7 Financial mathematics Series, time and investment Learning objectives After completing this chapter students should be able to: • Calculate the final sum, the initial sum, the time period and the interest rate for an investment. • Calculate the Annual Equivalent Rate for part year investments and compare this with the nominal annual rate of return. • Calculate the Net Present Value and Internal Rate of Return on an investment, constructing relevant spreadsheets when required. • Use the appropriate investment appraisal method to decide if an investment project is worthwhile. • Find the sum of finite and infinite geometric series. • Calculate the value of an annuity. • Calculate monthly repayments and the APR for a loan. • Apply appropriate mathematical methods to solve problems involving the growth and decline over discrete time periods of other economic variables, including the depletion of natural resources. 7.1 Discrete and continuous growth In economics we come across many variables that grow, or decline, over time. A sum of money invested in a deposit account will grow as interest accumulates on it. The amount of oil left in an oilfield will decline as production continues over the years. This chapter explains how mathematics can help answer certain problems concerned with these variables that change over time. The main area of application is finance, including methods of appraising different forms of investment. Other applications include the management of natural resources, where the implications of different depletion rates are analysed. The interest earned on money invested in a deposit account is normally paid at set regular intervals. Calculations of the return are therefore usually made with respect to specific time intervals.Forexample,Figure7.1(a)showstheamountofmoneyinadepositaccountatany given moment in time assuming an initial deposit of £1,000 and interest credited at the end of each year at a rate of 10%. There is not a continuous relationship between time and the total sum in the deposit account. Instead there is a ‘jump’ at the end of each year when the interest © 1993, 2003 Mike Rosser (a) Deposit account balance (b) Oil extraction 1,210 Million barrels £ 10 0 23456 Time (years) Time ( y ears) 123456 20 1,000 1,464 10 Figure 7.1 on the account is paid. This is an example of a ‘discrete’ function. Between the occasions when interest is added there is no change in the value of the account. A discrete function can therefore be defined as one where the value of the dependent variable is known for specific values of the independent variable but does not continuously change between these values. Hence one gets a series of values rather than a continuum. For example, teachers’ salaries are based on scales with series of increments. A hypothetical scale linking completed years of service to salary might be: 0 yrs = £20,000, 1 yr = £21,800, 2 yrs = £23,600, 3 yrs = £25,400 The relationship between salary and years of service is a discrete function. At any moment in time one knows what a teacher’s salary will be but there is not a continuous relationship between time and salary level. An example of a continuous function is illustrated in Figure 7.1(b). This shows the cumu- lative total amount of oil extracted from an oilfield when there is a steady 5 million barrels per year extraction rate. There is a continuous smooth function showing the relationship between the amount of oil extracted and the time elapsed. In this chapter we analyse a number of discrete-variable problems. Algebraic formulae are developed to solve some applications of discrete functions and methods of solution using © 1993, 2003 Mike Rosser spreadsheets are explained where appropriate, for investment appraisal analysis in particular. The analysis of continuous growth requires the use of the exponential function and will be explainedinChapter14. 7.2 Interest Time is money. If you borrow money you have to pay interest on it. If you invest money in a deposit account you expect to earn interest on it. From an investor’s viewpoint the interest rate can be looked on as the ‘opportunity cost of capital’. If a sum of money is tied up in a project for a year then the investor loses the interest that could have been earned by investing the money elsewhere, perhaps by putting it in a deposit account. Simple interest is the interest that accrues on a given sum in a set time period. It is not reinvested along with the original capital. The amount of interest earned on a given investment each time period will be the same (if interest rates do not change) as the total amount of capital invested remains unaltered. Example 7.1 An investor puts £20,000 into a deposit account and has the annual interest paid directly into a separate current account and then spends it. The deposit account pays 8.5% interest. How much interest is earned in the fifth year? Solution The interest paid each year will remain constant at 8.5% of the original investment of £20,000. Thus in year 5 the interest will be 0.085 × £20,000 = £1,700 Most investment decisions, however, need to take into account the fact that any interest earned can be reinvested and so compound interest, explained below, is more relevant. The calculation of simple interest is such a basic arithmetic exercise that the only mistake you are likely to make is to transform a percentage figure into a decimal fraction incorrectly. Example 7.2 How much interest will be earned on £400 invested for a year at 0.5%? Solution To convert any percentage figure to a decimal fraction you must divide it by 100. Therefore 0.5% = 0.5 100 = 0.005 and so 0.5% of £400 = 0.005 × £400 = £2 © 1993, 2003 Mike Rosser If you can remember that 1% = 0.01 then you should be able to transform any interest rate specified in percentage terms into a decimal fraction in your head. Try to do this for the following interest rates: (i) 1.5% (ii) 30% (iii) 0.075% (iv) 1.02% (v) 0.6% Now check your answers with a calculator. If you got any wrong you really ought to go back and revise Section 2.5 before proceeding. Converting decimal fractions back to percentage interest rates is, of course, simply a matter of multiplying by 100; e.g. 0.02 = 2%, 0.4 = 40%, 1.25 = 125%, 0.008 = 0.8%. Compound interest is interest which is added to the original investment every time it accrues. The interest added in one time period will itself earn interest in the following time period. The total value of an investment will therefore grow over time. Example 7.3 If £600 is invested for 3 years at 8% interest compounded annually at the end of each year, what will the final value of the investment be? Solution £ Initial sum invested 600.00 Interest at end of year 1 = 0.08 × 600 48.00 Total sum invested for year 2 648.00 Interest at end of year 2 = 0.08 × 648 51.84 Total sum invested for year 3 699.84 Interest at end of year 3 = 0.08 × 699.84 55.99 Final value of investment 755.83 Example 7.4 If £5,000 is invested at an annual rate of interest of 12% how much will the investment be worth after 2 years? Solution £ Initial sum invested 5,000 Year 1 interest = 0.12 × 5,000 600 Sum invested for year 2 5,600 Year 2 interest = 0.12 × 5,600 672 Final value of investment 6,272 © 1993, 2003 Mike Rosser The above examples only involved the calculation of interest for a few years and did not take too long to solve from first principles. To work out the final sum of an investment after longer time periods one could construct a spreadsheet, but an even quicker method is to use the formula explained below. Calculating the final value of an investment Consider an investment at compound interest where: A is the initial sum invested, F is the final value of the investment, i is the interest rate per time period (as a decimal fraction) and n is the number of time periods. The value of the investment at the end of each year will be 1 + i times the sum invested at the start of the year. For instance, the £648 at the start of year 2 is 1.08 times the initial investment of £600 in Example 7.3 above. The value of the investment at the start of year 3 is 1.08 times the value at the start of year 2, and so on. Thus, for any investment Value after 1 year = A(1 + i) Value after 2 years = A(1 + i)(1 + i) = A(1 + i) 2 Value after 3 years = A(1 + i) 2 (1 + i) = A(1 + i) 3 etc. We can see that each value is A multiplied by (1 + i) to the power of the number of years that the sum is invested. Thus, after n years the initial sum A is multiplied by (1 + i) n . The formula for the final value F of an investment of £A for n time periods at interest rate i is therefore F = A(1 + i) n LetusreworkExamples7.3and7.4usingthisformulajusttocheckthatwegetthesame answers. Example 7.3 (reworked) If £600 is invested for 3 years at 8% then the known values for the formula will be A = £300 n = 3 i = 8% = 0.08 Thus the final sum will be F = A(1 + i) n = 600(1.08) 3 = 600(1.259712) = £755.83 © 1993, 2003 Mike Rosser Example 7.4 (reworked) £5,000 invested for 2 years at 12% means that A = £5,000 n = 2 i = 12% = 0.12 F = A(1 + i) n = 5,000(1.12) 2 = 5,000(1.2544) = £6,272 Having satisfied ourselves that the formula works we can now tackle some more difficult problems. Example 7.5 If £4,000 is invested for 10 years at an interest rate of 11% per annum what will the final value of the investment be? Solution A = £4,000 n = 10 i = 11% = 0.11 F = A(1 + i) n = 4,000(1.11) 10 = 4,000(2.8394205) = £11,357.68 (ReferbacktoChapter2,Section8ifyoucannotrememberhowtousethe[y x ]functionkey on your calculator to work out large powers of numbers.) Sometimes a compound interest problem may be specified in a rather different format, but the method of solution is still the same. Example 7.6 You estimate that you will need £8,000 in 3 years’ time to buy a new car, assuming a reasonable trade-in price for your old car. You have £7,000 which you can put into a fixed interest building society account earning 4.5%. Will you have enough to buy the car? Solution You need to work out the final value of your savings to see whether it will be greater than £8,000. Using the usual notation, A = £7,000 n = 3 i = 0.045 F = A(1 + i) n = 7,000(1.045) 3 = 7,000(1.141166) = £7,988.16 © 1993, 2003 Mike Rosser So the answer is ‘almost’. You will have to find another £12 to get to £8,000, but perhaps you can get the dealer to knock this off the price. Changes in interest rates What if interest rates are expected to change before the end of the investment period? The final sum can be calculated by slightly adjusting the usual formula. Example 7.7 Interest rates are expected to be 14% for the next 2 years and then fall to 10% for the following 3 years. How much will £2,000 be worth if it is invested for 5 years? Solution After 2 years the final value of the investment will be F = A(1 + i) n = 2,000(1.14) 2 = 2,000(1.2996) = £2,599.20 If this sum is then invested for a further 3 years at the new interest rate of 10% then the final sum is F = A(1 + i) n = 2,599.20(1.1) 3 = 2,599.20(1.331) = £3,459.54 This could have been worked out in one calculation by finding F = 2,000(1.14) 2 (1.1) 3 = £3,459.54 Therefore the formula for the final sum F that an initial sum A will accrue to after n time periods at interest rate i n and q time periods at interest rate i q is F = A(1 + i n ) n (1 + i q ) q If more than two interest rates are involved then the formula can be adapted along the same lines. Example 7.8 What will £20,000 invested for 10 years be worth if the expected rate of interest is 12% for the first 3 years, 9% for the next 2 years and 8% thereafter? Solution F = 20,000(1.12) 3 (1.09) 2 (1.08) 5 = £49,051.90 © 1993, 2003 Mike Rosser Test Yourself, Exercise 7.1 1. If £4,000 is invested at 5% interest for 3 years what will the final sum be? 2. How much will £200 invested at 12% be worth at the end of 4 years? 3. A parent invests £6,000 for a 7-year-old child in a fixed interest scheme which guarantees 8% interest. How much will the child have at the age of 21? 4. If £525 is invested in a deposit account that pays 6% interest for 6 years, what will the final sum be? 5. What will £24,000 invested at 11% be worth at the end of 5 years? 6. Interest rates are expected to be 10% for the next 3 years and then to fall to 8% for the following 3 years. How much will an investment of £3,000 be worth at the end of 6 years? 7.3 Part year investment and the annual equivalent rate If the duration of an investment is less than a year the usual final sum formula does not always apply. It is usually the custom to specify interest rates on an annual basis for part year investments, but two different types of annual interest rates can be used: (a) the nominal annual interest rate, and (b) the Annual Equivalent Rate (AER). The ways that these annual interest rates relate to part year investments differ. They are also used in different circumstances. Nominal annual interest rates For large institutional investors on the money markets, and for some forms of individual savings accounts, a nominal annual interest rate is quoted for part year investments. To find the interest that will actually be paid, this nominal annual rate is multiplied by the fraction of the year that it is quoted for. Example 7.9 What interest is payable on a £100,000 investment for 6 months at a nominal annual interest rate of 6%? Solution 6 months is 0.5 of one year and so the interest rate that applies is 0.5 × (nominal annual rate) = 0.5 × 6% = 3% Therefore interest earned is 3% of £100,000 = £3,000 © 1993, 2003 Mike Rosser and the final sum is F = (1.03)100,000 = £103,000 If this nominal annual interest rate of 6% applied to a 3-month investment then the actual interest payable would be a quarter of 6% which is 1.5%. If it applied to an investment for one month then the interest payable would be 6% divided by 12 which gives 0.5%. The calculation of part year interest payments on this basis can actually give investors a total annual return that is greater than the nominal interest rate if they can keep reinvesting through the year at the same part year interest rate. The total final value of the investment can be calculated with reference to these new time periods using the F = A(1 +i) n formula as long as the interest rate i and the number of time periods n refer to the same time periods. For example, if £100,000 can be invested for four successive three month periods at a nominal annual interest rate of 6% then, letting i represent the effective quarterly interest rate and n represent the number of three month periods, we get A = £100,000 n = 4 i = 0.25 × 6% = 1.5% = 0.015 F = A(1 + i) n = 100,000(1.015) 4 = £106,136.35 This final sum gives a 6.13635% return on the initial £100,000 sum invested. (Although in practice interest rates are usually only specified to 2 dp.) The more frequently that interest based on the nominal annual rate is paid the greater will be the total annual return when all the interest is compounded. For example, if a nominal annual interest rate of 6% is paid monthly at 0.5% a month and £100,000 is invested for 12 months then A = £100,000 n = 12 i = 0.5% = 0.005 F = A(1 + i) n = 100,000(1.005) 12 = £106,167.78 This new final sum is greater than that achieved from quarterly interest payment and is equivalent to an annual rate of 6.17%. The Annual Equivalent Rate (AER) and Annual Percentage Rate (APR) Although some part year investments on money markets may earn a return which is not equivalent to the nominal annual interest rate, individual investors are usually quoted an annual equivalent rate (AER) which is an accurate reflection of the interest that they earn on investments. For example, interest on the money you may have in a building society will normally be worked out on a daily basis although you will only be told the AER and the interest on your account may only be credited once a year. For loan repayments the annual equivalent rate is usually referred to as the annual percentage rate (APR). If you take out a bank loan you will usually be quoted an APR even though you will be asked to make monthly repayments. The examples above have already demonstrated that the AER is not simply 12 times the monthly interest rate. To determine the relationship between part year interest rates and their true AER, consider another example. © 1993, 2003 Mike Rosser Example 7.10 If interest is credited monthly at a monthly rate of 0.9% how much will £100 invested for 12 months accumulate to? Solution Using the standard investment formula where the time period n is measured in months: A = £100 n = 12 i = 0.9% = 0.009 F = A(1 + i) n = 100(1.009) 12 = 100(1.1135) = £111.35 This final sum of £111.35 after investing £100 for one year corresponds to an annual rate of interest of 11.35%. This is greater than 12 times the monthly rate of 0.9%, since 12 ×0.9% = 10.8% The calculations in the above example that tell us that the ratio of the final sum to the initial sum invested is (1.009) 12 . Using the same principle, the corresponding AER for any given monthly rate of interest i m can be found using the formula AER = (1 +i m ) 12 − 1 Because (1 +i m ) 12 gives the ratio of the final sum F to the initial amount A the −1 has to be added to the formula in order to get the proportional increase in F over A. The APR on loans is the same thing as the annual equivalent rate and so the same formula applies. Example 7.11 If the monthly rate of interest on a loan is 1.75% what is the corresponding APR? Solution i m = 1.75% = 0.0175 APR = (1 +i m ) 12 − 1 = (1.0175) 12 − 1 = 1.2314393 − 1 = 0.2314393 = 23.14% If you have a credit card you can check out this formula by referring to the leaflet on inter- est rates that the credit card company should give you. For example, in October 2001 the © 1993, 2003 Mike Rosser [...]... 1 0.9 174 31 0.84168 0 .77 2183 0 .70 8425 0.649931 0.5962 67 0.5 470 34 0.501866 0.460428 0.422411 0.3 875 33 0.355535 1 0.961538 0.924556 0.888996 0.854804 0.8219 27 0 .79 0315 0 .75 9918 0 .73 069 0 .70 25 87 0. 675 564 0.649581 0.6245 97 1 0.952381 0.9 070 29 0.863838 0.82 270 2 0 .78 3526 0 .74 6215 0 .71 0681 0. 676 839 0.644609 0.613913 0.584 679 0.5568 37 1 0.943396 0.889996 0.839619 0 .79 2094 0 .74 7258 0 .70 4961 0.6650 57 0.6 274 12... 0.82 270 2 0 .78 3526 0 .74 6215 0 .71 0681 0. 676 839 0.644609 0.613913 0.584 679 0.5568 37 1 0.943396 0.889996 0.839619 0 .79 2094 0 .74 7258 0 .70 4961 0.6650 57 0.6 274 12 0.591898 0.558395 0.52 678 8 0.496969 1 0.934 579 0. 873 439 0.816298 0 .76 2895 0 .71 2986 0.666342 0.62 275 0.582009 0.543934 0.508349 0. 475 093 0.444012 1 0.925926 0.8 573 39 0 .79 3832 0 .73 503 0.680583 0.630 17 0.58349 0.540269 0.500249 0.463193 0.428883 0.3 971 14... 18 19 20 B RETURN -7 5000 15000 20000 20000 25000 25000 12000 14.14% C D Interest Rate 4% 5% 6% 7% 8% 9% 10% 11% 12% 13% 14% 15% 16% 17% 18% 19% 20% E NPV 270 96.19 23813.36 20686.58 177 06. 57 14864. 67 12152.86 9563.64 70 90.04 472 5.54 2464.06 299.94 -1 77 2.14 -3 75 7.14 -5 659 .71 -7 484.20 -9 234.68 -1 0914.99 Example 7. 30 A firm has to choose between projects A and B Project A involves an initial outlay of £18,000... 0.558395 0.52 678 8 0.496969 1 0.934 579 0. 873 439 0.816298 0 .76 2895 0 .71 2986 0.666342 0.62 275 0.582009 0.543934 0.508349 0. 475 093 0.444012 1 0.925926 0.8 573 39 0 .79 3832 0 .73 503 0.680583 0.630 17 0.58349 0.540269 0.500249 0.463193 0.428883 0.3 971 14 1 0.9 174 31 0.84168 0 .77 2183 0 .70 8425 0.649931 0.5962 67 0.5 470 34 0.501866 0.460428 0.422411 0.3 875 33 0.355535 © 1993, 2003 Mike Rosser Test Yourself, Exercise 7. 4 1 The... Adjusts the Excel NPV for project B Writes “PROJECT A” under relevant NPV Writes “PROJECT B” under relevant NPV Table 7. 5 A B C D E 1 Ex 7. 27 Interest rate = 10% 2 3 YEAR PROJECT A PV A PROJECT B PV B 4 0 -3 0000 -3 0000.00 -3 0000 -3 0000.00 5 1 6000 5454.55 8000 72 72 .73 6 2 10000 8264.46 8000 6611. 57 7 3 10000 75 13.15 8000 6010.52 8 4 10000 6830.13 8000 5464.11 9 5 8000 49 67. 37 8000 49 67. 37 10 11 NPV = 3029.66... For example, to calculate £525(1. 07) −8 enter 525 [÷] 1. 07 [yx ] 8 [=] or 525 [×] 1. 07 [yx ] 8 [+/−] [=] © 1993, 2003 Mike Rosser Table 7. 6 Discounting factors for Net Present Value Rate of interest i 4% 5% 6% 7% 8% 9% 10% 11% 12% 13% 14% 15% Year 0 1 2 3 4 5 6 7 8 9 10 11 12 1 0.961538 0.924556 0.888996 0.854804 0.8219 27 0 .79 0315 0 .75 9918 0 .73 069 0 .70 25 87 0. 675 564 0.649581 0.6245 97 1 0.952381 0.9 070 29... is added.) Table 7. 2 1 2 3 4 5 6 7 8 9 10 11 12 13 14 A Ex 7. 26 B C Interest rate= YEAR 0 1 2 3 4 5 RETURN -2 5000 5000 6000 10000 10000 10000 PV -2 5000 43 47. 83 4536.86 6 575 .16 571 7.53 4 971 .77 NPV = 1149.15 Excel NPV less cost = £26,149.15 £1,149.15 © 1993, 2003 Mike Rosser D 15% out the NPV for more than one project The following example shows how the spreadsheet created for Example 7. 26 can be extended... B5) =1/(1 + B6) =1/(1 + B7) =1/(1 + B8) =1/(1 + B9) Table 7. 13 A 1 Ex 7. 31 2 3 YEAR 0 4 1 5 2 6 3 7 4 8 5 9 10 11 B C D E F NPV WITH VARIABLE INTEREST RATES i 0 0.15 0.12 0.1 0.11 0.12 DISCOUNT 1 0.8695652 0.8928 571 0.9090909 0.9009009 0.8928 571 FACTOR 1 0.869565 0 .77 6398 0 .70 5816 0.635 87 0.5 677 41 RETURN -2 5000 6000 8000 8000 10000 6000 PV -2 5000.00 52 17. 39 6211.18 5646.53 6358 .70 3406.45 TOTAL NPV =... in decimal format Example 7. 13 If an annual discount rate of 4 7 % is quoted for 3-month Treasury Bills, what would it cost 8 to buy a tranch of these bills with redemption value of £100,000? What would be the annual equivalent rate of return on the sum paid for them? © 1993, 2003 Mike Rosser Solution A nominal annual rate of 4 7 % corresponds to a 3-month rate of 8 47 8 7 = 1 32 = 1.21 875 % 4 As this... discount on the maturity sum then the cost of 3-month Treasury Bills with redemption value of £100,000 of would be £100,000(1 − 0.0121 875 ) = £98 ,78 1.25 and the amount of the discount is £1,218 .75 Therefore, the rate of return on the sum of £98 ,78 1.25 invested for 3 months is 1,218 .75 = 0.012338 = 1.2338% 98 ,78 1.25 If this investment could be compounded for four 3-month periods at this quarterly rate of 1.2338% . by (1 + i) n . The formula for the final value F of an investment of £A for n time periods at interest rate i is therefore F = A(1 + i) n LetusreworkExamples7.3and7.4usingthisformulajusttocheckthatwegetthesame answers. Example. rate of return on the sum paid for them? © 1993, 2003 Mike Rosser Solution A nominal annual rate of 4 7 8 % corresponds to a 3-month rate of 4 7 8 4 = 1 7 32 = 1.21 875 % As this rate is actually. the rate of return on the sum of £98 ,78 1.25 invested for 3 months is 1,218 .75 98 ,78 1.25 = 0.012338 = 1.2338% If this investment could be compounded for four 3-month periods at this quarterly rate