Engineering Economics MODULE 5 Engineering Economics Engineering Economics Introduction

Why Engineering Economics? MODULE 5 Engineering Economics Why Engineering Economics? Accreditation requirement in Ontario Engineers have to understand financial implications Communicate with the bean counters

MODULE 5 Engineering Economics Method of Instruction 1 Lecture per week Lecture PLUS (Participation, Learning, Understanding, Success) Participation - in class Learning - together Understanding - individual Success - individual A B C

Demonstration - Lecture PLUS Quick Checks MODULE 5 Engineering Economics Demonstration - Lecture PLUS Quick Checks You are buying a new car and have three choices: Choice A B C Auto 57 Chevy 87 Honda 82 Mercedes Price ($) 12,000 7,000 20,000 Operation($) 200/mo. 50/mo. 150/mo. Resale($) 13,000 6,000 20,000 Which would you choose? Sur une base conomique seulement, en ignorant l’inflation et la possibilit d’un changement drastique dans le teux de rendement gnral, le Chevy et preferable. La raison pour ceci sera vident aprs la premire partie de ce cours. Cependent, on devrait aussi prendre en consideration d’autres facteurs tels: la fiabilit du Chevy la disponibilit des fonds quelle auto est le plus confortable quelle auto s’accorde le mieux avec votre mode de vie le Chevy pollue l’environnement beaucoup plus etc….

MODULE 5 Engineering Economics Module 4 Outline Intro to Engineering Economic Decisions Read Chapter 1 of the text Study the Chapter Summary (pp. 36) The time value of money Equivalence and Interest Formulas ( Chapter 2) Real world examples (Chapter 3) Analysis of independent Investments (Chapter 4)

MODULE 5 Engineering Economics Why Pay Interest? The borrower pays the lender for: administrative costs associated with the loan compensation for the risk of default opportunity cost associated with not investing the money elsewhere

MODULE 5 Engineering Economics Key Notation Interest ……………………………… I nominal interest rate……………… r effective interest rate (per period)… i Present Value (or Worth) ………… P Total Number of interest Periods…… N Future Value………….…………. F Annuity amount …………………….. A A discrete payment or receipt at the end of an interest period …………An Autre termes pour Principal: valeur presente, valuer actualise, valeur actuelle, valeur escompte. Autre termes pour annuit: rente (f), redevance, pension

Interest Rate Example (Quick Check) MODULE 5 Engineering Economics Interest Rate Example (Quick Check) If you borrow $1000 from me and agree to pay the $1000 plus $125 extra at the end of one year, what is the interest rate? A. 8 % B. 12.5 % C. .125 % Aprs le quick check: Le rapport de l’intrt payable la fin d’une priode de temps, (d’habitudeune anne ou moins) et l’argent d au dbut de la priode i = I/P ici, I=125 P=1000 i=125/1000 = 0.125 = 12.5 % donc B et correct.

MODULE 5 Engineering Economics Simple Interest: Interest is paid only on the original principal amount, not on any accumulated interest In = Pin Fn = P + In = P(1+in) Rarely used

Simple Interest:Example (QC) MODULE 5 Engineering Economics Simple Interest:Example (QC) How much interest is due for a loan of $10,000 at 10% simple interest: 1) after 1 yr? I = Pin = 10000(.10)(1) = $1000 2) after 2 yrs? A) $1,000 B) $1,100 C) $2,000 I = Pin = 10 000(.10)(2) = 2 000 C.

MODULE 5 Engineering Economics Compound Interest The interest accumulates interest F1= P(1+i) where F1 = future value an the end of period 1 F2= F1 (1+i) = P(1+i) (1+i) = P(1+i)2 F3= P(1+i)3 : Fn= P(1+i)n where, (1+i)n = (F/P,i,n),is called the compound amount factor

Compound Interest (QC) MODULE 5 Engineering Economics Compound Interest (QC) How much borrowed today will have to be repaid by $10,000 in three periods(i =10%) ? A) $9 000 B) $9 091 C) $7 513 F=P(1+i)n P=F/(1+i)n P = 10000/(1.10)3 = 10000/1.331 = 7513 C.

Compound Interest Factors MODULE 5 Engineering Economics Compound Interest Factors There are three categories of factor: those that convert a single payment to a present or future amount those that convert a series of payments to a present or future amount those that convert a series of payments that increase or decrease by some constant amount G (for Gradient series) every period to an equivalent uniform series

Single Payment Factors MODULE 5 Engineering Economics Single Payment Factors Factor Formula Name Description (P/F,i,n) Present worth Factor Converts a single Future amount to a Present amount (F/P,i,n) Compound amount factor present amount to a single future

MODULE 5 Engineering Economics Single Payment F i Yrs 1 2 3 4 5 P Two amounts P, F

MODULE 5 Engineering Economics Example (QC) F=? i = 6% yrs 1 2 3 4 5 P=200 A. F=200(1.06)5 = 267.6451 B. F=200(F/P,6,5)= 200(1.3382)= 267,64 C. errondissment pour les deux. Reponse, A,B ou C A. $267.65 B. $267.64 C. $268.00

Uniform Series Payment Factors MODULE 5 Engineering Economics Uniform Series Payment Factors Factor Formula Name Description Uniform series compound amount factor Converts a series of uniform payments to a single future amount (F/A,i,n) Sinking fund factor Converts a single future amount into a series of equal payments that would be necessary to accumulate F in n periods (A/F,i,n)

More Uniform Series Factors MODULE 5 Engineering Economics More Uniform Series Factors Factor Formula Name Description Series present worth factor Converts a series of equal payments to a present amount (P/A,i,n) Capital recovery factor Finds the series of payments necessary to recover (repay) a present amount in a fixed number of periods. (A/P,i,n)

Uniform Series (Annuity) MODULE 5 Engineering Economics Uniform Series (Annuity) A A A A A yrs 0 1 2 3 4 5 i F Example, RRSP

MODULE 5 Engineering Economics Uniform Series A A A A A yrs 1 2 3 4 5 i P Example - bank loan/mortgage

Example - Loan repayment MODULE 5 Engineering Economics Example - Loan repayment You take out a mortgage (150 000$) for a term of 20 yrs, with i=5.25% per yr. What are your annual payments? A A A A A 1 2 3 19 20 i = 5.25 P = 150 000

MODULE 5 Engineering Economics Solution by formula A A A A A 1 2 3 19 20 i = 5.25 P = 150 000 A = P (A/P,i,n) A = 12 292,83$

Solution by interpolation MODULE 5 Engineering Economics Solution by interpolation A = P (A/P,i,n) A = 150000 (A/P,5.25,20) (A/P, 5, 20) = 0.0802 (A/P, 6, 20) = 0.0872 0.0872 ? (A/P, 5.25%,20) 0.0802 5% 5.25% 6% (A/P, 5.25%,20)= =0.08195 A = 150000(0.08195) = 12 292,50$

Gradient present worth factor MODULE 5 Engineering Economics Gradient present worth factor Factor Formula Description Converts an increasing or decreasing series of payments by a constant amount G to a single present value. (P/G,i,n) Arithmetic gradient to annuity conversion factor (A/G,i,n)

MODULE 5 Engineering Economics Arithmetic Gradient 4G 3G 2G G Yrs 1 2 3 4 5 i P

MODULE 5 Engineering Economics Example (QC) Find the Present value P=? i = 10% 7 150 500 500/yr P= 150(P/A,10,3) + 500(P/G,10,5)(P/F,10,2) =150(2.4869)+500(6.8618)(0.8264) =373.035 + 2835.296 =3208.33 (C) Erreurs P=150(P/A,10,3) +500(P/G,10,5) = 3803.94 (B) P=150(P/A,10,3) + 500(P/G,10,4)(P/F,10,3) = 150(2.4869)+500(4.3781)(.7513) =2017.67 (A) Aussi, une a la fois, P = F/(1+i)n 2000 A. $2 018 B. $3 804 C. $3 208

Formula Summary See pp. 94 Complete Assignment 1(from Chapter 2)

Chapter 2 Review Problems The following questions from Chapter 2 are recommended: Level 1: 2.1 to 2.8 (answers on pp. 905), 2.20 to 2.22, 2.33 to 2.35 Level 2: 2.40 to 2.42, 2.50 to 2.51 Level 3: 2.65, 2.67 a only

Nominal and Effective Interest MODULE 5 Engineering Economics Nominal and Effective Interest Generally interest rates are quoted on an annual basis (annual percentage rate - APR); but the contract may specify that compounding occurs more frequently: monthly quarterly (every 3 months) daily etc

MODULE 5 Engineering Economics Nominal Vs. Effective What is the effect of compounding more frequently than once per yr? Given: r = nominal interest per yr (APR - annual percentage rate) m= number of times per year (sub-periods) that interest is compounded is = r/m = effective interest rate per period

MODULE 5 Engineering Economics Example $10 000 is deposited in the bank at a nominal rate of 10% per year, compounding is quarterly.. Given: r = 10% m = 4 periods per yr. Therefore: is = r/m = 10/4 = 2.5% per quarter

MODULE 5 Engineering Economics Quick Check Do you think the corresponding effective annual interest rate will be: A. Less B. Equal C. More than the nominal rate? A, a cause de l’intrt compose.

MODULE 5 Engineering Economics Example Cont’d We have P=10,000, is=2.5% Therefore: F3mo=P(1+ is) = 10 000(1.025)= 10 250 F6mo = F3mo(1+ is)= 10250(1.025)= 10 506 F9mo = F6mo(1+ is)= 10 506(1.025) = 10 769 F12mo= F9mo(1+ is) = 11 038 Therefore the accumulated interest at the end of the yr = $1,038 which is more than the $1,000 That we would receive with simple interest The Effective annual rate : ie =(F-P)/P = (11038-10000)/10000 = 10.38%

Formula for calculating ia for 1 yr MODULE 5 Engineering Economics Formula for calculating ia for 1 yr ia = (1 + r/m)m -1 where ia = effective annual rate r = nominal interest rate per year m = number of compounding periods per yr Note: 1. This formula applies when compounding is more frequent than once per yr. 2. If m=1, ia = r 3. The factor (1+r/m)m increases with m,therefore the more frequent the compounding, the more interest will accumulate.

MODULE 5 Engineering Economics Quick Check Choose between: A) borrowing at 12% compounded monthly B) borrowing at 13% compounded semi- annually C) borrowing at 11.5% compounded daily. ia = (1+0.12/12)12 - 1 = .1268 ib = (1+0.13/2)2 - 1 = .1342 ic = (1+0.115/365)365 - 1 = .1218 Alors,Choisir c

MODULE 5 Engineering Economics Example Would you prefer to receive: $200 after 1 yr (case 1) $150 in 1 yr and another $100$ after 2 yrs (case 2)? The interest rate at the bank is 15% compounded monthly.

MODULE 5 Engineering Economics Solution 200$ Case 1 i = 15/12 = 1.25% 12 P = 200 (P/F,1.25,12) = 172,30$ 150$ 100$ i = 15/12 = 1.25% Case 2 12 24 Or ieff = (1+0.15/12)12 - 1 = 0.16075 Cas 1 P= 200/1.16075 = 172,30 Cas 2 P = 150/1.16075 + 100/(1.16075)2 = 203,45$ P = 150 (P/F,1.25,12) + 100(P/F,1.25,24) = $203.45 By calculating a single present amount that is equivalent to each cash flow we can compare the 2 cases directly.

MODULE 5 Engineering Economics Quick Check A student borrowed $10,000 from the bank to buy a new car. The bank charges 10% compounded annually but give the student three repayment options. Which would you choose? End Yr Plan A Plan B Plan C 1 $2638 - $1800 2 2638 - 1800 3 2638 - 1800 4 2638 - 1800 5 2638 16105 6916 Total $13190 $16105 $14116 Plan A P=2638(P/A,10,5) F=2638(F/A,10,5) =2638(3.7908) = 10 000$ = 2638(6.1051)=16105 Plan B P=16105(P/F,10,5) F=16105 =16105(0.6209) = 10 000 Plan C P = 1800(P/A,10,$) + 6916(P/F,10,5) F=16105. = 1800(3.1699) + 6916(0.6209) = 10 000 Nous obtenons le mme rsultat c.a.d. les trois options sont equivalentes! Exercise: Trouvez Valeur Future et Annuite Equiv

Effective Interest rate – any payment period Effective interest are usually based on the payment period, but can Be calculated for any desired period. For example, if cash flow payments Occur quarterly, but interest is compounded monthly we may wish To calculate an effective quarterly interest rate. i = (1 + r/m)C -1 = (1 + r/CK)C -1 Where M=The number of compounding periods per year C= The number of compounding periods per payment period K=The number of payment periods per year Note that M=CK

Example – effective rate per period Suppose you make quarterly deposits in a savings account which earns 9% interest compounded monthly. Calculate the effective interest rate per quarter. r=9%, C = 3 compounding periods per quarter K=4 quarterly payments per year, and M=12 compounding periods per year. Or, M=CK = 3x4=12 i = (1 + r/m)C -1 i = (1 + 0.09/12)3 –1 = 2.27%

Example explained First Quarter Second Third Fourth r = 9% compounded monthly, i=9%/12 = 0.75% 0.75% 0.75% 0.75% 2.27% 0.75% 0.75% 0.75% First Quarter Second Third Fourth i = (1 + 0.09/12)3 –1 = 2.27% Effective interest rate per quarter ia = (1 + r/m)m –1 = (1+.0075)12-1 = (1+.0227)4-1 = 9.38%

Equivalence and demand loans MODULE 5 Engineering Economics Equivalence and demand loans A demand loan is a loan that either the lender or borrower can decide to have terminated by immediate payment of the outstanding balance. In practice, borrowers often choose to pay off a loan in order to save on interest charges. Using the principal of equivalence, the remaining balance of a loan can be determined in two ways. Example: Julie borrowed $1 000 from the bank at 9% compounded monthly. She agreed to repay the loan in 6 monthly payments but since this is a demand loan, she can clear the balance at any time. Immediately following her second payment, Julie wins big in the lottery and she decides to clear off her debt. How much must she pay to clear the loan?

MODULE 5 Engineering Economics Solution - Method 1 Payments: A = 1000(A/P,9/12,6) = 1000(.1711) = $171.10/mo We find an equivalent value of all preceding transactions just after the second payment: B2 = 1000(F/P,.75,2) - 171.1(F/A,.75,2) = 1000(1.015)-1.71(2.008) = $671.76 is the remaining balance after the second payment

MODULE 5 Engineering Economics Solution - Method 2 We simply discount the remaining payments to the time at which the loan is to be paid off. I.e., we calculate the PW of the remaining payments. B2 =171.1(P/A,.75,4) =171.1(3.9261) = $671.76

MODULE 5 Engineering Economics Loan Tables End Yr 1 2 3 4 5 6 A balance 1 000.00 836.40 671.57 505.51 338.20 169.63 0.00 B payments 171.10 A(.0075) interest 7.50 6.27 5.04 3.79 2.54 1.27 capital recovered 163.60 164.83 166.06 167.30 168.56 169.80

Calculations involving compound interest MODULE 5 Engineering Economics Calculations involving compound interest The frequency of payments is not always the same as the compounding period Usually we have 3 of {P,A,F,i,n} and we have to find the forth We generally assume that P occurs at the beginning of the first period (or at the end of period 0) and that F and A occur at the end of the period (end of period convention)

Example - Single Transaction MODULE 5 Engineering Economics Example - Single Transaction You have $1000 available to invest. You also know that 6 years from now you have a requirement for $1600. What rate of return i is necessary achieve $1600? P=1000 F=1600 n=6 i=?

Solution - single transaction MODULE 5 Engineering Economics Solution - single transaction F=1600 i=? n=5 P=1000 F = P(F/P,i,n) therefore, (F/P,i,6) = F/P = 1.6 = (1+i)6  (1+i) = 1.6(1/6) = 1.08148 therefore, i = 0.08148 = 8.15%/yr Therefore, we require an internal rate of return of 8.15%/ yr. To meet the $1600 requirement.

Multiple Transaction Quick Check MODULE 5 Engineering Economics Multiple Transaction Quick Check In general, most projects have a combination of individual transactions and annuities. Example: An investment pays $10,000 immediately and $1,000 at the end of each yr for a period of 5 yrs and also has an individual payment of $2,000 at the end of the 5th year. If i=7%, what is the present worth of the transactions?

MODULE 5 Engineering Economics Special Cases There are 3 special cases that prevent the direct application of the compound interest factors: Compounding is more frequent than payments The interest rate is not constant for the whole period Annuities occur at the start of the period

Compounding more frequent than payments MODULE 5 Engineering Economics Compounding more frequent than payments Example: three deposits of $2,500 are made every 2 years starting in 2 years. i = 10%/yr, how much will be in the bank at the end of 6 yrs? The series compound amount factor (F/A, i, n) cannot be used directly because the payment interval does not match the compounding period.

MODULE 5 Engineering Economics Solution F=? i = 10% 1 2 3 4 5 6 2500 2500 2500 Method 1: F = 2500 + 2500[(F/P, 10,2) + (F/P, 10,4)] = 2500 + 2500[(1.21) + (1.4641)] = $9,185 Method 2: we calculate an effective interest rate to match the compounding period. I.e. an effective 2 yr interest rate. ie = (1.10)2-1 = 0.21 = 21% F = 2500(F/A,21,3) = 2500(3.6741) = $9185

Variable interest rates (QC) MODULE 5 Engineering Economics Variable interest rates (QC) If the interest rate changes for different periods, the present worth has to be calculated in steps. Ex: Given the following CFD, calculate the PW 0 1 2 3 4 5 6 i = 5% i = 7% i=4% 30 20 10 A. $428 B. $89 C. $484

Annuities at the start of the period MODULE 5 Engineering Economics Annuities at the start of the period For annuities where the first payment is made today instead of at the end of the first period P = A + A(P/A,i,n-1) Ex: Jaqueline needs to rent a machine for 5 yrs. She has to choose between paying $10,000 up front or paying $2,100 at the start of each of the 5 yrs. Use i = 8%. P = 2100 + 2100(P/A,8,4) = 2100(1+3.3121) = $9,055 Conclusion: it is less expensive to pay monthly. The difference between the up front cost and the PW of the annuity is $945. How would you interpret this amount?

Interpretation - Present Worth MODULE 5 Engineering Economics Interpretation - Present Worth Yr 1 2 3 4 5 A Balance at the start of the yr. (C) - (D) 7900.00 6432.00 4846.56 3134.28 1285.03 B Interest (@8%) during the yr. 632.00 514.56 387.72 250.74 C Balance at the end of yr (A) + (B) 10000.00 8532.00 6946.56 5234.28 3385.03 D Payments at start of yr 2100

MODULE 5 Engineering Economics Quick Check The PW of the remaining balance after the last payment is… A) P = 1285.03(P/A,8%,5) B) P = 1285.03(P/F,8%,4) C) P= 1285.03(P/F,8%,5)

Chapter 3 Review Problems The following questions from Chapter 3 are recommended: Level 1: 3.2 to 3.9, 3.11 to 3.17 (answers on pp. 905) Level 2: 3.22, 3.26, 3.27 a and b only, 3.31, 3.38, 3.40, 3.43, 3.45, 3.54, 3.58, 3.59, 3.91 Level 3: 3.95. For part b, use Excel’s IRR function to find the effective interest rate.

Analysis of Independent Investments MODULE 5 Engineering Economics Analysis of Independent Investments

Outline Cash flow representation of projects Payback Period without interest with interest Decision Criteria NPW AW FW IRR Capitalized Value

The Payback Method Justification important to know when a project starts to make a profit traditional payback approach ignores the time value of money do not require a MARR for calculations Useful for project screening

Payback Method Without interest With interest (discounted payback) Given Ft the net sum of all payments up to period t, the payback period is the smallest value of n that satisfies: Without interest With interest (discounted payback)

MODULE 5 Engineering Economics Example (QC) $1 000 invested that returns $200 per yr for 10 yrs (i=8%) The payback period: without interest: 5 ans with interest: 1000=200(P/A,8,n) n = ? A. n=5 B. n=6 C. n=7 (P/A,8,n) = 1000/200 = 5 n=5, P/A = 3.9927 n=6, P/A = 4.6229 n=7, P/A = 5.2 Therefore between 6 and 7, therefore 7.

Disadvantages of payback period MODULE 5 Engineering Economics Disadvantages of payback period Suppose there were a second option in the previous example that returned the $1 000 after only 1 yr but had no subsequent receipts. The payback period would be 1 yr, but the investment would not earn any interest!!! Even though the first option earned: 1000=200(P/A,i,10)  i=15.1% (IRR) Because payback ignores transactions that occur after the payback has occurred it should only be used as supplemental information and never as the main decision.

Payback without interest Payback period Ex. 2 i=15% t 1 2 3 4 5 6 Ft -1000 -500 500 700 1000 1500 -1500 -300 2200 2700 -435 378 460 572 746 216 -1435 -1057 -597 -25 721 937 With interest = 5 ans Payback without interest = 4 ans

Typical Project Cash Flow Initial investment Followed by a series of expected profits 1 2 3 4 5 6 …..

Project Example New Factory Initial investment Expected profits construction costs investment in production equipment training Expected profits decision required: accept/reject the project need a decision criteria

Example The company Fevolard of Kingston is considering the purchase of a new packaging machine. One model has an initial cost of $120 000 and a salvage value of $5 000 after it’s 10 yr service life. Given an anticipated increase in sales revenue the machine is anticipated to result in a cost savings of $15 000$ the first yr. Increasing by $5 000 each of the following 9 yrs. Supplemental operating costs associated with this piece of equipment are $10 000 per yr. The company uses a required rate of return of 12%. Should they invest in the machine? 15 000 - 120 000 - 10 000 5 000 /yr 5 000 i = 12%

PW Criteria Conclusion: Accept the project because PW(@12%) > 0 15 000 - 120 000 - 10 000 5 000 /yr 5 000 i = 12% 5.6502 20.2541 0.3220 PW = -120 000+[ (15000-10000) (P/A,12,10)]+5000(P/G,12,10)+5000(P/F,12,10) PW = $11 132 Conclusion: Accept the project because PW(@12%) > 0

Quick Check If PW(@ MARR) = 0 we should: A. Accept the project B. reject the project

Annual Equivalent (AE) We could solve the same problem by calculating an single annuity amount that is equivalent to entire project cash flow as follows: 15 000 - 120 000 - 10 000 5 000 /yr 5 000 i = 12% 0.1770 3.5847 0.0570 AE = -120000(A/P, 12,10)+15000+5000(A/G,12,10)-10000+5000(A/F,12,10) = $1 968

Annual Equivalent (AE) Of course, if we already have the PW, we can calculate the AE directly: AE=PW(A/P,12,10) = 11 132(0.1770) = $1 970

Quick Check Do you believe that the PW criteria and the AE criteria will always yield the same decision? A. Yes B. No C. It depends

Future Equivalent The same reasoning applies if we chose to use FE as our decision criteria. The FW, PW and AE can be calculated directly from each other by multiplying by a positive constant amount. Therefore, all three methods will yield consistent results. PW1 FE1 AE1 Note: = = PW2 FE2 AE2

Present Worth Profile To help understand the relationship between the PW and the IRR, examine the PW as a function of the interest rate i. IRR, PW=0 = 13.6% MARR=12%

Comparison Example - Equipment Purchase Machine #1 $10,000 A = $1,000 S = $3,000 1 5 2 3 4 i=10% Machine #2 $6,000 A = $2,000 S = $1,500 1 5 2 3 4 i=10%

NPW for Machine 1 (check) MODULE 5 Engineering Economics NPW for Machine 1 (check) S = $3,000 i=10% 1 2 3 4 5 A = $1,000 $10,000 A. P = -10000 - 1000(P/F,10,5) + 3000(P/F,10,5) B. P = -10000 - 1000(P/A,10,5) + 3000(P/F,10,5) C. P = -10000 - 1000(A/P,10,5) + 3000(P/F,10,5)

NPW - Machine #1 NPW = -10000 - 1000[P/A,10,5] + 3000[P/F,10,5] $10,000 A = $1,000 S = $3,000 1 5 2 3 4 i=10% NPW = -10000 - 1000[P/A,10,5] + 3000[P/F,10,5] = $-11,928

AEC (Check) A. AEC = -10000(A/P,10,5) - 1000 + 3000[A/F,10,5] $10,000 A = $1,000 S = $3,000 1 5 2 3 4 i=10% A. AEC = -10000(A/P,10,5) - 1000 + 3000[A/F,10,5] B. AEC = -10000(P/A,10,5) - 1000 + 3000[A/F,10,5] C. AEC = $-11,928*(A/P,10,5) * equivalent NPW

Solution - Machine #1 NPW = -10000 - 1000[P/A,10,5] + 3000[P/F,10,5] $10,000 A = $1,000 S = $3,000 1 5 2 3 4 i=10% NPW = -10000 - 1000[P/A,10,5] + 3000[P/F,10,5] = $-11,928 AEC = -10000(A/P,10,5) - 1000 + 3000[A/F,10,5] = $-3,146.50 AEC = $11,928(A/P,10,5) = $-3,146.50

Solution - Machine #2 NPW = -6000 - 2000[P/A,10,5] + 1500[P/F,10,5] $6,000 A = $2,000 S = $1,500 1 5 2 3 4 i=10% NPW = -6000 - 2000[P/A,10,5] + 1500[P/F,10,5] = $-12,650 AEC = -6000(A/P,10,5) - 2000 + 1500[A/F,10,5] = $-3,337.10 AEC = NPW (A/P,10,5) = $-12,650(A/P,10,5) = $-3,337.10

MODULE 5 Engineering Economics Solution Summary Machine #1 NPW = $-11,928 AEC = $-3,146.50 Machine #2 NPW = $-12,650 AEC = $-3,337.10 With equal lives, the two methods yield consistent results.

The Net Present Worth Function PW(i) $ 5 10 15 20 i %

PW Function Check Suppose your MARR = 6%, would you: a. Select project A b. Select project B c. Select project A and B PW(i) $ i % 5 10 15 20 Project A Project B

Internal Rate of Return: IRR rendering revenue equivalent to expenses MODULE 5 Engineering Economics Internal Rate of Return: IRR rendering revenue equivalent to expenses Ex: Your company can buy a machine for $10,000 and then rent it out for 2500 per yr for it’s service life of 5 yrs. What is the is the interest rate (IRR) for this investment? P = 10 000 A = 2 500 n = 5 i =? Therefore, 10000=2500(P/A,i,5), (P/A,i,5 ) = 10000/2500 = 4 From the interest tables: (P/A,8,5) = 3.993 and (P/A,7,5) = 4.1001 By interpolation: i = .07+.01[(4.1001-4)/(4.1001-3.993)] = .07935 or IRR = 7.9% Using Excel 7.93: DEMO

MODULE 5 Engineering Economics Symbol Convention i* represents the interest rate that makes the NPV of the project equal to zero IRR represents the internal rate of return of the investment, for simple investments IRR=i*, and this is frequently referred to as the rate of return ROR

Chapter 4 Review Problems The following questions from Chapter 4 are recommended: Level 1: 4.1, 4.7 a and c only, 4.8,4.10, 4.18, 4.23, 4.28, 4.35 Level 2: 4.38, 4.43, 4.46, 4.62 Level 3: 4.67 use excel to find i*