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Engineering Economy Chapter 4: The Time Value of Money Lotto Example • If you win $5,000,000 in the California lottery, how much will you be paid each year? How much money must the lottery commission have on hand at the time of the award? Assume interest = 3%/year. • Given: Jackpot = $5,000,000, N = 19 years (1st payment immediate), and i = 3% year • Solution: A = $5,000,000/20 payments = $250,000/payment (This is the lottery’s calculation of A P = $250,000 + $250,000(P | A, 3%, 19) P = $250,000 + $3,580,950 = $3,830,950 2 Example: Uniform Series Compound Amount Factor • Assume you make 10 equal annual deposits of $2,000 into an account paying 5% per year. How much is in the account just after the 10th deposit? • Solution: • F= $2,000 (F|A, 5%, 10) = $25,156 3 Finding A Given F Example 2 How much would you need to set aside each year for 25 years, at 10% interest, to have accumulated $1,000,000 at the end of the 25 years? 4 Example - Uniform Series Capital Recovery Factor • Suppose you finance a $10,000 car over 60 months at an interest rate of 1% per month. How much is your monthly car payment? • Solution: A = $10,000 (A | P, 1%, 60) = $222 per month 5 Finding A Given P If you had $500,000 today in an account earning 10% each year, how much could you withdraw each year for 25 years? 6 An Example • You would need to deposit $22,100 today into an account paying 10% per year in order to have $1,000,000 40 years from now. Instead of the single deposit, what uniform annual deposit for 40 years would also make you a millionaire? • Solution: A = $1,000,000 (A | F, 10%, 40) = $1,000,000 x .0023 = $2,300 7 It can be challenging to solve for N or i. • We may know P, A, and i and want to find N. • We may know P, A, and N and want to find i. • These problems present special challenges that are best handled on a spreadsheet. 8 Finding i - Example Jill invested $1,000 each year for five years in a local company and sold her interest after five years for $8,000. What annual rate of return did Jill earn? So, Again, this can be solved using the interest tables and interpolation, but we generally resort to a computer solution. 9 Basic Setup for Interpolation •Work with the following basic relationships 10 Estimating for i = 7.3% • Form the following relationships 11 Finding N - Example Acme borrowed $100,000 from a local bank, which charges an interest rate of 7% per year. If Acme pays the bank $8,000 per year, now many years will it take to pay off the loan? So, This can be solved by using the interest tables and interpolation, but we generally resort to a computer solution. 12 There are specific spreadsheet functions to find N and i. The Excel function used to solve for N is NPER(rate, pmt, pv), which will compute the number of payments of magnitude pmt required to pay off a present amount (pv) at a fixed interest rate (rate). One Excel function used to solve for i is RATE(nper, pmt, pv, fv), which returns a fixed interest rate for an annuity of pmt that lasts for nper periods to either its present value (pv) or future value (fv). 13 Deferred Annuities • Deferred annuities are uniform series that do not begin until some time in the future. • If the annuity is deferred J periods then the first payment (cash flow) begins at the end of period J+1. 14 Deferred Annuities Diagram 15 Finding the value of a deferred annuity at time 0 1. Use (P/A, i%, N-J) find the value of the deferred annuity at the end of period J (where there are N-J cash flows in the annuity). 2. Use (P/F, i%, J) to find the value of the deferred annuity at time zero. 16 Deferred Annuities Example • A father, on the day his son is born, wishes to determine what amount should be deposited into an account with 12% annual interest to provide $2,000 on each of son’s 18th, 19th, 20th, and 21st birthdays. • P17 = $2,000 x (P/A,12%,4) • = $2,000 x 3.0373 = $6,074.60 • P0 = F17 x (P/F,12%,17) • = 6,074.60 x .1456 = $884.46 17 Interest Rates that vary over time • In practice – interest rates do not stay the same over time unless by contractual obligation. • There can exist “variation” of interest rates over time – quite normal! • Example: (e.g., a variable rate mortgage). • If required, how do you handle that situation? 18 Basic Cash Flow Rules • Rule 1. Cash flows cannot be added or subtracted unless they occur at the same point in time. • Rule 2. To move a cash flow forward in time by one time unit, multiply the magnitude of the cash flow by ( 1 + i), where i is the interest rate that reflects the time value of money. • Rule 3. To move a cash flow backward in time by one time unit, divide the magnitude of the cash flow by ( 1 + i). 19 Interest Rates Vary Over Time The present equivalent of a cash flow occurring at the end of period N can be computed with the equation below, where ik is the interest rate for the kth period. If F4 = $2,500 and i1=8%, i2=10%, and i3=11%, then 20 Multiple Interest Factors • Some situations include multiple unrelated sums or series, requiring the problem be broken into components that can be individually solved and then re-integrated. See page 145. • Example: Problem 4-16 21 Uniform Arithmetic Gradient Factors • An arithmetic (linear) Gradient is a cash flow series that either increases or decreases by a constant amount over n time periods. •A linear gradient is always comprised of TWO components: •The Gradient component •The base annuity component •The objective is to find a closed form expression for the Present Worth of an arithmetic gradient 22 Uniform Arithmetic Gradient We can model these situations as a uniform gradient of cash flows. The table below shows such a gradient. End of Period Cash Flows 1 0 2 G 3 2G : : N (N-1)G Linear Gradient Example A1+(n-1)G A1+(n-2)G • Assume the following: A1+2G A1+G 0 1 2 3 n-1 N This represents a positive, increasing arithmetic gradient 24 Present Worth: Gradient Component • General CF Diagram – Gradient Part Only (n-1)G (n-2)G 3G 2G 1G 0G We want the PW at time t = 0 (2 periods to the left of 1G) 0 1 2 3 4 ……….. n-1 n 25 To Begin- Derivation of P/G,i%,n P = G(p / F,i%,2) + 2G(P/F,i%,3) + … + [(n-2)G](P/F,i%,n-1) + [(n-1)G](P/F,i%,n) P = G{(p / F,i%,2) + 2(P/F,i%,3) + … + [(n-2)](P/F,i%,n-1)+[(n-1)](P/F,i%,n)} 1 2 n-2 n-1 P=G 2 3 ... n-1 n (1+i) (1+i) (1+i) (1+i) Multiply both sides by (1+i) 26 Subtracting [1] from [2]….. 1 2 n-2 n-1 1 P(1+i) =G ... 1 (1+i) (1+i) 2 (1+i) n-2 (1+i)n-1 2 - 1 2 n-2 n-1 P=G 2 3 ... n-1 n 1 (1+i) (1+i) (1+i) (1+i) G (1 i) 1 N N P= N i i(1 i) N (1 i) ( P / G, i%, N ) factor 27 The A/G Factor • Convert G to an equivalent A A G( P / G, i, n)( A / P, i, n) G (1 i) N 1 N i(1 i) N P= N i i(1 i) N (1 i) (1 i) 1 N 1 n A/G,i,n = G i (1 i) 1 N 28 Gradient Example $700 $600 $500 $400 $300 $200 $100 0 1 2 3 4 5 6 7 •PW(10%)Base Annuity = $486.84 •PW(10%)Gradient Component= $1,276.3 •Total PW(10%) = $ 486.84 + $ 1,276.3 29 •Equals $1,763.14 The annual equivalent of End of Year Cash Flows this series of cash flows can 1 2,000 be found by considering an 2 3,000 annuity portion of the cash flows and a gradient 3 4,000 portion. 4 5,000 End of Year Annuity ($) Gradient ($) 1 2,000 0 2 2,000 1,000 3 2,000 2,000 4 2,000 3,000 30 Geometric Gradients • An arithmetic (linear) gradient changes by a fixed dollar amount each time period. •A GEOMETRIC gradient changes by a fixed percentage each time period. •We define a UNIFORM RATE OF CHANGE (%) for each time period •Define “g” as the constant rate of change in decimal form by which amounts increase or decrease from one period to the next 31 Geometric Gradients: Increasing • Typical Geometric Gradient Profile •Let A1 = the first cash flow in the series 0 1 2 3 4 …….. n-1 n A1 A1(1+g) A1(1+g)2 A1(1+g)3 A1(1+g)n-1 32 Geometric Gradients: Starting • Pg = The Aj’s time the respective (P/F,i,j) factor •Write a general present worth relationship to find Pg…. A1 A1 (1 g ) A1 (1 g ) 2 A1 (1 g ) n 1 Pg ... (1 i )1 (1 i ) 2 (1 i) 3 (1 i) n Now, factor out the A1 value and rewrite as.. 33 Geometric Gradients 1 (1 g )1 (1 g ) 2 (1 g ) n 1 Pg A1 ... n (1) (1 i) (1 i) (1 i ) (1 i ) 2 3 (1+g) Multuply both sides by to create another equation (1+i) (1+g) (1+g) 1 (1 g )1 (1 g ) 2 (1 g ) n1 (2) Pg A1 ... (1+i) (1+i) (1 i) (1 i) 2 (1 i) 3 (1 i) n Subtract (1) from (2) and the result is….. 34 Geometric Gradients 1+g (1 g ) n 1 Pg 1 A1 n 1 1+i (1 i) 1 i Solve for Pg and simplify to yield…. 1 g n nA1 1 Pg Pg A1 1 i g i (1 i ) ig For the case i = g 35 Geometric Gradient: Example •Assume maintenance costs for a particular activity will be $1700 one year from now. •Assume an annual increase of 11% per year over a 6-year time period. •If the interest rate is 8% per year, determine the present worth of the future expenses at time t = 0. •First, draw a cash flow diagram to represent the model. 36 Geometric Gradient Example (+g) •g = +11% per period; A1 = $1700; i = 8%/yr 0 1 2 3 4 5 6 7 $1700 $1700(1.11)1 $1700(1.11)2 $1700(1.11)3 PW(8%) = ?? $1700(1.11)5 37 Nominal and effective interest rates. • More often than not, the time between successive compounding, or the interest period, is less than one year (e.g., daily, monthly, quarterly). • The annual rate is known as a nominal rate. • A nominal rate of 12%, compounded monthly, means an interest of 1% (12%/12) would accrue each month, and the annual rate would be effectively somewhat greater than 12%. • The more frequent the compounding the 38 greater the effective interest. Nominal and Effective Interest Rates • Nominal interest (r) = interest compounded more than one interest period per year but quoted on an annual basis. • Example: 16%, compounded quarterly • Effective interest (i) = actual interest rate earned or charged for a specific time period. • Example: 16%/4 = 4% effective interest for each of the four quarters during the year. 39 Relationship • Relation between nominal interest and effective interest: i=(1+r/M)M -1, where • i = effective annual interest rate • r = nominal interest rate per year • M = number of compounding periods per year • r/M = interest rate per interest period 40 Nominal and Effective Interest Rates –Examples • Find the effective interest rate per year at a nominal rate of 18% compounded (1) quarterly, (2) semiannually, and (3) monthly. • (1) Quarterly compounding; i=(1+0.18/4)4 - 1=0.1925 or 19.25% • (2) Semiannual compounding; i=(1+0.18/2)2 - 1=0.1881 or 18.81% • (3) Monthly compounding ... 41 Nominal and Effective Interest Rates –Example • A credit card company advertises an A.P.R. of 16.9% compounded daily on unpaid balances. What is the effective interest rate per year being charged? r = 16.9% M = 365 • Solution: ieff = (1+0.169/365)365 -1=0.184 or 18.4% per year 42 Nominal and Effective Interest Rates • Two situations we’ll deal with in Chapter 4: • (1) Cash flows are annual. We’re given r per year and M. Procedure: find i/yr = (1+r/M)M-1 and discount/compound annual cash flows at i/yr. • (2) Cash flows occur M times per year. We’re given r per year and M. Find the interest rate that corresponds to M, which is r/M per time period (e.g., quarter, month). Then discount/compound the M cash flows per year at r/M for the time period given. 43 Example: 12% Nominal No. of EAIR EAIR Comp. Per. (Decimal) (per cent) Annual 1 0.1200000 12.00000% semi-annual 2 0.1236000 12.36000% Quartertly 4 0.1255088 12.55088% Bi-monthly 6 0.1261624 12.61624% Monthly 12 0.1268250 12.68250% Weekly 52 0.1273410 12.73410% Daily 365 0.1274746 12.74746% Hourly 8760 0.1274959 12.74959% Minutes 525600 0.1274968 12.74968% seconds 31536000 0.1274969 12.74969% 12% nominal for various compounding periods 44 Interest Problems with Compounding more often than once per Year – Example A • If you deposit $1,000 now, $3,000 four years from now followed by five quarterly deposits decreasing by $500 per quarter at an interest rate of 12% per year compounded quarterly, how much money will you have in you account 10 years from now? r/M = 3% per quarter and year 3.75 = 15th Quarter P @yr. 3.75 = P qtr. 15 = 3000(P/A, 3%, 6) - 500(P/G, 3%, 6) = $9713.60 F yr. 10 = F qtr. 40 = 9713.60(F/P, 3%, 25) + 1000(F/P, 3%, 40) = = $23,600.34 45 Interest Problems with Compounding more often than once per Year – Example B • If you deposit $1,000 now, $3,000 four years from now, and $1,500 six years from now at an interest rate of 12% per year compounded semiannually, how much money will you have in your account 10 years from now? • i per year = (1+0.12/2)12-1 = 0.1236 • F = $1,000(F/P, 12.36%, 10) + $3,000(F/P, 12.36%, 6) +$1,500(F/P, 12.36%, 4) or r/M = 6% per half-year • F = 1000(F/P, 6%, 20) + 3000(F/P, 6%, 12)+ 1500(F/P, 6%, 8) • = $11,634.50 46 Interest can be compounded continuously. • Interest is typically compounded at the end of discrete periods. • In most companies cash is always flowing, and should be immediately put to use. • We can allow compounding to occur continuously throughout the period. • The effect of this compared to discrete compounding is small in most cases. 47 Derivation of Continuous Compounding • We can state, in general terms for the EAIR: r m i (1 ) 1 m Now, examine the impact of letting “m” approach infinity. 48 Derivation of Continuous Compounding • We re-define the general form as: r m r m 1 r 1 r (1 ) 1 m m •From the calculus of limits there is an important limit that is quite useful. h 1 lim 1 e 2.71828 h h m r r lim 1 e, ieff.= er – 1 m m 49 Derivation of Continuous Compounding • Example: • What is the true, effective annual interest rate if the nominal rate is given as: – r = 18%, compounded continuously Solve e0.18 – 1 = 1.1972 – 1 = 19.72%/year The 19.72% represents the MAXIMUM effective interest rate for 18% compounded anyway you choose! 50 Continuous compounding interest factors. The other factors can be found from these. 51 Example • An investor requires an effective return of at least 15% per year. What is the minimum annual nominal rate that is acceptable if interest on his investment is compounded continuously? • Solution: er – 1 = 0.15 er = 1.15 ln(er) = ln(1.15) r = ln(1.15) = 0.1398 = 13.98% 52 We can use the effective interest formula to derive the interest factors. As the number of compounding periods gets larger (M gets larger), we find that 53