Chapter 6endustri.eskisehir.edu.tr/onur_kaya/İKT356/icerik/Ch6...Unequal Project Lives 1. Required...

1

Chapter 6Annual Equivalence Analysis

Annual equivalent criterionApplying annual worth analysisMutually exclusive projects

2

Annual Worth Analysis

Principle: Measure investment worth on annualbasis

Benefit: • Annual reports, yearly budgets• Seek consistency of report format• Determine unit cost (or unit profit)• Facilitate unequal project lifecomparison

3

Annual Equivalent Worth

AE(i)=PW(i)*(A|P,i,N)

If AE(i) > 0 , acceptAE(i) < 0 , rejectAE(i) = 0 , remain indifferent

Since (A|P,i,N) > 0 for –1<i<∞AE(i) > 0 PW(i) > 0

4

Computing Equivalent Annual Worth

$100$50

$80$120

$700

2 3 4 5 61

A = $46.07

2 3 4 5 61

AE(12%) = $189.43(A/P, 12%, 6)= $46.07

$189.43

00

PW(12%) = $189.43

5

Annual Equivalent Worth - Repeating Cash Flow Cycles

$500$700 $800

$400 $400 $500$700

$800

$400 $400

$1,000 $1,000

Repeating cycle

6

• First Cycle:

PW(10%) = -$1,000 + $500 (P/F, 10%, 1)+ . . . + $400 (P/F, 10%, 5)

= $1,155.68AE(10%) = $1,155.68 (A/P, 10%, 5) = $304.87

• Both Cycles:

PW(10%) = $1,155.68 + $1,155.68 (P/F, 10%, 5)= $1,873.27

AE(10%) = $1,873.27 (A/P, 10%,10) = $304.87

7

Annual Equivalent Cost

When only costs are involved, the AE method is called the annual equivalent cost.Revenues must cover two kinds of costs: Operating costs and capital costs.

Capital costs

Operating costs

+

Ann

ual E

quiv

alen

t Cos

ts

8

Capital and Operating CostsCapital costs are incurred by purchasing assets tobe used in production and service. Normally, they arenonrecurring (one-time) costs.Operating costs are incurred by the operation of physical plant or equipment needed to provideservice (e.g. labor and raw materials). Normally, theyrecur for as long as an asset is owned.Operating costs are on annual basis anyway. Annualequivalent of a capital cost is calledcapital recovery cost, CR(i).

Remember: (A|P, i, N) is called the capital recovery factor.

9

Capital (Ownership) CostsDef: The cost of owning an

equipment is associated with two transactions—(1) its initial cost (I) and (2) its salvage value(S).Capital costs: Taking into these sums, we calculate the capital costs as:

since (A|F,i,N)=(A|P,i,N)-i

0 1 2 3 N

0N

I

S

CR(i)CR i I A P i N S A F i N

I S A P i N iS( ) ( / , , ) ( / , , )

( )( / , , )= −= − +

10

Example - Capital Cost Calculation

Given:I = $200,000N = 5 yearsS = $50,000i = 20%

Find: CR(20%) $200,000

$50,000

50

CR i I S A P i N iSCR A P

( ) = ( - ) ( / , , ) + ( = ($200, - $50, ) ( / , )

+ (0.20)$5 , = $60,

20%) 000 000 20%, 50 000

157Annual cost of owning the asset at 20%

11

Justifying an investment based on AE Method

Given: I = $20,000, S = $4,000, N = 5 years, i = 10%Find: see if an annual revenue of $4,400 is enough to cover the capital costs.Solution:CR(10%) = $4,620.76Conclusion: Need an additional annual revenue in the amount of $220.76.

12

Example - Capital Cost Calculation for Mini Cooper

Given:I = $19,800N = 3 yearsS = $12,078i = 6%

Find: CR(6%) $19,800

$12,078

30

( ) = ( - ) ( / , , ) + (6%) = ($19,800 - $12,078) ( / , 6%, 3)

+ (0.06)$12,078 = $3,613.55

CR i I S A P i N iSCR A P

13

Applying Annual Worth Analysis

•Unit Cost (Profit) Calculation

• Unequal Service Life Comparison

• Minimum Cost Analysis

14

Equivalent Worth per Unit of Time

0

1 2 3

$24,400$55,760

$27,340

$75,000 Operating Hours per Year

2,000 hrs. 2,000 hrs. 2,000 hrs.

• PW (15%) = $3553• AE (15%) = $3,553 (A/P, 15%, 3)

= $1,556

• Savings per Machine Hour= $1,556/2,000= $0.78/hr.

Note: 3553/6000=0.59/hour: instant savings in presentworth for each hourly use; does not consider the time over which the savings occur

15

Equivalent Worth per Unit of Time(cont’d)

0

1 2 3

$24,400$55,760

$27,340

$75,000 Operating Hours per Year

1,500 hrs. 2,500 hrs. 2,000 hrs.

•Let C denote the equivalent annual savings per machine hour• $1,556=[(C)(1500)(P|F,15%,1)

+(C)(2500)(P|F,15%,2)+(C)(2000) (P|F,15%,3)] (A|P,15%,3)

C=$0.79/hr

16

Breakeven Analysis

Problem:

At i = 6%, what should be the reimbursement rate per mile so that Sam can break even? $11,72539,000Total

$4,680$3,624$3,421

14,50013,00011,500

123

Total costs(Ownership

& Operating)

Miles Driven

Year(n)

1711,500 miles13,000 miles14,500 milesExpected miles driven

$3,421$3,624$4,680Total of all costs

$971$1,003$988Total operating cost

110125135Parking and tolls

10010080Oil

522650688Gasoline and taxes

121315Accessories

273035Replacement tires

2008535Nonscheduled repairs

$2,450$2,621$3,693Total ownership cost

505778Registration and taxes

635635635Insurance

220153100Scheduled maintenance

$1,545$1,776$2,879Depreciation

Third YearSecond YearFirst Year

Breakeven Analysis

18

• Equivalent annual cost of owning and operating the car[$4,680 (P/F, 6%, 1) + $3,624 (P/F, 6%, 2) + $3,421 (P/F, 6%, 3)] (A/P, 6%,3) = $3,933 per year

• Equivalent annual reimbursement Let X = reimbursement rate per mile [14,500X(P/F, 6%, 1) + 13,000X(P/F, 6%, 2) + 11,500 X (P/F, 6%, 3)] (A/P, 6%,3) = 13.058X

• Break-even value 13.058X = 3,933

X = 30.12 cents per mile

19

Annual equivalent reimbursement as a function of cost per mile

4000

3000

2000

1000

0 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40

Annual equivalentcost of owning andoperating ($3,933)

Loss

Annual reimbursementamount

Minimumreimbursementrequirement ($0.3012)

Gain

Reimbursement rate ($) per mile (X)

Ann

ual e

quiv

alen

t ($)

13,058X

20

Mutually Exclusive Alternativeswith Equal Project Lives

Standard PremiumMotor Efficient Motor

25 HP 25 HP$13,000 $15,60020 Years 20 Years$0 $018.65 kW per motor 18.65 kW per motor

89.5% 93%$0.07/kWh $0.07/kWh3,120 hrs/yr. 3,120 hrs/yr.

SizeCostLifeSalvageOutputEfficiencyEnergy CostOperating Hours

(a) At i= 13%, determine the operating cost per kWh for each motor.(b) At what operating hours are they equivalent?

21

Solution:

(a) Operating cost per kWh per unit

Determine total input power

Conventional motor:

input power = 18.65 kW/ 0.895 = 20.838kW

PE motor:

input power = 18.65 kW/ 0.930 = 20.054kW

Input power = output power% efficiency

22

Determine total kWh per year with 3120 hours of operation

Conventional motor:

3120 hrs/yr (20.838 kW) = 65,018 kWh/yr

PE motor:

3120 hrs/yr (20.054 kW) = 62,568 kWh/yr

Determine annual energy costs at $0.07/kwh: Conventional motor:

$0.07/kwh × 65,018 kwh/yr = $4,551/yrPE motor:

$0.07/kwh × 62,568 kwh/yr = $4,380/yr

Solution:

23

Capital cost:Conventional motor:

$13,000(A/P, 13%, 20) = $1,851PE motor:

$15,600(A/P, 13%, 20) = $2,221

Total annual equivalent cost:Conventional motor:

AE(13%) = $4,551 + $1,851 = $6,402Cost per kwh = $6,402/58,188 kwh = $0.1100/kwh

(where 58188=18.65*3120 is the total output power per year)PE motor:

AE(13%) = $4,380 + $2,221 = $6,601Cost per kwh = $6,601/58,188 kwh = $0.1134/kwh

Solution:

24

Savings from switching from conventional to PE motors?

Incremental capital cost required=2221-1851=370Incremental energy savings=4551-4380=171Hence, at 3120 annual operating hours, a loss of 199 for each motor.

Solution:

25

(b) Break-evenoperatinghours = 6,742

Solution:

26

Mutually Exclusive Alternatives with Unequal Project Lives

1. Required service period = Indefinite

2. Each alternative replaced by identical assetwith same costs and performance

Then,

We may solve for AE of each project based on its initial life span rather than on LCM of project lives.

27

Model A:0 1 2 3

$12,500$5,000 $5,000

$3,000

Model B:0 1 2 3 4

$15,000$4,000 $4,000 $4,000

$2,500

Mutually Exclusive Alternatives with Unequal Project Lives

Required servicePeriod = Indefinite

Analysis period =LCM (3,4) = 12 years

(Least common multiple)

28

Model A:

$12,500$5,000 $5,000

$3,000

0 1 2 3

• First Cycle:PW(15%) = -$12,500 - $5,000 (P/A, 15%, 2)

- $3,000 (P/F, 15%, 3)= -$22,601

AE(15%) = -$22,601(A/P, 15%, 3) = -$9,899• With 4 replacement cycles:

PW(15%) = -$22,601 [1 + (P/F, 15%, 3) + (P/F, 15%, 6) + (P/F, 15%, 9)]

= -$53,657AE(15%) = -$53,657(A/P, 15%, 12) = -$9,899

29

Model B:

$15,000$4,000 $4,000

$2,500

0 1 2 3 4

$4,000

• First Cycle:PW(15%) = - $15,000 - $4,000 (P/A, 15%, 3)

- $2,500 (P/F, 15%, 4)= -$25,562

AE(15%) = -$25,562(A/P, 15%, 4) = -$8,954• With 3 replacement cycles:

PW(15%) = -$25,562 [1 + (P/F, 15%, 4) + (P/F, 15%, 8)]= -$48,534

AE(15%) = -$48,534(A/P, 15%, 12) = -$8,954

30

IF:1. The service of the selected alternative required on a continuous basis2. Each alternative replaced by identical asset with same costs andperformance

31

SummaryAnnual equivalent worth analysis, or AE, is—along with present worth analysis—one of two main analysis techniques based on the concept of equivalence. The equation for AE is

AE(i) = PW(i)(A/P, i, N).AE analysis yields the same decision result as PW

analysis.

32

The capital recovery cost factor, or CR(i), is one of the most important applications of AE analysis in that it allows managers to calculate an annual equivalent cost of capital for ease of itemization with annual operating costs.

The equation for CR(i) isCR(i)= (I – S)(A/P, i, N) + iS,

where I = initial cost and S = salvage value.

Summary

33

AE analysis is recommended over NPW analysis in many key real-world situations for the following reasons:1. In many financial reports, an annual equivalent value

is preferred to a present worth value.2. Calculation of unit costs is often required to determine

reasonable pricing for sale items.3. Calculation of cost per unit of use is required to

reimburse employees for business use of personal cars.

4. Make-or-buy decisions usually require the development of unit costs for the various alternatives.

Summary

34

Example 1:Susan wants to buy a car that she will keep forthe next four years. She can buy a Honda Civicat 15000 and then sell it for 8000 after fouryears. If she bought this car, what would be her annual ownership cost (capital recovery cost)? Assume that her interest rate is 6%.

35

2.2500)8000)(06.0()4%,6,|()800015000(or

2.2500)4%,6,|(8000)4%,6,|(15000%)6(

=+−

=−=

PA

FAPACR

36

Example 2:The following cash flows represent the potential annualsavings associated with two different types of productionprocesses, each of which requires an investment of 12000. Assuming an interest rate of 15%, complete thefollowing tasks: (a) Determine the equivalent annualsavings for each process. (b) Determine the hourlysavings for each process assuming 2000 hours of operation per year. (c) Determine which process shouldbe selected.

635045603635022804

635068402635091201

-12000-120000BAn

37

( )

B.Select (c)

hour/07.12000

08.2147 :B Process

hour/95.02000

14.1893 :A Process

(b)

08.2147)4%,15,|()4%,15,|(635012000%)15(

14.1893)4%,15,|(15.1

228015.1

456015.1

684015.1

912012000%)15(

(a)

432

=

=

=+−=

=⎟⎠⎞

⎜⎝⎛ ++++−=

PAAPAE

PAAE

B

A

38

Example 3:Norton Auto Parts, Inc. is considering two different forklifttrucks for use in its assembly plant.

Truck A costs 15000 and requires 3000 annually in operating expenses. It will have a 5000 salvagevalue at the end of its 3-year service life. Truck B costs 20000 and requires 2000 annually in operating expenses. It will have a 8000 salvagevalue at the end of its 4-year service life.

The firm’s MARR is 12%. Assuming that the trucks areneeded for 12 years and no significant changes areexpected in the future price and functional capacity of both trucks, select the most economical truck based on AE analysis.

39

.economical more is BTruck

4.69102000)8000)(12.0()4%,12,|()800020000(%)12(

77633000)5000)(12.0()3%,12,|()500015000(%)12(

=++−=

=++−=

PAAE

PAAE

B

A

Capital Recovery Cost

40

Example 4:Your company needs a small front loader for handling bulk materials at the plant. It can be leased from the dealer for three years for $4050 per year including all maintenance. It can also be purchased for $14,000. You expect the loader to last for six years and to have a salvage value of $3000. You predict that maintenance will cost $400 the first year and increase by $200 per year in each year after the first. Your MARR is 15%. Use annual equivalence analysis to determine whether to lease or buy the loader. What is the shortest project life for which the annual worths you have calculated are exactly correct?

41

( )

6) of multipleinteger an (or years 6for correct Exactly

LEASE.

46.4175 )6%,15,|()6%,15,|(3000)6%,15,|(200)6%,15,|(40014000%)15(

=−++= PAFPGPAPAE

Present worth

42

Example 5:Consider the following cash flows for two types of models. Bothmodels will have no salvage value upon their disposal (at the end of their respective service lives). The firm’s MARR is known to be 15%. (a) Notice that models have different service lives. However, model A will be available in the future with the same cash flows. Model B is available now only. If you select model B now, you will have toreplace it with model A at the end of year 2. If your firm uses thepresent worth as a decision criterion, which model should be selectedassuming that your firm will need either model for an indefiniteperiod? (b) Suppose that your firm will need either model for only twoyears. Determine the salvage value of model A at the end of year twothat makes both models indifferent (equally likely).

3500310000350021000035001-15000-60000

BAnProject cash flow

4354.20721257)2%,15,|()2%,15,|(35006000

(b)

47.5653)2%,15,|(15.015.8721257%)15(

33.581415.015.872%)15(

(a)

18.773)2%,15,|(1257%)15(15.872)3%,15,|(2.1991%)15(

1257)2%,15,|(1000015000%)15(2.1991)3%,15,|(35006000%)15(

:cycle oneFor

foreverA Model and now B Model : 2Option foreverA ModelandnowA Model:1Option

2

1

==++−

=+=

==

====

=+−==+−=

SFPSAP

FPPW

PW

PAAEPAAE

APPWAPPW

Opt

Opt

B

A

B

A