© chia fah choy 2005 topic 4 – life cycle costing
TRANSCRIPT
© Chia Fah Choy 2005
Topic 4 – Life Cycle Costing
© Chia Fah Choy 2005
Learning Objectives
After studying this topic you will:• be able to define Life Cycle Costing (LCC)• understand the objectives of LCC• know the components of total cost of building• understand the LCC techniques• understand the difficulties in assessing LCC• know how to conduct LCC• understand the implementation problem
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Definition
• An economic evaluation which takes into account of all relevant costs such as capital, finance, energy, maintenance and salvage value over the client’s time horizon and adjusting to an equivalent time difference to give the total cost in present value terms.
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Example
Roof AInitial Construction
Cost PV
Replacement after 30 years
$12,000 $12,000
$12,000 $2,089.32
$24,000 $14,089.32Total Cost
Roof BInitial Construction $18,000 $18,000
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TOTAL COSTS(LIFE SPAN OF BUILDING)
MaintenanceOperating services
(operating & cleaning)Energy
RatesInsuranceModifications &AlterationsEstate control (management)
RUNNING COSTS OCCUPATIONAL CHARGES
LandConstruction
Professional fees
USER COSTSINITIAL COSTS
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MAINTENANCE
UNPLANNED MAINTENANCE
PLANNED MAINTENANCE
Schedule maintenance
Condition-based maintenance
Preventive maintenance
Corrective (including emergency maintenance)
Corrective (including emergency maintenance)
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Maintenance Costs
How can it be reached?How can it be reached?
How can it be cleaned?How can it be cleaned?
How long will it last?How long will it last?
How can it be replaced?How can it be replaced?
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Breakdown of typical total costs for various types of buildings
Type of annual cost Houses High Flats
Industrial Buildings
Schools Offices
Maintenance 14 12 18 16 13
Fuel and attendance for heating and lighting
24 24 30 18 29
Initial costs (amortized)
Building
Land and development
48
14
56
8
47
5
51
15
47
11
Total Costs 100 100 100 100 100
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Objectives
• To achieve a more cost effective design
• To achieve the desired function, quality and standard of works for the building as a whole
• Identify all relevant costs
• To achieve least total cost commitment
• Establish historical data
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Life Cycle Costing Techniques
• Life Cycle Cost Analysis (LCCA)
• Life Cycle Cost Management (LCCM)
• Life Cycle Cost Planning (LCCP)
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Life Cycle Cost Analysis
• The collection and analysis of historical data on the actual costs of occupying comparing buildings, having regard to running costs and performance
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Life Cycle Cost Management
• Identifies those areas in which the costs of using the building as detailed by the LCCA can be reduced.
• To assist client to compare building costs in a meaningful way and in assessing and controlling occupancy costs throughout the life of a building.
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Life Cycle Cost Planning
• Part of LCCM
• The prediction of total cost of building
• Planning the timing of work and expenditure on the building
• Updated as necessary
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InceptionInception
FeasibilityFeasibility
Outline Outline ProposalProposal
SchemeSchemeDesignDesign
Measurement information
Detailed Cost Plan
Budget Forecast
Level 1
Budget Estimate
Level 2
Budget Cost Plan and Price Prediction
Level 3
Client in conjunction with QS establishes Budget range for running cost targets for total building for investment appraisal
Establish tax aspects of the building
Identify historical data on running costs of homogeneous building
Modify running cost targets in light of further information. Breakdown running costs into RM/m2 items (Budget LCCP)
Establish use pattern of the building.Define discount rate and produce brief LCCP.
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Cost Checks
SchemeSchemeDesignDesign
Detailed Cost Plan
Measurement information
Level 3
Bill of Bill of QuantitiesQuantities
ConstructionConstruction
Detail Detail DesignDesign
Cost Checks Production Production InformationInformation
Detailed LCCA after 12 Months Occupancy
Check LCCA with LCCP
Assist client cash flow by producing details on Capital Allowances as cost is incurred
Produce detailed information on taxation cost plan
Compare capital cost plan with LCCP
Produce taxation cost plan
Undertake cost checks on LCCP as design develops
Evaluate alternative design options for running cost implications
Produce LCCP for individual elementsCompare LCCP with capital cost plan
Produce detailed LCCP for building
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Difficulties in Assessing Life Cycle Costs
• Accurately assessing the maintenance and running costs of different materials, processes and systems
• Initial, annual and periodic payments have to be related to a common basis for comparison purposes
• Taxation rates and allowances are subject to considerable variation over the life of the building
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Difficulties in Assessing Life Cycle Costs
• Selection of suitable interest rates for calculation involving periods of up to 60 years
• Inflationary tendencies may not affect all costs in a uniform manner
• Client’s interest on projects are to be sold as an investment on completion
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Difficulties in Assessing Life Cycle Costs
• The initial funds available to the building client are severely restricted
• Changes of taste and fashion, changing statutory requirements for buildings and the replacement of worn out components by superior updated items
• Forecasting lives of different types of buildings
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3 Key Concepts
• Time value of money
• Time horizon
• Relevant costs
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Time value for money
• Money is capable of growth
• The value of $1 today is definitely not equal to $1 in years to come
• Apply discounting techniques to enable comparison analysis
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Time horizon
• The time scale when the client is affected either by incurring expenses or through using the facility until such time his interest in the physical asset ceases
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Time horizon (contd.)
• Categories of obsolescence– Economic obsolescence– Functional obsolescence– Technological obsolescence– Social and legal obsolescence– Physical obsolescence
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Useful lives of buildings (years)
Departments 40 Banks 50
Dwelling 45 Factories 45
Farm buildings 25 Hotels 40
Machine shops 45 Office buildings 45
Stores 50 Theaters 40
Warehouses 60
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Useful lives of buildings components/equipments (years)
Large equipment 20 Medium equipment 15
Small equipment 10 Boilers/furnaces 20
Gas burners 16 Oil burners 10
Radiators 25 Awnings 5
Clocks 15 Elevators 25
Fire alarm systems 25 Incinerators 14
Lighting fixtures 15 Plumbing fixtures 25
Roofs 20 Transformers 25
Water tanks (metal) 25 Window screens 10
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Relevant costs
• Acquisition costs• Construction costs• Installation costs• Operation/running
costs• Maintenance costs
• Replacement costs• Fees/charges• Demolition, salvage or
resale value• Other costs such as
loan repayment, taxation, etc
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When to implement
• Impact during the early stages of the design phase i.e. – inception– schematic design– design development phase
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When to implement
• Important decisions during the early stages– To proceed or to abandon– Best/better ways of utilization of available
resources– Choice between alternative design solutions– Financial implications of the options and its
selection
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How to implement
• State the required objectives & constraints• Identify possible options • Collect the relevant data/information• Formulate parameters/assumptions• Identify all relevant costs• Compare options using appropriate appraisal
techniques• Apply sensitivity analysis• Report the recommendation
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LCC Techniques
• Selection depend on the following criteria– Project circumstances– Evaluation stage– Availability of time, resources and
expertise– Degree of accuracy required
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LCC Techniques
• Two categories– Non-discounting techniques– Discounting techniques
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Example 1
To find the PV of the running costs of a building with a life of 60 years.
Annual cleaning cost = $1,600
Annual decoration cost = $600
Annual repairs = $400
External painting = $4,000 every 5 years
A new roof every 30 years = $ 40,000.
Interest rate = 5%
Solution 1
0.051.05
1-1
$2,600 repairs and sdecoration cleaning, PV60
$ 49,216
3722.3$4,000 painting External $ 13,489
301.05
1 $40,000t replacemen Roof $ 9,255
$ 71,960Year 5%5 0.7835
10 0.613915 0.481020 0.376925 0.295330 0.231435 0.181340 0.142045 0.111350 0.087255 0.0683
Total 3.3722
nrPV
1
1
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Example 2
To find the annual equivalent cost over the life of the building.
Initial construction cost = $400,000Annual cleaning and minor repairs cost = $16,000Quinquennial repairs = $40,000Replacement costs at every 20 years = $80,000Demolition costs = $5,800Salvage value at the end of 60 years = $1,800Life of the building = 60 yearsInterest rate = 5%; ASF = 2 ½%
Solution 2
1025.01
025.0 1/2% 2at years 60for ASF 60
Year 5%5 0.7835
10 0.613915 0.481020 0.376925 0.295330 0.231435 0.181340 0.142045 0.111350 0.087255 0.0683
Total 3.3722 nr
PV
1
1
Annual equivalentBuildingInterest 5% = 0.05
= 0.007350.05735
$400,000 x 0.05735 $22,940
Cleaning and minor repairs $16,000
Large repairs $40,000 x 3.3722 = $134,888Replacement $80,000 x (0.3769+0.1420) = $ 41,512
1.05
1$1,800-$5,800 Value eSalvageabl-Demolition
60 = $ 214
= $176,6140.05735
$ 10,129
Annual equivalent of LCC $ 49,069
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Example 3
Compare the LCC of the following alternative building schemes.
Scheme A Scheme B
Cost of building 200,000 260,000
Site cost 40,000 40,000
Annual running cost 6,000 4,800
Replacement cost at every 20 years 24,000 16,000
Replacement cost at every 30 years 32,000 20,000
Life of building = 60 yearsInterest rate = 5%ASF = 2 ½%
Solution 3
Scheme ACost of site= $40,000Annual equivalent in perpetuity at 5% = $40,000 x 0.05 $ 2,000
Cost of building =$200,000
First replacement cost in 20 years 045,9$05.1
1000,24$
20
Second replacement cost in 40 years 409,3$05.1
1000,24$
40
Replacement cost in 30 years 404,7$05.1
1000,32$
30
$219,858PV of building and replacement costs
Annual equivalent over 60 years
Interest at 5% 0.05
ASF to replace $1 in 60 years at 2 ½% 0.00735 0.05735 $ 12,609
Annual running costs $ 6,000LCC $ 20,609
Solution 3
Scheme BCost of site= $40,000Annual equivalent in perpetuity at 5% = $40,000 x 0.05 $ 2,000
Cost of building =$260,000
First replacement cost in 20 years 027,6$05.1
1000,16$
20
Second replacement cost in 40 years 273,2$05.1
1000,16$
40
Replacement cost in 30 years 627,4$05.1
1000,20$
30
$272,927PV of building and replacement costs
Annual equivalent over 60 years
Interest at 5% 0.05
ASF to replace $1 in 60 years at 2 ½% 0.00735 0.05735 $ 15,652
Annual running costs $ 4,800LCC $ 22,452
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Example 4
A building which is to be demolished in 25 years times requires repainting now and will also require repainting every 5 years until demolition. The cost of each repainting is estimated at $1,200. In 10 years time $8,000 is to be spent on alterations, and $600 will be spent at the end of each year on sundry repairs. What sum must be set aside now to cover the cost of all work, assuming that the rate of interest obtainable on investment is 6%, and ignoring the effect of taxation?
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Solution 4
Cost of painting 1200Present repainting 1.00000PV of $1 in 5 years at 6% 0.74726PV of $1 in 10 years at 6% 0.55839PV of $1 in 15 years at 6% 0.41727PV of $1 in 20 years at 6% 0.31180
3.034723642
Cost of alterations 8000PV of $1 in 10 years at 6% 0.55839
4467Cost of sundry repairs 600PV of $1 pa for 24 years at 6% 12.55036
7530Sum to be set aside 15639
nr1
1
06.006.11
1 24
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Example 5
An electrically operated 8 person lift installation to serve 6 floors is required for a new building with planned life of 30 years. The initial cost of the lift installation is $42,000, and the running costs are made up of wiping down finishes 12 times a year at $1.60, vacuuming the floor 100 times a year at $0.12, replacing the carpet tile flooring and painting the lift car every 5 years at $300, replacing the installation after 20 years at a cost of $45,000 and allowing for a comprehensive maintenance contract at $920 p.a. (excluding the first year). Calculate the PV of the LCC for the lift installation at a compound rate of interest of 5%.
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Solution 5Initial cost 42000Annual costWipping down finishes: 12X$1.60 19.20Vaccuming floor: 100x$0.12 12.00
31.20PV of $1 pa for 30 years at 5% 15.3725
480Maintenance contract 920PV of $1 pa for 30 years at 5% 15.3725Less first year 0.9524
14.420113266
Replacing floor finish and repainting every 5 years 300PV of $1 in 5 years at 5% 0.7835PV of $1 in 10 years at 5% 0.6139PV of $1 in 15 years at 5% 0.4810PV of $1 in 25 years at 5% 0.2953
2.1738652
Replacing lift installation after 20 years 45000PV of $1 in 20 years at 5% 0.3769
16960PV of LCC of lift 73358
05.005.11
1 30
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Example 6
A temporary building is to be replaced in 15 years’ time by a new building which it is estimated will then cost $240,000. What sum must be set aside at the end of each year, if the interest rate on investment (after deducting for tax) is 3%, to accumulate to the building cost figure in 15 years?
Solution 6
Cost of new building in 15 years’ time $240,000
Sinking fund to provide $1 in 15 years at 3%
05377.0103.1
03.015
Sum to be set aside = 240,000 x 0.05377 = $12,905
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Example 7
Lift in government building consume 2M Kwh electricity p.a.Present cost of electricity = RM0.06/KwhExtensive overhaul & modification = RM100,000; reduce power
consumption by 10% p.a. Equipment life extended to 25 years.No overhaul, equipment last 6 years, after which, overhaul not
feasibleNew equipment = RM400,000 & cost RM20,000 to remove &
dispose old equipmentNew equipment is 25% more energy efficient than current without
overhaul & estimated life of exceeding 25 yearsMaintenance contract = RM1,000 p.a.Bank interest rate = 12% Inflation (general) = 4%Inflation (fuel) = 6%
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Solution – Example 7
Options available1. Overhaul & modification
2. No overhaul, but replace with new after 6 years
3. Replace with new equipment now
Solution – Example 7
%0566.0106.1
12.1
ratediscount inflation Fuel
%0769.0104.1
12.1
ratediscount General
n
n
ii
i
1
11p.a. RM1 PV
Factortion Multiplica
207.130566.010566.0
10566.01p.a. RM1 PV
ratediscount inflation for Factor tion Multiplica
25
25
964.100769.010769.0
10769.01p.a. RM1 PV
ratediscount generalfor Factor tion Multiplica
25
25
Solution – Example 7
p.a. RM120,000
RM0.062,000,000
n consumptioenergy ofCost
Cost Opeating
RM20,000
equipment old of dispose and Remove
Cost Disposal
Option 1 – Overhaul and modification
Capital cost for overhaul & modify RM 100,000
Operating costEnergy reduced by 10% annuallyTherefore, cost of energy consumption= RM120,000 x 90% = RM108,000
PV RM1 p.a. for energy cost over 25 years time horizon= RM108,000 x 13.207 RM1,426,356
Maintenance CostPV RM1 p.a. for maintenance= RM1,000 x 10.964 RM 10,964
Total Life Cycle Cost for Option 1 RM1,537,320
Solution – Example 7
97.40566.010566.0
10566.01p.a. RM1 PV
ratediscount inflation for Factor tion Multiplica
6
6
667.40769.010769.0
10769.01p.a. RM1 PV
ratediscount generalfor Factor tion Multiplica
6
6
236.80566.010566.0
10566.01
0566.1
1
years 6 deferred (fuel) p.a. RM1 PV
19
19
6
297.60769.010769.0
10769.01
0769.1
1
years 6 deferred (general) p.a. RM1 PV
19
19
6
6411.00769.01
1 PV 6
Option 2 – No Overhaul but replace with new after 6 years
Operating cost PV RM1 p.a. for energy consumption over 6 years= RM120,000 x 4.97 RM 596,400
Capital cost/ReplacementFor new lift equipment = RM400,000x1.046x0.6411
Disposal RM20,000 x 1.046 x 0.6411
RM 324,478
Maintenance costPV RM1 p.a. for maintenance cost over 25 years time horizon= RM1,000 x 4.667
RM 16,224Total Life Cycle Cost for Option 2 RM1,689,306
RM 4,667
Operating cost (thereafter)PV RM1 p.a. for fuel deferred 6 years= RM120,000 x 75% x 8.236 RM 741,240
Maintenance costPV RM1 p.a. for maintenance deferred 6 years= RM1,000 x 6.297
RM 6,297
Option 3 – Replace with new
Capital cost of new lift equipment RM 400,000
Disposal
Operating cost - energyPV RM1 p.a. for energy over 25 years time horizon= RM120,000 x 75% x 13.207
RM 20,000
Total Life Cycle Cost for Option 3 RM1,619,554
RM1,118,630
Maintenance costPV RM1 p.a. for maintenance cost over 25 years time horizon=RM1,000 x 10.964 RM 10,964
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Solution – Example 7
Option 1 - Overhaul & modification
Option 2 - No overhaul, but replace with new after 6 years
Option 3 - Replace with new equipment now 1. RM1,537,320
2. RM1,619,594
3. RM1,689,306
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Implementation Problems
• Data collection and its related problems
• Forecasting/accuracy
• Sensitivity analysis
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Data collection and its related problems
• LCC demands different types of data– Cost data– Performance data– Physical data– Quality data
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Data collection and its related problems
• Possible sources of data– Internal/records, etc– Published cost data– Financial institutions– Sub-contractors, suppliers,
specialists, etc– Property& professional bodies– Trade associations, government
bodies, others– BCIS
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Data collection and its related problems
• Reason for lack of data– Standard method of collection, classifying and recording– Many firms do not keep such record– Running cost and performance are suitable for accounting
purposes only– Long time lag between design stage and running cost
available– Frequency of alterations to fabric and occupancy pattern– Lack of understanding of interrelationship between
components of the system
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Data collection and its related problems
• Reason for lack of data– Value perceived for such data may not be
worth the time and money involved in its collection
– Data does not indicate expenditure, efficiency or effectiveness on performance of the work involved
– Cost consequences of delayed works are not ascertainable from the available data
– Poor coordination in organising, monitoring and feedback of data
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Data collection and its related problems
• Reason for lack of data– High variability in running costs data due to un-
identical nature of works– Lack of understanding of the importance and
utilisation of such data for LCC
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Forecasting/accuracy
• Forecasting of future costs is heavily dependent upon availability of historical data, professional judgments and expertise
• Historic data may be defective or mismatched• Use of assumptions or default values when
historic data is not available• Variability in cost data• Lack of knowledge over application of LCC
techniques• Lack of reliable cost data
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Sensitivity Analysis
• SA is capable of identifying the extent to which a particular result is dependent upon the parameters and assumption made in arriving at the result
• A range of value i.e. maximum and minimum values are incorporated into the suspected variables i.e. discount rate, time horizon, estimation of initial costs or ownership costs
• Re-evaluate the alternative options to determine if the upper and lower limits of the variable affect the original choices
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Learning Objectives
After studying this topic you will:• be able to define Life Cycle Costing (LCC)• understand the objectives of LCC• know the components of total cost of building• understand the LCC techniques• understand the difficulties in assessing LCC• know how to conduct LCC• understand the implementation problem