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%
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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
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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
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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