Ann Oper Res (2006) 143: 91–106

DOI 10.1007/s10479-006-7374-1

“Optimization of aircraft maintenance/supportinfrastructure using genetic algorithms—level ofrepair analysis”

Haritha Saranga · U. Dinesh Kumar

C© Springer Science + Business Media, Inc. 2006

Abstract Level of repair analysis (LORA) is an approach used during the design stage of

complex equipment for analysis of the cost effectiveness of competing maintenance strategies.

LORA is carried as a part of the life cycle cost and cost of ownership analysis and plays

a significant role in minimizing the life cycle cost and cost of ownership of the capital

equipment. Since many purchasing decisions of complex equipment are based on cost of

ownership, it has become essential to carry out LORA to compete in the market. In this

paper, we develop a mathematical model for LORA and propose a solution methodology

based on genetic algorithms. The concept is illustrated using a hypothetical aircraft engine.

Keywords Aircraft maintenance · Cost of ownership · Genetic algorithms · Level of repair

analysis · Life cycle cost · Multi echelon · multi indenture

From being a “necessary evil”, maintenance has become a strategic issue of equipment

and asset management. Concepts such as Reliability Centered Maintenance (RCM), Total

Productive Maintenance (TPM) and Business Centered Maintenance (BCM) all have the

same objective of improving the productivity of the equipment through effective maintenance.

Optimisation of maintenance tasks has attracted the attention of many researchers in the past

few decades (Park, 1979; Jayabalan et al., 1992; Zhang et al., 1993; Makis et al., 1993). Many

of these articles, however, deal with micro level issues related to maintenance such as optimal

replacement time and optimal preventive maintenance intervals based on reliability and cost

of repair, replacement etc. The secret of effective maintenance is not about the optimization

of maintenance tasks during operation of the equipment, but optimization of maintenance and

maintainability during the design stage of the equipment in order to achieve minimum life

cycle cost (LCC) or minimum total cost of ownership (TCO). In fact all the LCC and TCO

H. Saranga (�)Learning Resource Center, Indian School of Business, Gachibowli, Hyderabad, 500019, Indiae-mail: haritha saranga@isb.edu

U. D. KumarIndian Institute of Management Calcutta, India

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92 Ann Oper Res (2006) 143: 91–106

models stress the need to minimise maintenance costs to minimise the LCC/TCO (Asiedu

and Gu, 1998; Sherif and Kolarik, 1981; Taylor, 1981).

Equipment such as commercial aircraft would cost as high as $200 million and an addi-

tional $2 billion towards operation, maintenance and support throughout the economic life,

which is around 20 to 25 years. For most equipment, 80–85% of the total LCC is spent during

the operation and maintenance of the equipment. A significant part of the LCC is spent on

maintenance alone. Woodward (1997) claims that life cycle cost is all about operating phys-

ical assets at minimum cost. Accordingly, minimization of maintenance costs is essential to

minimize the total LCC of the asset. It is well established in the literature that the design

stage is the best period of the product life cycle, during which maintenance related issues

should be addressed.

The concept of maintainability (Blanchard et al, 1991; Knezevic, 2000) developed by the

defense and aerospace industries tries to address the maintenance related issues during the

design stage of an item. One of the most important tasks related to maintenance, which is

performed at this time, is called Level of Repair Analysis (LORA). LORA, is a systematic

procedure to determine the cost of alternative maintenance options and maintenance levels,

taking into account all cost drivers, such as, spare parts, support equipment, manpower,

training and transportation that would be required at each line of maintenance. LORA is used

to develop the initial maintenance concept based on economic and non-economic constraints

and readiness requirements, and to determine whether a particular maintenance action is

cost-effective. One of the main objectives of LORA is to minimize the life cycle cost of the

equipment. LORA is especially required for complex systems like aircraft and other defense

equipment, where maintenance requires vast numbers of complex support equipment and

highly skilled personnel and also for systems whose unavailability can result in a huge loss

of revenue.

Several researchers have discussed the importance of LORA in reducing the life cycle

cost (Blanchard et al., 1992 & 2000; Dinesh Kumar et al., 2000). Barros (1998) proposes a

pure integer programming formulation for LORA, which is effective whenever fixed costs are

significant. In order to include the fixed as well as the variable costs, Barros (2001) derived

a Level of Repair Optimization Model (LOROM) using the branch and bound technique,

which can obtain tight upper and lower bounds for life cycle cost estimates. Although,

several commercial software packages are available in the market for solving LORA, there

is huge gap between the real problem and the models available to solve the problem.

LORA demands an in-depth analysis of repair options, due to maintenance task complex-

ity at various maintenance levels, in practical military and industrial problems. Manpower

skill-level requirements, frequency of occurrence, special facility needs, turn around time,

reliability, cost of test programs, cost of technical documentation, transportation, holding

costs etc., dictate to a great extent the selection of a specific repair task to be accomplished

at each level of maintenance. Each option reflects the characteristics of system design which

is evaluated in terms of effectiveness criteria such as availability, reliability, maintainability,

supportability, total cost of ownership etc. The maximum benefits in implementing LORAare obtained by performing it at the early stages of system design and using the analysis

to change the design accordingly, to prepare maintenance plans and to determine logistic

resource allocation.

The main objective of this paper is to present a holistic approach to level of repair

analysis and develop optimisation models for LORA. The rest of the paper is organized as

follows. In section 1, we describe the issues involved in maintenance of complex systems.

Mathematical models for analysis of LORA are discussed in Section 2. These models can

be used when the system under consideration has very few components. Section 3 deals

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Ann Oper Res (2006) 143: 91–106 93

Aircraft

Engines

Modules

Parts

Fig. 1 Multi-indentureenvironment

with optimisation models for LORA. A case study based on an aircraft engine is presented

in Section 4 to illustrate the Genetic Algorithms (GA) approach. We finally conclude with

future research possibilities in Section 5.

1. Issues in level of repair analysis

The complexity of LORA lies in the fact that most of the complex equipment has modular

design (multi indenture design as shown in Figure 1) and maintenance is carried out at multiple

echelons. Multi-echelon refers to the hierarchy of operating locations and supporting depots.

For example, aircraft may be dispersed in squadrons (1st echelon) across a number of bases

(2nd echelon) such that each base may have one or more operational squadrons as illustrated

in Figure 2. Note that although a base may support a squadron, in practice they could be

thousands of miles apart (when the squadron is on a deployment). A 3rd echelon depot

or maintenance unit supports the bases. The contractor or original equipment manufacturer

(OEM) (4th echelon) may support all the echelons above it, thus, for example, aircraft engines

may be removed at first echelon and sent directly to the contractor for recovery. These

additional lines of communication are shown by dotted lines in Figure 2. Where there are

several bases situated in close proximity, it may prove cost-effective to use one of them to

provide the 2nd echelon support.

Aircraft Squadron

Base

Depot

Contractor

Fig. 2 Multi-echelon environment

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94 Ann Oper Res (2006) 143: 91–106

The level of repair analysis technique can be used to decide whether an item should be

repaired, reconditioned or discarded, and if repaired or reconditioned, to find the location

where this is best performed. Whenever a system fails, the faulty assembly is isolated and

replaced with a spare assembly if available. The removed assembly may be discarded, recon-

ditioned or repaired (One would not normally repair a burst tire; equally one would normally

not discard a faulty engine).

If it is decided to discard an assembly then all the modules and parts within the assembly

are also discarded. The decision to discard is generally made for one of the following two

reasons either it is simply not economical to repair it or; it is impractical/impossible to do

so. For example, the booster tanks on the space shuttle are normally repairable because they

are ejected at a relatively low altitude and the flight path is such that they would normally

fall into the shallow waters off the Florida coast, so are easily retrievable. The main rocket

motor, however, is not repairable, at least not after a normal flight, because it would normally

burn up on re-entry to the earth’s atmosphere, thus making it impossible. Another common

example for discardables are sparkplugs in a standard family car. These may be repairable

(by sand blasting the electrodes and re-setting the gap) but the cost of paying for a mechanic

to do this is likely to be several times the cost of a new plug.

If it has been decided that an assembly (1st indenture item) is repairable, the next decision

is where this should be done. In most cases it is either impractical or undesirable to perform

this task in situ. It will therefore need to be moved to one of the deeper echelons (2nd, 3rd

or 4th). Once there, the modules (2nd indenture items), which require maintenance will be

removed and replaced with serviceable ones (which could be the same ones after they have

been recovered). Those modules, which are considered repairable, may be recovered at the

same location as the assembly, or sent away to a still deeper echelon. In most cases this would

be achieved by the replacement of one, or more of the constituent parts (3rd indenture items).

These rejected parts may, in turn be repaired either at the same location as the module [from

which they were removed] or sent to an even deeper echelon.

Obviously, each of these decisions has a different economic impact. LORA attempts to

find the best combination of repair/discard decisions and the maintenance level that min-

imises the total support cost. For example, if an item is recommended to be repaired by the

manufacturer, then the operator need not procure logistic support resources such as test and

support equipment, facilities etc. However if the repair is to be carried out by the operator,

then the operator needs to procure all the support resources required.

For most systems, support is spread over three echelons viz. base, depot and contractor.

Thus, in the proposed model we also assume a three-echelon system. It is also assumed that

the system consists of 3 indentures. Without loss of generality, the first indenture item is called

‘assembly’, the second indenture item is called ‘module’ and the third indenture item is called

‘part’. Each item has three repair options. These assumptions are the closest representation

of the physics of the problem in terms of echelon and repair options. However, the indenture

levels can be more than three in a complex system, but for mathematical simplicity we assume

only three indenture levels. Thus we have the following assumptions:

Assumption 1: The system consists of three echelons (e). Without loss of generality, we

assume that e =1 refers to base, e = 2 refers to the intermediate and e = 3 refers to depot.

Assumption 2: There are three indentures (v), namely, assemblies (v =1) enclosed in the

system, modules (v = 2) enclosed in the assembly and parts (v = 3) enclosed in the

module.

Assumption 3: Each component can have at most three repair options r, namely, repair

(r =1), discard (r = 2) and replace (r = 3).

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Ann Oper Res (2006) 143: 91–106 95

Repair options are defined as follows:

1. Repair: Item, say an assembly is repaired at the current echelon; enclosed modules of

the assembly i.e., lower indenture level items are repaired at that echelon or sent to lower

echelon.

2. Discard: Item, say a module is removed from the assembly and replaced with a spare

item. The removed item including all the lower indenture enclosed parts is discarded. That

is no further maintenance is carried out on these items.

3. Replace: Item, say an assembly is replaced by a spare at the current echelon, and the

replaced assembly including all the lower indenture enclosed modules is sent to the lower

echelon for further decision making.

The following assumptions are made for modeling the optimization problem. These as-

sumptions are derived based on interviewing several logistics engineers who are involved in

maintaining complex systems such as aircraft and its components.

Assumption 1: At first echelon, only replacement or discard of assemblies is carried out.

No other maintenance is done at this echelon.

Assumption 2: At second echelon, all three maintenance activities are carried out for

assemblies, modules and parts.

Assumption 3: At third echelon maintenance of modules and parts is carried out with

repair, discard or replace options.

2. Mathematical analysis of LORA

One of the basic problems within LORA is to make a decision between repair, discard and

replace options. Ebling (1997) provides a simple model that compares the cost of repair

versus replace and recommends the one that is less. According to Degraeve et al. (1999) and

Plank et al. (2002) decision between repair and replace option should be based either on life

cycle cost or cost of ownership. In an LCC exercise, the cost is divided into capital cost and

revenue cost. In this paper we use the terminology, fixed cost and variable cost respectively.

The fixed cost consists of cost for maintenance equipment, facilities and technical manuals.

Variable cost consists of costs for material, manpower and spare parts associated with each

maintenance task. Variable cost mainly depends on the removal rate of the equipment from

the system for maintenance. In the following section we develop a mathematical model for

making the decision between repair and replace.

2.1. Repair versus replace

Any effective decision analysis with respect to repair versus replace should include the cost

comparison throughout the life cycle of the system as well as repair effectiveness. If the repair

is 100% effective (as good as new), then the removal rate will remain the same throughout

the life of the system, which in turn will be reflected in the variable cost. Thus, modelling

total cost of maintenance in terms of removal rate and optimising the resultant cost function

would ensure a cost effective repair versus replace decision analysis.

Assume that λR and λD denote the annual removal rate of the equipment under the

maintenance option repair and replacement respectively. Then, unless the repair is as-good-as-new, λR > λD . If repair is as good as new, then λR = λD .

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96 Ann Oper Res (2006) 143: 91–106

Since all the costs associated with a system throughout its life should be considered, we

use the Present Value, PV, for calculating cost of maintenance under different strategies. The

present value of the total cost of maintenance under the repair option, PVR , throughout the

life of the system, is given by:

PV R = FCR +N∑

t=1

λR × VCR

(1 + r )t(1)

where,r = Annual discount rate

N = Design life of the system in years

FCR = Fixed cost of maintenance associated with repair

V CR = Variable cost of maintenance associated with repair

The present value of the total cost of maintenance under the replacement option, PV D ,

throughout the life of the system, is given by:

PV D = FCD +N∑

t=1

λD × VCD

(1 + r )t(2)

Where FCD and VCD are the fixed and Variable costs associated with replacement.

We know that:

N∑t=1

1

(1 + r )t= 1 − (1 + r )−N

r

Substituting the above equation in equations (1) and (2), we get:

PV R = FCR + λR × VCR × (1 − (1 + r )−N )

r

PV D = FCD + λD × VCD × (1 − (1 + r )−N )

r

Comparing RHS of above two equations, repair option will be chosen if:

FCR + λR × VCR × (1 − (1 + r )−N )

r< FCD + λD × VCD × (1 − (1 + r )−N )

r

Multiplying both sides withr

(1 − (1 + r )−N )and rearranging we get:

VCR × λR − VCD × λD < (FCD − FCR) ×(

r

1 − (1 + r )−N

)(3)

Thus one can safely conclude that the repair option should be chosen when the condition in

equation (3) is satisfied.

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Ann Oper Res (2006) 143: 91–106 97

Fig. 3 (a) Single echelon maintenance/support system, (b) two-echelon maintenance/support system

2.2. Location of repair/replacement decision—multi echelon versus single echelon

Once the decision is made whether to repair or replace the equipment under consideration, the

next decision would be to decide on the location of the repair/replacement. In this section, we

look at two possible scenarios: (1) Repair/replacement at the base, and (2) repair/replacement

at depot, which supports ‘n’ bases. Figures 3a & 3b illustrate both the scenarios:

Cost effectiveness of the above two-maintenance/support designs can be analysed as

follows:

The cost drivers for single echelon maintenance/support design are: (1) Facility, (2)

Manpower, (3) Test and maintenance equipment, (4) spare parts, (5) spares holding, and

(6) technical documentation. The cost drivers for the two-echelon maintenance/support

design include all the cost drivers as in the first case and the transportation cost.

Assume that:λi : Demand for maintenance task at Base i.

Fc : Facility cost per unit.

Mc : Manpower cost per unit.

Ec : Test and maintenance equipment cost per unit.

Dc : Technical documentation cost.

Sc : Spare parts cost per unit part (including holding cost).

Ti : Transportation cost per unit between base and depot.

Li : Transport times between base i and the depot.

MTTR : Mean time to restore the system.

n : Number of bases (1st echelon units)

The costs associated with facility, equipment and technical documentation are fixed costs,

whereas, the costs associated with manpower, transportation and spare parts come under

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98 Ann Oper Res (2006) 143: 91–106

Fig. 4 Relationship between thefixed cost and the demand inmaintenance

variable cost. The investment in capital equipment, facility and manpower would depend on

the demand (required capacity) for maintenance tasks. The facility can accommodate up to a

certain demand, after which the capacity of the facility needs to be increased to accommodate

the next batch of demands and so on as shown in Fig. 4. We define the following function to

relate the investment in equipment, facility and manpower:

F(λi) = i × Fc, for(i − 1)λf < λi ≤ iλf

That is for every incremental increase of λf in the demand for maintenance, there is an

incremental increase of Fc in the facilities cost.

E(λi) = i × Ec, for (i − 1)λe < λi ≤ iλe

That is for every incremental increase of λe in the demand for maintenance, there is an

incremental increase of Ec in the test and equipment cost.

M(λi) = i × Mc, for (i − 1)λm < λi ≤ iλm

That is for every incremental increase of λm in the demand for maintenance, there is an

incremental increase of Mc in the manpower cost.

The total maintenance cost for the single echelon design is given by:

NPVSE =n∑

i=1

[F(λi ) + E(λi )] + n × Dc +n∑

i=1

N∑t=1

M(λi ) + λi × Sc

(1 + r )t(4)

λi × Sc is the spare parts cost at base i.

In the case of two-echelons, spare parts need to be maintained at each of the bases and

also at the depot. Depot will send a spare item to the base as soon as it receives the item for

maintenance from the bases. Thus, the expected number of spares to be stocked at base will

be λ × Li. At the depot the demand for spare will be equal to the sum of the demands for the

maintenance tasks form each of the bases, i.e.∑n

i=1 λi . Using Palm’s theorem (Sherbrooke,

1992), the expected number of spares to be stocked at depot is given by,∑n

i=1 λi × MTTR.

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Ann Oper Res (2006) 143: 91–106 99

The total maintenance cost for the two echelons design is given by:

NPVTE = F

(n∑

i=1

λi

)+ E

(n∑

i=1

λi

)+ Dc +

N∑t=1

M(n∑

i=1

λi)

(1 + r)t+

n∑i=1

N∑t=1

λixLi

(1 + r)t

+N∑

t=1

MTTR × ∑ni=1 λi

(1 + r )t

(5)

The difference between NPVSE and NPVTE should be considered while making decision

between single echelon and multi-echelon maintenance support system.

3. Mathematical programming model for LORA

The previous section deals with the repair/replacement decision analysis as well as location of

maintenance action for a single component considering various relevant factors. However, in

case of complex systems like aircraft, where thousands of components needs to be analysed,

it is necessary to optimise the maintenance decisions by considering all the components.

Thus, in this section we develop a general mathematical programming model for 3-indenture

3-echelon maintenance support design for commercial airline, whose maintenance/support

are 3-echelon, however the number of indentures can be more than three. In this model we

try to minimise the total cost of maintenance for the system under consideration, with fixed

and variable costs, and the rate of failure as the input parameters. Here the repair/replacement

decision analysis as well as location of repair is automatically considered while minimising

the objective function.

The following notations are used for this model.E = The total number of echelons for the system under

consideration

V = The total number of indenture levels for the system under

consideration.

Nl = Number of components at the indenture level l.(i, j, k) = An ordered triplet, where i denotes the item (LRU) i at

the first indenture, j denotes the enclosed item (SRU) j of

i , at the second indenture and k denotes the enclosed item

(part) k of j of i , at the third indenture.

(i, 0, 0) = The first indenture item, i.e, LRU i(i, j, 0) = The second indenture item, i.e., the SRU j enclosed in

LRU iR = Total number of repair options.

M(i, j,k)(T ) = Expected number of failures for item (i, j, k) during the

cumulative life (T ) of the fleet.

FC (i, j,k)r,e = Fixed cost associated with repair option ‘r ’ at echelon e,

for item (i, j, k).

λ(i, j,k) = Total number of maintenance tasks required in the whole

life time of the item (i, j, k).

V Cr,e(i, j, k) = Variable cost associated with repair option ‘r’ at echelon

level e, for item (i, j, k).

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100 Ann Oper Res (2006) 143: 91–106

Apart from the above notation, we also assume the following variables, which take the

values 1 or 0 depending on whether a particular repair option is selected or not for each item

at every indenture and echelon level.

zr,e(i, j, k) ={

1 if repair option r at echelon e is selected for item (i, j, k)

0, otherwise

The mathematical programming formulation for the Level of Repair Analysis is stated

below:

Minimise:∑(i, j,k)

3∑r=1

3∑e=1

[VCr,e(i, j, k)λ(i, j,k) + FCr,e(i, j, k)

]Zr,e(i, j, k) (6)

Subject to the constraints:

Single repair option constraints: Exactly one repair option is chosen in the first echelon

3∑r=1

Zr,1(i, j, k) = 1 ∀(i, j, k) (7)

Replace option constraints� If replace option is chosen in echelon e, then exactly one repair option is chosen in echelon

e + 1.� If not, no repair option is chosen in e + 1.

Zreplace, e(i, j, k) =3∑

r=1

Zr,e+1(i, j, k) ∀e and ∀(i, j, k) (8)

Compatibility constraintsIf an item is either discarded or replaced in echelon e, all the enclosed lower indenture items

are also discarded or replaced respectively in echelon e.

Zr,e(i, 0, 0) ≤ Zr,e(i, j, 0) ∀e and ∀(i, j, k) where r = discard, replace (9)

Zr,e(i, j, 0) ≤ Zr,e(i, j, k) ∀e and ∀(i, j, k) where r = discard, replace (10)

The objective function in equation (6) tries to minimize the total support cost per flying

hour associated with different repair options at various echelons. Constraint equation (7)

ensures that at least one repair option is chosen for each LRU in the system at echelon one.

Constraint option (8) makes sure that if replace option is chosen in the current echelon,

then at lease one repair option is chosen in the next echelon. Equations (9) and (10) ensure

the relationship between the parent (higher indenture item) children (lower indenture item)

with respect to the replace/discard options at different echelons. For example, if a SRU is

discarded in the second echelon, then all the constituent parts of the SRU are also discarded

in the second echelon. The above optimization is a 0-1 linear programming problem.

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Ann Oper Res (2006) 143: 91–106 101

For a system with “N” LRUs, “M” SRUs and “S” parts with n echelons and ri repair

options at echelon i, the number of possible solutions is equal to RN+M+S. Where R =∑ni=1 ri , ri is the number of repair options at echelon i .

For a system with 10 LRUs, 20 SRUs and 40 parts, the size of the solution space will be

as high as 1.21 × 1019. Modern aircraft will have around 1200 LRUs, 100,000 SRUs and

millions of parts. This is one of the reasons why this problem has never been solved or even

never attempted, at the aircraft level. In the next Section we apply genetic algorithms to solve

this problem for an aircraft engine.

4. Case study—LORA for an aircraft engine

An aircraft engine (LRU) is considered for the application of GA in our LORA model,

which consists of 10 modules (SRU) with 22 parts spread between these modules in the way

described in the engine breakdown structure shown in Figure 5. The solution space for this

system is 6.28 × 1010.

4.1. Genetic algorithms

Due to the complexity and the large number of variables involved in LORA (160 variables

with 225 constraints in the present problem alone, which consists of 10 modules and 22

parts), it is very difficult to optimize the cost of maintenance using traditional optimization

techniques. The number of variables and constraints depend on the number of indenture

and echelon levels and the maintenance strategy. Techniques like integer programming and

branch and bound method become difficult to use even for problems involving limited number

Part 1.1 Part 1.2

M-01

Part 2.1 Part 2.2Part 2.3

M-02

Part 3.1Part 3.2

M-03

Part 4.1Part 4.2Part 4.3

M-04

Part 5.1 Part 5.2

M-05

Part 6.1 Part 6.2

M-06

Part 7.1Part 7.2Part 7.3Part 7.4Part 7.5Part 7.6

M-07

Part 8.1

M-08

Part 9.1

M-09 M-10

Engine

Fig. 5 Engine breakdown structure

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102 Ann Oper Res (2006) 143: 91–106

Table 1 Cost heads for different indenture levels at different echelons

Echelon 1 Echelon 2 Echelon 3

Indenture 1 (LRU) 1. Facility Cost 1. Facility Cost None

2. Manpower Cost 2. Manpower Cost

3. Test and 3. Test and

Maintenance Maintenance

Equipment Cost Equipment Cost

4. Spare Part Cost 4. Spare Part Cost

(LRU) (SRU)

5. Spare holding Cost 5. Spare holding Cost

6. Technical 6. Technical

Documentation Cost Documentation Cost

7. Transportation Cost

Indenture 2 (SRU) None 1. Manpower Cost 1. Manpower Cost

2. Test Equipment Cost 2. Test Equipment Cost

3. Spare Part Cost 3. Spare Part Cost

(parts) (parts)

4. Spare holding Cost 4. Spare holding Cost

5. Technical 5. Technical

Documentation Cost Documentation Cost

6. Transportation Cost

Indenture 3 (Part) None 1. Manpower Cost 1. Manpower Cost

2. Test Equipment Cost 2. Test Equipment Cost

3. Spare Part Cost (spares 3. Spare Part Cost

for the parts) (spares for the parts)

4. Spare holding cost 4. Spare holding cost

5. Technical 5. Technical

Documentation Cost Documentation Cost

6. Transportation Cost

of variables. To overcome this problem, we focus on a more recently developed optimization

technique called Genetic Algorithms (GA), which is proved to be a robust tool to tackle

complex problems with a large number of parameters.

A Genetic Algorithm begins with a random initial population of chromosomes, which

correspond to the variables of the solution space. However, to fasten the evolution process,

there are ways to choose an effective initial population, in which case it is called a semi

random initial population. Next, the parameters of initial population are passed to the objective

function, which in case of GAs is more aptly called the “Fitness function” for evaluation.

Only the chromosomes with best fitness values are retained for the next generation, and

the rest are discarded. Genetic operations called “Cross Over” and “Mutation” are used to

generate new offspring from the survived chromosomes. The population is usually ranked

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Ann Oper Res (2006) 143: 91–106 103

according to the fitness values. The chromosomes with good fitness values are retained for

the next generation, and the rest are replaced with new offspring. According to the Schema

theorem (Holland, 1975) the subspaces with higher than average payoffs will be allocated

exponentially more trials over time, while those subspaces with below average payoffs will be

allocated exponentially fewer trials. This assumes that the effect of crossover and mutation

are not disruptive (Spears and De Jong, 1991). Thus, a new population is formed with

the survived parents from the initial population and their offspring, which will replace the

discarded chromosomes of the initial population. This iterative process will continue until a

population with the same chromosomes is being generated repeatedly, which means GA has

arrived at a solution (Haupt and Haupt, 1998).

4.2. Optimization using GA

We are considering three indenture levels in this example (the aircraft engine), namely, engine,

modules and constituent parts respectively. Here, engine is the first indenture item, modules

are the second indenture items, and parts fall under the third indenture level. For simplicity

sake as well as in compliance with the assumptions made in Section 2, we make the following

assumptions for the example under consideration.� At the first echelon, which is in general an airport, only the first indenture item, i.e., the

engine, is replaced and no other maintenance activity is carried out.� In second echelon, both modules as well as parts, are repaired, discarded or replaced.� In the third echelon, modules and parts are either repaired or discarded. No replacement

option is available.

The type of variable and fixed costs for various repair options at different echelons for all

indenture items are listed in Table 1 and the actual costs, along with failure or removal rate

for the example under consideration are given in Table 2.

Thus, even though we are considering E = 3 echelons, with R = 3 repair options, we

essentially end up with repair, discard and replace options at the base (2nd echelon) and

repair or discard options at the depot (3rd echelon) for all the modules and constituent parts

of the engine.

Using Evolver, the GA application software, we minimize the fitness function given by

equation (6) under the constraints (7)-(10) and obtain the optimum repair options for the

modules and constituent parts as given in Table 3. In general GA deals with constraints by

assigning penalty for any violation of the constraint in the fitness function.

Evolver is a genetic optimization package that functions as an add-in to Microsoft Excel.

One can enter the data into the Excel spreadsheet and maximize the fitness function under the

given constraints using Evolver. For the problem under consideration, we have 160 binary

variables (Zr,e(i, j, k)) and 225 constraints while modeling the optimization problem into a

GA program.

4.2.1. Treatment of constraints

Constraint (7) is entered as it is, by forcing the sum (Z1,1(i, j, k) + Z2,1(i, j, k) +Z3,1(i, j, k)) to be equal to 1 for each (i, j, k).

Constraints (8) are entered in the form of the equations:

Zreplace,2(i, j, k) −3∑

r=1

Zr,3(i, j, k) = 0 ∀ (i, j, k)

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104 Ann Oper Res (2006) 143: 91–106

Table 2 The fixed and variable costs of different modules and constituent parts for various repair options atdifferent echelons

VC e2 VC e2 VC e2 VC e3 VC e3 FC e2 FC e2 FC e2 FC e3 FC e3

repair Discard Replace Repair Discard repair Discard Replace repair Discard

Item (r1) (r2) (r3) (r1) (r2) (r1) (r2) (r3) (r1) (r2) λi, j,k

Module 1 400 1500 40 520 1550 150 80 350 720 25 0.11Part 1.1 400 900 10 300 902 700 20 350 720 25 0.067

Part 1.2 400 600 10 340 601 700 20 350 720 25 0.043

Module 2 1000 2500 45 1050 550 1000 85 200 1150 34 0.164

Part 2.1 300 1000 20 250 501 1000 85 50 1150 34 0.068

Part 2.2 250 800 20 200 201 1000 80 50 1150 34 0.058

Part 2.3 150 700 20 100 201 1000 80 50 1150 34 0.038

Module 3 1000 3000 60 1100 2020 150 90 900 1250 40 0.041Part 3.1 800 1500 30 500 1502 100 30 900 1250 40 0.004

Part 3.2 300 500 10 300 501 120 30 900 1250 40 0.037

Module 4 900 1000 20 950 1050 200 64 1000 1900 40 0.199Part 4.1 100 400 20 75 405 200 64 1000 1900 40 0.081

Part 4.2 75 300 10 60 302 200 34 1000 1900 40 0.066

Part 4.3 250 300 10 200 301 200 34 1000 1900 40 0.052

Module 5 200 1200 30 700 1250 200 22 600 750 28 0.217Part 5.1 500 1000 20 400 401 200 22 600 750 28 0.109

Part 5.2 100 200 10 75 150 150 22 600 750 28 0.108

Module 6 1500 1800 50 1504 500 2500 40 200 2700 50 0.193Part 6.1 500 1200 40 450 240 2500 40 50 2700 50 0.066

Part 6.2 400 600 30 300 130 2500 40 50 2700 40 0.127

Module 7 1200 2000 60 1250 2005 1800 40 600 2000 42 0.294Part 7.1 400 600 50 300 604 1800 40 600 2000 42 0.063

Part 7.2 200 400 40 190 405 1800 40 600 2000 42 0.037

Part 7.3 300 350 30 300 351 1800 40 600 2000 42 0.037

Part 7.4 250 250 20 200 250 1800 40 600 2000 42 0.055

Part 7.5 100 200 10 100 202 1800 40 600 2000 42 0.04

Part 7.6 75 200 10 50 201 1800 40 600 2000 42 0.062

Module 8 500 1000 30 505 600 2000 100 50 2100 50 0.075Part 8.1 500 1000 30 505 500 2000 46 40 2100 30 0.075

Module 9 500 2000 50 600 2050 1200 30 200 1250 31 0.036Part 9.1 500 2000 50 600 2050 1200 30 400 1250 31 0.036

Module 10 400 1000 40 450 1000 1500 20 350 1600 22 0.066

Constraints (9)-(10) are treated through the equations;

Zr,e(i, j, k) − Zr,e(i, j, 0) ≥ 0 ∀e and ∀(i, j, k) where r = discard, replace

Evolver randomly generates many organisms (possible solutions), and calculates the result

each produces. Entire “population” is ranked. It then selects good organisms, swaps their

variables (genes) using crossover and mutation to produce “offspring”. If offspring do not

produce a good result, two more parents are selected. If offspring organism is good, it is

re-inserted into the population. As the above steps are repeated, the population “evolves”

increasingly optimal solutions.

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Ann Oper Res (2006) 143: 91–106 105

Table 3 The optimum values of the decision variables after GA application

Z1,2(1,j,k) Z2,2(1,j,k) Z3,2(1,j,k) Z1,3(1,j,k) Z2,3(1,j,k)

Z1,2(1,1,0) 1 Z2,2(1,1,0) 0 Z3,2(1,1,0) 0 Z1,3(1,1,0) 0 Z2,3(1,1,0) 0

Z1,2(1,1,1) 0 Z2,2(1,1,1) 1 Z3,2(1,1,1) 0 Z1,3(1,1,1) 0 Z2,3(1,1,1) 0

Z1,2(1,1,2) 0 Z2,2(1,1,2) 1 Z3,2(1,1,2) 0 Z1,3(1,1,2) 0 Z2,3(1,1,2) 0

Z1,2(1,2,0) 0 Z2,2(1,2,0) 0 Z3,2(1,2,0) 1 Z1,3(1,2,0) 0 Z2,3(1,2,0) 1

Z1,2(1,2,1) 0 Z2,2(1,2,1) 0 Z3,2(1,2,1) 1 Z1,3(1,2,1) 0 Z2,3(1,2,1) 1

Z1,2(1,2,2) 0 Z2,2(1,2,2) 0 Z3,2(1,2,2) 1 Z1,3(1,2,2) 0 Z2,3(1,2,2) 1

Z1,2(1,2,3) 0 Z2,2(1,2,3) 0 Z3,2(1,2,3) 1 Z1,3(1,2,3) 0 Z2,3(1,2,3) 1

Z1,2(1,3,0) 1 Z2,2(1,3,0) 0 Z3,2(1,3,0) 0 Z1,3(1,3,0) 0 Z2,3(1,3,0) 0

Z1,2(1,3,1) 0 Z2,2(1,3,1) 1 Z3,2(1,3,1) 0 Z1,3(1,3,1) 0 Z2,3(1,3,1) 0

Z1,2(1,3,2) 0 Z2,2(1,3,2) 1 Z3,2(1,3,2) 0 Z1,3(1,3,2) 0 Z2,3(1,3,2) 0

Z1,2(1,4,0) 0 Z2,2(1,4,0) 1 Z3,2(1,4,0) 0 Z1,3(1,4,0) 0 Z2,3(1,4,0) 0

Z1,2(1,4,1) 0 Z2,2(1,4,1) 1 Z3,2(1,4,1) 0 Z1,3(1,4,1) 0 Z2,3(1,4,1) 0

Z1,2(1,4,2) 0 Z2,2(1,4,2) 1 Z3,2(1,4,2) 0 Z1,3(1,4,2) 0 Z2,3(1,4,2) 0

Z1,2(1,4,3) 0 Z2,2(1,4,3) 1 Z3,2(1,4,3) 0 Z1,3(1,4,3) 0 Z2,3(1,4,3) 0

Z1,2(1,5,0) 0 Z2,2(1,5,0) 1 Z3,2(1,5,0) 0 Z1,3(1,5,0) 0 Z2,3(1,5,0) 0

Z1,2(1,5,1) 0 Z2,2(1,5,1) 1 Z3,2(1,5,1) 0 Z1,3(1,5,1) 0 Z2,3(1,5,1) 0

Z1,2(1,5,2) 0 Z2,2(1,5,2) 1 Z3,2(1,5,2) 0 Z1,3(1,5,2) 0 Z2,3(1,5,2) 0

Z1,2(1,6,0) 0 Z2,2(1,6,0) 0 Z3,2(1,6,0) 1 Z1,3(1,6,0) 0 Z2,3(1,6,0) 1

Z1,2(1,6,1) 0 Z2,2(1,6,1) 0 Z3,2(1,6,1) 1 Z1,3(1,6,1) 0 Z2,3(1,6,1) 1

Z1,2(1,6,2) 0 Z2,2(1,6,2) 0 Z3,2(1,6,2) 1 Z1,3(1,6,2) 0 Z2,3(1,6,2) 1

Z1,2(1,7,0) 0 Z2,2(1,7,0) 1 Z3,2(1,7,0) 0 Z1,3(1,7,0) 0 Z2,3(1,7,0) 0

Z1,2(1,7,1) 0 Z2,2(1,7,1) 1 Z3,2(1,7,1) 0 Z1,3(1,7,1) 0 Z2,3(1,7,1) 0

Z1,2(1,7,2) 0 Z2,2(1,7,2) 1 Z3,2(1,7,2) 0 Z1,3(1,7,2) 0 Z2,3(1,7,2) 0

Z1,2(1,7,3) 0 Z2,2(1,7,3) 1 Z3,2(1,7,3) 0 Z1,3(1,7,3) 0 Z2,3(1,7,3) 0

Z1,2(1,7,4) 0 Z2,2(1,7,4) 1 Z3,2(1,7,4) 0 Z1,3(1,7,4) 0 Z2,3(1,7,4) 0

Z1,2(1,7,5) 0 Z2,2(1,7,5) 1 Z3,2(1,7,5) 0 Z1,3(1,7,5) 0 Z2,3(1,7,5) 0

Z1,2(1,7,6) 0 Z2,2(1,7,6) 1 Z3,2(1,7,6) 0 Z1,3(1,7,6) 0 Z2,3(1,7,6) 0

Z1,2(1,8,0) 0 Z2,2(1,8,0) 0 Z3,2(1,8,0) 1 Z1,3(1,8,0) 0 Z2,3(1,8,0) 1

Z1,2(1,8,1) 0 Z2,2(1,8,1) 0 Z3,2(1,8,1) 1 Z1,3(1,8,1) 0 Z2,3(1,8,1) 1

Z1,2(1,9,0) 0 Z2,2(1,9,0) 1 Z3,2(1,9,0) 0 Z1,3(1,9,0) 0 Z2,3(1,9,0) 0

Z1,2(1,9,1) 0 Z2,2(1,9,1) 1 Z3,2(1,9,1) 0 Z1,3(1,9,1) 0 Z2,3(1,9,1) 0

Z1,2(1,10,0) 0 Z2,2(1,10,0) 1 Z3,2(1,10,0) 0 Z1,3(1,10,0) 0 Z2,3(1,10,0) 0

Note that only modules 1, 3 and 5 have been opted for repair with consisting parts being

discarded in echelon two. Whereas, modules 2, 6 and 8 opted to be replaced in echelon two,

along with the constituent parts, which are then discarded in echelon three after a further

examination. The rest of the modules, namely 4, 7, 9 and 10 along with their parts are opted

for discard in echelon two itself.

5. Conclusions

An application of LORA using GA with actual data would give many insights into how to

allocate various repair/replacement options in different echelons, so that the total cost of

maintenance, and in turn the LCC can be minimized at the design stage.

A further mathematical analysis to incorporate the repair versus replace decision analysis

and location of repair/replacement decision of section 2 into the general mathematical model

of section 3 would result in a generic model with much wider scope.

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106 Ann Oper Res (2006) 143: 91–106

It should however be recognized that at this time, the time to failure distribution parameters

are more likely to be targets than accurate estimates and will generally be given as a single

value, the “MTBF” (mean time between failures).

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