3.7 / Notation and Cash Flow Diagrams /Tables 75 A= $2524 _| 2 3 4=N = 10% per year FIGURE 3-3 Cash Flow Diagram for Plan 3 of Table 3-1 (Lender's Viewpoint) P= $8,000 The cash flow diagram employs several conventions: The horizontal line is a time scale, with progression of time moving from left to right. The period (e.g., year, quarter, month) labels can be applied to intervals of time rather than to points on the time scale. Note, for example, that the end of Period 2 is coincident with the beginning of Period 3. When the end-of-period cash flow convention is used, period numbers are placed at the end of each time interval as illustrated in Figures 3-2 and 3-3. The arrows signify cash flows and are placed at the end of the period. If a distinction needs to be made, downward arrows represent expenses (negative cast flows or cash outflows) and upward arrows represent receipts (positive cash Hows or cash inflows). 3. The.cash How diagram is dependent on the point of view. For example, the Situations shown in Figures 3-2 and 4-4 were based on cash flow as seen by the ‘ender. If the directions of all arrows had been reversed, the problem would have been diagramme: the borrower's viewpoint. 7 EXAMPLE 3-1 Before evaluating the economic merits of a proposed investment, the XYZ Corpo- | ration insists that its engineers develop a cash flow diagram of the proposal. An | investment of $10,000 can be made that will produce uniform annual revenue of | | ¥ $5,310 for five years and then have a market (recovery) value of $2,0( df of year five. Annual expenses wil 000 at the end of each year for operating | and maintaining the project. Draw a cash flow diagram for the five-year life of the projéct. Use the corporation’s viewpoint. SOLUTION As shown in Figure 3-4, the initial investment of $10,000 and annual expenses of $3,000 are cash outflows, while annual revenues and the market value are cash inflows. Notice that the beginning of a given year is the end of the preceding year. For example, the beginning of year two is the end of year one. Example 3-2 presents a situation in which cash flows are represented in tabular form to facilitate the analysis of plans/designs. @ scanned with OKEN ScannerG7 76 FIGURE 3.4 CHAPTER 3 / MoNey~Time RELATIONSHIPS AND EQUIVALENCE $5310 $5310 $5,310 $5310 $5310 $3,000 $3,000 $3,000 $3,000 $3,600 Years 10,000 Cash Flow Diagram for Example 3-1 EXAMPLE 3-2 In a company's renovation of a small office building, two feasible alternatives {ot upgrading the heating, ventilation, and air conditioning (HVAC) system have been identified. Bither Alternative A or Altemative B muse be implemented. The costs are ‘ Alternative A Rebuild (overhaul) the existing HVAC system + Equipment, labor, and materials to upgrade... $18,000 + Annual cost of electricity... vores 32,000 * Annual maintenance expenses... eee 100 Alternative B_ Install a new HVAC system that utilizes existing ductwork * Equipment, labor, and materials to install. 560,000 * Annual cost of electricity... 9,000 * Annual maintenance expenses. : E 16,000 * Replacement of a major component 4 years hence. 9400 Atthe end of eight ‘Years, the estimated market value for Alternative A is $2,000, and for Alternative B it is $8,000. Assur me that both alternatives will provide mapharable service (comfort) over an eightsyear time period, and assume that the major component teplaced in. Alternative B will have no market value at the end of Year eight. (1) Use a cash flow table and end-of-year convention to tabulate the ret £28h flows for both alternatives. 2) Determine ae annual n between the alternatives (l ‘et cash flow difference B~ A). (3) Compute the cumul the end of year eight. (Th lative difference through © cumulative difference is the sum of differences, B - A, from year zero through year eight) Soturion The cash flow table (company’s viewpoint) for this exai imple is shown in Table 3-2 Based on these results, several points can be made (1) doing nothing is not an option—either A or B must be selected; (2) even though positive and negative cash flows are included in the table, on balance we are investigating two “cost-only” alternatives; (3) a decision between the two alternatives can be made just as easily @ scanned with OKEN Scanner7 native A Alternative B Difference cr astcash FLOW Net Cash Flow (BA) me $ 18,000 60,000 12000 =" 34400 25,000 32,600 = hoo ~ 25,000 9,400 = B20) = 34400 - 25,000 9,400 = 13,800 = 34400 = 25,000 - 9,400 0 ~ 13,800 = Hho - 25,000 9400 ~ 4400 = 3400 — 2,000 9400 5,000 = 400 = 25,000 9400 14400 = 3400 + 2,000 25,000-+ 8,000 15,400 29,800 $291,200 $261,400 IZ2235 is5 on the difirence in cashflows (i, 0” the avoidable difference) as it can on the ae tone net cash flows for Alternatives A and B; (4) Alternative B has cash flows identical to those of Alternative A except for the differences shown in the table, so if the avoidable difference can “pay its own way,” Alternative B is the verotamended choice; (6) cash flow changes caused by inflation or other suspected Fnfluences could have easily been inserted into the table and included in the analysis; and (6) it takes six years for the extra $42,000 investment in Alternative B to generate sufficient cumulative savings in annual expenses to justify the higher investment (this ignores the time value of money). So, which. alternative is better? Weill be able to answer this question later when we consider the time value of money in order to recommend choices between alternatives. hs It should be apparent that a cash flow table clarifies the timing 6f cash flows, the assumptions that are being made, and the data that are available. A cash flow table is often useful when the complexity of a situation makes it difficult to show all cash flow amounts on a diagram. xl Jopment and illustration assessing the economic les 3-1 and 3-2 , The remainder of Chapter 3 deals with the devel of equivalence (time value of money) principles for activeness of investments such as those proposed in Examp] Tae ‘ewpoint: In most examples presented in this chapter, the company’s investor's) e “ewpoint will be taken 2 ginba Bead Lip, 7 D> i Interest Formul oo : las Relating P1 it setergegmaiea ater So lows ~ Oy SCRBL Figure 3.5 2 a future coe cash flow diagram involving-a presént Sing é am, im, F, separated by N periods with interest at *% per period. @ scanned with OKEN Scanner«ce 7s cnapren 3. J) Movey=Te ReLarisuPs ano Eouv F = Future Equivalent Fing 1 1 je Interest Rate per Period FIGURE 3-5 General Cash Flow Diagram Relating Present Equivalent and Future Equivalent of Single Payments int (Given) p = Present Equivale shown in Figure 3-5, indicates lashed arrow, such as that shown 3 . re ned ‘Taro formulas relating a given P and its unknoyin 3-2and 33. ‘Throughout this chay the quantity to be det equivalent F are provided in Equations Finding F When Given P . dollars is invested at a point in time and ; % is the interest ; ; il] grow to a future amount of th) rate per period, the amount will gi ) (prof or grote a Pepi y teen of two periods the amour i grow to P(1 +i)(1 +1) (1 +i); by the end of three pet ods, the amount will grow TPC + ae(1 +i) = P+ and by theend of N periods the am: F=pa+i (32) If an amount of P EXAMPLE 3-3 Suppose that you borrow $8,000 now, promising to repay the loan principal plus SUPP ost ated interest in four years at i = 10% per year. How much would you repay at the end of four years? SOLUTION ‘Amount Owed Interest Owed ‘Amount Owed Total End-of- Year___at Start of Year for Each Year atEnd of Year___Year Payment 1 P = $8000 iP =$ 800 PU+i) =$ 8800 0 2 Pati =$8800 iP +i) 880 P+i =$ 9,680 0 3 PUFF = $9,680 iPU+I =$ 968 PL +i) = $10,648 0 4 P+)? = $10,648 PCL +i) = $1,065 POL +i)* = $1,713 F = $11,713 In general, we see that F = P(1 +)", and the total amount to be repaid is $11,713. This further illustrates Plan 4 in Table 3-1 in terms of notation that we shall be using throughout this book. The quantity (1 + i) in Equation 3-2 is commonly called the single paymen! compoun ar. Numerical values for this factor are given in the second column from the left in the tables of Appendix C for a wide range of values of! @ scanned with OKEN Scannerture Equivalent Values of Single Cash Flows \d Ful present an ; plating 38! 11 use the functional symbol (F/P, %,N) pook we shal nantewy oe ton oe can be expressed 3 ence E F = P(F/P,i%,N) oo ventheses i oad “find F given P at Me interest per is Bae ot Eatin 5.3, where the unknown quantity, F,is placed onthe ihe initial Fa Of the equation. ‘This sequencing of letters is true of all functional jeft-hand 5 in this book and makes them easy to remember. sybase eof Binding F when given P, together with a cash flow diagram Another x ein Table33. Noten Tablet for each ofthe six common and solution Tfand interest circumstances covered, two problem statements are discrete com: fogy and (b) in equivalence terminology —but rm orvaing-ending ter ) ation. Indeed, there are generally many Jepresentthesame cas love situ rer given cash flow situation can be expressed. an agood way to interpret a relationship such as Equation 3-3 is that thecalculated amount, F, at the point in time at which it occurs, is equionlent to (i.e, can be traded for) the known value, P, at the point in time at which it occurs, for the given interest or profit rate i. iven—(@) they both ways in W In genet Finding P When Given F 3-2, F = P(1 + i) 1 P= (45) =Fa+i® (3-4) ‘The quantity (1 4-1)" is called the single payment present worth factor. Numerical fhe third column of the tables in Appendix C for Values for this factor are given in # wide range of values of 'and N. We shall use the functional symbol (P/F, i%,N) for this factor. Hence From Equation N, Solving this for P gives the relationship P = F(P/F, i%,N) (5) EXAMPLE 3-4 \/ | An investor owner) has an option to purchase a tract of land that will Be worth each year, how much $10,000 in six years. If the value of the land ii % : “ars. I increases at 8% should the investor be willing to pay now for this property? SOLUTION The purchase pri : rchase price can be determined . Reece aie letermined from Equation 3-5 ai nd Table C-11 in P = $10,000(P/F, 8%, 6) P = $10,000(0.6302) = $6,302 @ scanned with OKEN Scannerrd Chap, ER 1 Money—Te Retarowsues ano EOUVALENSE Another example of this type of problem, together with a cash flow diagran, and " Solution, is given in Table 3-3- 39 Inter ang a Uniform Se est Formulas Relating 4 | and Fut ies (Annuity) to Its Present ‘ure Equivalent Values nvolvi jes of uniform Figure 3-6 showsa general cash flow digit” valving peri ay teceipts, each of amour! & Seeeee end of 220 Fried an annuity, tt sho A interest at M% per period. Such uniform reais often etch hou i“ be noted that the formulas and tables to be presen| occurs at the end of each period, an‘ thus: ; + period before the fi 1. P (present equivalent value) occurs one interest period eto! first A iform mount). . elast A, dN . 2 yar equivalent value) occurs at the same time aS the last A, and N periods fter P. a. 3. A (annual equivalent value) occurs at the end of periods | through N, inclu. sive. “The timing relationship for P,A,and F ean be observed in Figur’ 3-6. Four formulas relating A to F and P will be developed, » 3.9.1. Finding F When Given A at the end of each period for the amount of A dollars occurs ‘ is the interest (profit or growth) rate per period, the future nd of the Nth period is ‘obtained by summing the he cash flows. Thus, i%,N — 2) + A(E/P, i%,N -—3)+° Ifa cash flow i N periods and equivalent value, F, at the © feture equivalents of each of t f= A(E/P,i%,N ~ 1) + AE/P, “+ A(E/P,i%, 1) + AE/P, 1%, 0) rea+)h 240+) Wo pee t+) +I = Al( +) ‘A= Uniform Amounts (Given) jf i it i ! 1 2 3 N-1 | N=Number ' Period | of Interest ' | Periods t i= Interest Rate per Period ‘ . 1 P= Present Equival resent Equivalent (Find) F = Future Equivalent (Find) FIGURE 3-6 General Cash Flow Diagram Relatit if ing Uniform Series (Ordinary Annu Present Equivalent and Future Equivalent Values aaa raree @ scanned with OKEN Scannerso! ty) to Its Present and Future E iform Series (Annu ‘quivalent peiating 2 UN Values, es erms eometric sequence f - pracketed terms comprise @ quence havin is ry Recall that the sum of the first N terms (5.) of a go 8 8 common ratio of ‘Ometric sequence is (#1) ay isthe frst term in the sequence, a» is the last term, and bis the com, mon where " si see welet b= (11a = 1+ 9"% and ay = (1+ 2°, then 14 (1+i%) - (+a FrA i a5 which reduces to (1+ i%-1 Fe {een (3-6) The quantity {[(0 + — 1]/i} is called the uniform series compound amount factor, Tris the starting point for developing the remaining three uniform series interest factors, Numerical values for the uniform series compound amount factor are given in the fourth column of the tables in Appendix C for a wide range of values of i and N. We shall use the functional symbol (F/A, i%, N) for this factor. Hence Equation 3-6 can be expressed as F = A(F/A,i%,N) (3-7) Examples of this type of “wealth accumulation” problem based on the (F/A, i%, N) factor are provided here and in Table 3-3. EXAMPLE 3-5 {@) Suppose you make 15 equal annual deposits of $1,000 each into a bank account Paving 5% interest per year. The first deposit will be made one year from today. ow much money can be withdrawn from this bank account immediately after the 15th deposit? SoLurion merits of A 's $1,000, N equals 15 years, and i = 5% per year. Immediately Payment, the future equivalent amount is F = $1,000(F/4,5%, 15) = $1,000(21.5786) = $21,578.60 @ scanned with OKEN Scanner84 3.9.2 QUVALENCE ‘omapten 3 / Money—Tue Rearinsns 0 E coincident with Notice in the cash flow diagram below that the value of F is ¢ he in the cash flow last payment of $1,000. ae? End of Year (OY) (b) ‘oillustrate further the amazing effects of compound ee coi he 7 ani for lity of this statement: age an meer Ba Ife, you can become a millionaire.” Lets assume you live to age BD and the annual interest rate is 10% (i = 10%) Under these specific conditions, wre compute the future compound amount (F)t0 be F = $365/yr. (F/A, 10%, 60 years) = $365 (3,034.81) = $1,107,706 . sumptions given! The moral isto start saving | ‘Thus, the statement is true for the ass ding work on your behalf! early and let the “magic” of compoun‘ ney early and preserving resources through frugality avoiding waste) are extremely important ingredients of veal cretion in general. Often, being frugal means postponing the satisfaction of immediate material wants for the creation of a better tomorow. In this regard, be very cautious about spending tomorrow's cash today by undisciplined borrowing (e.g. with credit cards). The (F/A, i%, N) factor demonstrates how fast your debt can accumulate! A feco words to the wise: Saving mo Finding P When Given A From Equation 3-2, F = P(1+i)". Substituting for F in Equation 3-6, one determines that Pa +i = ajar] i Dividing both sides by (1+, = (+aN-1 u Ae : oS @ scanned with OKEN Scanner(Annuity) to Its Present and Future Equivalen 4g | Relating @ Uniform series ( t Values i ‘on 38s the relation for finding the present equival the beginning of the first period) of a uniform series of chloe bs of toe peeint for N periods. The quantity in brackets is called the unyjorn con® act wort aor. Numerical vals for this factor are given in he fh column rhe tables in Appendix C fora wide range of values of i and N. We shall earn functional symbol (P/A,i%, N) for this factor. Hence e the P = A(P/A,i%,N) (09) Thus, Equati EXAMPLE 3-6 Ifa certain machine undergoes a major overhaul now, its output can be increased by 20%-—which translates into additional cash flow of $20,000 atthe end of each year for five years. If | = 15% per year, how much can we afford to invest to overhaul this machine? SOLUTION The increase in cash flow is $20,000 per year, and it continues for five years at 15% annual interest. The upper limit on what we can afford to spend is P = $20,000(P/A, 15%, 5) te = $20,000(3.3522) h a = $67,044 Rx 5 yO - EXAMPLE 3-7 Suppose that your rich uncle fias $1,000,000 that he wishes to distribute to his heirs at the rate of $100,000 per year. If the $1,000,000 is deposited in a bank account that earns 6% interest per year, how many years will it take to completely deplete the account? How long will it take if the account earns 8% interest per year instead of 6%? ce SOLUTION From Table C-9, solve for N in the following equation: $1,000,000 = $100,000(P/A, 6%,N); N = 15.7 years. When the interest rate is increased to 8% per year, it will take 20.9 years to bring the account balance to zero, which is found by solving this | yy equation: 10 = (P/A, 8%, N). = “ Ue See ee 4,06) * 3.9.3. Finding A When Given F ye! n on Takin, i i a 7 hal ¢ ie faking Equation 3-6 and solving for A, one finds that jue t) ( y 05) a eae, (3-10) a=Flaae=al wh | ‘A,of a uniform series Thus, Equation 3-10 is the relation for finding the amount, A, of Sent Of cash flows occurring at the end of N interest periods that woul @ scanned with OKEN Scanner86 3.9.4 couarren 3 /- Money=Tive RevaTionsirs AND EaUvALENCE urring at the end of the sinking fund factor, Numeric, ann of the tables in Appendix C fo, inctional symbol (A/F, i%,N) alent value occ qui is called to (have the same value as) its future € xth colt last period. The quantity in bracket the values for this factor are given the Ful a wide range of values of and N. We shall use the for this factor. Hence Ae F(A/E,i%/N) (3-11) ersonal savings totaling $1,000,009 ld. If the annual interest rate will ‘account, what equal end-of-year EXAMPLE 3-3 ‘An enterprising student is a when she retires at age 65 the next 45 yee average 7% over BF must she save to accomP! lanning to have Ps fe is now 20 years ars on her savings ish her goal? amount must nual amount this student must 000,000. The equal anr 45 years at 7% annual interest so.urion 1, Be el $1,000,000 in ture amou oe fund that grows tO place in a sinking ‘A = $1,000,000 (A/F, 7%, 45) = $1,000,000(0.0035) = $3,500 (sce Table C-10) is together with a cash flow diagram ‘Another example of this type of problem, and solution, is given in Table 3-3. Finding A When Given P Taking Equation 3-8 and solving for A, one finds that = ia + aN usa ala + ON (3-12) Thus, Equation 3-12 is the relation for finding the amount, A, of a uniform series of cash flows occurring at the end of each of N interest periods that would be equivalent to, or could be traded for, the present equivalent P, occurring at the beginning of the first period. The quantity in brackets is called the capital recovery {factor.* Numerical values for this factor are given in the seventh column of the tables in Appendix C for a wide range of values of i and N. We shall use the functional symbol (A/P, i%, N) for this factor. Hence A = P(A/P,i%,N) (09) “The capital recovery factor is more convenien ; . hand-held calculator. more conveniently expressed as i/[1 ~ (1 + i)-] for computation witha @ scanned with OKEN ScannerJ 2 Uniform Series (Annuity) to Its Present and Future Equivalent Values fe 39 / Relatin An example that utilizes the equivalence between a present cunt and a series of equal uniform annual payments starting a ne and continuing through year four is provided in Taki 31 a fon 3-13 yields the equivalent value of A that Tepays the $8,000 interest per year over four years A = $8,000(4/P, 10%, 4) = $8,000(0:3155) = $2594 The entries in columns three and five of Plan 3 in Table 3-1 can now be better understood. Interest owed at the end of year one equals $8,00n¢ 10), and there- fore the principal repaid out of the total end-of-year Payment of §25594 je ge, difference, $1,724. At the beginning of year two, the amount : $8,000 ~ $1,724 = $6,276, Interest owed at the end $628, and the principal repaid at that time is $2,524. ~ $628 entries in Plan 3 are obtained by performing, these calcul and four. lump-sum Joan the end of year S Plan 3, Equa- loan plus 10% $1,896. The remaining lations for years three that 10% interest is being paid on the beginning-of-year amount owed year-end payments of $2,524, consisting of interest and principal owed to $0 at the end of the fourth year. (The exact value of an $2,523.77 and Produces an exact value of $0 at the end of four years, It is important t Another example of a problem where we desire to compute an equiv value for A, from a given value of P and a known interest rate and number of compounding periods, is given in Table 3-3, For an annual interest rate of 10%, the reader should now be convinced from Table 3-3 that $1,000 at the beginning of year one is equivalent to $187.45 athe end of years one through eight, which is then equivalent to $2,143.60 at the and year eight. co $6,903.60 ~ [Repay ‘Amount $2524 481756 mount _ “o sam esax “| Repay $2,524 = Ea a 2 Ey 4 End of Year FIGURE 3-7 _ Relationship of Cash Flows for Plan 3 of Table 3-1 to Repayment of the $8,000 Loan Principal @ scanned with OKEN Scanner3.11 / Deferred Annuities (Uniform Series) 89 yl aee eeede 1e Compounding Interest Factors and Symbols? Factor by Which to Factor To Find: Given: Multiply ‘Given’? Factor Name Functional Symbol? For single cash flows: : : P a+ay Single payment 1 compound amount (F/P, i%, N) 7. F a7 Single payment a+ present worth (P/E, i%,N) For uniform series (arnnuities): F 7 a+ i = Uniform series i compound amount (F/A, i%, N) p yn a+aN Uniform series i(1 + present worth (PLA, i%, N) i A F ae Sinking fund (A/E,i%, N) il + DN 7 A P apes Capital recovery (A/P, i%,N) per interest period; N;, number of interest periods; A, uniform series amount (occurs at the end F, future equivalent; P, present equivalent. ghout this book. +i, effective interest rate of each interest period); 'The functional symbol system is used throu; © scanned with OKEN Scanner7” 90 cMaten 3 J) MONE wt TTT ime Pre Period Defesred Annuity (Uniform Series) sdinary annuity has been ‘is. Remember that in The end of periag entire framed 0 “time 0," by J perio payment iS Lin length. tat t fan annuity with cash flows of at al 3.9, AP/A, M,N ~ P). The present equivalent ofthe a yas of time 0 will then be portray aved fora annuity d assuming, that nt oquivale fom Equation amount 4 is, 6 a amount A(P/A, M,N gle ; A(P/A, i%,N ~ (P/E i%,]) EXAMPLE 3-9 * a father, on the day his son strate the preceding discussion, suppose that 1 Tape wishes % amount would have to be paid into an is born, wishes to determine what Lut se orm pearing interest of 12% per year to provide withdrawals of $2,000 on each of the son's 18th, 19th, 20th, and 2ist birthdays. SOLUTION ; The problem is represented in Figure 3:9. One should first recognize that an ordinary annuity of four withdrawals of $2,000 each is involved, and that the present equivalent of this annuity occurs at the 17th birthday when a (P/A,i%, N — factor is utilized. In this problem, N = 21 and = 17. [tis often helpful to use a subscript with P ot F to denote the respective point in time. Hence Pir = A(P/A,12%,4) = $2,000(3.0373) = $6,074.60 Note the dashed arrow in Figure 3-9, denoting Pir. Now that Piz is known, the next step isto calculate Po, With respect to Pp, Pr7 i a future equivalent, and hence it could also be denoted Fir. Money at a given point in time, such as the end of A= $2,000 FIGURE 3-9 Cash Flow Diagram of the Deferred ‘Annuity Problem in Example 3-9 @ scanned with OKEN Scanner| PB \eollon. 1° it "daa eta! TI a 3.12 3.42 7 Equi 'quivalence Calculations Involving Multiple Interest Formulas 1 A= $2,000 er er ee ee Ye ' : ears | FIGURE 3-10 | Cash Flow Diagram for the Deferred Annuity Problem in Example 3-10 period 17, is the same regardless of whether itis called a present equivalent or a future equivalent. Hence Pp = Fir(P/F,12%, 17) = $6,074.60(0.1456) = $884.46 which is the amount that the father would have to deposit on the day his son iS born. EXAMPLE 3-10 ; ose that the father wishes to ‘As an addition to the problem in Example 3-9, supp Jeterine the equivalent worth ofthe four $2,000 withdrawals a5 of the son’s 24th .drawn or possibly birthday. This could mean that four amounts were nevet with that the son took them and immediately redeposited them in an account also earning interest at 12% per year. Using our subscript system, we wish to calculate Fy, as shown in Figure 3-10 above. SOLUTION One way to work thisis to calculate Fy = A(E/A,12%,4) = $2,000(4.7793) Fa, can now be denoted Pay, and Fog = Pal E/P,12%,3) = $9,558.60(1.4049) = $13,428.88, fo work the problem is to recognize that Pi7 ih equivalent to the four $2,000 withdrawal Using Po, we obtain = $13,424.86 = $9,558.60 To determine F2s, $6,074.60 . Hence one ‘Another, quicker way | and Po = $884.46 are ea gan find Fx, directly, given Pa ot Po Foy = Po(F/P,12%, 24) = $884.46(15.1786) the previous answer. The two numbers differ by $4.02, ates which closely approximé to round-off error in the interest factors. which can be attributed alence Calculations Involving ple Interest Formulas jer should now be comfortable with equivalence problems that involve if interest and discrete cash flows. All compounding of Equi Multi The read discrete compounding o' @ scanned with OKEN Scanner—_= 2 cuarren 3 Mover—Tive ReLanoNshs aX0 EQUVALENCE. interest takes place once per time period (e.g:, a year), and to this point cash flows falso occur once per time period. This section provides two examples involving two oor more equivalence calculations to solve for an unknown quantity. End-of-year cash flow convention is used. Again, the interest rate is constant over the N time periods. KK EXAMPLE 3-11 igure +11 depicts an example problem with a series of year-end cash flows | extending over eight years. The amounts are $100 for the first year, $200 for | the second year, $500 for the third year, and $400 for each year from the fourth through the eighth. These could represent something like the expected maintenance expenditures for a certain piece of equipment or payments into a fund. Note that the payments are shown at the end of each year, whichis a standard assumption (convention) for this book and for economic analyses in general unless one has information to the contrary. It is desired to find the (a) present equivalent expenditure, Pr; (b) future equivalent expenditure, Fs; and (c) annual equivalent expenditure, A, of these cash flows ifthe annual interest rate is 20%, SOLUTION (a) To find the equivalent Pp, one needs to sum the equivalent values of all payments sof the beginning of the first year (time zero). The required movements of money through time are shown graphically in Figure 3-11a. Py = Fi(P/F, 20%, 1) = $100(0.8333) =$ 83.33 + Fy(P/F,20%, 2) + $200(0.6944) + 138.88 + Fs(P/F, 20%, 3) + $500(0.5787) + 289.35 +A(P/A,20%,5) x (P/F, 20%,3) + $400(2.9900) x (0.5787) + 692.26 $1,203.82 (0) To find the equivalent Fs, one can sum the equivalent values of all payments as of the end of the eighth year (time eight). Figure 3-11b indicates these movements of money through time. However, since the equivalent Po is already known to be $1,203,82, one can calculate directly Fg = Po(F/P,20%,8) = $1,203.82(4.2998) = $5,176.19 (c) The equivalent A of the irregular cash flows can be calculated directly from either Po or Fy as follows: A = Po(A/P, 20%, 8) = $1,203.82(0.2606) = $313.73 or A = Fe(A/F,20%,8) = $5,176.19(0.0606) = $313.73 The computation of A from Pp and Fs is shown in. Figure 3-11c. Thus, one finds that the irregular series of payments shown in Figure 3-11 is equivalent to $1,203.82 at | time zero, $5,176.19 at time eight, or a uniform series of $313.73 at the end of each | of the eight years, ! @ scanned with OKEN Scanner3.12 J Equivalence Calculations Involving Multiple Interest Formulas 93 @ ry $100 be (PM, 20% 1) $200 _— x16, 20%, 2) ——an er po Y Py= $1,208.82 z ee 7 ws = $400 $400 $400 $400 400 $500 oo x X(E/A, 20%, 5) 3 = 85,176.19 © A x(AP, Or) ee x (AFF, 20%, 8) ee i Fg (Grom b) (from a) FIGURE 3-11 Example 3-11 for Calculating the Equivalent P, F, and A Values * EXAMPLE 3-12 Transform the cash flows on the left-hand side of Figure 3-12 to their equivalent cash flows on the right-hand side. That is, take the left-hand quantities as givens and determine the unknown value of Q in terms of H in Figure 3-12. The interest rate is 10% per year. (Notice that “<=>” means “equivalent to.”) @ scanned with OKEN ScannerCHAPTER 3 J Money—Twwe ReLaTionsiiPs AND EQUIVALENCE peel eo ‘Trrtse ty Endoof year(EOY) | 12345678 End of year (EOY) 1 Q FIGURE 3-12 Cash Flow Diagrams for Example 3-12 soturion If all cash flows on the left are discounted to year zero, we have P, = 2H{P/A,10%, 4) + H(P/A,10%.3\P/F,10%,5) = 7.8839H. When cash flows on the right are also discounted to year zero, we can solve for Qin terms of H. [Notice that Q at the end of year (FOY) two is positive, Q at EOY seven is negative, ang the two Q values must be equal in amount] 7.8839H = Q(P/F, 10%,2) ~ Q(P/F, 10%, 7) Q = 25.172H __ | EXAMPLE 3-13 Suppose you start a savings plan in which you save $500 each year for 15 years, You make your first payment at age 22 and then leave the accumulated sum in the savings plan (and make NO more annual payments) until you reach age 65, at which time you withdraw the total accumulated amount. The average annual interest rate you'll earn on this savings plan is 10%, A friend of yours (exactly your age) from Minnesota State University waits 10 years to start her savings plan (ie., she is age 32). She decides to save $2,000 each year in an account earning interest at the rate of 10% per year. She will make these annual payments until she is 65 years old, at which time she will withdraw the total accumulated amount, How old will you be when your friend’s accumulated savings amount (including interest) exceeds yours? State any assumptions you think are necessary. or SOLUTION Creating cash flow diagrams for Example 3-13 is an important first step in solving for the unknown number of years, N, until the future equivalent values of both savings plans are equal. The two diagrams are shown in Figure 3-13, The future equivalent (F) of your plan is $500(F/A, 10%, 15)(F/P,10%,N ~ 36), and that of your friend is F’ = $2,000(F/4, 10%,N — 31). It is clear that N, the age at which F = F', is greater than 32. Assuming that the interest rate remains constant @ scanned with OKEN Scannerlents 95 3.13 / Relating a Uniform Gradient of Cash Flows to Its Annual and Present Equival Your savings plan: ' End of year (BOY) id of year ( a A= $500/year Your friend's savings plan: i End of year (OY) on 5 36 ' B38 5 A’ = $2,000/ year FIGURE 9-13 Cash Flow Diagrams for Example 3-13 at 10% per year, the value of Ncan be determined by tral and error: 's F_Friend’s F N__ Your Pla 36 $15,886 «$12,210 38 $19,222 $18,974 39 $21,145 $22,872 40 $23,259 $27,159 end’s accumulated savings will exceed yours. £$500, you would be over 76 years old when foral: Start saving early!) By the time you reach age 39, your fi (if you had deposited $1,000 instead o your friend’s plan surpassed yours. Interest Formulas Relating a Uniform Gradient of Cash Flows to its Annual and Present Equivalents Some problems involve receipts or expenses that are projected to increase or decrease by a uniform amount each period, thus constituting an arithmetic sequence of cash flows. For example, because of leasing a certain type of equipment, maintenance and repair savings relative to purchasing the equipment may increase by a roughly constant amount each period. This situation can be modeled with a uniform gradient. Figure 3-14 is a cash flow diagram of a sequence of end-of-period cash flows increasing by a constant amount, G, in each period. The G is known as the uniform gradient amount. Note that the timing of cash flows on-which the derived formulas @ scanned with OKEN Scanner96 CHAPTER 3-1 Mowey—Tiwe Revanionstirs ano EouvaLehce i= Effective Interest Rate per Period FIGURE 3-14 Cash Flow Diagram for a Uniform Gracient increasing by G Dollars per Period and tabled values are based is as follows: Endof Period __Cash Flows 1 0 2 G 3 2G N-1 (N-2)G N (W=DG we (w-2)6 (-3)G 3G I { o | ft f fs eee | 3 4 N-2 0 N-1 N End of Period Notice that the first flow occurs at the end of period two. ‘The future equivalent, F, of the arithmetic sequence of cash flows shown in Fig- F = G(F/A,i%,N — 1) + G(F/A,i%,N = 2) +++ + G(F/A, i%, 2) + G(F/A, i%, 1) 3.13.1. Finding F When Given G ure 3-14 is or Nat F= [ote N-2 +Otn a, +OePt, Gey et = Sia +N 14 (1+ NP 4 +(1+ 9? + (14H) +1) - NG i @ scanned with OKEN Scanner3.43 / Relat ‘ating a Uniform Gradient of Cash Flows to Its Annual and Present Equivalents 97 Nel NG = >a + at] > = & = Srya,i%,N)- Me. (220) ; Instead of working with future equivalent values, itis usually more practical to deal with annual and present equivalents in Figure 3-14. 3.13.2 Finding A When Given G From Equation 3-20, itis easy to develop an expression for A as fol A= F(A/E,i,N) . [Sevaam - NSlare, iN) lows: = $-amriny ol Stee i! N 4 otal > ‘The term in braces in Equation 3-21 is called the gradient to uniform series conversion factor. Numerical values for this factor are given on the right side of Appendix C for a range of iand N values. We shall use the funétional symbol (A/G, #%,N) for this factor. Thus A = G(A/G,i%,N) (3-22) 3.13.3 Finding P When Given G We may now utilize Equation 3-21 to establish the equivalence between P and G: P = A(P/A,i%,N) e(i Ne de =¢[} all i +i | (1+) -1-Ni Sear | : of aaet _ oN I (22) Se i] a+) (1+ 9h The term in braces in Equation 3-28 is called the gradient to present equivalent conversion factor. It can also be expressed as (1/i)[(P/A, i%,N) — N(P/F, i%,N)]- @ scanned with OKEN Scanner3.13.4 [xaeree 3-15 Asa further example of the use of arithm CHAPTER 3 / Money—Tiwe ReLaTionshiPs AND EQUIVALENCE Numerical values for this factor are given in column 8 of Appendix C for a wide assortment of i and N’ values. We shall use the functional symbol (P/G, 1%, N) for this factor. Hence P = G(P/G.i%.N) G24 Computations Using G Be sure to notice that the direct use of gradient conversion factors applies when there is no cash flow at the end of period one, as in Example 3-14. There may be an A amount at the end of period one, but it is tr ated separal tely, ee isuated in Examples 3-13 and 3-16. A major advantage of using gradient conversion factors (.e., computational time savings) is realized when N becomes large. EXAMPLE 3-14 As an example of the straightforward use of the gradient conversion factors, suppose that certain end-of-year cash flows are expected to be $1,000 for the second year, $2,000 for the third year, and $3,000 for the fourth year, and that if interest is 15% per year, it is desired to find the (a) present equivalent value at the beginning of the first year, and (b) uniform annual equivalent value at the end of each of the four years.” SOLUTION Observe that this schedule of cash flows fits the model of the arithmetic gradient formulas with G = $1,000 and N = 4 (see Figure 3-14). Note that there is no cash flow at the end of the first period. (a) The present equivalent can be calculated as Py = G(P/G, 15%,4) = $1,000(3.79) = $3,790 (b) The annual equivalent can be calculated from Equation 3-22 as A = G(A/G, 15%,4) = $1,000(1.3263) = $1,326.30 Of course, once Py is known, the value of A can be calculated as A = Po(A/P. 15%, 4) = $3,790(0.3503) = $1,326.30 etic gradient formulas, suppose that one has cash flows as follows: EndofYear___ Cash Flows (S) 1 5,000 2 6,000 3 ~7, 000 4 8,000 y @ scanned with OKEN Scanner3.43 / Relatin, ga Uni a Horm Gradient of Cash Flows to Ite Annual and Present Equivalents 99 and that one wis Pate alent ati = 15% per year using Sriten y ca alent ati ery s to calculate their present equ interest formulas SOLUTION Aue schedule of cash flows is depicted in the top di fom two diagrams of Figure 3-15 show how the original schedule can be broken into two separate sets of cash flows, a uniform series of $5,000 payments, plus an arithmetic gradient payment of $1,000 that fits the general gradient model for which factors are tabled. The summed present equivalents of these two separate sets of payments equal the present equivalent of the original problem. Thus, using the symbols shown in Figure 3-15, we have Por iagram of Figure 3-15. The Poa + Pog = ~A(P/A, 15%, 4) ~ G(P/G, 15%, 4) —$5,000(2.8550) — $1,000(3.79) = ~$14,275 - 3,790 = ~$18,065 be calculated with the : The annual equivalent of the original cash flows could aid of Equation 3-22 as follows Ar = A+Ac $5,000 — $1,000(A/G, 15%, 4) = $6,326.30 Ar is equivalent to Por because ~$6,326.30(P/A, 15%, 4) = — $18,061, whi same value obtained previously (subject to round-off error). End of Year 1 2 3 4 — i= 15% ' $5,000 000 $6000 Pars? $7,000 $8,000 Equals End of Year 1 2 3 4 f — i= 19% t | ’ er A= $5,000 Plus End of Yea 1 2 3 4 ee : $1,000 Pon? a an $3,000 FIGURE 3-15 Example 3-15 Involving an Increasing Arithmetic Gradient @ scanned with OKEN Scanner100 CHAPTER 3 / MoNey=Tiwe RELATIONSHIPS AND EQUIVALENCE and of Year : 1 2 2 | ' | ‘ ua ' A=$8,000 } ne \ {sign of Gis positive) $3,000 } Pos $2,000 1 $1000 Coie = 15% 1 7 4 End of Year FIGURE 3-16 Example 9-16 Involving a Decreasing Arithmetic Gradient EXAMPLE 3-16 ithmeti formulas, suppose that one For another example of the use of arithmetic gradient form has cash flows that are timed in exact reverse of the situation depicted in Example 3-15, The top diagram of Figure 3-16 shows the following sequence of cash flows: Endof Year___Cash Flows (S) 1 8,000 2 7,000 3 6,000 4 -5,000 Calculate the present equivalent at i = 15% per year using arithmetic grat interest factors. SOLUTION ‘The bottom two diagrams of Figure 3-16 show how these cash flows can be bro! into two separate sets of cash flows. It must be remembered that the arithmetic gradient formulas and tables provided are for increasing gradients only. Hence one must subtract an increasing gradient of payments that did not occur. Thus @ scanned with OKEN Scanner3.14 / Relating a Geometric Sequence of Cash Flows to Its Present and Annual Equivalent alents 101 Por = Poa ~ Pog = —A(P/A,15%, 4) + G(P/G, 15%, 4) = —$8,000(2.8550) + $1,000(3.79) = —$22,840 + $3,790 = —$19,050 Again, the annual equivalent of the ori; be calculated by the same rationale: A=A-Ac = —$8,000 + $1,000(A/G, 15%, 4) = ~$6,673.70 Note from Examples 3-15 and 3-16 that the present equivalent of —$18,065 for an increasing arithmetic gradient series of payments is different from the present equivalent of — $19,050 for an arithmetic gradient of payments of identical amounts but reversed timing. This difference would be even greater for higher interest rates and gradient amounts and exemplifies the marked effect of timing of cash flows on equivalent values. It is also helpful to observe that the sign of G corresponds to the general slope of the cash flows over time. For instance, in Figure 3-15 the slope of the cash flows is negative (G is negative), whereas in Figure 3-16 the slope is positive (G is positive). ginal decreasing series of cash flows can ® scanned with OKEN ScannerCastiriee = = EXAMPLE 9-17 eometric sequence of cash flows in Figure 3-19 a ider the end-of-year § \ » of increase is Consi the P. A, Ao, and F equivalent values. The rate of increase is 29%, a 4 eal the first year, and the interest rate is 25% per year. year d , goLUTION $833.33(P/A, 4.167%, 4) $1,000 25% — 20% 4 (rvs 1.20 ° 3 (1.04167)* = 1 = $8335. 0:04167(1.04167)* = $833.33(3.6157) = $3,013.08 A = $3,013.08(A/P, 25%, 4) = $1,275.86 ‘Ag = $3,013.08(A/P, 4.167%, 4) 4 0.04167(1.04167) = $833.34 = ssi] abaent = 1 F = $3,013.08(F/P, 25%, 4) = $7,356.15 EXAMPLE 3-18 Suppose that the geometric gradient in Example 3-17 begins with $1,000 at the end of year one and decrenses by 20% per year after the first year. Determine P, A, Ag, and F under this condition. ‘ SOLUTION The value off is -20% in this case and icg = [(1+4)/(1+f)]-1 = (1.25/0.80)-1 = 0.5625, or 56.25% per year. The desired quantities are as follows: ey $1,000 = “Fg (P/A,56.25%,4) = $1,250(1.4795) = $1,849.38 P @ scanned with OKEN Scanner345 1 Interest Rates That Vary with Time j0F A = $1,849,38(A/P,25%, 4) = $783.03 Ag = $1,849,38(A /P,56,25%, 4) = $1,250.00 F = $1,849,38(F/P, 25%, 4) = $4,515.08 arr 3.15 Interest Rates That Vary with Time eserve When the interest rate on a loan can vary with, for example, the Federal R Board’s discount rate, it is necessary to take this into account when determining the future equivalent value of the loan. It is becoming common to see interest-rate “escalation riders” on some types of loans. Example 3-19 demonstrates how this situation is treated. EXAMPLE 3.19 A person has made‘an arrangement to borrow $1,000 now and another $1,000 two years hence. The entire obligation is to be repaid at the end of four years. If the Projected interest rates in years one, two, three, and four are 10%, 12%, 12%, and 14%, respectively, how much will be repaid as a lump-sum amount at the end of four years? SOLUTION This problem can be solved by compounding the amount owed at the beginning of each year by the interest rate that applies to each individual year and repeating this process over the four years to obiain the total future equivalent value: F, = $1,000(F/P, 10%,1) = $1,100 F, = $1,100(F/P, 12% 1) = $1,232 Fs = ($1,232 + $1,000)(F/P, 12%, 1) = $2,500 Fy = $2,500(F/P, 14%,1) = $2,850 ‘To obtain the present equivalent of a series of future cash flows subject to varying interest rates, a procedure similar to the preceding one would be utilized with a sequence of (P/F, ic%, ) factors. In general, the present equivalent value of a cash flow occurring at the end of period N can be computed with Equation 3.32, where i isthe interest rate forthe kth period (the symbol |] means “the product of"): p=— fy TE iy (3-32) For instance, if Fy = $1,000 and i, = 10%, iy = 12% j = 13%, and is = 10% 1,000{(P/F, 10%, 1)(P/F, 12%, 1)(P/F, 139 13%, 1)(P/E, 10%, 7, 000|(0.9091)(0.8629)(0.8850)(0.9091)] = i St p= @ scanned with OKEN ScannerN EXAMPLE 3-20 | A credit card company charges an interest rate of 1.375% per month on the unpaid i hey claim, is 12(1.375%) = 16.5%, balance of all accounts. The annual interest rate, t i What is the effective rate of interest per year being charged by the company? ict d on time periods that may be annual, Interest tables in Appendix C are based on e ual, quarterly, monthly, and so on. Because we have no 1.375% tables (or 16.5% tables), Equation 3-33 must be used to compute the effective rate of interest in this example: 12, i= ( + a -1 = 0.1781, or 17.81% /year | Note that r = 12(1.375%) = 16.5%, which is the APR. It is true that r = M(r/M), as seen in Example 3-20, where r/M is the interest rate per period. t | 1 ® scanned with OKEN Scanner317.2 ’ anyorest Preptea th compounding MOHe OA TaN One per, 1 Yoar 109 EXAMPLE 9:21 — ana SUD Tans HHO AVNET AOE 1D sapposetnatasty for VO yearnaia vs sores ped ter HW URE GT Swamnatinte “ Wem OH We tent ‘ Oe as SOLUTION are four co" pounding periods Pe YN, OLA AHL OEE 1 ty riod is 0/4 = L5% When the vahten mercy’ ae seta There The interest rate Per interest pel Equation 33, 07€ finds that = SWOOSH) = Ststao 40) = $100.00(1L0) p= PE/P.1S e effective interest rate front ‘Alternatively, the ft F ene P.6.14 10) = $100.00(1.0614)"" = a3 0: | Torte vas quation INI is 1.1%, Therefor, | nd Gradient Series When there is more than one compounded interest period per year, the formula: and tables for uniform series and gradient series can be used as Lg as there a ar ow atthe end of each interest period, as shown in Figures 3-6 and 314 fora cos inna series and a uniform gradient series respectively. ° Uniform Series a EXAMPLE 3- 22 stone has a bank loan for $10,000, which is to be repaid in equal endo Vi cont for five years with a nominal interest rate of 12% compounded itis the amount of each payment? a Suppose thi monti instal monthly. Whi SOLUTION The number of install month is 12%/12 = 1%. When these ), and the interest rate per one finds fe rp. A ‘A = P(A/P, 1%, 60) = $10,000(0.0222) = $222 \ ra h month (interest period), including | Qn ent payments is 5 X 12 = 60, values are used in Equation 3 1268 that Notice that there is a cash flow at the end of eacl tmonth 60, in this example. EXAMPLE 3.23 stain operating savings are expected to be Oat the end of the first six months, at the end bi Sonat he eu ofthe second ix months, at increase by $1,000 re pind heer fo total of four years, It is desired to find the cauicakn nm amour, athe endl of each of the eight six-month periods if jerest rate is 20% compounded semiannually. @ scanned with OKEN Scanneruy ws saa som and regulations of siggy n2s/80” these methods ave often specified States, and are used for d, fee te and municipal governments in the United i rie Poses in other countries. In addition, as We do not discuss the applicati Internal Revenue Service the Modified Accelerated RS) non Of ACRS in the chapter, but eadily available Cost Re lications describe its use.* Selected parts of “covery System, however, are described and 'ystem applies to d future engineering projects, ©preciable property in present and fone Additional Definitions any terms that are not generally included in the lucation and practice, an abbreviated set of definitions i iginal cost basis of the asset, adjusted by allowable increases or decreases, is used to compute depreciation and depletion deduc- tions! For example, the cost of any improvement to a capital asset with a useful life greater than one year increases the original cost basis, and a casualty or theft loss decreases it, If the basis is altered, the depreciation deduction may need to be adjusted. Basis, or cost basis The initial cost of acquiring an asset (purchase price plus any sales taxes), including transportation expenses and other normal costs of making the asset serviceable for its intended use. This amount is also called the unadjusted cost basis.) Book Value (BV) The worth of a depreciable property as shown on the accounting records of a company. It is the original cost basis of the property, including any adjustments, less all allowable depreciation ot depletion deductions. It thus represents the amount of capital that remains invested in the property and must be recovered in the future through the accounting process. The BV. of a property may not be a useful measure of its market value.(In general, the book value of a property at the end of year k is k _ABook value), = adjusted cost basis — >” (depreciation deduction), (6-1) ja Iseful references on material in this chapter, available from the Intemal Revenue Service in an annually updated version, are Publication 534 (Depreciation), Publication 334 (Tay Guide for Small Business), Publication 542 (Tix Information on Corporations), andl Publication 544 (Sales and Other Dispositions of Assets). @ scanned with OKEN Scanner248 S 634 capren 6) Devneonox avo Weowe TES willing buyer £04 Willing der no compuleya Market value aay) Theamownt thal gage and is gh fora prope ae Tee the present valle of vat lle Teeeige buy or el The MY PT operty nell the time va mnt through ownership abasis of a property is ree as arsover which the bast ; Recovery period The number bees Tor the classical methods of eprecation Hs through the accounting PrP MACRS, this period is the prop, eriod is ee mm (GDS), ee is the class life for 4! Glass for the General D syste (etn “Alternative Dep! : System (1 form) for each year of the Macy, Recovery APSTS pats utized te an annual depreciation dedy i i to compute a a oe recovery period ie ofa property at the end of is useful life, ene it ic Then the asset can no longer be ys; i ice of a property W ; “ Fa Te a atone pro i i nd these cash outflow: ill ‘osing of the property, and t ot ell incur expenses in disp tae to obtain a final net SV. When the classi ted from the cash inflo' a ‘ veto of depreciation are applied, an estimated salvage value is initia mMesblished and used in the depreciation ‘calculations. Under MACRS, the gy "depreciable property is defined to be 2er0, i pat tte The expected (est ‘od of time that a property will be use mated) peri ina trade or business or come.jit is not how long the propery ts to productively use it. Useful lie, ual useful life of an asset, howeve, to produce in ‘will ast but how long the owner expec Sometimes referred to as depreciable life. Ac may be different than its depreciable ife. The Classical (Historical) Depreciation Methods ‘This section describes and illustrates the straight-line, declining balance, and sun. of-the-years-digits methods of calculating depreciation deductions. As mentionsi jin Section 6.2, these historical methods continue to apply, directly and indirect to the depreciation of property. Also included is a discussion of the units o production method. Straight-Line (SL) Method Straightline depreciation)is the simplest depreciation method. Itjassumes thts constant amount is depreciated each year over the depreciable (useful) lft oth asset The following definitions are used in the equations below. If we define” N = depreciable life of the asset in years cost basis, including allowable adjustments dy = annual depreciation deduction in year k (1 = k = N) “We often use the term Market Value (MY) in place of Salvage Value (SV). @ scanned with OKEN Scannerm\ 6.3 / The Classical (Historical) Depreciation Methods 247 BV; = book value at end of year k SVw = estimated salvage value at end of year N dj = cumulative depreciation through year k then d, = (B-SVy)/N (6-2) dj = kd forl sk =N 63) BV; = B-d; 4) Note for this method you must have an estimate of the final SV, which will also be the final book value at the end of year N. In some cases, the estimated SViw may not equal an asset's actual terminal MV. EXAMPLE 6-1 ‘Anew electric saw for cutting small pieces of lumber in a furniture manufacturing plant has a cost basis of 4,000 and a 10-year depreciable life. The estimated SV of the saw is zero at the end of 10 years, Determine the annual depreciation amounts using the straight-line method. Tabulate the annual depreciation amounts and the book value of the saw at the end of each year. oe | BV SOLUTION The depreciation amount, cumulative depreciation, and book value for each year are obtained by applying Equations 6-2, 6-3, and 6-4. Sample calculations for year five are shown below. = $2,000 BV; = $4,000 — a0 —9) = $2,000 ‘ A a4 The depreciation and book value amounts for each year are shown below. s\° * * 9 EOY,kK dr BV wee : .) argon - OSD 0 1 s4o0 (3,6 2 400 3,200 3 400 2,800" 4 400 2,400 - 5 400 2,000 6 400 1,600 7 400. 1,200 8 400 '800 9 400 400 0 400 0 @ scanned with OKEN Scanner