ENGR 3360U Engineering Economics: Lecture 4

• ENGR 3360U Engineering Economics covers time cash flow modeling, compound interest and series, sinking fund, and capital recovery.
• The course uses material from "Engineering Economics" by Fraser N. et al., 3rd Edition.

Subjects Covered

• Macro & Micro Economics are introduced in relation to Engineering Economics
• Understanding and Estimating Models of Supply and Demand
• Role of the BANK of Canada is studied
• Balance Sheet Accounting and Financial RATIOS are covered
• Time value of money is explored
• Cash Flow Analysis (C.F.) focuses on interest rates and cash flow diagrams
• MARR & IRR / ERR & Comparisons are discussed
• Uncertainty & Risk, Probability Analysis & Decision Trees are examined
• Market Failures & Remedies are covered
• Students will learn about Depreciation and Taxes & Cash Flow
• Replacement Analysis & Decisions, Inflation & Price Change, and Public Decision Making are also covered
• The course culminates in a FINAL Exam

Lecture 4: Outline

• Comparisons between projects
• Compound interest and annuities are defined
• Present worth factor, uniform series, arithmetic series, and geometrical series are covered
• Capital recovery factor and sinking fund

Goals of Lecture 4

• Review and understand the concepts of compounding & annuities, capital recovery & sinking fund, and gradient series & applications.

Time and Interest Rate Importance

• Time is a critical factor in engineering economics, with a preference for consuming goods and services sooner rather than later.
• The stronger the preference for current consumption, the greater the importance of time in investment decisions.
• The amount and timing of a project's cash flows are critical to a project's worth and doability.

Project Comparison Example

• Projects A and B have identical initial costs ($10,000), duration (5 years), and cumulative cash flows over 5 years ($15,000) but different cash flow patterns.
• If the discount rate i = 0%, then A and B are equivalent
• If i ≠ 0%, then A and B are not equivalent

Rate of Interest

• Rate of return is based on lending money
• Rate of return is paid by use of a lender for the use of fund

Simple Interest

• Interest payment calculation: Multiply the interest rate by the principal each year
• Accumulated value under simple interest: FN=8 = P(1 + in)*
• $1000 invested at 9% simple interest for 8 years yields $1720 (F = $1000[1+0.09(8)] = $1720)
• Simple interest is not commonly used.

Compound Interest

• Interest each year is found by multiplying the interest rate by the accumulated value of money (principal and interest.)

Compound Interest Example

• Value for amount P invested for n periods at i rate: F = P(1+i)^N
• $1000 invested at 9% compound interest for 8 years yields $1992.6 (F8 = $1000 (1 + 0.09)8 = $1992.6)
• Compound interest is the basis for practically all monetary transactions
• $1000 invested at 9% compound interest for 4 years yields $1411.6 (F4 = $1000 (1 + 0.09)4 = $1411.6)

Present/Future Value

• Discounting is the process of translating a future value (or cash flows) to a present value (PV = FV / (1+i)^N) or (P = F / (1+i)^N and F = P(1+i)^N
• A promise of $450,000 in 50 years, assuming 6% interest, has a value today of $24,429.7

Present/Future Value Example

• An amount of deposit today is (PV): $60,000
• An an interest rate of: 12%
• Compounded Annually
• Over a period of 5 years (5 periods)
• The future value (FV) of this single sum is $105,740.5

Present/Future Value Example

• An amount of deposit is at the end (FV): $105,740.5
• At an interest rate of: 12%
• Compounded Annually
• Over a period of 5 years (5 periods)
• The present value (PV) of this single sum is $60001.1

Compound & P. Worth Factors

• Compounding is: F= P(1+i)^N
• Present Worth Factor is: 1/(1+i)^N

Present/Future Value Example

• F = P(1+i)^n is the single payment future value factor.
• The future value in 7 years of $500 today at an interest rate of 8 1/2% is about $885.07
• P = F(1/(1+i)^n) = F(1+i)^-n is the single payment present value factor.
• The value today of $500 in 7 years at 8 1/2% interest is approx. $282.46

Present/Future Value & Compounding Example

• An interest rate is at 8% compounded quarterly
• Nominal rate is 8% (per year)
• The periodic rate (per quarter) is 8%/4 = 2%
• The effective rate is [1+(8%/4)]^4 – 1 = 8.2432%
• Investing $1 at 2% per quarter is equivalent to investing $1 at 8.2432% annually.

Annuities & Annuity Computation

• Defined as a series of equal dollar payments coming at the end of a certain time period for a specified number of time periods
• Examples are lifetime insurance benefits, lottery payments, and retirement payments
• Periodic payments or receipts should always be the same amount
• Interval between such payments or receipts should always be the same, and the interest should always be compounded once each interval.

Annuity Example

• Saving $1,500 a year at the end of every year for three years in an account earning 7% interest, compounded annually leads to a final amount of $4,822.35 at the end of the third year
• FVYear3 = $1,500 (1.07)² + $1,500(1.07)1 + $1,500(1.07)° = $4,822.35

Annuity Equation

• Total Future Value (FV) = A + A(1+i) + A(1+i)² + ... + A(1+i)^n-1 , where n is the year
• Using multiplication: F(1+i) = A(1+i) + A(1+i)² + ... + A(1+i)^n
• Uniform Series of Annuity: F = A((1+i)^N -1) / i

Annuities Future Value Example 1

• A deposit made at the end of each period :$5,000
• Compounded Annually
• Number of periods: Five
• Interest rate: 12%
• Future value (FV) means: (F = FV and A= Annuity)
• Using F = A[(1+i)^n - 1] / i provides $31,764.25

Annuities Present Value Example 1

• A deposit made at the end of each period is: $5,000 compounded annually
• For a Number of periods: Five
• And an interest rate of: 12%
• Formula: PV = FV_n/ (1 + i)^n
• The Present value (PV) is: $18,023.87 ($31,764.25 / (1.76234)

Annuities Future Value Example 2

• Deposit made at the begining of each of the the period
• Value: $8,000. Compounded Annually
• Number of periods: N = 8 (at the end)
• Interest rate: 12%

Annuities Present value Example 2

• Ordinary ANNUITY FACTOR (for 8 years) if START at the end of year 1
• [(1+i)^n - 1] / i = 12.2996 (i = 12% or 0.12)
• Conversion: 1.12 x 12.2996 = 13.7756 (the END of the year)
• FV = $8000 x (13.7756) = $110,205.2

Uniform Series

• Series Discount Amount (PV)
• Transformations
• Series Compound Amount (FV)

Uniform Series Concepts

• A series is the sum of the terms of a sequence
• A formula: F=A ((1+i)^N −1)/ i
• Important Factors: Sinking Factor (Uniform Series), Capital Recovery Factor and Series Present Worth

Compound Amount Factor (Uniform or equal Payment Series)

• Finds future value, (F) of a uniform series of equal annual payments,(A), over n periods at i rate
• F = A ((1+i)^n −1)/ i
• Example: 6 year worth of $800 at 12%: Finds F given A

Sinking Fund Factor

• Determines needed deposit each period with interest rate per period to yeild a sum
• A= F(i/ ((1+i)^n -1))
• Example: $1.2 million bond retired in 20 years, finds amount for: Finds A given F

Capital Recovery Factor

• Finds an annuity
• Uniform series of payments over n periods at an interest rate that is equivelant of a present value

Present Worth Factor

• Finds present value of the series of period payments
• P = A ((1+i)^n -1)/ (i(1+i)^n)

Capital Recovery & Sinking F. Factors

• Formula: F = A(1+i)^N -1/ i
• Capital Recovery Factor * (1+i)^N -1/ i*
• Sinking Fund Factor is i/ * (1+i)^N -1*

Capital Recovery & Sinking F. Factors

• Sinking Fund: interest-bearing account into which uniform payment or deposits are placed to accumulate some amount.
• Capital Recovery Factor*: value (A) of equal periodic payments equivalent to a present amount P (interest rate 𝑖, N periods). Interest in understanding Arithmetic Gradient Series and Geometric Gradient Series.

Uniform Series Example

• Find the balance over ten years of annual deposits that pays interest of 8% compounded annually
• F = $1500 (((1+0.08)^10 -1)// 0.08

Arithmetic Series

• P = G ((1+i)^N -iN-1)/ i^2(1+i)^N)
• N = number of deposits
• Cash flows: 1G, 2G, …,(N-1)G at the end of periods 1, 2, …N

Homework Topics for next class

• Read and study CHAPT. 3 & CHAPT. 4 (from Fraser / Annual Worth Comparison) or CHAPT. 4 Newnan (Gradients) CHAPT. 5 (PW Analysis) and CHAPT. 6 (Annual Cash Flow Analysis) from Newnan.