engineering economy chapter 5

1Blank & Tarquin: 5th Edition. Ch. 5 Authored By: Dr. Don Smith, Texas A&M University.

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

CHAPTER V

PRESENT WORTH ANALYSIS

Mc

GrawHill

ENGINEERING ECONOMY Fifth Edition

Blank and Tarquin

2


Blank & Tarquin: 5th Edition. Ch. 5 Authored By: Dr. Don Smith, Texas A&M University.

LEARNING OBJECTIVES

PURPOSE OF THIS CHAPTER FORMULATION OF MUTUALLY

EXCLUSIVE ALTERNITIVES PROPER COMPARISON/ANALYSIS OF

MUTUALLY EXCLUSIVE ALTERNATIVES PRESENT WORTH METHOD EXTENSIONS OF THE PRESENT

WORTH METHOD

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CHAPTER TOPICS

Formulating AlternativesPW of Equal-Life AlternativesPW of Different-Life alternativesFuture Worth AnalysisCapitalized Cost AnalysisPayback PeriodLife-Cycle CostsPW of BondsSpreadsheet Applications

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5.1 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

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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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5.1 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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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.

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5.1 Type of Alternatives

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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5.1 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!

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5.1 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

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5.1 Alternatives

Problem

DoNothing

Alt.1

Alt.2

Alt.m

Analysis

Selection

Execution

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5.1 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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5.2 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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5.2 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

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P(i%) = P(+) – P(-).

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

OR, . . .

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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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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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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:

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5.2 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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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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5.2 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

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5.2 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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5.2 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

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5.2 PRESENT WORTH: Example

Consider: Machine A Machine BFirst 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?

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5.2 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=$200

A = $700

MB

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5.2 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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5.3 PRESENT WORTH: Different Lives

Present Worth of Alternatives with Different Lives

Comparison must be made over equal time periods

Compare over the least common multiple, LCM, for their lives

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5.3 PRESENT WORTH: Unequal Lives

Present Worth of Alternatives with Different LivesRemember – if the lives of the alternatives are not equal, one must create or force a study period where the life is the same for all of the alternatives

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5.3 Present Worth with Unequal Lives: The Rule

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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5.3 PRESENT WORTH: Lowest Common Multiple 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 is 12 years.Evaluate all over 12 years for a PW analysis.

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5.3 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

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5.3 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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5.3 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

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5.3 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

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5.3 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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5.3 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

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5.3 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

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5.3 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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5.3 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

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5.3 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.

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5.3 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!

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5.4 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

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5.4 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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5.4 Future Worth Example (Figure 5.3)

See Example 5.3Calculate the Future Worth of determining the selling price in order to earn exactly 25% on the investmentDraw the cash-flow diagram!!

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5.5 CAPITALIZED COST

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

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5.5 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

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5.5 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

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5.5 Derivation - Continued

1 AP A

i i

Or,

CC(i%) = A/i

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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 = ?

…………………..

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1 2 3 4 5 ..

N=inf.

A=$50/yr

P = ? Find the PW of an infinite annuity flow

…………………..

P0 = A[P/A,i%,N]

(1 ) 1, let N

(1 )

(1 ) 1 1lim

(1 )

N

N

N

N N

iP A

i i

i

i i i

P0=A(1/i)

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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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5.5 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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5.5 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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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.

5.5 Capitalized Cost Example

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$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.

5.5 Cash Flow Diagram

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5.5 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

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5.5 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.

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5.5 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 ……..

…….

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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) = $5,315

Recurring costs = $5,315/i = 5,315/0.15 =$3,443/yr

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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).

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5.6 Payback Period Analysis

Two forms for this method Discounted Payback Period (uses an

interest rate) Conventional Payback Period (does not

use an interest rate)

Payback is the period of time it takes for the cash flows to recover the initial investment.

62



5.6 Payback Period Analysis

Discounted Payback Approach

Find the value of np such that:

1

0 ( / , , )pt n

tt

P NCF P F i t

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5.6 Payback Period Analysis - Example

Example 5.8

Machine 1: N=7

Machine 2: N=14

i = 15%

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5.6 Payback Period Analysis - Example 5.8

Tabular Format: Machine 1

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5.6 Payback Period Analysis- Machine A

Payback is

between 6 and 7

yeas(6.57 yrs)

PW(15%)= +$481.00

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5.6 Non-Discounted Analysis – Machine A

At a “0” interest rate the PB time is seen to equal 4 years!

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5.6 Payback for Machine B at 15%, N = 14 yrs

Payback for B is between 9 and 10 years!Longer time

period to recover the investment.

9.52 years

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5.6 Payback at “0”% for Machine B

Payback for B at 0% is 6

years!

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5.6 Payback for Example 5.8

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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5.6 Payback Method Summarized

Payback is only a rough estimator of desirabilityUse as an initial screening methodAvoid using this method as a primary analysis technique for selection projectsTotally avoid the no-return payback period

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5.6 Payback Method Summarized

The “No-return” method 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 to more technically correct methods like present worth!

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5.7 Life Cycle Costs (LCC)

Extension of the Present Worth methodUsed for projects over their entire life span where cost estimates are employedUsed for: Military/Defense Projects New Product Lines Large construction projects

73



5.7 Life Cycle Defined – Detailed Phases

Needs Assessment PhaseConceptual Design PhaseDetailed Design PhaseProduction/Construction PhaseOperation – (upgrading to extend) PhaseRetirement/Disposal Phase

The life can be for years into the future

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5.7 Life Cycle: Two General Phases

TIME

Cost-$

Acquisition Phase

Operation Phase

Cumulative Life Cycle Costs

75



5.7 Life-Cycle Costs: Impact of Design Changes

Cost of a design change tends to multiply by 10 with each phaseAny design changes that might occur late in the life cycle drastically increase the total life cycle costs!

76



5.7 Life-Cycle Costs: Acquisition Phase

Needs Assessment

Conceptual Design

Detailed Design

Acquisition Phase

Costs - $

Rule: About 80% of LCC are locked in by the end of the Acquisition Phase.

Emphasis is on good design!

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5.7 Life-Cycle Costs – Purpose

Make explicit as possible the relationship of costs over the total life span of a product/systemDesign Process Objective Minimize the life-cycle costs And meet other performance

requirements By making correct trade-offs between

costs in the acquisition phase and costs during the operations phase

78



5.7 Life-Cycle Costs – Warning

Beware of introducing certain cost- cutting measures in the acquisition phase and early production phaseSuch cost-cutting measures could impact the future operations and degrade safety or require modifications later onThese cost-cutting measures can be misleading and dangerous!

79



5.7 Life-Cycle Costs – Warning

Engineers have a ethical and moral responsibility to ensure that designs are: Economically sound Functional Safe Perform as expected

80



5.8 Present Worth of Bonds

Bonds represent a source of funds for the firm.Bonds are sold (floated) by investment banks for firms in order to raise additional debt capitalA bond is similar to an IOUBonds are evidence of Debt

81



5.8 Bond Types – Treasury Bonds

Treasury bonds Issued by Federal Government Full backing of the Government 1 year or less; 2-10 year issues; and

10-30 year issues Conservative-type investment

82



5.8 Bond Types – Municipal Bonds

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 interest and face value

83



5.8 Bond Types – Mortgage Bonds

Mortgage Bonds Issued by Corporations Secured by the firm’s 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 the firm

84



5.8 Bond Types – Debentures

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

85



5.8 Present Worth of Bonds – Overview

The Firm

Investment Bankers

Commissions/Fees

Proceeds fromThe sale

Sell the Bonds to The lending public

Bondholders

86



5.8 Bonds – Basics

Bonds are negotiable instrumentsCan be traded by the current bondholderSource of funds to the firmDebt capitalBondholders are loaning $$ to the firmEarn periodic interestSell the bonds at any time

87



5.8 Bonds – Firm’s View

Firm authorizes a bond saleBonds are sold by an outside agencyFirm pays a commission to the selling agencyThe firm receives the proceeds from the saleThis is now DEBT capital to the firm

88



5.8 Bond Basics – Continued

The bondholders are not ownersThey are lendorsThe firm pays periodic interest payments to the current bond holdersAt the end of the bond’s life, the bonds are redeemed (bought back) from the current bond holder

89



5.8 Bond Basics - Continued

The bond itself is just a piece of paperEvidence of the debt the firm has incurredThe firm may be able to “call” the bonds back by paying the current bondholder a calculated sum

90



5.8 Bonds – Notation

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

r – The nominal annual bond interest rate

91



5.8 Bonds – Notation and Example

Given the nominal annual bond interest rate, the payment frequency of the interest (monthly, quarterly semi-annually, etc.) is also statedExample: V = $5,000 (face value) r = 4.5% per year paid semiannually N = 10 years

92



5.8 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

93



5.8 Bonds – Example 5.11

Given V = $10,000 (Face value of the bond) r = 4.5% paid semiannually N = 10 years or 20 interest periods $I/6 months = $5,000(0.045/2) =

$112.50 paid to the current bondholder

Bonds are bought at sold in a bond market. Thus the price of the bond is subject to the

pressures of the bond market.

94



5.8 Example 5.11

Key Point – The purchase price of the bond can be considered a value that is determined by a willing buyer and a willing seller.Assume the potential buyer of this bond requires a interest rate of no less than 8%/year compounded quarterly.

95



5.8 Example 5.11 – Continued

The purchaser will consider this bond if he/she can earn 8%/yr c.q.What is fixed? The future interest payments are fixed The future face value of the bond in

fixed

What can vary? The purchase price such that the buyer

can earn at least the 8%/yr c.q.

96



5.8 Example 5.1 – Continued

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 buyer’s required

interest rate

97



5.8 Example 5.11 – Solving

The objective is to determine the purchase price of this bond discounted at the buyer’s required rate of 4.04% per 6 monthsDraw the cash-flow diagramWork the problem with N = 20 (not 10) We have 20 interest payments (every 6

months) = 10 years

98



5.8 Example 5.11 – Cash-Flow Diagram

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 cash flows to the potential bond buyer

99



5.8 Example 5.11 – Solving

P = $112.50(P/A,4.04%,20)+ $5,000(P/F,2%,40)

P = $3,788IF the buyer can buy this bond for $3,788 or less, he/she will earn at least the 8% c.q. rate.

100



Summary: Present Worth

• Present Worth is the basic analysis approach for most engineering economy studies.

• It also forms a basis for the Internal Rate of Return method to be presented later

• Requires knowledge of the discount rate as part of the analysis

101



Summary: Present Worth

• PW represents a family of methods

• Annual worth

• Future Worth

• Capitalized Cost

• Life-cycle cost analysis – application

• Bond Problems – application

102



End of Chapter 5 Lecture Set

engineering economy chapter 5

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