Basic Mathematics for Economists phần 5 ppt

Solution A nominal annual rate of 4 7 8 % corresponds to a 3-month rate of 4 7 8 4 = 1 7 32 = 1.21875% As this rate is actually the 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 − 0.0121875) = £98,781.25 and the amount of the discount is £1,218.75. Therefore, the rate of return on the sum of £98,781.25 invested for 3 months is 1,218.75 98,781.25 = 0.012338 = 1.2338% If this investment could be compounded for four 3-month periods at this quarterly rate of 1.2338% then the annual equivalent rate calculated using the standard formula would be AER = (1.012338) 4 − 1 = 1.050273 − 1 = 0.050273 = 5.0273% Test Yourself, Exercise 7.2 1. If £40,000 is invested at a monthly rate of 1% what will it be worth after 9 months? What is the corresponding AER? 2. A sum of £450,000 is invested at a monthly interest rate of 0.6%. What will the final sum be after 18 months? What is the corresponding AER? 3. Which is the better investment for someone wishing to invest a sum of money for two years: (a) an account which pays 0.9% monthly, or (b) an account which pays 11% annually? 4. If £1,600 is invested at a quarterly rate of interest of 4.5% what will the final sum be after 18 months? What is the corresponding AER? 5. How much interest is earned on £50,000 invested for three months at a nominal annual interest rate of 5%? If money can be reinvested each quarter at the same rate, what is the AER? 6. If a credit card company charges 1.48% a month on any outstanding balance, what APR is it charging? 7. A building society pays an AER of 5.5% on an investment account, calculated on a daily basis. What daily rate of interest will it pay? 8. If 3-month government Treasury Bills are offered at an annual discount rate of 4 7 16 %, what would it cost to buy bills with redemption value of £500,000? What would the AER be for this investment? © 1993, 2003 Mike Rosser 7.4 Time periods, initial amounts and interest rates The formula for the final sum of an investment contains the four variables F, A, i and n. So far we have only calculated F for given values of A, i and n. However, if the values of any three of the variables in this equation are given then one can usually calculate the fourth. Initial amount A formula to calculate A, when values for F, i and n are given, can be derived as follows. Since the final sum formula is F = A(1 + i) n then, dividing through by (1 +i) n , we get the initial sum formula F (1 + i) n = A or A = F(1 + i) −n Example 7.14 How much money needs to be invested now in order to accumulate a final sum of £12,000 in 4 years’ time at an annual rate of interest of 10%? Solution Using the formula derived above, the initial amount is A = F(1 + i) −n = 12,000(1.1) −4 = 12,000 1.4641 = £8,196.16 What we have actually done in the above example is find the sum of money that is equivalent to £12,000 in 4 years’ time if interest rates are 10%. An investor would therefore be indifferent between (a) £8,196.16 now and (b) £12,000 in 4 years’ time. The £8,196.16 is therefore known as the ‘present value’ (PV) of the £12,000 in 4 years’ time. We shall come back to this concept in the next few sections when methods of appraising different types of investment project are explained. Time period Calculating the time period is rather more tricky than the calculation of the initial amount. From the final sum formula F = A(1 + i) n © 1993, 2003 Mike Rosser Then F A = (1 + i) n If the values of F, A and i are given and one is trying to find n this means that one has to work out to what power (1 + i) has to be raised to equal F/A. One way of doing this is via logarithms. Example 7.15 For how many years must £1,000 be invested at 10% in order to accumulate £1,600? Solution A = £1,000 F = £1,600 i = 10% = 0.1 Substituting these values into the formula F A = (1 + i) n we get 1,600 1,000 = (1 + 0.1) n 1.6 = (1.1) n (1) If equation (1) is specified in logarithms then log 1.6 = n log 1.1 (2) since to find the nth power of a number its logarithm must be multiplied by n. Finding logs, this means that (2) becomes 0.20412 = n 0.0413927 n = 0.20412 0.0413927 = 4.93 years If investments must be made for whole years then the answer is 5 years. This answer can be checked using the final sum formula F = A(1 + i) n = 1,000(1.1) 5 = 1,610.51 If the £1,000 is invested for a full 5 years then it accumulates to just over £1,600, which checks out with the answer above. A general formula to solve for n can be derived as follows from the final sum formula: F = A(1 + i) n F A = (1 + i) n © 1993, 2003 Mike Rosser Taking logs log  F A  = n log (1 +i) Therefore the time period formula is log (F/A) log (1 + i) = n (3) An alternative approach is to use the iterative method and plot different values on a spreadsheet. To find the value of n for which 1.6 = (1.1) n this entails setting up a formula to calculate the function y = (1.1) n and then computing it for different values of n until the answer 1.6 is reached. Although some students who find it difficult to use logarithms will prefer to use a spreadsheet, logarithms are used in the other examples in this section. Logarithms are needed to analyse other concepts related to investment and so you really need to understand how to use them. Example 7.16 How many years will £2,000 invested at 5% take to accumulate to £3,000? Solution A = 2,000 F = 3,000 i = 5% = 0.05 Using these given values in the time period formula derived above gives n = log (F/A) log (1 + i) = log 1.5 log 1.05 = 0.1760913 0.0211893 = 8.34 years Example 7.17 How long will any sum of money take to double its value if it is invested at 12.5%? © 1993, 2003 Mike Rosser Solution Let the initial sum be A. Therefore the final sum is F = 2A and i = 12.5% = 0.125 Substituting these value for F and i into the final sum formula F = A(1 + i) n gives 2A = A(1.125) n 2 = (1.125) n Taking logs of both sides log 2 = n log 1.125 n = log 2 log 1.125 = 0.30103 0.0511525 = 5.9 years Interest rates A method of calculating the interest rate on an investment is explained in the following example. Example 7.18 If £4,000 invested for 10 years is projected to accumulate to £6,000, what interest rate is used to derive this forecast? Solution A = 4,000 F = 6,000 n = 10 Substituting these values into the final sum formula F = A(1 + i) n Gives 6,000 = 4,000(1 + i) 10 1.5 = (1 + i) 10 1 + i = 10  (1.5) = 1.0413797 i = 0.0414 = 4.14% © 1993, 2003 Mike Rosser A general formula for calculating the interest rate can be derived. Starting with the familiar final sum formula F = A(1 + i) n F A = (1 + i) n n  (F/A) = 1 + i n  (F/A) − 1 = i (4) This interest rate formula can also be written as i =  F A  1/n − 1 Example 7.19 At what interest rate will £3,000 accumulate to £10,000 after 15 years? Solution Using the interest rate formula (4) above i = n   F A  − 1 = 15   10,000 3,000  − 1 = 15  (3.3333 − 1 = 1.083574 − 1 = 0.083574 = 8.36% Example 7.20 An initial investment of £50,000 increases to £56,711.25 after 2 years. What interest rate has been applied? Solution A = 50,000 F = 56,711.25 n = 2 Therefore F A = 56,711.25 50,000 = 1.134225 © 1993, 2003 Mike Rosser Substituting these values into the interest rate formula gives i = n   F A  − 1 = 2  (1.13455) − 1 = 1.065 − 1 = 0.065 i = 6.5% Test Yourself, Exercise 7.3 1. How much needs to be invested now in order to accumulate £10,000 in 6 years’ time if the interest rate is 8%? 2. What sum invested now will be worth £500 in 3 years’ time if it earns interest at 12%? 3. Do you need to invest more than £10,000 now if you wish to have £65,000 in 15 years’ time and you have a deposit account which guarantees 14%? 4. You need to have £7,500 on 1 January next year. How much do you need to invest at 1.3% per month if your investment is made on 1 June? 5. How much do you need to invest now in order to earn £25,000 in 10 years’ time if the interest rate is (a) 10% (b) 8% (c) 6.5%? 6. How many complete years must £2,400 be invested at 5% in order to accumulate a minimum of £3,000? 7. For how long must £5,000 be kept in a deposit account paying 8% interest before it accumulates to £7,500? 8. If it can earn 9.5% interest, how long would any given sum of money take to treble its value? 9. If one needs to have a final sum of £20,000, how many years must one wait if £12,500 is invested at 9%? 10. How long will £70,000 take to accumulate to £100,000 if it is invested at 11%? 11. If £6,000 is to accumulate to £10,000 after being invested for 5 years, what rate must it earn interest at? 12. What interest rate will turn £50,000 into £60,000 after 2 years? 13. At what interest rate will £3,000 accumulate to £4,000 after 4 years? 14. What monthly rate of interest must be paid on a sum of £2,800 if it is to accumulate to £3,000 after 8 months? 15. What rate of interest would turn £3,000 into £8,000 in 10 years? 16. At what rate of interest will £600 accumulate to £900 in 5 years? 17. Would you prefer (a) £5,000 now or (b) £8,000 in 4 years’ time if money can be borrowed or lent at 11%? 7.5 Investment appraisal: net present value Assume that you have £10,000 to invest and that someone offers you the following proposal: pay £10,000 now and get £11,000 back in 12 months’ time. Assume that the returns on this investment are guaranteed and there are no other costs involved. What would you do? Perhaps © 1993, 2003 Mike Rosser you would compare this return of 10% with the rate of interest your money could earn in a deposit account, say 4%. In a simple example like this the comparison of rates of return, known as the internal rate of return (IRR) method, is perhaps the most intuitively obvious method of judging the proposal. This is not the preferred method for investment appraisal, however. The net present value (NPV) method has several advantages over the IRR method of comparing the project rate of return with the market interest rate. These advantages are explained more fully in the following section, but first it is necessary to understand what the NPV method involves. We have already come across the concept of present value (PV) in Section 7.4. If a certain sum of money will be paid to you at some given time in the future its PV is the amount of money that would accumulate to this sum if it was invested now at the ruling rate of interest. Example 7.21 What is the present value of £1,500 payable in 3 years’ time if the relevant interest rate is 4%? Solution Using the initial amount investment formula, where F = £1,500 i = 0.08 n = 3 A = F(1 + i) −n = 1,500 (1.04) 3 = 1,500(1.04) −3 = 1,500 1.124864 = £1,333.49 An investor would be indifferent between £1,333.49 now and £1,500 in 3 years’ time. Thus £1,333.49 is the PV of £1,500 in 3 years’ time at 4% interest. In all the examples in this chapter it is assumed that future returns are assured with 100% certainty. Of course, in reality some people may place greater importance on earlier returns just because the future is thought to be more risky. If some form of measure of the degree of risk can be estimated then more advanced mathematical methods exist which can be used to adjust the investment appraisal methods explained in this chapter. However, here we just assume that estimated future returns and costs, are correct. An investor has to try to make the most rational decision based on whatever information is available. The net present value (NPV) of an investment project is defined as the PV of the future returns minus the cost of setting up the project. Example 7.22 An investment project involves an initial outlay of £600 now and a return of £1,000 in 5 years’ time. Money can be invested at 9%. What is the NPV? © 1993, 2003 Mike Rosser Solution The PV of £1,000 in 5 years’ time at 9% can be found using the initial amount formula as A = F(1 + i) −n = 1,000(1.09) −5 = £649.93 Therefore NPV = £649.93 − £600 = £49.93. This project is clearly worthwhile. The £1,000 in 5 years’ time is equivalent to £649.93 now and so the outlay required of only £600 makes it a bargain. In other words, one is being asked to pay £600 for something which is worth £649.93. Another way of looking at the situation is to consider what alternative sum could be earned by the investor’s £600. If £649.93 was invested for 5 years at 9% it would accumulate to £1,000. Therefore the lesser sum of £600 must obviously accumulate to a smaller sum. Using the final sum investment formula this can be calculated as F = A(1 + i) n = 600(1.09) 5 = 600(1.538624) = £923.17 The investor thus has the choice of (a) putting £600 into this investment project and securing £1,000 in 5 years’ time, or (b) investing £600 at 9%, accumulating £923.17 in 5 years. Option (a) is clearly the winner. If the outlay is less than the PV of the future return an investment must be a profitable ven- ture. The basic criterion for deciding whether or not an investment project is worthwhile is therefore NPV > 0 As well as deciding whether specific projects are profitable or not, an investor may have to decide how to allocate limited capital resources to competing investment projects. The rule for choosing between projects is that they should be ranked according to their NPV. If only one out of a set of possible projects can be undertaken then the one with the largest NPV should be chosen, as long as its NPV is positive. Example 7.23 An investor can put money into any one of the following three ventures: Project A costs £2,000 now and pays back £3,000 in 4 years Project B costs £2,000 now and pays back £4,000 in 6 years Project C costs £3,000 now and pays back £4,800 in 5 years The current interest rate is 10%. Which project should be chosen? © 1993, 2003 Mike Rosser Solution NPV of project A = 3,000(1.1) −4 − 2,000 = 2,049.04 − 2,000 = £49.04 NPV of project B = 4,000(1.1) −6 − 2,000 = 2,257.90 −2,000 = £257.90 NPV of project C = 4,800(1.1) −5 − 3,000 = 2,980.42 −3,000 =−£19.58 Project B has the largest NPV and is therefore the best investment. Project C has a negative NPV and so would not be worthwhile even if there was no competition. The investment examples considered so far have only involved a single return payment at some given time in the future. However, most real investment projects involve a stream of returns occurring over several time periods. The same principle for calculating NPV is used to assess these projects, the initial outlay being subtracted from the sum of the PVs of the different future returns. Example 7.24 An investment proposal involves an initial payment now of £40,000 and then returns of £10,000, £30,000 and £20,000 respectively in 1, 2 and 3 years’ time. If money can be invested at 10% is this a worthwhile investment? Solution PV of £10,000 in 1 year’s time = £10,000 1.1 = £9,090.91 PV of £30,000 in 2 years’ time = £30,000 1.1 2 = £24,793.39 PV of £20,000 in 3 years’ time = £20,000 1.1 3 = £15,026.30 Total PV of future returns £48,910.60 less initial outlay −£40,000 NPV of project £8,910.60 This NPV is greater than zero and so the project is worthwhile. At an interest rate of 10% one would need to invest a total of £48,910.60 to get back the projected returns and so £40,000 is clearly a bargain price. The further into the future the expected return occurs the greater will be the discounting factor.ThisismadeobviousinExample7.25below,wherethereturnsarethesameeach time period. The PV of each successive year’s return is smaller than that of the previous year because it is multiplied by (1 + i) −1 . © 1993, 2003 Mike Rosser [...]... 0.698924 0.68 658 2 0.674463 0.66 256 4 0. 650 880 0.639409 0.6281 45 0.617084 0.606224 0 .59 556 0 0 .58 5089 0 .57 4807 0 .56 4711 0 .55 4797 0 .54 5061 0 .53 550 1 0 .52 6114 0 .51 68 95 0 .50 7842 0.498 953 0.490223 0.741372 0.71 957 4 0.698427 0.677911 0. 658 008 0.638699 0.619966 0.601791 0 .58 4 157 0 .56 7049 0 .55 0449 0 .53 4343 0 .51 8717 0 .50 355 4 0.488842 0.47 456 7 0.4607 15 0.4472 75 0.434232 0.42 157 7 0.4092 95 0.397378 0.3 858 13 0.37 459 0 0.363699... 0.7084 25 0.649931 0 .59 6267 0 .54 7034 0 .50 1866 0.460428 0.422411 0.38 753 3 0. 355 5 35 1 0.96 153 8 0.92 455 6 0.888996 0. 854 804 0.821927 0.7903 15 0. 759 918 0.73069 0.70 258 7 0.6 755 64 0.64 958 1 0.62 459 7 1 0. 952 381 0.907029 0.863838 0.822702 0.78 352 6 0.7462 15 0.710681 0.676839 0.644609 0.613913 0 .58 4679 0 .55 6837 1 0.943396 0.889996 0.839619 0.792094 0.747 258 0.704961 0.6 650 57 0.627412 0 .59 1898 0 .55 83 95 0 .52 6788... = 300 0 .50 0 .55 0.60 0. 65 0.70 0. 75 0.80 0. 85 0.90 0. 95 1.00 1. 05 1.10 1. 15 1.20 1. 25 1.30 1. 35 1.40 1. 45 1 .50 1 .55 1.60 1. 65 1.70 1. 75 1.80 1. 85 1.90 1. 95 2.00 0.9419 05 0.936300 0.930731 0.9 251 97 0.919700 0.914238 0.908811 0.903418 0.898061 0.892738 0.887449 0.882194 0.876972 0.871784 0.866630 0.86 150 8 0. 856 419 0. 851 363 0.846339 0.841347 0.836387 0.831 459 0.82 656 2 0.821696 0.816861 0.812 057 0.807284... 0.863838 0.822702 0.78 352 6 0.7462 15 0.710681 0.676839 0.644609 0.613913 0 .58 4679 0 .55 6837 1 0.943396 0.889996 0.839619 0.792094 0.747 258 0.704961 0.6 650 57 0.627412 0 .59 1898 0 .55 83 95 0 .52 6788 0.496969 1 0.93 457 9 0.873439 0.816298 0.7628 95 0.712986 0.666342 0.622 75 0 .58 2009 0 .54 3934 0 .50 8349 0.4 750 93 0.444012 1 0.9 259 26 0. 857 339 0.793832 0.7 350 3 0.68 058 3 0.63017 0 .58 349 0 .54 0269 0 .50 0249 0.463193 0.428883... 0.80 254 1 0.797828 0.793146 0.788493 0.8871 85 0.876 658 0.866260 0. 855 991 0.8 458 48 0.8 358 31 0.8 259 37 0.8161 65 0.80 651 4 0.796981 0.78 756 6 0.778266 0.769081 0.760008 0. 751 048 0.742197 0.733 454 0.724819 0.716290 0.7078 65 0.69 954 3 0.691324 0.683204 0.6 751 85 0.667263 0. 659 438 0. 651 708 0.644073 0.63 653 1 0.629080 0.621721 0.8 356 44 0.8208 15 0.806 255 0.791961 0.777927 0.764148 0. 750 621 0.737339 0.724299 0.7114 95. .. 0 .59 1898 0 .55 83 95 0 .52 6788 0.496969 1 0.93 457 9 0.873439 0.816298 0.7628 95 0.712986 0.666342 0.622 75 0 .58 2009 0 .54 3934 0 .50 8349 0.4 750 93 0.444012 1 0.9 259 26 0. 857 339 0.793832 0.7 350 3 0.68 058 3 0.63017 0 .58 349 0 .54 0269 0 .50 0249 0.463193 0.428883 0.397114 1 0.917431 0.84168 0.772183 0.7084 25 0.649931 0 .59 6267 0 .54 7034 0 .50 1866 0.460428 0.422411 0.38 753 3 0. 355 5 35 © 1993, 2003 Mike Rosser Test Yourself, Exercise... function key For example, to calculate 52 5(1.07)−8 enter 52 5 [÷] 1.07 [yx ] 8 [=] or 52 5 [×] 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 0.96 153 8 0.92 455 6 0.888996 0. 854 804 0.821927 0.7903 15 0. 759 918 0.73069 0.70 258 7 0.6 755 64 0.64 958 1 0.62 459 7 1 0. 952 381... 0.363699 0. 353 130 0.342873 0.332918 0.323 257 0.313881 0.304782 0.302096 0.268103 0.237949 0.211199 0.187467 0.166412 0.147731 0.131 154 0.116444 0.103390 0.0918 05 0.08 152 3 0.072397 0.064296 0. 057 1 05 0. 050 721 0.0 450 53 0.040022 0.0 355 54 0.03 158 6 0.028064 0.0249 35 0.022 156 0.019689 0.017497 0.0 155 50 0.013820 0.012284 0.010919 0.009706 0.008628 0.2239 65 0.192920 0.166190 0.143174 0.123 355 0.106287 0.09 158 8 0.078927... 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 0.86 956 52 0.892 857 1 0.9090909 0.9009009 0.892 857 1 FACTOR 1 0.86 956 5 0.776398 0.7 058 16 0.6 358 7 0 .56 7741 RETURN - 250 00 6000 8000 8000 10000 6000 PV - 250 00.00 52 17.39 6211.18 56 46 .53 6 358 .70 3406. 45 TOTAL NPV = 1840. 25 © 1993, 2003 Mike Rosser Test Yourself, Exercise 7 .5 1 Calculate the IRR for. .. year Thus 8,000 8,000 6,000 + + 1. 15 1. 15 × 1.12 1. 15 × 1.12 × 1.1 10,000 6,000 + + 1. 15 × 1.12 × 1.1 × 1.11 1. 15 × 1.12 × 1.1 × 1.11 × 1.12 NPV = − 25, 000 + © 1993, 2003 Mike Rosser 6,000 8,000 8,000 10,000 6,000 + + + + 1. 15 1.288 1.4168 1 .57 2648 1.7613 657 = − 25, 000 + 5, 217.39 + 6,211.18 + 5, 646 .53 + 6, 358 .70 + 3,406. 45 = − 25, 000 + = − 25, 000 + 26,840. 25 = £1,840. 25 This is positive and so the investment . 0.428883 0.38 753 3 0.64 958 1 0 .58 4679 0 .52 6788 0.4 750 93 0.428883 0.38 753 3 12 0.62 459 7 0 .55 6837 0.496969 0.444012 0.397114 0. 355 5 35 0.62 459 7 0 .55 6837 0.496969 0.444012 0.397114 0. 355 5 35 © 1993, 2003. 0.644609 0 .59 1898 0 .54 3934 0 .50 0249 0.460428 10 0.6 755 64 0.613913 0 .55 83 95 0 .50 8349 0.463193 0.422411 0.6 755 64 0.613913 0 .55 83 95 0 .50 8349 0.463193 0.422411 11 0.64 958 1 0 .58 4679 0 .52 6788 0.4 750 93 0.428883. 0.7903 15 0.7462 15 0.704961 0.666342 0.63017 0 .59 6267 0.7903 15 0.7462 15 0.704961 0.666342 0.63017 0 .59 6267 7 0. 759 918 0.710681 0.6 650 57 0.622 75 0 .58 349 0 .54 7034 0. 759 918 0.710681 0.6 650 57 0.622 75 0 .58 349