1

CE 561 Lecture Notes

Set 2

Engineering Economic Analysis

Time value of money– Inflation– Opportunity cost

Cash Flow DiagramP A A P=Investment

A=Yearly Return0 N N=No. of Years

Interest– Profit Motive MARR– Public Project Opportunity Cost– Stable Economy 5-8%– Developing Countries 10-15%

2

Types of Compounding and Interest Rates

Interest

SimpleCompound

ContinuousDiscreet

VariableFixed

Example of Compounding(Int. Rate 10%)

Year Principal Interest

1 1000 100 1100

2 1100 110 1210

3 1210 121 1331

4 1331 133.1 1464.1

P=1000 F=1464.1

0 1 2 3 4

*Compounding vs. Discounting

Amt. at the end of the year

Nominal and Effective Rates

Compounded interest rate for periods less than a yearIf nominal rate = 6% per year, compounded every 3 months – 4 periods at 1.5% per periodEffective Rate = %100

PPF

−

3

P=100 F=?

0 3 6 9 12

P=100

100 (1 + 0.015) = 101.50

101.5 (1.015) = 103.02

103.02 (1.015) = 104.57

104.57 (1.015) = 106.14 = F

Effective Rate

=

= 6.14%

Effective Rate = SPCAFi/m,m-1

m = Number of Periods in a Year

100100

10014.106×

−

Interest Equations: Equivalence

P i/period F=?

0 1 2 N-1 N

Compounding

At the end of the 1st period = P+Pi = P(1+i)

At the end of the 2nd period = P(1+i) + P(1+i)i = P(1+i)2

At the end of the Nth period = P(1+i)N

Single Payment Compound Amount Factor

(SPCAFi,N) = (1+i)N

4

Discounting

P=? i/period F

0 1 2 N

P = F/(1+i)N

Single Payment Present Worth Factor

( )NN,i i11SPPWF+

=

Examples

1. How much will be accumulated in a fund at the end of 25 years if $2,000 is invested now with 6% compounded annually?

F = 2000 SPCAF6%,25

= 2000 x (1 + .06)25

= 2000 x 4.2919

= $8584

2. How much money should be invested now at 6% compounded annually so that $8584 can be received 25 years hence?

P = 8584 SPPWF6%,25

= 8584 x 0.2330

= $2000

5

Interest Equations

Three sets– Uniform Periodic Sums– Uniformly Increasing Periodic Sums-Linear

Gradient– Exponentially Increasing Periodic Sums-

Geometric

Uniform Series Compound Amount Factor

A A A A F=?

0 1 2 N-1 N

( ) ( ) ( ) Ai1Ai1Ai1AF 2N1N +++⋅⋅⋅++++= −−

Multiplying by (1 +i)

( ) ( ) ( ) ( )i1Ai1Ai1Ai1F 1NN ++⋅⋅++++=+ −

Subtracting, ( )[ ]1i1AFi N −+=

( )[ ] i/1i1USCAF NN,i −+=∴

Example

1. How much money will be accumulated in a fund at the end of 25 years from now if $2000 is invested at the end of each year (interest 6% year)?

F = A x USCAF6%,25

= 2000 x (1 + .06)25-1

.06

= 2000 x 54.865

= $109,729

6

Sinking Fund Deposit Factor

( )

−+=

1i1iFA N

A A=? A A A

0 1 2 N-1 N

The amount of uniform series of end-of-period deposits for N periods to provide F at the end of N periods.

( ) 1i1iSFDF Nn,i−+

=∴

F

Capital Recovery Factor

( )Ni1PF +=

[ ]( )1i1/iFA N −+=

( )( )

−+

+=

1i1i1iP N

N

P A A=? A A A

Given P, find A

( )( )

−+

+=∴

1i1i1iCRF N

N

N,i

Examples

1. A county expects that a bridge reconstruction project 16 years from now will require $35 million. What annual payment should be made to a fund with 6% rate?

A = F x SFDF6%,16

= 35 x 0.06

(1.06)16-1

= 35 x 0.03895

= $1.36 million/year

(0.97 million/year if interest rate = 10%)

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1. If an agency borrows $20 million to build a toll road, what must be the prospective annual toll income for 20 years to pay back with 7% interest?

A = P x CRF7%,20

= 20 x 106 x 0.07(1.07)20

(1.07)20-1

= 20 x 106 x 0.09439

= $1.9 million

Uniform Series Present Worth Factor

Given A, find P

Reciprocal of CRF

( )( )N

N

n,i i1i1i1USPWF

+−+

=

For i = 9%

Year

1

2

3

4

5

6

SPCAF

1.090

1.188

1.295

1.412

1.538

1.677

SPPWF

0.917

0.842

0.772

0.708

0.650

0.596

CRF

1.09

0.568

0.395

0.309

0.257

0.223

USPWF

0.917

1.759

2.531

3.24

3.89

4.49

SFDF

1.0

0.478

0.305

0.218

0.167

0.133

USCAF

1.0

2.09

3.28

4.57

5.98

7.52

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A county decided to raise its share of a construction project through local resources. The Bd. of Commissioners decided on Jan. 1, 1989 to invest an assured annual income in a special fund starting Jan. 1, 1990 until the start of the project. The last investment will be on Jan. 1, 2005. The fund was expected to earn an interest of 10% per annum and the county would be able to withdraw $10 million each year for 4 years as from Jan. 1, 2005, the start of the project. How much is being deposited each year?

$10m

Jan ’90 95 2000 2005 2008

A

Example

Present Worth = 10 million (USPWF10%,4)(SPCAF10%,1)

On Jan. 1, 2005 = 34.87 million

Annual Amt (A) so that 34.87 million is available at 2005 is:

A = 34.87 (SFDF10%,16)

= $970,000/yr for 16 years (1990-2005)

Present Worth of Periodic Payments in Perpetuity

( ) ( )K+

++

+= N2N i1

Ri1

RPW

( ) ( )

+

++

+= KN2N i1

1i1

1R

( )

−

+−

= 1

i111

1R

N

R R R R R

N N N N N

( ) 1i1R

N −+=

αK

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Continuous Compounding

( )mNm/i1PF +=

iNi

m

mi1P

+=

emi1

im

=

+

infinite number of periods

i = nominal rate, m = no. of periods

iNPeF =

Limα→m

Continuous Compound Amount Factor

CCAF = eiN

Effective Annual Interest Rate

( ) 1eP

PePP

PF ii

−=−

=−

Computation of PW of Continuous Benefit/Cost Flows

RR2

R3Rn

R1r = rate of growth

i = interest

Amounts compounded continuously

Interest compounded continuously

0 n

10

Bring all future streams to the present

R PW of R = R

R1=Rer “ R1 = Rer/ei

R2=R1er “ R2 = Re2r/e2i

Take summation,

+⋅⋅+++=Σ ni

nr

i2

r2

i

r

ee

ee

ee1R

( )

−−

=−

ir1eR

irn

Computation of PW of Benefits/Costs

Annual growth rate

Y = Period of the estimate

Ylogr α

=

=α Future Annual Value Estimate (F)

First Year Value Estimate (P)

Assuming continuous compounding,rYPeF =

α== P/FerY

α= logrY

Ylogr α

=∴

logF

logP

Y

r

YP/Flog

YPlogFlogr =

−=

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Example

If the current user benefit from a road improvement project is $5 million (1st yr) and the total project cost is $80 million, what must be the minimum rate of growth of benefit (continuously increasing)? Assume the project life is 15 years and interest rate of 5% per year.

PW of Benefit must be at least equal to PW of cost∴ PW of Benefit = $80 million

PW of Benefit = 1st Yr Benefit

*Present Worth Factor (PWF)

( )

ir1ePWF

nir

−−

=−

( )16

580

05.r1e 1505.r

==−

−−

Try r = 6%( )

1605.06.

1e 1505.06.

≈−

−−

minimum rate of growth = 6%∴

ir >

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Methods of Economic Analysis

Equivalent Uniform AC MethodPresent Worth of CostsEquivalent Uniform Annual Net ReturnNet Present ValueBenefit-Cost RatioInternal Rate of Return

Formulas Related

SPPWF1

USPWF1

SPCAF =

USCAF =

CRF =

SFDF1

CRF = SFDF + i

Annual Cost Method

KSFDF*TCRF*PAC n,in,i +−=

( ) Ki*TCRF*TP n,i ++−=

P = Initial Cost

T = Salvage

K = Annual Maint. & Op. Cost

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Example

100,00050,00050,00050,00060,00060,00060,000

0 1 2 3 4 5 6A

T=40,000

i = 15%

B80,000

55,00055,000

55,00065,000

65,00065,000

T=30,0000 1 2 3 4 5 6

ACA = 100,000 CRF(15%,6) – 40,000 SFDF(15%,6)

+ 50,000 USPWF(15%,3)

+ 60,000 USPWF(15%,3) SPPWF(15%,3) CRF(15%,6)

= 75,850

ACB = 76,680

System A is economically more desirable

Present Worth of Cost

4%,84%,8 USPWF800SPPWF1001000 +−=

312.38007350.01001000 ×+×−=

A1000 K=800/yr T=100

4 yrs

B800 K=900/yr T=0

4 yrsPresent Worth of A (8% interest)

576,3=

14

PW of B = 800 + 900USPWF8%,4

= 3780

Advantage of A = (3780 – 3576)

= $204

Choose A∴

Equivalent Uniform Net Annual Return

ENAR = Annual Income (Benefit) – Annual Cost

Initial Inv. (P)

Terminal Value (T)

Annual Maint. (K)

Annual Benefit (R)

125,000

170,000

185,000

30,000

50,000

70,000

10,000

13,000

14,000

60,000

75,000

85,000

A

B

C

i = 10%, Proj. Pd = 12 yrs

( ) ( )[ ]12%,10SFDF*TK12%,10CRF*PRENARA −+−=

( )[ ]4676.0*000,30000,1014676.0000,125000,60 −+×−=

058,23=

389,39ENARB =

122,47ENARC = Highest Annual Net Return

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Internal Rate of Return

IRR is the interest rate which makes the costs equivalent to thebenefits.

P TK/yr

0 nR/yr

R = Benefit/yr

i = unknown

( ) ( ) RKn,iSFDF*Tn,iCRF*P =+−

Find i by trial and error

Example

P=30,000 T=15,000R=5,000/yr

0 10K=2,000/yr

Find the internal rate of return000,5000,2SFDF000,15CRF000,30 10,i10,i =+×−×

First trial, i = 0%

Selecting zero shows whether the income is actually sufficient to recover the costs

000,5000,2101000,15

101000,30 ≅+×−×

000,53500 ≅500,10 ≅

∴

At 5%, Annual Cost > Annual Benefit by 1,500

An approximate rate of return

100000,30

500,1×=

%5≈

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Try 6%,000,5000,2SFDF000,15CRF000,30 10%,610%,6 ≅+×−×

000,5938,4 ≅620 ≅

Try 7%,000,5000,2SFDF000,15CRF000,30 10%,710%,7 ≅+×−×

000,5186,5 ≅

%25.6%118662

62%6i =×+

+=

By interpolation,

Direction finder approximate rate of return after 2nd trial

%21.6100000,30

62%6i ≅×+≅

Net Present Value

( ) ( ) ( )[ ]n,iUSPWFKn,iSPPWFTPn,iUSPWF*R ×+×−−0≥

[ ]10%,610%,610%,6 USPWF000,2SPPWF000,15000,30USPWF000,5NPW +−−=

PW of Benefit > PW of CostT=15,000P=30,000 K/yr=2,000

R/yr=5,000 N=10

[ ]3601.7000,255839.0000,15000,303601.7000,5 ×+×−−×=

[ ]720,14376,8000,30805,36 +−−=

0461<−= Therefore, the project is not feasible at 6%.

i = 6%

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Benefit-Cost Ratio Method

US Flood Control Act of 1936—”benefit to whomsoever it may accrue” should exceed “estimated costs.”B/C ratio expresses the ratio of equivalent uniform annual benefit (or its present worth) to the equivalent uniform annual cost (or its present worth).Same proposed project may indicate different B/C ratios depending upon whether certain items are considered “disbenefits” or “costs”

Example

100K 54K 58K

105K 61K 65K

111K 62K 71K

Alt 1

Alt 2

Alt 3

Alt 1

PW of Cost = 100K i = 10%

PW of Benefit ( ) ( )21.1

581.1

54+

++

=

93.4701.49 +=94.96=

97.0100/94.96C/B 1 ==

Similarly, 04.1C/B 2 =036.1C/B 3 =

∴ Alt 2 is best

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Benefit-Cost Ratio Method-Mutually Exclusive Projects

B/C = Benefit

Additional Cost

= RA – RB

ACB – ACA

= Ann. Rd. User Cost in A – Ann. Rd. User Cost in B

Ann. Cost of B – Ann. Cost of A

Choosing from several mutually exclusive projects—A, B, C

∴

∴

Comparing B with A, Ben/Cost = 1.2

So, B is better than A

C with A, Ben/Cost = 1.1

C is better than A

C with B, Ben/Cost = 0.6

C is not better than B

Most economic solution is B

∴

Example: Incremental Analysis

Loc H J K

Construction Cost $110,000 622,000 1,158,000

Ann. Maint. 35,000 21,000 17,000

Ann. Road User Cost 266,500 199,900 173,200

i = 7%, 20 yrs

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Rate of Return Computations

( ) ( )[ ] ( )20%,iUSPWF000,21000,35900,199500,266 −+−

000,110000,622 −=

Balance the prospective savings in road user and maintenance costs to recover the extra construction outlay.

MARR = 7%

J compared to H

%8.12*i =

Rate of Return Computations

J compared to H will yield 12.8%

K compared to J will yield 3%

location J is economically superior∴

Benefit-Cost Ratio

Benefit—savings in road user costs

Costs—Initial outlay plus maintenance

Compare J with H:

B = (266,000 – 199,900)USPWF(7%, 20)

C = (622,000 – 110,000) – (35,000 – 21,000)USPWF(7%, 20)

= 364,000

NPV = B-C = 342,000 > 0

B/C = 706,000 or 66,600

364,000 34,400=1.94 J is better than H∴

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Compare K with J:

NPV = B-C = -211,000 < 0

B/C = 282,000 or 26,700

493,000 46,500

favorable location is J

=0.57

∴

Discussion of Various Methods

AC Method– Benefits are not considered– Can be used in project formulation phase– Not useful in evaluation– Alternatives must have the same level of service

Net Present Value– Profitability expressed as a lump sum, not a rate– Unambiguous, direct index– Can be used in project evaluation as well as in

project formulation– Standard procedure-AASHTO

Rate of Return– Benefit or Income is considered– Solution is not unique– Wrong assumptions when terminal dates are

different-reinvestment of funds– Used by World Bank

Benefit-Cost Ratio– Concern over significance in relative values of

B/C ratio– Definition of benefits and costs– Policy-maker’s lack of understanding

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Other Methods– Capitalized Cost--PW of costs with the analysis

period of infinity– Payback Period--period required to accumulate

savings to equal investment– Breakeven Analysis--example