Equipment is a critical resource in the execution of most construction projects. The equipment fleet may represent the largest long-term capital investment in many construction companies. Consequently, equipment management decisions have significant impacts on the economic viability of construction firms.
Equipment must pay for itself by earning more for the contractor than it costs to purchase, own, and use it. Idle equipment is a drain on income - operating costs incur only when the equipment is used, but ownership costs incur irrespective of frequency of use. Contractors must continually evaluate their equipment fleets to determine when to acquire additional items, when to replace items and when to dispose of items that is underutilized.
One of the key decisions in planning and executing a construction project is the selection of equipment to use on the project. The type of equipment chosen will determine how the work will be done, the time required to complete the work, and the cost of construction. Therefore, it is important that construction managers understand what type of construction equipment is most appropriate for each construction task and how to estimate equipment productivity and costs. Success in construction is greatly influenced by the selection of equipment for the tasks to be performed.
The capabilities of construction equipment are described in manufacturers' literature and can be used to estimate equipment productivity. The costs to be considered are the cost of owning, leasing, or renting the equipment and the costs of operating, maintaining, and repairing it. The effectiveness of a contractor s preventive maintenance program will significantly influence equipment operating and repair costs.
Most construction operations can be performed by more than one type 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.
The following factors should be considered in selecting equipment for a project: Cost effectiveness. This means not only.selecting the appropriate type of equipment for the task, but also selecting an appropriate-sized machine. This involves comparison of the increased production rates of larger machines with their increased ownership and operating costs. Where possible, contractors should select the size of equipment that minimizes the unit cost (e.g., dollars per cubic meter of performing the construction task. The soil conditions of the job site may dictate the type of equipment that should be selected. Tracked equipment usually is selected when the surface condition of the job site is soft or wet, because they exert less ground pressure and generally have better traction than wheeled equipment under such conditions. Construction site access or working-area restrictions may also limit the types and sizes of equipment that can be used on a construction site.
Versatility. To control total project costs and minimize equipment transportation costs, equipment should be selected that can perform multiple tasks on a given project site. Using a tractor to excavate for a foundation, backfill the completed foundation, and grade around the newly constructed building is usually more efficient than using a different type of equipment for each task. The project must be analyzed in its entirety to select the most cost-effective set of construction equipment to be used on the project.
Contractors continually analyze their equipment fleets to ensure that none of their equipment is losing money for them. Major company decisions include purchasing, leasing, depreciating, repairing, and replacing equipment. These management decisions are based on economic analysis of each alternative course of action. The time value of money must be considered in order to make the best decisions.
Equivalence Concept The concept of equivalence means that payments that differ in magnitude but are made at different time periods may be equivalent to one another. The cash flow factors can be used to determine the equivalent value of money at a time period different from the one in which the money is paid or received. This involves consideration of time and the interest rate. For example, a contractor might be interested in purchasing a truck in five years and wants to determine how much he or she should invest today to have sufficient funds at the end of the five-year period. Another example might be a contractor who is considering either purchasing or leasing a crane.
Each alternative has differing costs that are incurred at different times. To compare the two, the contractor decides to determine an equivalent cost for each based on its present worth, which means determining an equivalent cost at today's value. To be meaningful, any economic comparison must be based on equivalent costs at the same point in time. In other words, comparing a future cost of one alternative with the present worth cost of a second alternative is not valid, and therefore not meaningful.
Single Payments CONSTRUCTION EQUIPMENT 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 F is used 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 i is the interest rate. For n periods, the formula becomes F = P (1+ i) n
The term (1+ i) n is called the single payment compound amount factor, 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 present worth factor which 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 each factor 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.
The shorthand notation for the single payment compound amount factor is written as (F/P, i, n) which means find a future sum given a present value at i interest for n time periods. A similar shorthand notation for the single payment present worth factor would be (P/F i, n). This means: Find the present worth of a given future sum received or paid at the end of n periods at an effective interest rate of i.
Example: A contractor plans to purchase a pickup truck in 5 years. How much should the contractor invest at 6% interest today to have the $30,000 needed to purchase the truck at the end of the 5 years?
Solution CONSTRUCTION EQUIPMENT In this problem, the purchase price is a known future value, and the unknown is the present worth amount. Mathematically, this can be written as P = 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 shorthand notation (P/F), and the known is the denominator. Looking at tables, we find the factor value to be 0.747. Solving the equation yields the following answer: P = ($30,000) (0.747) = $22,410
Example A contractor is considering the purchase of a new pump that will be used to remove storm runoff from open excavations. The pump will cost $15,000 and have au expected life of 10 years. After 10 years of 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%?
Solution In this problem, the purchase price is a known present worth cost and the salvage value is a future receipt. To determine the present worth of the total cost, we subtract the present worth of the salvage value from the initial cost. Mathematically, this is written as P = $15000 - $4,000 / (1+0.08) 10 Using our shorthand notation, it is written as 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
Uniform Series of Payments In some situations, it is desirable to determine the present worth or future worth of a uniform series of payments or receipts. In other situations, it is necessary to determine a series of equal payments or receipts. To accomplish these analyses, we will introduce A, which is defined as a series of equal payments or receipts that occur at the end of each period for n periods. It is important that you learn this definition and understand that A is not used for payments or receipts mode or received at the beginning of each time period.
The uniform series compound amount factor is used to determine the future worth of a series of equal payments or receipts. Mathematically, it is written as [(1 + i) n -1]/ i, and the shorthand notation is (F/A,i,n). The uniform series present worth factor is 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 the shorthand notation is (P/A,i,n).
The uniform series sinking fund factor is used to determine a series 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 shorthand notation is (A/F,i,n). The uniform series capital recovery factor is used to determine a series of equal payments or receipts that is equivalent to a given present worth sum. Mathematically, it is written as [i (1 + i) n ] / [(1 + i) n - 1], and the shorthand notation is (A/P, i, n).
Example A contractor is investing $5,000 per year in savings certificates at an interest rate of 6% and plans to continue the investment program for 6 years. He is doing this so he will have a down payment for some new construction equipment. What will the value of the contractor's investment be at the end of 6 years?
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 Using our shorthand notation, it is written as F = ($5,000) (F/A, 6%, 6) Inserting the factor value from table yields the following: F = ($5,000) (6.975) = $34,875
Example 2.4 A contractor has purchased a new truck for $125,000 and plans to use the truck for 6 years. After 6 years of use, the estimated salvage value for the truck will be $30,000. What is the contractor's annual cost (annual uniform series) for the truck at an interest rate of 10%?
Solution In this problem, the purchase price is given as a present value and the salvage value as a future value. The unknown is a series of equal annual 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] Using our shorthand notation, it is written as 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
Cash Flow Diagrams Cash flow diagrams are used to analyze economic alternatives. Although they are not always necessary in simple problems, such diagrams allow the student to better visualize each of the individual sums and uniform series involved in the alternative. The following conventions are used to standardize cash flow diagrams: The horizontal (time) axis is marked off in equal increments, one per interest period, up to the end of the time period under consideration (period of ownership). The interest period may be years, months, days, or any other equal time period. Receipts are represented by arrows directed up and payments are represented by arrows pointing down. Two or more receipts or payments in the same period are placed end-toend, 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.
Example A contractor purchased a small used tractor for $20,000 that she intends to use for landscaping around newly constructed houses. Maintenance costs for the tractor are esti-mated to be $1,000 per year. The contractor plans to dispose of the tractor after 5 years and realize a salvage value of $7,000. Annual income generated by the tractor is esti-mated to be $5,000 per year. Draw the cash flow diagram.
Solution Arrows representing the initial purchase price and the annual maintenance costs will be drawn down in accordance with our convention, since they are payments. The salvage value and the income will be represented by arrows pointing up, because they are receipts. The resulting cash flow diagram is shown below.
Alternative Analysis When two or more alternatives are capable of performing the same function, the economically superior alternative will be the one with the least present worth cost. This present worth method of alternative comparison should be restricted to evaluating alter-natives with equal life spans. Alternatives that accomplish the same function but have unequal lives must be compared using the annual cost method of comparison. The annual cost method assumes that each alternative will be replaced by an identical twin at the end of its useful life (infinite renewal).
The first step in comparing economic alternatives is to construct a cash flow diagram for each alternative. Then a common basis (either P, F, or A) is selected for comparing the alternatives, and an equivalent sum or uniform annual series is determined for each. Using the common basis, the alternatives are compared to select the one that is most favorable. Contractors are usually interested in earning more from their equipment investment than simply the cost of money. They often use a minimum attractive rate of return to perform cash flow analysis. The minimum attractive rate of return usually includes the cost of money (interest), taxes on the equipment, and equipment insurance costs. It is used as the effective interest rate in cash flow analysis. These concepts are illustrated in the following examples.
Example A contractor is considering purchasing a used tractor for $180,000 that she could use for 10 years and then sell for an estimated salvage 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 tractor from an equipment dealer? Annual operating cost is approximately the same for both alternatives. Use a minimum attractive rate of return of 12%.
Solution Since the rental alternative is known on an annual cost basis, we will compare the alter-natives on an annual cost basis. The annual cost for the rental alternative is A = (12 months) ($4,000/month) = $48,000 Following is a cash flow diagram for the purchase alternative:
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 a lower annual cost.
Example A contractor has decided to add a grader to his equipment fleet. He could purchase either a new or a used one. Interest, insurance, and taxes total about 12%, and the contractor anticipates using the grader about 2,000 hours per year. Which of the following alternatives should the contractor select?
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 4,000 hours of use) and major repairs will be needed after 8,000 hours of operation at a cost of $6,000. Fuel, oil, and minor maintenance cost about $15.25 for each hour the grader is used. Estimated salvage value at the end of 16,000 hours of opertion is $10,000.
b) The used grader costs $75,000 to purchase and is expected to have a useful life of 8,000 hours of operation. Tires cost $5,000 to replace (estimated to occur after every 4,000 hours of use). Fuel, oil, and minor maintenance cost about $18.25 for each hour the grader is used. Estimated salvage value at the end of 8,000 hours of use is $8,000.
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.
The cash flow diagram for the used grader alternative is shown below. Annual fuel, oil, and minor maintenance cost is (2,000 hr.) ($18.25/hr.) or $36,500.
Because the two alternatives have different lives, an annual cost comparison will be used.
The annual cost for the new grader alternative can be determined with the following 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 to one end or the other (P or F) prior to applying a cash flow factor to determine on equivalent 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
The annual cost for the used grader alternative can be determined with the fol-lowing 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
The contractor should purchase the new grader because it has the lower annual cost.
Rate of Return Analysis Contractors often want to estimate the prospective rate of return on an investment or compare anticipated rates of return for several alternative investments. The rate of return is the annual interest rate at which the sum of investment and expenditures equals total income from the investment. Rate of return analysis involves setting receipts equal to expenditures and solving for the interest rate. Sometimes the interest rate can be solved for directly, but in most cases it can be found only through a trial-and-error solution. In these cases, two or more interest rates are assumed, equivalent present worth or annual costs are calculated, and the rate of return is found by interpolation. Both types of problems are illustrated in the following two examples.
Solution Following is a cash flow diagram for the investment:
This problem could be solved either on a present worth or an annual cost basis. Let s use the annual cost basis. Setting income equal to expenditures yields the following equation: $115,000 = $60,000 + $300,000 (A/P i, 8) Solving for the uniform series capital recovery factor yields the following: (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)]
Example A contractor is considering the purchase of a new dump truck at a cost of $85,000. Annual maintenance and repair costs are estimated to be $4,000 per year. The truck would be used for 8 years and then sold for an estimated salvage value of $10,000. The contractor estimates that the annual income generated by the truck will be $16,000 per year. What is the prospective rate of return for this investment?
Solution Following is the cash flow diagram for this investment:
Let s 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 both sides 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.
Let s try an interest rate of5%. 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)
Next, let s try an interest rate of4%. P = [($12,000) (P/A, 4%, 8)] + [($10,000) (P/F, 4%, 8)] Substituting the factor values from Appendix C yields P = [($12,000) (6.733)] + [($10,000) (0.731)] P = $80,796 + $7,310 = $88,106 (which is greater than the $85,000)
Now we can determine the prospective rate of return by interpolation. i = 4 + [($88,106 85,000)/($88,106 84,326)] = = 4 + [($3,106 / 3,780] = 4.8%