Engineering EconomicsIntroduction

Why Engineering Economics?

• Accreditation requirement in Ontario

• Engineers have to understand financial implications

• Communicate with the bean counters

Method of Instruction

• 1 Lecture per week

• Lecture PLUS (Participation, Learning, Understanding, Success)– Participation - in class

– Learning - together

– Understanding - individual

– Success - individual

ABC

Demonstration - Lecture PLUSQuick 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?

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)

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

Key NotationInterest ……………………………… Inominal interest rate……………… reffective interest rate (per period)… i Present Value (or Worth) ………… PTotal Number of interest Periods…… NFuture Value………….………….FAnnuity amount …………………….. AA discrete payment or receipt at

the end of an interest period …………An

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 %

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)

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

Compound InterestThe 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)

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

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

Factor Formula Name Description

(P/F,i,n) ni)1(1

Present worthFactor

Converts a singleFuture amount to a Present amount

(F/P,i,n)ni)1( Compound

amountfactor

Converts a single present amount toa single futureamount

Single Payment

P

F

1 2 3 4 5Yrs

i

Two amounts P, F

Example (QC)

P=200

F=?

1 2 3 4 5yrs

i = 6%

A. $267.65B. $267.64C. $268.00

Uniform Series Payment Factors

Factor Formula Name Description

(F/A,i,n) ii n 11 Uniform series

compound amountfactor

Converts a series of uniform payments to a single future amount

(A/F,i,n) 1)1( nii

Sinking fund factor

Converts a single future amount into a series of equal payments that would be necessary to accumulate F in nperiods

More Uniform Series Factors

Factor Formula Name Description

(P/A,i,n)n

n

iii

)1(1)1(

Series present

worth factorConverts a series of equalpayments to a presentamount

(A/P,i,n) 1)1()1(

n

n

iii

Capital recoveryfactor

Finds the series of payments necessary to recover (repay) a presentamount in a fixed numberof periods.

Uniform Series (Annuity)

F

0 1 2 3 4yrs

i

Example, RRSP

5

A A A A A

Uniform Series

P

1 2 3 4yrs

i

Example - bank loan/mortgage

5

A A A A A

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?

P = 150 000

1 2 3 19 20

i = 5.25

A A A A A

Solution by formula

P = 150 000

1 2 3 19 20

i = 5.25

A A A A A

A = P (A/P,i,n)

1)1(

)1(150000

n

n

iii

A

1)0525.1()0525.1(0525.

15000020

20

A

A = 12 292,83$

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

5% 6%5.25%0.0802

0.0872

? (A/P, 5.25%,20)

0802.00872.000.500.600.525.5

0802.0

(A/P, 5.25%,20)= =0.08195

A = 150000(0.08195) = 12 292,50$

Gradient present worth factor

Factor Formula Description

(P/G,i,n)n

n

iiini)1(

1)1(2

Converts an increasing ordecreasing series of payments by a constant amount G to a singlepresent value.

(A/G,i,n) Arithmetic gradient to annuity conversionfactor

1)1(

1

ni

n

i

Arithmetic Gradient

P

1 2 3 4Yrs

i

5

G

2G

3G

4G

Example (QC)

Find the Present value

150 500

500/yr2000

P=?

A. $2 018B. $3 804C. $3 208

i = 10%7

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

• 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

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

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

Quick Check

Do you think the corresponding effective annual interest rate will be:

A. Less

B. Equal

C. More

than the nominal rate?

Example Cont’dWe 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 038Therefore 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

ia = (1 + r/m)m -1

where ia = effective annual rater = nominal interest rate per yearm = 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 = r3. The factor (1+r/m)m increases with m,therefore the more frequent the compounding, the more interest will accumulate.

Quick Check

Choose between:

A) borrowing at 12% compounded monthly

B) borrowing at 13% compounded semi-annually

C) borrowing at 11.5% compounded daily.

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.

Solution

Case 1200$

0 12

i = 15/12 = 1.25%

P = 200 (P/F,1.25,12) = 172,30$

Case 2150$ 100$

0 12 24

i = 15/12 = 1.25%

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.

Quick Check

A student borrowed $10,000 from the bank to buy a new car. The bank charges 10% compounded annually but give the studentthree 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

Effective Interest rate – any payment period

Effective interest are usually based on the payment period, but canBe calculated for any desired period. For example, if cash flow paymentsOccur quarterly, but interest is compounded monthly we may wishTo calculate an effective quarterly interest rate.

i = (1 + r/m)C -1

= (1 + r/CK)C -1Where

M=The number of compounding periods per yearC= The number of compounding periods per payment periodK=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

r = 9% compounded monthly, i=9%/12 = 0.75%

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%

0.75% 0.75% 0.75%

2.27%

0.75% 0.75% 0.75%

2.27%

0.75% 0.75% 0.75% 0.75% 0.75% 0.75%

FirstQuarter

SecondQuarter

ThirdQuarter

FourthQuarter

2.27% 2.27%

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?

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) = $671.76is the remaining balance after the second

payment

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

Loan Tables

EndYr

0123456

Abalance

1 000.00836.40671.57505.51338.20169.630.00

Bpayments

171.10171.10171.10171.10171.10171.10

A(.0075)interest

7.506.275.043.792.541.27

capitalrecovered

163.60164.83166.06167.30168.56169.80

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

You have $1000 available to invest. You also know that6 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

P=1000

F=1600

n=5

i=?

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 TransactionQuick 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?

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

• 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.

Solution

1 2 3 4 5 6

i = 10%

F=?

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)

If the interest rate changes for different periods, the present worthhas 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. $428B. $89C. $484

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

Yr

012345

ABalance at the start

of the yr.

(C) - (D)

7900.006432.004846.563134.281285.03

BInterest

(@8%) duringthe yr.

632.00514.56387.72250.74

CBalance at

the end of yr

(A) + (B)

10000.008532.006946.565234.283385.03

DPayments

at start of yr

21002100210021002100

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

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

Given Ft the net sum of all payments up to period t, the paybackperiod is the smallest value of n that satisfies:

00

n

ttF

Without interest

With interest (discounted payback)

0)1(0

tn

tt iF

Example (QC)

$1 000 invested that returns $200 per yr for 10 yrs (i=8%)

The payback period: without interest: 5 answith interest: 1000=200(P/A,8,n) n = ?

A. n=5B. n=6C. n=7

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 period Ex. 2i=15%

t

0123456

Ft

-1000-500500700

10001500500

n

ttF

0

-1000-1500-1000-300700

22002700

tt iF )1(

-1000-435378460572746216

tn

tt iF

0)1(

-1000-1435-1057-597-25721937

Payback without interest= 4 ans

With interest= 5 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

• construction costs• investment in production equipment• training

– Expected profits– decision required:

• accept/reject the project• need a decision criteria

ExampleThe 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

15 000

- 120 000

- 10 000

5 000 /yr

5 000

5.6502 20.2541 0.3220PW = -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

i = 12%

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

AE = -120000(A/P, 12,10)+15000+5000(A/G,12,10)-10000+5000(A/F,12,10) = $1 968

0.1770 3.5847 0.0570

i = 12%

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.

Note:PW1 FE1 AE1

PW2 FE2 AE2

= =

Present Worth ProfileTo help understand the relationship between the PW and the IRR,

examine the PW as a function of the interest rate i.

($50.0)

$0.0

$50.0

$100.0

$150.0

$200.0

0 5 10 15 20 25

Interest rate %

Pre

se

nt

Wo

rth

($0

00

)

IRR, PW=0= 13.6%MARR=12%

Comparison Example - Equipment Purchase

Machine #1

$10,000

A = $1,000

S = $3,000

1 52 3 4

i=10%

Machine #2

$6,000

A = $2,000

S = $1,500

1 52 3 4

i=10%

NPW for Machine 1 (check)

$10,000

A = $1,000

S = $3,000

1 52 3 4

i=10%

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]= $-11,928

$10,000

A = $1,000

S = $3,000

1 52 3 4

i=10%

AEC (Check)

A. AEC = -10000(A/P,10,5) - 1000 + 3000[A/F,10,5]

C. AEC = $-11,928*(A/P,10,5) * equivalent NPW

$10,000

A = $1,000

S = $3,000

1 52 3 4

i=10%

B. AEC = -10000(P/A,10,5) - 1000 + 3000[A/F,10,5]

Solution - Machine #1

NPW = -10000 - 1000[P/A,10,5] + 3000[P/F,10,5] = $-11,928

$10,000

A = $1,000

S = $3,000

1 52 3 4

i=10%

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

$6,000

A = $2,000

S = $1,500

1 52 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

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)

$

0

i %

5 10 15 20

PW Function CheckP

W(i

)$

0

i %

5 10 15 20 Project A

Project B

Suppose your MARR = 6%, would you:

a. Select project Ab. Select project Bc. Select project A and B

Internal Rate of Return: IRRrendering 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

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*