Engineering Economics Fundamentals

Engineering Economics – Fundamentals John P. Greaney, PE Copyright 2004 All rights reserved (Revised 02/19/2004) 1.0. Time Value of Money Understanding the time value of money is the central theme of Engineering Economics. Interest is the cost of borrowing money from a lender (e.g., a bank or other financial institution, or an individual) Even if an owner doesn’t take on debt to finance a project, the opportunity cost of not investing the funds elsewhere is considered in the economic analysis of the project. This interest rate equivalent is often called the discount rate. 1.1. Future Value With Compound Annual Interest Calculation Virtually every commercial lending transaction makes use of compound interest. The general form of the Future Value equation takes this into consideration FV = P (1+i/t)nt Where FV = Future Value P = initial Principal i = effective interest rate n = term of the investment in years t = number of times interest is compounded during year Example 1: What is the Future Value of $1,000 invested at 8% interest compounded annually for 20 years? FV = P (1+i/t)nt = $1,000 (1 + (0.08/1))(20 x 1)

= $4,660.96

Interest may be compounded monthly, daily, hourly, or even continuously Compounded monthly: Compounded daily: Compounded hourly:

FV = $1,000 (1 + (0.08/12))(20 x 12) FV = $1,000 (1 + (0.08/365))(20 x 365) FV = $1,000 (1 + (0.08/8760))(20 x 8760)

= $4,926.80 = $4,952.16 = $4,953.00

As the period for compounding becomes infinitely small (i.e. continuous compounding), the Future Value formula is expressed as: FV = Pein Where FV = Future Value

P = initial Principal e = constant (2.71828183) i = effective interest rate n = term of the investment in years (Note: e is a constant and the base of the natural logarithm. Mathematicians often call it Euler’s (sounds like “oiler”) Number in honor of the Swiss mathematician Leonhard Euler. Financial types usually just call it the “compound interest constant”.) Example 2: What is the Future Value of $1,000 invested at 8% interest compounded continuously for 20 years? FV = Pein

= $1,000 (2.71828)(0.08 x 20)

= $4,953.03

The effect of increasing the frequency of compounding is depicted graphically in the following chart. Over a 20-year term, there’s little increase in the end value of the investment if interest is compounded more frequently than daily.

Compound Interest $1,000 invested @ 8.00% for 20 years

Value After 20 Years

$5,000 $4,900 $4,800 $4,700 $4,600

Cont.

Hourly

Daily

Monthly

Yearly

$4,500

Frequency of Compounding

1.2. Nominal and Effective Rates of Interest The phenomenon of compound interest may also be considered by calculating the effective rate of interest. In Examples 1 and 2 above, the nominal rate is

8%. The effective rate of interest grows as the period of compounding shortens. The formula for the effective rate of interest is as follows. i = (1+ ((i(t))/t))t)-1 Where i = effective interest rate i(t) = nominal interest rate t = number of times interest is compounded during year Note that in the term i(t) , (t) is a superscript, not “i” to the “t” power. The variable for the nominal rate of interest with monthly compounding would be denoted i(12). For daily compounding, the variable would be i(365) Example 3: What is the effective rate of interest for an 8% nominal annual rate of interest compounded annually? i = (1+ ((i(t))/t))t)-1 = (1+ ((0.08)/1))1)-1 = 0.0800000 = 8.00% In the case of annual compounding, the nominal and effective rates of interest are the same. Example 4: What is the effective rate of interest for an 8% nominal annual rate of interest compounded monthly? i = (1+ ((i(t))/t))t)-1 = (1+ ((0.08)/12))12)-1 = 0.0829995 = 8.30% As we increase the frequency of compounding from daily to hourly to an infinite number of periods (i.e. continuous compounding), the effective interest rate equation becomes

Where i = effective interest rate i(t) = nominal interest rate (where t approaches infinity) e = constant (2.71828183) t = number of times interest is compounded during year Example 5: What is the effective rate of interest for an 8% nominal annual rate of interest compounded continuously?

i = (2.71828183)0.08) -1 = 0.083287 = 8.33%

1.3. Average Annual Return vs. Average Annualized Return Many people don’t realize that there’s a big difference between average annual returns and average annualized returns. Average annual return usually provides a poor description of investment performance. Average annual return is simply the sum of the annual change in the value of an investment divided by the number of years. Rave = ( ∑ Rn) / n Where Rave = Average Annual Return Rn = Annual Return for Year (n) n = term of the investment in years Let’s look at two examples. Example 6: What is the average annual return for a $1,000 investment that gains 30% the first year, loses 30% the second year, and then gains 30% the third year? How much money do you have at the end of three years? Rave = ( ∑ Rn) / n = (30%+(-30%)+30%)/3 = 10%

Year 1 Year 2 Year 3

Starting Annual Balance Return (%) $ 1,000 30% $ 1,300 -30% $ 910 30%

Annual Year End Return ($) Balance $ 300 $ 1,300 $ (390) $ 910 $ 273 $ 1,183

Example 7: What is the average annual return for a $1,000 investment that gains 10% per year for three years? How much money do you have at the end of three years? Rave = ( ∑ Rn) / n = (10%+10%+10%)/3 = 10%

Year 1 Year 2 Year 3

Starting Annual Balance Return (%) $ 1,000 10% $ 1,100 10% $ 1,210 10%

Annual Year End Return ($) Balance $ 100 $ 1,100 $ 110 $ 1,210 $ 121 $ 1,331

1.4. Compound Annual Growth Rate (CAGR)

Because of the inconsistencies illustrated in the two examples above (i.e., an investment with the same average annual return had very different values after three years), Annualized Annual Return (often referred to as Compound Annual Growth Rate (CAGR)) is the preferred measure of investment performance. CAGR = (FV / P)(1 / n) - 1 Where CAGR = Compound Annual Growth Rate FV = Future Value P = initial Principal n = term of the investment in years Example 8: What is the CAGR for the $1,000 investment described in Examples 3 and 4 above? For Example 3: CAGR = (FV / P)(1 / n) – 1 = ($1,183 /$1,000)(1 / 3) – 1 = 5.67% For Example 4: CAGR = (FV / P)(1 / n) – 1 = ($1,331 /$1,000)(1 / 3) – 1 = 10.0% 1.5. Present Value of a Series of Uniform Payments Present Value is usually expressed as a factor for a series of uniform payments of $1. PV = ((1+i)n – 1) / (i(1+i)n ) Where PV = Present Value i = interest rate n = term of the investment in years Example 9: What is the Present Value for a uniform series of 10 annual payments of $1,000 at 7% interest? PV = ((1+i)n – 1) / (i(1+i)n ) = ((1+0.07)10 – 1) / (0.07(1+0.07)10 ) = 7.024 PV= $1,000 x 7.024 = $7,024 You can also use the Present Value function in Microsoft Excel to make this calculation PV(rate,nper,pmt) = PV(0.07,10,1000) = 7,024 1.6. Amortizing a loan The word “amortize” is derived from “mors”, the Latin word for death. Amortizing a loan literally mean “killing” the loan or “working it to death.”

We can use the Present Value formula to calculate the annual payment required to retire a debt. Example 10: What is the year-end uniform annual payment for a $100,000 loan at 8% interest, compounded annually, that must be repaid over 5 years? PV = ((1+i)n – 1) / (i(1+i)n ) = ((1+0.08)5 – 1) / (0.08(1+0.08)5 ) = 3.993 Annual loan payment = Loan amount/PV factor = $100,000/3.993 = $25,046 You can also use Payment function in Microsoft Excel for this calculation: PMT(rate,nper,pv) = PMT(0.08,5,100000) = $25,046 If the interest on a loan is compounded more frequently than annually, simply calculate the effective rate of interest and then use the effective rate in the Present Value formula. Example 11: What is the year-end uniform annual payment for a $100,000 loan at 8% interest, compounded daily, to be repaid over 5 years? First, calculate the effective interest rate. i = (1+ ((i(t))/t))t)-1 = (1+ ((0.08)/365))365)-1 = 0.0832776 = 8.33% Then input the result in the Present Value formula. PV = ((1+i)n – 1) / (i(1+i)n ) = ((1+0.0832776)5 – 1) / (0.0832776(1+0.0832776)5 ) = 3.9584553 Annual loan payment = Loan amount/PV factor = $100,000/3.9584553 = $25,262.38 You can also use Payment function in Microsoft Excel for this calculation: PMT(rate,nper,pv) = PMT(0.0832776,5,100000) = $25,262.38 1.7 Loan Amortization Tables It is customary to provide a loan amortization table detailing the split between interest and repayment of principal for each loan payment. This is necessary because the interest payments on an investment loan are tax deductible, but the repayment of principal is not.

There are several different loan amortization methods. The example we’ll consider below is based on the actuarial or “normal” method. The actuarial method is also called the “US Simple Rule”. Federal law requires that the amortization method be used for consumer loans with durations of 61 months or longer. Other amortization methods include the “maximum yield method”, “the Merchant’s rule”, “the constant ratio method” and the “direct ratio method” The direct ratio method is also called the “Rule of 78” or the “Sum of the Digits” method. This method front loads the interest charges on the amortization table and penalizes borrowers who pay off the debt early. That’s why lenders are prohibited from using the Rule of 78 to amortize loans with a term of longer than 5 years. Constructing a loan amortization table is a very tedious calculation best organized in tabular format. A Microsoft Excel spreadsheet is an ideal tool for this purpose. First we calculate the periodic (usually monthly or annual) loan payment as we did in Example 10 above. For a $100,000 loan at 8% annual interest, compounded annually, to be repaid over 5 years we get an annual payment of $25,045.65. Next we calculate the first year interest charge using the formula, It = (i)(B(t-1)) Where It = interest paid for that year i = the effective interest rate B(t-1) = the outstanding balance at the end of the previous year It = (i)(B(t-1)) = (.0.08)($100,000) = $8,000 Then the principal repaid during the first year Pr = PMT - It Where Pr = is the principal repaid in that year PMT = the periodic (e.g., annual or monthly) loan payment It = interest paid for that year Pr = PMT - It = $25,045.65 - $8,000 = $17,045.65 And finally, the outstanding principal remaining at the end of the first year Bt = B(t-1) - Pr Where It = interest paid for that year

Pr = is the principal repaid in that year B(t-1) = the outstanding balance at the end of the previous year Bt = B(t-1) - Pr = $100,000.00 - $17,045.65 = $82,954.35 In the second year, we get the following: Interest paid = It = (i)(B(t-1)) = (.0.08)($82,954.35) = $6,636.35 Principal repaid = Pr = PMT - It = $25,045.65 - $6,636.35 = $18,409.30 Outstanding balance = Bt = B(t-1) - Pr = $82,954.35 - $18,409.30 = $64,545.06 Here’s the complete amortization table. Duration Years t 0 1 2 3 4 5

Annual Payment PMT $25,045.65 $25,045.65 $25,045.65 $25,045.65 $25,045.65

It = (i)(B(t-1)) Interest Paid It

Pr = PMT - It Principal Repaid Pr

$ $ $ $ $

$17,045.65 $18,409.30 $19,882.04 $21,472.60 $23,190.41

8,000.00 6,636.35 5,163.60 3,573.04 1,855.23

Bt = B(t-1) - Pr Outstanding Principal Bt $100,000.00 $ 82,954.35 $ 64,545.06 $ 44,663.02 $ 23,190.41 $ 0.00

Example 12: Construct a loan amortization table for a $50,000 loan at a 6% annual rate of interest, compounded monthly, to be repaid in 3 years with equal year-end payments. First we calculate the effective interest rate. i = (1+ ((i(t))/t))t)-1 = (1+ ((0.06)/12))12)-1 = 0.0616778 = 6.17% Then the present value factor PV = ((1+i)n – 1) / (i(1+i)n ) = ((1+0.0616778)3 – 1) / (0.0616778(1+0.0616778)3 ) = 2.664736 The annual loan payment = Loan amount/PV factor = $50,000/2.664736 = $18,763.58 The calculated loan amortization table appears below.

Duration Years t 0 1 2 3

2.0.

Annual Payment PMT $18,763.58 $18,763.58 $18,763.58

It = (i)(B(t-1)) Interest Paid It

Pr = PMT - It Principal Repaid Pr

$ $ $

$15,679.69 $16,646.78 $17,673.52

3,083.89 2,116.80 1,090.06

Bt = B(t-1) - Pr Outstanding Principal Bt $ 50,000.00 $ 34,320.31 $ 17,673.52 $ 0.00

Discount Rate

The discount rate is usually taken as the interest rate on the debt used to finance a project, the organization’s cost of capital, or some specified “hurdle rate” that a company uses to screen projects for approval. Cost of Capital is what an organization pays to raise funds (i.e., the interest rate on its debt.) The US Government typically has the lowest cost of capital as measured by the yields on Treasury securities. Large financially sound corporations like ExxonMobil or General Electric would pay a small premium over the Treasury rate. A small, untested company might pay 5% or 10% more than the US Treasury rate, if they could sell debt at all. States and municipalities can issue debt that is exempt from Federal income taxes (e.g., tax-free municipal bonds.) The tax-free advantage can give these securities yields even lower than Treasury securities. Companies can also raise capital by selling equity (stock) in the company. Many companies specify a “hurdle rate”, which covers their cost of capital plus a risk premium. Projects normally would not be funded unless the expected return on investment exceeds the hurdle rate. 3.0. Inflation Inflation should be considered for projects with longer useful lives. Over the past 50 years inflation as measured by the Consumer Price Index (CPI) has increased at an annualized rate of 3.8% per annum. A basket of goods that cost $100 fifty years ago would cost about $670 today. Over shorter periods of time inflation varies greatly. The worst ten-year period (1973-1983) saw inflation rising at an 8.7% annualized rate. The most benign ten-year period (1953-1963) saw prices increase at only a 1.3% annualized rate. The most commonly used inflation measure is the Consumer Price Index – All Urban Consumers (CPI-U) The US Bureau of Labor Statistics (BLS) updates this

index monthly. The latest figures are published on their website at http://www.bls.gov/cpi/home.htm Example 13: In January 1993 the CPI-U was 142.6 and rose to 181.7 by January 2003. What was the average annualized inflation rate for that ten-year period? You can use the CAGR formula to calculate the annualized inflation rate. CAGR = (FV / P)(1 / n) – 1 = (181.7 /142.6)(1 / 10) – 1 = 2.45% 4.0. Depreciation There are two definitions of depreciation. 1) Depreciation is the reduction in the value of an asset over time due to wear, deterioration, or obsolescence. 2) Deprecation is the reduction in the value of an asset as prescribed by the relevant tax authority. Wear and tear on equipment is an expense just like electricity or labor, so depreciation must be considered in evaluating the economics of a project. 4.1. Depreciation Methods: There are many methods for calculating depreciation, but the three most common are: 1) Straight-Line 2) Declining Balance 3) Modified Accelerated Cost Recovery System (MACRS) Straight-line depreciation is the easiest to calculate. Under this method the asset is assumed to depreciate in equal amounts each year. In order to calculate straight-line depreciation you need to know (or assume) three things; the acquisition cost of the item, its useful life, and its salvage value at the end of its useful life. DSL = (A – S)/n Where DSL = Annual Depreciation (Straight-line) A = Acquisition Cost S = Salvage Value

n = useful life in years Example 14: A company purchases a machine for $100,000 with an expected 10-year useful life and a $20,000 salvage value after 10 years. What is the annual depreciation using the straight-line method? Using the Straight-Line depreciation formula we get: DSL = (A - S)/n = ($100,000 – $20,000) / 10 = $8,000 per year Declining balance depreciation is somewhat more cumbersome to calculate. Under this method the asset is assumed to depreciate at a faster rate than the equal annual amounts used in the straight-line method. This is done by applying a multiple to the acquisition cost (book value in subsequent years) and ignoring the salvage value. The annual depreciation factor is calculated with the formula Df = m/n Where Df = Annual Depreciation Factor (Declining balance) m = declining balance multiple n = useful life in years The first year depreciation is calculated using this formula: DDB = ADf Where DDB = Annual Depreciation (Declining balance) A = Acquisition Cost Df = Annual Depreciation Factor (Declining balance) In subsequent years depreciation is calculated using this formula: DDB = BDf Where DDB = Annual Depreciation (Declining balance) B = Book Value Df = Annual Depreciation Factor (Declining balance) And Book Value B equals Acquisition Cost minus the sum of the annual depreciation charges accumulated to date. B = A - ( ∑ DDB)

Multiples of 125%, 150%, and 200% are commonly used in the declining balance depreciation calculations. The case where a multiple of 200% is used is often referred to as the double-declining balance method. Example 10: A company purchases a machine for $100,000 with an expected 10-year useful life and a $20,000 salvage value after 10 years. What is the annual depreciation in the 10th year using the 200% declining balance method? First we calculate the Annual Depreciation Factor. Df = m/n = 200%/10 years = 20.00% per year The easiest way to organize a declining balance depreciation problem is in tabular form. Year 1 2 3 4 5 6 7 8 9 10

$100,000 x 20% $80,000 x 20% $64,000 x 20% $51,200 x 20% $40,960 x 20% $32,768 x 20% $26,214 x 20% $20,972 x 20% $16,777 x 20% $13,422 x 20%

Annual Depreciation $20,000 $16,000 $12,800 $10,240 $ 8,192 $ 6,554 $ 5,243 $ 4,194 $ 3,355 $ 2,684

Book Value $ 100,000 $100,000-$20,000 $ 80,000 $80,000-$16,000 $ 64,000 $64,000-$12,800 $ 51,200 $51,200-$10,240 $ 40,960 $40,960-$8,192 $ 32,768 $32,768-$6,554 $ 26,214 $26,214-$5,243 $ 20,972 $20,972-$4,194 $ 16,777 $16,777-$3,355 $ 13,422 $13,422-$2,684 $ 10,737

Modified Accelerated Cost Recovery System (MACRS) is the third system. It’s basically a variant of the declining balance method. MACRS uses a variety of conventions for when the property is placed in service, different recovery periods, and several depreciation methods depending on the type of property being considered. The IRS requires that MACRS be used for most business and investment property placed in service after 1986. The IRS helpfully publishes a hundred-page document explaining this complex subject (Publication 946 How to Depreciate Property http://www.irs.gov/pub/irspdf/p946.pdf.) Fortunately, once you’ve identified the appropriate depreciation schedule to use, it’s a simple matter of applying some percentages from a table to calculate the annual depreciation.

Modified Accelerated Cost Recovery System (MACRS). Depreciation schedules run from 3 to 20 years depending on the type of property. Here’s an example for property with a three-year recovery period. Annual Depreciation for 3-Year Recovery Period (Half-Year Convention, Table A-1, IRS Publication 946 http://www.irs.gov/pub/irs-pdf/p946.pdf ) Year 1 2 3 4

5.0.

Depreciation 33.33% 45.45% 14.81% 7.41%

Project Cash Flows

The first step in economic analysis is defining the project’s cash flows. A project might be the purchase of a machine or construction of a building to perform some task or provide some benefit. This usually entails an up front purchase or construction cost, periodic annual expenses during the life of the project such as labor & materials, fuel, debt service, etc., and some residual or resale value that can be recouped at the end of the project. Example 11: Company A plans to purchase a new machine with a 5-year useful life for $100,000 by taking out a 5-year loan at 8% interest, compounded daily. Annual operating costs are estimated at $2,000 per year for labor & materials, escalating at 3% per annum. Fuel costs are estimated at $1,000 the first year, increasing at 5% per annum in years 2 through 5. The salvage or resale value for the machine at the end of 5 years is expected to be $20,000. Straight-line depreciation is assumed. Prepare a tabulation of the project cash flows. Solution: First, we need to calculate the annual payment on the $100,000 loan. Since the interest is compounded daily, we need to find the effective interest rate. i = (1+ ((i(t))/t))t)-1 = (1+ ((0.08)/365))365)-1 = 0.0832776 = 8.33% Where i = effective interest rate i(t) = nominal interest rate t = number of times interest is compounded during year Then input the result in the Present Value for a Uniform Series of Payments formula.

PV = ((1+i)n – 1) / (i(1+i)n ) = ((1+0.0832776)5 – 1) / (0.0832776(1+0.0832776)5 ) = 3.9584553 Annual loan payment = Loan amount/PV factor = $100,000/3.9584553 = $25,262.38 You can also use Payment function in Microsoft Excel for this calculation: PMT(rate,nper,pv) = PMT(0.0832776,5,100000) = $25,262.38 Next calculate the annual depreciation using the Straight-line method. DSL = (A – S)/n Where DSL = Annual Depreciation (Straight-line) A = Acquisition Cost S = Salvage Value n = useful life in years

DSL = (A – S)/n = ($100,000-$20,000)/5 = $16,000/year Third, we calculate the inflated values for labor & materials and fuel costs. Labor & Materials: $2,000 (Year 1) x 1.03 = $2,060 (Year2) x 1.03 = $2,122 (Year3) x 1.03 = $2,185 (Year4) x 1.03 = $2,251 (Year5) Fuel: $1,000 (Year 1) x 1.05 = $1,050 (Year2) x 1.05 = $1,103 (Year3) x 1.05 = $1,158 (Year4) x 1.05 = $1,216 (Year5) Once the annual charges for each item are determined, we put it in a table.

Project Cash Flows Year 0 Purchase Price $ 100,000 Debt Service @ 8% Labor & Materials Fuel Depreciation ----Subtotal---Salvage Value Sum of Cash flows $ 100,000

1

2

3

4

$ (25,262) $ (25,262) $ (25,262) $ (25,262) $ $ (2,000) $ (2,060) $ (2,122) $ (2,185) $ $ (1,000) $ (1,050) $ (1,103) $ (1,158) $ $ (16,000) $ (16,000) $ (16,000) $ (16,000) $ $ (44,262) $ (44,372) $ (44,486) $ (44,605) $ $ $ (44,262) $ (44,372) $ (44,486) $ (44,605) $

Note: Labor & Materials inflated at 3% per annum, Fuel at 5% per annum

5

(25,262) (2,251) (1,216) (16,000) (44,728) 20,000 (24,728)