Simple and Compound Interest PDF

## LECTURE 2 ### SIMPLE AND COMPOUND INTEREST #### 2.1 Introduction Everybody uses money. Sometimes you work for your money and other times your money works for you. In the case of money placed into a savings account at your local bank, your money is affected positively by the interest the bank aff...

## LECTURE 2 ### SIMPLE AND COMPOUND INTEREST #### 2.1 Introduction Everybody uses money. Sometimes you work for your money and other times your money works for you. In the case of money placed into a savings account at your local bank, your money is affected positively by the interest the bank affords and negatively by the inflation rate. If i is the given interest rate per period, then ₦1 now will be worth 1+i naira one period hence - A point of departure is whether or not the interest computation for periods 2, 3, ... involves the interest earned in previous periods. This leads to two computation formulae: simple interest and compound interest. i.e. #### 2.2 Simple Interest Interest on loans in a year or less is frequently calculated as simple interest. This type of interest charged/or paid only on the amount borrowed/invested and not on past interest. In another explanation, the simple interest is only applied to the original sum of money invested/borrowed. For example: an amount P accumulates to $P + Pc = P(1 + i)$ at the end of one period, and further accumulates to $P + 2Pc = P(1 + 2i)$ at the end of the two periods, and on, and P accumulates to $P(1 + in)$ at the end of n periods. ##### Example 1: Mrs Davide borrowed ₦125,000 at 8% simple interest for 11 months. How much interest will she pay? **Solution:** Since 8% is the interest rate per year, then 11 months is converted to a year as = 11/12 yrs. rate, $r = 8% = 0.08$ Then, the total interest Mrs Davide will pay is $I = Prt$ $=125,000 × 0.08 × 11/12$ $= ₦9,166.67$ #### Future Value A deposit of ₦P today at a rate of interest r for time t in years produces interest of I = prt. By adding the interest to the original principal P gives: $A = p + prt = p(1 + rt)$ The amount is known as the future value of ₦P at an interest rate r for time t in years, but we are actually dealing with loan, the future value is then called the maturity value of the loan. #### Example 2: Mr Uno took a loan of ₦250,000 from Alabuto Cooperative Society (ACS), and the loan was granted at the interest rate of 4.3%, to be repaid within 8 months. What is the maturity value of the loan. **Solution:** $A = p (1 + rt)$ $P = ₦ 250,000$ $r = 4.3% = 0.043$ $t = 8/12 = 2/3$ Then, $A = 250,000 × \left[ 1 + (0.043 × 2/3) \right]$ $A = 250,000 × (1 + 0.02867)$ $A = 250,000 × 1.02867$ $A = ₦257,167.50$ #### Example 3: Alice wants to borrow ₦8000 from Dar Yu. She agrees to payback ₦8180 in 6 months, what interest rate will she pay? **Solution:** $A = P(1 + rt)$ $A = 8180$ $t = 6/12 = 0.5$ $P = 8000$ ⇒ $8180 = 8000 × (1 + 0.5r)$ ⇒ $\frac{8180}{8000} = 1+0.5r$ ⇒ $0.5r = \frac{8180}{8000}-1$ ⇒ $0.5r = 1.0225 -1$ ⇒ $0.5r = 0.0225$ ⇒ $r = \frac{0.0225}{0.5}$ ⇒ $r = 0.045$ Interest rate = $r = 4.5%$ #### Compound Interest For compound interest, the interest computation at the end of a given period involves the interest already accrued or earned in previous periods as part of the balance. Thus, the principal ₦P today is worth: $A = P(1 + i)ⁿ = P(1 + i)ⁿ⁻¹ +P(1 + i)ⁿ⁻²$ in periods hence. The factor 1+i is called the accumulation factor in that multiplying a balance P by $(1 + i)ⁿ$ allows us to accumulate P forward in time in n years. #### Derivation of Compound Interest Formula Firstly, suppose that ₦P is deposited at a rate of interest (r) per year. The amount on deposit at the end of the first year is found by the simple interest formula with t = 1: $A = P (1 + r (1))$ $= P(1 + r)$ If the deposit earns a compound interest, the interest is paid on the total amount deposited on the total amount invested after the first year. Then: $A = P(1+r) \left\{1 + r(1) \right\}$ $= P(1 +r)²$ This is the interest at the end of the second year. Over and over after the third year ending, $A = P(1 + r)³$ By generalizing this, at the end of t years, the total amount on deposit is: $A = P (1 + r)t$ Called the compounded amount. By comparison: * Compound interest $A = P (1 + r) t$ * Simple interest $A = P (1 + rt)$ Interest can be compounded more than once in a year. take it semi-annually (twice a year), quarterly (four times a year), monthly (12 times a year), etc. The interest rate per period, i, is found by dividing the annual interest rate, r, by the number of compounding period, n per year. To find the total number of compounding period, we multiply the number of years t by the number of compounding periods per year, m. **Compounded Amount:** $A = pC1 + i)ⁿ$ where $i = \frac{r}{m}$ and $n = mt$ * $A$ is the future (maturity value) * $P$ is the principal * $r$ is the annual interest rate * $m$ is the number of compounding period per year * $t$ is the number of years * $n$ is the number of compounding period * $i$ is the interest rate per period. #### Example 4: If $3500 is invested at 9% compounded monthly, what will the future value be in four years? **Solution:** $A = p(1+i)ⁿ$ $P = $3500$ $r = 9% = 0.09$ Clearly, this interest has to be paid every month, and if it is compounded, then, $i = \frac{r}{m} = \frac{0.09}{12} = 0.0075$ and n = 48, the number of compounding periods. $A = 3500 (1 + 0.0075)^{48}$ $=3500 × 1.0075^{48}$ $= $5009.92$ ∴ $3500 invested at 9% compounded monthly will yield $5009.92 in four years. #### Example 5: What principal invested at 9% compounded monthly will yield $5000 in four years? **Solution:** $A = $5000$ $r = 9% = 0.09$ $t = 4$ years $m = 12$ months Then, $n = mt = 4 × 12 = 48$ $i = \frac{r}{m} = \frac{0.09}{12} = 0.0075$ $A = P(1+i)ⁿ$ ⇒ $5000 = P (1+0.0075)^{48}$ ⇒ $P = \frac{5000}{(1+0.0075)^{48}}$ $P = \frac{5000}{1.5882250716}$ $P = $3188.32$ #### Example 6: If $4000 is invested at 4% compounded annually, how long will it take to accumulate to $6000? **Solution:** $A = P(1 + i)ⁿ$ $i = \frac{r}{m} = \frac{r}{1} = r = 4.2% = 0.04$ $P = $4000$ $A = $6000$ ∴ $6000 = 4000 (1 + 0.04)ⁿ$ $1.5 = (1.04)ⁿ$ $\frac{6000}{4000} = (1.04)ⁿ$ $log_{1.04} 1.5 = log_{1.04} (1.04)ⁿ$ $log_{1.04} 1.5 = n$ $n = 10.33$ years #### Effective Interest Rate Banks are required to state their interest rate in terms of an Effective yield or effective interest rate for comparison purposes. The effective rate is also called the Annual Percentage Yield (APY) or Annual Percentage Rate (APR). The effective rate is the interest rate compounded annually, it would be equivalent to the stated rate and compounding periods. For example, if Bank A pays 7.2% interest compounded monthly, what is the effective interest rate? If Bank B pays 7.25% interest compounded semi-annually, what is the effective interest rate? Which Bank pays more interest? **Solution:** * **Bank A**: Suppose we wish to deposit $1 in this bank and leave it for a year, then the accrued amount will be: $1(1+0.072) ^{12 × 1}$ =$(1+0.006)^{12}$ $= 1.006^{12}$ $= 1.0744$ Thus, interest of $(1.0744 - 1) = $0.0744 is earned while investing $1. $r_{EFF} = $ the effective interest rate = $\frac{0.0744}{1} × 100$ $r_{EFF} = 7.449%$ * **Bank B**: Interest is paid semi-annually; then the amount accrued for a year in the same $1 is: $1 (1 + \frac{0.0725}{2})^{2}$ $= (1 + 0.03625)^{2}$ $= 1.03625^{2}$ $= 1.073814$ $r_{EFF}$ Interest, $m = 1.073814 - 1$ $= 0.0738$ Thus $0.0738 is earned on every $1. $r_{EFF} = \frac{0.0738 }{1} × 100$ $= 7.138%$ So, Bank A pays slightly higher interest. #### The Theory of Interest A nominal interest rate j refers to an annual interest rate where interest calculation is done more frequently than once a year, in a compounding rule. When the annual rate j is given as a nominal rate compounded mthly, what is meant is that interest is compounded at the rate i/m in every mth of the years and this i/m is the effective rate for each mth. So, at the end of k number of mth of a year, P accumulates to $P(1 + i/m)^{mk}$ by the formula for compounded interest. By assuming that j is a nominal rate with mthly compounding, and if j is equivalent to an effective rate i, then, $1 + i = (1 + \frac{j}{m})^m$ where i is the effective rate at an annual rate. Solving for j in the above leads to, $j = m\left\{(1+i)^{1/m }- 1 \right\}$ Conversely, if $j = m\left\{(1+i)^{1/m }- 1 \right\}$, then j satisfies $1+i = (1+\frac{j}{m})^m$ and ₦1 accumulates to $(1+i)^t$ after t years. $1 + i = (1 + \frac{j}{m})^{mt}$ Note, if you start with ₦P and apply the effective rate i, then the balance at the end of n years is $P(1 + i)^n$. Conversely, if you have ₦Q for n years after depositing ₦P, then, $P(1+i)ⁿ = Q$ So that $P = \frac{Q}{(1 + i)^n} = v^n Q$ Where $v = \frac{1}{1+i}$. We name v = $\frac{1}{1+i}$ the Discount factor corresponding to i. The accumulation factor 1 + i allows investors to more money forward, while the discount factor $v = \frac{1}{1+i}$ allows one to draw money back in time. Recall, 1 + i is the accumulation factor. The number $δ = lim_{m→∞} δ (m)$ has a meaning, i.e., interest rate assuming continuous compounding, corresponding to the effective rate i. The δ is the force of interest. By L'Hospital Rule, $δ(m) = \frac{(1+i/m)^m -1}{1/m}$ and it converges to $δ = ln(1+i) = -lnv$. Thus, the accumulation factor is also derived as $1+i = e^δ$ And the accumulated savings after t years from a principal P is then put as, $P(1 + i)t = Pe^{δt}$. #### Revision Question 1. Given that j being the nominal rate and i being the effective rate at an annual rate, show that $j = m[(1+i)^{1/m} - 1]$. 2. For a principal P, compounded monthly for t years, show that the amount accrued A is the same as the amount accrued at i effective rate, and $P = \frac{A}{(1+i)^{n}}$. 3. Given the effective interest rate i in $1+i = (1+\frac{j}{m})^{m}$, show that by L'Hospital rule, the above equality holds, and that $δ' = ln(1+i)=-lnv$ hence $1+i=e^δ$. 4. Give detailed explanation: Annuities (certain) - Annuities - Due and Annuities - Immediates.