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GENG 360:EngineeringEconomics
Chapter 5. Present Worth
(PW) Analysis
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Contents
Formulating Alternatives PW of Equal-Life Alternatives PW of Different-Life alternatives Future Worth Analysis Capitalized Cost Analysis Payback Period
Life-Cycle Costs PW of Bonds Spreadsheet Applications
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5.1. Formulating Mutually Exclusive
Alternatives
Firms have the capability to generatepotential projects for investment
Two types of investment categories
Mutually Exclusive Set
Independent Project Set
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Mutually Exclusive set
It is a set where a candidate set ofalternatives exist (more than one)
Objective: Pick one and only one from theset.
Once selected, the remaining alternativesare excluded.
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Independent Project Set
Given a set of alternatives (more than one)
Objective is to:
Select the best possible combination of projects fromthe set that will optimize a given criteria.
Subjects to constraints
More difficult problem than the mutually exclusive
approach
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Mutually Exclusive Vs.Independent Alternatives
Mutually exclusive alternatives compete witheach other.
Independent alternatives may or may notcompete with each other
The independent project selection problem dealswith constraints and may require a mathematicalprogramming or bundling technique to evaluate
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Type of Alternatives
(Revenue based or service-based)Cash Flows determine whether the alternativesare revenue based or service based. Revenue/Cost the alternatives consist of cash
inflow and cash outflows
Select the alternative with the maximumeconomic value
Service the alternatives consist mainly of costelements
Select the alternative with the minimumeconomic value (min. cost alternative)
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Problem
DoNothing
Alt.1
Alt.2
Alt.m
Analysis
Selection
Execution
Selection depends upon data, life, discount rate, and assumptions made.
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5.2 Present Worth Approach with Equal-Lives
Simple: Transform all of the current and future
estimated cash flow back to a point in time (time t = 0)
Result is in equivalent dollars now!
P(i%) = P(+cash flows) +P(-cash flows)
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If P(%) is:
> 0 then the project is deemed acceptable.
< 0 the project is usually rejected= 0 (Present worth of costs = Present
worth of revenues) Indifferent!If P(%) = 0 that meansthe project earned exactlythe discount rate that wasused to discount the cash
flows!
The interest rate that causesa cash flows NPV to equal 0is called the Rate of Returnof the cash flow!
The net present worth is purely a function of theMARR (the discount rate one uses).
If one changes the discount rate, a different NPV willresult.
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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).
Acceptance or rejection of a project is afunction of the timing and magnitude of theproject's cash flows, and the choice of thediscount rate.
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PRESENT WORTH: Special Applications
Present Worth of Equal Lived Alternatives
Alternatives with unequal lives: Beware
Capitalized Cost Analysis
Both Calculations Require knowledge of the discount rate before
we conduct the analysis
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Present Worth of Equal Lived
Alternatives Straightforward: Compute the PW of each
alternative and select the best,
i.e., smallest if cost and largest if profit
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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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Solution: Cash Flow Diagrams:
0 1 2 3 4 5
$2,500A = $900
F5=$200
MA
0 1 2 3 4 5
$3,500
F5=$350
A = $700
MB
PA = 2,500 + 900 (P|A, 0.10, 5) 200 (P|F,0.10, 5)= 2,500 + 900 (3.7908) - 200 (0.6209)= 2,500 + 3,411.72 - 124.18 = $5,788
PB = 3,500 + 700 (P|A, 0.10, 5) 350 (P|F,0.10, 5)= 3,500 + 2,653.56 - 217.31 = $5,936
Which alternativeshould we select?
SELECTMACHINE A:Lower PW cost!
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Present Worth of Alternativeswith Different Lives
BASIC RULE: In an analysis one cannot effectivelycompare the PW of one alternative with a study period
different from another alternative that does not have the
same study period.
Comparison must be made over equal time periods
Compare over the Least Common Multiple (LCM) for their lives
Example:You have different projects over{3,4, and 6}yearsThe LCM life is 12 years (6x2) Evaluate all over 12 years for aPW analysis.
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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 apositive value to cost and negative to inflows
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LCM(6,9) = 18 year study period will apply for present worth
A commonmistake isto computethe presentworth of
the 6-yearproject andcompare it
to thepresentworth of
the 9-yearproject.
i = 15% per year
0 1 2 3 4 5 6
$11,000
F6=$1,000
A 1-6
=$3,500
Machine A
0 1 2 3 4 5 6 7 8 9
F6=$2,000
A 1-9
=$3,100
$18,000
Machine B
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i = 15% per year
Machine A
LCM(6,9) = 18 year study period will apply for present worth
Cycle 1 for A Cycle 2 for A Cycle 3 for A
6 years 6 years 6 years
Cycle 1 for B Cycle 2 for B
18 years
9 years 9 years
Machine B
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First, Calculate the PW of Machine A:
Machine A
Cycle 1 for A Cycle 2 for A Cycle 3 for A
6 years 6 years 6 years
PA6 = 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
0 1 2 3 4 5 6
$11,000
F6=$1,000
A 1-6 =$3,500
Machine A
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0 6 12 18
$23,813 $23,813 $23,813
Machine A
PA18=23,813 +23,813 (P|F, 0.15, 6) +23,813 (P|F, 0.15, 12)
=23,813 +10,294 +4,451 =$38,558
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Calculate the Present Worth of a 9-year cycle for B
0 1 2 3 4 5 6 7 8 9
F6=$2,000
A 1-9
=$3,100
$18,000
PB9 = 18,000+3,100(P|A, .15, 9) 2,000(P|F, .15, 9)= 18,000 + 3,100(4.7716) - 2,000(.2843)= $32, 223 which occurs at time 0 and 9
0 9 18
$32,508 $32,508
PB18 = 32,508 + 32,508 (P|F, .15, 9)= 32,508 + 32,508(.2843) = $41,750
Choose Machine A because it costs less
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11000
18 years
Alternative (A) cash flow:
3500 35003500
11000 11000
1000 10001000
Alternative (B) cash flow:
18000
3100 3100
18000
20002000
18 years
PWA = - 11,000 3,500 (P/A, 15%, 18) 10,000 (P/F, 15%, 6)-10,000 (P/F, 15%, 12) + 1,000 (P/F, 15%, 18)= - 38,559
PWB = - 18,000 3,100 (P/A, 15%, 18) 16,000 (P/F, 15%, 9)
+ 2,000 (P/F, 15%, 18) = -41,384
We Select (A)
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5.4. FUTURE WORTH APPROACH
Compound all cash flows forward in time to somespecified time period using (F/P), (F/A), factors or,
Given P, the F = P(1+i)N
Applications of FW Analysis: 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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FW Exercise See Example 5.3
Calculate the Future Worth of determiningthe selling price in order to earn exactly25% on the investment
Draw the cash-flow diagram!!
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5.5 Capitalized Cost
Capitalized Cost: the PW of a project that lastsforever, such as:Government Projects
Roads, Dams, Bridges (projects that possessperpetual life)
Infinite analysis period
Perpetual Investments: Investments that lastfor such a long time (n ~ )
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Derivation for Capitalized Cost:
Start with the closed form for the P/A factor:
Next, let N approach infinity and divide thenumerator and denominator by (1+i)N
(1 ) 1
(1 )
N
N
iP A
i i
1
1 (1 )NiP A
i
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Now, let n approach infinity and the right hand sidereduces to.
iAPn
1CostdCapitalize
Or,
CC(i%) = A/i
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Example:Calculate the Capitalized Cost of a project
that has an initial cost of $150,000. Theannual operating cost is $8,000 for thefirst 4 years and $4,000 thereafter. Thereis an recurring $15,000 maintenancecost each 15 years. Interest is 15% per
year.
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Cash Flow Diagram:
$4,000
0 1 2 3 4 5 6 7 15 30
$150,000
$8,000
$15,000 $15,000 $15,000 $15,000How much $$ at t = 0is required to fundthis project?
The capitalized cost is thetotal amount of $ at t = 0,when invested at theinterest rate, will provideannual interest that coversthe future needs of theproject.
i=15%/YR
Solution:
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1. Consider $4,000 of the $8,000 cost for thefirst four years to be a one-time cost, leaving
a $4,000 annual operating cost forever:
P0= 150,000 + 4,000 (P|A, .15, 4) = $161,420
Recurring annual cost is $4,000 plus theequivalent annual of the 15,000 end-of-cyclecost.
.0 15 30 45 60 ..
Take any 15-year period and find the equivalentannuity for that period using the F/A factor.
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Take any 15-year period and find the equivalentannuity for that period using the F/A factor
$15,000
A for a 15-year period
0 15 30 45 60 ...
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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) = $4,315/yr
Recurring costs = $4,315/i = 4,315/0.15
=$28,767
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Capitalized Cost = 161,420 + 4315/.15
= $190,187
Thus, if one invests $190,187 at time t = 0,then the interest at 15% will supply the end-of-year cash flow to fund the project so longas the principal sum is not reduced or theinterest rate changes (drops).
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Example: Find the capitalized cost for the following project (use i =15%).
Time $
Initial Cost At end of year 0 150,000
Additional Cost At end of year 10 50,000
Annual Cost From end of year 1 to end of year 4 5000
Annual Cost From end of year 5 to 8000
Operation Cost End of every13 years 15000
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Solution:
8 15 ( / ,15%,13)150 50 ( / ,15%,10) 5 8 ( / ,15%,4)
0.15 0.15
15 0.029118150 50 0.1229 3 5.8474
0.15 0.15
$194.847K
K K A FP K K P F K K P A
KKK K K
5000
150000
8000
5000015000 15000 15000
0 1 4 5 10 13 26 39
13 years
8 15150 50 ( / ,15%,10) 5 8 / ,15%,40.15
8 15150 50 0.1229 3 5.8474
0.15 5.1528
$194.847K
K KP K K P F K K P A
K KK
i
K K
OR:
i13 years =(1+0.15)13 1 =5.1528
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Example:
38Which one do you choose?
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Incremental Analysis:The value of the present worth can be
calculated by using either incrementalmethod or individual method.
Both methods should lead to the sameanswer.
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Example:
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5.6 Payback Period Analysis Payback period is the period of time it
takes for the cash flows to recover theinitial investment.
Two forms for this methodDiscounted Payback Period (uses an interest
rate)
Conventional Payback Period (does not use
an interest rate)
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Payback period is only a rough estimator of desirability
Use it as an initial screening method
Avoid using this method as a primary analysis techniquefor selection projects
Totally avoid the no-return payback period
The No-return method (example follows):
Does not employ the time value of money
Disregards all cash flows past the payback time period
If used, can lead to conflicting selections when compared tomore technically correct methods like present worth!
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Discounted Payback Approach: Find the value of np such that:
1
0 ( / , , )pt n
tt
P NCF P F i t
NCFt = Net CashFlow at time t
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Example 5.8 (Discounted Analysis)Machine A: N=7 i = 15%
Payback
isbetween6 and 7yeas(6.57yrs)
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Non-Discounted Analysis
At a 0interest ratethe PB timeis seen to
equal 4years!
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Machine B (Discounted):Machine B: N=14 i = 15%
Payback for B isbetween 9 and
10 years!Longer time
period torecover theinvestment.
9.52 years
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Machine B (Undiscounted):
Payback for B at0% is 6 years!
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Example Summary
Discounted
Machine A: 6.57 years
Machine B: 9.52 years
Undiscounted
Machine A: 4.0 years
Machine B: 6.0 years Go with Machine A lower time period payback
to recover the original investment
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5.8 Present Worth of Bonds
Bonds represent a source of funds for thefirm.
Bonds are sold (floated) by investmentbanks for firms in order to raise additionaldebt capital
A bond is similar to an IOU
Bonds are evidence of Debt
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Bond Types: Treasury bonds
Issued by Federal Government
Full backing of the Government
Conservative-type investment
State and Municipal Bonds
Issued by states and local governments
Generally tax-exempt by the Federal Government
Used to finance state and local projects
Backed by future tax and user fees to pay the interestand face value
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Mortgage Bonds Issued by Corporations
Secured by the firms assets Money received by the firm is used to fund projects
Referred to a Debt Capital
Buyers of these bonds are not owners they are lenders to thefirm
Debenture Bond Issued by Corporations
Not backed by specific assets
Backing good faith of the firm
Pays higher interest rates
Higher risks involved Bond interest rate may float
Could be convertible to common stock
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PW of Bonds - Overview
The Firm
Investment Bankers
Proceeds from
The sale
Sell the Bonds toThe lending public
Bondholders
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Basics of Bonds: The bond itself is just a piece of paper Bonds are negotiable instruments
Can be traded by the current bondholder
Source of funds to the firm (Debt capital)
Bondholders are loaning $$ to the firm, they Earn periodicinterest
They can sell the bonds at any time
The bondholders are lenders not owners
The firm pays periodic interest payments to the currentbond holders
At the end of the bonds life, the bonds are redeemed(bought back) from the current bond holder
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Bonds Notations: P0:The time (t = 0) selling price of the bond the cost to
the buyer of the bond
V:The face value of the bond
The value printed on the bond
Face values are usually: $100, $1,000, $5,000, $10,000 increments
N:The life of the bond in years (Defining the Maturity Date)
r: The nominal annual bond interest rate
Given the nominal annual bond interest rate, the payment
frequency of the interest (monthly, quarterly semi-annually, etc.) is
also stated(Facevalue)(Bondinerestrate)
Numberof paymentperiodperyearI I: Amount of interest
paid per period
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Example: Calculate the Bond Semiannual interest if:
V = $5,000 (face value)
r = 4.5% per year paid semiannually
N = 10 years
Solution:
0.045$5,000( ) $5,000(0.0225)
2$112.50 every 6 months
I
I
The bondholder,buys the bondand will receive
$112.50 every 6months for thelife of the bond
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Example 5.11
Given V = $5,000 (Face value of the bond)
r = 4.5% paid semiannually
N = 10 years or 20 interest periods
$I per 6 months = $5,000(0.045/2) = $112.50 paid tothe current bondholder
The buyer will consider buying this bond if he canearn a nominal rate of return of 8%/yr c.q.
(compounded quarterly)
Calculate the acceptable purchase price
Bonds are bought and sold in a bond market. Thus the priceof the bond is subject to the pressures of the bond market.
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Solution: What is fixed?
The future interest payments are fixed
The future face value of the bond is fixed
What can vary?
The purchase price such that the buyer can earn at least the 8%/yr c.q.
8% c.q is the same as
0.08/4 =0.02 =2% per quarter.
Bond interest flows every 6 months
Need an effective 6-month rate
The effective 6-month rate is then
(1.02)2 1 =0.0404 =4.04%/6 months
This is the potential buyers required interest rate
The objective is to determine the purchase price of this bonddiscounted at the buyers required rate of 4.04% per 6 months
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Draw the cash-flow diagram
Work the problem with N = 20 (not 10): We have 20 interest payments (every 6 months for 10 years)
A = 112.50/6 months
0 1 2 3 4 . .. 19 20
P=??
$5,000
i=4.04%/6 months
Find the PW(4.04%) of the future cashflows to the potential bond buyer
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P = $112.50(P/A,4.04%,20)+
$5,000(P/F,4.04%,20)= $3,788
IF the buyer can buy this bond for $3,788 orless, he/she will earn at least the 8% c.q. rate.
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Example:
A 15-year bond with a face value of$10,000 is offered for sale.
The rate of interest on the bond is 6%,paid semi-annually.
Assume that MARR = 8% nominal
compounded semi-annually, what would be the selling price of the
bond?
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Solution:( )( ) (10000)(0.06) $300
2Facevalue Bond inerest rateI
Numberof payment period per year
I =$300
Price of bond =?
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Face value =$ 10,000
P =300(P/A, 4%, 30) +10,000(P/F, 4%, 30) =$8,270.6