CHAPTER 5 Engineering Economic

(1)

EVALUATING SINGLE

PROJECT

CHAPTER 5

(2)

Introduction

•

Capital projects should consider return on theCapital projects should consider return on the

capital that is sufficiently attractive in view of the

risks involved and potential alternative uses.

•

We will discuss 5 methods for evaluating theWe will discuss 5 methods for evaluating the

economic profitability of a project:

1 1.. PPrreesseennt t wwoorrtth h ((PPWW)) 2 2.. FFuuttuurre e wwoorrtth h ((FFWW)) 3 3.. AAnnnnuuaal l wwoorrtth h ((AAWW)) 4 4.. IInntteerrnnaal l RRaatte e oof f RReettuurrn n ((IIRRRR)) 5 5.. EExxtteerrnnaal l RRaatte e oof f RReettuurrn n ((EERRRR))

Convert cash flows into equivalent worth at some point

Convert cash flows into equivalent worth at some point (s)(s)

in time using an interest rate

in time using an interest rate known asknown as M i n i m u mM i n i m u m

A

(3)

Introduction

•

Capital projects should consider return on theCapital projects should consider return on the

capital that is sufficiently attractive in view of the

risks involved and potential alternative uses.

•

We will discuss 5 methods for evaluating theWe will discuss 5 methods for evaluating the

economic profitability of a project:

1 1.. PPrreesseennt t wwoorrtth h ((PPWW)) 2 2.. FFuuttuurre e wwoorrtth h ((FFWW)) 3 3.. AAnnnnuuaal l wwoorrtth h ((AAWW)) 4 4.. IInntteerrnnaal l RRaatte e oof f RReettuurrn n ((IIRRRR)) 5 5.. EExxtteerrnnaal l RRaatte e oof f RReettuurrn n ((EERRRR))

Convert cash flows into equivalent worth at some point

Convert cash flows into equivalent worth at some point (s)(s)

in time using an interest rate

in time using an interest rate known asknown as M i n i m u mM i n i m u m

A

(4)

The

The basic basic question question to to be be addressed:addressed:

is the proposed

is the proposed c a p i t ac a p i t al il in vn v e s t me s t me n t e n t and its associated and its associated e x p e n d i t u r e s

e x p e n d i t u r e s ( ( cash cash out out flow)flow)

can be recovered by

r e v e

r e v en un u e (e (s a v i n gs a v i n g s ) s ) over time over time (cash in flow)(cash in flow)

in addition to a

in addition to a r e tr e tu r n u r n o n o n t h e c a p i t at h e c a p i t al l (rate of return ~ MARR)(rate of return ~ MARR)

that is sufficiently attractive in view of the risk involved.

(5)

MINIMUM ATTRACTIVE RATE OF

RETURN ( MARR )

•

• An interest

An interest

rate used

to convert

cash flo

ws into

equivalent worth at some point(s) in time

•

• Usually a policy issue based on:

Usually a policy issue based on:

- amount, source and cost of money available for

investment

- number and purpose of good projects available for

investment

- amount of perceived risk of investment

opportunities and estimated cost of administering

projects ove

projects over short and long runr short and long run

- type of organization involved

MARR is sometimes referred to as hurdle rate

(6)

CAPITAL RATIONING

•

Establishing MARR involves opportunity cost viewpoint results from phenomena of CAPITAL

RATIONING

•

Exists when management decides to restrict the

total amount of capital invested, by desire or limit of available capital

•

Select only those projects which provide annual rate of return in excess of MARR

•

As amount of investment capital and opportunities

available change over time, a firm’s MARR will also

change

(7)

PRESENT WORTH METHOD ( PW )

• Based on concept of equivalent worth of all

cash flows relative to the present as a base

• All cash inflows and outflows discounted to

present at interest -- generally MARR

• PW is a measure of how much money can be

afforded for investment in excess of cost

• PW is positive if dollar amount received for

investment exceeds minimum required by

investors

(8)

FINDING PRESENT WORTH

•

Discount future amounts to the present by using the interest rate over the appropriate study period

PW =

F

_k

( 1 +

i

)

- k

–

i = effective interest rate, or MARR per compounding period

–

k = index for each compounding period

–

F_k = future cash flow at the end of period k

–

N = number of compounding periods in study period

•

interest rate is assumed constant through project

•

The higher the interest rate and further into future a cash flow occurs, the lower its PW

k = 0 N

(9)

FUTURE WORTH METHOD (FW )

•

FW is based on the equivalent worth of all cash inflows and outflows at the end of the planning

horizon at an interest rate that is generally MARR

•

The FW of a project is equivalent to

FW = PW ( F / P, i %, N )

•

If FW > 0, it is economically justified

FW (

i

% ) =



_F

k

( 1 +

i

)

N - k k = 0 N

–i = effective interest rate

–k = index for each compounding period

–F_k = future cash flow at the end of period k

(10)

ANNUAL WORTH METHOD ( AW )

•

AW is an equal annual series of dollar amounts, over a stated period ( N ), equivalent to the cash inflows and outflows at interest rate that is generally MARR

•

AW is annual equivalent revenues ( R ) minus annual equivalent expenses ( E ), less the annual equivalent capital recovery (CR)

AW ( i % ) = R - E - CR ( i % )

•

AW = PW ( A / P, i %, N )

•

AW = FW ( A / F, i %, N )

•

If AW > 0, project is economically attractive

(11)

A piece of new equipment has been proposed by engineers to increase the productivity of a certain manual welding operation. The investment cost $25,000, and the equipment will have a salvage

value $5,000 at the end of a five year study period. Increased productivity attributable to this equipment will amount to $8,000 per year after extra operating costs have been subtracted from the revenue

generated by the additional production. The firm MARR is 20% per year.

a) Draw the cash flow diagram

(12)

CAPITAL RECOVERY ( CR )

• CR is the equivalent uniform annual cost of

the capital invested

• CR is an annual amount that covers:

–

Loss in value of the asset

–

Interest on invested capital ( i.e., at the MARR )

CR (

i

% ) = I ( A / P,

i

%, N ) - S ( A / F,

i

%, N )

I = initial investment for the project

S = salvage ( market ) value at the end of the

study period

(13)

Example of capital recovery calculation

Consider a machine that cost $10,000, last for five years, and have a salvage (market) value $2,000.

Thus, the loss in value of this asset over five years is $8,000. Additionally, the MARR is 10% per year. Thus CR (i%)=I ( A/P ,i%,N )

–

S( A/F,i%,N )

Solution:

CR (10%) = $10,000( A/P,10%,5 ) - $2,000( A/F,10%,5 )

= $10,000 (0.2638) - $2,000 (0.1638) = $2,310.40

(14)

Example of Annual Worth Method:

A piece of new equipment has been proposed by engineers to increase the productivity of a certain manual welding operation. The investment cost $25,000, and the equipment will have a market value $5,000 at the end of a study period of five years. Increased productivity attributable to this equipment will amount to $8,000 per year after extra operating costs have been

subtracted from the revenue generated by the additional

production. If the firm MARR is 20% per year, is this proposal a sound one? Use the AW method.

Solution:

AW (i%) = R - E – CR (i%)

AW (20%) = $8,000 – [$25,000(A/P,20%,5) - $5,000(A/F,20%,5)] = $8,000 – [$8360 - $672]

= $312

Conclusion: because AW (20%) is positive, the equipment more than pays for itself over the planning horizon.

(15)

Problem 4.19a

A certain service can be performed satisfactorily by process X, which has a capital investment cost of $8,000, an estimated life of 10 years, no salvage value, an annual net receipts (revenue

– expenses) of $2,400. Assuming a MARR of 18% before

income taxes, find the AW of this process and specify whether you would recommend it.

Solution:

AW (i%) = R - E – CR (i%)

AW (18%) = $2,400 – [$8,000(A/P,18%,10) - $0(A/F,18%,10)] = $2,400 – [$1,780 - $0]

= $620

Conclusion: because AW (18%) is positive, the process X is recommended.

(16)

CAPITAL RECOVERY ( CR)

• CR is also calculated by adding sinking fund

amount (i.e., deposit) to interest on original

investment

CR (

i

% ) = ( I - S ) ( A / F,

i

%, N ) + I (

i

% )

• CR is also calculated by adding the equivalent

annual cost of the uniform loss in value of the

investment to the interest on the salvage value

(17)

• 5 methods for evaluating the

economic profitability of a project:

1. Present worth (PW)

√

2. Future worth (FW)

√

3. Annual worth (AW)

√

4. Internal Rate of Return (IRR) 5. External Rate of Return (ERR)

(18)

INTERNAL RATE OF RETURN METHOD ( IRR )

• IRR solves for the interest rate that equates the

equivalent worth of an alternative’s cash

inflows (receipts or savings) to the equivalent

worth of cash outflows (expenditures)

• Also referred to as:

– investor’s method

–

discounted cash flow method

–

profitability index

• IRR is positive for a single alternative only if:

–

both receipts and expenses are present in the cash flow pattern

(19)

INTERNAL RATE OF RETURN METHOD ( IRR )

• IRR is

i

’

%, using the following PW formula:



_R

k

( P / F,

i

’

%, k ) =



E

_k

( P / F,

i

’

%, k )

R

_k

= net revenues or savings for the kth year

E

_k

= net expenditures including investment

costs for the kth year

N = project life ( or study period )

• If

i

’

> MARR, the alternative is acceptable

• To compute IRR for alternative, set net PW = 0

PW =  R _k ( P / F, i ’ %, k ) -  E _k ( P / F, i ’ %, k ) = 0

•

i ’ is calculated on the beginning-of-year unrecovered

investment through the life of a project N k = 0 N k = 0 N k = 0 N k = 0

(20)

DISADVANTAGES OF IRR

• The IRR method assumes recovered funds, if not

consumed each time period, are reinvested at i ‘ %, rather than at MARR

• The computation of IRR may be unmanageable

• Multiple IRR’s may be calculated for the same problem

• The IRR method must be carefully applied and interpreted in the analysis of two or more alternatives, where only one is acceptable

• Basic IRR method cannot rank mutually exclusive projects, the project with higher IRR potentially having lower net

(21)

ADVANTAGES OF IRR

• IRR can be calculated without having to estimate cost of capital or MARR

• IRR, in the form of rate of return is more appealing to evaluate investment

• IRR is more appealing to communicate profitability

• When unique (not multiple IRR’s), it provides valuable

information about the return on the investment and is often viewed as a measure of efficiency

(22)

Using Multiple Internal Rates of Return: The IRR Parity Technique

PW vs rate of return -250 -200 -150 -100 -50 0 50 0 8 1 6 2 6 3 6 4 5 5 4 6 4 7 4 8 4 9 4 1 0 4 1 1 4 1 2 4 1 3 4 1 4 4 1 5 4 rate of return (%) P W ( $ )

Multiple IRRs @ 30% and 60%

EOY Cash flows ($) 0 -1000 1 2900 2 -2080

(23)

Using Multiple Internal Rates of Return: The IRR Parity Technique According to IRR parity technique proposed by Zhang (2005)

If the number of real IRRs which is greater than the MARR is: • Even (including zero), reject the project;

• Odd, accept the project.

Referring to the earlier chart of PW vs. rate of return

• If 30% < MARR < 60%, there is one real IRR (60%) that is greater than

MARR; therefore the project should be accepted. Then we identify the relevant IRR as 60%.

• If MARR < 30%, there are two real IRRs (both 30% and 60%) that are

greater than MARR. Hence, the project should be rejected.

• If MARR > 60%, there is no real IRR that is greater than MARR;

therefore the project should be rejected.

(24)

Example of Internal Rate of Return method

A capital investment of $10,000 can be made in a project that will produce a uniform annual income revenue of $5,310 for five years and then have a salvage value of $2,000. Annual expenses will be $ 3,000. The company is willing to accept any project that will earn at least 10% per year on all invested capital. Determine whether it is acceptable by using IRR method.

Solution:

Set net PW = 0

PW = 0 = -$10,000 + ($5,310 – $3,000)(P/A,i’%,5 ) + $2,000(P/F,i’%,5 ) ; i’ ’% = ?

By trial-and-error process; at i = 5%; PW = $1,568 at i = 15%; PW = -$1,262 By linear interpolation; i’ = 10.5%.

(25)

Problem 4.4 b

Determine the IRR of the following engineering project when the MARR is 15% per year.

Investment cost $10,000

Expected life 5 years

Salvage value -$ 1,000

Annual receipts $ 8,000

Annual expenses $ 4,000

Solution:

PW = 0 = -$10,000 – $1,000(P/F,i’%,5 ) + (8,000-$4,000)(P/A,i’%,5 ) By trial and error process, at i’ = 20%, PW = $1560.50

at i’ = 25%, PW = $ 429.50

(26)

Example of IRR Method:

A piece of new equipment has been proposed by engineers to

increase the productivity of a certain manual welding operation. The investment cost $25,000, and the equipment will have a market value $5,000 at the end of a study period of five years. Increased

productivity attributable to this equipment will amount to $8,000 per year after extra operating costs have been subtracted from the

revenue generated by the additional production. If the firm MARR is 20% per year, is this proposal a sound one? Use the IRR method. Solution:

PW = 0 = -$25,000 + $8,000(P/A,i’%,5 ) + $5,000(P/F,i’%,5 ) By trial and error process, at i’ = 20%, PW = $ 934.30

at i’ = 25%, PW = -$1847.10 By linear interpolation, i’ = 22 %.

(27)

Example of IRR method

Barron Chemical uses a thermoplastic polymer to enhance the

appearance of certain RV panels. The first cost of one process was $126,000 with annual costs of $49,000 and revenues of $88,000. A salvage value of $33,000 was realized when the process was

discontinued after 8 years.

What rate of return did the company make on the process ?

Solution: Set FW = 0

FW = 0 = -$126,000 (F/P,i’%,8 _{) + ($88,000 - 49,000)}

(F/A,i’%,8 _{) + $33,000}

By trial and error process, at i’ = 25%_{, FW = -$55,811}

at i’ = _{30%, FW = $64,370}

(28)

THE EXTERNAL RATE OF RETURN METHOD

( ERR )

• ERR directly takes into account the

interest rate (



) external to a project at

which net cash flows generated over the

project life can be reinvested (or

borrowed ).

• If the external reinvestment rate, usually

the firm’s MARR, equals the IRR, then

ERR method produces same results as

IRR method

(29)

CALCULATING EXTERNAL RATE OF

RETURN ( ERR )

1. All net cash outflows are discounted to the present (time 0) at _{% per compounding period.}

2. All net cash inflows are discounted to period N at _%.

3. ERR -- the equivalence between the discounted cash inflows and cash outflows -- is determined.

The absolute value of the present equivalent worth of the net cash outflows at  _{% is used in step 3.}

•

A project is acceptable when i ‘ % of the ERR

(30)

CALCULATING EXTERNAL RATE OF

RETURN ( ERR )



_E

k

( P / F,



%, k )( F / P,

i

‘

%, N )

=



_R

k

( F / P,



%, N - k )

R_k = excess of receipts over expenses in period k E_k = excess of expenses over receipts in period k N = project life or period of study

_{= external reinvestment rate per period}

N k = 0 N k = 0 i ‘ %= ? Time N 0  R_k ( F / P,  %, N - k ) N k = 0 _{E ( P / F,} _{%, k )( F / P,} _i‘ %, N ) N

(31)

ERR ADVANTAGES

• ERR has two advantages over

IRR:

1. It can usually be solved for

directly, rather than by trial and

error.

2. It is not subject to multiple rates

of return.

(see slide Problem 4.38)

(32)

Example of ERR Method:

A piece of new equipment has been proposed by engineers to increase the productivity of a certain manual welding

operation. The investment cost $25,000, and the equipment will have a market value $5,000 at the end of a study period of five years. Increased productivity attributable to this

equipment will amount to $8,000 per year after extra

operating costs have been subtracted from the revenue generated by the additional production. If the external investment rate = MARR = 20% per year, what is the

alternative’s ERR, and is this proposal a sound one?

Solution:

$25,000(F/P,i’%,5) = $8,000(F/A,20%,5) + $5,000

(F/P,i’%,5) = $64,532.80/$25,000 = 2.5813

i’ = 20.88%

(33)

Problem 4.32

Summary of the projected costs and annual receipts for a new product line is presented as follows:

End of Year Net Cash Flow

0 - $450,000 1 - 42,500 2 92,800 3 386,000 4 614,600 5 - 202,200

The company’s external reinvestment rate per year = MARR = 10%

per year Solution: [$450,000 + $42,500(P/F,10%,1 ) + $202,200(P/F,10%,5 )](F/P,i’%,5 ) = $92,800(F/P,10%,3 ) + $386,000(F/P,10%,2 ) + $614,600(F/P,10%,1 ) $614,182.73(F/P,i’%,5 ) = $1,265,544 (F/P,i’%,5 ) = 2.0622

(34)

Problem 4.34

Given a cash flow as stated below:

End of Year Net Cash Flow

0 - $10,000

1 – 7 1,400

7 10,000

The external rate of reinvestment for the company = 8% per year. Solution:

$10,000(F/P.i’%,7 ) = $1,400(F/A,8%,7 ) + $10,000

(F/P,i’%,7 ) = $22,491.92 / 10,000 = 2.2492

(35)

Problem 4.48:

A certain project has a net receipts equaling $1,000 now, has

costs of $5,000 at the end of the first year, and earns $6,000 at the end of the second year.

If the external reinvestment rate of 10% is available, what is the rate of return for this project using the ERR method ?

Solution:

$5,000(P/F,10%,1 )(F/P,i’%,2 ) = $1,000(F/P,10%,2 ) + $6,000

(F/P,i’%,2 ) = $7,210 / $4,545.50 = 1.5862

(36)

PAYBACK PERIOD METHOD

•

Sometimes referred to as simple payout method

•

Indicates liquidity (riskiness) rather than profitability

•

Calculates smallest number of years (  ) needed for

cash inflows to equal cash outflows -- break-even life

•

 ignores the time value of money and all cash flows

which occur after 



_{( R}

k

-E

k

) - I > 0

•

If  is calculated to include some fraction of a year, it

is rounded to the next highest year k = 1

(37)

PAYBACK PERIOD METHOD

•

The payback period can produce misleading results, and should only be used with one of the other

methods of determining profitability

•

A discounted payback period _‘ ( where _‘ < N )

may be calculated so that the time value of money is considered

i ‘ is the MARR

I is the capital investment made at the present time ( k = 0 ) is the present time

_‘ is the smallest value that satisfies the equation



_{( R}

k

- E

k

) ( P / F, i %, k ) - I > 0

k = 1

(38)

Simple Payback Period vs. Discounted Payback Period

EOY Net c / flow Cum PW @ i=0% PW @ i=20% Cum PW @ MARR 20%

0 - $25,000 - $25,0000 - $25,000 - $25,000 1 8,000 - 17,000 6,667 - 18,333 2 8,000 - 9,000 5,556 - 12,777 3 8,000 - 1,000 4,630 - 8,147 4 8,000 7,000 3,858 - 4,289 5 13,000 20,000 5,223 934

The payback period is at 4th years, because the cumulative balance turns positive at EOY 4

[ t i m e v al u e o f m o n e y is n o t c o n s i d e r e d ]

The payback period is at 5th

year, because the cumulative discounted balance turns

positive at EOY 5

[ t i m e v al u e o f m o n e y is c o n s i d e r e d ]

(39)

INVESTMENT-BALANCE

DIAGRAM

Describes how much money is

tied up in a project and how the

recovery of funds behaves over

its estimated life.

(40)

INTERPRETING IRR USING

INVESTMENT-BALANCE DIAGRAM

• downward arrows represent annual returns (R_k - E_k) : 1 < k < N

• dashed lines represent opportunity cost of interest, or interest on BOY investment balance

• IRR is value i ‘ that causes unrecovered investment balance to

equal 0 at the end of the investment period.

0 1 2 3 N $0 Unrecovered Investment Balance, $ 1 + i‘ 1 + i‘ 1 + i‘ 1 + i‘ P (1 + i‘) [ P (1 + i‘) - (R ₁ - E₁) ] (1 +i‘) (R₁ - E₁) (R₂ - E₂) (R₃ - E₃) (R_N-1 - E_N-1) (R_N - E_N) Initial investment = P

(41)

INVESTMENT-BALANCE

DIAGRAM EXAMPLE

• Capital Investment ( I ) = $10,000

• Uniform annual revenue = $5,310

• Annual expenses = $3,000

• Salvage value = $2,000

(42)

0 1 2 3 I n v e s t m e n t B a l a n c e , $ 4 5 5,000 - 5,000 - 10,000 -$10,500 - $2,310 - $2,310 - $2,310 - $2,310 - $2,310 - $8,190 - $6,290 - $4,294 - $2,199 - $8,600 - $6,604 - $4,509 + $4,310 $2,001 ( = FW ) Years MARR = 5% Area of Negative Investment Balance _’

(43)

WHAT INVESTMENT-BALANCE

DIAGRAM PROVIDES

• Discounted payback period (

_‘

) is 5 years

• FW is $2,001

• Investment has negative investment balance

until the fifth year

• Investment-balance diagram provides

additional insight into worthiness of proposed

capital investment opportunity and helps

(44)

INVESTMENT-BALANCE DIAGRAM EXAMPLE 4-17

Construct an investment balance diagram for the

following cash flow table. The MARR is 10% per year.

End of Year Net Cash Flow Three sign Changes

0 - $5,000

-1 6,000 negative to positive

2 - 1,000 positive to negative