CHAPTER 4 APPLICATIONS IN THE CONSTRUCTION INDUSTRY

CHAPTER 4 APPLICATIONS IN THE CONSTRUCTION INDUSTRY This chapter introduces some topics that are related to the economic evaluation of alternatives such as the depreciation, breakeven analysis and the sensitivity analysis. These topics are very important for the economic evaluation of projects. Also, the use of investment laws and economic evaluation are presented and applied for the construction projects such as: calculation of the renting cost of equipment, equipment replacement analysis and the evaluation of bids. 4.1 Depreciation The depreciation in defined as the decrease in market value of an asset over time through wear, deterioration or obsolescence. A machine may depreciate (decline in value) because it is wearing out and no longer performing its function as well as when it was new. Many kinds of machinery require increased maintenance as they age, reflecting a slow but continuing failure of individual parts. Also, the quality of outputs may decline due to wear in components. Another aspect of depreciation is that caused by obsolescence. Obsolescence occurs when the asset is no longer technologically superior to available alternatives. A machine is described as obsolete when the function it performs can be done in some better manner. A machine may be in excellent working condition, yet may still be obsolete. For example, electronic machines, computers, etc. As asset always has different values: initial value (P), book value (BV), salvage value (F) and market value (MV). The initial value represents the purchase price of an asset. Salvage value represents the expected price for selling the asset at the end of its useful life. The book value represents the current value in the accounting systems. The book value equals the initial value of the asset minus all the depreciation costs till given time. Engineering Economy 82 Dr. Emad Elbeltagi

The book value is always calculated at the end of each year. The market value, on the other hand, represents the value of the asset if it is sold in the free market. It is not necessary that the book value equals the market value. Depreciation is an accounting charge that allows for the recuperation of capital that was used to procure equipment or other physical assets. There are three common methods for calculating depreciation expense for financial accounting purposes: straight-line, sum-ofyears digits and the sinking fund method. Each method involves the spreading of the amount to be depreciated over the recovery life of an asset in a systematic manner. Each depreciation method selected produces different patterns of depreciation expense per period. The straight-line method assumes linear depreciation or the depreciation cost is allocated equally over the asset useful life. The sum-of-years digits assumes high rate of depreciation at the early age of an asset and decreasing rate at its aged life. The Third method assumes lower rate at the early ages and faster rate at the late age. 4.1.1 Straight-Line method The simplest and best known of the various depreciation methods is the straight-line depreciation method. In this method a constant depreciation charge is made. To obtain the annual depreciation charge at any year, D n, the total amount to be depreciated (initial value, P salvage value, F) is divided by the useful life in years, N. (Annual depreciation charge) D n = (P F) / N (4.1) Where: D n = Annual depreciation P = Initial value of the asset (include purchase price, delivery cost, installation cost and other equipment related costs). P = Salvage value of the asset (the net value after dismantling or removal costs have been subtracted from the actual monetary value). N = Expected depreciable life of the asset. Engineering Economy 83 Dr. Emad Elbeltagi

The book value at any time, n, could be calculated as given in the following equation: BV(n) = P nd n (4.2) Example 4.1: If an asset has a initial value of LE50,000 with LE10,000 salvage value after five years. Calculate the annual depreciation and calculate the book value of the asset after each year. Solution: Annual depreciation: D n = (P - F) / N = 50,000-10,000 / 8 = LE8,000 per year Book value of the asset after each year: BV(n) = P nd n (n = 1, 2, 3, 4, 5) BV(1) = 50,000 (1) 8,000 = LE42,000 BV(2) = 50,000 (2) 8,000 = LE34,000 BV(3) = 50,000 (3) 8,000 = LE26,000 BV(4) = 50,000 (4) 8,000 = LE18,000 BV(5) = 50,000 (5) 8,000 = LE10,000 = F Example 4.2: If the purchase price of an equipment is LE60,000 and its salvage value after 8 years is LE6,000, calculate the annual depreciation and the book value of the equipment each year. Solution: P = 60,000; F = 6,000; N = 8; Total depreciation = 60000 6000 = LE54,000 Annual depreciation = 54000 / 8 = LE6,750 Notice that the book value of the equipment equals its salvage value at the end of its useful life as shown in Table 4.1. Engineering Economy 84 Dr. Emad Elbeltagi

Table 4.1: Straight-line depreciation of Example 4.2 Year Annual depreciation Book value 0 0 60,000 1 6,750 53,250 2 6,750 46,500 3 6,750 39,750 4 6,750 33,000 5 6,750 26,250 6 6,750 19,500 7 6,750 12,750 8 6,750 6,000 4.1.2 Sum-of-years digits method Another method for allocating the cost of an asset minus its salvage value over its useful life is called sum-of-years digits depreciation method. The sum of the year digits is a rapid write-off technique by which most of the value of the asset is written off in the first one-third of its life. This method results in faster depreciation at the early life of an asset. Larger depreciation charges than straight-line depreciation during the early years of an asset and smaller charges as the asset nears the end of its estimated life. Each year, the depreciation charge is computed as the remaining useful life at the beginning of the year divided by the sum of the years digits for the total useful life, with this ratio multiplied by the total amount of depreciation (P F). The mechanics of the method involve finding the sum of the year digits from 1 through N. The depreciation charge for any given year is then obtained by multiplying total amount of depreciation by a ratio of the number of years remaining in the life of the asset to the sum of year digits. Thus means that the depreciation is calculated as the percentage of the remaining life to the original life. For an asset with useful life N, to obtain the annual depreciation charge, D n, at any year n, can be calculated as follows: D n = (Remaining useful life at beginning of a year / Sum of years digits) (P F) Engineering Economy 85 Dr. Emad Elbeltagi

Sum of years digits = N (N + 1) / 2 D n Ν - n 1 ( P F) N( N 1)/2 (4.3) Example 4.3: Resolve Example 4.2 using the straight-line depreciation method. Solution: P = 60,000; F = 6,000; N = 8 Sum-of-years digits = 8 (8 + 1) / 2 = 36 years The calculations are shown in the following table (Table 4.2). Table 4.2: Sum-of-years depreciation of Example 4.3 Year Remaining life / Annual sum-of-years depreciation Book value 0 0 0 60,000 1 8/36 12,000 48,000 2 7/36 10,500 37,500 3 6/36 9,000 28,500 4 5/36 7,500 21,000 5 4/36 6,000 15,000 6 3/36 4,500 10,500 7 2/36 3,000 7,500 8 1/36 1,500 6,000 Example 4.4: Calculate the depreciation charge for the first three years and the book value for year three for an asset which had a first cost of LE25,000, and a salvage value of LE4,000 and a life of eight years. Solution: The sum of years digits must be calculated first: Sum of Years Digits = 8 (8 + 1) / 2 = 36 D 1 = [(8 1 + 1) / 36] (25,000 4,000) = LE4667 Engineering Economy 86 Dr. Emad Elbeltagi

D 2 D 3 = [(8 2 + 1) / 36] (25,000 4,000) = LE4083 = [(8 3 + 1) / 36] (25,000 4,000) = LE3500 The book value for year three: BV(3) = 25,000 [4667 + 4083 + 3500] = LE12,750 4.1.3 Sinking fund method This method assumes that a uniform series of end-of-payments are deposited into an imaginary sinking fund at a given interest rate i. The amount of the annual deposit is calculated so that the accumulated sum at the end of the asset life, and at the stated interest rate, will just equal the value of the asset depreciated (i.e., P F). The amount of yearly depreciation is invested in a compound manner for the remaining period as a uniform series of payments using Eq. 4.4 as follows: A i ( P F) n (1 ) 1 i (4.4) Then the depreciation value, D n, at any year n is calculated using the following equation. D n = A (1 + i) n-1 ; n = 1, 2, 3,..., N (4.5) Example 4.5: Resolve Example 4.2 using the sinking fund depreciation method, assuming that the interest rate is 10%. Solution: P = 60,000; F = 6,000; N = 8; i =10% A = (60000 6000) [(0.1) / (1.1 8 1)] = LE4,722 Accordingly, the annual depreciation could be calculated as follows: At the first year: D 1 = LE4,722 At the second year: D 2 = 4722 (1.1) = LE5,194 At the third year: D 3 = 4722 (1.1) 2 = LE5,714.. At the eighth year: D 8 = 4722 (1.1) 7 = LE9,202 The results of the depreciation calculations are summarized in Table 4.3. Engineering Economy 87 Dr. Emad Elbeltagi

Book value Table 4.3: Sinking fund depreciation of Example 4.5 Year Annual depreciation Book value 0 0 60,000 1 4,722 55,278 2 5,194 50,084 3 5,714 44,370 4 6,285 38,085 5 6,913 31,172 6 7,605 23,567 7 8,365 15,202 8 9,202 6,000 After studying depreciation calculations from the previous listed three methods, Figure 4.1 illustrates the difference between the three methods. The figure shows that the sum-of-year digits method gives an accelerated depreciation compared to the straight-line method. On the other hand, the sinking fund is a decelerated method compared with the straight-line method. However, the straight-line method is the commonly used for calculating asset depreciation. Initial value Sinking fund Straightline Salvage value Sum-of years Age Figure 4.1: Comparison among the three depreciation methods Engineering Economy 88 Dr. Emad Elbeltagi

4.2 Estimating Equipment Costs (Rentals) The cost per unit of time of owning an item of equipment has to be determined. Costs associated with owing equipment called the ownership costs. Estimating equipment cost involves identifying the ownership and operating costs. Ownership costs include: initial cost, financing (investment) costs, depreciation costs and taxes and insurance costs. The operating costs include: maintenance and repair costs, storage costs and fuel and lubrication costs. 4.2.1 Initial cost The initial cost is the total cost required to purchase a piece of equipment. This initial cost is the basis for determining other costs related to ownership as well as operating costs. Generally, initial cost is made up of: price at the factory or used equipment price, extra options and accessories, sales tax, freight and assembly or setup charges. The initial cost is very straight forward, whereas the other costs require more analysis and computation. Te annual depreciated cost of the equipment should be calculated as described in the previous section. 4.2.2 Investment cost The purchase of construction equipment requires a significant investment of money. This money either be borrowed from a lender, or it will be taken from reserve fund of the contractor. Either the lender will charge an interest rate for the borrowed money, or the contractor will lose any interest money that could be gained if the contractor invest that amount of money used to purchase a piece of equipment. In order to calculate the cost of finance (or investment cost) of an equipment, both the purchase price, P, and the salvage value, F, should be converted into uniform annual values. In this situation, the purchase price is considered as a present value invested for n yeas as a series of uniform payments (equipment useful life) and the salvage value is considered as a future sum of money to be discounted for n years as a series of uniform payments. Engineering Economy 89 Dr. Emad Elbeltagi

Annual cos t of finance n i(1 i) (1 ) 1 P i P F n i n (1 i) n F 1 n (4.6) Example 4.6: An excavator purchase price is LE460,000 and its salvage value is Solution: LE40,000 after 10 years of useful life. Find the annual cost of finance of this excavator if the annual interest rate is 15%. P = 460,000; F = 40,000; n = 10; i = 15% Annual cost of finance = 10 0.15(1.15) 460000 0.15 40000 460000 40000 10 10 (1.15) 1 10 (1.15) 1 10 = LE47,684/year 4.2.3 Operating costs Operating cost accrue only when the unit of equipment is used, whereas ownership costs accrue whether or not the equipment is used. Operating costs include maintenance and repairs, fuel, oil and lubricants. The amounts consumed by a piece of equipment vary with the type and size of equipment, the conditions under which it is operated. An equipment is seldom used its total horse power and also it is seldom to work for 60 minute/hour. Thus, the fuel consumed should be based on the actual operating conditions. Perhaps the average demand on an engine might be 50 percent of its maximum power for an average 45 minutes/hour. Maintenance and repair costs: The cost for maintenance and repairs include the expenditures for replacement parts and the labor required to keep the equipment in good working condition. Historical cost records of maintaining and servicing equipment are the most reliable guide in estimating maintenance and repair cost. The manufacturers of construction equipment provide information showing recommended costs for maintenance and repairs for the equipment they Engineering Economy 90 Dr. Emad Elbeltagi

manufacture. The annual cost of maintenance and repairs is often expressed as a percentage of purchase prices or as a percentage of the straight-line depreciation costs. Fuel consumption: When operating under standard conditions, a gasoline engine will consume approximately 0.06 gallon of fuel for each horsepower-hour developed. A diesel engine will consume approximately 0.04 gallon of fuel for each horsepower-hour developed. Lubricating oil consumption: The quantity of lubricating oil consumed by an engine varies with the size of the engine, the capacity, the equipment condition and the number of hours between oil changes. Cost of rubber tires: Many types of construction equipment use rubber tires, whose life usually will not be the same as the equipment on which they are used. For example, a unit of equipment may have an expected useful life of six years, but the tires on the equipment may last only for two years. Therefore, a new set of tires must be placed on the equipment every two years, which would require three sets of tires during the six years the equipment will be used. Thus, the cost of depreciation and repairs for tires should be estimated separately from the equipment. Example 4.7: Calculate the ownership cost per hour for an excavator powered by a 250-hp engine based on the following data: - Purchase price (P) = LE420,000 - Salvage value (F) = LE250,000 - Operation factor = 50% - Useful life (N) = 6 years - Working hours per year = 2000 - Maintenance and repair costs = 110% of annual depreciation - Diesel fuel price = 3.8/gallon Engineering Economy 91 Dr. Emad Elbeltagi

- Fuel consumption = 0.04 gallon/hp/hr - Lube oil cost = 10% of fuel - Interest rate (i) = 10% Solution: Depreciation (assume straight-line) = (420000 250000) / 6 = LE28333.33/year Investment annual cost is calculated as follows: 6 0.1 (1.1) 420000 0.1 420000 250000 6 6 (1.1) 1 6 (1.1) 1 250000 6 Annual investment = (420000 0.2296 70000) (250000 0.1296 41666.67) = 26432 (- 9264.82) = LE35696.82/year Maintenance and repair cost = 1.1 28333.33 = LE31166.67/year Then, the total yearly costs = 28333.33 + 35696.82 + 31166.67 = LE95196.81/year Accordingly, the hourly cost = 95196.81 / 2000 = LE47.6/hr Fuel consumption = 250 0.04 0.5 = 5 gallon/hr Fuel cost = 5 3.8 = LE19/hr Lubricate oil cost = 19 0.1 = LE1.9/hr Finally, the total hourly cost = 47.6 + 19 + 1.9 = LE68.5/hr Example 4.7: Calculate the hourly rate of equipment based on the following data: - Purchase price (P) = LE460,000 - Salvage value (F) = LE40,000 - Useful life (N) = 10 years - Working hours per year = 2000 years - Annual maintenance costs = 10% of purchase price - Annual operating costs = LE47,000 - Interest rate (i) = 15% Engineering Economy 92 Dr. Emad Elbeltagi

Solution: Depreciation (assume straight-line) = (460000 40000) / 10 = LE42000/year Investment annual cost is calculated as follows: 10 0.1 5(1.15) 460000 0.15 40000 460000 40000 10 10 (1.15) 1 10 (1.15) 1 10 Annual investment = LE47684/year Maintenance and repair cost = 0.1 460000 = LE46000/year Operating costs = LE47000/year Then, the total annual costs = 42000 + 47684 + 46000 + 47000 = LE182684/year Accordingly, the hourly cost = 182684/ 2000 = LE91.34/hr 4.2.4 Calculate the equipment rentals using cash flows In this method, the cash flow for both the costs and the revenues are calculated and the net present worth is calculated so that the net present worth equal zero taking into account the time value of money. Example 4.8: Resolve example 4.7 using the cash flows method. Solution: Let s assume that the annual return from renting the equipment is x, accordingly, the cash flows could be summarized as given in Table 4.4. Annual costs = annual maintenance + operating cost = 0.1 460000 + 47000 = = 46000 + 47000 = LE93,000 Engineering Economy 93 Dr. Emad Elbeltagi

Table 4.4: Cash flows of Example 4.8 Year Initial cost Salvage value Annual costs Cash in Cash out Net cash 0 460000 0 0-460000 -460000 1-93000 x -93000 x - 93000 2-93000 x -93000 x - 93000 3-93000 x -93000 x - 93000 4-93000 x -93000 x - 93000 5-93000 x -93000 x - 93000 6-93000 x -93000 x - 93000 7-93000 x -93000 x - 93000 8-93000 x -93000 x - 93000 9-93000 x -93000 x - 93000 10 40000-93000 x + 40000-93000 x - 53000 By equating the PV for both the costs and revenues (i.e., NPV =0), as follows: 0 = -460000 + (x 93000) (P/A, 15%, 10) + 40000 (P/F, 15%, 10) = -460000 + (x 93000) (5.0188) + 40000 (0.2472) x 93000 = 89685 Then, x = LE182,685 / year The hourly rent = 182685 / 2000 = LE91.34/hr 4.3 Sensitivity Analysis All the previous analysis assumes that the data used for an economic decision is deterministic and certain. Also, since many data gathered in solving a problem represents projections of future consequences, there may be considerable uncertainty regarding the accuracy of that data. As the desired results of the analysis is decision making, as appropriate question is, to what extent do variations in the data affect my decision? What variations in a particular estimate would change selection of the alternative? In such case, the decision is said to be sensitive to the estimate. Engineering Economy 94 Dr. Emad Elbeltagi

The sensitivity analysis, as such, aims to check if the economic decision changes as one or more of the data used change such as the MARR, service life, annual maintenance cost, etc. This section is dedicated to show how the change of these values may change the economic decisions. To perform sensitivity analysis, follows the following steps: - Identify the factors(s) that values may change than that has/have been identified. - Identify the range of values change for this/these factor(s). - Identify the economic decision criterion (e.g., NPW, EUAW, IRR, etc.) that will be used to perform the economic analysis. - Calculate the values of this economic criterion at different levels of the changeable factor. - Draw a diagram for the obtained results from the previous step to study the effect of these changes on the economic criterion under study. Example 4.9: The purchasing cost of a given equipment is LE90,000 and its return for the first year is LE30,000 decreasing annually by LE3,000. If the investment rate changes between 10% and 25% and the equipment age ranges from 8 to 12 years. It is required to study the sensitivity of the decision considering the effect of the change of the investment rate and the age using the EUAW method. Neglect the equipment salvage value. Solution: The cash flow could be represented as follows: G = 3,000 LE30,000 LE90,000 Engineering Economy 95 Dr. Emad Elbeltagi

The sensitivity of the decision to the change is the investment rate: In this case, let s assume that the average age of the equipment is 10 years. Then, calculate the EUAW at different values for i, (10, 15, 20 and 25%) i = 10% EUAW = -90000 (A/P, 10%, 10) + 30000 3000(A/G, 10%, 10) = -90000 (0.16274) + 30000 3000 (3.7255) = LE4177 Similarly, The EUAW could be calculated at different i as follows: i = 15% EUAW = LE1919 i = 20% EUAW = - LE687 i = 25% EUAW = - LE3600 The sensitivity of the decision to the change is the equipment age: In this case, let s assume that the average investment rate is 15%. Then, calculate the EUAW at different age values (8, 10 and 12 years). n = 8 EUAW = LE1601 n = 10 EUAW = LE1919 n= 12 EUAW = LE1673 The previous results show that all the values of the EUAW remain positive for all values of n (in the specified range). Accordingly, the decision does not affected with the change of the equipment life. On the other hand, the change of the investment rate changes the values of the EUAW from being positive to negative. Accordingly, the decision changes with the change of the i values. Example 4.10: A test vehicle has a cost of LE100,000 with a life of 10 years. Additional revenues of LE17,500 per year are expected and the required MARR is 15%. If the additional revenues are only estimates and might fluctuate 15%. Evaluate the sensitivity of the analysis. Engineering Economy 96 Dr. Emad Elbeltagi

Solution: EUAW= -100,000(A/P, 15%, 10) + Revenue = - 19,925 + Revenue Then, calculate the EUAW at different values for the revenue within the range of ±15% (14875 to 20125). Revenue EUAW 14,875-5,050 15,750-4,175 16,625-3,300 17,500-2,425 18,375-1,550 19,250-675 20,125 200 Example 4.11: A company wants to purchase a new core driller for information gathering. It is expected that the machine can be purchased for LE275,000 it will last 8 years, expenses will total LE50,000 per year and that the revenues will be LE100,000 per year. The company requires a MARR of 10%. Engineering Economy 97 Dr. Emad Elbeltagi

Solution: a. Determine the sensitivity of the present worth to a change in the MARR. b. What would be the effects of a 10% range to all other parameters? a. PW = -275,000-50,000(P/A, i%, 8) + 100,000(P/A, i%, 8) = -275,000 + 50(P/A, i%, 8) MARR (P/A, i%, 8) PW 6 6.46321 48160.5 7 5.9713 23565 8 5.74664 12332 9 5.53482 1741 10 5.33493-8253.5 b. By changing the cost, expenses and revenues by 10% (-10% to +10%) at the interval of 5%, the effect on the present worth would be as shown in the following table. Engineering Economy 98 Dr. Emad Elbeltagi

Range Cost PW Cost Exp. PW Exp. Rev. PW Rev. -10 247500 39832 45000 41065.2 90000-45134.4-5 261250 26082 47500 26698.6 95000-16401.2 0 275000 12332 50000 12332 100000 12332 5 288750-1418 52500-2034.6 105000 41065.2 10 302500-15168 55000-16401 110000 69798.4 4.4 Breakeven Analysis A fundamental of accounting is that all revenues and costs must be accounted for and the difference between the revenues and costs is the profit, or loss, of the business. Costs can be classified as either a fixed cost or a variable cost. A fixed cost is one that is independent of the level of sales; rather, it is related to the passage of time. Examples of fixed costs include rent, salaries and insurance. A variable cost is one that is directly related to the level of sales, such as cost of goods sold and commissions. In planning and managing your business it is important to know what level of sales must be achieved in order to meet total costs. Every LE of sales above this will contribute to profits. Advantages of Break-even Analysis - It is simple to conduct and understand. - It shows profit and loss at different levels of output. - It can cope with changing circumstances, e.g. the following changes in the business environment can be shown in a break even chart. Disadvantages of Break-even Analysis - It assumes that all output is sold at the given price (this may well be untrue). - Although it can cope with changes in circumstances, these factors change regularly reducing its usefulness as a forecasting tool. - The model assumes that costs increase constantly and do not benefit from economies of scale. If the firm obtains purchasing economies of scale then its total cost line will no longer be straight. Engineering Economy 99 Dr. Emad Elbeltagi

- Break-even analysis is only as good as the data upon which it is based. Poor quality data will lead to inaccurate conclusions being drawn. Example 4.12: A factory produces concrete blocks units. Each unit sells for LE15 and costs LE5. The annual maintenance and operation costs are LE75,000. Calculate the number of blocks that should be produced to justify keep this business running. Solution: Let s assume that number of units produced is x. Accordingly, the cost of blocks equal 5x and the annual revenue is 15x. The annual cost = 75000 + 5x At breakeven point when total costs equal total revenue: 75000 + 5x = 15x Then, x = number of units produced = 7500 blocks The investment in this project would be acceptable when the production increases than 7500 units per year, otherwise it is rejected. Cost Revenue Total cost Fixed cost Units Engineering Economy 100 Dr. Emad Elbeltagi

In case of evaluating alternatives using the breakeven analysis, the breakeven point represents the point at which the total cost for alternatives are equal considering that all alternatives have equal revenue. In case of having alternatives with different revenues, the breakeven for each alternative is identified first, then the breakeven between each two alternatives is identified as follows: - Identify the common criterion (IRR, NPW, etc.) among alternatives. - Calculate the cost in one of these criteria. - For each two alternatives, find the value at which the two alternatives have equal cost. - Represent the results graphically and compare between alternatives on the breakeven point. Example 4.13: There are two proposals to buy a new production line for brick manufacturing. The information related to both alternatives is shown in the following table. If the interest rate is 10%, find: a. The volume of production that justifies the purchase of Alternative A. b. If the demand on the production of this factory is 2000 ton annually, which alternative do you recommend for purchasing? Alternative A B Initial cost LE230,000 LE80,000 Annual costs LE35,000 LE15,000 Salvage value LE40,000 - Labor cost (EL/ton) 15 40 Age (years) 10 5 Solution: First, let s calculate the EUAW for both alternatives: EUAW A = 230000(A/P, 10%, 10) + 35000 40000(A/F, 10%, 10) = LE69,922 EUAW B = 80000(A/P, 10%, 5) + 15000 Engineering Economy 101 Dr. Emad Elbeltagi

= LE36,103 Assume that x represents the annual production (ton/year), the direct costs for both alternatives could be represented as: Direct cost A = 15x Direct cost B = 40x Accordingly, the total cost for both alternatives are: Total cost A = 69922 + 15x Total cost B = 36103 + 40x By equating the total cost for both alternatives, then, the breakeven point equals; x = 1353 ton/year a. Thus means that, if the annual production is more than 1353 ton, then alternative A is better. b. In case of the required production is 2000, then alternative A is better. As shown in the figure below, if the production is less than 1353 ton annually then it is better to use alternative B. While alternative A is better in case of production is more that 1353 ton annually. A B Engineering Economy 102 Dr. Emad Elbeltagi

Example 4.14: It is required to calculate the breakeven point for the three options listed in the table below. Which alternative do you recommend? Annual fixed Variable cost Sell price Alternative cost (LE/ unit) (LE/unit) A LE30,000 17 27 B LE50,000 12 35 C LE60,000 10 40 Solution: First, calculate the breakeven point for each alternative (the point at which the total costs and the revenue are equal). Alternative A: 30000 + 17x = 27x; then x = 3000 units Alternative B: 50000 + 12 x = 35x; then x = 2174 units Alternative C: 60000 + 10 x = 40z; then x = 2000 units Second, draw the relation between the total cost and the number of units produced. A B C Engineering Economy 103 Dr. Emad Elbeltagi

To find the points of intersection, you may solve each two equations together or you may find it from the graph shown above. The above shows the following: - If the production is less than 4000 units, then alternative A is the best. - If the production is greater than 4000 and less than 5000, then alternative B is the best. - Finally, if the production is greater than 5000 units, then alternative C is the best. 4.5 Exercises 1. A company purchased a piece of equipment 3 years ago with an initial value of LE15,000, salvage value of LE3,000, annual operating cost of LE2,000, and estimated life of 10 years. Calculate the book value of the machine now using the straight-line, sum-of years digits and sinking fund depreciation method. Assume interest rate 10%. 2. A backhoe will be purchased for a cost of LE109,750. After a useful life of 5 years, it is assumed the equipment will be sold for LE35,000. Assume interest of 8% for borrowing money, 4% for risk and 2% for taxes, insurance and storage. Calculate the annual ownership cost and the cost per hour assuming the equipment will be used 1800 hr/year. 3. Calculate the ownership cost per hour for a dump truck powered by a 120-hp gasoline engine based on the following data: - Purchase price = LE175,000 - Freight charges = LE2,000 - Estimated salvage value = LE57,500 - Operation factor = 40% - Useful life = 5 years - Hours used per year = 1800 Engineering Economy 104 Dr. Emad Elbeltagi

- Maintenance and repair = 130% of annual depreciation - Tire cost = LE5,000 - Tire life = 4,000 hours - Maintenance and repairs (tires) = 15% of tire depreciation - Gasoline fuel price =LE4.0/gallon - Fuel consumption = 0.06 gallon/hp/hr - Lube oil cost = 10% of fuel - Interest rate (i) = 10% 4. When studying the different alternative air conditioning systems for a building, there were two available systems with their information as shown in the table below. It is required to calculate the decision of selecting any of the two systems to if the interest rate ranges from 8% to 15%. Use the EUAW method. Alternative A B Initial cost LE100,000 LE150,000 Annual costs LE2,000 LE1,500 Salvage value LE5,000 LE10,000 Maintenance at mid-age LE20,000 - Age (years) 8 12 5. Find the breakeven point for the following two alternatives. Assume that the investment rate is 10%. Which one do you select if the expected production is 2000 m 3 /year? Alternative Equipment A Equipment B Initial cost LE23,000 LE8,000 Maintenance cost/year LE3,500 LE1,500 Salvage value LE4,000 - Labor cost/hr LE12 24 Age (years) 10 5 Production (m 3 /hr) 8 6 Engineering Economy 105 Dr. Emad Elbeltagi

REFERENCES Newman, D.G., and Lavelle, J.P., Engineering Economic Analysis, 7 th Engineering Press, Austin, Texas, 1998. edition, Ammar, M., Principles of Engineering Economy, Lecture Notes, Tanta University, 2008. Griffis, F.H., Farr, J.V., and Morris, M.D., Construction Planning for Engineers, McGraw-Hill, Inc., New York, 2000. Engineering Economy 106 Dr. Emad Elbeltagi