Mathematical Interest Theory-Week 1

Transcription

1 Mathematical Interest Theory-Week 1 Jonathan Curtis September 2016 Contents 0.1 Introduction Chapter 1: The Growth of Money Section 1.3: Accumulation and Amount Functions Section 1.4: Simple Interest Section 1.5: Compound Interest Section 1.6: Effective Discount Rates/ Interest in Advanced

2 0.1 Introduction Although most will be using the free book that can be found on-line, I have already bought the Second Edition copy of Mathematical Interest Theory by Vaaler and Daniel, and this is where all the problems I do will be coming from unless otherwise stated. This book suggests that in order to pass that FM/2 Exam for Actuary the sections that need to be covered are , 1.14, , , 3.11, 3.13, , , , 6.9, 7.1, 8.3, and My goal then for the semester will be to at least roughly complete these sections first, and hopefully more, if time allows. This means I will have to cover about 3-4 sections per week in order to make it through all the sections. Every 2-3 weeks I will also check the sample exams to see if I can complete any of the problems and also complete those on here. From each section I will complete about 3 problems in varying difficult to prove to myself that I understand the material on that section and can move on. In this document I have shown the solutions to the problems that I have done from sections since 1.2 does not have any problems in the book. 2

3 1 Chapter 1: The Growth of Money 1.1 Section 1.3: Accumulation and Amount Functions Problem 1: Given that A K (t) = 1, t for 0 t < 100, find K and a(20). We know the following from a given equation in the book: From this we can find K: K = A K (0) = A K (0) = K 1, 000 1, 000 = = 10 With the value for K we can now find the value of a(20): a(20) = A 1,000 K(20) K = 10 = = 1.25 Problem 5: It is known that for each positive integer k, the amount of interest earned by an investor in the k th period is k. Find the amount of interest earned by the investor from time 0 to n, n a fixed positive integer. Since we know that at every time k the investor earns the interest amount k, which means at k = 1 the investor earns an interest of 1, and at k = 2, an interest of 2 and so on. This leads to the following equation for the amount of interest from time 0 to n: n We know this is equal to n(n+1) 2 so the amount of interest earned from time 0 to n is n(n+1) 2. Problem 6: It is known that for each positive integer k, the amount of interest earned by an investor in the k th period is 2 k. Find the amount of interest earned by the investor from time 0 to n, n a fixed positive integer. This amount of interest for every k th period from 0 to n leads to the following equation: n We know that this is equal to 2(2 n 1), which is the equation for the amount of interest. 3

4 1.2 Section 1.4: Simple Interest Problem 1: How much interest is earned in the fourth year if $1,000 is invested under simple interest at an annual rate of 5%? What is the balance at the end of the fourth year? With simple interest, the amount of interest gained every year is the same regardless of what the new value is after the interest is added, therefore the interest gained in only the fourth year is the same as the interest gained in the first year, which is the following: A $1,000 (t) = $1, 000( t) A $1,000 (1) = $1, 000( (1)) = $1, 000(1.05) = $1, 050 The amount of interest earned in every year= $1, 050 $1, 000 = $50. From the equation above we can also find the balance at the end of 4 years: A $1,000 (4) = $1, 000( (4)) = $1, 000(1.2) = $1, 200 Problem 3: The monthly simple interest rate is 0.5%. simple interest rate? What is the yearly The monthly simple interest rate is 1 12 of the yearly simple interest rate. This means to find the yearly simple interest rate we must multiply the monthly simple interest rate by 12: 0.5%(12) = 6% Problem 5: At a particular rate of simple interest, $1,200 invested at time t = 0 will accumulate to $1,320 in T years. Find the accumulated value of $500 invested at the same rate of simple interest and again at t = 0, but this time for 2T years. The first step in solving this problem is the solve for the interest rate, i, in terms of T from the following: $1, 320 = $1, 200(1 + it ) $1, 320 it = 1 = = 0.1 $1, 200 i = 0.1 T Now we have to use this value for the simple interest rate to solve for the accumulated value of $500 at a time of 2T : A $500 (2T ) = $500( (2T )) = $500( ) = $500(1.2) = $600 T 4

5 1.3 Section 1.5: Compound Interest Problem 3: Horatio invests money in an account earning compound interest at an unknown annual effective interest rate i. His money doubles in 9 years. Find i. To solve this problem, we can set up the compound interest equation for $1 becoming $2 and then solve for i: (1 + i) 9 = 2 i = = = % Problem 6: Sean deposits $826 in a savings account that earns interest at Increasing Rates Bank. For the first three years the money is on deposit, the annual effective interest rate is 3%. For the next two years the annual effective interest rate is 4%, and for the following five years the annual effective interest rate is 5%. What is Sean s balance at the end of ten years? To find the value of the balance at the end of the full ten years we must calculate the balance at each sub-interval where the interest changes and then using the new balance calculated to find the balance after the next sub-interval of years. The following equation expresses this: A $826 (10) = (A $826 (0))(1.03) 3 (1.04) 2 (1.05) 5 This calculates the balance at each sub-interval, and then multiplies it by the next amount of interest, so we can get the balance from this, which is this: A $826 (10) = $826(1.03) 3 (1.04) 2 (1.05) 5 = $1, Since the balance in the account cannot have any more decimal points after the second, there is nothing smaller than a penny, the balance after the ten years will be $1, Problem 7: For a fourteen-year investment, what level annual effective rate of interest gives the same accumulation as an annual effective rate of interest of 5% for eight years followed by a monthly effective rate of interest 0.6% for six years? To find this annual effective interest rate we must solve for i in the following equation: (1 + i) 14 = (1.05) 8 (1.006) 6(12) 1 + i = ((1.05) 8 (1.006) 72 ) 1 14 i = ((1.05) 8 (1.006) 72 ) = This means that in order to have an annual effective rate that equals the same as the tiered compound rate above, the annual effective rate of interest must be around %. 5

6 Problem 9: In 1963, an investor opened a savings account with $K earning simple interest at an annual rate of 2.5%. Four years later, the investor closed the account and invested the accumulated amount in a savings account earning 5% compound interest. Determine the number of years (since 1963) necessary for the balance to reach $3K. The first step to solve this problem is determine what the accumulated total of the investors savings in the simple interest account was after the first four years: A $K (4) = $K(1 + 4(0.025)) = $K( ) = $1.1K The next step to solve this problem is to solve for the amount of time it takes this accumulated value to reach the $3K that is desired from the following equation: $3K = $1.1K(1.05) t = (1.05)t ln(1.05) t = ln( ) t(ln(1.05)) = ln( ) t = ln( ln(1.05) = Since you need the full year it must be rounded up to 21 years and the 4 years from before must be added as well, which means the total number of years, since 1963, that it takes for the $K to grow to $3K in this circumstance is 25 years. 6

7 1.4 Section 1.6: Effective Discount Rates/ Interest in Advanced Problem 3: Jonathan borrows $1,450 for one year at a discount rate of D. He has the use of an extra $1,320. Find D and the annual interest rate that this is equivalent to. To find the discount rate we must use the following: $1, 320 = $1, 450(1 D) 1 D = D = = = % 1450 We then use this discount rate to find the interest rate it is equivalent to: i = (1 D) 1 1 = ( ) 1 1 = = = % 1320 Problem 5: A savings account earns compound interest at an annual effective interest rate i. Given that i [2,4.5] = 20% find d [1,3]. When we try to solve for any d between any time intervals, we need a (1 + i), which is what we will solve for first given the i [2,4.5] = 20%: 1.2 = 1 + i [2,4.5] = (1 + i) = (1 + i) 2.5 ) (1 + i) = (1.2) = (1.2) 0.4 Now that we have a value for (1 + i) we can solve for the d [1,3] : d [1,3] = 1 (1 + i [ 1, 3]) 1 = 1 (1 + i) 2( 1) = 1 (1 + i) 2 = 1 (1.2) 0.4( 2) = 1 (1.2) 0.8 = = % 7