CE 533 - ECONOMIC DECISION
ANALYSIS IN CONSTRUCTION
CHP IV-PRESENT WORTH ANALYSIS
By
Assoc. Prof. Dr. Ahmet ÖZTAŞ
Gaziantep University
Department of Civil Engineering
CHP IV-PRESENT WORTH ANALYSIS
TOPICS
Formulating Alternatives
PW of Equal-Life Alternatives
PW of Different-Life alternatives
Future Worth Analysis
Capitalized Cost Analysis
Independent projects
Payback Period
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4.1 FORMULATING MUTUALLY EXCLUSIVE
ALTERNATIVES
Viable firms/organizations have the
capability to generate potential
beneficial projects for potential
investment
Two types of investment categories


Mutually Exclusive Set
Independent Project Set
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4.1 FORMULATING MUTUALLY EXCLUSIVE
ALTERNATIVES
Mutually Exclusive set is where a
candidate set of alternatives exist
(more than one)
Objective: Pick one and only one from
the set.
Once selected, the remaining
alternatives are excluded.
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4.1 INDEPENDENT PROJECT SET
Given a set of alternatives (more than
one)
The objective is to:



Select the best possible combination of
projects from the set that will optimize a
given criteria.
Subjects to constraints
More difficult problem than the mutually
exclusive approach
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4.1 FORMULATING MUTUALLY EXCLUSIVE
ALTERNATIVES
Mutually exclusive alternatives compete
with each other.
Independent alternatives may or may
not compete with each other
The independent project selection
problem deals with constraints and may
require a mathematical programming or
bundling technique to evaluate.
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4.1 Type of Alternatives
Alternative’s CF are classified as
revenue-based or cost-based
Revenue/Cost – the alternatives consist
of cash inflow and cash outflows

Select the alternative with the maximum
economic value
Service – the alternatives consist mainly
of cost elements

Select the alternative with the minimum
economic value (min. cost alternative)
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4.1 Evaluating Alternatives
Part of Engineering Economy is the
selection and execution of the best
alternative from among a set of
feasible alternatives
Alternatives must be generated
from within the organization

One of the roles of engineers!
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4.1 Evaluating Alternatives
In part, the role of the engineer to
properly evaluate alternatives from
a technical and economic view
Must generate a set of feasible
alternatives to solve a specific
problem/concern
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4.1 Alternatives
Do
Nothing
Analysis
Alt.
1
Problem
Alt.
2
Alt.
m
If there are m investment proposals,
we can form up to 2m mutuallyCE533
exclusive
alternatives.
- PW Analysis
This includes DN option.
Selection
Execution
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4.1 Alternatives: The Selected Alternative
Problem
Alt.
Selected
Execution
Audit and Track
Selection is dependent upon the data, life,
discount rate, and assumptions made.
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4.2 Present Worth Approach Equal-Lifes
Simple – Transform all of the current
and future estimated cash flow back to
a point in time (time t = 0)
Have to have a discount rate before the
analysis in started
Result is in equivalent dollars now!
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4.2 THE PRESENT WORTH METHOD
A process of obtaining the
equivalent worth of future cash
flows to some point in time
– called the Present Worth
At an interest rate usually equal to or
greater than the Organization’s
established MARR.
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4.2 THE PRESENT WORTH METHOD
P(i%) = P( + cash flows) +
P( - cash flows)
OR, . . .
P(i%) = P(+) – P(-).
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4.2 THE PRESENT WORTH METHOD
If P(i%) > 0 then the project is
deemed acceptable.
If P(i%) < 0
the project is usually
rejected.
If P(i%) = 0 Present worth of costs = Present
worth of revenues –
Indifferent!
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4.2 THE PRESENT WORTH METHOD
If the present worth of a project turns out to =
“0,” that means the project earned exactly the
discount rate that was used to discount the cash
flows!
The interest rate that causes a cash flow’s NPV
to equal “0” is called the Rate of Return of the
cash flow!
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4.2 THE PRESENT WORTH METHOD
For P(i%) > 0, the following holds true:
A positive present worth is a dollar
amount of "profit" over the minimum
amount required by the investors
(owners).
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4.2 THE PRESENT WORTH METHOD –
Depends upon the Discount Rate Used
The present worth is purely a
function of the MARR (the
discount rate one uses).
If one changes the discount rate, a
different present worth will result.
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4.2 THE PRESENT WORTH METHOD
For P(i%) > 0, the following
holds true:
Acceptance or rejection of a
project is a function of the timing
and magnitude of the project's
cash flows, and the choice of the
discount rate.
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4.2 PRESENT WORTH: Special
Applications
Present Worth of Equal Lived
Alternatives
Alternatives with unequal lives: Beware
Capitalized Cost Analysis
Require knowledge of the discount rate
before we conduct the analysis
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4.2 PRESENT WORTH: Equal Lives
Present Worth of Equal Lived
Alternatives – straightforward
Compute the Present Worth of each
alternative and select the best, i.e.,
smallest if cost and largest if profit.
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4.2 Equal Lives – Straightforward!
Given two or more alternatives with
equal lives….
Alt. 1
Alt. 2
N = for all
alternatives
Alt. N
Find PW(i%) for each alternative then compare
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4.2 PRESENT WORTH: Example
Consider:
Machine A Machine B
First Cost
$2,500
$3,500
Annual Operating Cost
900
700
Salvage Value
200
350
Life
5 years
5 years
i = 10% per year
Which alternative should we select?
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4.2 PRESENT WORTH: Cash Flow Diagram
F5=$200
MA
0
1
2
3
4
A = $900
$2,500
F5=$350
MB
0
$3,500
5
1
2
3
4
5
A = $700
Which alternative should we select?
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4.2 PRESENT WORTH: Solving
PA = 2,500 + 900 (P|A, .10, 5) –
200 (P|F, .01, 5)
= 2,500 + 900 (3.7908) - 200 (.6209)
= 2,500 + 3,411.72 - 124.18 = $5,788
PB = 3,500 + 700 (P|A, .10, 5) –
350 (P|F, .10, 5)
= 3,500 + 2,653.56 - 217.31 = $5,936
SELECT MACHINE A: Lower PW cost!
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4.2 Present Worth of Bonds
Often corporations or government obtain
investment capital for projects by selling
bonds.
A good application of PW method is the
evaluation of a bond purchase alternative.
If PW < 0 at MARR, do-nothing alternative
is selected.
A bond is similar to an IOU for time periods
such as 5, 10, 20 or more years.
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4.2 Present Worth of Bonds
Each bond has a face value V of $100,
$1000, $5000 or more that is fully
returned to the purchaser when the
bond maturity is reached.
Additionaly, bond provide the
purchaser with periodic interest
payments I (bond dividends) using the
bond coupon (interest) b, and c, the
number of payment periods per year.
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4.2 Bonds – Notation and Example
(Bond face value)(bond coupon rate) V.b
I = ------------------------------------------ = ---number of payments per year
c
At the time of purchase, the bond may sell for
more or less than face value.
Example:



V = $5,000 (face value)
b = 4.5% per year paid semiannually
c = 10 years
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4.2 PW Bonds – Example – Continued
The interest the firm would pay to the
current bondholder is calculated as:
0.045
I  $5, 000(
)  $5, 000(0.0225)
2
I  $112.50 every 6 months
The bondholder, buys the bond and will receive
$112.50 every 6 months for the life of the bond
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4.2 Example 4.2
Ayşe has some extra money, requires safe
investment. Her employer is offering to
employees a generous 5% discount for 10year 5000 YTL bonds tat carry a coupon rate of
6% paid semiannually.
The expectation is to match her return on
other safe investments, which have averaged
6.7% per year compounded semiannually.
(This is an effective rate of 6.81% per year).
Should Ayşe buy the bond?
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4.2 Example 4.2 – Cash-Flow Diagram
$5,000
i=3.35%
A = 150
0
P=??
1
2
3
4
….
…..
19
20
Find the PW(3.35%) of the future cash
flows to the potential bond buyer
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4.2 Example 4.2 – Solving
I = (5000)(0.06)/2 = 150 YTL every 6
months for a total of n=20 dividend
payments.
The semiannual MARR is 6.7/2 =
3.35%, and the purchase price now is –
5000(0.95)= -4750 YTL.
Using PW evaluation:
PW = -4750 + 150(P/A, 3.35%, 20) +
5000(P/F, 3.35%, 20) = - 2.13 YTL <0
Effective rate is slightly less than 6.81%
per year since PW<0.
She sould not buy.
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4.3 Present Worth Analysis of Different-Life
Alternatives
In an analysis one cannot effectively
compare the PW of one alternative with
a study period different from another
alternative that does not have the same
study period.
This is a basic rule!
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4.3 PRESENT WORTH: Unequal Lives
If the alternatives have different lifes,
there are 2 ways to use PW analysis to
compare alternatives:
A) The lowest common Multiple (LCM)
B) Study period (planning horizon)
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4.3 PRESENT WORTH: Lowest Common
Multiple (LCM) of Lives
If the alternatives have different study
periods, you find the lowest common
life for all of the alternatives in
question.
Example: {3,4, and 6} years. The
lowest common life (LCM) is 12 years.
Evaluate all over 12 years for a PW
analysis.
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4.3 PRESENT WORTH: Example Unequal
Lives
EXAMPLE
Machine A Machine B
First Cost
$11,000
$18,000
Annual Operating Cost
3,500
3,100
Salvage Value
1,000
2,000
Life
6 years
9 years
i = 15% per year
Note: Where costs dominate a problem it is customary to assign a
positive value to cost and negative to inflows
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4.3 PRESENT WORTH: Example Unequal
Lives
A common mistake is to
compute the present
worth of the 6-year
project and compare it to
the present worth of the
9-year project.
NO! NO! NO!
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4.3 PRESENT WORTH: Unequal Lives
Machine A
0
1
2
F6=$1,000
3
4
5
6
A 1-6
0
1
F6=$2,000
=$3,500
$11,000
2
Machine B
3
4
A
5
6
7
8
9
1-9
=$3,100
i = 15% per year
$18,000
LCM(6,9) = 18 year study period will apply for present worth
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4.3 Unequal Lives: 2 Alternatives
Machine A
6 years
6 years
Cycle 1 for A
Cycle 2 for A
6 years
Cycle 3 for A
Machine B
9 years
9 years
Cycle 1 for B
Cycle 2 for B
18 years
i = 15% per year
LCM(6,9) = 18 year study period will apply for present worth
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4.3 Example: Unequal Lives Solving
LCM = 18 years
Calculate the present worth of a 6-year cycle
for A
PA = 11,000 + 3,500 (P|A, .15, 6) –
1,000 (P|F, .15, 6)
= 11,000 + 3,500 (3.7845) – 1,000 (.4323)
= $23,813, which occurs at time 0, 6 and 12
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4.3 Example: Unequal Lives
Machine A
0
6
$23,813
12
$23,813
18
$23,813
PA= 23,813+23,813 (P|F, .15, 6)+
23,813 (P|F, .15, 12)
= 23,813 + 10,294 + 4,451 = 38,558
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4.3 Unequal Lives Example: Machine B
Calculate the Present Worth of a 9-year
cycle for B
F6=$2,000
0
1
2
3
4
5
6
7
8
9
A 1-9
=$3,100
$18,000
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4.3 9-Year Cycle for B
Calculate the Present Worth of a 9-year cycle
for B
PB = 18,000+3,100(P|A, .15, 9) –
1,000(P|F, .15, 9)
= 18,000 + 3,100(4.7716) - 1,000(.2843)
= $32,508 which occurs at time 0 and 9
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4.3 Alternative B – 2 Cycles
Machine A: PW =$38,558
0
9
$32,508
18
$32,508
PB = 32,508 + 32,508 (P|F, .15, 9)
= 32,508 + 32,508(.2843)
PB = $41,750
Choose Machine A
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4.3 Unequal Lives – Assumed Study
Period
Study Period Approach


Assume alternative: 1 with a 5-year life
Alternative: 2 with a 7-year life
Alt-1: N = 5 yrs
LCM = 35 yrs
Alt-2: N= 7 yrs
Could assume a study period of, say, 5 years.
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4.3 Unequal Lives – Assumed Study
Period
Assume a 5-yr. Study period
Estimate a salvage value for the 7-year
project at the end of t = 5
Truncate the 7-yr project to 5 years
Alt-1: N = 5 yrs
Now, evaluate both
over 5 years using
the PW method!
Alt-2: N= 7 yrs
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FUTURE WORTH APPROACH
FW(i%) is an extension of the present
worth method
Compound all cash flows forward in
time to some specified time period
using (F/P), (F/A),… factors or,
Given P, the F = P(1+i)N
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Applications of Future Worth
Projects that do not come on line until
the end of the investment period




Commercial Buildings
Marine Vessels
Power Generation Facilities
Public Works Projects
Key – long time periods involving
construction activities
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Life Cycle Costs (LCC)
Extension of the Present Worth method
Used for projects over their entire life
span where cost estimates are
employed
Used for:





Buildings (new construction or purchase)
New Product Lines
Commercial aircraft
New automobile models
Defense systems
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Life Cycle: Two General Phases
Cost-$
Cumulative Life Cycle
Costs
Acquisition Phase
Operation Phase
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TIME
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4.4 CAPITALIZED COST ANALYSIS
CAPITALIZED COST- the present worth
of a project that lasts forever.
Government Projects
Roads, Dams, Bridges (projects that
possess perpetual life)
Infinite analysis period
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4.4 Derivation for Capitalized Cost
Start with the closed form for the P/A factor
 (1  i)  1 
P  A
N 
 i(1  i) 
N
Next, let N approach infinity and divide the
numerator and denominator by (1+i)N
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4.4 Derivation - Continued
Dividing by (1+i)N yields
1 

1

 (1  i ) N 
P  A

i




Now, let n approach infinity and the
right hand side reduces to….
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4.4 Derivation - Continued
1  A
P  A  
i  i
Or,
CC(i%) = A/i
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4.4 CAPITALIZED COST
Assume you are called on to maintain a
cemetery site forever if the interest rate
= 4% and $50/year is required to
maintain the site.
1
2
3
4
5 ..
…………………..
N=inf.
A=$50/yr
P=?
Find the PW of an infinite annuity flow
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4.4 CAPITALIZED COST
P0 = $50[1/0.04]
P0 = $50[25] = $1,250.00
Invest $1,250 into an account that
earns 4% per year will yield $50 of
interest forever if the fund is not
touched and the i-rate stays constant.
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4.4 CAPITALIZED COST: Endowments
Assume a wealthy donor wants to endow a
chair in an engineering department.
The fund should supply the department with
$200,000 per year for a deserving faculty
member.
How much will the donor have to come up
with to fund this chair if the interest rate =
8%/yr.
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4.4 CAPITALIZED COST: Endowed Chair
The department needs $200,000 per year.
P = $200,000/0.08 = $2,500,000
If $2,500,000 is invested at 8% then the
interest per year = $200,000
The $200,000 is transferred to the department,
but the principal sum stays in the investment to
continue to generate the required $200,000
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4.4 Capitalized Cost Example
EXAMPLE
Calculate the Capitalized Cost of a
project that has an initial cost of
$150,000. The annual operating cost is
$8,000 for the first 4 years and $5000
thereafter. There is an recurring
$15,000 maintenance cost each 15
years. Interest is 15% per year.
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4.4 Cash Flow Diagram
“i”=15%/YR
0
1
2
3
4
5
6
7
15
30
………
$4,000
$8,000
$150,000
$15,000
$15,000
$15,000
$15,000
N=
How much $$ at t = 0 is required to fund this
project?
The capitalized cost is the total amount of $ at t
= 0, when invested at the interest rate, will
provide annual interest that covers the future
needs of the project.
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4.4 CAPITALIZED COST - Example
Continued
1. Consider $4,000 of the $8,000 cost
for the first four years to be a one-time
cost, leaving a $4,000 annual operating
cost forever.
2.855
P0= 150,000 + 4,000 (P|A,
.15, 4) =
$161,420
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4.4 CAPITALIZED COST - Continued
Recurring annual cost is $4,000 plus the
equivalent annual of the 15,000 end-ofcycle cost.
0
15
30
45
…….
60 ……..
Take any 15-year period and find the equivalent
annuity for that period using the F/A factor.
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4.4 CAPITALIZED COST: One Cycle
Take any 15-year period and find the equivalent
annuity for that period using the F/A factor
0
15
30
45
…….
60 ……..
$15,000
A for a 15-year period
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4.4 CAPITALIZED COST
2. Recurring annual cost is $4,000 plus
the equivalent annual of the 15,000
end-of-cycle cost.
A= 4,000 + 15,000 (A|F, .15, 15)
= 4,000 + 15000 (.0210) = $5,315
Recurring costs = $5,315/i =
5,315/0.15 =$35,443/yr
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4.4 CAPITALIZED COST
Capitalized Cost = 161,420 + 5315/.15
= $196,853
Thus, if one invests $196,853 at time t
= 0, then the interest at 15% will
supply the end-of-year cash flow to
fund the project so long as the principal
sum is not reduced or the interest rate
changes (drops).
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Another Example
A wealthy businessman wants to start a
permanent fund for supporting research
directed toward sustainability. The donor
plans to give equal amounts of money for
each of the next 5 years, plus one now(i.e six
donations) so that $100,000 per year can be
withdrawn each year forever, beginning in
year 6. If the fund earns interest at a rate of
8% per year, how much money must be
donated each time?
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4.6 USING EXCEL FOR PW ANALYSIS
General format to determine the PW is;
PW = P – PV (i%,n,A,F)
When different-life alternatives are
evaluated using LCM; develop NPV function
PW = P +
NPV(i%,year_1_CF_cell:last_year_CF_cell)
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Summary: Present Worth
• PW represents a family of methods
• Annual worth
• Future Worth
• Capitalized Cost
• Life-cycle cost analysis – application
• Bond Problems – application
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End of Chapter 4
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