econ construction equipment

7/30/2019 Econ Construction Equipment

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CONSTRUCTION EQUIPMENT

Equipment is a critical resource in the execution of most

construction projects.

The equipment fleet may represent the largest long-term capitalinvestment in many construction companies. Consequently,

equipment management decisions have significant impacts on theeconomic viability of construction firms.


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Equipment must pay for itself by earning more for the contractorthan it costs to purchase, own, and use it. Idle equipment is a drainon income - operatingcosts incur only when the equipment is used,

but ownershipcosts incur irrespective of frequency of use.

Contractors must continually evaluate their equipment fleets todetermine when to acquire additional items, when to replace itemsand when to dispose of items that is underutilized.


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The capabilities of construction equipment are described inmanufacturers' literature and can be used to estimate equipmentproductivity.

The costs to be considered are the cost of owning, leasing, orrenting the equipment and the costs of operating, maintaining, andrepairing it.

The effectiveness of a contractors preventive maintenance programwill significantly influence equipment operating and repair costs.


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Most construction operations can be performed by more than onetype of equipment.

The equipment selected should complete the work in accordance

with the project plans and specifications, in the required time frame,and at the least overall cost.


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The following factors should be considered in selecting equipment fora project:

Cost effectiveness. This means not only .selecting theappropriate type of equipment for the task, but also selecting anappropriate-sized machine. This involves comparison of theincreased production rates of larger machines with theirincreased ownership and operating costs. Where possible,

contractors should select the size of equipment that minimizesthe unit cost (e.g., dollars per cubic meter of performing theconstruction task. The soil conditions of the job site may dictatethe type of equipment that should be selected. Trackedequipment usually is selected when the surface condition of the

job site is soft or wet, because they exert less ground pressureand generally have better traction than wheeled equipment undersuch conditions. Construction site access or working-arearestrictions may also limit the types and sizes of equipment thatcan be used on a construction site.


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Versatility. To control total project costs and minimize

equipment transportation costs, equipment should be selectedthat can perform multiple tasks on a given project site. Using atractor to excavate for a foundation, backfill the completedfoundation, and grade around the newly constructed building isusually more efficient than using a different type of equipment

for each task. The project must be analyzed in its entirety toselect the most cost-effective set of construction equipment tobe used on the project.


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Contractors continually analyze their equipment fleets to ensure thatnone of their equipment is losing money for them.

Major company decisions include purchasing, leasing, depreciating,repairing, and replacing equipment. These management decisionsare based on economic analysis of each alternative course of action.

The time value of moneymust be considered in order to make thebest decisions.


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Equivalence Concept

The concept of equivalencemeans that payments that differin magnitude but are made at different time periods may beequivalent to one another.

The cash flow factors can be used to determine the equivalentvalue of money at a time period different from the one in whichthe money is paid or received.

This involves consideration of time and the interest rate. Forexample, a contractor might be interested in purchasing atruck in five years and wants to determine how much he/sheshould invest today to have sufficient funds at the end of thefive-year period. Another example might be a contractor whois considering either purchasing or leasing a crane.


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Each alternative has differing costs that are incurred at differenttimes. To compare the two, the contractor decides to determine an

equivalent cost for each based on its present worth, which meansdetermining an equivalent cost at today's value.

To be meaningful, any economic comparison must be based onequivalent costs at the same point in time.

In other words, comparing a future cost of one alternative with thepresent worth cost of a second alternative is not valid, and thereforenot meaningful.


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Single Payments

Single payments may occur either today or at some time in the

future. P is used to indicate a sum paid or received today, and Fisused to indicate a future sum.

Let's determine the future value of $10 invested at 6% for one year.

$10 (1 + 0.06) = $10.60

This can be written symbolically as

F = P (1 + i)

where iis the interest rate. For nperiods, the formula becomes

F = P (1+ i)n


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The term (1+ i)n is called the single payment compound amountfactor, which is used to determine the future worth of a present sum

of money.

The reciprocal, or 1/ (1+ i)n is called the single payment presentworth factorwhich is used to determine the present worth of a

future sum of money. In solving economic analysis problems,students may use either their calculators or the formulas for eachfactor or a shorthand notation and the interest tables.

In following, we will set up the example problems both ways, but we

will use the shorthand notation for problem solution.


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The shorthand notation for the single payment compound amountfactor is written as (F/P, i, n)which means find a future sum given a

present value at iinterest for ntime periods.

A similar shorthand notation for the single payment present worthfactor would be (P/F i, n).

This means: Find the present worth of a given future sum receivedor paid at the end of nperiods at an effective interest rate of i.


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Example:

A contractor plans to purchase a pickup truck in 5 years. How muchshould the contractor invest at 6% interest today to have the$30,000 needed to purchase the truck at the end of the 5 years?


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Solution

In this problem, the purchase price is a known future value, and theunknown is the present worth amount. Mathematically, this can be

written asP = F/ (1+ i)n = $30,000 / (1 + 0.006)5

Using our shorthand notation, it is written as

P = F (P/F, i, n) =($30,000) (P/F, 6%, 5)

Note that the unknown is always the numerator in the shorthandnotation (P/F), and the known is the denominator.

Looking at the tables, we find the factor value to be 0.747. Solvingthe equation yields the following answer:

P = ($30,000) (0.747) = $22,410


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Example

A contractor is considering the purchase of a new pump that will beused to remove storm runoff from open excavations. The pump willcost $15,000 and have au expected life of 10 years. After 10 yearsof use, the contractor estimates the pump salvage value will be

$4,000. What is the contractor's total cost (on a present worth basis)of owning the pump, if the effective interest rate is 8%?


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Solution

In this problem, the purchase price is a known present worth costand the salvage value is a future receipt. To determine the presentworth of the total cost, we subtract the present worth of the salvagevalue from the initial cost. Mathematically, this is written as

P = $15000 - $4,000 / (1+0.08)10


P = $15,000 - [($4,000) (P/F, 8%,10)]

Inserting the factor value from table yields the following:

P = $15,000 - [($4,000) (0.463)] = $15,000 - $1,852 = $13,148


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UniformSeries of Payments

In some situations, it is desirable to determine the present worth orfuture worth of a uniform series of payments or receipts. In othersituations, it is necessary to determine a series of equal payments orreceipts.

To accomplish these analyses, we will introduce A, which is definedas a series of equal payments or receipts that occur at the endofeach period for nperiods. It is important that you learn thisdefinition and understand thatA is not used for payments or

receipts mode or received at the beginning of each time period.


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The uniform series compound amount factoris used todetermine the future worth of a series of equal payments or receipts.

Mathematically, it is written as [(1 + i)n-1]/ i, and the shorthandnotation is (F/A,i,n).

The uniform series present worth factoris used to determine the

present worth of a series of equal payments or receipts.Mathematically, it is written as [(1 + i)n-1]/ [i(1 + i)n] and theshorthand notation is (P/A,i,n).


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The uniform series sinking fund factoris used to determine aseries of equal pay-ments or receipts that is equivalent to a stated or

required future sum.Mathematically, it is written as i / [(1 + i)n-1], and the shorthandnotation is (A/F,i,n).

The uniform series capital recovery factoris used to determine a

series of equal payments or receipts that is equivalent to a givenpresent worth sum.Mathematically, it is written as [i (1 + i)n] / [(1 + i)n - 1], and theshorthand notation is (A/P, i, n).


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Example

A contractor is investing $5,000 per year in savings certificates at aninterest rate of 6% and plans to continue the investment program for6 years. He/she is doing this so he will have a down payment forsome new construction equipment.

What will the value of the contractor's investment be at the end of 6years?


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Solution

In this problem, the annual investment is an annual uniform series,and the unknown is the future worth.

Mathematically, this can be written as follows:

[A(1 + i)n-1]/ i = [$5,000)(1 + i)6-1]/ 0.06


F = ($5,000) (F/A, 6%, 6)

Inserting the factor value from table yields the following:

F = ($5,000) (6.975) = $34,875


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Example

A contractor has purchased a new truck for $125,000 and plans touse the truck for 6 years. After 6 years of use, the estimated salvagevalue for the truck will be $30,000.

What is the contractor's annual cost (annual uniform series) for thetruck at an interest rate of 10%?


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Solution

In this problem, the purchase price is given as a present value and thesalvage value as a future value. The unknown is a series of equalannual payments.

Mathematically, this can be written as

A = [($125,000)(0.1)(1+0.1)6/(1+0.1)6-1] - ($30,000)(0.01)/[(1+0.1)6-1]


A = [($125,000) (A/P, 10%, 6)] - [($30,000) (A/F, 10%, 6)]

Inserting the factor values from table yields the following:

A = [($125,000) (0.230)] - [($30,000) (0.130)] = $28,750 - $3,900 =

$24,850


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Cash Flow Diagrams

Cash flow diagrams are used to analyze economic alternatives.Although they are not always necessary in simple problems, suchdiagrams allow the student to better visualize each of the individual

sums and uniform series involved in the alternative. The followingconventions are used to standardize cash flow diagrams:

The horizontal (time) axis is marked off in equal increments, one perinterest period, up to the end of the time period under consideration

(period of ownership). The interest period may be years, months, days, orany other equal time period. Receiptsare represented by arrows directed up and paymentsare

represented by arrows pointing down. Two or more receipts or payments in the same period are placed end-to-

end, and these may be combined. All cash that flows during an interest period is considered to flow at the end

of the period. This is known as the year-end, convention.

These conventions are illustrated in the following example.


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Example

A contractor purchased a small used tractor for $20,000 that he/sheintends to use for landscaping around newly constructed houses.Maintenance costs for the tractor are estimated to be $1,000 peryear.

The contractor plans to dispose of the tractor after 5 years andrealize a salvage value of $7,000.

Annual income generated by the tractor is estimated to be $5,000per year.

Draw the cash flow diagram.


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Solution

Arrows representing the initial purchase price and the annualmaintenance costs will be drawn down in accordance with ourconvention, since they are payments. The salvage value and theincome will be represented by arrows pointing up, because they arereceipts. The resulting cash flow diagram is shown below.


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Alternative Analysis

When two or more alternatives are capable of performing the samefunction, the economically superior alternative will be the one withthe least present worth cost.

This present worth methodof alternative comparison should berestricted to evaluating alter-natives with equal life spans.

Alternatives that accomplish the same function but have unequallives must be compared using the annual cost methodofcomparison. The annual cost method assumes that each alternativewill be replaced by an identical twin at the end of its useful life(infinite renewal).


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The first step in comparing economic alternatives is to construct acash flow diagram for each alternative.

Then a common basis (either P, F, or A) is selected for comparing thealternatives, and an equivalent sum or uniform annual series isdetermined for each. Using the common basis, the alternatives arecompared to select the one that is most favorable.

Contractors are usually interested in earning more from theirequipment investment than simply the cost of money. They often usea minimum attractive rate of returnto perform cash flow analysis.

The minimum attractive rate of return usually includes the cost ofmoney(interest), taxes on the equipment, and equipment insurancecosts. It is used as the effective interest rate in cash flow analysis.

These concepts are illustrated in the following examples.


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Example

A contractor is considering purchasing a used tractor for $180,000that he/she could use for 10 years and then sell for an estimatedsalvage value of $10,000.

Annual maintenance and repair costs for the used tractor are

estimated to be $15,000 per year.As an alternative, the contractor could lease a similar tractor for$4,000 per month.

Should the contractor purchase the used tractor or lease the tractorfrom an equipment dealer?

Annual operating cost is approximately the same for bothalternatives. Use a minimum attractive rate of return of 12%.


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Solution

Sincethe rental alternative is known on an annual cost basis, we willcompare the alter-natives on an annual cost basis. The annual cost for

the rental alternative isA = (12 months) ($4,000/month) = $48,000

Following is a cash flow diagram for the purchase alternative:


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The annual cost can be determined using the following equation:

A=[($180,000) (A/P,12%,10)] + $15,000-[($10,000) (A/F,12%,10)]

Substituting factor values from tables yields the following:

A=[($180,000) (0.177)] + $15,000 - [($10,000) (0.057)] =

$31,860 + $15,000-$570 = $46,290

The contractor should purchase the used tractor, because it has alower annual cost.


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Example

A contractor has decided to add a grader to his equipment fleet. Hecould purchase either a new or a used one.

Interest, insurance, and taxes total about 12%, and the contractoranticipates using the grader about 2,000 hours per year.

Which of the following alternatives should the contractor select?


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a) The new grader costs $120,000 to purchase and is expected to

have a useful life of 16,000 hours of operation.

Tires cost $5,000 to replace (estimated to occur after every 4,000hours of use) and major repairs will be needed after 8,000 hours ofoperation at a cost of $6,000.

Fuel, oil, and minor maintenance cost about $15.25 for each hourthe grader is used.

Estimated salvage value at the end of 16,000 hours of operation is$10,000.


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b) The used grader costs $75,000 to purchase and is expected tohave a useful life of 8,000 hours of operation.

Tires cost $5,000 to replace (estimated to occur after every 4,000hours of use).

Fuel, oil, and minor maintenance cost about $18.25 for each hourthe grader is used.

Estimated salvage value at the end of 8,000 hours of use is $8,000.


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Solution

Following is the cash flow diagram for the new grader alternative.Annual fuel, oil, and minor maintenance cost is (2,000hr.)($15.25/hr.)or $30,500.


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The cash flow diagram for the used grader alternative is shownbelow. Annual fuel, oil, and minor maintenance cost is (2,000 hr.)

($18.25/hr.) or $36,500.


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Because the two alternatives have different lives, an annual cost

comparison will be used.



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The annual cost for the new grader alternative can be determined with thefollowing equation:

A = [($120,000) (A/P, 12%, 8)] + $30,500

+ [($5,000) (P/F, 12%, 2) (A/P, 12%, 8)]+ [($11,000) (F/F, 12%, 4) (A/P, 12%, 8)]+ [($5,000) (F/P, 12%, 2) (A/F, 12%, 8)]- [($10,000) (A/F, 12%, 8)]

Note that the single sums that occur within the analysis period must be moved toone end or the other (P or F) prior to applying a cash flow factor to determine onequivalent annual cost.

Substituting the cash flow factors from tables yields the following:A = [($120,000) (0.201)] + $30,500

+ [($5,000) (0.797) (0.201)]+ [($11,000) (0.636) (0.201)]+ [($5,000) (1.254) (0.081)]- [($10,000) (0.081)]

= $24,120 + $30,500 + $801 + $1,406 + $508 - $810 = $56,525


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The annual cost for the used grader alternative can be determined withthe following equation:

A = [($75,000) (A/P, 12%, 4)] + $36,500+ [($5,000) (P/F, 12%, 2) (A/P, 12%, 4)]

- [($8,000) (A/F, 12%, 4)]

Substituting the cash flow factors from tables yields the following:

A = [($75,000) (0.329)] + $36,500

+ [($5,000) (0.797) (0.329)]

- [($8,000) (0.209)]

= $24,675 + $36,500 + $1,311 - $1,672 = $60,814


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The contractor should purchase the new grader because it has the

lower annual cost.


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Rate of Return Analysis

Contractors often want to estimate the prospective rate of return onan investment or compare anticipated rates of return for severalalternative investments.The rate of return is the annual interest rate at which the sum ofinvestment and expenditures equals total income from the

investment.Rate of return analysis involves setting receipts equal toexpenditures and solving for the interest rate.Sometimes the interest rate can be solved for directly, but in mostcases it can be found only through a trial-and-error solution. In these

cases, two or more interest rates are assumed, equivalent presentworth or annual costs are calculated, and the rate of return is foundby interpolation.

Both types of problems are illustrated in the following two examples.


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Example:

A contractor is considering making a $300,000 investment in usedconstruction equipment. He estimates that his annual maintenanceand repair costs for the equipment will be $60,000 and that hisannual income from the equipment will be $115,000. He estimates

that he can get 8 years of use out of the equipment, but that therewill be no salvage value at the end of the 8 years. What would bethe contractors prospective rate of return from this investment?


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Solution

Following is a cash flow diagram for the investment:



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This problem could be solved either on a present worth or an annualcost basis. Lets use the annual cost basis. Setting income equal toexpenditures yields the following equation:

$115,000 = $60,000 + $300,000 (A/P i, 8)

Solving for the uniform series capital recovery factor yields thefollowing:

(A/P,i,8) = $115,000-$60,000/ $300,000 = $55,000 / $300,000 =0.183

Now we must find the appropriate interest rate from the interest tables.Examining the interest tables reveals the following:

(A/P, 9%, 8) = 0.181

(A/P, 10%, 8) = 0.187

The prospective rate of return can now be found by interpolation.i= 9 + [(0.183 - 0.181)/(0.187 - 0.181)]


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Example

A contractor is considering the purchase of a new dump truck at acost of $85,000.

Annual maintenance and repair costs are estimated to be $4,000per year.

The truck would be used for 8 years and then sold for an estimatedsalvage value of $10,000.

The contractor estimates that the annual income generated by thetruck will be $16,000 per year.

What is the prospective rate of return for this investment?

O O


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Solution

Following is the cash flow diagram for this investment:



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Lets use a present worth analysis and set total costs equal to total

receipts to find the prospective rate of return.

$85,000+[($4,000)(P/A,i,8)]=[($16,000)(P/A,i,8)]+[($10,000)(P/E i,8)]

Simplifying the equation by subtracting ($4,000) (P/A, i, 8) from bothsides of the equation yields the following:

$85,000 = [($12,000) (P/A, i, 8)] + [($10,000) (P/E i, 8)]

Now we must use a trial-and-error method to find the solution.


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Lets try an interest rate of 5%.

P = [($12,000) (P/A, 5%, 8)] + [($10,000) (P/F, 5%, 8)]

Substituting the factor values from tables yields

P = [($12,000) (6.463)] + [($10,000) (0.677)]

P = $77,556 + $6,770 = $84,326 (which is less than the $85,000)


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Next, lets try an interest rate of 4%.

P = [($12,000) (P/A, 4%, 8)] + [($10,000) (P/F, 4%, 8)]

Substituting the factor values from tables yields

P = [($12,000) (6.733)] + [($10,000) (0.731)]P = $80,796 + $7,310 = $88,106

(which is greater than the $85,000)


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Now we can determine the prospective rate of returnby interpolation.

i = 4 + [($88,106 85,000)/($88,106 84,326)] =

= 4 + [($3,106 / 3,780] = 4.8%

econ construction equipment

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