PRESENT WORTH ANALYSIS Anastasia L. Maukar 1. 2 Formulating Mutually Exclusive Alternatives One of the important functions of financial management and.

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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

cap

ital

Ris

kp

rem

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 13.1594

0.85 Required total kWh per year

13.1594kW 4,000 hours/year =52,638

kWkW

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

PRESENT WORTH ANALYSIS Anastasia L. Maukar 1. 2 Formulating Mutually Exclusive Alternatives One of the important functions of financial management and.

Documents

exclusive alternatives

creation of alternatives

multiple alternatives

set of feasible alternatives

project cash flows

given problem slide

largest net present

present worth techniques