Part II: Evaluating business & engineering assets

Ch 5: Present worth analysis

Ch 6: Annual equivalence analysis

Ch 7: Rate-of-return analysis

Net present worth measure

Principle: Compute the equivalent net surplus at n = 0 for a given interest rate of i.

Decision rule: Accept the project if the net surplus is positive.

2 3 4 50 1

inflow

outflow

0PW(i)inflow

PW(i)outflow

net surplus

PW(i) > 0

Minimum attractive rate of return (MARR)

Also called “required rate of return”Used to evaluate a single project – must have a positive

present worthAlso used to compare alternatives – choose one with

greatest positive present worthSteps in project evaluation:

1. determine i, or MARR2. estimate project service life3. estimate cash inflow for each period over service life4. estimate case outflow for each period over service life5. determine net cash flows for each period over service life6. find net PW of each net cash flow at the MARR & sum7. decide to accept or reject the investment based on PW

Example 5.3 - Tiger Machine Tool Company

$75,000

$24,400 $27,340$55,760

01 2 3

inflow

outflow

PW (15%) inflow P F P F$24 , ( / , ) $27 , ( / , )400 15%,1 340 15%, 2= +

760$55 P F, ( / , )15%, 3+ $78,553=

PW ( $75,15%) 000outflow =

PW ( $78, $75,15%) 553 000= - $3, ,553 0 Accept= >

Excel solution – example 5.3

Net present value

Year Cash flow PW

0 -75,000 -75,000

1 24,400 21,217

2 27,340 20,673

3 55,760 36,663

Sum 32,500 3,553

Ex 5.3: Present worth amounts at varying interest rates

i ,% PW(i)0 $32,500

2 27,743

4 23,309

6 19,169

8 15,296

10 11,670

12 8,270

14 5,077

16 2,076

17.46 0

18 -751

i, % PW(i)

20 -$3,412

22 -5,924

24 -8,296

26 -10,539

28 -12,662

30 -14,673

32 -16,580

34 -18,360

36 -20,110

38 -21,745

40 -23,302

Break even interest rate

(internal rate of return)

Present worth profile

40

$3553 17.46%

Break even interest rate(or rate of return)

Accept Reject30

20

PW

(i) (

$ th

ousa

nds) 10

0

-10

-20

-300 5 10 15 20 25 30 35 40

i = MARR (%)

Future worth criterionGiven: cash flows & MARR (i)Find: net equivalent worth at

end of project life

$75,000

$24,400 $27,340$55,760

01 2 3

project life

Should the firm leave the $75,000 in its investment pool to earn 15% over 3 years?

$75,000(F/P,15,3) = $114,066

Or should the firm invest the $75,000 in the project?

$24,400(F/P,15,2) + $27,340(F/P,15,1) + $55,760(F/P,15,0) = $119,470

Note: $5,404(P/F,15,3) = $3,553Net future worth: $119,470 - $114,066 = $5,404

Meaning of net present worth

$75,000

$24,400$27,340

$55,760

Investment pool How much would you have after 3 years if the Investment is made?

$119,470

0 1 2 3Project

Return

to investment pool How much would you have if the investment was not made?

$114,066

What is the net gain from the investment?

$5,404

There is a return of $5,404 to investment pool, above the 15% interest

Project balance on borrowed funds

NN 00 11 22 33Beginningbalance

Interest

Payment

Project balance

-$75,000

-$75,000

-$75,000

-$11,250

+$24,400

-$61,850

-$61,850

-$9,278

+$27,340

-$43,788

-$43,788

-$6,568

+$55,760

+$5,404

Net future worth, FW(15%)

If a firm has no investment pool but can borrow funds at 15% the result is the same

Project balance over 3 years60,000

40,000

20,000

0

-20,000

-40,000

-60,000

-80,000

-100,000

-120,0000 1 2 3

-$75,000-$61,850

-$43,788

$5,404

Year(n)

Terminal project balance(net future worth, or

project surplus)

Discounted payback period

Pro

ject

bal

ance

($)

Capitalized equivalent worth

Principle: PW for a project with an annual receipt of A over infinite service life

Equation: CE(i) = A(P/A, i, ∞) = A/i

A0

P = CE(i)

Ex 5.4: A = $2 million, i = 8%, N = ∞

A/i = CE(i) = $2 million/0.08 = $25 million

Practice problem

$2,000

10

$1,000

0

Given: i = 10%, N = ∞Find: P or CE (10%)

∞

P = CE (10%) = ?

Practice problem solution

10

$1,000$2,000

0

Given: i = 10%, N = ∞Find: P or CE (10%)

∞

P = CE (10%) = ?

CE (10%) =$1,0000.10

$1,0000.10

+ (P/F,10%,10)

= $10,000(1+0.3855) = $13,855

Bridge construction projectConstruction cost = $2,000,000Annual maintenance cost = $50,000Renovation cost = $500,000 every 15 yearsPlanning horizon = infinite periodInterest rate = 5%

$500,000 $500,000 $500,000 $500,000

$2,000,000

$50,000

0 15 30 45 60Years

Bridge project – solution

Construction cost: P1 = $2,000,000Maintenance costs P2 = $50,000/0.05 = $1,000,000Renovation costs: P3 = $500,000(P/F, 5%, 15)

+ $500,000(P/F, 5%, 30)+ $500,000(P/F, 5%, 45)+ $500,000(P/F, 5%, 60)…= {$500,000(A/F, 5%, 15)}/0.05 = $463,423

Total present worth: P = P1 + P2 + P3 = $3,463,423

Alternate way to calculate P3

Concept: Find the effective interest rate per payment period

Effective interest rate for a 15-year cycle i = (1 + 0.05)15 - 1 = 107.893%

Capitalized equivalent worth P3 = $500,000/1.07893 = $463,423

150

$500,000

30 45 60

$500,000 $500,000 $500,000

Practice problem – electric motor

Want to purchase an electrical motor rated at 15 HP 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 cost of owning and operating the motor at

10% interest.

Solution – electric motor

1 HP = 0.7457 kW

15 HP = 150 × 7457 = 11.1855 kW

Required input power at 85% efficiency rating:

= 13.1594 kW11.18550.85

kW

Required total kWh per year13.1594 kW × 4,000 hr/yr = 52,638 kWh/yr

Total annual energy cost to operate the motor52,638 kWh × $0.08/kWh = $4,211/yr

Total cost of owning and operating the motorPW(10%) = $1,000 + $4,211(P/A,10%,10) = $26,875

0 1 2 3 4 5 6 7 8 9 10

$4,211

$1,000

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

Comparing mutually exclusive projects (alternatives)

PrinciplePrinciple: projects must be compared over an equal time : projects must be compared over an equal time spanspan

Rule of thumbRule of thumb: If the required service period is given, the : If the required service period is given, the analysis period should be the sameanalysis period should be the same

Analysis period: time span over which the economic effects of an investment will be evaluated (study period or planning horizon)

Required service period: time span over which the service of an equipment (or investment) will be needed

Do nothing is an alternative!

Revenue projects: projects whose revenues depend on the choice of alternatives – maximize PW

Service projects: projects whose revenues do not depend on the choice of alternative – minimize PW

Case 1: analysis period equals project lives

Compute the PW for each project over its life

$1,400$2,075

$2,110

$450$600

$5000

$1,000A B$4,000

PW (10%)A = $283 PW (10%)B = $579

$1,000

$450$600

$500

Project A

$1,000

$600$500$450

$3,000

3,993

$4,000

$1,400

$2,075

$2,110

Project B

Comparing projects requiring different levels of investment – assume that theunused funds will be invested at MARR.

PW(10%)A = $283PW(10%)B = $579

This portionof investmentwill earn 10%return on investment.

ModifiedProject A

Case 2: analysis period shorter than project lives

Estimate the salvage value at the end of the required service period

Compute the PW for each project over the required service period

Required Service Period = 2 years

Ex. 5.6

PW(15%)B = -$364,000PW(15%)A = -$362,000

Ex. 5.6

Case 3: Analysis period longer than project lives

Come up with replacement projects that match or exceed the required service period

Compute the PW for each project over the required service period

Required service period = 5 yrs

Model A

Model B

$15,000

$12,500

$4,000 $4,500 $5,000$5,500

$1,500

$5,000 $5,500 $6,000

$2,000

40 1 2

3

1 2 34

5

05

Lease model A to serve the remaining service life

PW(15%)A = -$35.929

PW(15%)B = -$33,173

Summary – vocabulary

Present worth is an equivalence method of analysis in which a project’s cash flows are discounted to a lump sum amount at present time.

The MARR or minimum attractive rate of return is the interest rate at which a firm can always earn or borrow money.

MARR is generally dictated by management and is the rate at which NPW analysis should be conducted.

Two measures of investment, the net future worth and the capitalized equivalent worth, are variations to the NPW criterion.

The term mutually exclusive means that, when one of several alternatives that meet the same need is selected, the others will be rejected.

Summary – vocabulary (cont.)

Revenue projects are those for which the income generated depends on the choice of project.

Service projects are those for which income remains the same, regardless of which project is selected.

The analysis period (study period) is the time span over which the economic effects of an investment will be evaluated.

The required service period is the time span over which the service of an equipment (or investment) will be needed.

The analysis period should be chosen to cover the required service period.

When not specified by management or company policy, the analysis period to use in a comparison of mutually exclusive projects may be chosen by an individual analyst.