BES 513 Engineering Economy and Accounting (Weeks 10-13) PDF

0 TECHNOLOGICAL UNIVERSITY OF THE PHILIPPINES VISAYAS Capt. Sabi St., City of Talisay, Negros Occidental College of Engineering Office of the Program Coordinator LEARNING MODULE Subject Code: BES 513: ENGINEERING E...

0 TECHNOLOGICAL UNIVERSITY OF THE PHILIPPINES VISAYAS Capt. Sabi St., City of Talisay, Negros Occidental College of Engineering Office of the Program Coordinator LEARNING MODULE Subject Code: BES 513: ENGINEERING ECONOMY AND ACCOUNTING DEPARTMENT: COLLEGE OF ENGINEERING COMPILED BY: JOEL B. SUMUGAT, MT, PME 2020 This module is a property of Technological University of the Philippines Visayas and intended for EDUCATIONAL PURPOSES ONLY and is NOT FOR SALE NOR FOR REPRODUCTION.  1 LEARNING GUIDE Week No.: __10__ TOPIC/S 1. Rate of Return Analysis 2. Interpretation of a Rate of Return Value LEARNING OUTCOMES Upon completing this Learning Module, you will be able to: 1. State and understand the meaning of rate of return 2. Use a PW or AW relation to determine the ROR of a series of cash flow 3. State the difficulties of using the ROR method, relative to the PW and AW methods 4. Determine the maximum number of possible ROR values and their values for a specific cash flow series. 5. Determine the external rate of return using the techniques of modified ROR and return on invested capital. CONTENT/TECHNICAL INFORMATION Rate of Return Analysis The most commonly quoted measure of economic worth for a project or alternative is its rate of return (ROR). Whether it is an engineering project with cash fl ow estimates or an investment in a stock or bond, the rate of return is a well accepted way of determining if the project or investment is economically acceptable. Compared to the PW or AW value, the ROR is a generically different type of measure of worth, as is discussed in this chapter. Correct procedures to calculate a rate of return using a PW or AW relation are explained here, as are some cautions necessary when the ROR technique is applied to a single project’s cash flows. The ROR is known by other names such as the internal rate of return (IROR), which is the technically correct term, and return on investment (ROI). Interpretation of a Rate of Return Value From the perspective of someone who has borrowed money, the interest rate is applied to the unpaid balance so that the total loan amount and interest are paid in full exactly with the last loan payment. From the perspective of a lender of money, there is an unrecovered balance at each time period. The interest rate is the return on this unrecovered balance so that the total amount lent and the interest are recovered exactly with the last receipt. Rate of return describes both of these perspectives. This module is a property of Technological University of the Philippines Visayas and intended for EDUCATIONAL PURPOSES ONLY and is NOT FOR SALE NOR FOR REPRODUCTION.  2 Rate of return (ROR) is the rate paid on the unpaid balance of borrowed money, or the rate earned on the unrecovered balance of an investment, so that the final payment or receipt brings the balance to exactly zero with interest considered. Rate of Return Calculation Using a PW or AW Relation The ROR value is determined in a generically different way compared to the PW or AW value for a series of cash flows. For a moment, consider only the present worth relation for a cash fl ow series. Using the MARR, which is established independent of any particular project’s cash flows, a mathematical relation determines the PW value in actual monetary units, say, dollars or euros. For the ROR values calculated in this and later sections, only the cash flows themselves are used to determine an interest rate that balances the present worth relation. Therefore, ROR may be considered a relative measure, while PW and AW are absolute measures. Since the resulting interest rate depends only on the cash flows themselves, the correct term is internal rate of return (IROR); however, the term ROR is used interchangeably. Another definition of rate of return is based on interpretations of PW and AW. The rate of return is the interest rate that makes the present worth or annual worth of a cash fl ow series exactly equal to 0. To determine the rate of return, develop the ROR equation using either a PW or AW relation, set it equal to 0, and solve for the interest rate. Alternatively, the present worth of cash outflows (costs and disbursements) PWo may be equated to the present worth of cash inflows (revenues and savings) PWi. PWo = PWi or AWo = AWi The guideline is as follows: If i* => MARR, accept the project as economically viable. If i* < MARR, the project is not economically viable. The purpose of engineering economy calculations is equivalence in PW or AW terms for a stated i => 0%. In rate of return calculations, the objective is to find the interest rate i* at which the cash flows are equivalent. Example 1. if you deposit Php 1000 now and are promised payments of 500 three years from now and 1500 five years from now, the rate of return relation using PW factors is: This module is a property of Technological University of the Philippines Visayas and intended for EDUCATIONAL PURPOSES ONLY and is NOT FOR SALE NOR FOR REPRODUCTION.  3 Solution: Draw the cash flow diagram: PWo = PWi 1000 = 500 ( P/F , i*,3) + 1500( P/F , i*,5) 0 = -1000 + 500( P/F , i*,3) +1500( P/F , i*,5) i* = 16.9% It should be evident that rate of return relations are merely a rearrangement of a present worth equation. That is, if the above interest rate is known to be 16.9%, and it is used to find the present worth of 500 three years from now and 1500 five years from now, the PW relation is PW= 500( P/F ,16.9%,3) + 1500( P/F ,16.9%,5) =1000 The general procedure: 1. Convert all disbursements into either single amounts ( P or F ) or uniform amounts ( A ) by neglecting the time value of money. For example, if it is desired to convert an A to an F value, simply multiply the A by the number of years n. The scheme selected for movement of cash flows should be the one that minimizes the error caused by neglecting the time value of money. That is, if most of the cash flow is an A and a small amount is an F , convert the F to an A rather than the other way around. 2. Convert all receipts to either single or uniform values. 3. Having combined the disbursements and receipts so that a P/F , P/A , or A/F format applies, use the interest tables to find the approximate interest rate at which the P/F , P/A , or A/F value is satisfied. The rate obtained is a good estimate for the first trial. This module is a property of Technological University of the Philippines Visayas and intended for EDUCATIONAL PURPOSES ONLY and is NOT FOR SALE NOR FOR REPRODUCTION.  4 Example 2. An investment of Php 200,000 now that will reduce the maintenance amount in a manufacturing facility. Estimated savings are 15,000 per year for each of the next 10 years and an additional savings of 300,000 at the end of 10 years in facility and equipment upgrade costs. Determine the rate of return. Solution: Draw the cash flow diagram Use Equation format for the ROR equation: 0 = -200,000 + 15,000( P/A , i *,10) + 300,000( P/F , i *,10) Solve for i* i*= 10.58% ROR EVALUATION OF TWO OR MORE MUTUALLY EXCLUSIVE ALTERNATIVES When selecting from two or more mutually exclusive alternatives on the basis of ROR, equal- service comparison is required, and an incremental ROR analysis must be used. The incremental ROR value between two alternatives (B and A) is correctly identified as Δi*B→A, but it is usually shortened to Δi*. Guidelines: Select the alternative that: 1. requires the largest initial investment, and 2. has a Δi* ≥ MARR, indicating that the extra initial investment is economically justified. Before conducting the incremental evaluation, classify the alternatives as cost or revenue alternatives. The incremental comparison will differ slightly for each type. Cost: Evaluate alternatives only against each other. Revenue: First evaluate against do-nothing (DN), then against each other. This module is a property of Technological University of the Philippines Visayas and intended for EDUCATIONAL PURPOSES ONLY and is NOT FOR SALE NOR FOR REPRODUCTION.  5 The following procedure for comparing multiple, mutually exclusive alternatives, using a PW- based equivalence relation, can now be applied. 1. Order the alternatives by increasing initial investment. For revenue alternatives add DN as the first alternative. 2. Determine the incremental cash flow between the first two ordered alternatives (B – A) over their least common multiple of lives. (For revenue alternatives, the first ordered alternative is DN.) 3. Set up a PW-based relation of this incremental cash flow series and determine Δi*, the incremental rate of return. 4. If Δi*≥ MARR, eliminate A; B is the survivor. Otherwise, A is the survivor. 5. Compare the survivor to the next alternative. Continue to compare alternatives using steps (2) through (4) until only one alternative remains as the survivor. Example 3. Given the data for four machines. Select which machine to recommend base on the internal rate of return. Data currency in Peso. MARR = 13.5%. Solution: 1. These are cost alternatives and are arranged by increasing first cost. 2. The lives are all the same at n= 8 years. The B→A incremental cash flows are indicated in Table below. The estimated salvage values are shown. Incremental Cash Flow for Comparison of Machine B→A: Take the PW relation for B→A: 0 = - 1500 + 300(P/A, Δi*,8) + 400(P/F, Δi*,8) Solving the , Δi* : Δi* = 14.57%. This module is a property of Technological University of the Philippines Visayas and intended for EDUCATIONAL PURPOSES ONLY and is NOT FOR SALE NOR FOR REPRODUCTION.  6 Since Δi* = 14.57%. exceeds MARR = 13.5%, A is eliminated and B is the survivor. Year Cash Flow Cash Flow Incremental Machine B Machine C Cash Flow (B-C) 0 -6500 -15000 -3500 1-8 -3200 -3000 +200 8 +900 +700 -200 Take the PW relation for C→B: 0 = 3500 + 200(P/A, Δi*,8) - 200(P/F, Δi*,8) Solving the , Δi* : Δi* = - 18.77% Since Δi* = -18.77%. is less the MARR = 13.5%, then C is eliminated and B is the survivor. At this point its between B and D: Year Cash Flow Cash Flow Incremental Machine B Machine D Cash Flow (B-D) 0 -6500 -10000 -3500 1-8 -3200 -1400 +1800 8 +900 +1000 +100 Take the PW relation for D→B: 0 = - 8500 + 1800(P/A, Δi*,8) + 100(P/F, Δi*,8) Solving the , Δi* : Δi* = 13.60% With Δi* = 13.60%, machine D is the overall, though marginal, survivor of the evaluation; therefore machine D is recommended for purchase. This module is a property of Technological University of the Philippines Visayas and intended for EDUCATIONAL PURPOSES ONLY and is NOT FOR SALE NOR FOR REPRODUCTION.  7 PROGRESS CHECK Solve the following problems: 1. Supply can be provided by either gasoline or diesel-powered engines. The costs for the gasoline engines in pesos are as follows: The incremental PW cash flow equation associated with (diesel – gasoline) is: 0 = -40,000 + 11,000( P/A , i* , 15) + 16,000( P/F , i* , 15) Determine the following: (a) First cost of the diesel engines (b) Annual M&O cost of the diesel engines (c) Salvage value of the diesel engines 2. Drilling a new well at a cost of Php 1,000,000 or by installing a tank and self- cleaning screen ahead of the desalting equipment. The tank and screen will cost 230,000 to install and 61,000 per year to operate and maintain. A new well will have a pump that is more efficient than the old one, and it will require almost no maintenance, so its operating cost will be only 18,000 per year. If the salvage values are estimated at 10% of the first cost, use a present worth relation to ( a ) calculate the incremental rate of return and ( b ) determine which alternative is better at a MARR of 6% per year over a 20-year study period. 3. Determine which of the two types should be selected by calculating the rate of return on the incremental investment. Assume the company’s MARR is 21% per year. 4. Two additives for improving the dry-weather stability of its low cost acrylic paint. Additive A has a first cost of Php 110,000 and an annual operating cost of 60,000. Additive B has a first cost of 175,000 and an annual operating cost of 35,000. If the company uses a 3-year recovery period for paint products and a MARR of 20% per year, which process is economically favored? Use an incremental ROR analysis. This module is a property of Technological University of the Philippines Visayas and intended for EDUCATIONAL PURPOSES ONLY and is NOT FOR SALE NOR FOR REPRODUCTION.  8 5. The company uses a 3-year planning period and a MARR of 15% per year. Determine which copier the company should acquire on the basis of an incremental rate of return analysis. REFERENCE/S Basics of Engineering Economy by Leland Blank, P. E. and Anthony Tarquin, P. E. This module is a property of Technological University of the Philippines Visayas and intended for EDUCATIONAL PURPOSES ONLY and is NOT FOR SALE NOR FOR REPRODUCTION.  9 LEARNING GUIDE Week No.: __11__ TOPIC/S 1. Benefit/ Cost Analysis 2. Benefit/ Cost Analysis of a Single Project LEARNING OUTCOMES Upon completing this Learning Module, you will be able to: 1. Explain some of the fundamental differences between private and public sector projects 2. Calculate the benefit/cost ratio and use it to evaluate a single project 3. Select the better of two alternatives using the incremental B/C ratio method CONTENT/TECHNICAL INFORMATION Benefit/Cost Analysis The benefit/cost (B/C) ratio introduces objectivity into the economic analysis of public sector evaluation, thus reducing the effects of politics and special interests. The different formats of B/C analysis, and associated dis benefits of an alternative, are discussed here. The B/C analysis can use equivalency computations based on PW, AW, or FW values. Performed correctly, the benefit/cost method will always select the same alternative as PW, AW, and ROR analyses. Benefit/Cost Analysis of a Single Project: The benefit/cost ratio is relied upon as a fundamental analysis method for public sector projects. The B/C analysis was developed to introduce greater objectivity into public sector economic. There are several variations of the B/C ratio; however, the fundamental approach is the same. All cost and benefit estimates must be converted to a common equivalent monetary unit (PW, AW, or FW) at the discount rate (interest rate). The B/C ratio is then calculated using one of these relations: Present worth and annual worth equivalencies are preferred to future worth values. The sign convention for B/C analysis is positive signs; costs are preceded by a sign. Salvage values and additional revenues to the government, when they are estimated, are subtracted from costs in the denominator. Disbenefits are considered in different ways depending upon the model used. Most commonly, disbenefits are subtracted from benefits and placed in the numerator. This module is a property of Technological University of the Philippines Visayas and intended for EDUCATIONAL PURPOSES ONLY and is NOT FOR SALE NOR FOR REPRODUCTION.  10 The decision guideline is simple: If B/C ≥ 1.0, accept the project as economically justified for the estimates and discount rate applied. If B/C ≤ 1.0, the project is not economically acceptable. If the B/C value is exactly or very near 1.0, noneconomic factors will help make the decision. The conventional B/C ratio, probably the most widely used, is calculated as follows: disbenefits are subtracted from benefits, not added to costs. The B/C value could change considerably if disbenefits are regarded as costs. For example, if the numbers 10, 8, and 5 are used to represent the PW of benefits, disbenefits, and costs, respectively, the correct procedure results in B/C =(10 - 8)/5 =0.40. The incorrect placement of disbenefits in the denominator results in B/C =10/(8 + 5)= 0.77, which is approximately twice the correct B/C value of 0.40. Clearly, then, the method by which disbenefits are handled affects the magnitude of the B/C ratio. However, regardless of whether disbenefits are (correctly) subtracted from the numerator or (incorrectly) added to costs in the denominator, a B/C ratio of less than 1.0 by the first method will always yield a B/C ratio less than 1.0 by the second method, and vice versa. The modified B/C ratio includes all the estimates associated with the project, once operational. Maintenance and operation (M&O) costs are placed in the numerator and treated in a manner similar to disbenefits. The denominator includes only the initial investment. Once all amounts are expressed in PW, AW, or FW terms, the modified B/C ratio is calculated as: Salvage value is usually included in the denominator as a negative cost. The modified B/C ratio will obviously yield a different value than the conventional B/C method. However, as with disbenefits, the modified procedure can change the magnitude of the ratio but not the decision to accept or reject the project. The benefit and cost difference measure of worth, which does not involve a ratio, is based on the difference between the PW, AW, or FW of benefits and costs, that is, B − C. If ( B − C )≥ 0, the project is acceptable. This method has the advantage of eliminating the discrepancies noted above when disbenefi ts are regarded as costs, because B represents net benefits. Thus, for the numbers 10, 8, and 5, the same result is obtained regardless of how disbenefits are treated. This module is a property of Technological University of the Philippines Visayas and intended for EDUCATIONAL PURPOSES ONLY and is NOT FOR SALE NOR FOR REPRODUCTION.  11 Before calculating the B/C ratio by any formula, check whether the alternative with the larger AW or PW of costs also yields a larger AW or PW of benefits. It is possible for one alternative with larger costs to generate lower benefits than other alternatives, thus making it unnecessary to further consider the larger-cost alternative. By the very nature of benefits and especially disbenefits, monetary estimates are difficult to make and will vary over a wide range. The extensive use of sensitivity analysis on the more questionable parameters helps determine how sensitive the economic decision is to estimate variation. This approach assists in determining the economic and public acceptance risk associated with a defined project. Also, the use of sensitivity analysis can alleviate some of the public’s concerns commonly expressed that people (managers, engineers, consultants, contractors, and elected officials) designing (and promoting) the public project are narrowly receptive to different approaches to serving the public’s interest. Example 1. Use the conventional and modified B/C methods to determine if this grant proposal is economically justified over a 10-year study period. The foundation’s discount rate is 6% per year. Award amount: Php 20 million (end of) first year, decreasing by 5 million per year for 3 additional years; local government will fund during the first year only Annual costs: 2 million per year for 10 years, as proposed Benefits: Reduction of$8 million per year in health-related expenses for citizens Disbenefits: 0.1 to 0.6 million per year for removal of arable land and commercial districts Solution: Determine the AW for each parameter over 10 years. In Php 1 million units. Award: 20 − 5(A/G,6%,4) = 12.864 per year Annual costs: 2 per year Benefits: 8 per year Disbenefits: Use 0.6 for the first analysis Using conventional B/C analysis: This module is a property of Technological University of the Philippines Visayas and intended for EDUCATIONAL PURPOSES ONLY and is NOT FOR SALE NOR FOR REPRODUCTION.  12 The modifiied B/C analysis: The proposal is not justified economically since both measures are less than 1.0. If the low disbenefits estimate of 0.1 million per year is used, the measures increase slightly, but not enough to justify the proposal. It is possible to develop a direct formula connection between the B/C of a public sector and B/C of a private sector project that is a revenue alternative; that is, both revenues and costs are estimated. Further, we can identify a direct correspondence between the modified B/C relation in Equation [9.3] and the PW method we have used repeatedly. (The following development also applies to AW or FW values.) Let’s neglect the initial investment in year 0 for a moment, and concentrate on the cash flows of the project for year 1 through its expected life. For the private sector, the PW for project cash flows is : PW of project = PW of revenue - PW of costs Since private sector revenues are approximately the same as public sector benefits minus disbenefits (B - D), the modified B/C is: This relation can be slightly rewritten to form the profitability index (PI), which can be used to evaluate revenue projects in the public or private sector. For years t 1, 2,... , n ,: Note that the denominator includes only first cost (initial investment) items, while the numerator has only cash flows that result from the project for years 1 through its life. The PI measure of worth provides a sense of getting the most for the investment. That is, the result is in PW units per PW of money invested at the beginning. This is a “bang for the buck” measure. When used solely for a private sector project, the disbenefits are usually omitted, whereas they should be estimated and included in the modified B/C version of this measure for a public project. The evaluation guideline for a single project using the PI is the same as for the conventional B/C or modified B/C. This module is a property of Technological University of the Philippines Visayas and intended for EDUCATIONAL PURPOSES ONLY and is NOT FOR SALE NOR FOR REPRODUCTION.  13 If PI ≥ 1.0, the project is economically acceptable at the discount rate. If PI ≤ 1.0, the project is not economically acceptable at the discount rate. Remember, the computations for PI and modified B/C are essentially the same, except the PI is usually applied without dis benefits estimated. The PI has another name: the present worth index (PWI). It is often used to rank and assist in the selection of independent projects when the capital budget is limited. Example 1 In our first example, a financial technology startup is expanding and adding two new programmers. The CEO of the company decides to run a cost benefit analysis to determine whether the decision will be beneficial to the company - and to what degree. The company is analyzing a time horizon of one year, and estimates that revenue would increase some 50% if the two programmers were hired. On the cost side of the equation, the CEO must examine the cost of the two programmer's salaries - estimated at Php 75,000. Additionally, there is the cost of recruitment, which might be around 3,000. Training could add an additional 4,000. Additionally, there is the cost of new work areas and computers, totaling 5,000, and the cost of additional licensing for software and the like, around 2,000. The total direct and indirect costs would total around 89,000. When calculating benefits, the CEO would examine the benefit of additional revenue within a 12 month period, estimated around 100,000. Additionally, the increase in product quality resulting from the new programmers (and therefore presumed customer satisfaction) would increase by 10%, adding an estimated 10,000 in value to the company. In total, the benefits would be 110,000. Using the benefit-cost ratio equation, that would be BCR = 110,000/89,000, or 1.24. Given that the value is positive (and the total benefits are greater than the total costs), the cost benefit analysis indicates the decision to hire two additional programmers would be a beneficial move for the company. Alternative Selection Using Incremental B/C Analysis The technique to compare two mutually exclusive alternatives using benefit/cost analysis is virtually the same as that for incremental ROR. The incremental (conventional) B/C ratio, which is identified as ΔB/C, is determined using PW, AW, or FW calculations. The higher- cost alternative is justified if ΔB/C is equal to or larger than 1.0. The selection rule is as follows: This module is a property of Technological University of the Philippines Visayas and intended for EDUCATIONAL PURPOSES ONLY and is NOT FOR SALE NOR FOR REPRODUCTION.  14 If ΔB/C ≥ 1.0, choose the higher-cost alternative, because its extra cost is economically justified. If ΔB/C < 1.0, choose the lower-cost alternative. To perform a correct incremental B/C analysis, it is required that each alternative be compared only with another alternative for which the incremental cost is already justified. This same rule was used for incremental ROR analysis. There are two dimensions of an incremental B/C analysis that differ from the incremental ROR method in Chapter 8. We already know the first, all costs have a positive sign in the B/C ratio. The second, and significantly more important, concerns the ordering of alternatives prior to incremental analysis. Alternatives are ordered by increasing equivalent total costs, that is, PW or AW of all cost estimates that will be utilized in the denominator of the B/C ratio. When not done correctly, the incremental B/C analysis may reject a justified higher-cost alternative. Follow these steps to correctly perform a conventional B/C ratio analysis of two alternatives. Equivalent values can be expressed in PW, AW, or FW terms. 1. Determine the equivalent total costs for both alternatives 2. Order the alternatives by equivalent total cost: first smaller, then larger. Calculate the incremental cost (ΔC) for the larger-cost alternative. This is the denominator in ΔB/C. 3. Calculate the equivalent total benefits and any disbenefits estimated for both alternatives. Calculate the incremental benefits (ΔB) for the larger-cost alternative. This is Δ(B - D) if disbenefits are considered. 4. Calculate the ΔB/C ratio. 5. Use the selection guideline to select the higher-cost alternative if ΔB/C ≥1.0. When the B/C ratio is determined for the lower-cost alternative, it is a comparison with the do nothing (DN) alternative. If B/C < 1.0, then DN should be selected and compared to the second alternative. If neither alternative has an acceptable B/C value and one of the alternatives does not have to be selected, the DN alternative must be selected. In public sector analysis, the DN alternative is usually the current condition. Example This module is a property of Technological University of the Philippines Visayas and intended for EDUCATIONAL PURPOSES ONLY and is NOT FOR SALE NOR FOR REPRODUCTION.  15 The patient usage copay is an estimate of the amount paid by patients over the insurance coverage generally allowed for a hospital room. The discount rate is 5%, and the life of the building is estimated at 30 years. (a) Use incremental B/C analysis to select design A or B. (b) Once the two designs were publicized, the privately owned hospital in the directly adjacent city lodged a complaint that design A will reduce its own municipal hospital’s income by an estimated 500,000 per year because some of the day-surgery features of design A duplicate its services. Subsequently, the association argued that design B could reduce its annual revenue by an estimated 400,000, because it will eliminate an entire parking lot used by their patrons for short-term parking. The city financial manager stated that these concerns would be entered into the evaluation as disbenefits of the respective designs. Redo the B/C analysis to determine if the economic decision is still the same as when disbenefits were not considered. Solution: (a) Since most of the cash flows are already annualized, the incremental B/C ratio will use AW values. No disbenefit estimates are considered. Follow the steps of the procedure above: 1. The AW of costs is the sum of construction and maintenance costs. AWA = 10,000,000(A/P,5%,30) + 35,000 AWA =685,500 AWB = 15,000,000(A/P,5%,30) + 55,000 AWB = 1,030,750 2. Design B has the larger AW of costs, so it is the alternative to be incrementally justified. The incremental cost is : ΔC = AWB - AWA ΔC = 1,030,750 - 685,500 ΔC = 345,250 per year 3. The AW of benefits is derived from the patient usage copays, since these are consequences to the public. The benefits for the ΔB/C analysis are not the estimates themselves, but the difference if design B is selected. The lower usage copay is a positive benefit for design B. ΔB = usageA - usageB ΔB = 450,000 - 200,000 ΔB = 250,000 per year 4. The incremental B/C ratio is: ΔB/C = 250,000/345,250 ΔB/C = 0.72 This module is a property of Technological University of the Philippines Visayas and intended for EDUCATIONAL PURPOSES ONLY and is NOT FOR SALE NOR FOR REPRODUCTION.  16 5. The B/C ratio is less than 1.0, indicating that the extra costs associated with design B are not justified. Therefore, design A is selected. (b) The revenue loss estimates are considered disbenefits. Since the disbenefits of design B are 100,000 less than those of A, this positive difference is added to the 250,000 benefits of B to give it a total benefit of 350,000, thus ΔB/C = 350,000/345,250 ΔB/C = 1.01 Design B is slightly favored. In this case the inclusion of disbenefits has reversed the previous economic decision. This has probably made the situation more difficult politically. Incremental B/C Analysis of Multiple, Mutually Exclusive Alternatives Choose the largest-cost alternative that is justified with an incremental B/C ≥ 1.0 when this selected alternative has been compared with another justified alternative. The procedure for incremental B/C analysis of multiple alternatives is as follows: 1. Determine the equivalent total cost for all alternatives. Use AW, PW, or FW equivalencies. 2. Order the alternatives by equivalent total cost, smallest first. 3. Determine the equivalent total benefits (and any disbenefits estimated) for each alternative. 4. Direct benefits estimation only: Calculate the B/C for the first ordered alternative. If B/C < 1.0, eliminate it. By comparing each alternative to DN in order, we eliminate all that have B/C < 1.0. The lowest-cost alternative with B/C 1.0 becomes the defender and the next higher-cost alternative is the challenger in the next step 5. Calculate incremental costs (ΔC ) and benefits (ΔB ) using the relations ΔC = challenger cost - defender cost ΔB = challenger benefits - defender benefits If relative usage costs are estimated for each alternative, rather than direct benefits, ΔB may be found using the relation: ΔB defender usage costs - challenger usage costs Calculate the ΔB/C for the first challenger compared to the defender. ΔB/ΔC This module is a property of Technological University of the Philippines Visayas and intended for EDUCATIONAL PURPOSES ONLY and is NOT FOR SALE NOR FOR REPRODUCTION.  17 6. If ΔB/C ≥ 1.0, the challenger becomes the defender and the previous defender is eliminated. Conversely, if ΔB/C < 1.0, remove the challenger and the defender remains against the next challenger. 7. Repeat steps 5 and 6 until only one alternative remains. It is the selected one. In all the steps above, incremental disbenefits may be considered by replacing ΔB with Δ( B - D ). This module is a property of Technological University of the Philippines Visayas and intended for EDUCATIONAL PURPOSES ONLY and is NOT FOR SALE NOR FOR REPRODUCTION.  18 PROGRESS CHECK Solve the following problems: 1. The project cost is Php 300,000. It will be maintained at a cost of 25,000 per year. Even though the new road is not very smooth, it allows access to an area that previously could only be reached with off-road vehicles. The improved accessibility has led to a 150% increase in the property values along the road. If the previous market value of a property was 900,000, calculate the B/C ratio using an interest rate of 6% per year and a 20-year study period. 2. An irrigation canal for an area that was recently cleared of trees and large weeds is projected to have a capital cost of Php 2,000,000. Annual maintenance and operation costs will be 100,000 per year. Annual favorable consequences to the general public of 820,000 per year will be offset to some extent by annual adverse consequences of 400,000 to a portion of the general public. If the project is assumed to have a 20-year life, what is the B/C ratio at an interest rate of 8% per year? 3. A proposal to reduce traffic congestion has a B/C ratio of 1.4. The annual worth of benefits minus disbenefits is Php 560,000. What is the first cost of the project if the interest rate is 6% per year and the project is expected to have a 20-year life? REFERENCE/S Basics of Engineering Economy by Leland Blank, P. E. and Anthony Tarquin, P. E. This module is a property of Technological University of the Philippines Visayas and intended for EDUCATIONAL PURPOSES ONLY and is NOT FOR SALE NOR FOR REPRODUCTION.  19 LEARNING GUIDE Week No.: __12__ TOPIC/S 1. Breakeven Analysis 2. Payback Analysis LEARNING OUTCOMES Upon completing this Learning Module, you will be able to: 1. Determine the breakeven point for one parameter 2. Calculate the breakeven point of a parameter and use it to select between two alternatives CONTENT/TECHNICAL INFORMATION Breakeven Analysis Breakeven analysis is performed to determine the value of a variable or parameter of a project or alternative that makes two elements equal, for example, the sales volume that will equate revenues and costs. A breakeven study is performed for two alternatives to determine when either alternative is equally acceptable. Breakeven analysis is commonly applied in make-or- buy decisions when a decision is needed about the source for manufactured components, services, etc. Breakeven Analysis for a Single Project: The analysis is based on the relationship: Profit = revenue – total cost Profit = R – TC Where: R=Revenue TC=Total Cost At breakeven, there is no profit or loss, hence, revenue = total cost R = TC Costs, which may be linear or non-linear, usually include two components: This module is a property of Technological University of the Philippines Visayas and intended for EDUCATIONAL PURPOSES ONLY and is NOT FOR SALE NOR FOR REPRODUCTION.  20 Fixed costs (FC) – Includes costs such as buildings, insurance, fixed overhead, equipment capital recovery, etc. These costs are essentially constant for all values of the decision variable. Variable costs (VC) – Includes costs such as direct labor, materials, contractors, marketing, advertisement, etc. These costs change linearly or non-linearly with the decision variable, e.g. production level, workforce size, etc. For the analysis to be followed here, the variation will generally be assumed to be linear. TC = FC + VC Where: TC= total cost Revenue also changes with the decision variable. Again, for the analysis, the variation will generally be assumed to be linear. The following diagram illustrates the basics of the breakeven analysis. It can be seen that we have profit if the production level is above the breakeven quantity and loss if it is below. Examples: 1. The fixed costs of JJJ company are Php 1 million annually. The main product has revenue of 8.90 per unit and 4.50 variable cost. (a) Determine the breakeven quantity per year, and (b) Annual profit if 200000 units are sold. Solution: Let: QBE = Breakeven quantity or breakeven volume of production Given: This module is a property of Technological University of the Philippines Visayas and intended for EDUCATIONAL PURPOSES ONLY and is NOT FOR SALE NOR FOR REPRODUCTION.  21 FC = Php 1,000,000 Variable cost per unit = 4.5 Revenue per unit = 8.9 A.) Determine the breakeven quantity To break even: revenue = total cost R = FC + VC The variable Cost = Variable cost per unit x QBE Revenue = Revenue per unit x QBE Therefore: 8.9QBE = 1,000,000 + 4.5 QBE QBE = 1,000,000 / (8.9-4.5) QBE = 227,272 units B.) Annual Profit if 200,000 units are sold Q=200,000 Profit = R – TC = 8.90Q – 1,000,000 - 4.5Q Profit = 8.90(200,000) – 1,000,000 - 4.50(200,000) = Php -120,000 (loss) Examples: 2. The current price of a certain is Php 12 per unit. The variable costs are 4 per unit, and 10,000 units are sold every year and a profit of Php 30,000 is realized per year. A new design will increase the variable costs by 20% and Fixed Costs by 10% but sales will increase to 12,000 units per year. (a) At what selling price do we break even, and (b) If the selling price is to be kept same (12/unit) what will the annual profit be? Solution: Given: Price per unit = 12 This module is a property of Technological University of the Philippines Visayas and intended for EDUCATIONAL PURPOSES ONLY and is NOT FOR SALE NOR FOR REPRODUCTION.  22 Variable cost per unit = 4 Number of units sold every year = 10,000 units Profit = Php 30,000 per year For the new design: Variable costs increased by 20% Fixed Costs by 10% Sales increase of 12,000 units per year A.) Selling Price to breakeven B.) At 12/unit of selling price, solve for the annual profit Profit = Revenue – Costs 30000 = 10000(12) – [10000(4) + FC] Solving for the Fixed Cost (FC) FC = Php 50,000 A.) New variable cost = 4(1.2) = 4.8 per unit. New fixed costs = 50000(1.1) = 55,000 Let x = breakeven selling price per unit, then 12000x = 55000 + 12000(4.8) x = Php 9.38/unit (b) Profit = 12000(12) – 12000(4.8) - 55000 Profit = Php 31,400 Example 3.The total annual fixed costs is Php 100,000 and the total variable cost per unit of production as 33. (a) If only 5000 units is all that is expected to sell this year what should the per unit selling price be to make a 25% profit this year? (b) If sales of 3000 units per year is to be added to the 5000 units and a 25% profit is acceptable for this, what could be the new selling price per unit? This module is a property of Technological University of the Philippines Visayas and intended for EDUCATIONAL PURPOSES ONLY and is NOT FOR SALE NOR FOR REPRODUCTION.  23 Solution: Given: FC= Php 100,000 Variable cost per unit = 33 A.) Unit selling price at 5,000 units and 25% profit B.) Selling price at 3,000 additional units and a 25% profit a) Total costs = 100000 + 5000(33) = 265,000 % profit = 100(revenue – cost)/cost 25 = 100(revenue – 265000)/265000 revenue = 265000(1.25) = 331250 Selling price = revenue/number of units Selling price = 331250/5000 Selling price = Php 66.25 / unit. b) Total Cost = FC + VC VC = total number of units x variable cost per unit VC = (3000+5000)33 VC = 8000(33) Total cost = 100000 + 8000(33) Total cost = 364,000 Revenue for 25% profit = 364000(1.25) = 455,000 This module is a property of Technological University of the Philippines Visayas and intended for EDUCATIONAL PURPOSES ONLY and is NOT FOR SALE NOR FOR REPRODUCTION.  24 Therefore; New selling price = 455000/8000 New selling price = Php 56.875 per unit. Two or more Alternatives: This is commonly applied to between alternatives that serve the same purpose. As a result, breakeven analysis is carried out between the costs of the alternatives. It involves the determination of a common variable between two or more alternatives. The procedure to follow for two alternatives is as follows: Define the common variable and its dimensional units. Use AW or PW analysis to express the total cost of each alternative as a function of the common variable. (Use AW values if lives are different). Equate the two relations and solve for the breakeven value of the variable. If the anticipated level is below the breakeven value, select the alternative with the higher variable cost (larger slope). If the level is above the breakeven point, select the alternative with the lower variable cost. The same type of analysis can be performed for three or more alternatives. Then, compare the alternatives in pairs to find their respective breakeven points. The results are the ranges through which each alternative is more economical. Example 4. A company is evaluating the purchase of an automatic cloth-cutting machine. The machine will have a first cost of Php 22,000, a life of 10 years, and a Php 500 salvage value. The annual maintenance cost of the machine is expected to be 2000 per year. The machine will require one operator at a total cost of 24 an hour. Approximately 1500 meters of material can be cut each hour with the machine. Alternatively, if human labor is used, five workers , each earning 10 an hour , can cut 1000 meters per hour. If the company’s MARR is 8% per year , and 180,000 meters of material is to be cut every year should the company buy the automatic machine or use human labor instead? At how many meters cloth-cutting per year will the two alternatives breakeven? Solution: Given: For automatic cloth-cutting machine.: P= Php 22,000 This module is a property of Technological University of the Philippines Visayas and intended for EDUCATIONAL PURPOSES ONLY and is NOT FOR SALE NOR FOR REPRODUCTION.  25 n=10 years SV = 500 AOC = 2,000 Operator salary = 24/hr. Length that can be cut per hour = 1,500 meters If human labor is used: Number of workers = 5 Salary per worker = 10/hr. Length that can be cut per hour = 1,000 meters MARR = 8% Total length to cut per year = 100,000 meters Let: x = meters of material to be cut For Automatic machine: Solving for the Total annual cost: Let: ACA = Annual Cost of Automatic machine ACA = -22000(A/P,8%,10) + 500(A/F,8%,10) – 2000 – (x/1500)(24) = = -5244.15 – x/62.5 This module is a property of Technological University of the Philippines Visayas and intended for EDUCATIONAL PURPOSES ONLY and is NOT FOR SALE NOR FOR REPRODUCTION.  26 For Manual Labor: Let: ACM = Annual Cost of Manual, Labor ACM = -(x/1000)(5)(10) = -x/20 At breakeven, ACA = , ACM -5244.15 – x/62.5 = - x/20 Solving for x: X= 154,240 m Therefore, at 180,000 m, select the automatic machine. (profit occurs if quantity is above the breakeven). Payback Analysis Payback analysis is another use of the present worth technique. It is used to determine the amount of time, usually expressed in years, required to recover the fi rst cost of an asset or project. Payback is allied with breakeven analysis; this is illustrated later in the section. The payback period, also called payback or payout period, has the following definition and types. The payback period np is an estimated time for the revenues, savings, and any other monetary benefits to completely recover the initial investment plus a stated rate of return i. There are two types of payback analysis as determined by the required return. No return; i=0%: Also called simple payback, this is the recovery of only the initial investment. Discounted payback; i> 0%: The time value of money is considered in that some return, for example, 10% per year, must be realized in addition to recovering the initial investment. To calculate the payback period for i= 0% or I > 0%, determine the pattern of the NCF series. Note that np is usually not an integer. For t 1, 2,... , np This module is a property of Technological University of the Philippines Visayas and intended for EDUCATIONAL PURPOSES ONLY and is NOT FOR SALE NOR FOR REPRODUCTION.  27 After n p years, the cash flows will recover the investment in year 0 plus the required return of i %. If the alternative is used more than n p years, with the same or similar cash flows, a larger return results. If the estimated life is less than n p years, there is not enough time to recover the investment and i % return. It is important to understand that payback analysis neglects all cash flows after the payback period of n p years. Consequently, it is preferable to use payback as an initial screening method or supplemental tool rather than as the primary means to select an alternative. The reasons for this caution are that: No-return payback neglects the time value of money, since no return on an investment is required. Either type of payback disregards all cash flows occurring after the payback period. These cash flows may increase the return on the initial investment. Payback analysis utilizes a significantly different approach to alternative evaluation than the primary methods of PW, AW, ROR, and B/C. It is possible for payback analysis to select a different alternative than these techniques. However, the information obtained from discounted payback analysis performed at an appropriate I > 0% can be very useful in that a sense of the risk involved in undertaking an alternative is provided. For example, if a company plans to utilize a machine for only 3 years and payback is 6 years, indication is that the equipment should not be obtained. Even here, the 6-year payback is considered supplemental information and does not replace a complete economic analysis. Example: An Php 18 million design contract was approved. The services are expected to generate new annual net cash flows of 3 million. The contract has a repayment clause of 3 million at any time that the contract is canceled by either party during the 10 years of the contract period. (a) If i =15%, compute the payback period. (b) Determine the no-return payback period and compare it with the answer for i = 15%. Solution: (a) The net cash flow each year is 3 million. The single 3 million payment (call it CV for cancellation value) could be received at any time within the 10-year contract period. This module is a property of Technological University of the Philippines Visayas and intended for EDUCATIONAL PURPOSES ONLY and is NOT FOR SALE NOR FOR REPRODUCTION.  28 0 = - P + NCF(P/A,i,n) + CV(P/F,i,n) 0 = - 18,000,000 + 3,000,000(P/A,15%, np) + 3(P/F,15%, np) Solving for np, np = 15.3 years During the period of 10 years, the contract will not deliver the required return, since np = 15.3 years is greater than 10 years. (b) If no return on its Php 18 million investment, results in np = 5 years, as follows (in $ million) 0 = -18 + 5(3) + 3 There is a very significant difference in np for 15% and 0%. At 15% this contract would have to be in force for 15.3 years, while the no-return payback period requires only 5 years. A longer time is always required for I > 0% for the obvious reason that the time value of money is considered. This module is a property of Technological University of the Philippines Visayas and intended for EDUCATIONAL PURPOSES ONLY and is NOT FOR SALE NOR FOR REPRODUCTION.  29 PROGRESS CHECK Solve the following problems: 1. Suppose a firm is considering manufacturing a new product and the following data have been provided: Sales price Php 12.50 per unit Equipment cost Php 200,000 Overhead cost 50,000 per year Operating and maintenance cost 25 per operating hour Production time 0.1 hours per unit Planning period 5 years MARR 15% Assuming a zero salvage value for all equipment at the end of five years, determine the number of unit to be produced to break even. 2. A company is planning to convert a plant from producing low value cars to producing high end cars. The initial cost for equipment conversion will be Php 200 million with a 20% salvage value anytime within a 5-year period. The cost of producing a car will be 21,000, and it will be sold for 33,000. The production capacity for the first year will be 4000 units. At an interest rate of 12% per year, by what uniform amount will production have to increase each year in order for the company to recover its investment in 3 years? 3. Two types of pumps are available. Pump X costs Php 800 and has a life of 3 years. It also requires rebuilding after 2000 operating hours at a cost of Php 300. Pump Y costs 1,900 and is expected to last 5 years. It also requires overhaul after 8000 hours of operation at a cost of 700. If the operating cost of each pump is 1 per hour, how many hours per year must the pump be required to justify the purchase of pump Y? (Interest rate = 10% per year). 4. Your Company is faced with three proposed methods for making one of their products. Method A involves the purchase of a machine for Php 5,000. It will have a seven-year life, with a zero salvage value at that time. Using Method A involves additional costs of 0.20 per unit of product produced per year. Method B involves the purchase of a machine for 10,000. It will also have a seven-year life, with 2,000 salvage value at that time. Using Method B involves additional costs of 0.15 per unit of product produced per year. Method C involves the purchase of a machine for 8,000. It will have a 2,000 salvage value when disposed of in seven years. Additional costs of 0.25 per unit of product per year arise when Method C is used. An 8% interest rate is used by the ABC Company in evaluating investment alternatives. For what range of annual production volume values is each method preferred? This module is a property of Technological University of the Philippines Visayas and intended for EDUCATIONAL PURPOSES ONLY and is NOT FOR SALE NOR FOR REPRODUCTION.  30 5. Three options are considered for an engine part: Option A – complete in-house manufacturing, with initial equipment cost of Php 50 000, labor cost of 26, 000 per year, and material cost of 10 per engine part. Option B – partial manufacture, (i.e. partially finished engine parts are purchased), with initial equipment cost of 35 000, labor cost of 10 000 per year, material cost of 3 per engine part, and an additional cost of 40 per the partially finished engine part. Option C – purchase from outside at a cost of 120 per engine part. Any equipment purchased will have a life of 6 years. If the MARR is 10% per year, determine the number of engine parts that must be manufactured to justify (a) complete in-house manufacture and (b) partial manufacture. (c) Plot the total cost lines for all three options, and state the ranges of engine parts for which each option will have the lowest cost. REFERENCE/S Basics of Engineering Economy by Leland Blank, P. E. and Anthony Tarquin, P. E. This module is a property of Technological University of the Philippines Visayas and intended for EDUCATIONAL PURPOSES ONLY and is NOT FOR SALE NOR FOR REPRODUCTION.  31 LEARNING GUIDE Week No.: __13__ TOPIC/S 1. Depreciation 2. Purpose of Charging Depreciation 3. Factors affecting the amount of Depreciation 4. Sum-of-Years-Digits (SYD) 5. Sinking Fund method 6. Modified Accelerated Cost Recovery System (MACRS) LEARNING OUTCOMES Upon completing this Learning Module, you will be able to: 1. Define Depreciation 2. Know the Purpose of Charging Depreciation 3. Know the Factors affecting the amount of Depreciation 4. Define Sinking Fund Method 5. Calculate the SYD and MACRS CONTENT/TECHNICAL INFORMATION Depreciation The term depreciation refers to fall in the value or utility of fixed assets which are used in operations over the definite period of years. In other words, depreciation is the process of spreading the cost of fixed assets over the number of years during which benefit of the asset is received. The fall in value or utility of fixed assets due to so many causes like wear and tear, decay, effluxion of time or obsolescence, replacement, breakdown, fall in market value etc. Depreciation: Depreciation is treated as a revenue loss which is recorded when expired utility fixed assets such as plant and machinery, building and equipment etc. Depletion: The term depletion refers to measure the rate of exhaustion of the natural resources or assets such as mines, iron ore, oil wells, quarries etc. While comparing with depreciation, depletion is generally applied in the case of natural resources to ascertain the rate of physical shrinkage but in the case of depreciation is used to measure the fall in the value or utility of fixed assets such as plant and machinery and other general assets. Amortization: The term Amortization is applied in the case of intangible assets such as patents, copyrights, goodwill, trade marks etc., Amortization is used to measure the reduction in value of intangible assets. This module is a property of Technological University of the Philippines Visayas and intended for EDUCATIONAL PURPOSES ONLY and is NOT FOR SALE NOR FOR REPRODUCTION.  32 Obsolescence: Obsolescence means a reduction of usefulness of assets due to technological changes, improved production methods, change in market demand for the product or service output of the asset or legal or other restrictions. Purpose of Charging Depreciation The following are the purpose of charging depreciation of fixed assets: (1) To ascertain in the true profit of the business. (2) To show the true presentation of financial position. (3) To provide fund for replacement of assets. (4) To show the assets at its reasonable value in the balance sheet. Factors Affecting the Amount of Depreciation The following factors are to be considered while charging the amount of depreciation: (1) The original cost of the asset. (2) The useful life of the asset. (3) Estimated scrap or residual value of the asset at the end of its life. (4) Selecting an appropriate method of depreciation. (1) Straight Line Method This method is also termed as Constant Charge Method. Under this method, depreciation is charged for every year will be the constant amount throughout the life of the asset. Accordingly depreciation is calculated by deducting the scrap value from the original cost of an asset and the balance is divided by the number of years estimated as the life of the asset. D=P-F/N Where: D= depreciation per year P= first cost SV=salvage value N= life in years BVn = P-Dn BVn = book value on the nth year Dn = total depreciation on the nth year Example. The asset costs Php10,000 upon purchase. The Salvage value is estimated at 1,000 at he end of its useful life which is 10 years. Compute for the yearly depreciation and the book value on the fifth year of its life. This module is a property of Technological University of the Philippines Visayas and intended for EDUCATIONAL PURPOSES ONLY and is NOT FOR SALE NOR FOR REPRODUCTION.  33 Solution: Given: P=Php 10,000 SV= 1,000 N=10 years A.) Solving for the depreciation per year, D: D=P-SV/N D = (10,000 – 1000)/10 D = Php 900 B.) Solving for the Book Value on year 5 (BV5 ): BVn = P-Dn BV5 = P-D5 But D5 =nD D5 = 5(900) D5 = 4,500 So, BV5 = 10,000 – 4,500 BV5 = 5,500 (2.) Declining Balance Method The declining balance method or double declining balance method is an accelerated system to record depreciation over a period by multiplying an asset's beginning book value by a depreciation rate. Declining balance method is considered an accelerated depreciation method because it depreciates assets at higher rates in the beginning years and lower rates in the later years. In this sense, the declining balance method frontloads the depreciation expense on the first half of the asset's life opposed to the straight line method that evenly distributes depreciation expense over the asset's life. The maximum annual depreciation rate for the DB method is twice the straight line rate double declining balance (DDB). If n = 10 years, the DDB rate is 2/10 = 0.2; so 20% of the book value is removed annually. Another commonly used percentage for the DB method is 150% of the SL rate, where d = 1.5/n. The depreciation for year n is the fixed rate d times the book value at the end of the previous year. This module is a property of Technological University of the Philippines Visayas and intended for EDUCATIONAL PURPOSES ONLY and is NOT FOR SALE NOR FOR REPRODUCTION.  34 Dn=d(BVn-1) The actual depreciation rate for each year n , relative to the basis P , is: dn = d(1-d)n-1 Book value in year t is determined in one of two ways: by using the rate d and basis P or by subtracting the current depreciation charge from the previous book value. BVn=P(1-d)n or BVn= BVn-1 - Dn BVn = P (1 - d )n d = 1 - ( SV/P )1/n Example. The asset costs Php10,000 upon purchase. The Salvage value is estimated at 1,000 at the end of its useful life which is 10 years. Compute for the a.) depreciation and the book value on the fifth year of its life, b.) depreciation and the book value on the fifth year of its life, using double declining. Solution: Given: P=Php 10,000 SV= 1,000 N=10 years A.) D5 Solving for the depreciation rate, d: d = 1 - ( SV/P) 1/n d = 1 – (1000/10000)1/10 d = 0.2056 BVn = P (1 - d )n BV4 = 10000 (1 – 0.2056 )4 BV4 = 3982.5 Dn=d(BVn-1) D5=d(BV4) D5=0.2056(3982.5) D5=818.8 BV5 = 10000 (1 – 0.2056 )5 This module is a property of Technological University of the Philippines Visayas and intended for EDUCATIONAL PURPOSES ONLY and is NOT FOR SALE NOR FOR REPRODUCTION.  35 BV5 = 3163.7 Total Depreciation ay year 5 ( D5T) D5T = P - BV5 D5T = 10,000 – 3,163.7 D5T = 6,836.3 B.) Using double declining. d=2/N d=2/10 d= 0.2 Dn=d(BVn-1) D5=d(BV4) BVn = P (1 - d )n BV4 = 10000 (1 – 0.2)4 BV4 = 4096 D5=0.2(4096) D5= 819.2 BV5 = 10000 (1 – 0.2 )5 BV5 = 3276.8 Sum-of-Years-Digits (SYD) The sum of the years' digits method is used to accelerate the recognition of depreciation. Doing so means that most of the depreciation associated with an asset is recognized in the first few years of its useful life. This method is also called the SYD method. The method is more appropriate than the more commonly-used straight-line depreciation if an asset depreciates more quickly or has greater production capacity in the earlier years than it does as it ages. The total amount of depreciation is identical no matter which depreciation method is used - the choice of depreciation method only alters the timing of depreciation recognition. This module is a property of Technological University of the Philippines Visayas and intended for EDUCATIONAL PURPOSES ONLY and is NOT FOR SALE NOR FOR REPRODUCTION.  36 A problem with using this or any other accelerated depreciation method is that it artificially reduces the reported profit of a business over the near term. The result is excessively low profits in the near term, followed by excessively high profits in later reporting periods. Use of the method can have an indirect impact on cash flows, since accelerated depreciation can reduce the amount of taxable income, thereby deferring income tax payments into later periods. Use the following formula to calculate it: Number of years of estimated life Applicable remaining at the beginning of the percentage = year SYD n(n + 1) SYD = where n = useful life 2 Example of Sum of the Years' Digits Depreciation ABC Company purchases a machine for 100,000. It has an estimated salvage value of 10,000 and a useful life of five years. The sum of the years' digits depreciation calculation is: Compute for the depreciation on year three and the book value at year three. 5(5 + 1) SYD = = 15 2 This formula yields the sum of each year of the estimated useful life: 1 + 2 + 3 + 4 + 5 = 15 Remaining estimated useful Applicable Annual Year life at beginning of year SYD percentage depreciation 1 5 5/15 33.33% 30,000 2 4 4/15 26.67 24,000 3 3 3/15 20.00 18,000 4 2 2/15 13.33 12,000 5 1 1/15 6.67 6,000 Totals 15 100.00% $90,000 𝑥 BVn = P-(P-F)𝑆𝑌𝐷 Solution: Given: This module is a property of Technological University of the Philippines Visayas and intended for EDUCATIONAL PURPOSES ONLY and is NOT FOR SALE NOR FOR REPRODUCTION.  37 P= 100,000 SV=10,000 n= 5 years Required: D3 and BV3 At year 3, x=5+4+3 X=12 𝑥 D3 = (P-F)𝑆𝑌𝐷 12 D3 = (100,000-10,000)15 D3 = 72,000 𝑥 BVn = P-(P-F)𝑆𝑌𝐷 12 BV3 =100,000 - (100,000-10,000) 15 BV3 = 28,000 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: Where: F=salvage value Then the depreciation value, Dn, at any year n is: Dn = A × (1 + i)n-1 BVn=P-Dn This module is a property of Technological University of the Philippines Visayas and intended for EDUCATIONAL PURPOSES ONLY and is NOT FOR SALE NOR FOR REPRODUCTION.  38 Example: using the sinking fund depreciation method, assuming that the interest rate is 10%. Solution: Given the following: P = 60,000; F = 6,000; N = 8; i =10% A = (60000 – 6000) × [(0.1) / (1.18 – 1)] = 4,722 Accordingly, the annual depreciation could be calculated as follows: At the first year: D1 = 4,722 At the second year: D2 = 4722 × (1.1) = 5,194 At the third year: D3 = 4722 × (1.1)2 = 5,714 and so on up to the 8th year. At the eighth year: D8 = 4722 × (1.1)7 = 9,202 The results are tabulated at the table below: Sinking fund Annual Book value depreciation depreciation of Example 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 Modified Accelerated Cost Recovery System (MACRS) Over the years the US Government has used each of these systems in turn to determine allowable depreciation for tax purposes. MACRS is now the system used for most depreciable property. Many aspects of MACRS deal with the specific depreciation accounting aspects of tax law. This section covers only the elements that materially affect after-tax economic analysis. Additional information on how the DDB, DB, and SL methods are embedded into MACRS and how to derive the MACRS depreciation rates. This module is a property of Technological University of the Philippines Visayas and intended for EDUCATIONAL PURPOSES ONLY and is NOT FOR SALE NOR FOR REPRODUCTION.  39 MACRS determines annual depreciation amounts using the relation Dt = dt B The Accelerated Cost Recovery System (ACRS) replaced historical depreciation methods in 1981. In 1986, ACRS was modified (MACRS) and it remains in effect. The advantages of MACRS over historical methods are: (a) an asset may be fully depreciated before the end of its service life, (b) a zero salvage value is assumed, and (c) tables make the system simple to apply. Under MACRS both a general depreciation system (GDS) and alternative depreciation system (ADS) are available. GDS is based on declining balance converted to straight line depreciation. ADS is less desirable for profitable companies and is not discussed here. Notation: B = asset cost basis = asset initial cost plus other costs to make the asset operational rt = appropriate MACRS % rate for year t. dt = depreciation allowance for year t BVt = book value EOY t (BV0 = B) Procedure: 1. Determine the asset's property class, which defines the recovery period. Definitions of property classes will be given in most engineering economic analysis textbooks, and are available from the IRS. These are in the chapter on Depreciation in the text. For personal property (not real estate) there are six property classes (in years of life): 3, 5, 7, 10, 15, and 20. 2. For the asset's recovery period, obtain the appropriate annual % rates. The rates will be given in most engineering economic analysis textbooks, and are available from the IRS. Note there are four rates given for 3-year property and six rates given for 5-year property. This is due to the requirement that only a half-year of depreciation may be taken in the first and last recovery years. Thus 3-year property, for example, is actually depreciated over four years. Special rates apply if more than 40% of the year's MACRS property is placed in service in the last quarter. This case is not discussed here. 3. Calculate depreciation and book value for each year by: dtt = (B)(rt) BVt = B - (accumulated depreciation through year t) This module is a property of Technological University of the Philippines Visayas and intended for EDUCATIONAL PURPOSES ONLY and is NOT FOR SALE NOR FOR REPRODUCTION.  40 Note that the salvage value is not considered. The book value at the end of the recovery period will be zero. Also, although MACRS is based on the double-declining-balance method, the percentages in the tables are always applied to the original basis value, never the book value. Depreciation Rates dt Applied to the Basis B for the MACRS Method Example: An asset with a 3-year recovery period has an initial cost (basis) of Php 10,000. Calculate the annual depreciation and book value for this asset using MACRS. calculation: d1 = (B)(r1) = (10,000)(33.33%) = 3,333 d2 = (B)(r2) = (10,000)(44.45%) = 4,445 d3 = (B)(r3) = (10,000)(14.81%) = 1,481 d4 = (B)(r3) = (10,000)(7.41%) = 741 Depreciation schedule: EOY t dt (Php ) BVt (Php) This module is a property of Technological University of the Philippines Visayas and intended for EDUCATIONAL PURPOSES ONLY and is NOT FOR SALE NOR FOR REPRODUCTION.  41 0 --- 10,000 1 3,333 6,667 2 4,445 2,222 3 1,481 741 4 741 0 This module is a property of Technological University of the Philippines Visayas and intended for EDUCATIONAL PURPOSES ONLY and is NOT FOR SALE NOR FOR REPRODUCTION.  42 PROGRESS CHECK Solve the following Problems: 1. ABC corporation purchased a machine that had a first cost of Php 40,000, an expected useful life of 8 years, a recovery period of 10 years, and a salvage value of 10,000. The operating cost of the machine is expected to be 15,000 per year. The inflation rate is 6% per year and the company’s MARR is 11% per year. Determine ( a ) the depreciation charge for year 3, ( b ) the present worth of the third-year depreciation charge in year 0, the time of asset purchase, and ( c ) the book value for year 3 according to the straight line method. 2. Equipment for cooling has an installed value of Php 182,000 with an estimated trade- in value of 40,000 after 15 years. For years 2 and 10, use DDB book depreciation to determine ( a ) the depreciation charge and ( b ) the book value. 3. Exactly 10 years ago, you purchased Php 100,000 in depreciable assets with an estimated salvage of 10,000. For tax depreciation, the SL method with n=10 years was used, but for book depreciation, the DDB method with n= 7 years and neglected the salvage estimate. The company sold the assets today for 12,500. (a) Compare the sales price today with the book values using the SL and DDB methods. (b) If the salvage of 12,500 had been estimated exactly 10 years ago, determine the depreciation for each method in year 10. 4. Tabulate the results of the following, given useful life of 10 years , present cost of Php 125,000 and a salvage value of 12,500 with i=10% using the sum of the years digit method. Draw the graph using book value vs life in years. 5. Tabulate the results of the following, given useful life of 10 years , present cost of Php 125,000 and a salvage value of 12,500 with i=10% using the sinking fund method. Draw the graph using book value vs life in years. REFERENCE/S Basics of Engineering Economy by Leland Blank, P. E. and Anthony Tarquin, P. E. This module is a property of Technological University of the Philippines Visayas and intended for EDUCATIONAL PURPOSES ONLY and is NOT FOR SALE NOR FOR REPRODUCTION.  43 List of References This module is a property of Technological University of the Philippines Visayas and intended for EDUCATIONAL PURPOSES ONLY and is NOT FOR SALE NOR FOR REPRODUCTION.  44 1. References: Basics of Engineering Economy by Leland Blank, P. E. and Anthony Tarquin, P. E. About the Author/s Joel B. Sumugat is a Professional Mechanical Engineer by profession, with a Master`s Degree in Masters of Technology, and an Outstanding Mechanical Engineer in the Philippines in year 2002 in Research, Invention and Innovation. This module is a property of Technological University of the Philippines Visayas and intended for EDUCATIONAL PURPOSES ONLY and is NOT FOR SALE NOR FOR REPRODUCTION.