Simple, Compound Interest, and Annuity - Free Electrical Engineering Board Exam Reviewer

LessonAbout 1.5 h1 of 2 in Engineering Economics

Engineering Economics

Engineering Economics is the application of economic principles and mathematical techniques to evaluate engineering projects and financial decisions.

1. Interest and Discount

1.1 Simple Interest

Simple interest is calculated solely on the original principal amount.

Principal: initial amount borrowed or invested
Ordinary interest: based on 360 days per year
Exact interest: based on 365 or 366 days

Simple Interest and Total Amount

\[ I = Pin \]

\[ F = P(1 + in) \]

ME Board April 1995

Example. A sum of \(4,000\ \text{pesos}\) is borrowed for \(75\ \text{days}\) at \(16\%\) per annum simple interest. Find the total amount due at the end of \(75\ \text{days}\).

Solution.

\[ F = P\left(1 + i\dfrac{d}{360}\right) = 4000\left[1 + 0.16\left(\dfrac{75}{360}\right)\right] = 4,133.33\ \text{pesos} \]

Answer: \(4,133.33\ \text{pesos}\)

ME Board April 1998

Example. A bank charges \(12\%\) simple interest on a \(300.00\ \text{pesos}\) loan. How much will be repaid if the loan is paid back in one lump sum after \(3\ \text{years}\)?

Solution.

\[ F = P(1 + in) = 300[1 + 0.12(3)] = 408.00\ \text{pesos} \]

Answer: \(408.00\ \text{pesos}\)

1.2 Compound Interest

Compound interest is interest calculated on both initial principal and accumulated past interest.

Nominal rate: basic annual interest rate
Effective rate: actual annual rate earned
Compounding period: frequency interest is added

Compound Amount and Effective Rate

\[ F = P(1 + i)^n \]

\[ ER = \left(1 + \dfrac{r}{m}\right)^m - 1 \]

Where: \(m\) = number of interest periods per year

ECE Board April 1998

Example. Find the accumulated amount of \(12,800\ \text{pesos}\) in \(4\ \text{years}\) at \(5\%\) interest compounded quarterly.

Solution.

\[ F = P\left(1 + \dfrac{r}{m}\right)^{mt} = 12800\left(1 + \dfrac{0.05}{4}\right)^{4(4)} = 15,614.59\ \text{pesos} \]

Answer: \(15,614.59\ \text{pesos}\)

ECE Board November 1998

Example. At an interest rate of \(10\%\) compounded annually, how much will a deposit of \(1,500\ \text{pesos}\) be in \(15\ \text{years}\)?

Solution.

\[ F = P(1 + i)^n = 1500(1 + 0.10)^{15} = 6,265.87\ \text{pesos} \]

Answer: \(6,265.87\ \text{pesos}\)

1.3 Continuous Compounding

Continuous compounding occurs when interest is compounded continuously as periods approach infinity.

Continuous Compounding Amount

\[ F = P e^{rt} \]

\[ ER = e^r - 1 \]

ME Board April 1998

Example. A nominal interest of \(3\%\) compounded continuously is given on an account. Find the accumulated amount of \(10,000\ \text{pesos}\) after \(10\ \text{years}\).

Solution.

\[ F = P e^{rt} = 10000 e^{0.03(10)} = 13,498.60\ \text{pesos} \]

Answer: \(13,498.60\ \text{pesos}\)

Example. An account earns \(3\%\) nominal interest compounded continuously. Find the present worth if the account accumulates to \(6,749.30\ \text{pesos}\) in \(10\ \text{years}\).

Solution.

\[ P = F e^{-rt} = 6749.30 e^{-0.03(10)} = 5,000.00\ \text{pesos} \]

Answer: \(5,000.00\ \text{pesos}\)

1.4 Discount and Inflation

Discount is interest deducted in advance from a loan or payment.

Bank discount: interest collected upfront
Proceeds: net amount received
Inflation: decline in money purchasing power

Discount and Inflation Adjustment

\[ D = F d n \]

\[ P' = \dfrac{F}{(1 + f)^n} \]

Where: \(d\) = rate of discount, \(f\) = annual inflation rate

ME Board April 1993

Example. A loan of \(20,000\ \text{pesos}\) is granted with an interest of \(6\%\) for \(180\ \text{days}\) collected in advance. Find the proceeds of the note.

Solution.

\[ D = F d n = 20000(0.06)\left(\dfrac{180}{360}\right) = 1,200.00\ \text{pesos} \]

\[ \text{Proceeds} = F - D = 20000 - 1200 = 18,800.00\ \text{pesos} \]

Answer: \(18,800.00\ \text{pesos}\)

ME Board October 1995

Example. In year zero, you invest \(10,000.00\ \text{pesos}\) in a \(15\%\) security for \(5\ \text{years}\). During that time, the average annual inflation is \(6\%\). How much, in terms of year zero pesos, will be in the account at maturity?

Solution.

\[ F = P(1 + i)^n = 10000(1 + 0.15)^5 = 20,113.57\ \text{pesos} \]

\[ P_{\text{adjusted}} = \dfrac{F}{(1 + f)^n} = \dfrac{20113.57}{(1 + 0.06)^5} = 15,030.00\ \text{pesos} \]

Answer: \(15,030.00\ \text{pesos}\)

2. Annuities

2.1 Ordinary Annuity

Ordinary annuity consists of equal periodic payments made at the end of each period.

Uniform payment: equal payment \(A\) each period
Payment period: interval between consecutive payments
Term: duration from start to end

Ordinary Annuity Future and Present Worth

\[ F = \dfrac{A[(1 + i)^n - 1]}{i} \]

\[ P = \dfrac{A[(1 + i)^n - 1]}{(1 + i)^n i} \]

ECE Board November 1998

Example. Find the accumulated amount of a \(5\ \text{-year}\) annuity paying \(6,000\ \text{pesos}\) at the end of each year, with interest at \(15\%\) compounded annually.

Solution.

\[ F = \dfrac{A[(1 + i)^n - 1]}{i} = \dfrac{6000[(1 + 0.15)^5 - 1]}{0.15} = 40,454.29\ \text{pesos} \]

Answer: \(40,454.29\ \text{pesos}\)

ECE Board April 1998

Example. How much must you invest today in order to withdraw \(2,000\ \text{pesos}\) annually for \(10\ \text{years}\) if the interest rate is \(9\%\)?

Solution.

\[ P = \dfrac{A[(1 + i)^n - 1]}{(1 + i)^n i} = \dfrac{2000[(1 + 0.09)^{10} - 1]}{(1 + 0.09)^{10}(0.09)} = 12,835.32\ \text{pesos} \]

Answer: \(12,835.32\ \text{pesos}\)

2.2 Annuity Due

Annuity due consists of equal periodic payments made at the beginning of each period.

Annuity Due Present Worth

\[ P = A + \dfrac{A[(1 + i)^{n-1} - 1]}{(1 + i)^{n-1} i} \]

Example. A washing machine costs \(12,000\ \text{pesos}\) cash or can be bought in \(5\ \text{years}\) installment. If money is worth \(8\%\) compounded annually, find the yearly amortization paid at the beginning of each year.

Solution.

\[ 12000 = A + \dfrac{A[(1 + 0.08)^4 - 1]}{(1 + 0.08)^4(0.08)} \implies A = 2,400.00\ \text{pesos} \]

Answer: \(2,400.00\ \text{pesos}\)

ME Board April 1998

Example. How much must be deposited at \(6\%\) each year beginning on January 1, year 1, in order to accumulate \(5,000\ \text{pesos}\) on the date of the last deposit, January 1, year 6?

Solution.

\[ F = \dfrac{A[(1 + i)^n - 1]}{i} \implies 5000 = \dfrac{A[(1 + 0.06)^6 - 1]}{0.06} \implies A = 717.00\ \text{pesos} \]

Answer: \(717.00\ \text{pesos}\)

2.3 Deferred Annuity

Deferred annuity is an annuity where the first payment begins after a specified number of periods.

Deferred Annuity Present Worth

\[ P = \dfrac{A[(1 + i)^n - 1]}{(1 + i)^n i (1 + i)^m} \]

Where: \(m\) = deferment period

ECE Board April 1998

Example. A land is bought for \(100,000\ \text{pesos}\) down payment and \(10\) deferred semi-annual payments of \(8,000\ \text{pesos}\) each, starting \(3\ \text{years}\) from now. Find the present value if interest is \(12\%\) compounded semi-annually.

Solution.

\[ P_1 = \dfrac{8000[(1 + 0.06)^{10} - 1]}{(1 + 0.06)^{10}(0.06)} = 58,880.69\ \text{pesos} \]

\[ P_2 = \dfrac{58880.69}{(1 + 0.06)^5} = 43,999.08\ \text{pesos} \implies \text{Total} = 100000 + 43999.08 = 143,999.08\ \text{pesos} \]

Answer: \(143,999.08\ \text{pesos}\)

ME Board April 1998

Example. A house and lot can be acquired by a downpayment of \(500,000\ \text{pesos}\) and a yearly payment of \(100,000\ \text{pesos}\) at the end of each year for \(10\ \text{years}\), starting at the end of \(5\ \text{years}\) from purchase date. If money is worth \(14\%\) compounded annually, find the cash price.

Solution.

\[ P_1 = \dfrac{100000[(1 + 0.14)^{10} - 1]}{(1 + 0.14)^{10}(0.14)} = 521,611.56\ \text{pesos} \]

\[ P_2 = \dfrac{521611.56}{(1 + 0.14)^4} = 308,811.23\ \text{pesos} \implies \text{Total} = 500000 + 308811.23 = 808,811.00\ \text{pesos} \]

Answer: \(808,811.00\ \text{pesos}\)

2.4 Perpetuity and Capitalized Cost

Perpetuity is a series of uniform payments that continue indefinitely.

Perpetuity: endless stream of periodic payments
Capitalized cost: initial cost plus present worth of perpetual replacements

Perpetuity and Capitalized Cost

\[ P = \dfrac{A}{i} \]

\[ CC = FC + \dfrac{FC - SV}{(1 + i)^n - 1} \]

Where: \(FC\) = first cost, \(SV\) = salvage value

CE Board November 1995

Example. Find the present value of a perpetuity of \(15,000\ \text{pesos}\) payable semi-annually if money is worth \(8\%\) compounded quarterly.

Solution.

\[ 1 + i_{\text{semi}} = \left(1 + \dfrac{0.08}{4}\right)^2 = 1.0404 \implies i = 0.0404 \]

\[ P = \dfrac{A}{i} = \dfrac{15000}{0.0404} = 371,287.13\ \text{pesos} \]

Answer: \(371,287.13\ \text{pesos}\)

CE Board November 1996

Example. At \(6\%\), find the capitalized cost of a bridge whose cost is \(250\ \text{M pesos}\) and life is \(20\ \text{years}\), if the bridge must be partially rebuilt at a cost of \(100\ \text{M pesos}\) at the end of each \(20\ \text{years}\).

Solution.

\[ CC = FC + \dfrac{RC}{(1 + i)^n - 1} = 250 + \dfrac{100}{(1 + 0.06)^{20} - 1} = 295.30\ \text{M pesos} \]

Answer: \(295.30\ \text{M pesos}\)

Formula Summary

Formula Name	Formula
Simple Interest	\(I = Pin\)
Simple Interest Total Amount	\(F = P(1 + in)\)
Compound Amount	\(F = P(1 + i)^n\)
Effective Interest Rate	\(ER = \left(1 + \dfrac{r}{m}\right)^m - 1\)
Continuous Compounding Amount	\(F = P e^{rt}\)
Continuous Effective Rate	\(ER = e^r - 1\)
Bank Discount	\(D = F d n\)
Inflation Adjusted Present Worth	\(P' = \dfrac{F}{(1 + f)^n}\)
Ordinary Annuity Future Worth	\(F = \dfrac{A[(1 + i)^n - 1]}{i}\)
Ordinary Annuity Present Worth	\(P = \dfrac{A[(1 + i)^n - 1]}{(1 + i)^n i}\)
Annuity Due Present Worth	\(P = A + \dfrac{A[(1 + i)^{n-1} - 1]}{(1 + i)^{n-1} i}\)
Deferred Annuity Present Worth	\(P = \dfrac{A[(1 + i)^n - 1]}{(1 + i)^n i (1 + i)^m}\)
Perpetuity Present Worth	\(P = \dfrac{A}{i}\)
Capitalized Cost	\(CC = FC + \dfrac{FC - SV}{(1 + i)^n - 1}\)