PRESENT WORTH ANALYSIS

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PRESENT WORTH ANALYSISAnastasia L. Maukar

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Formulating Mutually Exclusive Alternatives

• One of the important functions of financial management and engineering is the creation of “alternatives”

• If there are no alternatives to consider then there really is no problem to solve!

• Given a set of “feasible” alternatives, engineering economy attempts to identify the “best” economic approach to a given problem

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Types of Economic Projects• Mutually exclusive alternatives

– From a set of feasible alternatives, pick one and only one to execute

– Mutually exclusive alternatives compete against each other

• Independent projects– From a set of feasible alternatives select as many as can be

funded in the current period

• The “Do Nothing” (DN) alternative should always be considered

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Economic Criteria

• The decision process requires that the outcomes of feasible alternatives be arranged so that they may be judged for economic efficiency in terms of the selection criterion.

• Equivalence provides the logic by which we may adjust the cash flow for a given alternative into some equivalent sum or series

• How should they be compared?• Learn how analysis can resolve alternatives into equivalent

present consequences, referred to simply as present worth analysis.

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APPLYING PRESENT WORTH TECHNIQUES

• One of the easiest ways to compare mutually exclusive alternatives is to resolve their consequences to the present time

• The three criteria for economic efficiency are restated in terms of present worth analysis :

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Chapter Opening Story – GE’s Healthymagination Project

GE Unveils $6 Billion Health-Unit Plan:• Goal: Increase the market share in the healthcare sector.• Strategies: Develop products that will lower costs, increase access and improve health-care quality.• Investment required: $6 billion over six years• Desired project outcome: Would help GE’s health-care unit grow at least twice as fast as the broader economy.

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Ultimate Questions

• GE’ s Point of View:– Would there be enough demand for their

products to justify the investment required in new facilities and marketing?

– What would be the potential financial risk if the actual demand is far less than its forecast or adoption of technology is too slow?

– If everything goes as planned, how long does it take to recover the initial investment?

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Bank Loan vs. Project Cash Flows

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Example 3.1 Describing Project Cash Flows – A Computer-Process Control Project

Year(n)

Cash Inflows(Benefits)

Cash Outflows(Costs)

NetCash Flows

0 0 $650,000 -$650,000

1 215,500 53,000 162,500

2 215,500 53,000 162,500

… … … …

8 215,500 53,000 162,500

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Cash Flow Diagram for the Computer Process Control Project

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Net Present Worth Measure Principle: Compute the equivalent net surplus at n = 0 for a given

interest rate of i. Decision Rule for Single Project Evaluation: Accept the project if the

net surplus is positive. Decision Rule for Comparing Multiple Alternatives: Select the

alternative with the largest net present worth.

2 3 4 50 1

Inflow

Outflow

0PW(i) inflow

PW(i) outflow

Net surplus

PW(i) > 0

0

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Example 3.2 - Tiger Machine Tool Company

inflow

outflow

(12%) $35,560( / ,12%,1) $37,360( / ,12%,2)$31,850( / ,12%,3) $34,400( / ,12%,4)$

(12%) $76,000(12%) $106,065 $76,000

$30,065 0, Accept

PW P F P FP F P F

PWPW

$76,000

$35,560 $37,360$34,400

01 2 3

outflow

inflow

$31,850

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Present Worth Amounts at Varying Interest Rates

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Can you explain what $30,065 really means?

1. Project Balance Concept

2. Investment Pool Concept

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Project Balance Concept

• Suppose that the firm has no internal funds to finance the project, so will borrow the entire investment from a bank at an interest rate of 12%.

• Then, any proceeds from the project will be used to pay off the bank loan.

• Then, our interest is to see if how much money would be left over at the end of the project period.

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Calculating Project Balances

End of Year (n ) 0 1 2 3 4

Beginning Project Balance (76,000)$ (49,560)$ (18,147)$ 11,525$

Interest Charged (12%) (9,120)$ (5,947)$ (2,178)$ 1,383$

Payment (76,000)$ 35,560$ 37,360$ 31,850$ 34,400$

Ending Project Balance (76,000)$ (49,560)$ (18,147)$ 11,525$ 47,308$

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Project Balance Diagram – Four Pieces of Information

The exposure to financial risk

The discounted payback period

The profit potential

The net future worth

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Useful Lives Equal the Analysis Period

Examples 3.3:

• A firm is considering which of two mechanical devices to

install to reduce costs in a particular situation. Both devices

cost $1000 and have useful lives of 5 years and no salvage

value. Device A can be expected to result in $300 savings

annually. Device B will provide cost savings of $400 the first

year but will decline $50 annually, making the second-year

savings $350, the third-year savings $300, and so forth. With

interest at 7%, which device should the firm purchase?

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• Answer:

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Examples 3.4:• A purchasing agent is considering the purchase of some new

equipment for the mail room. Two different manufacturers have provided quotations. An analysis of the quotations indicates the following:

• The equipment of both manufacturers is expected to perform at the desired level of (fixed)output. For a 5-year analysis period, which manufacturer's equipment should be selected? Assume 7% interest and equal maintenance costs

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• answer

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• A firm is trying to decide which of two weighing scales it should install to check a package-filling operation in the plant. The ideal scale would allow better control of the filling operation and result in less overfilling. If both scales have lives equal to the 6-year analysis period, which one should be selected? Assume an 8% interest rate

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• answer

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Useful Lives Different from the Analysis Period

• A diesel manufacturer is considering the two alternative production machines which are specific data are as follows:

• The manufacturer uses an interest rate of 8% and wants to use the PW method to compare these alternatives over an analysis period of 10 years.

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Infinite Analysis Period: Capitalized Cost

• Another difficulty in present worth analysis arises when we encounter an infinite analysis period (n= ~)

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Example:• A city plans a pipeline to transport water from a

distant watershed area to the city. The pipeline will cost $8 million and will have an expected life of 70 years. The city anticipates it will need to keep the water line in service indefinitely. Compute the capitalized cost, assuming7% interest

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Multiple Alternatives

• The minimum required interest rate for invested money is called the minimum attractive rate of return, or MARR

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Investment Pool Concept• Suppose the company has $76,000. It has two options.

(1)Take the money out and invest it in the project or (2) leave the money in the pool and continue to earn a 12% interest.

• If you take Option 1, any proceeds from the project will be returned to the investment pool and earn 12% interest yearly until the end of the project period.

• Let’s see what the consequences are for each option.

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If $76,000 were left in the investment poolfor 4 years

$76,000(F/P,12%,4) $119,587

If $76,000 withdrawalfrom the investmentpool were invested inthe project

• AmountYear• $35,6501• $37,3602

• $31,8503• $34,4004

$35,650(F/P,12%,3)

$37,360(F/P,12%,2)

$31,850(F/P,12%,1)

$34,400(F/P,12%,0)

1 • $49,959

2 • $46,864

3 • $35,672

4• $34,400

$166,896

$166,896

$119,587

$47,309

Option A

Option B

The net benefit of investing in the project

Investment Pool

PW(12%) = $47,309(P/F,12%,4) = $30.065

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What Factors Should the Company Consider in Selecting a MARR in Project Evaluation?

• Cost of capital– The required return necessary to

make an investment project worthwhile.

– Viewed as the rate of return that a firm would receive if it invested its money someplace else with a similar risk

• Risk premium– The additional risk associated with

the project if you are dealing with a project with higher risk than normal project

MA

RR

Cos

t of c

apita

lR

isk

prem

ium

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WEEK 4

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Practice Problem• An electrical motor rated at 15HP needs to be

purchased for $1,000. • The service life of the motor is known to be 10 years

with negligible salvage value.• Its full load efficiency is 85%.• The cost of energy is $0.08 per Kwh.• The intended use of the motor is 4,000 hours per

year.• Find the total present worth cost of owning and

operating the motor at 10% interest.

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Solution

1HP=0.7457kW 15HP = 15 0.7457 = 11.1855kW Required input power at 85% efficiency rating:

11.1855 13.15940.85

Required total kWh per year 13.1594kW 4,000 hours/year =52,638

kW kW

kWh/yr Total annual energy cost to operate the motor

52,638kWh $0.08/kWh =$4,211/yr The total present worth cost of owning and operating the motor

(10%) $1,000 $4,211( / ,10PW P A %,10) = $26,875

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Exercise 1

• A piece of land may be purchased for $610,000 to be strip-mined for the underlying coal. Annual net income will be $200,000 per year for 10 years. At the end of the 10 years, the surface of the land will be restored as required by a federal law on strip mining. The reclamation will cost $1.5 million more than the resale value of the land after it is restored. Using a 10%interest rate, determine whether the project is desirable

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Exercise

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Exercise 2

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Exercise 3

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Exercise 4

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Exercise 5

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PRESENT WORTH ANALYSIS