CHAPTER 5 Engineering Economic

EVALUATING SINGLE PROJECTCHAPTER 5

IntroductionCapital 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 the economic profitability of a project:Present worth (PW)Future worth (FW)Annual worth (AW)Internal Rate of Return (IRR)External Rate of Return (ERR)Convert cash flows into equivalent worth at some point (s) in time using an interest rate known as Minimum Attractive rate of Return (MARR)

The basic question to be addressed:is the proposed capital investment and its associated expenditures (cash out flow)

can be recovered by

revenue (savings) over time (cash in flow)in addition to a return on the capital (rate of return ~ MARR) that is sufficiently attractive in view of the risk involved.

MINIMUM ATTRACTIVE RATE OF RETURN ( MARR )An interest rate used to convert cash flows into equivalent worth at some point(s) in timeUsually 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 over short and long run - type of organization involvedMARR is sometimes referred to as hurdle rate

CAPITAL RATIONINGEstablishing MARR involves opportunity cost viewpoint results from phenomena of CAPITAL RATIONINGExists when management decides to restrict the total amount of capital invested, by desire or limit of available capitalSelect only those projects which provide annual rate of return in excess of MARRAs amount of investment capital and opportunities available change over time, a firms MARR will also changeSee slide

PRESENT WORTH METHOD ( PW )Based on concept of equivalent worth of all cash flows relative to the present as a baseAll cash inflows and outflows discounted to present at interest -- generally MARRPW is a measure of how much money can be afforded for investment in excess of costPW is positive if dollar amount received for investment exceeds minimum required by investors

FINDING PRESENT WORTHDiscount future amounts to the present by using the interest rate over the appropriate study periodPW = Fk ( 1 + i ) - k i = effective interest rate, or MARR per compounding periodk = index for each compounding periodFk = future cash flow at the end of period kN = number of compounding periods in study periodinterest rate is assumed constant through projectThe higher the interest rate and further into future a cash flow occurs, the lower its PWk = 0N

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 MARRThe FW of a project is equivalent to FW = PW ( F / P, i%, N )If FW > 0, it is economically justifiedFW ( i % ) = Fk ( 1 + i ) N - kk = 0Ni = effective interest ratek = index for each compounding periodFk = future cash flow at the end of period kN = number of compounding periods in study period

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 MARRAW 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 attractiveAW = 0 : annual return = MARR earned

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 diagramb) Determine the equivalent present worth (PW)

CAPITAL RECOVERY ( CR )CR is the equivalent uniform annual cost of the capital investedCR is an annual amount that covers:Loss in value of the assetInterest 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 N = project study period

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

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.

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.

CAPITAL RECOVERY ( CR)CR is also calculated by adding sinking fund amount (i.e., deposit) to interest on original investmentCR ( 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 valueCR ( i % ) = ( I - S ) ( A / P, i %, N ) + S ( i % )

5 methods for evaluating the economic profitability of a project:

Present worth (PW) Future worth (FW) Annual worth (AW) Internal Rate of Return (IRR)External Rate of Return (ERR)

INTERNAL RATE OF RETURN METHOD ( IRR )IRR solves for the interest rate that equates the equivalent worth of an alternatives cash inflows (receipts or savings) to the equivalent worth of cash outflows (expenditures)Also referred to as:investors methoddiscounted cash flow methodprofitability indexIRR is positive for a single alternative only if:both receipts and expenses are present in the cash flow patternthe sum of receipts exceeds sum of cash outflows

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 acceptableTo compute IRR for alternative, set net PW = 0PW = R k ( P / F, i %, k ) - E k ( P / F, i %, k ) = 0i is calculated on the beginning-of-year unrecovered investment through the life of a project (slide)Nk = 0Nk = 0Nk = 0Nk = 0

DISADVANTAGES OF IRRThe IRR method assumes recovered funds, if not consumed each time period, are reinvested at i %, rather than at MARRThe computation of IRR may be unmanageableMultiple IRRs may be calculated for the same problemThe IRR method must be carefully applied and interpreted in the analysis of two or more alternatives, where only one is acceptableBasic IRR method cannot rank mutually exclusive projects, the project with higher IRR potentially having lower net present value . . . inconsistent ranking.

ADVANTAGES OF IRRIRR can be calculated without having to estimate cost of capital or MARRIRR, in the form of rate of return is more appealing to evaluate investment IRR is more appealing to communicate profitabilityWhen unique (not multiple IRRs), it provides valuable information about the return on the investment and is often viewed as a measure of efficiency

Using Multiple Internal Rates of Return: The IRR Parity TechniqueMultiple IRRs @ 30% and 60%

EOYCash flows ($)0-1000129002-2080

Chart3

-180

-167.7286540535

-156.093810073

-134.6153846154

-115.3435386258

-98.0795610425

-82.6446280992

-68.8775510204

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-45.7788347206

-36.19649526

-27.7777777778

-20.4246170384

-14.0478668054

-8.5663895188

-3.90625

0

3.2139577594

5.7919358432

7.785467128

9.2417559336

10.2040816327

10.7121602857

10.8024691358

10.7015457788

10.5085381873

9.8612125639

8.8888888889

7.6177285319

6.0718502277

4.2735042735

2.2432302516

0

-2.4386526444

-5.0565139798

-7.8385832487

-10.7709750567

-13.8408304498

-17.0362358031

-20.3461487647

-23.7603305785

-27.2692841813

-30.8641975309

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-70

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-94.8736205055

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-111.5702479339

-115.7373589806

-119.8979591837

-124.0504346464

-128.1932902432

-132.3251417769

-136.4447086801

-140.5508072175

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-148.7183108538

-152.7777777778

-156.8198893518

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-172.8

-176.7447719829

-180.6683613367

-184.5703125

-188.4502133285

-192.3076923077

-196.1424159431

rate of return (%)

PW ($)

PW vs rate of return

Sheet1

k%NPV (k)EOYCash flowsk%NPV (k)EOYCash flows(2-1)Alt 1Alt 2(2-1)

0-1800-10000-950-1200-200-180-9585

1-168129001-8513200300-168-8582

2-1562-20802-762-2095-15-156-7680

4-1354-60-135-6075

6-1156-46-115-4670

8-988-33-98-3365

10-8310-22-83-2260

12-6912-13-69-1356

14-5714-5-57-552

15-5115-2-51-250

16-46162-46247

18-36187-36743

20-282012-281240

22-202215-201536

24-142418-141832

26-92620-92029

28-42821-42125

300302202222

323322232219

346342162116

368362082012

389381991910

4010401710177

4211421511154

4411441211121

451145101110-0

4611469119-2

4810486106-4

50950292-7

52852-28-2-9

54654-56-5-12

56456-104-10-14

58258-142-14-16

60060-180-18-18

62-262-23-2-23-21

64-564-28-5-28-23

66-866-33-8-33-25

68-1168-38-11-38-27

70-1470-43-14-43-29

72-1772-48-17-48-31

74-2074-53-20-53-33

76-2476-58-24-58-34

78-2778-63-27-63-36

80-3180-69-31-69-38

82-3582-74-35-74-40

84-3884-80-38-80-41

86-4286-85-42-85-43

88-4688-91-46-91-45

90-5090-96-50-96-46

92-5492-102-54-102-48

94-5894-107-58-107-49

96-6296-113-62-113-51

98-6698-118-66-118-52

100-70100-124-70-124-54

102-74102-129-74-129-55

104-78104-135-78-135-57

106-82106-140-82-140-58

108-87108-146-87-146-59

110-91110-151-91-151-61

112-95112-157-95-157-62

114-99114-162-99-162-63

116-103116-168-103-168-64

118-107118-173-107-173-66

120-112120-178-112-178-67

122-116122-184-116-184-68

124-120124-189-120-189-69

126-124126-194-124-194-70

128-128128-199-128-199-71

130-132130-205-132-205-72

132-136132-210-136-210-73

134-141134-215-141-215-75

136-145136-220-145-220-76

138-149138-225-149-225-77

140-153140-230-153-230-78

142-157142-235-157-235-79

144-161144-240-161-240-80

146-165146-245-165-245-81

148-169148-250-169-250-81

150-173150-255-173-255-82

152-177152-260-177-260-83

154-181154-265-181-265-84

156-185156-270-185-270-85

158-188158-274-188-274-86

160-192160-279-192-279-87

162-196162-284-196-284-88

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rate of return (%)

PW ($)

PW vs rate of return

Sheet3

Using Multiple Internal Rates of Return: The IRR Parity TechniqueAccording 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 returnIf 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.

D Zhang (2005), Engineering Economist, Vol. 50, Issue 4, pp 237-335

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%. Therefore, the project is minimally acceptable (i% > MARR)

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 B linear extrapolation, i = 27%, hence the project is acceptable; > MARR

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 %. Therefore, the project is acceptable (i% > MARR)

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 By linear interpolation, i = 27.3%

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 firms MARR, equals the IRR, then ERR method produces same results as IRR method

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 method is greater than or equal to the firms MARR

CALCULATING EXTERNAL RATE OF RETURN ( ERR ) Ek ( P / F, %, k )( F / P, i %, N ) = Rk ( F / P, %, N - k )Rk = excess of receipts over expenses in period kEk = excess of expenses over receipts in period kN = project life or period of study = external reinvestment rate per periodNk = 0Nk = 0i %= ?TimeN0 Rk ( F / P, %, N - k )Nk = 0 Ek ( P / F, %, k )( F / P, i %, N )Nk = 0

ERR ADVANTAGESERR 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)

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 alternatives 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%

Since i > MARR, the alternative is minimally justified.

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 companys external reinvestment rate per year = MARR = 10% per yearSolution:[$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) = 2.0622By interpolation, i% , ERR = 15.6%

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

By interpolation, ERR = i% = 12.3%

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

By interpolation, ERR = i% = 25.9%

PAYBACK PERIOD METHODSometimes referred to as simple payout methodIndicates liquidity (riskiness) rather than profitabilityCalculates 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 ( Rk -Ek) - I > 0If is calculated to include some fraction of a year, it is rounded to the next highest yeark = 1

PAYBACK PERIOD METHODThe payback period can produce misleading results, and should only be used with one of the other methods of determining profitabilityA 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 ( Rk - Ek) ( P / F, i %, k ) - I > 0k = 1

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[time value of money is not considered]The payback period is at 5th year, because the cumulative discounted balance turns positive at EOY 5[time value of money is considered]

INVESTMENT-BALANCE DIAGRAM Describes how much money is tied up in a project and how the recovery of funds behaves over its estimated life.

INTERPRETING IRR USING INVESTMENT-BALANCE DIAGRAMdownward arrows represent annual returns (Rk - Ek) : 1 < k < Ndashed lines represent opportunity cost of interest, or interest on BOY investment balanceIRR is value i that causes unrecovered investment balance to equal 0 at the end of the investment period.0123N$0UnrecoveredInvestmentBalance, $1 + i1 + i1 + i1 + iP (1 + i)[ P (1 + i) - (R1 - E1) ] (1 +i)(R1 - E1)(R2 - E2)(R3 - E3)(RN-1 - EN-1)(RN - EN)Initial investment= P

INVESTMENT-BALANCE DIAGRAM EXAMPLE Capital Investment ( I ) = $10,000Uniform annual revenue = $5,310Annual expenses = $3,000Salvage value = $2,000MARR = 5% per year

0123

InvestmentBalance, $455,000- 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 )YearsMARR = 5%Area of NegativeInvestmentBalance

WHAT INVESTMENT-BALANCE DIAGRAM PROVIDESDiscounted payback period ( ) is 5 yearsFW is $2,001Investment has negative investment balance until the fifth yearInvestment-balance diagram provides additional insight into worthiness of proposed capital investment opportunity and helps communicate important economic information

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 3 4,000 negative to positive

0123

InvestmentBalance, $- $5,000-$5,500+ $500+$6,000+ $4,000$3505 ( = FW )End of yearMARR = 10%-$495-$1,000+$550-$450Example 4-17 Investment Balance Diagram1.10