#set document(title: "6.4 Compound Interest", author: "OpenStax / XYZ Homework") #set page(width: 8.5in, height: auto, margin: 1in) #import "@preview/cetz:0.5.2" #set text(font: ("STIX Two Text", "Libertinus Serif", "New Computer Modern"), size: 10.5pt, lang: "en") #show math.equation: set text(font: ("STIX Two Math", "New Computer Modern Math")) #set par(justify: true, leading: 0.62em, spacing: 0.9em) #set enum(spacing: 1.1em) // room between list items so tall inline fractions don't collide #set list(spacing: 1.1em) #set table(stroke: 0.5pt + rgb("#c7ccd3")) #let BLUE = rgb("#183B6F") // brand navy — section bars + example/solution labels (white on navy 11.09:1) #let ORANGE = rgb("#A94509") // brand primary-700 — AA-safe deep orange for TEXT (5.93:1 on white; raw brand #F37021 is 2.94:1 and must never carry text) #let RED = rgb("#DC2626") // brand error-600 #let GREEN = rgb("#059669") // brand success-600 (decoration only; small green text uses green-text #007942) #show heading.where(level: 1): it => block(width: 100%, above: 0pt, below: 16pt, fill: gradient.linear(BLUE, rgb("#2C5AA0")), inset: (x: 14pt, y: 12pt), radius: 3pt, text(fill: white, weight: "bold", size: 19pt, it.body)) #show heading.where(level: 2): it => block(width: 100%, above: 18pt, below: 10pt, fill: BLUE, inset: (x: 10pt, y: 6pt), radius: 2pt, text(fill: white, weight: "bold", size: 12pt, it.body)) #show heading.where(level: 3): it => text(fill: ORANGE, weight: "bold", size: 12.5pt, it.body) #show heading.where(level: 4): it => text(fill: BLUE, weight: "bold", size: 10.5pt, it.body) #let examplebox(label, title, body) = block(width: 100%, breakable: true, fill: rgb("#EFF1F5"), stroke: 0.5pt + rgb("#CFDDF0"), radius: 4pt, inset: 10pt, above: 12pt, below: 12pt)[ #block(below: 6pt)[#box(fill: BLUE, inset: (x: 6pt, y: 2pt), radius: 2pt, text(fill: white, weight: "bold", size: 8.5pt, label)) #h(0.4em) #strong[#title]] #body] // rail = decorative left rule (raw brand token); labelcolor = AA-safe label text shade #let notebox(label, rail, labelcolor, tint, body) = block(width: 100%, breakable: true, fill: tint, stroke: (left: 3pt + rail), inset: (left: 10pt, rest: 8pt), radius: (right: 4pt), above: 11pt, below: 11pt)[ #text(fill: labelcolor, weight: "bold", size: 7.5pt, tracking: 0.5pt)[#upper(label)] #linebreak() #body] #let solutionbox(body) = block(above: 4pt, below: 8pt)[ #text(fill: BLUE, weight: "bold", size: 8.5pt)[Solution] #linebreak() #body] #let figph(msg) = block(width: 100%, height: 60pt, fill: rgb("#f6f7f9"), stroke: (paint: rgb("#c7ccd3"), dash: "dashed"), radius: 4pt, inset: 10pt)[ #align(center + horizon, text(fill: rgb("#889"), style: "italic", size: 9pt, msg))] // Standardize inlined figure sizes: measure the natural CeTZ canvas, then scale to a // consistent envelope (aspect-aware; see build_typst.py FIG_* constants). Unlike the // print preamble, dimensions are FLOORED: in an editor a user can trim a figure to a // degenerate 1-D shape (a bare line), and w/h or tw/w would then divide by zero. #let _STD_W = 3.5 #let _WIDE_W = 5.6 #let _MAX_H = 3.4 #let _ASPECT_WIDE = 2.2 #let _UPSCALE_MAX = 1.15 #let stdfig(body) = context { let m = measure(body) let w = calc.max(m.width / 1in, 0.01) let h = calc.max(m.height / 1in, 0.01) let tw = if w / h > _ASPECT_WIDE { _WIDE_W } else { _STD_W } let s = calc.min(tw / w, _MAX_H / h, _UPSCALE_MAX) align(center, box(scale(x: s * 100%, y: s * 100%, reflow: true, body))) } #show figure: set block(breakable: false) #set figure(gap: 8pt) #show figure.caption: set text(size: 8.5pt, fill: rgb("#555")) == 6.4#h(0.6em)Compound Interest #figure(figph[Coins are placed in stacks ordered from smallest to largest.], alt: "Coins are placed in stacks ordered from smallest to largest.", caption: [The impact of compound interest]) === Learning Objectives After completing this section, you should be able to: + Compute compound interest. + Determine the difference in interest between simple and compound calculations. + Understand and compute future value. + Compute present value. + Compute and interpret effective annual yield. For a very long time in certain parts of the world, interest was not charged due to religious dictates. Once this restriction was relaxed, loans that earned interest became possible. Initially, such loans had short terms, so only simple interest was applied to the loan. However, when loans began to stretch out for years, it was natural to add the interest at the end of each year, and add the interest to the principal of the loan. After another year, the interest was calculated on the initial principal plus the interest from year 1, or, the interest earned interest. Each year, more interest was added to the money owed, and that interest continued to earn interest. Since the amount in the account grows each year, more money earns interest, increasing the account faster. This growth follows a geometric series (Geometric Sequences). It is this feature that gives compound interest its power. This module covers the mathematics of compound interest. === Understand and Compute Compound Interest As we saw in Simple Interest, an account that pays simple interest only pays based on the original principal and the term of the loan. Accounts offering #strong[compound interest] pay interest at regular intervals. After each interval, the interest is added to the original principal. Later, interest is calculated on the original principal plus the interest that has been added previously. After each period, the interest on the account is computed, then added to the account. Then, after the next period, when interest is computed, it is computed based on the original principal AND the interest that was added in the previous periods. The following example illustrates how compounded interest works. #examplebox("Example 1")[Interest Compounded Annually][ Abena invests \$1,000 in a CD (certificate of deposit) earning 4% compounded annually. How much will Abena’s CD be worth after 3 years? #solutionbox[ Since the interest is compounded annually, the interest will be computed at the end of each year and added to the CD’s value. The interest at the end of the following year will be based on the value found form the previous year. #strong[Step 1:] After the first year, the interest in Abena’s CD is computed using the interest formula #math.equation(block: false, alt: "I equals P times r times t")[$I = P × r × t$]. The principal is #math.equation(block: false, alt: "P")[$P$] = 1,000, the rate, as a decimal, is 0.04, and the time is one year, so #math.equation(block: false, alt: "t")[$t$] = 1. Using that, the interest earned in the first year is #math.equation(block: false, alt: "I equals P times r times t equals 1,000 times 0.04 times 1 equals 40")[$I = P × r × t = 1,000 × 0.04 × 1 = 40$], so the interest earned in the first year was \$40.00. This is added to the value of the CD, making the CD worth #math.equation(block: false, alt: "$ 1,000 plus $ 40 equals $ 1,040")[$"$" 1,000 + "$" 40 = "$" 1,040$]. #strong[Step 2:] At the end of the second year, interest is again computed, but is computed based on the CD’s new value, \$1,040. Using this new value and the interest formula (#math.equation(block: false, alt: "r")[$r$] and #math.equation(block: false, alt: "t")[$t$] are still 0.04 and 1, respectively), we see that the CD earned #math.equation(block: false, alt: "I equals P times r times t equals 1,040 times 0.04 times 1 equals 41.6")[$I = P × r × t = 1,040 × 0.04 × 1 = 41.6$], or \$41.60. This is added to the value of the CD, making the CD now worth #math.equation(block: false, alt: "$ 1,040.00 plus $ 41.60 equals $ 1,081.60")[$"$" 1,040.00 + "$" 41.60 = "$" 1,081.60$]. #strong[Step 3:] At the end of the third year, interest is again computed, but is computed based on Abena’s CD’s new value, \$1,081.60. Using this value and the interest formula (#math.equation(block: false, alt: "r")[$r$] and #math.equation(block: false, alt: "t")[$t$] are still 0.04 and 1, respectively), we see that the CD earned #math.equation(block: false, alt: "I equals P times r times t equals 1,081.60 times 0.04 times 1 equals 43.264")[$I = P × r × t = 1,081.60 × 0.04 × 1 = 43.264$], or \$43.26 (remember to round down). This is added to the value of the CD, making the CD now worth #math.equation(block: false, alt: "$ 1,081.60 plus $ 43.26 equals $ 1,124.86")[$"$" 1,081.60 + "$" 43.26 = "$" 1,124.86$]. After 3 years, Abena’s CD is worth \$1,124.86. ] ] === Determine the Difference in Interest Between Simple and Compound Calculations It is natural to ask, does compound interest make much of a difference? To find out, we revisit Abena’s CD. #examplebox("Example 2")[Comparing Simple to Compound Interest on a 3-Year CD][ Abena invested \$1,000 in a CD that earned 4% compounded annually, and the CD was worth \$1,124.86 after 3 years. Had Abena invested in a CD with simple interest, how much would the CD have been worth after 3 years? How much more did Abena earn using compound interest? #solutionbox[ Had Abena invested \$1,000 in a 4% simple interest CD for 3 years, her CD would have been worth #math.equation(block: false, alt: "P plus P times r times t equals 1,000 plus 1,000 times 0.04 times 3 equals 1,120")[$P + P × r × t = 1,000 + 1,000 × 0.04 × 3 = 1,120$], or \$1,120.00. With interest compounded annually, Abena’s CD was worth \$1,124.86. The difference between compound and simple interest is #math.equation(block: false, alt: "$ 1,124.86 minus $ 1,120.00 equals $ 4.86")[$"$" 1,124.86 − "$" 1,120.00 = "$" 4.86$]. So compound interest earned Abena \$4.86 more than the simple interest did. ] ] #notebox("Video", rgb("#DC2626"), rgb("#DC2626"), rgb("#f7f8fa"))[ #link("https://openstax.org/r/compound_interest_beginners")[Compound Interest] ] === Understand and Compute Future Value Imagine investing for 30 years and compounding the interest every month. Using the method above, there would be 360 periods to calculate interest for. This is not a reasonable approach. Fortunately, there is a formula for finding the future value of an investment that earns compound interest. #notebox("Formula", rgb("#059669"), rgb("#007942"), rgb("#EAF3EC"))[ The future value of an investment, #math.equation(block: false, alt: "A")[$A$], when the principal #math.equation(block: false, alt: "P")[$P$] is invested at an annual interest rate of #math.equation(block: false, alt: "r")[$r$] (in decimal form), compounded #math.equation(block: false, alt: "n")[$n$] times per year, for #math.equation(block: false, alt: "t")[$t$] years, is found using the formula #math.equation(block: false, alt: "A equals P open parenthesis 1 plus the fraction r over n close parenthesis to the power n t")[$A = P attach(( 1 + frac(r, n) ), t: n t)$]. This is also referred to as the future value of the investment. ] #notebox("Checkpoint", rgb("#059669"), rgb("#007942"), rgb("#EAF3EC"))[ #emph[Note, sometimes the formula is presented with the total number of periods, #math.equation(block: false, alt: "n")[$n$], and the interest rate per period, #math.equation(block: false, alt: "r")[$r$]. In that case the formula becomes] #math.equation(block: false, alt: "A equals P open parenthesis 1 plus r close parenthesis to the power n")[$A = P attach(( 1 + r ), t: n)$]. ] #examplebox("Example 3")[Computing Future Value for Compound Interest][ In the following, compute the future value of the investment with the given conditions. + Principal is \$5,000, annual interest rate is 3.8%, compounded monthly, for 5 years. + Principal is \$18,500, annual interest rate is 6.25%, compounded quarterly, for 17 years. #solutionbox[ + The principal is #math.equation(block: false, alt: "P")[$P$] = \$5,000, interest rate, in decimal form, #math.equation(block: false, alt: "r")[$r$] = 0.038, compounded monthly so #math.equation(block: false, alt: "n")[$n$] = 12, and for #math.equation(block: false, alt: "t")[$t$] = 5 years. Substituting these values into the formula, we find #math.equation(block: true, alt: "A, equals, P open parenthesis 1 plus the fraction r over n close parenthesis to the power n t equals 5 , 000 open parenthesis 1 plus the fraction 0.038 over 12 close parenthesis to the power 12 times 5; equals, 5 , 000 open parenthesis 1 plus 0.0031 6 ¯ close parenthesis to the power 60; equals, 5 , 000 open parenthesis 1.0031 6 ¯ close parenthesis to the power 40; equals, 5 , 000 times 1.20888663572; equals, 6 , 044.4332")[$A & = & P attach(( 1 + frac(r, n) ), t: n t) = 5 , 000 attach(( 1 + frac(0.038, 12) ), t: 12 × 5) \ & = & 5 , 000 attach(( 1 + 0.0031 overline(6) ), t: 60) \ & = & 5 , 000 attach(( 1.0031 overline(6) ), t: 40) \ & = & 5 , 000 × 1.20888663572 \ & = & 6 , 044.4332$] #linebreak() The future value of the investment is \$6,044.43. #linebreak() + The principal is #math.equation(block: false, alt: "P")[$P$] = \$18,500, interest rate, in decimal form, #math.equation(block: false, alt: "r")[$r$] = 0.0625, compounded quarterly so #math.equation(block: false, alt: "n")[$n$] = 4, and for #math.equation(block: false, alt: "t")[$t$] = 17 years. Substituting these values into the formula, we #math.equation(block: true, alt: "A, equals, P open parenthesis 1 plus the fraction r over n close parenthesis to the power n t equals 18 , 500 open parenthesis 1 plus the fraction 0.0625 over 4 close parenthesis to the power 4 times 17; equals, 18 , 500 open parenthesis 1 plus 0.015625 close parenthesis to the power 68; equals, 18 , 500 open parenthesis 1.015625 close parenthesis to the power 68; equals, 18 , 500 times 2.86992151999; equals, 53 , 093.5481")[$A & = & P attach(( 1 + frac(r, n) ), t: n t) = 18 , 500 attach(( 1 + frac(0.0625, 4) ), t: 4 × 17) \ & = & 18 , 500 attach(( 1 + 0.015625 ), t: 68) \ & = & 18 , 500 attach(( 1.015625 ), t: 68) \ & = & 18 , 500 × 2.86992151999 \ & = & 53 , 093.5481$] The future value of the investment is \$53,093.54. ] ] #examplebox("Example 4")[Interest Compounded Quarterly][ Cody invests \$7,500 in an account that earns 4.5% interest compounded quarterly (4 times per year). Determine the value of Cody’s investment after 10 years. #solutionbox[ Cody’s initial investment is \$7,500, so #math.equation(block: false, alt: "P")[$P$] = \$7,500. The annual interest rate is 4.5%, which is 0.045 in decimal form. Compounding quarterly means there are four periods in a year, so #math.equation(block: false, alt: "n")[$n$] = 4. He invests the money for 10 years. Substituting those values into the formula, we calculate #math.equation(block: true, alt: "A, equals, P open parenthesis 1 plus the fraction r over n close parenthesis to the power n t equals 7 , 500 open parenthesis 1 plus the fraction 0.045 over 4 close parenthesis to the power 4 times 10; equals, 7 , 500 open parenthesis 1 plus 0.01125 close parenthesis to the power 40; equals, 7 , 500 open parenthesis 1.01125 close parenthesis to the power 40; equals, 7 , 500 times 1.564376865391; equals, 11 , 732.8265")[$A & = & P attach(( 1 + frac(r, n) ), t: n t) = 7 , 500 attach(( 1 + frac(0.045, 4) ), t: 4 × 10) \ & = & 7 , 500 attach(( 1 + 0.01125 ), t: 40) \ & = & 7 , 500 attach(( 1.01125 ), t: 40) \ & = & 7 , 500 × 1.564376865391 \ & = & 11 , 732.8265$] After 10 years, Cody’s initial investment of \$7,500 is worth \$11,732.82. ] ] #examplebox("Example 5")[Interest Compounded Daily][ Kathy invests \$10,000 in an account that yields 5.6% compounded daily. How much money will be in her account after 20 years? #solutionbox[ Kathy’s initial investment is \$10,000, so #math.equation(block: false, alt: "P")[$P$] = \$10,000. The annual interest rate is 5.6%, which is 0.056 in decimal form. Compounding daily means there are 364 periods in a year, so #math.equation(block: false, alt: "n")[$n$] = 365. She invests the money for 20 years, so #math.equation(block: false, alt: "t")[$t$] = 20. Substituting those values into the formula, we calculate #math.equation(block: true, alt: "A, equals, P open parenthesis 1 plus the fraction r over n close parenthesis to the power n t equals 10,000 open parenthesis 1 plus the fraction 0.056 over 365 close parenthesis to the power 365 times 20; equals, 10,000 open parenthesis 1 plus 0.000153424657534 close parenthesis to the power 7300; equals, 10,000 open parenthesis 1.000153424657534 close parenthesis to the power 7300; equals, 10,000 times 3.06459091598; equals, 30,645.909")[$A & = & P attach(( 1 + frac(r, n) ), t: n t) = 10,000 attach(( 1 + frac(0.056, 365) ), t: 365 × 20) \ & = & 10,000 attach(( 1 + 0.000153424657534 ), t: 7300) \ & = & 10,000 attach(( 1.000153424657534 ), t: 7300) \ & = & 10,000 × 3.06459091598 \ & = & 30,645.909$] After 20 years, Kathy’s initial investment of \$10,000 is worth \$30,645.90. ] ] #notebox("Video", rgb("#DC2626"), rgb("#DC2626"), rgb("#f7f8fa"))[ - #link("https://openstax.org/r/compare_simple_compound_interest1")[Compare Simple Interest to Interest Compounded Annually] - #link("https://openstax.org/r/compare_simple_compound_interest2")[Compare Simple Interest and Compound Interest for Different Number of Periods Per Year] ] #notebox("Work It Out", rgb("#059669"), rgb("#007942"), rgb("#EAF3EC"))[ To truly grasp how compound interest works over a long period of time, create a table comparing simple interest to compound interest, with different numbers of periods per year, for many years would be useful. In this situation, the principal is \$10,000, and the annual interest rate is 6%. + Create a table with five columns. Label the first column YEARS, the second column SIMPLE INTEREST, the third column COMPOUND ANNUALLY, the fourth column COMPOUND MONTHLY and the last column COMPOUND DAILY, as shown below. #figure(table( columns: 5, align: left, inset: 6pt, table.header([YEARS], [SIMPLE INTEREST], [COMPOUND ANNUALLY], [COMPOUND MONTHLY], [COMPOUND DAILY]), [], [], [], [], [], [], [], [], [], [], [], [], [], [], [], )) + In the years column, enter 1, 2, 3, 5, 10, 20, and 30 for the rows. + Calculate the account value for each column and each year. + Compare the results from each of the values you find. How do the number of periods per year (compoundings per year) impact the account value? How does the number of years impact the account value? + Redo the chart, with an interest rate you choose and a principal you choose. Are the patterns identified earlier still present? ] === Understand and Compute Present Value When investing, there is often a goal to reach, such as “after 20 years, I’d like the account to be worth \$100,000.” The question to be answered in this case is “How much money must be invested now to reach the goal?” As with simple interest, this is referred to as the present value. #notebox("Formula", rgb("#059669"), rgb("#007942"), rgb("#EAF3EC"))[ The money invested in an account bearing an annual interest rate of #math.equation(block: false, alt: "r")[$r$] (in decimal form), compounded #math.equation(block: false, alt: "n")[$n$] times per year for #math.equation(block: false, alt: "t")[$t$] years, is called the present value, #math.equation(block: false, alt: "P V")[$P V$], of the account (or of the money) and found using the formula #math.equation(block: false, alt: "P V equals the fraction A over open parenthesis 1 plus the fraction r over n close parenthesis to the power n times t")[$P V = frac(A, attach(( 1 + frac(r, n) ), t: n × t))$], where #math.equation(block: false, alt: "A")[$A$] is the value of the account at the investment’s end. Always round this value up to the nearest penny. ] #examplebox("Example 6")[Computing Present Value][ Find the present value of the accounts under the following conditions. + #math.equation(block: false, alt: "A")[$A$] = \$250,000, invested at 6.75 interest, compounded monthly, for 30 years. + #math.equation(block: false, alt: "A")[$A$] = \$500,000, invested at 7.1% interest, compounded quarterly, for 40 years. #solutionbox[ + To reach a final account value of #math.equation(block: false, alt: "A")[$A$] = \$250,000, invested at 6.75% interest, in decimal form #math.equation(block: false, alt: "r")[$r$] = 0.0675 (decimal form!), compounded monthly, so #math.equation(block: false, alt: "n")[$n$] = 12, for 30 years, substitute those values into the formula for present value. Calculating, we find the present value of the \$250,000. #linebreak() #math.equation(block: true, alt: "P V, equals, the fraction A over open parenthesis 1 plus the fraction r over n close parenthesis to the power n times t equals the fraction 250,000 over open parenthesis 1 plus the fraction 0.0675 over 12 close parenthesis to the power 12 times 30; equals, the fraction 250,000 over open parenthesis 1 plus 0.005625 close parenthesis to the power 360; equals, the fraction 250,000 over open parenthesis 1.005625 close parenthesis to the power 360; equals, the fraction 250,000 over 7.5332454772; equals, 33,186.2277")[$P V & = & frac(A, attach(( 1 + frac(r, n) ), t: n × t)) = frac(250","000, attach(( 1 + frac(0.0675, 12) ), t: 12 × 30)) \ & = & frac(250","000, attach(( 1 + 0.005625 ), t: 360)) \ & = & frac(250","000, attach(( 1.005625 ), t: 360)) \ & = & frac(250","000, 7.5332454772) \ & = & 33,186.2277$] #linebreak() In order for this account to reach \$250,000 after 30 years, \$33,186.23 needs to be invested. #linebreak() + To reach a final account value of #math.equation(block: false, alt: "A")[$A$] = \$500,000, invested at 7.1% interest, in decimal form #math.equation(block: false, alt: "r")[$r$] = 0.071, compounded quarterly, so #math.equation(block: false, alt: "n")[$n$] = 4, for 40 years, substitute those values into the formula for present value. Calculating, we find the present value of the \$500,000. #linebreak() #math.equation(block: true, alt: "P V, equals, the fraction A over open parenthesis 1 plus the fraction r over n close parenthesis to the power n times t equals the fraction 500 , 000 over open parenthesis 1 plus the fraction 0.071 over 4 close parenthesis to the power 4 times 40; equals, the fraction 500 , 000 over open parenthesis 1 plus 0.01775 close parenthesis to the power 160; equals, the fraction 500 , 000 over open parenthesis 1.01775 close parenthesis to the power 160; equals, the fraction 500 , 000 over 16.6946672846; equals, 29 , 949.6834")[$P V & = & frac(A, attach(( 1 + frac(r, n) ), t: n × t)) = frac(500 "," 000, attach(( 1 + frac(0.071, 4) ), t: 4 × 40)) \ & = & frac(500 "," 000, attach(( 1 + 0.01775 ), t: 160)) \ & = & frac(500 "," 000, attach(( 1.01775 ), t: 160)) \ & = & frac(500 "," 000, 16.6946672846) \ & = & 29 , 949.6834$] #linebreak() In order for this account to reach \$500,000 after 40 years, \$29,949.69 needs to be invested. ] ] #examplebox("Example 7")[Investment Goal with Compound Interest][ Pilar plans early for retirement, believing she will need \$1,500,000 to live comfortably after the age of 67. How much will she need to deposit at age 23 in an account bearing 6.35% annual interest compounded monthly? #solutionbox[ Knowing how much to deposit at age 23 to reach a certain value later is a present value question. The target value for Pilar is \$1,500,000. The interest rate is 6.35%, which in decimal form is 0.0635. Compounded monthly means #math.equation(block: false, alt: "n")[$n$] = 12. She’s 23 and will leave the money in the account until the age of 67, which is 44 years, making #math.equation(block: false, alt: "t")[$t$] = 44. Using this information and substituting in the formula for present value, we calculate #math.equation(block: true, alt: "P V, equals, the fraction A over open parenthesis 1 plus the fraction r over n close parenthesis to the power n times t equals the fraction 1,500,000 over open parenthesis 1 plus the fraction 0.0635 over 12 close parenthesis to the power 12 times 44; equals, the fraction 1,500,000 over open parenthesis 1 plus 0.005291 6 ¯ close parenthesis to the power 528; equals, the fraction 1,500,000 over open parenthesis 1.005291 6 ¯ close parenthesis to the power 528; equals, the fraction 1,500,000 over 16.226302189; equals, 92,442.5037")[$P V & = & frac(A, attach(( 1 + frac(r, n) ), t: n × t)) = frac(1","500","000, attach(( 1 + frac(0.0635, 12) ), t: 12 × 44)) \ & = & frac(1","500","000, attach(( 1 + 0.005291 overline(6) ), t: 528)) \ & = & frac(1","500","000, attach(( 1.005291 overline(6) ), t: 528)) \ & = & frac(1","500","000, 16.226302189) \ & = & 92,442.5037$] Pilar will need to invest \$92,442,51 in this account to have \$1,500,000 at age 67. ] ] === Compute and Interpret Effective Annual Yield As we’ve seen, quarterly compounding pays interest 4 times a year or every 3 months; monthly compounding pays 12 times a year; daily compounding pays interest every day, and so on. #strong[Effective annual yield] allows direct comparisons between simple interest and compound interest by converting compound interest to its equivalent simple interest rate. We can even directly compare different compound interest situations. This gives information that can be used to identify the best investment from a yield perspective. Using a formula, we can interpret compound interest as simple interest. The effective annual yield formula stems from the compound interest formula and is based on an investment of \$1 for 1 year. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Effective annual yield is #math.equation(block: false, alt: "Y equals open parenthesis 1 plus the fraction r over n close parenthesis to the power n minus 1")[$Y = attach(( 1 + frac(r, n) ), t: n) − 1$] where #math.equation(block: false, alt: "Y")[$Y$] = effective annual yield, #math.equation(block: false, alt: "r")[$r$] = interest rate in decimal form, and #math.equation(block: false, alt: "n")[$n$] = number of times the interest is compounded in a year. #math.equation(block: false, alt: "Y")[$Y$] is interpreted as the equivalent annual simple interest rate. ] #examplebox("Example 8")[Determine and Interpret Effective Annual Yield for 6% Compounded Quarterly][ Suppose you have an investment paying a rate of 6% compounded quarterly. Determine and interpret that effective annual yield of the investment. #solutionbox[ Here, #math.equation(block: false, alt: "n")[$n$] = 4 (quarterly) and #math.equation(block: false, alt: "r")[$r$] = 0.06 (decimal form). Substituting into the formula we find that the effective annual yield is #math.equation(block: true, alt: "Y, equals, open parenthesis 1 plus the fraction 0.06 over 4 close parenthesis to the power 4 minus 1; equals, open parenthesis 1.015 close parenthesis to the power 4 minus 1; equals, 1.06136 minus 1; equals, 0.0614; equals, 6.14 %")[$Y & = & attach(( 1 + frac(0.06, 4) ), t: 4) − 1 \ & = & attach(( 1.015 ), t: 4) − 1 \ & = & 1.06136 − 1 \ & = & 0.0614 \ & = & 6.14 %$] Therefore, a rate of 6% compounded quarterly is equivalent to a simple interest rate of 6.14%. ] ] #examplebox("Example 9")[Determine and Interpret Effective Annual Yield for 5% Compounded Daily][ Calculate and interpret the effective annual yield on a deposit earning interest at a rate of 5% compounded daily. #solutionbox[ In this case, the rate is #math.equation(block: false, alt: "r")[$r$] = 0.05 and #math.equation(block: false, alt: "n")[$n$] = 365 (daily). Using the formula #math.equation(block: false, alt: "Y equals open parenthesis 1 plus the fraction r over n close parenthesis to the power n minus 1")[$Y = attach(( 1 + frac(r, n) ), t: n) − 1$], we have #math.equation(block: true, alt: "Y, equals, open parenthesis 1 plus the fraction 0.05 over 365 close parenthesis to the power 365 minus 1; equals, open parenthesis 1.0001369863 close parenthesis to the power 365 minus 1; equals, 1.051267 minus 1; equals, 0.0513")[$Y & = & attach(( 1 + frac(0.05, 365) ), t: 365) − 1 \ & = & attach(( 1.0001369863 ), t: 365) − 1 \ & = & 1.051267 − 1 \ & = & 0.0513$] This tells us that an account earning 5% compounded daily is equivalent to earning 5.13% as simple interest. ] ] #examplebox("Example 10")[Choosing a Bank][ Minh has a choice of banks in which he will open a savings account. He will deposit \$3,200 and he wants to get the best interest he can. The banks advertise as follows: #figure(table( columns: 2, align: left, inset: 6pt, table.header([Bank], [Interest Rate]), [ABC Bank], [2.08% compounded monthly], [123 Bank], [2.09% compounded annually], [XYZ Bank], [2.05% compounded daily], )) Which bank offers the best interest? #solutionbox[ To compare these directly, Minh could change each interest rate to its effective annual yield, which would allow direct comparison between the rates. Computing the effective annual yield for all three choices gives: ABC Bank: #math.equation(block: false, alt: "Y equals open parenthesis 1 plus the fraction 0.0208 over 12 close parenthesis to the power 12 minus 1 equals 0.0210 equals 2.10 %")[$Y = attach(( 1 + frac(0.0208, 12) ), t: 12) − 1 = 0.0210 = 2.10 %$] 123 Bank: #math.equation(block: false, alt: "Y equals open parenthesis 1 plus the fraction 0.0209 over 1 close parenthesis to the power 1 minus 1 equals 0.0209 equals 2.09 %")[$Y = attach(( 1 + frac(0.0209, 1) ), t: 1) − 1 = 0.0209 = 2.09 %$] XYZ Bank: #math.equation(block: false, alt: "Y equals open parenthesis 1 plus the fraction 0.0205 over 365 close parenthesis to the power 365 minus 1 equals 0.0207 equals 2.07 %")[$Y = attach(( 1 + frac(0.0205, 365) ), t: 365) − 1 = 0.0207 = 2.07 %$] ABC Bank has the highest effective annual yield, so Minh should choose ABC bank. ] ] === Key Terms - Compound interest - Effective annual yield === Key Concepts - Compound interest means that the interest earned during one period will earn interest in later periods. Essentially, the amount of the principal grows from period to period. - The important values in computing compound interest are the interest rate, the principal, the length of time the investment, and the number of times the investment is compounded. - Compound interest has minimal impact early, but later has a very large impact. - You can determine how much to invest today in order to reach a goal for some time later. - Compound interest can be translated into an effective annual yield, which allows for comparison between investment options. === Videos - #link("https://openstax.org/r/compound_interest_beginners")[Compound Interest] - #link("https://openstax.org/r/compare_simple_compound_interest1")[Compare Simple Interest to Interest Compounded Annually] - #link("https://openstax.org/r/compare_simple_compound_interest2")[Compare Simple Interest and Compound Interest for Different Number of Periods Per Year] === Formulas #math.equation(block: true, alt: "A equals P open parenthesis 1 plus the fraction r over n close parenthesis to the power n t")[$A = P attach(( 1 + frac(r, n) ), t: n t)$] #math.equation(block: true, alt: "P V equals the fraction A over open parenthesis 1 plus the fraction r over n close parenthesis to the power n times t")[$P V = frac(A, attach(( 1 + frac(r, n) ), t: n × t))$] #math.equation(block: true, alt: "Y equals open parenthesis 1 plus the fraction r over n close parenthesis to the power n minus 1")[$Y = attach(( 1 + frac(r, n) ), t: n) − 1$]