1. Topic 16 Exponential and Log Functions and Applications III

2. Application of exponential and logarithmic functions (SLEs 2,5,7,10,11,13) 2 hours Applications of geometric progressions to compound interest including present and future values (SLEs 18,19,24) 4 hours Applications of geometric progressions to annuities and amortization of a loan (SLEs 20-23,25-27,29) 4 hours

3. Simple vs Compound Interest.xls

5. Amount borrowed = 25000 – 2000 = 23000 Amount of interest= 13% X 23000 X 2 = 5980 Amount to be repaid= 23000 + 5980 = 28980 Monthly repayment= 28980 ÷ 24 = $1207.50

6. Amount borrowed = 9500 Total amount being repaid= 24 x 480 = 11520 Interest paid in 2 years= 11520 – 9500 = 2020 Interest paid in 1 year= 1010 Interest rate p.a.= 1010 ÷ 9500 x 100% = 10.63% p.a.

7. What are the: i) Growth Factor ii) Amount and iii) Simple Interest On $5000 at 15.5% for 2 years? (i) Growth Factor (G) (ii) Amount (iii) Simple Int

9. If we invested $10000 at 10% p.a. compound interest, the amount after the 1 st year = 1.1 x 10000 =11000 the amount after the 2 nd year = 1.1 x (the amount after the 1 st year) = 1.1 x (1.1 x 10000) = 12100 the amount after the 3 rd year = 1.1 x (the amount after the 2 nd year) = 1.1 x (1.1 x 1.1 x 10000) = 13310 the amount after the n th year = 1.1 n x 10000

10. If we invested $10000 at 6% p.a. compound interest, the amount after the 1 st year = 1.06 x 10000 =10600 the amount after the 2 nd year = 1.06 x (the amount after the 1 st year) = 1.06 x (1.06 x 10000) = 11236 the amount after the 3 rd year = 1.06 x (the amount after the 2 nd year) = 1.06 x (1.06 x 1.06 x 10000) = 111910.16 the amount after the n th year = 1.06 n x 10000 In general, if P = the principal, i % = the interest rate, n = the term A = P x (1 + i ) n

11. A = ? P = 35000 i =.07 n = 8 A = P x (1 + i) n = 35000 x (1.07) 8 = $60136.51

12. It is often necessary to use half-yearly, quarterly, monthly or even daily rates of interest. To find half-yearly rates, divide annual rates by 2 To find monthly rates, divide annual rates by 12 To find daily rates, divide annual rates by 365 etc

13. Daily interest rate = 15%  365 = 0.15  365 Number of days = 21 Interest = PRT = 1200 x 0.15x 21 365 = $10.36

14. A = P = $10 000 i = 0.09/12 n = 8 A = P x (1 + i) n = 10 000 x 1.0075 8 = $10615.98

15. Model: It is expected that inflation over the next 10 years will be at 5% p.a. What would you expect the price of a 4WD costing $45 000 now to be in 10 year’s time? P = 45 000, i=0.05, n=10 A = P(1+i) n = 45000(1.05) 10 = $73300

17. A geometric progression (GP) is a series of numbers in which the ratio between successive terms is constant: e.g. 1, 3, 9, 27, …. Common ratio = 3 e.g. 32, -16, 8, -4, 2, …Common ratio = -½ The first term is denoted by a and the common ratio is denoted by r Note r = T n T n-1

18. So the terms of a GP are in the form a, ar, ar 2, ar 3, ar 4, ….. T 1 = a T 2 = ar T 3 = ar 2 T 4 = ar 3.. T n = ar n-1 r 0 1

19. (a) a = 3, r = 2 T 1 = 3, T 2 = 6, T 3 = 12, T 4 = 24 (b) ar 3 = 24 …. (1) ar 5 = 96 …. (2) (2) = ar 5 = 96 (1) ar 3 24 r 2 = 4 r =  2 If r = 2 then a = 3 if r = -2 then a = -3 If r = 2, a = 3  3, 6, 12, 24,… If r = -2, a = -3  -3, 6, -12, 24,… (c) ar 5 = ¼  a(½) 5 = ¼ a = 8  8,4,2,1,….

21. Let S n denote the sum of n terms of a GP S n = a + ar + ar 2 + …… + ar n-3 + ar n-2 + ar n-1 …(1) S n = a + ar + ar 2 + …… + ar n-3 + ar n-2 + ar n-1 …(1)also r x S n = ar + ar 2 + …… + ar n-3 + ar n-2 + ar n-1 + ar n …(2) Subtracting (1) and (2) S n -r x S n = a – ar n S n - r x S n = a – ar n (1-r)S n = a(1-r n ) S n = a(1-r n ) S n = a(1-r n ) 1-r 1-r

22. a = 2, r = 3, n=10 S n = a(1-r n ) 1-r = 2(1-3 10 ) 1-3 = 59 048

23.  a = 3r = 2 find n T n = ar n-1 384 = 3  2 n-1 128 = 2 n-1 2 7 = 2 n-1 n = 8

24.  a = 2¼r = 4/3

26. If you were able to invest $1 at 10% p.a. compound interest, its value (v) would increase so that after n years, its value would be given by: n123456789101112 v1.101.211.331.461.611.771.952.142.362.592.853.14 13141516171819202122232425 3.453.804.184.595.055.566.126.737.408.148.959.85 10.83 i.e. the future value of $1 in 19 years at 10% is $6.12 The present value of $1 in 19 years at 10% is 1/6.12 = $0.16

27. Press Finance TVM Solver Enter the data: APPS N=19 I%=10 PV=0 PMT=0 FV=1 P/Y=1 C/Y=1 PMT:END BEGIN Place the cursor on the 0 in PV = 0 Press ALPHA SOLVE

28. N=18 I%=8 PV=0 PMT=0 FV=30000 P/Y=1 C/Y=1 PMT:END BEGIN Using with the cursor on PV=0,  PV = $7507.47SOLVE

29. N=8  12 I%=10 PV=-10 000 PMT=0 FV= Alpha Solve P/Y=12 C/Y=12 PMT:END Using with the cursor on FV=0,  FV = $22 181. 76 SOLVE

31. An annuity is a series of fixed payments made regularly over a long period of time. Each payment is made up of the interest earned by the investment together with a small part of the capital invested. The initial capital is calculated in such a way that the final annuity payment will completely exhaust the fund.

32. To collect $10 000 in 1 year’s time  $9259.26 To collect $10 000 in 2 year’s time  $8573.39 To collect $10 000 in 3 year’s time  $7938.32  Amount to invest now is $25770.97

33. To collect $10 000 in 1 year’s time  $9259.26 To collect $10 000 in 2 year’s time  $8573.39 To collect $10 000 in 3 year’s time  $7938.32  Amount to invest now is $25 770.97 Another way to do this N=3 I% = 8 PV = 0 PMT = 10 000 FV = 0 P/Y = 1 PMT : END Place cursor on PV and solve

34. Find the present value of payments of $58 per week over 5 years at 8.4% interest. N= 52  5 I%=8.4 PV= 0 PMT= -58 FV= Alpha Solve P/Y=52 C/Y=52 PMT:END FV = $18 722.39 Approach 1: (i) what will this annuity amount to? (ii) what one off investment is needed now to have $18 722.39 in 5 years under the same conditions? N= 52  5 I%=8.4 PV= Alpha Solve PMT= 0 FV= $18 722.39 P/Y=52 C/Y=52 PMT:END PV = $12 305.66

35. Find the present value of payments of $58 per week over 5 years at 8.4% interest. N= 52  5 I%=8.4 PV= Alpha Solve PMT= 58 FV= 0 P/Y=52 C/Y=52 PMT:END Approach 2: Think of investing an unknown amount of money now under the same conditions and receiving $58 from that investment every week until there is nothing left in the account. PV = $12 305.66

39. John is 20, plans to retire at 55 and is likely to live to be 85. In his retirement he would like to receive an indexed annuity of $1000 per month. How much should he be contributing each month during his working life to afford this? Assume he can invest during his working life at 7%pa and during retirement at 8.5%.