STR 402: Quantity Surveying & Cost Control of Construction Projects

Transcription

1 Time Value of Money Depreciation Accounting Techniques STR 402: Quantity Surveying & Cost Control of Construction Projects Dr. Ahmed Saad Eldin Eldieb Time Value of Money Money has a time value. The value of money increases with time because of the interest rate, which represents the cost of borrowing the money or the return obtainable by investing it. Because of the cost of the money, interest must be considered by contractors when making decisions regarding their equipment. This requires a cash flow analysis, which recognizes that money has a different economic value depending on when it is received or paid. Major construction company decisions include purchasing, leasing, depreciating, repairing, and replacing equipment. These management decisions are based on economic analysis of each alternative course of action. The time value of money must be considered in order to make the best decisions. 1

2 Equivalence Concept The concept of equivalence means that payments that differ in magnitude but are made at different time periods may be equivalent to one another. The cash flow factors can be used to determine the equivalent value of money at a time period different from the one in which the money was paid or received. This involves consideration of time and interest rate. Single Payments Single payments may occur either today or at some time in the future. P is used to indicate a sum paid or received today, and F is used to indicate a future sum. Single payment compound amount factor The term (1 + i) n is used to determine the future worth of a present sum of money F = P (1 + i) n F = P (F/P,i,n) i = interest rate n = number of time periods 2

3 Single payment present worth factor 1 The term (1+i) n is used to determine the present worth of a present sum of money i = interest rate n = number of time periods Single Payments P = F 1 (1+i) n P = F (P/F,i,n) Example A contractor plans to purchase a pickup truck in 5 years. The estimated price of the truck at that time is 300,000 EGP. How much should the contractor invest today at a 6% annual interest rate to purchase the truck at the end of the 5 years? P = F (P/F,i,n) P = (300,000) (P/F,6%,5) P = (300,000) (0.7473) P = 224,190 EGP 300,000 3

4 Example A contractor is considering the purchase of a new de-watering pump. The pump will cost 15,000 EGP and will have an expected life of 10 years. After 10 years of use, the contractor estimates the pump salvage value to be 4,000 EGP. What is the contractor s total cost (on present worth basis) of owning the pump, if the effective interest rate is 8%? 4, ,000 P = 15,000-4,000 (P/F,i,n) P = 15,000-4,000 (P/F,8%,10) P = 15,000-4,000 (0.463) P = 13,148 EGP 4

5 Uniform Series Payments In some situations it is desirable to determine the present worth or future worth of a uniform series of payments or receipts. A is used to indicate a series of equal payments or receipts that occur at the end of each period for n periods. Uniform series compound amount factor The term [ 1+i n 1] is used to determine the future worth of a i series of equal payments or receipts i = interest rate n = number of time periods F = A [ 1+i n 1] i F = A (F/A,i,n) Uniform series present worth factor The term [ 1+i n 1] is used to determine the present worth of a [i 1+i n ] series of equal payments or receipts i = interest rate n = number of time periods Uniform Series Payments P = A [ 1+i n 1] [i 1+i n ] P = A (P/A,i,n) 5

6 Uniform Series Payments Uniform series sinking fund factor The term i [ 1+i n 1] is used to determine a series of equal payments or receipts that is equal to a stated or required future sum i = interest rate n = number of time periods A = F i [ 1+i n 1] A = F (A/F,i,n) Uniform series capital recovery factor The term [i 1+i n ] [ 1+i n 1] is used to determine a series of equal payments or receipts that is equal to a given present worth sum i = interest rate n = number of time periods Uniform Series Payments A = P [i 1+i n ] [ 1+i n 1] A = P (A/P,i,n) 6

7 Example Using an interest rate of 12%, find the equivalent uniform annual cost for a piece of construction equipment that has an initial purchase cost of 30,000 EGP, an estimated economic life of 6 years, and an estimated salvage value of 10,000 EGP. Annual maintenance will amount to 600 EGP per year and periodic overhauls costing 1,000 EGP each will occur at the end of the 10,000 second and fourth years ,000 1,000 1,000 Equivalent uniform annual cost = ,000 (A/P,12%,6) - 10,000 (A/F,12%,6) + 1,000 (P/F,12%,2) (A/P,12%,6) + 1,000 (P/F,12%,4) (A/P,12%,6) = ,000 (0.243) 10,000 (0.123) + 1,000 (0.797)(0.243) + 1,000 (0.636)(0.243) = 7, EGP/year Alternative Analysis When two or more alternatives are capable of performing the same function, the economically superior alternative will be the one with the least present worth cost or the most present worth return. This present worth method of alternative comparison should be restricted to evaluating alternatives with equal life spans. Alternatives that accomplish the same function but have unequal lives must be compared using the annual worth method of comparison. The annual worth method assumes that each alternative will be replaced by an identical alternative at the end of its useful life (infinite renewal). 7

8 Alternative Analysis Steps for alternative analysis: Construct a cash flow diagram for each alternative A common basis (either P, F or A) is selected for comparison An equivalent sum or uniform series is determined for each Using the common basis, the alternatives are compared to select the most favorable Example A contractor is considering purchasing a used tractor for 180,000 EGP which he will use for 10 years, and then sell for an estimated salvage value of 10,000 EGP. Annual maintenance and repair costs for the used tractor are estimated to be 15,000 EGP per year. As an alternative the contractor could lease a similar tractor for 4,000 per month. Should the contractor purchase the used tractor or lease the tractor from the equipment dealer? Assume that the annual operating cost are approximately the same for both alternatives. Use a minimum attractive rate of return of 12%. We can compare the alternatives on annual cost basis Lease alternative A lease = (4,000 EGP/month)(12 months/year) A lease = 48,000 EGP/year 8

9 10, ,000 15,000 Purchase alternative A purchase = 180,000 (A/P,12%,10) + 15,000 10,000 (A/F,12%,10) A purchase = 180,000 (0.177) + 15,000 10,000 (0.057) A purchase = 46,290 EGP/year The contractor should purchase the used tractor, because it has a lower annual cost Example A contractor has decided to add a grader to his equipment fleet. He could purchase either a new or a used one. Interest, insurance, and taxes total about 12% and the contractor expects using the grader about 2,000 hours per year. Which of the following alternatives should the contractor select? 1. The new grader costs 120,000 EGP to purchase and is expected to have a useful life of 16,000 hours of operation. Tires cost 5,000 EGP to replace every 4,000 hours of use and major repairs will be needed after 8,000 hours of operation at a cost of 6,000 EGP. Fuel, oil and minor maintenance cost about EGP for each hour the grader is used. Estimated salvage value at the end of 16,000 hours of operation is 10,000 EGP. 2. The used grader costs 75,000 EGP to purchase and is expected to have a useful life of 8,000 hours of operation. Tires cost 5,000 EGP to replace every 4,000 hours of use. Fuel, oil and minor maintenance cost about EGP for each hour the grader is used. Estimated salvage value at the end of 8,000 hours of operation is 8,000 EGP. 9

10 Alternative (1): New Grader Annual fuel, oil, and minor maintenance cost = (2000 hrs)(15.25 EGP/hr) = 30,500 EGP/year ,000 30, ,000 5,000 5,000 5,000 Annual cost for the new grader A = 120,000 (A/P,12%,8) + 30,500 + [5,000 (P/F,12%,2)(A/P,12%,8)] + [11,000 (P/F,12%,4)(A/P,12%,8)] + [5,000 (P/F,12%,6)(A/P,12%,8)] - [10,000 (A/F,12%,8) = 56,525 EGP/year 6,000 Alternative (2): Used Grader Annual fuel, oil, and minor maintenance cost = (2000 hrs)(18.25 EGP/hr) = 36,500 EGP/year 8, ,000 36,500 Annual cost for the used grader A = 75,000 (A/P,12%,4) + 36,500 + [5,000 (P/F,12%,2)(A/P,12%,4)] - [8,000 (A/F,12%,4) = 60,814 EGP/year 5,000 The contractor should purchase the new grader because it has the lower annual cost 10

11 Rate of Return Analysis Contractors often want to estimate the prospective rate of return on an investment or compare anticipated rates of return for several alternative investments. The rate of return is the annual interest rate at which the sum of investments and expenditures equals total income from the investment. Rate of return analysis involves setting receipts (revenues) equal to expenditures and solving for the interest rate. Example A contractor is considering making a 300,000 EGP investment in used equipment. He estimates that his annual maintenance and repair costs for the equipment will be 60,000 EGP and his annual income from the equipment will be 115,000 EGP. He estimates that he can get 8 years of use out of the equipment, but there will be no salvage value at the end of the 8 years. What would be the contractor s prospective rate of return from this investment? 115,000 Using the annual worth method Income = Expenditures 115,000 = 60, ,000 (A/P,i,8) 300,000 (A/P,i,8) = From interest tables (A/P,9%,8) = (A/P,10%,8) = Using interpolation i = 9 + = = 9.3% ,000 11

12 Example A contractor is considering the purchase of a new dump truck at the cost of 85,000 EGP. Annual maintenance and repair costs are estimated to be 4,000 EGP. The truck is estimated to be used for 8 years and then sold for a salvage value of 10,000 EGP. The contractor estimates that the annual income from the truck will be 16,000 EGP. What would be the contractor s prospective rate of return from this investment? Using the present worth method Income = Expenditures 16,000 (P/A,i,8) + 10,000 (P/F,i,8) = 85, ,000 (P/A,i,8) 12,000 (P/A,i,8) + 10,000 (P/F,i,8) = 85,000 Using the trial and error method Assume i= 5%, LHS = 84,326 EGP < 85,000 EGP Assume i= 4%, LHS = 88,106 EGP > 85,000 EGP Using interpolation 88,106 85,000 i = ,106 84,326 = 4.8% ,000 16,000 4,000 10,000 Depreciation Accounting Techniques Construction equipment loses value with age. This loss in value is called depreciation. It occurs because of wear and tear from use and/or obsolescence which leads to decrease in the market value of the used equipment. The value of the equipment at the time of purchase is called the purchase price (P), and the value of the equipment at the time of disposal or retirement is the salvage value (S). Contractors generally use the estimated annual loss value of their equipment to offset income for tax purposes. To accomplish this, they use one of the approved methods of depreciation accounting to determine the annual depreciation and a residual book value for their equipment. Depreciation is an artificial expense that spreads the depreciable value of the equipment over the depreciation period. 12

13 The depreciable value is the purchase price less the estimated salvage value. The depreciable period (n) is the number of years over which the equipment is owned. The book value (BV t ) is the equipment value at the end of year t. The annual depreciation rate (R t ) is the annual rate by which a piece of equipment is depreciated. The annual depreciation amount (D t ) is the amount by which a piece of equipment is depreciated. The Salvage Value (S) is the estimated trade-in or market value at the end of the asset's useful life. Depreciation accounting methods generally used and approved by the tax regulations are: 1. Straight-Line method 2. Sum-of-Years method 3. Declining-Balance method 13

14 Straight-Line Method The straight-line method of depreciation accounting depreciates the equipment value equally in each of the years the equipment is owned. The annual depreciation rate is R = 1 n The annual depreciation amount is D = R (P S) The book value is BV t = BV t 1 D t The book value at the end of the depreciation period must be the salvage value Example A contractor purchased a tractor at a cost of 135,000 EGP and plans to use it for 5 years, and estimates the salvage value of the tractor to be 65,000 EGP at the end of the 5 years. Using the straight-line method of depreciation accounting, what is the book value of the tractor at the end of each of the 5 years? The annual depreciation rate is R = 1 n = 1 5 = 0.2 The annual depreciation amount is D = R P S = ,000 65,000 = 14,000 EGP The book value can be determined as follows Year Book Value (EGP) 0 135, , , , , ,000 Purchase price Salvage value 14

15 Sum-of-Years Method In the Sum-of-Years method, the annual depreciation rate is not the same for each year. The annual depreciation rate is n t + 1 R t = SOY SOY is the sum of years = n+(n-1)+(n-2)+.+1 = n(n+1) 2 The annual depreciation amount is D t = R t (P S) The book value is BV t = BV t 1 D t Example Solve the previous example using the sum-of-years method of depreciation accounting. SOY = = 15 Year Depreciation Rate (R t ) Depreciation Amount (D t ) Book Value (BV t ) , /15 23, , /15 18,667 93, /15 14,000 79, /15 9,333 69, /15 4,667 65,000 Purchase price Salvage value 15

16 Declining-Balance Method Declining balance depreciation is also known as the fixed percentage or uniform percentage method. DB depreciation is determined by multiplying the book value by a fixed percentage R expressed in decimal form. If R = 0.1, then 10% of the book value is removed each year. Therefore, the depreciation amount decreases each year. The maximum annual depreciation rate for the DB method is twice the SL rate; R max = 2 n In this case, the method is called double declining balance (DDB). For example, if n = 8 years, then the DDB rate = 2/8 = 0.25, so 25% of the book value is removed annually. The depreciation for year t is D t = R. BV t 1 The depreciation in year t, relative to the first cost is D t = P. R. (1 R) t 1 The book value for year t is BV t = BV t 1 D t The book value for year t relative to the first cost is BV t = P. (1 R) t 16

17 The book value of the asset for the DB method never goes to zero, since the book value is always decreased by a fixed percentage. The implied salvage value after n years is the BV n amount, that is, Implied S = BV n = P.(1-R) n P Note: The estimated S is not used in the DB method to calculate annual depreciation If the implied S < estimated S, it is correct to stop charging further depreciation when the book value is at or below the estimated S Example Solve the previous example using (a) 1.5 declining-balance method and (b) double-declining-balance method of depreciation accounting. a) 1.5 declining-balance method R = 1.5/5 = 0.3 Year Depreciation Amount (D t ) Book Value (BV t ) , ,500 94, ,350 66, ,150 65, , ,000 Purchase price Salvage value Note that the full depreciation amount (19,845) computed at the end of year 3 could not be taken, because it would reduce the residual book value below the salvage value (65,000) 17

18 Book Value Example b) Double-declining-balance method R = 2/5 = 0.4 Year Depreciation Amount (D t ) Book Value (BV t ) , ,000 81, ,000 65, , , ,000 Purchase price Salvage value 135,000 65, Time (years) 18