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Chapter I

Chapter I. Mathematics of Finance. I-1 Interest. I-1-01: Simple Interest Let: p = Principal in Riyals r =Interest rate per year t = number of years → The accumulated amount A after t years: A = p + prt = p ( 1 + rt). Graphing the straight line segment: A(t) = p + prt ; t ≥ 0.

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Chapter I

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  1. Chapter I Mathematics of Finance

  2. I-1 Interest I-1-01: Simple Interest Let: p = Principal in Riyals r =Interest rate per year t = number of years → The accumulated amount A after t years: A = p + prt = p ( 1 + rt)

  3. Graphing the straight line segment:A(t) = p + prt ; t ≥ 0

  4. Example (1) What are the interest and the total accumulated amount after 10 years on a deposit of 2000 Riyals at a simple interest rate of 1% per year? Solution: The accumulated amount A = p + prt The interest paid I = prt where p = 2000, r = 1/100 = 0.01 and t = 10. A = 2000 + 2000(1/100)(10) = 2000 + 200 = 2200 I = prt = 200 Textbook: Example 1 page 186

  5. Graphing the straight line:A(t) = 2000 + 2000(0.01)t ; t ≥ 0

  6. Homework Exercises 4.1 Page: 197 All odd numbered exercises from 1 to 9

  7. Homework 1.a. What are the interest I and the total accumulated amount A after 2 years on a deposit of 500 Riyals at a simple interest rate of 8% per year? Answer: I = 80 Riyals and A = 580 Riyals 2.b.c. If theaccumulated amountafter 2 years on a deposit at a simple interest rate of 8% per year is 580, then what’s the principal P (the deposit)? Answer: P = 500 Riyals 3.d. How many years t would it take a deposit of 500 Riyals at a simple interest rate of 8% per year to grow to an accumulated amount of 580 Riyals? Answer: t = 2 years 3.f. If a deposit of 500 Riyals grows to an accumulated amount of 580 Riyals in 2 years, then what’s the simple interest rate? Answer: r = 0.08 (8% per year )

  8. Homework 2.a. What are the interest I and the total accumulated amount A after 9 months on a deposit of 800 Riyals at a simple interest rate of 6% per year? Answer: I = 36 Riyals and A = 836 Riyals 2.b. If theaccumulated amountafter 9 months on a deposit at a simple interest rate of 6% per year is 836, then what’s the principal P (the deposit)? Answer: P = 800 Riyals 3.c. How long would it take a deposit of 800 Riyals at a simple interest rate of 6% per year to grow to an accumulated amount of 836 Riyals? Answer: t = 0.75 year = (3/4)(12 months) = 9 months 3.d. If a deposit of 800 Riyals grows to an accumulated amount of 836 Riyals in 9 months, then what’s the simple interest rate? Answer: r = 0.06 = 6% per year

  9. I-1-02: Compound Interest Let: p = Principal in Riyals r =Interest rate per year t = number of years → The accumulated amount A1 after 1 year: A1 = p + pr(1) = p (1 + r) The accumulated amount A2 after 2 years: A2 = A1 + A1 r(1) = A1 (1 + r) = p(1 + r) (1 + r)= p(1 + r)2

  10. The accumulated amount A3 after 3 years: A3 = A2 + A2 r(1) = A2 (1 + r) = p(1 + r)2 (1 + r)=p (1 + r)3 The accumulated amount A after 4 years: A4 = A3 + A3 r(1) = A3 (1 + r) = p(1 + r)3 (1 + r)= p(1 + r)4 We conclude that the accumulated amount after t years At = p(1 + r)t

  11. Example (2) What are the interest and the total accumulated amount after 2 years on a deposit of 2000 Riyals at a compound interest rate of 10% per year? Solution: The accumulated amount A = p(1+ r)t The interest paid I = A – p Where, p = 2000, r = 10/100 = 1 /10 = 0.1 and t = 2 A = 2000[1 + (1/10)]2 The interest paid = 2420 – 2000 = 420

  12. I-1-03: Interest compounded m times a year Let: p = Principal in Riyals r = Interest rate per year m = number of times a year the interest is compounded Conversion period = the period of time between successive interest calculations The interest rate per conversion period = i = r / m t = the number of years (term) n = Number of periods in t years = mt → The accumulated amount A1 after 1 period: A1 = p + pi = p (1 + i) The accumulated amount A2 after 2 periods: A2 = A1 + A1 i = A1 (1 + i) = p(1 + i) (1 + i)= p(1 + i)2

  13. The accumulated amount A3 after 3 periods: A3 = A2 + A2 i = A2 (1 + i) = p(1 + i)2 (1 + i)=p (1 + i)3 The accumulated amount A after 4 periods: A4 = A3 + A3 i = A3 (1 + i) = p(1 + i)3 (1 + i)= p(1 + i)4 We conclude that the accumulated amount after n periods An = p(1 + i)n = p (1+i)mt = p (1 + r/m )mt Where, p = the principal r = the interest per year m = the number of times (periods) in a year the interest is compounded

  14. Example (3) What is the total accumulated amount after 3 years on a deposit of1000 Riyals at interest rate of 10% per year compounded: 1. semiannually ( 2 periods in a year) 2. quarterly ( 4 periods in a year ) 3. monthly ( 12 periods in a year) 4. daily ( 365 periods in a year) 5. every 4 months ( 3 periods in a year ) 6. every two months ( 6 periods in a year ) 7. annually Solution: In all of these cases, we use the formula An = p(1 + i)n = p (1+i)mt = p (1 + r/m )mt

  15. 1. semiannually ( 2 periods in a year) A = 1000[1 + (0.1)/2 ]2(3 2. quarterly ( 4 periods in a year ) A = 1000[1 + (0.1)/4 ]4(3) 3. monthly ( 12 periods in a year) A = 1000[1 + (0.1)/12 ]12(3)

  16. 4. daily ( 365 periods in a year) A = 1000[1 + (0.1)/365 ]365(3) 5. every 4 months ( 3 periods in a year ) A = 1000[1 + (0.1)/3 ]3(3) 6. every two months ( 6 periods in a year ) A = 1000[1 + (0.1)/6 ]6(3) 7. annually ( 1 periods in a year ) A = 1000[1 + 0.1 ]3 Textbook: Example 3 page 189

  17. Homework Exercises 4.1 Page: 197 All odd numbered exercises from 11 to 19

  18. Homework What is the total accumulated amount after tyears on a deposit of P Riyals at interest rate of r per year, compounded: 1. annually, if P= QR1000, t = 8 years and r = 7% Answer: A = QR 1718.19 2. quarterly , if P= QR12000, t = 10.5 years and r = 8% Answer: A = QR 27566.93 3. monthly , if P= QR150000, t = 4 years and r = 14% Answer: A = QR 261751.04 4. daily , if P= QR150000, t = 3 years and r = 12% Answer: A = QR 214986.69 4. semiannually, if P= QR2500, t = 10 years and r = 7% Answer: A = QR 4974.47

  19. Homework What is the total accumulated amount after tyears on a deposit of P Riyals at interest rate of r per year, compounded: 1. annually, if P= QR1000, t = 8 years and r = 7% Answer: A = QR 1718.19 2. quarterly , if P= QR12000, t = 10.5 years and r = 8% Answer: A = QR 27566.93 3. monthly , if P= QR150000, t = 4 years and r = 14% Answer: A = QR 261751.04 4. daily , if P= QR150000, t = 3 years and r = 12% Answer: A = QR 214986.69 4. semiannually, if P= QR2500, t = 10 years and r = 7% Answer: A = QR 4974.47

  20. I-1-04: Continuous Compounding of Interest“ If the interest is compounded more and more frequently” We have: A = p (1 + r/m )mt = p (1 + r/m )\(m/r) rt = p [1 + (1/u)]u(rt) , by letting m/r = u and so r/m = 1/u = p [1 + (1/u)]u(rt) = p [[1 + (1/u)]u](rt) Notice that u →0 as m→∞ and so as you are supposed to know: [[1 + (1/u)]u → e And thus, p [[1 + (1/u)]u](rt) → p ert Thus the amount accumulated if the interest is compounded continuously is: A = p ert Where p is the principal, r the interest rate and t the term ( number of years)

  21. Example (4) What is the total accumulated amount after 3 years on a deposit of1000 Riyals at interest rate of 10% compounded continuously? Solution: The amount accumulated: A = p ert , where p=1000, r = 0.1 and t = 3. Thus, A = 1000 e(0.1)(3) Assignment: 1. Find the answer approximated to 2 decimal digits (Accuracy of one Dirham) 2. Use the appropriate formula to calculate the amount accumulated if the interest is compounded daily ( with the same accuracy) and compare the that with the answer to question (1). Textbook: Example 4 page 191

  22. Homework Exercises 4.1 Page: 198 Exercise 29

  23. Homework Find, using the continuous compounding formula: the total accumulated amount A after 4 years on a deposit of 5000 Riyals at interest rate of 8% per year. Answer: A = QR 6885.64

  24. I-1-05: Effective Rate of Interest( the annual percentage yield) reff = [1 + (r/m)]m - 1 Example (5): What’s the effective rate if the rate if the nominal rate of interest is 10% per year Compounded: 1. quarterly 2. monthly 3. annually

  25. Solution We use the formula reff = [1 + (r/m)]m – 1, Where r = 10/100 = 1/10 = 0.1 1. quarterly → m = 4 reff = [1 + (0.1)/4)]4 – 1 2. monthly→ m = 12 reff = [1 + (0.1)/12)]12 – 1 3. annually → m = 1 reff = [1 + (0.1)/1)]1– 1 = 0.1 = the nominal interest Textbook: Example 5 page 192

  26. Deducing the Formula of the Effective rate of Interest We let: p(1+ reff ) = p[1+ (r/m)]m Where, r = the nominal yearly interest rate m= the number of conversion periods per year →1+ reff = [1+ (r/m)]m →reff = [1+ (r/m)]m - 1

  27. Homework Exercises 4.1 Page: 197 Exercises 21 and 23

  28. Homework What is the effective rate of interest reff corresponding to the following nominal interest rate: 1. 10% per year compounded semiannually Answer: reff = 10.25 % per year 2. 8% per year compounded monthly Answer: reff = 8.3 % per year

  29. I-1-06: Present Value What’s the principal (present value) that should be deposited in a bank paying an Interest at a rate of r compounded m times a Year such that at the end of t years the accumulated amount will be A. We had A = p [ 1 +(r/m) ]mt →p =A / [ 1 +(r/m) ] mt →p =A [ 1 +(r/m) ] - mt

  30. Example (6) What’s the principal (present value) that should be deposited in a bank paying an Interest at a rate of 10% compounded 3 times a year such that at the end of 4 years the accumulated amount will be 5000. Solution: p =A [ 1 +(r/m) ] - mt We have A=5000, r =10/100=1/10= 0.1, m=3 and t=4 →p =5000 [ 1 +(0.1)/3 ] –3(4) Textbook: Examples 6 & 7 page 194

  31. Homework Exercises 4.1 Page: 197 - 198 Exercises 25, 27

  32. Homework 1. What’s the present value( money already in an account) in a bank paying an Interest at a rate of 6% compounded 2 times a year such that at the end of 4 years the accumulated amount will be QR 40000. Answer: P = 31576.37 Riyals 2. What’s the present value( money already in an account) in a bank paying an Interest at a rate of 7% compounded monthly such that at the end of 4 years the accumulated amount will be QR 40000. Answer: P = 30255.95 Riyals

  33. I-1-07: using Logarithms to Solve Problems in Finance a. Finding the term number of years) t needed for a principal p to grow to an accumulated amount A if invested at rate r(per year) compounded m times a year. We have: A = p [ 1 + (r/m) ]mt Hence, A / p = [ 1 + (r/m) ]mt , dividing both sides of the equation by p Taking the natural logarithm for both sides, we get: ln(A/p) = ln [ 1 + (r/m) ]mt = mt ln [ 1 + (r/m) ] Thus, t = ln(A/p) / m ln [ 1 + (r/m) ]

  34. Example (7)-a How long it will take a principal of 1000 Riyals to grow to 1270.24 if deposited in a bank paying an Interest at a rate of 8% compounded monthly ? Solution: t = ln(A/p) / m ln [ 1 + (r/m) ] We have: p=1000, A=1270.24, r=0.08 and m=12 We can either use the deduced formula or do the calculation from scratch: 1. Using the deduced formula: t = ln(A/p) / m ln [ 1 + (r/m) ] → t = ln(1270.24/1000) / 12 ln [ 1 + (0.08/12) ] Assignment find t ! Textbook: Example 8 page 194

  35. 2. Solving from scratch: We have t = ln(1270.24/1000) / 12 ln [ 1 + (0.08/12) ] A = p [ 1 + (r/m) ]mt Thus, 1270.24 = 1000[ 1 + (0.08/12) ]12t → 1270.24/1000 = [ 1 + (0.08/12) ]12t ( dividing both sides by 1000) → ln(1270.24/1000) = ln [ 1 + (0.08/12) ]12t ( applying the logarithm to both sides) → ln(1270.24/1000) =12t ln [ 1 + (0.08/12) ] ( using a logarithmic property) →12t = ln(1270.24/1000) / ln [ 1 + (0.08/12) ] →t = ln(1270.24/1000) / 12 ln [ 1 + (0.08/12) ] Textbook: Example 8 page 194

  36. Homework Exercises 4.1 Page: 198 Exercises: 31, 33 and 37

  37. Homework 1. How long it will take an investment of P Riyals to grow to A Riyals in a bank paying an Interest at a rate of r, compounded monthly: a. If P = QR 5000, r= 0.12 per year and A = QR6500 Answer: 2.2 years b. If P = QR 2000, r= 0.09 per year and A = QR4000 Answer: 7.7 years 2. How long it will take an investment of 6000 Riyals to grow to 7000 Riyals in A bank paying an Interest at a rate of 7.5%, compounded continuously. Answer: 2.06 years

  38. b. Finding the interest rate needed for a principal p to grow to an accumulated amount A if invested for t years and the Interest is compounded m times a year. We have: A = p [ 1 + (r/m) ]mt Hence, A / p = [ 1 + (r/m) ]mt , dividing both sides of the equation by p Taking the natural logarithm for both sides, we get: ln(A/p) = ln [ 1 + (r/m) ]mt = mt ln [ 1 + (r/m) ] Thus, ln [ 1 + (r/m) ] = ln(A/p) / mt → 1 + (r/m) = e ln(A/p) / mt → r/m = e ln(A/p) / mt – 1 → r = m [ e ln(A/p) / mt – 1]

  39. Example (7)-b Find the interest rate needed a principal of 1000 Riyals to grow to 1270.24 if deposited in a bank for 3 years and the interest is compounded monthly ? Solution: We have: p=1000, A=1270.24, t=3 and m=12 We can either use the deduced formula or do the calculation from scratch: 1. Using the deduced formula: r = m [ e ln(A/p) / mt – 1] → r = 12 [ e ln(1270.24/1000) / 12(3) – 1] Assignment find r !

  40. 2. Solving from scratch: We have: A = p [ 1 + (r/m) ]mt →1270.24= 1000 [ 1 + (r/12) ]12(3) Hence, 1270.24 / 1000 = [ 1 + (r/12) ]36 , dividing both sides of the equation by p Taking the natural logarithm for both sides, we get: ln(1270.24 /1000) = ln [ 1 + (r/12) ]36 = 36ln [ 1 + (r/12) ] Thus, ln [ 1 + (r/12) ] = ln(1270.24 /1000) / 36 → 1 + (r/12) = e ln(1270.24/1000) / 36 → r/12= e ln(A/p) / 36 – 1 → r = 12 [ e ln(A/p) / 36 – 1] Textbook: Example 9 page 195

  41. Homework Exercises 4.1 Page: 198 Exercise 35

  42. Homework 1. Find the interest rate needed for an investment of 5000 Riyals to grow to 6500 Riyals in 2.2 years If interest is compounded monthly Answer: 0.12 = 12% 1. Find the interest rate needed for an investment of 5000 Riyals to grow to 6000 Riyals in 3 years If interest is compounded continuously Answer: 0.0608 = 6.08%

  43. I.2. Progressions

  44. I-2-01: Arithmetic Progression An arithmetic progression is a sequence of numbers of the form: a, a+d, a+2d, a+3d,…… Where the difference between any term and the one preceding it is a constant d (called the difference). Examples (8) Consider the following sequence: 5, 7, 9, 11, 13, 15, Here the first term a = 5 and the difference d = 2 a2 =a1 + d = 5 + 2 = 7 a3 = a2 + d = 7 + 2 = 9 = 5 + 2(2) = a + 2d a4 = a3 + d = 9 + 2 = 11= 5 + 3(2) = a + 3d a5 = a4 + d = 11 + 2 = 13= 5 + 4(2) = a + 4d In general: an = an-1+ d = a + (n-1)d For this progression: an = 5 + 2 (n-1) For example: a21 = 5 + (21-1)(2) = 5 + 40 = 45

  45. Examples (9) Consider the following sequence: 3, 13, 23, 33, 43, 53, Here the first term a = 3 and the difference d =10 a2 =a1 + d = 3 + 10 = 13 a3 = a2 + d = 13 + 10 = 23 = 3 + 2(10) = a + 2d a4 = a3 + d = 23 + 10 = 33= 3 + 3(10) = a + 3d a5 = a4 + d = 33 + 10 = 43= 3 + 4(10) = a + 4d In general: The n-th term Formula an = a + (n-1)d For this progression: an = 3 + 10 (n-1) For example: a21 = 3 + (21-1)(10) = 3 + 200 = 203 Textbook: Example 1 page 229

  46. Example (10) Find the arithmetic progression with the 21st term is being 45 and the 23rd term 49. Solution: We have: 45 = a + 20d and 49 = a + 22d Subtracting the first equation from the second, we get: 4 = 2d → d =2 Substituting d = 2 n the first equation, we get: 45 = a + 20(2) → a = 45 – 40 = 5 Thus the progression is the one given in example (8), namely: 5, 7, 9, 11, 13, 15,……., 5+2(n-1), ……….. Textbook: Example 2 page 229

  47. Sum of the first n terms of an arithmetic progression Consider the arithmetic progression a, a+d, a+2d, a+3d, a+4d, ….a+(n-2)d, a+(n-1)d, ……. Let s be the sum of the first n terms of the progression. Then: s = a + (a+d) + (a+2d) + (a+3d)+…….+[a + (n-2]d + an s = an + (an -d) + (an -2d) + (an -3d)+…….+[an - (n-2)d] + a → 2s = ( a + an ) + ( a + an ) + ( a + an ) +……+ ( a + an ) = n ( a + an ) = n [a + a + (n-1)d] = n [ 2a + (n-1)d] → s = n [ 2a + (n-1)d] / 2 The sum formula for the first n terms of the arithmetic progression: sn = n [ 2a + (n-1)d] / 2

  48. Example (11) Consider the arithmetic progression of example (8) Find: 1. The the sum of its first five terms and the first ten terms 2. the sum of the term from the sixth to the tenths Solution: 1. We use the formula Sn = n [ 2a + (n-1)d] / 2 From example (8), we have: a=5 and d=2 Thus: S10 = 10 [ 2(5) + (10 -1)2] / 2 = 140 And S5 = 5[ 2(5) + (5 -1)2] / 2 = 45 2. The sum of the terms from a6 to a10 = S10– S5 = 140 – 45 = 95 Textbook: Example 3 page 229, Applied Example (4)

  49. Homework Exercises 4.4 Page: 234 - 235 1. All odd numbered exercise from 1 to 11 And exercise 13* 2. Exercise from 17 and 19

  50. Homework 1. For each of the following arithmetic progressions, find the 31st term and the sum of the first 31 terms: a. 2, 5, 8, 11, 13…. Answers: a31 = 94 and s31 = 1457 b. 1, 2, 3, 4, 5, … Answers: a31 = 32 and s31 = 496 c. -5, -4 ½ , -4, -3 ½ , -3 Answers: a31 = 5 and s31 = 77.5 d. -5, -7, -9, -11,… Answers: a31 = - 70 and s31 = - 1085 e. -5, 5 ½ , -6, -6½ , -7,… Answers: a31 = - 25 and s31 = - 387.5

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