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## Chapter I

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**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)**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**Graphing the straight line:A(t) = 2000 + 2000(0.01)t ; t**≥ 0**Homework**Exercises 4.1 Page: 197 All odd numbered exercises from 1 to 9**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 )**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**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**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**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**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**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**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**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)**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**Homework**Exercises 4.1 Page: 197 All odd numbered exercises from 11 to 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**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**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)**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**Homework**Exercises 4.1 Page: 198 Exercise 29**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**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**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**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**Homework**Exercises 4.1 Page: 197 Exercises 21 and 23**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**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**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**Homework**Exercises 4.1 Page: 197 - 198 Exercises 25, 27**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**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) ]**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**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**Homework**Exercises 4.1 Page: 198 Exercises: 31, 33 and 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**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]**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 !**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**Homework**Exercises 4.1 Page: 198 Exercise 35**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%**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**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**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**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**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)**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**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