chapter 4 sept 3 - purdue engineeringce561/classnotes/chapter 4.pdf · result in decreasing value...

Economic Analysis of Transportation Investments 97

Chapter 4 Economic Analysis of Transportation Improvements


Topics: Interest Equations and Equivalencies, Methods of Economic Analysis, An Application of Economic Analysis Methods: Optimal Timing of Investments,

4.1 Interest Equations and Equivalencies

4.1.1 Engineering Economics 4.1.2 Cash Flow Illustrations 4.1.3 The Concept of Interest 4.1.4 Types of Compounding and Interest Rates 4.1.5 Interest Equations 4.1.6 Special Cases of Interest Equations 4.1.7 Estimation of the Rate of Growth of Benefits or Costs

4.2 Methods of Economic Analysis

4.2.1 Introduction 4.2.2 Equivalent Uniform Annual Cost Method 4.2.3 Present Value of Costs 4.2.4 Equivalent Uniform Annual Return Method 4.2.5 Net Present Value Method 4.2.6 Internal Rate of Return 4.2.7 Benefit Cost Ratio Method 4.2.8 Evaluation Methods Using Incremental Attributes

4.3 An Application of Economic Analysis Methods: Optimal Timing of Investments

4.3.1 Introduction 4.3.2 Economic Effect of Postponement 4.3.3 Effect of Multi-Year Implementation of Investments 4.3.4 Multi-stage Implementation of Investments 4.3.5 Optimization of Stage Implementation of Investments 4.3.6 Application of Optimization of Stage Implementation: Optimal Time for Road Paving

4.1 Interest Equations and Equivalencies

4.1.1 Engineering Economics

The fundamental principle underlying all engineering economic analysis is that the value of money is

directly related to the time at which the value is considered. A given amount of money at the current time is not

equivalent to the same amount at a future year. Typically, the combined forces of inflation and opportunity cost

result in decreasing value of money over time. Inflation refers to the increase in the prices of goods and services

with time, and is reflected by a decrease in the purchasing power of a given sum of money with time. Opportunity

cost is the income that is forgone at a later time by not investing a given sum of money at a current period.


4.1.2 Cash Flow Illustrations

The time stream of amounts of money that occur within a given time period can be displayed either as a

cash flow table or a cash flow diagram. On a cash flow table, there are two columns: one for time and the other for

amount. On a cash flow diagram, time is represented on a horizontal axis, while vertical arrows depict the inflow or

outflow of money at various points in time. The sign of the amount and the direction of the arrows in cash flow

tables and figures respectively indicate the movement of the amount: A popular convention is to represent money

“coming in” (i.e., deposits or benefits) by positive signs and downward arrows pointing towards the horizontal time

line in the cash flow table and figure respectively. This convention also stipulates that money “going out” (i.e.,

disbursements or costs) is represented by negative signs and upward arrows in the cash flow table and diagram

respectively. The entire payment period (often referred to as the contract period, payment interval, planning horizon,

or planning period) is represented by the interval between the present period (denoted by time = 0), and the end of

that period (denoted by time = N). The planning period is typically divided into a number of equal periods called

compounding periods, or simply referred to as period, which is often taken as one year. In economic analysis, it is

conventional to assume that all transactions that occur within the compounding period occur at the end of that

period.

4.1.3 The Concept of Interest

The amount by which a given sum of money differs from its future value is typically represented as

interest. In this respect, a borrower of money from a financial institution at a current time, owes the initial amount

plus interest when the loan is due, to reflect for the fact that the value of the initial amount is not the same as the

value of the amount at the time of payback. Typically, lending institutions also add a small margin for profit.

The ratio of the interest paid at the end of the contract period (the originally agreed time between loan

acquisition and final payment) to the original amount invested at the beginning of the period, is termed the interest

rate. Therefore, a 10% interest rate, for example, indicates that for every dollar borrowed at the initial year, 10 cents

must be paid as interest at the end of each year. Central banks typically control of interest rates to remedy current or

expected economic problems. For instance, in a sluggish economy, the United States Federal Reserve Board

decreases interest rates to discourage saving and encourage individual spending and business investments, while

interest rates are increased when the economy is overheated. The interest rate typically undergoes several changes to

reflect the state of the economy. Therefore, the relationship between interest rate and the supply and demand for

money is a simultaneous one: interest rates are both a cause and effect of the market price of money.

In a stable economy such as those of developed countries, interest rates are typically lower than in

economies that have high inflation and are associated with uncertainty of investment returns. In developed countries,

the Minimum Attractive Rate of Return (MARR) typically ranges from 5-8%, while for most developing countries it

is estimated to be in the order of 10-15%.


The concept of interest rate is the underlying rationale for transforming amounts of or series of money from

one time period to another. This is addressed in Section 4.1.5 of this Chapter.

4.1.4 Types of Compounding and Interest Rates

Figure 4-1 illustrates the various categories of interest rates. Such rates could be simple or compound,

discrete, continuous, fixed or variable.

Interest

Compound Simple (Interest is earned on (Interest calculation is based only principal plus the interest on the original principal amount) earned earlier)

Discrete Continuous (Interest is calculated (Interest is calculated

and added to the existing and added to the existing principal and interest at finite time principal and interest at infinitesimally intervals) short time intervals)

Fixed Variable Fixed Variable (Interest rate (Interest rate (Interest rate (Interest rate is constant across changes after one or is constant changes compounding more compounding over time) over time) periods) periods)

Figure 4-1: Types of Interest Rates.

Interest may either be simple or compounded. In simple interest computations, the amount of interest at the

end of each period is the same as each of such amounts are a percentage of the initial amount, or principal. The

amount of compound interest in a given period is the interest charged on the total amount owed at the end of the

previous period, i.e., the sum of the principal and the previous period’s amount of interest. Therefore, amounts

borrowed on compound interest involve higher payments for amortization. Interests are typically computed using

compound interest rates.


Interests that are computed only at the end of each compounding period and with a constant interest rate are

typically referred to as fixed periodic rates, and the compounding period is 1 period for such cases, e.g., a fixed

annual rate, refers to the period of compounding as one year. Very often the compounding period is less than one

year, e.g., quarterly, monthly, or weekly. It is customary to quote interest rates on an annual basis, followed by the

compounding period if different from one year in length. For example, if the interest rate is 6% per interest period

and the interest period is six months, it is customary to speak of this rate as “12% compounded semiannually.” Here

the annual rate of interest is known as the nominal rate, 12% in this case. A nominal interest rate is represented by r.

But the actual annual rate on the principal is not 12% but something greater, because compounding occurs twice

during the year. The actual or exact rate of interest earned on the principal during one year is known as the effective

interest rate. The effective interest rate is computed by incorporating the effect of the multiplicity of compounding

periods. It should be noted that effective interest rates are always expressed on an annual basis unless specifically

stated otherwise. Table 4-1 shows how to compute effective annual interest rates for various scenarios of the

frequency of compounding within a year.

Table 4-1: Effective Annual Interest Rates for Various Compounding Frequency Scenarios

Frequency of Compounding Effective Annual Interest Rate Notation

Once a year = r

Twice a year = (1 + r/2)2 – 1

Three times a year = (1 + r/3)3 – 1

Every quarter = (1 + r/4)4 – 1

Six times a year = (1 + r/6)6 – 1

Every month = (1 + r/12)12 – 1

m times a year = (1 + r/m)m – 1

r = Nominal

Annual

Interest Rate

m = Number of

Compounding

Periods per year

4.1.5 Interest Equations

Interest equations, also referred to as equivalence equations, are relationships between amounts of money

that occur at different points in time, and are necessary to transform amounts or series of amounts of money from

one time period to another taking due cognizance of the time value of money. The vital links in such relationships

are the interest factors, which are functions of the interest rate and the payment period. Interest factors are expressed

as a formula (see Tables 4-2 to 4-4), or as values derived from such formula (see Appendix 1).

Interest equations typically involve the following five key variables as shown:

P- The initial amount

F- The amount at a specified future period

A- Periodic (typically yearly) amount

i- Effective Interest rate for compounding period

N- The number of compounding periods, or the analysis period


In a typical problem, three of these variables are given and it is required to find the fourth. The initial

amount, P, refers to the money borrowed or sunk into an investment at the initial period (represented by time 0 on a

cash flow diagram), while F refers to a payment or amount received at a future date. The periodic payment is the

amount paid at the end of every compounding period either to amortize a loan or to set up a sinking fund to achieve

a target amount after a number of years. Periodic payments could be a uniform series, or they can be increasing or

decreasing in a systematic manner (most commonly, arithmetic or geometric). i is the effective annual interest rate

for each period, while N is the number of compounding periods or the payment period.

Tables 4-2 to 4-4 present the formulae, cash flow diagram and notation for various cases of interest

equivalencies and relationships. This is done for discrete compounding and continuous compounding of the interest

rate.


Table 4-2: Interest Formulae for Discrete and Continuous Compounding

Description Cash Flow Diagram

Computational Formula

Factor Computation

Finding the future compounded amount (F) at the end of a specified period, given the initial amount (P) and interest rate.

P F? 0 1 … … N

F = P * SPCAF The Single Payment Compound Amount Factor, SPCAF (i%, N) may be read off from Appendix 1, or may be computed as shown.

NiSPCAF )1( +=

iNeSPCAF *=

Finding the initial amount (P) that would yield a given future amount (F) at the end of a specified period, given the interest rate.

P? F 0 1 … … N

P = F * SPPWF The Single Payment Present Worth Factor, SPPWF (i%, N) may be read off from Appendix 1, or may be computed as shown.

NiSPPWF

)1(1+

=

iNeSPPWF *

1=

Finding the uniform yearly amount (A) that would yield a given future amount (F) at the end of a specified period, given the interest rate.

A? A? A? A? 0 1 2 … N F

A = F * USSFDF The Uniform Series Sinking Fund Deposit Factor, USSFDF (i%, N) may be read off from Appendix 1, or may be computed as shown.

1)1( −+= Ni

iUSSFDF

11

* −−

= iN

i

eeUSSFDF

Finding the future compounded amount (F) at the end of a specified period due to annual payments (A), given the interest rate.

A A A A 0 1 2 … N F

F = A * USCAF The Uniform Series Compounded Amount Factor, USCAF (i%, N) may be read off from Appendix 1, or may be computed as shown.

iiUSCAF

N 1)1( −+=

11*

−−

= i

iN

eeUSCAF

Finding the initial amount (P) that is equivalent to a series of uniform annual payments (A), given the interest rate.

P? A A A A 0 1 2 … N

P = A * USPWF The Uniform Series Present Worth Factor, USPWF (i%, N) may be read off from Appendix 1, or may be computed as shown.

N

N

iiiUSPWF

)1(*1)1(

+−+

=

11 *

−−

=−

i

iN

eeUSPWF

Finding the amount of uniform yearly payments (A) that would completely recover an initial amount (P) at the end of a specified period, given the interest rate.

P A? A? A? A? 0 1 2 … N

A = P * USCRF The Uniform Series capital Recovery Factor, USCRF (i%, N) may be read off from Appendix 1, or may be computed as shown.

1)1()1(*−+

+= N

N

iiiUSCRF

iN

i

eeUSCRF *1

1−−−

=

Note: 1) In column 4, upper and lower formulae are for discrete and continuous compounding, respectively.

2) For fixed discrete compounding yearly, i = nominal interest rate and N represents years. 3) When there is more than one compounding period per year, the formulae and tables can be used as long as there is a cash flow at the end of each interest period. i represents the interest rate per period and N is the number of periods. 4) When the compounding is more frequent than a year, but the cash flows are annual, the formulae can be used with N as number of years and i as the effective annual interest rate.


Table 4-3: Interest Formulae for Arithmetic Gradient series with discrete compounding

Description Cash Flow Diagram


Factor Computation

Finding the future compounded amount (F) at the end of a specified period due to linearly increasing annual payments (G), given the interest rate.

(N-1)G 2G 1G 0G 0 1 2 … N F?

F = G * GSCAF The Gradient Series Compounded Amount Factor, GSCAF (i%, N) may be computed as shown.

2

1)1(iiGSCAF

N −+=

iN

−

Finding the amount of linearly increasing annual payments (G) that would yield a future compounded amount (F) at the end of a specified period, given the interest rate.

(N-1)G? 2G? 1G? 0G? 0 1 2 … N F

G = F * GSSFDF The Gradient Series Sinking Fund Deposit Factor, GSSFDF (i%, N) may be computed as shown.

GSSFDF

iNii

N *1)1(

2

−−+=

Finding the initial amount, (P) that would be equivalent to specified linearly increasing annual payments (G), given the interest rate.

(N-1)G P? 2G 1G 0G 0 1 2 … N

P = G * GSPWF The Gradient Series Present Worth Factor, GSPWF (i%, N) may be read off from Appendix 1, or may be computed as shown.

GSPWF

NN

iiN

ii ]1[/]1)1([ 2 +−

−+=

Finding the amount of linearly increasing annual payments (G) that would completely recover an initial amount, (P) at the end of a specified period, given the interest rate.

P (N-1)G? 2G? 1G? 0G? 0 1 2 … N

G = P * GSCRF The Gradient Series Capital Recovery Factor, GSCRF (i%, N) may be computed as shown.

GSCRF

iNiii

N

N

*1)1(*)1( 2

−−++

=

Note: 1) In column 4, upper and lower formulae are for discrete and continuous compounding, respectively.

2) For fixed discrete compounding yearly, i = nominal interest rate and N represents years. 3) When there is more than one compounding period per year, the formulae and tables can be used as long as there is a cash flow at the end of each interest period. i represents the interest rate per period and N is the number of periods. 4) When the compounding is more frequent than a year, but the cash flows are annual, the formulae can be used with N as number of years and i as the effective annual interest rate.


Table 4-4: Conversion Factors between Uniform Annual Series and Selected Non-Uniform Series

Type of

Compounding

Direction of Conversion


Linear Gradient Series (G) to Equivalent Uniform Series (A)

Discrete Compounding

Geometric Series1 (M) to Equivalent Uniform Series (A)

Linear Gradient Series (G) to Equivalent Uniform Series (A)

Continuous Compounding

Geometric Series (M) to Equivalent Uniform Series (A)

a GSUAF, Gradient Series Uniform Amount Factor, can also be read off from Appendix 1. 1 The cash flow patterns are changing at a constant rate of t% per period. The initial cash flow in this series, M, occurs at the end

of period 1. The cash flow at the end of period 2 is M(1+t) and at the end of period N is M(1+t)N-1.

4.1.6 Special Cases of Interest Equations

4.1.6.1 Present Worth of Periodic Payments in Perpetuity

In some cases of investment economic evaluation, it may be more appropriate to consider the analysis

period as not finite, but perpetual in time. This is particularly the case where an investment is expected to recur

forever, such as most civil engineering structures (construction of bridges, highways, etc.) that will probably be

needed as long as there is a need for land transportation by humans. Consider the general case of an infrastructure

system that has periodic payments, R, made after every period, N (Figure 4-2). The period N is described as the life-

cycle or service life of that part of the infrastructure system that needs to be provided every Nth year. Such a

provision may involve removal of the existing component and its replacement, or the addition of a new component

on the existing component. The period N is assumed to be constant for this analysis, but it should be noted that N

could be increasing or decreasing as time goes on, depending on the level of usage and technological advances.

−−

−+−+

=11*

1)1(* Ni

i

i

NiN

ee

etetMA

−−

−=

111* *iNi e

Ne

GA

−+

+

−

−

++

=1)1(

)1(*1

11

* N

N

N

iii

itit

MA

aN GSUAFG

iN

iGA *

1)1(1* =

−+

−=


Increasing levels of usage, all else being constant would translate to decreasing values of N as time goes on,

and vice versa. Also, increasing quality of materials and other inputs would lead to increasing values of N, all else

being the same. In Figure 4-2, it is assumed that the initial investment, P is not the same as the periodic investments,

as recurring investments typically cover a lesser scope than the initial one. A case in point is highway construction,

where the initial investment includes right-of-way acquisition, embankment construction, relocation of utilities,

wetlands restoration and other costs aspects that are typically not found in recurring investments such as pavement

resurfacing or reconstruction.

R = Compounded amount of all cash flows within a life-cycle N = Life-cycle of the project that recurs every N years.

Figure 4-2: Present Worth of Periodic Payments (R) in Perpetuity.

Present Worth all R payments made every N years in perpetuity is given by:

P R R R R

Time

∞ N N N N

...)1()1()1( 32, +

++

++

+=∞ NNNR i

RiR

iRPW

+

++

++

+= ...

)1(1

)1(1

)1(1

32 NNN iiiR

−

+−

= 1

)1(11

1

Ni

R 1)1( −+= Ni

R


4.1.6.2 Effect of Infinite Number of Compounding Periods in a Year

In some cases of economic evaluation, not only are there several times the interest rate is compounded

within a year, but it is possible for the frequency of compounding of such rates (and consequently, their periods) to

be so many that the number of compounding periods can be considered infinite. Consider the general case of an

investment where r is the nominal interest rate per year and m is the number of interest periods in a year. This means

that the interest rate per compounding period is given by r/m. The continuously compounded value of a single

amount P, after n years is given by:

But

Therefore,

er*n is defined as the continuously compounded amount factor for the case of infinitely multiple compounding

periods. Similarly, the continuously discounted value of a single future amount F, after n years is given by:

1/er*n is defined as the continuously discounted factor for the case of infinitely multiple compounding periods.

For the case of an infinite number of compounding periods in a year, the effective annual interest rate is given by:

4.1.6.3 Present Worth of Continuously Compounded Payments with Continuously Compounded Interest

In yet another instance of economic evaluation, the cash stream may involve continuous compounding of

payments with continuously compounded interest within a year. Consider a general case where Rj payments are

made each year j. i is the interest rate per year, and r is the rate of growth of payments (expressed as a percentage)

per year (Figure 4-3).

nrrm

nm mrPmrPF**

* )/1()/1(

+=+=

emr r

m

m=

+

∞→1lim

nrPeF *=

nreFP *=

1)(−=

−=

− rr

eP

PePP

PF


Rj = Cash flow at the end of year j Figure 4-3: Present Worth of Continuously Compounded Payments with Continuously Compounded Interest.

Bringing all future streams to the present gives:

R0: Present worth of R0 = R0 = R

R1: Present worth of R1 = (R × er)/ei

R2: Present worth of R2 = (R × e2r)/e2i

Rn: Present worth of Rn = (R × enr)/eni

Summing up the values of all present worths yields:

PWCCP, CCI = present worth of continuously compounded payments with continuously compounded interest.

4.1.7 Estimation of the Rate of Growth of Benefits or Costs

Assuming continuous compounding,

F = P × ern

=> ern = F/P = α, say

r × n = loge α

r = (loge α)/n

The slope measured from a log-normal plot of payments vs. time (Figure 4-4), is r.

Where α is the ratio of future annual value estimate to first year value estimate, F/P,

And n is the number of years (period of estimate).

R0 R1 R2 R3

1st year 2nd year 3rd year

Rn-1

nth year

Rn

−−

×=

++++×= −

−+

11...1

))(1(

2

2

, ir

irn

ni

nr

i

r

i

r

CCICCP eeR

ee

ee

eeRPW

nnPF

nPFr αlog)/log(loglog

==−

=


Figure 4-4: Estimation of the Rate of Growth of Benefits or Costs.

4.2 Methods of Economic Analysis

4.2.1 Introduction

The evaluation of engineering projects is typically carried out using a broad range of criteria that may be

monetary or non-monetary or both. While non-monetary criteria involve effectiveness, the monetary criteria are

based on dollar values, and are therefore associated with project efficiency. In this regard, the fundamental principle

underlying all engineering economic analysis is that the value of money is directly related to the time at which the

value is considered. Therefore, the interest equations studied in Section 4.1.5 are a key aspect of economic analysis.

For economic analysis of engineering projects, benefits and costs are translated to their common dollar

values. For public projects such as non-toll road construction, there are not direct benefits, and therefore reduction in

road user costs is considered a benefit. There are six computational methods for economic analysis as identified and

discussed below. The first two methods are applicable only when all alternatives are associated with the same level

of benefits, and therefore cost minimization is the sole criterion.

1 Equivalent Uniform Annual Cost (EUAC)

2 Present Value of Costs

3 Equivalent Uniform Annual Return

4 Net Present Value

5 Internal Rate of Return

6 Benefit-Cost Ratio

4.2.2 Equivalent Uniform Annual Cost Method

This method combines all initial costs, subsequent maintenance and operation costs and terminal salvage

costs, if any, of an infrastructure system into a single annual cost over the analysis period of n years. Consider the

general case of a system that is associated with an initial investment P, maintenance costs Mj for each year j,

operation costs Rj for each year j, and salvage value S at the end of its service life. The equivalent annual cost EUAC

is given by the following expression:

[ ] ),(*),(*),(*)(),(*1

niUSSFDFSniUSCRFjiSPPWFRMniUSCRFPEUACn

jjj −++= ∑

=

r

n years

log F

log P Time

log ($Value)


4.2.3 Present Value of Costs

This method converts all initial investments, subsequent maintenance and operation costs and terminal

salvage costs, if any, of an infrastructure system into an equivalent single cost incurred at the beginning of the

analysis period (time zero). The equivalent present value of costs, or Present Value of Costs, PVC, is given by the

following expression:

The use of EUAC and PVC for economic evaluation is associated with the assumption that benefits across

alternatives are similar or their monetary values are difficult to determine. However, using costs alone for economic

analysis is inadequate if there are measurable and significant differences in values of benefits from one alternative to

another.

4.2.4 Equivalent Uniform Annual Return Method

The EUAR method combines all initial costs, subsequent maintenance and operation costs and terminal

salvage costs, if any, and annual, periodic, and terminal benefits of an infrastructure system into a single annual

value of return (benefits less costs) over the analysis period. Consider the general case of a system that is associated

with an initial investment P, maintenance costs Mj for each year j, operation costs Rj for each year j, annual benefits

Bj for each year j, and salvage value S at the end of its service life. The equivalent uniform annual return, EUAR, is

given by the following expression:

4.2.5 Net Present Value Method

The NPV of an investment is the difference between the present value of benefits and the present value of

costs. The NPV method combines all initial costs, subsequent maintenance and operation costs and terminal salvage

costs, if any, and annual, periodic, and terminal benefits of an infrastructure system into an equivalent single value

of return (benefits less costs) at the beginning of the analysis period. Consider the general case of a system that is

associated with an initial investment P, maintenance costs Mj for each year j, operation costs Rj for each year j,

annual benefits Bj for each year j, and salvage value S at the end of its service life. The equivalent net present value,

NPV, is given by the following expression:

NPV = Present worth of Benefits – Present Worth of Costs

[ ] [ ] ),(*),(*),(*11

niSPPWFSjiSPPWFRjiSPPWFMPPVCn

jj

n

jj −++= ∑∑

==

[ ]∑=

−−+−=n

jjjj niUSCRFjiSPPWFRMBniUSCRFPEUAR

1),(*),(*)(),(*

),(* niUSSFDFS+


Investments associated with the highest NPV are considered the best investments.

4.2.6 Internal Rate of Return

An economic rate of return is defined as the vestcharge, that is, interest rate at which the net present worth

or equivalent uniform annual return is equal to zero. The Internal Rate of Return (IRR) method determines the

interest rate that is associated with a zero NPV and is consequently associated with an equilibration of present value

of benefits and present value of costs. Then the IRR is compared to the Minimum Attractive Rate of Return

(MARR). If IRR exceeds MARR, then the investment is considered worthwhile.

Consider the general case of a system that is associated with an initial investment P, maintenance costs Mj

for each year j, operation costs Rj for each year j, annual benefits bj for each year j, and salvage value S at the end of

its service life. The IRR value is found as follows:

Present worth of Benefits = Present Worth of Costs

Alternatively, IRR value can be found as follows:

Equivalent Uniform Annual Benefits = Equivalent Uniform Annual Costs

The value of the interest rate that satisfies the above equations can be found using trial and error, or a

computer spreadsheet. If several alternative investments are being evaluated, the best investment is one that has the

highest IRR.

4.2.7 Benefit Cost Ratio Method

The benefit cost ratio (BCR) is a ratio of the equivalent uniform annual value (or net present value) of all

benefits to that of all costs incurred over the analysis period. An alternative investment with a BCR exceeding 1 is

[ ] [ ] [ ] ),(*),(*),(*),(*111

niSPPWFSjiSPPWFRjiSPPWFMPjiSPPWFBNPVn

jj

n

jj

n

jj +−−−= ∑∑∑

===

[ ] [ ] [ ]∑∑∑===

++=+n

jj

n

jj

n

jj jiSPPWFRjiSPPWFMPniSPPWFSjiSPPWFB

111),(*),(*),(*),(*

[ ]

[ ]∑

∑

=

=

++=

+

n

jjj

n

jj

niUSCRFjiSPPWFRMniUSCRFP

niUSSFDFSniUSCRFjiSPPWFB

1

1

),(*),(*)(),(*

),(*),(*),(*


considered to be economically feasible, and the alternative with the highest BCR value is considered the best

alternative. The BCR of an investment is calculated as follows:

Where EUAB = Equivalent Uniform Annual Benefits

EUAC = Equivalent Uniform Annual Costs

NPB = Net Present Benefits

NPC = Net Present Costs

The U.S. Flood Control Act of 1936, in making a reference to the use of BCR for evaluating economic

feasibility, was probably the first organization to use this method of evaluation.

4.2.8 Evaluation Methods Using Incremental Attributes

The foregoing discussion has been on determining the values of an evaluation attribute (benefit, cost, IRR,

benefit/cost ratio, etc.) associated with each individual investment. The best investment is considered as that with the

“most desired” value of the attribute. However, to obviate certain problems associated with such individualized

approached, evaluation of investments are increasingly been carried out using an incremental approach. In the

incremental approach, a particular investment is taken as the base case or base alternative. The attributes of all other

investment alternatives are compared to that of the base alternative (the difference between such is termed the

incremental value of the attribute). An example is found in highway pavement maintenance: For a given family of

pavements, there are several maintenance and rehabilitation scenarios over pavement life. Each scenario is

associated with costs (contract sums, or force expenditure on labor, equipment, material, etc.) and benefits (decrease

in vehicle operating costs arising from decreased surface roughness). The base case could be the scenario where the

pavement does not receive any maintenance. For each alternative scenario, the incremental benefits and costs would

be the benefits and costs of that scenario with respect to those of the zero maintenance alternative.

Table 4-5: Economic Evaluation Methods and Criteria for Choice of Best Investment

Criteria for Choice of Best Investment Method of Economic Evaluation

Individual Approach Incremental Approach

Equivalent Uniform Annual Cost (EUAC) Lowest EUAC Lowest (EUAC - EUACBase Case)

Net Present Cost (NPC) Lowest NPC Lowest (NPC - NPCBase Case)

Equivalent Uniform Annual Return (EUAR) Highest EUAR Highest (EUAR - EUARBase Case)

Net Present Value (NPV) Highest NPV Highest (NPV - NPVBase Case)

Internal Rate of Return (IRR) Highest IRR that exceeds MARR Highest (IRR - IRRBase Case)

Benefit Cost Ratio (BCR) Highest BCR that exceeds 1 Highest (BCR - BCRBase Case)

NBCNPB

EUACEUABBCR ==


In infrastructure management, a frequently proposed alternative is to improve an existing facility. The

procedure discussed above involves the comparison of evaluation attributes with the base case, which is also

referred to as the “null” or “do nothing” alternative, strategy, or scenario. This formulation is particularly useful in

instances where it is sought to determine the viability of improvements to an existing facility, for instance “Is it

worthwhile to keep a freeway section operating at a certain level of congestion? Are the suggested improvements to

a transit system economically viable, or should the system be left as in its current form?”

Each economic analysis method has its unique logic, merits and demerits. The equivalent uniform annual

cost and present value of costs methods are applicable only when all competing alternatives are associated with the

same level of service, and therefore the monetary value of benefits do not change from one alternative to another.

The equivalent uniform annual return and net present value methods, are a step above the cost-only methods, as they

consider the benefits, and are therefore appropriate where competing alternatives have significant and very different

levels of service. NPV, which is expressed as a lump sum, and not a rate, ratio, or index, provides a readily

comprehensible magnitude of the net benefit of an investment. For this reason, NPV is the method recommended by

AASHTO (1997). Like the NPV and EUAR, the IRR method considers both benefits and costs. Also, no assumption

need be made about the interest rate, although the minimum attractive rate of return must be known. The main

disadvantage of the IRR method is the mathematical problems associated with the computation of IRR: a unique

solution is not always guaranteed. Many multilateral agencies, including the World Bank, use the IRR method for

project appraisal. The benefit cost ratio duly considers both benefits and costs, but the use of this method is

susceptible to any ratio-based index: different values of BCR may be obtained depending on the definition, units,

and dimensions of benefits and costs. Most importantly, BCR does not provide any indication of the total extent of

benefit. The use of benefit cost ratio has been common, but is no longer recommended as a desirable approach.

Other evaluation criteria were identified by Winfrey (1970) who stated that the degree of economic

desirability could also be determined using any of the following approaches:

• Determination of the willingness of users in the community to pay for the facility,

• Comparison of the costs associated with the absence of the facility (for example, if bridge

replacement is being considered, evaluation of this activity could be carried out on the basis of the

extra costs incurred by road users on lengthy detours if the bridge were put out of action for any

reason,

• Determination of the social welfare function served by the facility.

4.3 An Application of Economic Analysis Methods: Optimal Timing of Investments

4.3.1 Introduction

In the previous chapter, the economic evaluation of an alternative was made assuming a given project was

implemented immediately or in year 0. However, varying the time of implementing a project can result in very

different economic returns. Suppose a state government plans to build a four-lane highway through a region. Some

pertinent questions that may be posed include: “Should the highway be built now or some years in the future? If it is


to be delayed, what is the desirable year to start construction?” In the case where the location is such that the

highway will have little traffic if it is built now, the user benefit will not be sufficient compared to the construction

cost. In such cases, the government may prefer to postpone the construction by a certain number of years until such

a time that the new road will have enough traffic to produce sufficient user benefit. This type of decision is

particularly common in many developing countries. Finding a point in time when traffic grows high enough to

justify the project is considered as the optimal timing for the improvement project. The underlying assumption for

such analysis is that there is a growing demand for road services with time, reflected by increasing traffic levels.

This section discusses the procedures for determining optimal timing for transportation investments.

4.3.2 The Economic Effect of Postponement

Consider a project that seeks to improve an existing road. Figures 4-5A and 4-5B illustrate the cash flows

associated with each timing scenario: implementing the project immediately, and postponing the project by one year,

respectively. A comparison of the two cash flow diagrams shows that the difference between them includes the loss

of the first year’s benefit by postponing the project by one year and the gain by not paying interest by one year on

the capital investment. In addition to that, the salvage value of the facility constructed a year later (Case B) may be

slightly higher than that constructed at the current year (Case A), even though the gain due to higher salvage will be

relatively small. Salvage differences can therefore be ignored in such cases where there is relatively little timing

difference in the initial investments.

A: Implementing the Project immediately

B: Postponing Project Implementation by One Year

Figure 4-5: Cash Flow Illustration of the Effect of Investment Postponement.

C

1 2 n-1 n 3 4 5 6

b1 b5 b6 bn-1 bn b2 b3 b4

C

1

2 n-1 n 3 4 5 6

b5 b6 bn-1 bn b2 b3 b4

0

0

Year

Year


In Figure 4-5 the initial investment cost is denoted by C, while the benefit in each year is denoted by bt. Let

NPVt be the Net Present Value in the general case where the initial investment occurs in year t. Assuming that the

benefits associated with the occurrence of an initial investment start accruing in the following year, the Net Present

Value for Case A (ignoring the salvage value) is:

NPV0 = -C + b1*SPPWF (i, 1) +b2 * SPPWF (i, 2) + … + bn*SPPWF (i, n)

Where i is the interest rate.

In Case B (postponement by one year), the net present value is:

NPV1 = -C*SPPWF (i, 1) +b2 * SPPWF (i, 2) + … + bn*SPPWF (i, n)

The algebraic difference between the above two equations for NPV0 and NPV1 yields the saving obtained

due to the delay of the postponement by one year, is given as follows:

S(1) = NPV1 – NPV0 = (-C-b1)*SPPWF (i, 1) + C

The numerator in the above equation is the interest on initial construction cost minus the first year’s benefit

and the denominator is simply SPCAF. If S(1) is discounted to Year 1, the NPV0 equation 1 ends up with interest ion

initial capital minus the first year’s benefit. For the postponement by one year to be justifiable, it is obvious that first

year’s benefit should be less than i*C and it should be tested if additional postponement is also worthwhile.

Additional postponement by another year is worthwhile if NPV2 is greater then NPV1. In other words, S(2) must be

positive in order to justify postponement of the implementation by an additional year. Generally,

NPV2 = -C* SPPWF (i, 2) + b3*SPPWF (i, 3) + …

S(2) = NPV2 - NPV1 = (-C – b2)* SPPWF (i, 2) + C*SPPWF (i,1)

)1(* 1

ibCi

+−

=

22

)1(*

ibCi

+−

=


S(2) must generally be positive to make the postponement of the implementation by one additional year

justifiable. Generally,

The timing of the initial investment can be delayed as long as the NPV at year t is equal to or greater than

that of the previous year (i.e., as long as S(t) ≥ 0). The year at which NPV is just less than NPV at the previous year

(i.e., when S(t) becomes < 0) is the optimal year for implementing the project by carrying out the initial investment.

In other words, the optimal timing of the investment is the year at which bt is greater than i*C. Therefore the

decision criterion for deferring (or implementing) a project can be stated as follows:

1. If S(t) ≥ 0, defer the project

2. If S(t) < 0, implement the project so that the benefit for the first year after the initial investment is

greater than the annual interest on the capital.

This criterion is called the First Year Benefit (FYB) criterion. Whenever S(t) is less than 0, the interest on

the initial capital investment, C, is less than the first year benefit, bt. Therefore the FYB criterion can be restated as

follows:

1. If the ratio of the first year benefit to the initial cost, bt /C) is smaller than or equal to the interest rate,

defer the project,

2. If the ratio of the first year benefit to the initial cost, bt /C) is greater than the interest rate, then

implement the project.

The ratio bt /C is the first year rate of return. So the second criterion above is referred to as the First Year

Rate of Return criterion. As soon as the project implementation is justified by the criterion, it is imperative that the

implementation begins early enough to open the facility in year t, so that there is full benefit bt obtained by the end

of the first year. If the project implementation period is expected to last for a long time, the project must be started

well in advance, so that the implementation is completed by the start of the year t.

Figure 4-6 shows the gains from delaying a project by one year, S(t), and NPV of the project over a period

of time. NPV keeps increasing and reaches the point at which NPV becomes positive for the first time. This point of

time is called the breakeven year, nBE. From this point of time, the starting of the project is justifiable. From this

breakeven point, postponement of the project by one year will produce a net gain because the benefit increases and

NPV will grow until it reaches the maximum value. The point at which NPV is a maximum is the optimal time, nOPT

for implementing the project. As shown in Figure 4-6, NPV will be a maximum when S(t) is zero. After nOPT, the

gain by postponement by another year becomes negative and NPV starts to decrease.

tt

ibCi

tS)1(

*)(

+−

=


Figure 4-6: Net Present Value and Gain due to Postponement of Implementation.

Example 4.3-1

A county highway agency intends to build a bridge. The data below shows the cash flow of the project for a

period of 10 years. There will be a major improvement after 10 years. Find the optimal timing for the bridge project.

Assume MARR is 10%.

Year 0 1 2 3 4 5 6 7 8 9 10

Construction Cost ($103) -1000

Benefits ($103) -- 83 101 124 138 155 170 188 200 210 215

Year 11 12 13 14 15 16 17 18 19 20

Construction Cost ($103)

Benefits ($103) 225 232 241 255 270 284 300 311 325 329

Solution

Five investment scenarios are considered: Construction in each year from year 0 to year 4. For each

scenario, the cash flows and present (at Year 0) worth of all such payments are calculated, and the net present value

is found (Table 4-6). The table shows that opening the facility in Year 1 produces an NPV of 300 when discounted

to Year 0. Since NPV is greater than 0, the project is feasible. The table also presents the NPV values if the project

were deferred. It can be seen that NPV values keep increasing with postponement of up to 2 years. After 2 years,

further postponement does not yield an increase in NPV. Therefore, the optimal time of implementing the project is

Year 3. Applying the First Year Benefit or the FYRR criterion would yield the same result as shown.

Time of Implementation

Time of Implementation

S(t)

NPV(t)

nBE nOPT


From the FYB criterion,

bt = 101

Ci = 1000* 0.1= 100

The first time bt exceeds 100 is the second year.

From the FYRR criterion,

b t /C = 0.101

MARR = 0.1

The first time bt/C exceeds MARR is the second year.

Table 4-6: Cash Flows and Present Worths for Different Project Timing for Example 4.3-1

Construction Starts in Year 0





Year Cash Flow

PW year 0

Cash Flow

PW year 1

Cash Flow

PW year 2

Cash Flow

PW year 3

Cash Flow

PW year 4

0 -1000 -1000

1 83 74 -1000 -893

2 101 81 101 81 -1000 -797

3 124 88 124 88 124 88 -1000 -712 -1000 -636

4 138 88 138 88 138 88 138 88 138 88

5 155 88 155 88 155 88 155 88 155 88

6 170 86 170 86 170 86 170 86 170 86

7 188 85 188 85 188 85 188 85 188 85

8 200 81 200 81 200 81 200 81 200 81

9 210 76 210 76 210 76 210 76 210 76

10 215 69 215 69 215 69 215 69 215 69

11 225 65 225 65 225 65 225 65 225 65

12 232 60 232 60 232 60 232 60 232 60

13 241 55 241 55 241 55 241 55 241 55

14 255 52 255 52 255 52 255 52 255 52

15 270 49 270 49 270 49 270 49 270 49

16 284 46 284 46 284 46 284 46 284 46

17 300 44 300 44 300 44 300 44 300 44

18 311 40 311 40 311 40 311 40 311 40

19 325 38 325 38 325 38 325 38 325 38

20 339 35 339 35 339 35 339 35 339 35

NPV +300 +333 +348 +338 +250


4.3.3 Effect of Multi-year Implementation of Investments

So far, it has been assumed that the construction cost was accrued in one year. Such as assumption may be

unduly restrictive in many actual cases. If the implementation period (time taken for construction) exceeds 1 year,

then implementation costs spread over more then one year, and consequently a separate set of computations may be

necessary to determine the optimal timing for construction. This can be illustrated through the use of an example

below:

Example 4.3-2

Consider a project that would take 4 years to complete. Figure 4-7A show the cash flow diagram if the

project is started immediately, and Figure 4-7-B shows the cash flows if the beginning of the construction is

postponed by one year. C1, C2, C3 and C4 are the costs for the first, second, third and fourth years of completion

respectively. The benefit streams start at the completion of the project, and they are counted at the end of each year.

The net present worth values for the two cases can be computed as follows:

The saving from deferring the project by one year, S(1), is obtained by subtracting NPV0 from NPV1, as

shown below:

Where,

C* represents the sum of the discounted values of all the construction costs in the year preceding the

opening of the project. To justify one year delay of the project, S(t) should be greater then 0, or i*C* should be

greater than b4. The First Year Benefit criterion applied to this case in the same way as in the case of one-year

construction except that the construction costs should be compounded in the year preceding the first benefit year.

…55

44

34

232

10 )1()1()1()1(1 ib

ib

iC

iC

iCCNPV

++

++

+−

+−

+−−=

…55

44

33

221

1 )1()1()1()1(1 ib

iC

iC

iC

iCNPV

++

+−

+−

+−

+−=

44

4443

22

31

)1(*

)1())1()1()1((

)1(ibiC

ibCiCiCiCi

S+−

=+

−++++++=

432

23

1 )1()1()1(* CiCiCiCC ++++++=


A: Case A- Implementing the Project “Immediately”

B: Case B- Postponing Project Implementation by One Year

Figure 4-7: Cash Flows for Investments with Multiyear Implementation Periods.

Example 4.3-3

Assume a 3-year construction period for a project; benefits begin accruing in the fourth year. The

construction costs spread over three years, and $50,000, $250,000 and $200,000 are spent in the first, second and

third year, respectively. Yearly benefits for four years are tabulated in Table 4-7. The interest rate is assumed to be

6%.

Table 4-7: Yearly Benefits for Example 4.3-3

Year Benefits (in $103)

1998 31

1999 34

2000 37

2001 40

Solution

Table 4-8 shows costs discounted to the year preceding the first benefit year. Using FYB criterion, the first

year S(t) becomes negative is 1999, (b3 = 34, which is more than 521*6%). Since the construction takes 3 years, the

construction should start in 1996.

1 2 34 5 6 N-1 N

0

C1 C2 C3 C4

b4 b5 b6 bN-1 bN

1 2 35 6 N-1 N

0

C1 C2 C3

b5 b6 bN-i bN

4

C4


Table 4-8: Yearly Construction Costs for Example 4.3-3

Year Costs (in $103) Cost Compounded in Year 3

1 50 56

2 250 265

3 200 200

Total 521

4.3.4 Multi-stage Implementation of Investments

When considering a construction of transportation facility, the following question is often raised, “Should

we build the facility now at the full service level to serve the future demand, or construct in stages as the demand

develops?” This problem can be illustrated by considering the following example.

Example 4.3-4

Consider the following situation where a four-lane roadway is to be constructed to full capacity now or in

two stages (two lane road way is built now and additional two lanes are built n years from now). Construction costs

for the two cases are the following:

Single Stage Construction

Construct four-lane roadway now $1,866,000

Two Stage Construction

Construct two-lanes now $1,200,000

Construct additional two-lanes n years from now $1,866,000

For the purpose of simplifying the example, annual benefits for both cases are assumed to be the same and zero

salvage value at end of the service life. Assuming the service life is infinite, find which option is economically more

viable? Interest rate is 7%.

Solution

Since the annual benefits for both cases are the same, the Net Present Cost method can be used. Present

worth of cost for the single stage construction is:

PW1 = $1,866,000

PW of cost for the two-stage construction case is:

PW2 = $1,200,000 + $1,500,000*SPPWF(7%, n)

Sample calculations of PW2 for several values of n are

n = 5 $1,200,000 + $1,500,000*SPPWF(7%, 5) = $2,269,500

n = 10 $1,200,000 + $1,500,000*SPPWF(7%, 10) = $1,962,450

n = 12 $1,200,000 + $1,500,000*SPPWF(7%, 12) = $1,896,000

n = 20 $1,200,000 + $1,500,000*SPPWF(7%, 20) = $1,587,600

n = 30 $1,200,000 + $1,500,000*SPPWF(7%, 30) = $1,397,100


A plot of PW of costs for the two cases is shown in Figure 4-8. The x-axis variable is the time the second

stage construction is conducted. PW of the single stage construction remains constant regardless of the second

construction time of the two-stage construction alternative. The present worth of the two stage construction

alternative decreases as the second construction is done 12 years from now. Until year 12, PW of cost for the two-

stage construction alternative is greater than that of single stage construction alternative. If the second stage is

needed before year 12, PW of cost for the single stage is smaller then PW of the two stage alternative. This implies

that if, in the two-stage construction, the second stage should be constructed before year 12, the single stage

construction is preferred. On the other hand, if the second stage construction is not needed until year 12, it would be

better to wait and build the facility in two stages.

4.3.5 Optimization of Stage Implementation of Investments

Stage construction, an example of stage implementation of investments, is the process of making

progressive improvements to a facility over an extended period of time, rather than constructing or improving it

initially to some high standard (World Bank, 1990). Stage construction can be economically desirable even though

the total of several stage construction costs may be higher than the single stage construction cost because of the time

value of money. If stage construction is preferred, the optimal timing of each construction stage should be

determined. Methods of determining the optimal timing of successive stage are introduced in the section.

Consider an existing roadway that currently provides a low level of service. Improvements are being

considered for the road. Instead of improving the road to achieve the full standard level of service, a two-stage

construction is proposed. Annual costs including user costs and maintenance costs are assumed to increase linearly

with time. In other words, the recurring functions are of the form: a + b*t. After each improvement, it is expected

that these annual costs will be significantly reduced and the form of the cost functions will change even though the

linear increase over time is assumed. Figure 4-9 shows the pattern of costs with the two-stage construction. The

period from now to the first upgrading time is defined as Stage 0. For upgrading the roadway from Stage 0 to Stage

1, the construction cost is C1 incurred in year t1. After the first upgrading of the road, there is a sharp drop in annual

cost line. Following the drop, cost start to grow linearly again until the next upgrading. This so-called saw-tooth

pattern continues every time a stage upgrading is conducted.


Figure 4-8: Present Worth of Costs for Single Stage and Two-Stage Construction.

Figure 4-9: Pattern of Costs with Two-Stage Construction.

1.0

2.0

3.0

10 20 25 15 30 5

Single Stage

2-Stage

Present Worth of Costs ($*106)

Construction Year

Total Costs

Year

a0+b0*t

C1

(Existing) t1 t2

Stage 0 Stage 1 Stage 2

C2

a1+b1*t a2+b2*t


The variables in Figure 4-9 are defined as:

tj = optimal timing of upgrading from Stage j to j+1,

aj, bj = coefficients of annual costs as a linear function of time during Stage j.

The optimal timing of the two improvements in the example should be determined so that the total costs are

minimized. Assuming each stage construction is completed in one year, the present worth (PW) of all costs with

interest rate is:

It can be noticed that continuous discounting is used for the above equation.

To obtain a set of timings, t1, and t2 that minimizes the present worth value, the PW expression is

differentiated with respect to t1 and t2, and the resulting partial derivatives are set to zero. After simplifying, optimal

values of t1 and t2 can be obtained as shown below:

The generalized form of optimal stage construction timing can be stated as:

Where tj = optimal timing for any sequential upgrading from stage j to j+1

Cj = upgrading cost

aj, bj = coefficients of linear yearly cost functions

dtetbaeCdtetbaeCdtetbaPW ti

t

titit

t

titit

*22

*2

*11

*1

*

000

2

2

2

1

1

1

)*(*)*(*)*( −∞

−−−− ∫∫∫ +++++++=

10

1011

)(*bb

aaiCt−

−−=

21

2122

)(*bb

aaiCt−

−−=

jj

jjjj bb

aaiCt

−

−−=

−

−

1

1 )(*


…

Figure 4-10: Pattern of Costs with Two-Stage Construction.

Example 4.3-5

A developing country is planning to upgrade an existing 14.3 mile gravel road first by stabilizing it in a few

years from now and then providing an asphalt surface several years after the first improvement. The project will thus

have three stages and the base year is 1995. A strategy is needed for implementation of the project. The data on the

type of construction and unit cost for each upgrading are as follows:

1. From gravel surface to stabilized surface: $320,000/mile

2. From stabilized surface to asphaltic concrete surface: $400,000/mile

From previous similar type of projects in this particular region, the continuously increasing recurring cost

functions (user and maintenance) are estimated as follows: Stage 0 (from 1995 to year of stabilizing): 2.93 + 3.41t

Stage 1 (from year of stabilizing to year of asphaltic concrete surfacing): 2.59 + 3.41t

Stage 2 (after asphaltic concrete surfacing): 2.17 + 2.88t

Where t is the number of years from the base year.

What would be the optimal year for successive regarding? Assume that traffic growth is linear and the MARR is

5.5%.

Solution:

Construction cost for each upgrade is as follows:

C1 = $320,000 * 1.43 = $4,576,000

C2 = $400,000 * 1.43 = $5,720,000

Total Costs

Year

a0+b0*ekt

C1

(Existing) t1 t2

Stage 0 Stage 1 Stage 2

C2

a1+b1*ekt a2+b2*ekt


Cost functions are defined as aj + bj *t,

Where aj’s and bj’s are as follows:

Stage j aj bj

0 2.93 3.41

1 2.59 3.14

2 2.17 2.88

Applying the general equation for optimal stage construction timing yields:

So, the stabilizing should be done in 8 years from 1995 and then in another two years the road should be

upgraded to an asphalt surface.

The above example assumes linear increasing function of recurring costs. However, in practice, annual

recurring costs can be rather exponentially increasing (Figure 4-10) because the traffic increase is exponential and

expressed in percent per year. In this case the recurring costs can be expressed as ai + bi*ekt, where k is the annual

traffic growth rate. The optimal timing (tj) for stage construction with exponentially increasing cost function can be

derived in a similar manner, and is given by:

Example 4.3-6

A two-lane 12.5-mile highway section has a current Average Daily Traffic of 6,500 vehicles. The highway

is in a poor condition causing excessive user and maintenance costs each year. Due to the high demand of this

highway, the county has decided to do a major improvement. It will include the reconstruction of the existing

pavement in addition to building two new lanes. However, because of high cost and funding problems, the county is

planning to do the reconstruction in stages. The county will undertake the setting up of a sinking fund to finance the

project where an annual deposit will be made until the completion of the project. The fund is expected to earn 10%

per year. The recurring annual cost functions (in $105) are exponential and of the form: a + b*ekt where k is traffic

growth rate and t is time.

06.810*)14.341.3(

10*)59.293.2(055.0*000,576,4)(*5

5

10

1011 =

−−−

=−

−−=

bbaaiCt

48.1010*)88.214.3(

10*)17.259.2(055.0*000,270,5)(*5

5

21

2122 =

−−−

=−

−−=

bbaaiCt

kbb

aaiC

t jj

jjje

j

−

−−

= −

−

1

1 )(*log


The plan is to implement the project in the following stages and it is expected that the construction in each

stage can be completed in one year.

Table 4-9: Costs for Stage Construction for Example 4.3-6

Coefficients of Recurring Cost Function

Stage Work to be

Accomplished

Construction Cost

(in $105) a b

Traffic Growth Rate

0 Current

-

0.62

0.20 -

1

Reconstruction of the old pavement

4.72

0.413

0.14

5.1

2 New one-lane widening

2.5

0.317

0.07

5.7

3 New one-lane widening overlaying all 3 lanes

4.12

0.167

0.005

7.0

From the given data, we will determine the optimal timing to start each stage. Knowing that the depositing

in the fund will start at the beginning of the project, how much should be deposited annually for each stage? Assume

MARR = 6.5 %.

Applying the equation for tj,

t1 = 3

t2 = 5

t3 = 7

4.3.6 Application of Optimization of Stage Implementation: Optimal Timing for Paving Low-Volume Gravel

Roads (Bhandari and Sinha, 1985)

Gravel surface roads are generally adequate for most situations of low volume traffic. However, as traffic

increases, the maintenance and vehicle operating costs also increase. Therefore, it is eventually necessary to

consider paving the gravel surface. In order for the paving the road to be economically viable, savings in

maintenance cost and user cost should be substantial enough to justify the construction cost. If maintenance cost is

assumed to be constant over time, traffic volume, which benefits by the improvement, should be large enough to

produce much savings in comparison to improvement cost. In most cases of gravel low-volume road, the base year

volume is low so that investing the capital is hardly returned if construction is conducted in base year (n0). In that

case, it is preferred postponing the construction until it is, at least, economically feasible to pave the road. When the

base year volume is low but increasing with time, it is possible to invest the capital somewhere else in the economy

to obtain returns that are in excess of the benefits foregone during the years that construction is deferred. If the

concept of break-even analysis is applied to determine the cut-off volume (and the corresponding year) above which

it is economically feasible, this volume represents the minimum volume, above which the net present value of

paving is in excess of zero. But, it is more advantageous to postpone construction beyond the break-even volume


until such time when the net present value is maximized. The net present value by paving the road in the year in

which first year benefits are equal to the interest rate by First Year Benefit criterion described in the previous

section. The volume of traffic in this year is referred to as the optimal volume. If the road is paved a year before the

optimal volume, it would mean foregoing excess benefits that could be obtained from an alternative investment in

the economy, while paving the road a year after, would mean losing excess benefits that could have accrued had the

road been paved a year earlier. This typical situation is illustrated in Figure 4-11. The top portion of the figure shows

the stream of annual net benefits arising from paving the road in the year, say n1.

Compared to the benefit patterns in Figure 4-11, annual net benefits is increasing somewhat exponentially

rather than linearly increasing. The net benefits of paving the road in any other year should be the same as shown,

except for the years before paving when the net benefits are zero. For example, the net benefits corresponding to

construction in the break-even year (nbe) and the optimal year (nopt) would be correspond to the lines 0nbepq and

0noptq, respectively. The lower portion of the figure shows the net present value of the entire project, given the year

of construction. The points nbe and nopt correspond to zero and maximum net present values, respectively.

Figure 4-11: Net Present Value as a Function of the Year of Construction [Bhandari and Sinha, 1985]


In order to appropriately analyze the project, it is necessary to properly define important factors such as

service life, costs and benefits. For most road project it is true that a newly constructed facility will have a fixed

physical life varying from 15 to 49 years. However, generally, or often the case, major rehabilitation will be done at

the end of this period and in subsequent periods to perpetuate the useful life of that facility. For road projects, this is

often done in the form of major reconstruction and overlay, whenever the serviceability of the pavement has fallen

below an acceptable level. Under this assumption, benefits may be assumed to accrue indefinitely, for all practical

purposes. Thus, service life is assumed be infinite. The frequency of major rehabilitation is assumed to be planned in

advance, even though the physical life of a pavement is a function of the design standard and the traffic volume that

it is subjected to. Gain in salvage values being zero by deferring construction a few years can be also explained

using the concept of this indefinite service life. The primary costs associated with road projects consist of the

construction, routine maintenance, planned rehabilitation and vehicle operating costs. Construction costs are

assumed to occur only in the first year of construction while routine maintenance costs are assumed to remain

constant over time. Major rehabilitation costs are planned to occur every N years, where N is presumably close to the

physical life of the pavement.

The benefits from paving a gravel road comprise largely of the reduction in the total vehicle operating and

routine maintenance costs. Net benefits are then obtained after allowing for the construction and rehabilitation costs.

Let the variables relevant to a gravel road being considered for paving be defined as follows.

Q0: volume of traffic in the base year, in vehicles per day

m,m': uniform equivalent annual routine maintenance costs per km, before and after paving, respectively

c,c': average operating costs per veh-km on gravel and paved surfaces, respectively

C: fixed construction costs per km of paving

N: frequency of major rehabilitation, in years

R: cost of rehabilitation, in dollars per km

i: opportunity cost of capital (interest rate)

k: traffic growth rate per annum (generally less than i)

n: the year in which the gravel road is paved

The present value (PV) of various cost elements per km of roadway are obtained as follows:

PV of construction costs

Ci n *)1(

1+

=


P.V. of rehabilitation costs

It should be noted that the road is kept in service indefinitely through periodic rehabilitation.

PV of savings in routine maintenance costs

PV pf savings in vehicle operating costs

for k < i,

Where

Qn = Q0*(1+k)n, and

(1+k)/(i-k) = the discount factor for present value of a geometric series with growth rate k and

discount rate i, over an indefinite period (for k < i).

The net present value of paving is then obtained as follows:

If the breakeven year is now denoted by n, then the break-even volume, Qbe, may be obtained by setting the

NPV equal to zero in the above equation.

1)1(*

)1(1

−++= Nn i

Ri

imm

i n

)'(*)1(

1 −+

=

−+

−+

=kikccQ

i nn

1)'(*365**)1(

1

−+

−−

−

+

−+

−+

=1)1(

'1)'(*365**)1(

1Nnn i

RCi

mmkikccQ

iNPV

−+

−

−−

−++

=

kikcc

imm

iRC

QN

be 1*)'(*365

)'(1)1(


Provided the base year volume, Q0, is less than the volume in the break-even year (Qbe), the break-even

year, nbe, may then be obtained as follows:

for Q0 < Qbe, otherwise zero.

The optimal year for paving the gravel road is obtained by maximizing the NPV equation with respect to n.

However, as seen in Figure 4-12, with the discount rate equal to the opportunity cost of capital, the net present value

is maximum when the undiscounted net benefits in the year of paving are equal to zero. For such maximum to exist,

it is necessary that the benefits increase monotonically with time. This is ensured as long as the savings in the

vehicle operating costs and traffic growth rate both non-negative. The net benefits in the first year after paving are

given as follows:

To obtain the optimal volume Qopt, Qn is substituted by Qopt in the above expression and set it equal to zero.

Again, if Q0 is less than Qopt, the optimal year of paving, nopt, is obtained as follows:

for Q0 < Qopt, otherwise zero.

The break-even and optimal years obtained from the nbe and nopt equations are rounded to the nearest whole

numbers, in line with the usual assumption of year-end cost outlays.

benbe kQQ )1(0 +=

( )1)1(

')'(*365*)1(*−+

−−−+−+= Nn iRiCimmcckQNPV

)1(*)'(*365

)'(1)1(

*

kcc

mmi

iRCiQ

N

opt +−

−−−+

+=

)1(log)/(log 0

kQQ

ne

opteopt +

=

)1(log)/(log 0

kQQ

ne

beebe +=


Figure 4-12: Net Present Value as a Function of the Year of Construction.

Example 4.3-7

A 40-km gravel surface road on a reasonably good alignment is now carrying an annual average daily

traffic of 100 (10% truck). The traffic is expected to grow at an annual rate of 5% compounded (for each road type

and all vehicles). The minimum attractive rate of return is 10% per annum. It is proposed to pave the road with

minor reconstruction and the rest of the project details are shown below:

Item Value

1. Road Paving or Reconstruction Cost $50,000 per km

2. Road Maintenance a) Existing Road b) Paved Road

$600 per km/year $450 per km/year

3. Vehicle Operating Costs on a Gravel Surfaced-Road

Cars- 10 cents per day Trucks- 12 cents per day

4. Vehicle Operating Costs on a Bituminous Surfaced-Road

Cars- 7 cents per day Trucks- 9 cents per day

5. Rehabilitation/Resealing Cost $25,000 per km every 10 years

Determine:

1. The optimal year of paving and the break-even year of paving

2. The traffic volumes at the optimal year of paving and the break-even year of paving

3. The first year benefits and costs after the year of paving

4. The sensitivity of break even and optimal traffic volumes (and their corresponding years) with respect

to the interest rate.

Solution

1. Average operating cost per veh-km on gravel surface is:

c = 0.9 × 0.1 + 0.1 × 0.12 = $0.102/veh-km

Net Present Value (NPV)

Year of Implementation(Construction) nopt nbe

0


2. Average operating cost per veh-km on bituminous surface is:

c’ = 0.9 × 0.07 + 0.1 × 0.09 = $0.072/veh-km

Using the equations for break-even traffic volume and year, and the optimal volume and year are obtained,

as shown below:

After break-even year:

1st year benefits = Qbe × (k+1) × 365 × (c – c’) + (m-m’)

= 279 × 1.05 × 365 × (0.102 – 0.072) + 150

= $3357.8

1st year costs

dayvehsQbe /279

05.01.005.1*)072.0102.0(365

1.0450600

11.12500050000 10

=

−−

−−

−+

=

yearsne

ebe 21

05.1log)100/279(log==

dayvehsQopt /558)05.01)(072.0102.0(365

)450600(11.110.0*2500010.0*50000 10

=+−

−−−

+=

yearsne

eopt 2.35

05.1log)100/558(log==

1)1(*

−++= Ni

iRiC

63.6568$1)10.01(

10.0*2500010.0*50000 10 =−+

+=


After the optimal year:

1st year benefits = Qopt × (1+k) × 365 × (c – c’) + (m-m’)

= 586 × 1.05 × 365 × (0.102 – 0.072) + 150

= $6565.61

1st year costs

3. The sensitivity of the break-even volumes and their corresponding years is presented as follows:

Interest Rate Qbe nbe Qopt nopt

5% 0 0 369 28.2

8% 182 12.3 509 33.4

10% 279 21.0 558 35.2

12% 369 26.8 664 38.8

15% 498 32.9 784 42.2

Figure 4-13: Sensitivity of Breakeven Volume to Interest Rate.

63.6568$1)10.01(

10.0*2500010.0*50000 10 =−+

+=

1)1(*

−++= Ni

iRiC

0

200

400

600

800

1000

0 5 10 15 20

Interest Rate (%)

Traf

fic V

olum

e

Qbe Qopt


Figure 4-14: Sensitivity of Breakeven Years to Interest Rate.

EXERCISES

1) What is the difference between fixed annual rate and effective annual interest rate?

2) Under what assumption can ‘Equivalent Uniform Annual Cost’ or ‘Present value of cost’ method be used

for evaluation of engineering projects?

3) What is the First Year Benefit criterion? What is the First Year Rate of Return criterion? What are they

used for?

4) What is breakeven year? How is it related to NPV? What is the optimal year?

5) What fixed annual interest rate compounded monthly yields an effective annual interest rate of 19.56%?

6) Derive an expression for calculating the present worth of a geometric gradient series with an interest rate of

i% compounded continuously over a period of n years. The series starts with an amount of M at the end of

first year and grows at the rate of r% per year.

7) Derive a generalized expression to compute the present worth of a geometric gradient series with an

interest rate of i% per year over a period of N years (compounded annually). Assume that the cash flow at

the end of year one is M which grows at f% per year.

8) The rate of growth of traffic on a newly constructed bridge is 3% per year. By the end of first year 500,000

vehicles would have used it. Determine the number of vehicles using the bridge in the tenth year of service.

Assuming that a toll of $0.75 is collected per vehicle, calculate the present worth of the total toll collections

during the ten-year period using the equation derived in problem-6. Assume an interest rate of 10% with

continuous compounding. For simplicity cash flows can be considered discrete, occurring at the end of each

year.

0

10

20

30

40

50

0 5 10 15 20

Interest Rate (%)

Yea

rs

nbe nopt


9) Calculate the present worth of the cash flow diagram shown below. Assume interest is compounded

monthly.

P=? $1000 $5000 $1000 $10,000

0 1 2 3 4 5 (Years)

5% 7% 10% (Nominal Interest

Rate per year)

10) 20 equal annual payments, of $2000 each, are made starting one year from today. What is the future

equivalent value of this series at the end of 20 years? Assume 10% interest per year.

11) Consider the following two alternatives for a transportation project:

Alternative A Alternative B

Initial investment $ 50,000 $ 100,000

Annual maintenance cost $ 15,000 $ 10,000

Annual User Benefits $ 20,000 $ 25,000

Useful life (years) 10 10

Salvage Value $ 20,000 $ 25,000

Which alternative would you recommend based on:

(i) Equivalent Uniform Annual Return method?

(ii) Net Present Value method?

(iii) Internal Rate of Return method?

(iv) Benefit-Cost Ratio method?

Assume 10% interest per year and MARR of 8%.


12) Three alternative designs are being considered for a potential improvement project related to the operation

of an engineering department. The prospective net cash flows for these alternatives are shown below.

MARR is 15% per year.

End of year A B C

0 -$200,000 -$230,000 -$212,500

1 $90,000 $108,000 -$15,000

2 $90,000 $108,000 $122,500

3 $90,000 $108,000 $122,500

4 $90,000 $108,000 $122,500

5 $90,000 $108,000 $122,500

6 $90,000 $108,000 $122,500

Apply the IRR method and the NPV method using incremental analysis procedure to select an

alternative. (Source: Engineering Economy, Degarmo, E. P. et al.)

13) The alternatives for an engineering project are reduced to three designs. The estimated capital investment

amounts and annual savings are shown in the table below. Assume that the MARR is 12% per year, the

study period is six years, and the market value is zero for all three designs. Apply the NPV method to

determine the preferred alternative. (Source: Engineering Economy, Degarmo, E. P. et al.)

Hint: Use the expression derived in Problem-7 for Design-1.

End of year Design-1 Design-2 Design-3 0 -$100,000 -$80,000 -$100,000 1 $25,000 $30,000 $20,000 2 3 4 5 6 7 8 9

10

After year one, the annual savings are estimated to increase at the rate of 6% per year.

After year one, the annual savings are estimated to increase $150 per year.

Uniform sequence of annual savings.

14) You loaned out an amount of $12,500 at an interest rate of 13% per year for a period of six years. How

much total interest would you owe at the beginning of the 4th year? Assume that you make no payments

until the end of loan period and interest is compounded annually.

15) Draw a cash flow diagram for a $15,000 loan which needs to be repaid over a 5 year period in equal

uniform annual amounts. What is the amount of annual repayment if the interest rate is 12%?


16) Answer the following questions with reference to the cash flow diagram shown below:

(i) What future sum can be accumulated in 5 years with equal annual payments of $1,000 if the interest

rate is 10%?

(ii) What is the amount of uniform annual payment required to provide a future sum of $10,000 in 5 years

from now if the interest rate is 10%?

(iii) You want to accumulate a future sum of $15,000, but you can afford to pay only $1000 per year. If the

interest rate is 12%, what is the minimum number of years you will take to accumulate that money?

(iv) You make uniform annual payments of $1,000 each for 10 years. If you accumulate an amount of

$20,000 at the end of 10 years, what is the interest rate?

A A A A 0 1 2 … N

F

17) The initial investment for constructing a road is $500,000. The maintenance is expected to cost $15,000 per

year for the first 5 years of service, followed by $30,000 per year for the next 5 years. A rehabilitation

program is scheduled at the end of 10th year costing $150,000 after which the maintenance costs are

expected to reduce to $20,000 per year. What is the equivalent uniform annual cost over a 20 year period of

service if money is worth 8% per year? What is the equivalent present worth of all the road expenses?

18) A rehabilitation program is scheduled 4 years from now. If the expenditure is expected to be $50,000 then,

what amount should be set aside now to provide for that expenditure? Assume that funds earn 8% interest

per year.

19) A bridge project is considered for which two alternatives are being studied. Under the first alternative, a 4-

lane bridge will be constructed now for $5,000,000. The maintenance cost would be $30,000 for the first

year which is expected to increase by $2500 per year. Economic salvage value at the end of 20 years is

$1,000,000. The second alternative is to construct a 2-lane bridge now costing $3,000,000. Maintenance

would be $30,000 per year for 10 years. At the 10th year 2 more lanes would be added to accommodate for

the expected growth in traffic. This would cost $4,000,000 after which the maintenance cost for the new

facility would be $60,000 per year. Economic salvage value at the end of year 20 is $1,500,000. If the

interest rate is 10%, should the deferment of 2-lanes be preferred?

20) The cash flow table below shows expenditures and benefits for the construction of a parking facility

assuming that the construction takes place now. Find the optimal timing for the project. MARR is 9%.

Year 0 1 2 3 4 5

Construction Cost $100,000

Benefits - - $5000 $8000 $10,000 $13,000 $15,000


21) A transportation facility has a linear cost function given by: $(30,000 + 15,000*t ) where t is in years. A

rehabilitation program is planned in two stages. The maintenance cost function after stage-1 is expected to

be linear given by: $(25,000 + 10,000*t)

And the cost function after stage-2 is given by: $(20,000 + 8000*t)

The costs of implementing stages 1 and 2 are $300,000 and $200,000 respectively. Find the optimal timing

for each stage. MARR is 8%.

22) Find the present worth of the following cash flow diagram if i=18%. (Hint: The given series can be looked

at as a cumulative sum of various uniform and gradient series, positive and negative).

100 100

150 150

200 200

250 250

300 300

350

23) You deposit funds which represent an arithmetic gradient series. The first deposit is at the end of month 2

of amount G. The amount increases by G every month. You deposit till the end of 2 years. What is the

value of G if you want to accumulate $100,000 at the end of 2 years? The interest rate is 12% per year

compounded monthly?

24) The capital investment for a project is $100,000 now. This is to be repaid by a uniform annual series in 5

years. What is the annual repayment amount if the interest rate is 12% per year compounded monthly?

25) What is the effective annual interest rate for a nominal annual interest rate of 13% compounded

continuously?

26) A highway improvement is scheduled every 5 years in perpetuity. Each improvement costs $50,000. What

is the present worth of all the costs of the improvement project if the interest rate is 8% per year?

27) If the current year (beginning of year 1) user benefit from a road improvement is $5 million and the total

project cost is $80 million, what must be the minimum rate of growth of benefit in % per year, for the

project to be economically viable? Assume that the project life is 25 years and the benefit grows

continuously. The interest rate is 8% per year compounded continuously.


REFERENCES

1. AASHTO (1997) 2. Winfrey (1970)

3. World Bank, 1990

4. Bhandari and Sinha, 1985

5. Engineering Economy, Degarmo, E. P. et al

(Full detail of incomplete references will be provided to the students in due course.)

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