Chapter 1 - Construction Economics

8/9/2019 Chapter 1 - Construction Economics

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CONSTRUCTIONECONOMICS


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Time Value of Money

The amount of money left in a saving account for aperiod of time may be calculated by using the equationof:

F=(1+i)n

Where F= value at the end of n periods (future value)P= Present valuei= interest rate per period

n = number of periods

Often (1+i)n is called the single-payment compoundinterest factor.


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The present value of some future amountcan be calculates as:

The expression is called as single-payment present worth factor.

P =

F

(1 + i) n

1

1( + i)n


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Equipment Cost

Elements of Equipment Cost

To tender it is necessary to know thecost of a unit production (money/m3excavation, Money/m2 of top soilstripping, etc.)


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To find the cost of operating of an equipment,it is essential to estimate many factors, suchas, fuel consumption, tire life, and so on. The

best basis for estimating such factors is touse the historical data (=records). If this datais not available, the equipment manufacturercan be consulted.


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Equipment Cost Cont..

Owning and operating cost (O & Ocosts) are composed of owning costsand operating costs.

Owning Costs are fixed costs that areincurred each year whether theequipments is operated or not.

Operating Costsare incurred only whenthe equipment is used.


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Owning Costs

Owning cost is made up by;

- Depreciation

- Investment (interest) cost- Insurance cost

- Taxes

- Storage cost


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Depreciation Costs

Depreciation represents the decline in market value of an item ofequipment due to age, wear, deterioration, and obsolescence.

Depreciation is used for two purposes.- Evaluating tax liability.

- Obtaining depreciation component of hourly equipment costs.

For tax liability many contractors prefer to depreciate the equipmentas rapidly as possible to obtain the maximum reduction in tax liabilityduring the first few years of equipment life.

For rubber-tired equipment, the value of tires should be subtractedfrom the amount to be depreciated because tire cost will becomputed separately as an element of operating cost.


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Depreciation Costs Cont.

Equipment salvage value should be estimated asrealistically as possible based on historical data.

Equipment economical life should also be determined.

For many construction equipment, the economical lifetime is accepted to be 5 years.

The common depreciation methods are:

- Straight line method- Sum of the years digits method

- Double declining balance method


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Straight Line Method

Straight line method produces a uniformdepreciation for each year of equipment life.The amount to be depreciated annually isequipment's initial cost less salvage value (and

less tire cost for rubber-tired equipment).

Where: N= equipment life (years)n = year of life (1, 2, 3 etc.)

D= Depreciation value

DCost - Salvage - (tires)

Nn =


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Example

Using the straight-line method ofdepreciation, find the annualdepreciation and book value at the end

of each year for a track loader having aninitial cost of $50000, a salvage value of$5000, and expected life of 5 years.


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Solution

D $90001,2,3,4,5 =-

=50000 50005

Year Depreciation($)

Book Value ($)(End of Period)

0 0 50000

1 9000 41000

2 9000 32000

3 9000 23000

4 9000 14000

5 9000 5000


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Sum-of-the-Years-Digits Method

The sum of the years digits method ofdepreciation produces a non-uniformdepreciation which is the highest in the firstyear of life and gradually decreases thereafter.The amount to be depreciated is the same asthat used in the straight-line method. Thedepreciation for a particular year is calculatedby multiplying the amount to be depreciated by

a depreciation factor.

ddepreciatebeAmount todigitsyears'ofSum

digitYearDn


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Example

For the loader of Example 3.1, find theannual depreciation and book value atthe end of each year using the sum-of-

the-years'-digits method.


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Solution

Using above equation

15000=5000)-(50000)54321(

5D1

12000=5000)-(5000015

4D2

9000=5000)-(5000015

3D3

6000=5000)-(5000015

2D4

3000=5000)-(5000015

1D5


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Year Depreciation($)

Book Value ($)(End of Period)

0 0 50000

1 15000 35000

2 12000 23000

3 9000 14000

4 6000 8000

5 3000 5000


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Double-Declining-Balance Method

This method produces its maximumdepreciation in the first year of life. Thedepreciation for a particular year is found bymultiplying a depreciation factor by theequipment's book value at the beginning of theyear. The annual depreciation factor is foundby dividing 2 by the equipment life in years.Care must be taken not to reduce the book

value below the salvage value of theequipment.

yearofbeginningat theBook valueN

2Dn


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Example

For the loader of example 3.1, find theannual depreciation and book value atthe end of each year using the double-

declining-balance method.


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Solution

Annual depreciation factor =25=0 4.

20000=500004.0D1

12000=300004.0D2

7200=180004.0D3

4320=100004.0D4

*

5 $1480use2592=64804.0D


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Year Depreciation ($) Book Value ($) (Endof Period)

0 0 50000

1 20000 300002 12000 18000

3 7200 10800

4 4320 6480

5 1480 5000

* Because a depreciation of $2592 in the fifth year would reduce the bookvalue to less than $5000, only $1480 ($6480-5000) may be taken asdepreciation.


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Investment Cost

Investment cost represents the annual cost ofthe capital invested in a machine. It is simplythe interest charge on these funds. Investmentcost is computed by multiplying the interestrate by the value of the equipment. For aspecific year investment cost is calculated byusing the book value of the equipment in thatyear.

However, the simple way to find it is:

Average investment =Initial cost + Salvage value

2


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Insurance, Tax and Storage

Insurance costrepresents the cost of fire, theft,accident, and liability insurance for the equipment.

Tax costrepresents the cost of property tax andlicenses for the equipment

Storage costrepresents the cost of storage yard,wages of guards and employees involved in handlingequipment in and out of storage yard and associateddirect overhead.

Owning CostOwning cost is computed as the sum of depreciation,investment, insurance, tax and storage costs.


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Operating Costs

Operating costs are incurred only whenequipment is operated.

- Fuel cost- Service cost

- Repair cost

- Tire cost

- Cost of special items

- Operators' wages


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L d C diti *


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Type of Equipment

Load Conditions *

Low Average Severe

Clamshell and dragline 0.024 0.030 0.036

Self propelled Compactors 0.038 0.052 0.060

Crane 0.018 0.024 0.060

Excavator Hoe or Shovel 0.035 0.040 0048

Loader - Track 0.030 0.042 0.051

LoaderWheel 0.024 0.036 0.047

Motor Grader 0.025 0.035 0.047

Scraper 0.026 0.035 0.044

TractorCrawler 0.028 0.037 0.046

TractorWheel 0.028 0.038 0.052

Wagon 0.029 0.037 0.046

*: Low, light, or considerable idling; average, normal load and operatingconditions; severe, heavy work, little idling.


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Service Cost

Service cost representsthe cost of oil, hydraulicfluids, grease, and

filters as well as thelabour required toperform routinemaintenance service.

The manufacturersdata or Table 1.2 canbe used.

Operating

ConditionsService Cost

Factor

Favourable 20

Average 33

Severe 50


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Repair Cost

Repair cost represents all repair and maintenance cost except for tirerepair or replacement, routine services, and replacement of high-wear items such as ripper teeth. It is the largest item of operatingexpenses.

Lifetime repair is estimated as a percentage of initial cost. See Table3.2.

However repair cost is lower in the first year than the last years.Therefore, for an accurate estimate during a year the followingequation is used.

operatedHours

costrepairLifetime

digits-years-of-Sum

digitYear=costrepairHourly


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Example

Estimate the hourly repair cost for thefirst year of operation of a crawler tractorcosting $136000 and having a 5-year

life. Assume average operatingconditions and 2000 hours of operationduring the year.


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Solution

Lifetime repair cost factor = 0.90 (Table 3.2)

$122400=$1360000.90=costrepairLifetime

$4.08

2000

122400

15

1=costrepairHourly


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Tire Cost

Tire cost represents the cost of tire andreplacement. Table 3.4 may be used toapproximate tire life if the data is not

available. For replacement of tire, a 15%of repair cost is to be added.

lifetireExpected($)tireofsetaofCost1.15=costTire


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Type of Equipment

Operating Conditions

Favourable Average Severe

Dozers and Loaders 3200 2100 1300

Motor Graders 5000 3200 1900

Conventional Scrapers 4600 3300 2500

Twin-engine Scrapers 4000 3000 2300

Push-pull and elevating Scrapers 3600 2700 2100

Trucks and Wagons 3500 2100 1100


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Operating Costs (cont.)

Special ItemsIt includes the cost of replacing high-wear items such asdozer, grader, and scraper blade cutting edges and endbits, ripper tips, shanks etc. The unit cost is calculated bydividing expected cost in lifetime by working hours in a

life time.Operator Cost

It include the operator's wages. Wages should include allinsurances, social security, taxes, overtime etc.

Total Owning and Operating CostsTotal owning and operating cost is used for tenderingpurposes. However, it does not include overhead andprofit.


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Example

Calculate the expected hourly owning and operating cost for thesecond year of the operation of the twin-engine scraper describedbelow.

Cost delivered=$152,000

Tire cost=$12,000Estimated life=5 years (2000 hours operation per year)Salvage value=$16,000Depreciation method=sum-of-the-years-digitsInvestment (interest) rate=10%Tax, insurance, and storage rate=8%

Operating conditions=averageRated power=465 hpFuel price=$0.40/galOperator's wages=$8.00/h


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Solution

Owning Cost

Depreciation cost:

$33,067=)1200016000152000(15

4=D2

$16.53/hr=2000

33067

=onDepreciati

I t t t i d


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Investment, tax, insurance, andstorage cost

Cost rate = Investment + tax, insurance, andstorage= 10+8= 18%

Average investment =

152000 + 16000

= $840002

Investment, tax, insurance, and storage =84000 0.18

= $7.56 / hr

2000

$24.09/hr=7.56+16.53=costowningTotal


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Operating cost

Fuel cost:

Service Cost:

gal/hr16.3=4650.035=nconsumptioEstimated

Fuel cost =16.3 0.40 = $6.52 / hr

Service cost = 0.33 6.52 =$2.15/ hr


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Repair cost:

Tire cost:

Estimated tire life: 3000 hrs

Operators wages = $ 8.00/hr

Lifetime repair cost = 0.90 (152000 -12000) = $126000

Repair cost =2

15= $8.40 / hr

126000

2000

Tire cost =1.15 $120003000

= $4.60 / hr


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Total opearting cost = 6.52 + 2.15+ 8.40+ 4.60+ 8.00=$29.67 / hr

Total O & O Cost:

53.76/hr$=29.67+24.09=costoperatingandOwning


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Calculate the expected hourly owning cost for a third year ofoperation of a wheel mounted loader with 465 hp. For the thirdyear the loader is scheduled to work at different sites for 5 days aweek and 8 hours a day and it remains idle for 28 days in thethird year for maintenance, servicing and holidays.. Initial cost of loader = $265,000

. Estimated salvage value = $20,000

. Tax cost for third year = $5000

. Insurance cost for third year = 1250

. Storage cost for third year = 1500

. Estimated depreciation for third year = $25,000

The contractor borrowed money from bank to buy the loader andhe pays back his loan on equal instalment of $50,000 per year.The bank rate for borrowed capital is 12% per year.


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Calculate the expected hourly owning cost for a second year of operationof wheel mounted hydraulic shovel. For the second year the shovel isscheduled to work at different sites for 6 days a week and 9 hours a day.Initial cost = 175,000Tax for second year = 4200Storage cost for 2nd year = 1200Estimated equipment life = 7 yearsTire cost = $10,000Estimated salvage value = 16,000Insurance cost for 2nd year = 1100Depreciation method = sum of the years digits

The contractor borrows money from bank to buy the shovel and he paysback his loan on equal instalment of 40,000 per year. The bank rate forborrowed capital is 15% per year.

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Chapter 1 - Construction Economics