Chapter 9 engineering economics

Chapter 9 Engineering Economic Analysis

Copyright R.Turton and J. Shaeiwitz - 2012

Chemical Engineering Department

West Virginia University

1

Definitions

• P – Principal or Present Value (of an investment)

• Fn – Future Value (of an investment)

• n – Years (or other time unit) between P and F

• i – Interest Rate (based on time interval of n) per anum

Basic premise: Money when invested earns money

$1 today is worth more than $1 in the future

Copyright R.Turton and J. Shaeiwitz - 2012 2

Interest

• Simple Interest – Annual Basis

– Interest paid in any year = Pis

• Pis – Fraction of investment paid as interest per year

– After n years total interest paid = Pisn

– Total investment is worth = P + Pisn

– Total investment after 1 year (n = 1) = P (1+is)

– What is the drawback of simple interest?


We can earn interest on earned interest

3

Interest

• Compound Interest

At time 0 we have P

At the end of Year 1, we have F1 = P (1 + i )

At the end of Year 2, we have F2 = P (1 + i )2

At the end of Year n, we have Fn = P (1 + i )n

or P = Fn / (1 + i )n


Example

• How much would I need to invest at 8 % p.a. to yield $5000 in 10 years?


10

10

0.08

10

5000

5000$2315.97

1 0.08

i

n

F

P

5

What if Interest Rate Changes with Time?


1 21(1 ) 1 1 ....... 1

n

n j nj

F P i P i i i

Eq. (7.7)

6

Different Time Basis for Interest Calculations

• Relates to statement “Your loan is 6 % p.a., compounded monthly”

• Define actual interest rate per compounding period as r

– inom = Nominal annual interest rate

– m = Number of compounding periods per year (12)


Different Time Basis for Interest Calculations cont.

– ieff = effective annual interest rate

• Look at condition after 1 year


nomir

m

1 1 1

mnom

eff

iF P i P

m

1 1

mnom

eff

ii

m

8

Example

• I invest $1000 at 10 % p.a. compounded monthly. How much do I have in 1 year, 10 years?


12

1

12

10

10

0.101 1000 1 $1104.71

12

0.101 1 0.1047

12

1 $2707.04

mnom

eff

eff

iF P

m

i

F P i

9

Example cont.

• As m decreases ieff increases

• Is there a limit as m goes to infinity

– Yes – continuously compounded interest

– Derivation – pp. 265-266

– ieff (continuous) = e inom – 1


Cash Flow Diagram (CFD)

• Represent timings and approximate magnitude of investment on a cfd

– x-axis is time and y-axis is magnitude

– both positive and negative investments are possible.

• In order to determine direction (sign) of cash flows, we must define what system is being considered.


Consider a Discrete Cash Flow Diagram

• Discrete refers to individual CFDs that are plotted


Example

• I borrow $20 K for a car and repay as a $400 monthly payment for 5 years.


For Bank For Me

1 2 3

$400

60 3 2 1

$20,000

60

$400

$20,000

13

Cumulative CFD


Cumulative CFD

14

Annuities


1 2 3 n

Uniform series of equally spaced, equal value cash flows Note: The first payment is at the beginning of year 1 not

at t = 0

15

Annuities

• What is future value Fn?

• Geometric progression


1 2

1 1 .....n n

nF A i A i A

1 1n

n n

iF S A

i

16

Discount Factors

• Just a shorthand symbol for a formula in i and n


1, ,

1 1

1, ,

1

1 1, ,

1

n n

n

n

n

F PP i n

Fi i

PP F i n F

F i

iPA P i n

A i i

See Table 9.1

17

Discount Factors

n

n

n

n

ii

ini

A

P

ini

F

P

i

ini

A

F

1

11,,

therefore

1

1,,

11,,


Table 9.1 has six versions of these Three are reciprocals of the other three

18

Calculations with Cash Flow Diagrams

• Invest 5K, 1K, 2K at end of Years 0, 1, 3, and take 3K at end of Year 4

• Note that annuity payments are all at the end of the year


0

$3,000

$2,000 $1,000

$5,000

3

7 4

1

19

Example 1

• How much in account at end of Year 7 if i = 8% p.a.?

• What would investment be at Year 0 to get this amount at Year 7?


7 6 47

3

7

5,000 1 0.08 1000 1 0.08 2000 1 0.08

3000 1 0.08

$9097.84

F

F

7

9097.845308.50

1.08P

20

Example 2

• What should my annual monthly car payment be if interest rate is 8% p.a. compounded monthly?


$20,000

A

21

Example 2 (cont’d) • Compare at n = 60


60

60

60

60

0.081 1

1273.47

0.08

12

0.0820,000 1 29,796.90

12

73.47 29,796.90 0

$405.53

F A A

F

A

A

Interest paid = $4,331.20

22

Example 2 (cont’d)

• Another method

20.4331$000,20)52.405(60 paidinterest

52.405$)020276.0(000,2060,08.0,

020276.0

112

08.01

12

08.01

12

08.0

11

1

60,08.0,60

60

P

APA

m

i

m

i

m

i

P

Amn

mn


Example 3

• You buy a house where you finance $200 K at 6% p.a. interest, compounded monthly. What is your monthly payment, and how much interest do you pay over the lifetime of the loan for a 15-year and a 30-year mortgage in current dollars?



30,12mortgageyear -30

15,12mortgageyear -15

11

1

1)1(

)1(

nm

nm

m

i

m

i

m

i

i

ii

P

Amn

mn

n

n



• For 15-year mortgage – $1687.71/month

– total of $303,788 paid

– $103,788 interest

• For 30-year mortgage – $1199.10/month

– total of $431,676 paid

– $231,676 interest


Example 4

• You invest $5000/year (the maximum, for now) in a Roth IRA, starting at age 25 for 40 years. Assuming a return of 8% p.a., how much will you have at age 65 in future dollars?



283,295,1$

5000

06.259

08.0

1)08.01(

1)1(

40

F

A

A

F

A

F

i

i

A

F n


Example 5

• Repeat the previous calculation, assuming that you do not start investing until age 35 or age 45.

800,228$76.4520if

400,566$28.11330if

5000

08.0

1)08.01(

1)1(

FA

Fn

FA

Fn

A

A

F

i

i

A

F

n

n


Depreciation

• Total Capital Investment = Fixed Capital + Working Capital

– Fixed Capital – All costs associated with new construction, but Land cannot be depreciated

– Working Capital – Float of material to start operations cannot depreciate


LTCI FCI Land WC

30

Definitions

• Salvage Value, S – Value of FCIL at end of project

– Often = 0

• Life of Equipment – n – set by IRS

• Not related to actual equipment life

• Total Capital for Depreciation – FCIL - S


4 Basic Methods for Depreciation

• Straight Line

• Sum of Years Digits (SOYD)

• Double Declining Balance (DDB)

• Modified Accelerated Cost Recovery System (MACRS)


Straight Line


SL Lk

FCI Sd

n

n = # of years over which depreciation is taken

33

Sum of Years Digits (SOYD)


1

11

2

LSOYDk

n k FCI Sd

n n

SOYD

34

Double Declining Balance (DDB)


1

0

2k

DDBk L j

j

d FCI dn

35

MACRS

• Current IRS-approved method


Year

1

2

3

4

5

6

Depreciation Percentage

20.00

32.00

19.20

11.52

11.52

5.76

See Chapter 9

Based on combination of DDB and SL

36

Example 9.21


6

6

$150 10

$10 10

7

1 ?

L

stSL

FCI

S

n

Year d

Same for Years 1-7

150 1020

7SLd

37

Example 9.21 (cont’d)


,1

,2

7 1 1 7150 10 150 10 35

1 287 82

7 1 2 6150 10 150 10 30

1 287 82

SOYD

SOYD

d

d

Sum of Year’s Digits

38



,1

,2

2150 42.9

7

2150 42.9 30.6

7

DDB

DDB

d

d

Double Declining Balance

39


64.8)150(0576.0

28.17)150(1152.0

28.17)150(1152.0

8.28)150(192.0

48)150(32.0

30)150(20.0

6,

5,

4,

3,

2,

1,

yrMACRS

yrMACRS

yrMACRS

yrMACRS

yrMACRS

yrMACRS

d

d

d

d

d

d


MACRS

40

Taxation, Cash Flow, and Profit

• Tables 9.3 – 9.4

• Expenses = COMd + dk

• Income Tax = (R – COMd - dk)t

• After Tax (net)Profit =

(R – COMd –dk)(1 – t)

• After Tax Cash Flow =

(R – COMd – dk)(1 – t) + dk (+ other cash flows)

• Other cash flows might include working capital return, salvage value, etc.


Inflation

• $ Net Worth Now vs. $ Next Year

• f = Average inflation rate between years j and n


1n

CEPCI j n f CEPCI j

42

Inflation

• Example


10

0.1

1993 359

2003 402

4021

359

4021 0.0114 or 1.14%

359

CEPCI

CEPCI

f

f

43

Inflation

• What is inflation rate since 2003? The current CEPCI is 600 (2011).

%13.5

0513.014925.1

4925.1402

600)1(

125.0

8

f

f


Inflation • Effect of inflation on interest rate

f affects the purchasing power of the $

• Look at the purchasing power of future worth, then

• If this future worth was obtained by investing at a rate i, then the inflation adjusted interest rate, i ’ is given by


'

(1 )n

FF

f

' '(1 ) 1(1 )

1(1 ) (1 )

nnn

n n

F i iF P P P i

ff f

' 11

1

ii i f

f

45