(2)
Where:
Ct = the cash flow in year t t = the year
r = discount rate / weighed average capital cost
It is necessary to discount to present value to ensure that inflation is taken into account, but also to realise a minimum return on investment, in this case for an action. This is the so-‐called Weighted
Average Cost of Capital (WACC). Usually this can be requested at the financial department of an organisation.
To discount the value of a future cash flow in year t to present value, the following formula can be used.
(3)
PV = Present Value; Current value of future cash flow
To determine the NPV both incoming and outgoing cash flows are necessary. In this analysis we assume that, regardless of the action, the incoming cash flows will remain equal. In practise, these cash flows
will increase due to the fact that an asset will attain a higher availability. For this analysis this is not relevant.
In this analysis we will thus work solely with outgoing cash flows. Moreover, we will only work with
specific consequence costs (deprivation costs) and the costs of action. We assume that the other outgoing cash flows will remain the same, despite the action.
Please be aware that we will be using positive values in our calculations, where outgoing cash flows and
thus negative values should be used.
By using the Net Present Value, the formula for critical action (1) can be rewritten:
NPVA + NPVNC <NPVC (4)
NPVa = Cost of the action
NPCNC = Non critical risk cost NPVC = Critical risk cost
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In this calculation the Net Present Value of the action is also taken into account, as actions do not always have to take place straight away and can be planned in future years. The action can be taken in a
specific year, or can be spread over several years, which means that depreciation has to be taken into account.
On the basis of the total NPV, the left side of the formula (4) can show the optimal timing for the
implementation of a single action. This is the year that shows the lowest NPV. Several scenarios thus need to be calculated, where the year the action is taken varies.
Example: If the remaining life cycle of an asset is another 20 years, 20 scenarios’ can be calculated,
where the cost of the action is billed in different years. The scenario with the lowest total NPV is the year that the action should be implemented, according to cost strategy. In the period of time before the implementation of the action, the ‘old’ risk cost has to be taken into account; while the ‘new’ risk cost
has to be taken into account after the action is completed.
4. The notion “risk cost”
As has been mentioned before, risk cost is the product of consequence cost and probability. Both
parameters are elements of the risk matrix. The probability can often be found immediately, and the consequences can then be adjusted to costs.
The probability that something will happen is a function of the failure pattern of the asset. The determination of this failure pattern will be described extensively in this document.
At this point, the determination of the specific costs that are used in the calculations are key, as well as
the year in which these costs are effectively booked or charged.
4.1. Costs that are charged
The following costs are components of the total factor costs:
Direct costs = maintenance costs:
• Direct labour costs as a result of yearly maintenance activities • Direct material costs as a result of yearly maintenance activities
There is a chance that these costs will be different before and after the action has been taken. An
example would be when a maintenance plan is altered of made more specific, which could lead to the addition or removal of maintenance activities.
Indirect cost = deprivation cost:
• The consequence costs that are coupled to the failure of the asset. This could be losses due to loss of production, costs coupled to an accident etc.
Here it is also possible that these costs differ before and after taking the action.
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4.2. Determination of the year where the costs are charged
The direct costs for a specific year can be determined and assigned very easily. This a summation of the maintenance costs, and possible labour-‐ and material costs of longer shutdowns and revisions.
In the example shown below, the yearly maintenance costs remain the same. The exception is a 3-‐
yearly revision which leads to higher costs in those specific years (3, 6, 9, 12 …).
Total maintenance costs per year
Year 1 Year 2 Year 3 Year 4 Year 5
Direct labour 6.200 € 6.200 € 9.400 € 6.200 € 6.200 € Direct material 1.300 € 1.300 € 2.900 € 1.300 € 1.300 € Total maintenance cost 7.500 € 7.500 € 12.300 € 7.500 € 7.500 € PV maintenance cost @ WACC 7%
7.009,35 € 6.550,8 € 10.040,5 € 5.721,75 € 5.347,4 €
Table 1: Total yearly maintenance costs (PV: Present Value; WACC: Weighted Average Cost of Capital)
Remark: The current costs (year 0) are not taken into account in this calculation. The option exists to incorporate these costs. When they are incorporated they should not be discounted as they are
(logically) at present value.
The difficulty in this case is the determination of the year in which the deprivation costs should be charged, as this is a function of the failure pattern of the asset.
There are two ways in which this can be determined:
• A deterministic method: Simple, but not very accurate • A stochastic method: A little more complex, but more accurate
In both cases the starting point is the Mean Time To Failure (MTTF) or Mean Time Between Failure
(MTBF). The theoretical discussion of the difference between the two is not shown here.
4.2.1. Determination of the failure pattern according to the deterministic method
In this case the deterministic method assumes that we know when an asset will fail. In practise this is not realistic, as we will never know when an asset is going to fail. This is impossible to predict.
We assume that the value of the MTTF is the moment that the asset will actually fail, even though this
is an average value in reality. This is the reason that the deterministic method is the least accurate method.
This method is very suitable to give a simple and fast indication.
The MTTF can help determine the number of failures per year. The table shows these values for different values of the MTTF.
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Number of failures per year
Year 1 Year 2 Year 3 Year 4 Year 5 Year 6 Year 7
MTTF = 1 year 1 1 1 1 1 1 1 MTTF = 3 year 0 0 1 0 0 1 0 MTTF = 0,5 year 2 2 2 2 2 2 2 MTTF = 1,2 year 0 1 1 1 1 1 0
Table 2: Examples of a failure pattern in the function of the MTTF
A deprivation cost can now be assigned to every failure. The table shown below shows the complete
cost overview per year, taking into account the direct and indirect costs. In this example we use a deprivation cost of € 25.000 and an MTTF of 1,2 years.
Total costs per year Year 1 Year 2 Year 3 Year 4 Year 5 Direct labour 6.200 € 6.200 € 9.400 € 6.200 € 6.200 € Direct material 1.300 € 1.300 € 2.900 € 1.300 € 1.300 € Total maintenance costs 7.500 € 7.500 € 12.300 € 7.500 € 7.500 € Deprivation costs @ MTTF = 1,2
0 € 25.000 € 25.000 € 25.000 € 25.000 €
Total cost 7.500 € 32.500 € 37.300 € 32.500 € 32.500 € PV cost @ WACC 7% 7.009,35 € 28.386,76 € 30.447,91 € 24.794,09 € 23.172,05 € NPV 113.810,16 €
Table 3: Total cost per year with a deterministic failure pattern and an MTTF of 1,2 years
Pay attention: This calculated Net Present Value shows only the outgoing cash flow, more specifically
the direct costs related to maintenance and the deprivation costs related to the failure of the asset.
If the MTTF, the costs of the action and the direct and indirect costs are known for the situation before taking the action and after taking the action, one can compare the NPV’s of both situations.
On the basis of these calculated NPV’s one can decide whether the action should be implemented. In
the example shown in the table above an MTTF of 1,2 years was used. If we label that example as the current situation, and assume that the action ensures that the MTTF has increased to 4 years. We
assume that the direct cost and the deprivation costs remain unchanged after taking the action. The situation after taking the action is shown in the next table.
Total costs per year Year 1 Year 2 Year 3 Year 4 Year 5 Direct labour 6.200 € 6.200 € 9.400 € 6.200 € 6.200 € Direct material 1.300 € 1.300 € 2.900 € 1.300 € 1.300 € Total maintenance costs 7.500 € 7.500 € 12.300 € 7.500 € 7.500 € Deprivation costs @ MTTF = 1,2
0 € 0 € 0 € 25.000 € 0 €
Total cost 7.500 € 7.500 € 12.300 € 32.500 € 7.500 € PV cost @ WACC 7% 7.009,35 € 6.550,8 € 10.040,5 € 24.794,09 € 5.347,4 € NPV 53.742,14 €
Table 4: Total cost per yearwith a deterministic failure pattern and an MTTF of 4 years
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Comparing the NPV’s provides the following results:
NPV before taking the action: 113.810,16 € NPV after taking the action: 53.742,14 € Result: IMPLEMENT ACTION
Table 5:Comparison of NPV’s before and after taking the action
When we want to execute a single action in this example, such as a modification, with a cost of €
12.500, we can use the data to assess several scenarios. In the scenario’s the action will be taken in different years. The scenario with the lowest total NPV shows the year that the action can best be implemented.
NPV scenario year 1: 65.424,33 € NPV scenario year 2: 86.496,04 € NPV scenario year 3: 116.229,69 € NPV scenario year 4: 115.562,16 € NPV scenario year 5: 132.762,95 €
Table 6:Comparison NPV’s ‘best implementationyear’ of single action
Graph 1: Comparison NPV’s ‘best implementationyear’ of single action according to the deterministic method
We clearly see that the single action can best be implemented in the first year. We can even seen that if the action is taken in year 3 or later, it does not make sense, as the NPV is larger after that year, rather
than smaller.
In other words, if the practical implementation of this single action will not be possible before year 3, due to budgeting problems for example, there is no added value to reserving the budget or putting
energy into preparing the action. In this case we can attempt to find an alternative, or accept the current risk cost.
€ 0.00
€ 20,000.00
€ 40,000.00
€ 60,000.00
€ 80,000.00
€ 100,000.00
€ 120,000.00
€ 140,000.00
Year 1: Year 2: Year 3: Year 4: Year 5:
Recommended year of implementation of action
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If the action is simply changing the maintenance plan of the asset, then the direct cost before and after the action will be different. Even if this is the case it is possible that the choice is made for an additional
single action. In the function of the calculated NPV it can be decided that the altered maintenance plan alone will be sufficient.
4.2.2. Determination of the failure pattern according to the stochastic method.
When using the deterministic method for our calculation, we have used the premise that we know exactly when a failure will take place, using the average value MRRF. We are aware that in real life this
will not be possible.
By viewing the failure behaviour of the asset stochastically, we attempt to incorporate a certain amount of accuracy.
The failure behaviour of an asset can be viewed stochastically in several ways. Here we use the most
common probability (distribution) functions. In the world of asset management we usually work with the following distribution functions:
• Exponential distribution
• Standard distribution (Bell curve) • Weibull distribution
The way that the failure behaviour of the asset is distributed and which probability function we thus
may use depends on several factors.
Every probability function has its specific shape and parameters that are necessary to perform the
essential calculations.
Exponential distribution Standard distribution 2 parameter Weibull distribution
Curve shape
Parameters λ = constant failure rate µ = average
σ = standard deviation β = shape factor η = scale factor
Probability distribution-‐ function f(t) =
Reliability-‐ function R(t) =
Table 7:Functionshape and parameters of the most commonly used probability distributions in asset management
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The standard distribution is the most commonly used probability distribution in nature. Failure behaviour of assets that has a standard distribution has increasing failure intensity. The degradation
symptoms, amongst other, can thus be quickly described according to a standard distribution function.
The Weibull distribution can be used as a 2 parameter of 3 parameter distribution. The 2 parameter distribution is shown in table 7. When using a 3 parameter distribution, the third parameter is ‘minimal
lifetime’, the lifetime in which a component will not fail.
The Weibull distribution provides a good demonstration of situation in which several failure mechanisms, individually, can be the cause of failure of a component of an asset.
In the function of the value of the shape factor β, the Weibull distribution contains several other
probability distribution functions, such as: β = 1: Exponential distribution
β = 2: Raleigh distribution β = 3,5: Normal distribution
The exponential distribution is the most simple to use and to perform the calculations. This is why this
function is often used to simulate failure behaviour. The exponential distribution is based on ‘random’ failure behaviour. Due to the mathematical simplicity of the exponential distribution, we will use the exponential distribution in this document. Please note that the working method for other probability
functions identical.
Based on the specific parameters of the probability distributions numerous failure patterns can be set up.
Similar to the approach in the deterministic method, we will need the number of failures per years to
set up the failure pattern.
The probability that the component will still fulfil its function at a certain time is the reliability R(t) of
this component on that specific time. The probability for the exponential distribution in the function of the constant failure rate λ is equal to:
(5)
Where: t = a chosen time expressed in X year
λ = the (constant) failure rate, equal to (6)
To map a certain failure pattern, we will attempt to determine certain chosen failure times, where it is
of essence that these failure patterns still meet the exponential distribution with a predetermined identical constant failure rate λ.
This can be done by choosing random values of reliability R(t) and calculating the failure time several
times:
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(7)
In this equation t is time, expressed in year, where a failure occurs. In table 8, shown on the next page, 7 of these random times of failure are illustrated.
Random Fail generator
1 2 3 4 5 6 7
Fail after x years 0,37196 1,12248 2,52626 0,06852 2,91534 5,07631 1,04147
Table 8: Random times of failure, exponentially distributed with MTTF=1,2 (λ = 0,833)
If these failures are shown cumulatively, we can see a very specific failure pattern, where we can notice how this failure pattern changed over time. The table below shows the amount of time, in years, has passed after failure number 7.
Random Fail generator
1 2 3 4 5 6 7
Failure pattern 0,37196 1,49444 4,02070 4,08922 7,00456 12,08087 13,12234
Table 9: Cumulative failure pattern over time, exponentially distributedwitha MTTF =1,2 (λ = 0,833)
We can see in this table that the first failure occurs after 0,37 years. The second failure occurs 1,49 years from now, etc. At the occurrence of failure number 7, more than 13 years have passed.
On the basis of this information we can determine the number of failures that will yearly take place,
according to the specific failure pattern.
Failure pattern on a yearly basis
Year 1 Year 2 Year 3 Year 4 Year 5 Year 6 Year 7
# failures per year 1 1 0 0 2 0 0
Table 10: Possible failure pattern, exponentially distributed with MTTF =1,2 (λ = 0,833)
Using the data of the cumulative failure pattern, the table shown above could be expanded with an
additional failure in year 13 and 14 (number 6 and 7).
If we now choose alternative random values for R(t), we will find another failure pattern. It is important to know that all these failure patterns are still valid for the exponential distribution according to the
chosen value of λ.
The failure pattern in the table has been set up with an MTTF of 1,2 year (λ = 1/1,2 = 0,83333). What immediately attracts our attention is that the same MTTF of 1,2 years, a clearly different failure pattern
is found using this method compared to the deterministic method. This is logical, as there are an infinite number of failure patterns possible with an MTTF of 1,2 years, and we have shown 2 random patterns.
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We now have sufficient data as input for the table with the various yearly costs. The procedure of filling this table is identical to the procedure used in the deterministic method.
Total costs per year Year 1 Year 2 Year 3 Year 4 Year 5 Direct labour 6.200 € 6.200 € 9.400 € 6.200 € 6.200 € Direct material 1.300 € 1.300 € 2.900 € 1.300 € 1.300 € Total maintenance costs 7.500 € 7.500 € 12.300 € 7.500 € 7.500 € Deprivation costs @ MTTF = 1,2
25.000 € 25.000 € 0 € 0 € 50.000 €
Total cost 32.500 € 32.500 € 12.300 € 7.500 € 57.500 € PV cost @ WACC 7% 30.373,83 € 28.386,76 € 10.040,46 € 5.721,71 € 40.996,71 € NPV 115.519,47 €
Table 11: Total cost per year using the stochastic determined failure pattern and MTTF of 1,2 years
If we would generate an alternative failure pattern, the total NPV, as well as the decision whether action would be taken, would be different.
The total NPV and the decision to implement an action, is thus dependent on the randomly generated
failure pattern. This can be an aggravating situation as a basis for a decision. If the NPV’s that we have calculated for the situation before and after implementing an action are very close, one failure pattern could show a decision should be implemented, where another failure pattern could show that it should
not.
This is the uncertainty that rears its head again, due to the fact that we cannot predict the exact time of the failure of an asset.
To lower this uncertainty we will generate several different failure patterns by using a Monte Carlo
Simulation. For each failure pattern we will decide whether the action should be implemented or not.
Through this method it could be the case that 200 different failure patterns within an exponential
distribution with the same MTTF, the results show that 146 times a decision should be made to implement the action, and 54 times a decision should be made to not implement the action. In other words, when we take into account the uncertainty of the exact failure pattern, we have a 73& chance
that the action leads to positive results and a 27% chance that it does not.
0.00%
100.00%
OK Not ok
Decision Action / No Action
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Graph 2:The probability of a positive or negative effect on the decision to implement an action
On the basis of these numbers a decision can be taken. Every organisation will have their own decision
restrictions. One organisation may implement a decision only if the probability lies above 80%, where another organisation may decide to take action if they are 55% certain.
When the NPV values are divergent, the percentages will also lie further apart, which makes taking a
decision easier. An example would be 99% positive and 1& negative.
The greater the number of failure patterns that are simulated, the more accurately the percentages will approach reality.
To determine in what year a single action can best be implemented, a similar approach to the
deterministic method is used. Once again the function of the generated failure pattern will show the ‘best year for implementation’. By using the Monte Carlo Simulation for every failure pattern, and
registering in what year the single action can best be implemented, one can see how many times a specific year is uncovered as the best implementation year. Using this data one can then make the decision when the action should be implemented. In the example shown below it is recommended to
take the single action in year one, as this year scored best in 86 of the 200 simulations.
Graph 3:Comparison NPV’s ‘best implementation year’of single action according to the stochastic method
5. Decision
Risk management does not only imply the negation of risks, but it can also mean that some risks need
to be accepted.
This is what can be done using the methodology shown in this paper. The method can support and verify whether an action should be implemented or not. In other words, we attempt to decrease the
risk as much as possible, or we accept the risk as the costs or action does not outweigh the added value of the action. The method can also be used to compare different potential actions with reference to one another.
0 20 40 60 80 100
Year 1 Year 2 Year 3 Year 4 Year 5
Recommeded year of action implementation
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Every organization attempts to work in a cost efficient way. The methods that have been described take into account the cost aspects in decisions. The taking of actions to lower risks is expressed in a ‘financial
value’, which supports and simplifies the budget justification of asset managers.
Risk analysis and cost driven decision making will become increasingly important for asset managers to optimise the improvement plans and the performance of their assets.
In closing a last remark is made to emphasise the importance and the use of the methods. The methods
should be seen as providing guidance. It will never be possible to determine the exact failure pattern of an asset, and we can thus never take a decision that we are absolutely certain about.
The use of the stochastic method to determine failure behaviour, and the use of Monte Carlo
Simulation to generate several failure pattern, can give us more substantive guidance than the use of the deterministic method which is solely based on the failure pattern according the MTTF.