By Assoc. Prof. Dr. Ahmet ÖZTAŞ Gaziantep University Department of Civil Engineering CHP IV-PRESENT WORTH ANALYSIS CE 533 - ECONOMIC DECISION ANALYSIS

ByAssoc. Prof. Dr. Ahmet ÖZTAŞ

Gaziantep UniversityDepartment of Civil Engineering

CHP IV-PRESENT WORTH ANALYSIS

CE 533 - ECONOMIC DECISION ANALYSIS IN CONSTRUCTION

2 / 68CE533 - PW Analysis

TOPICS

Formulating AlternativesPW of Equal-Life AlternativesPW of Different-Life alternativesFuture Worth AnalysisCapitalized Cost AnalysisIndependent projectsPayback Period

CHP IV-PRESENT WORTH ANALYSIS


4.1 FORMULATING MUTUALLY EXCLUSIVE ALTERNATIVES

Viable firms/organizations have the capability to generate potential beneficial projects for potential investmentTwo types of investment categories Mutually Exclusive Set Independent Project Set



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.


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



Mutually exclusive alternatives compete with each other.Independent alternatives may or may not compete with each otherThe independent project selection problem deals with constraints and may require a mathematical programming or bundling technique to evaluate.


4.1 Type of Alternatives

Alternative’s CF are classified as revenue-based or cost-basedRevenue/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)


4.1 Evaluating Alternatives

Part of Engineering Economy is the selection and execution of the best alternative from among a set of feasible alternativesAlternatives must be generated from within the organization One of the roles of engineers!


4.1 Evaluating Alternatives

In part, the role of the engineer to properly evaluate alternatives from a technical and economic viewMust generate a set of feasible alternatives to solve a specific problem/concern


4.1 Alternatives

Problem

DoNothing

Alt.1

Alt.2

Alt.m

Analysis

Selection

Execution

If there are m investment proposals, we can form up to 2m mutually exclusive alternatives.This includes DN option.


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.


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!


4.2 THE PRESENT WORTH METHOD

At an interest rate usually equal to or greater than the Organization’s established MARR.

A process of obtaining the equivalent worth of future cash flows to some point in time

– called the Present Worth



P(i%) = P(+) – P(-).

P(i%) = P( + cash flows) + P( - cash flows)

OR, . . .



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!



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!



A positive present worth is a dollar amount of "profit" over the minimum amount required by the investors (owners).

For P(i%) > 0, the following holds true:


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.



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.


4.2 PRESENT WORTH: Special Applications

Present Worth of Equal Lived AlternativesAlternatives with unequal lives: BewareCapitalized Cost AnalysisRequire knowledge of the discount rate before we conduct the analysis


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.


4.2 Equal Lives – Straightforward!

Given two or more alternatives with equal lives….

Alt. 1

Alt. 2

Alt. N

N = for all alternative

s

Find PW(i%) for each alternative then compare


4.2 PRESENT WORTH: Example

Consider: Machine AMachine B

First Cost $2,500 $3,500Annual Operating Cost 900

700Salvage Value 200 350Life 5 years 5 years

i = 10% per year

Which alternative should we select?


4.2 PRESENT WORTH: Cash Flow Diagram

Which alternative should we select?

0 1 2 3 4 5

$2,500

A = $900

F5=$200MA

0 1 2 3 4 5

$3,500

F5=$350

A = $700

MB


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 = $5,936SELECT MACHINE A: Lower PW cost!


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.


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.


4.2 Bonds – Notation and Example

(Bond face value)(bond coupon rate) V.bI = ------------------------------------------ = ---- 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


4.2 PW Bonds – Example – Continued

The interest the firm would pay to the current bondholder is calculated as:

0.045$5,000( ) $5,000(0.0225)

2$112.50 every 6 months

I

I

The bondholder, buys the bond and will receive $112.50 every 6 months for the life of the bond


4.2 Example 4.2

Ayşe has some extra money, requires safe investment. Her employer is offering to employees a generous 5% discount for 10-year 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?


4.2 Example 4.2 – Cash-Flow Diagram

A = 150

0 1 2 3 4 …. ….. 19 20

P=??

$5,000

i=3.35%

Find the PW(3.35%) of the future cash flows to the potential bond buyer


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 <0Effective rate is slightly less than 6.81% per year since PW<0.She sould not buy.


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!


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)


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.


4.3 PRESENT WORTH: Example Unequal Lives

EXAMPLEMachine A

Machine BFirst Cost $11,000 $18,000Annual Operating Cost 3,500

3,100Salvage Value 1,000 2,000Life 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


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!


4.3 PRESENT WORTH: Unequal Lives

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

LCM(6,9) = 18 year study period will apply for present worth


4.3 Unequal Lives: 2 Alternatives

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 yearsMachine B


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


4.3 Example: Unequal Lives

PA= 23,813+23,813 (P|F, .15, 6)+

23,813 (P|F, .15, 12) = 23,813 + 10,294 + 4,451 = 38,558

0 6 12 18

$23,813 $23,813 $23,813

Machine A


4.3 Unequal Lives Example: Machine B

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


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


4.3 Alternative B – 2 Cycles

PB = 32,508 + 32,508 (P|F, .15, 9)

= 32,508 + 32,508(.2843)

PB = $41,750

Choose Machine A

0 9 18

$32,508 $32,508

Machine A: PW =$38,558


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

Alt-2: N= 7 yrs

LCM = 35 yrs

Could assume a study period of, say, 5 years.


4.3 Unequal Lives – Assumed Study Period

Assume a 5-yr. Study periodEstimate a salvage value for the 7-year project at the end of t = 5Truncate the 7-yr project to 5 years

Alt-1: N = 5 yrs

Alt-2: N= 7 yrs

Now, evaluate both over 5 years using the PW method!


FUTURE WORTH APPROACH

FW(i%) is an extension of the present worth methodCompound 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


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


Life Cycle Costs (LCC)

Extension of the Present Worth methodUsed for projects over their entire life span where cost estimates are employedUsed for: Buildings (new construction or

purchase) New Product Lines Commercial aircraft New automobile models Defense systems


Life Cycle: Two General Phases

TIME

Cost-$

Acquisition Phase

Operation Phase

Cumulative Life Cycle Costs


4.4 CAPITALIZED COST ANALYSIS

CAPITALIZED COST- the present worth of a project that lasts forever.Government ProjectsRoads, Dams, Bridges (projects that possess perpetual life)Infinite analysis period


4.4 Derivation for Capitalized Cost

Start with the closed form for the P/A factor

Next, let N approach infinity and divide the numerator and denominator by (1+i)N

(1 ) 1

(1 )

N

N

iP A

i i


4.4 Derivation - Continued

Dividing by (1+i)N yields

Now, let n approach infinity and the right hand side reduces to….

11

(1 )NiP A

i


4.4 Derivation - Continued

1 AP A

i i

Or,

CC(i%) = A/i


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.

Find the PW of an infinite annuity flow

1 2 3 4 5 ..

N=inf.

A=$50/yr

P = ?

…………………..



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.


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.


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


EXAMPLECalculate 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.

4.4 Capitalized Cost Example


$4,000

0 1 2 3 4 5 6 7 15 30 ………

$150,000

$8,000 $15,000 $15,000 $15,000 $15,000

“i”=15%/YR

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.

4.4 Cash Flow Diagram


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.P0= 150,000 + 4,000 (P|A, .15, 4) =

$161,420

2.855


4.4 CAPITALIZED COST - Continued

Recurring annual cost is $4,000 plus the equivalent annual of the 15,000 end-of-cycle cost.

…….0 15 30 45 60 ……..

Take any 15-year period and find the equivalent annuity for that period using the F/A factor.


4.4 CAPITALIZED COST: One Cycle

Take any 15-year period and find the equivalent annuity for that period using the F/A factor

$15,000

A for a 15-year period

0 15 30 45 60 ……..

…….



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



Capitalized Cost = 161,420 + 5315/.15

= $196,853Thus, 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).

Another ExampleA 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?



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 functionPW = P + NPV(i%,year_1_CF_cell:last_year_CF_cell)


Summary: Present Worth

• PW represents a family of methods

• Annual worth

• Future Worth

• Capitalized Cost

• Life-cycle cost analysis – application

• Bond Problems – application


End of Chapter 4

Documents

By Assoc. Prof. Dr. Ahmet ÖZTAŞ Gaziantep University Department of Civil Engineering CHP IV-PRESENT WORTH ANALYSIS CE 533 - ECONOMIC DECISION ANALYSIS