I have developed a maintenance cost optimisation model and utilised simulation technique which basically imitates the operation of the real-world processes and system over time. Simulations require the use of models that represents the key characteristics or behaviours of the selected system or processes, whereas the simulation represents the evolution of the model over time. Simulation is used in many contexts, such as simulation of technology for performance tuning or optimising, safety engineering, testing, training, education, and video games. Simulation is also used with scientific modelling of natural systems or human systems to gain insight into their functioning, as in econimics. Simulation can be used to show the eventual real effects of alternative conditions and courses of action. Simulation is also used when the real system cannot be engaged, because it may not be accessible, or it may be dangerous or unacceptable to engage, or it is being designed but not yet built, or it may simply not exist.

1. Asset Maintenance Cost
Optimisation Model
Arash Riazifar
September 2014

2. Objective
Develop asset maintenance cost
model to optimise total cost of
maintenance for complex and
critical operational assets and
recommend a suitable asset
through life support services
procurement and supply chain
management strategies.

3. Key Cost Components
• Preventive maintenance
• Corrective maintenance
• Periodic and ad hoc inspections
• Unmitigated risks
• Penalty cost
• Failures and downtimes

4. Definitions
• Corrective maintenances (CM) are unscheduled actions intended to
restore the system/asset to its operational state through corrective actions
after occurrence of failures (defects).
• Planned preventive maintenance (PM) actions are carried out to reduce
the likelihood of failures or to prolong the reliability of the
asset/equipment and/or to reduce the risk of failures.

5. Correlations
The below graph shows there is an exponential correlation between failure
rate, age of an asset, PM and CM activities.
t
a
b
c
Time / Age / Usage
Failure
probability

6. Correlations
• The asset operability or contract
duration for an asset to run
without any failure terminates
when it reaches a certain time or
usage level (L)
• CM and PM measures prolong
system reliability
• PM actions are carried out at
constant intervals of (x)
• Each PM restores the reliability of
the asset to a pre-defined extent
• Between two successive
preventive maintenance there
could be one or more corrective
maintenance needs

7. Developed Maintenance Cost Model
Total cost of asset maintenance over the period of (L) can be expressed as:
CT = Cm +Ci +Cr + p
Where,
CT : Total cost of asset maintenance
Cm: Cost of maintenance over the life of asset or maintenance contract (CM + PM)
Ci: Inspection cost
Cr: Cost of risk associated with accidents
p: Penalty Costs for not conforming to the safety, reliability and availability standards or agreed
contractual obligations

8. Estimating Maintenance Cost (Cm)
• Expected total cost of maintenance service =
𝐶𝐶𝑀 + 𝐶𝑃𝑀
(1+r)
𝑛
Where,
𝐶𝐶𝑀: Cost of corrective maintenance
𝐶𝑃𝑀: Cost of preventive maintenance
r: is the discount rate over Year (n)
n: 1, 2, 3

9. Corrective Maintenance Cost (CM)
• Expected cost of corrective maintenance and repairs:
𝐶𝑚𝑟.
𝑘=0
𝑁+1
𝑘𝑥
𝑘+1 𝑥
Λ 𝑡 − 𝑘. 𝛼. 𝑥 𝑑𝑡
Where,
𝑁: number of times maintenance is performed
𝑥: fixed time interval of maintenance
𝑘: kth PM , k = 1,2,3,…
𝛼𝑥: restoration out of maintenance action. α is the quality of the maintenance

10. Preventative Maintenance Cost (PM)
• Expected cost of preventive maintenance during the life of the asset or
intended contract period:
𝑁. 𝐶𝑝𝑚
Where,
𝑁: number of times preventative maintenance is performed
𝐶𝑝𝑚: Cost of preventative maintenance

11. Inspection Cost(Ci)
• Total inspection cost (Ci) over the contract can be calculated by :
𝐶𝑖 =
𝑗=0
𝑁𝑖
𝐶𝑖
(1 + 𝑟𝑖)𝑗
.
𝑟
1 −
1
(1 + 𝑟)𝑛
Where,
𝑁𝑖: Integer
𝐿
𝐼𝑓
, the expected number of inspection during the asset life or contract term
𝑟𝑖: Discount rate associated with inspection interval
𝐼𝑓: Inspection interval
𝑟: Annual discount rate

12. Risk Cost (Cr)
• The risk cost associated with system failure and accident is based on the
probability of inspection detecting potential failures and failures not being
detected by inspection, which can be calculated as follows:
𝐶𝑟 =
𝑛=0
𝐿
𝐸 𝑁. (𝑡𝑛, 𝑡𝑛+1) . 𝑃𝑛 𝐵 . 𝑏 + 1 − 𝑃𝑛 𝐵 . 𝑃𝑛 𝐴 . 𝑎
(1 + 𝑟)𝑛
Where,
a: expected cost per accident;
b: expected cost of repairing a failure based on non-destructive testing
𝑃𝑛(B): probability of detecting a potential failure using non-destructive testing
𝑃𝑛(A): probability of undetected potential failure leading to an accident between 𝑛 and
𝑛 +1 period
𝐸 𝑁 𝑡𝑛+1 , 𝑡𝑛 : expected number of failures between 𝑛 and 𝑛 +1 years

13. Total cost of maintenance (CT)
𝐶𝑇 = [𝐶𝑚𝑟. 𝑘=0
𝑁+1
𝑘𝑥
𝑘+1 𝑥
Λ 𝑡 − 𝑘. 𝛼. 𝑥 𝑑𝑡 + 𝑁. 𝐶𝑝𝑚]/ (1 + 𝑟)𝑛
+
𝑗=0
𝑁𝑖
𝐶𝑖
(1 + 𝑟𝑖)𝑗
.
𝑟
1 −
1
(1 + 𝑟)𝑛
+
𝑛=0
𝐿
𝐸 𝑁. (𝑡𝑛, 𝑡𝑛+1) . 𝑃𝑛 𝐵 . 𝑏 + 1 − 𝑃𝑛 𝐵 . 𝑃𝑛 𝐴 . 𝑎
(1 + 𝑟)𝑛
+ 𝑝
Where,
𝑝: is penalty cost for not conforming to the contractual obligations or failure to meet mandatory
safety, reliability and availability standards or conditions

14. Maintenance Service Provider Price (𝑃𝑐)
Maintenance service provider’s price per unit time can be calculated by:
𝑃𝑐 =
𝐶𝑇 + 𝑃
𝐿
Where,
L: Asset maintenance contract period or lifecycle
Ρ: Profit margin of the maintenance service provider

15. Simulation
• The following slides shows the number of failures (defect occurrences) for
a rail track asset achieved based on simulating the real life system when
for a period of time that the rail track asset reaches a usage level of ‘L’
Millions of Gross Tonnes (MGT). The results of the simulation is used in
the total cost of the asset maintenance model parameters as presented
previously.

16. Simulating the Real-Life System
• The next slides demonstrate the system simulation step by step

17. Simulation
The following table shows results generated by the simulated model and
assisted to identify the probability distribution function of the system failure
behaviour.
Index Ranges of MGT failiures Mean of the range
Vi
Frequesncy Probibility Vi
Cummulative
Probility
The intercept point of
the regression line and
the y axis
The slope of the
regression line
Weibull Probibility χ^2
χ^2
consolidated
ranges with
frquencies <5
i Bin Bin Average fi p(Vi) P(V) Y = ln(− ln(1 − P(V ))) X = ln Vi Pw (Vi) (Oi-Ei)2/Ei (Oi-Ei)2/Ei
1 0.5 - 1.5 1 15 0.00100963 0.00101 -6.897671187 0 0.0171176 0.0151579 0.0151579
2 1.5 - 2.5 2 110 0.00740392 0.00841 -4.773691083 0.693147181 0.0495494 0.0358479 0.0358479
3 2.5 - 3.5 3 3757 0.25287743 0.26129 -1.194513763 1.098612289 0.0873856 0.3134104 0.3134104
4 3.5 - 4.5 4 2862 0.19263647 0.45393 -0.502521166 1.386294361 0.1215021 0.0416462 0.0416462
5 4.5 - 5.5 5 4838 0.32563775 0.77957 0.413534639 1.609437912 0.1435289 0.2310588 0.2310588
6 5.5 - 6.5 6 1164 0.07834691 0.85791 0.668500617 1.791759469 0.1482912 0.0329905 0.0329905
7 6.5 - 7.5 7 795 0.05351013 0.91142 0.885367211 1.945910149 0.1356503 0.0497383 0.0497383
8 7.5 - 8.5 8 483 0.03250993 0.94393 1.058204365 2.079441542 0.1103809 0.0549360 0.0549360
9 8.5 - 9.5 9 288 0.01938480 0.96332 1.195569783 2.197224577 0.0799678 0.0458973 0.0458973
10 9.5 - 10.5 10 279 0.01877903 0.98210 1.391960803 2.302585093 0.0515178 0.0208050 0.0208050
11 10.5 - 11.5 11 85 0.00572121 0.98782 1.483359678 2.397895273 0.0294419 0.0191112 0.0191112
12 11.5 - 12.5 12 73 0.00491351 0.99273 1.594140539 2.48490665 0.0148789 0.0066744 0.0066744
13 12.5 - 13.5 13 40 0.00269233 0.99542 1.683936426 2.564949357 0.0066253 0.0023347 0.0023347
14 13.5 - 14.5 14 22 0.00148078 0.99690 1.753985778 2.63905733 0.0025894 0.0004746 0.0004746
15 14.5 - 15.5 15 24 0.00161540 0.99852 1.874135447 2.708050201 0.0008847 0.0006036 0.0006036
16 15.5 - 16.5 16 10 0.00067308 0.99919 1.963093068 2.772588722 0.0002631 0.0006387 0.0006387
17 16.5 - 17.5 17 5 0.00033654 0.99953 2.036053271 2.833213344 0.0000679 0.0010640 0.0010640
18 17.5 - 18.5 18 1 0.00006731 0.99960 2.055976752 2.890371758 0.0000151 0.0001804 0.0132632
19 19.5 - 20.5 20 3 0.00020193 0.99980 2.140961542 2.995732274 0.0000005 0.0856820
20 20.5 - 21.5 21 1 0.00006731 0.99987 2.187519775 3.044522438 0.0000001 0.0683362
21 24.5 - 25.5 25 1 0.00006731 0.99993 2.262411473 3.218875825 0.0000000 961
22 25.5 - 26.5 26 1 0.00006731 1.0 2.779942594 3.258096538 0.0000000 16229
Total 14,857 1.0 1.0 17,191 0.89

18. Simulation
By measuring the frequency of the system failure occurrences it was
determined that the system failure behaviour follows a Weibull probability
distribution function. This helped to develop a non-linear regression model
for the occurrence of the failures function.
0.00
0.02
0.04
0.06
0.08
0.10
0.12
0.14
0.16
0
1000
2000
3000
4000
5000
6000
0.5
-
1.5
1.5
-
2.5
2.5
-
3.5
3.5
-
4.5
4.5
-
5.5
5.5
-
6.5
6.5
-
7.5
7.5
-
8.5
8.5
-
9.5
9.5
-
10.5
10.5
-
11.5
11.5
-
12.5
12.5
-
13.5
13.5
-
14.5
14.5
-
15.5
15.5
-
16.5
16.5
-
17.5
17.5
-
18.5
19.5
-
20.5
20.5
-
21.5
24.5
-
25.5
25.5
-
26.5
Observed
frequency
𝑃(𝑉)= 𝐾/𝐶 〖.[𝑉/𝐶]〗^(𝐾−1) 〖𝑒𝑥𝑝〗^(−[𝑉/𝐶]^𝐾 )

19. Regression Model
Through regression technique the Weibull probability distribution function
parameters were calculated as follows:
• Shape parameter = 2.5814
• Scale parameter = 6.9625
-8
-6
-4
-2
0
2
4
0 0.5 1 1.5 2 2.5 3 3.5
Y
=
ln(−
ln(1
−
P(V
)))
X = ln Vi
y = 2.5814x - 5.0093
A=K=2.5814
B=-5.0093
C=𝑒^(−[𝐵/𝐴])=6.9625

20. Simulation Results Goodness-of-Fit Test
In the performed simulation calculated𝜒2
was equal to 0.89 which is less
than 𝜒2
critical at certainty level of 95%. This means that the actual system of
the asset has been correctly and properly simulated which reflects the real
system behaviour with 95% certainty.