solving typical fe problems

7/23/2019 Solving Typical FE Problems

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Solving Typical FE Problems

What to do: Find P or F from Single Amounts

How to do it:The simplest problems to solve in engineeringeconomic analysis are those which involve findingthe value of a single amount of money at an earlieror later date than that which is given. Suchproblems involve finding the future worth (F) of aspecified present amount (P), or vice versa. Theseproblems involve using the equations

F ! P(" # i)n or P ! F$"% (" # i)n&

'n terms of standard factor notation, the equationon the left is represented as F!P(F%P,i,n) and the

equation on the right is represented as P !F(P%F,i,n).

Eample: !": person deposits *,+++ into amoney maret account which pays interest at a rateof - per year. The amount that would be in theaccount at the end of ten years is most nearly

a. /,01/

b. 1,+++

c. "+,01*

d. "/,"2*

Solution: The *,+++ represents a present amount,P. The future amount, F, is

Eample !#: small company wants to deposit asingle amount of money now so that it will haveenough to purchase a new truc costing *+,+++five years from now. 'f the money can be depositedinto an account which earns interest at "+ peryear, the amount that must be deposited is most



nearly

a. "+,+++

b. 3",+*+

c. 33,/*+

d. 3"1,"2+

Solution: The *+,+++ is a future amount in yearfive that must be moved to the present.


What to do: Find P from a $niform Series %A& and 'ice'ersa

How to do it:

4niform series cash flows are represented by thesymbol . uniform series refers to cash flowswhich (") occur in consecutive interest periods, and

(/) are the same amount each time. To solve for Pfor these types of problems, the following equationis used

P !

'n standard factor notation, the equation is P !(P%,i,n). 't is important to note in using thisequation that the present worth, P, is located one

interest period ahead of the first . 't is alsoimportant to remember that n must be equal to thenumber of values and the interest rate, i, must bee5pressed in the same time units as n. For e5ample,if n is in months, i must be an effective interest rateper month.

This standard equation can be used in reverse toconvert a present worth into a uniform seriesamount using the form ! P(%P,i,n). This, fore5ample, is used to determine the monthly payment



associated with a car purchase or house loan for acompound interest rate of i.

Eample !(: company e5pects the material costof a certain manufacturing operation to be /+,+++per year. t an interest rate of - per year, thepresent worth of this cost over a five year pro6ectperiod is closest to

a. /1,3-2

b. *2,//+

c. 01,-*7

d. ""0,33/

Solution:

Eample !) piece of machinery has a first cost

of 3",+++ with a monthly operating cost of

"+,+++. 'f the company wants to recover itsinvestment in five years at an interest rate of "per month, the monthly income must be closest to

a. *,71-

b. 2,3-2

c. -,/1*

d. "+,2--

Solution: The value is per month.




What to do: Find F from a $niform Series %A& and 'ice'ersa

How to do it:

'n the previous problem type, the procedure forconverting a uniform series into anequivalent present amount was discussed. 8ere auniform series is converted into a future amountinstead of a present one. The equation for doing sois

F !

The standard notation form is F! (F%,i,n). 't isimportant to remember that the F occurs inthe sameperiod as the last . s before, the n isequal to the number of values and the i used inthe calculation must be e5pressed over the sametime units as n.

Eample !*: 'f a person deposits "++ per monthinto an account which pays interest at a rate of 2per year compounded monthly, the amount in theaccount at the end of five years would be nearestto

a. *27

b. 3,21

c. 2,100

d. 0,11/

Solution Since the cash flow (i.e., values) occursover monthly interest periods, the n and i musthave monthly time units.



s in the previous problem type, the standardequation can be set up and solved in reverse to findan value from a given future worth, F, using !F(%F,i,n).

Eample !+ small company wants to haveenough money saved to purchase a new /++,+++warehouse in five years. 'f the company can investmoney at "- per year, the amount that must beinvested each year is closest to

a. /0,12+

b. 32,1/+

c. 71,2*+

d. 23,12+

Solution


What to do: Find P, A, or F from $niform -radient .ashFlows

How to do it:

uniform gradient cash flow is one wherein thecash flow changes (increases or decreases) bythe same amount in each payment period. Fore5ample, if the cash flow in period " is -++, and inperiod two it is 1++, with amounts increasing by"++ in each subsequent period, this is a uniformgradient cash flow series with the gradient, 9, equal

to "++. The standard factor equation to find thepresent worth of the gradient is represented as P !9(P%9,i,n). This equation finds the value of only thegradient , not the amount of money that thegradient was :built on: (i.e., the base amount) inperiod one. The base amount in period one must behandled separately as a uniform series cash flow.Thus, the general equation to find the present worthof a uniform gradient cash flow series is

P ! (P%,i,n) # 9(P%9,i,n)



'f the gradient is negative, the total cash flowdecreases from one period to the ne5t. The onlydifference is that the plus sign becomes a minussign in the equation above.

Eample !/: company e5pects the cost ofequipment maintenance to be *,+++ in year one,*,*++ in year two, and amounts increasing by*++ per year through year "+. t an interest rateof "+ per year, the present worth of themaintenance cost is nearest to

a. 3-,//+

b. 7/,"0+

c. 72,22+

d. *",01+

Solution: This is an increasing gradient (use #sign) with 9 ! *++ and base amount ! *,+++

0ore to do: Find A from $niform -radient .ashFlows

;onvert a gradient cash flow into an equivalentuniform series, , in one of two ways

". Find the present worth, P, of the cash flowsas discussed above and then convert the Pvalue to an value using the !P(%P,i,n)factor, or

/. 4se the uniform gradient annual worthfactor, (%9,i,n), directly in ! 9(%9,i,n).

<hen using (%9,i,n), it is important to rememberthat this factor converts only the gradient into an



value, 6ust lie the P%9 factor affects only thegradient. The base amount in year one is simplyadded to the value obtained from ! 9(%9,i,n).The general equation for the total value, T, is

T ! " # 9

The future worth, F, of a gradient cash flow is foundby either first finding P and then using the F%Pfactor, or first finding and then using the F%factor.

Eample !1 The cash flow associated with astripper oil well is e5pected to be 3,+++ in monthone, /,1*+ in month two, and amounts decreasingby *+ each month through year five. t an interestrate of "/ per year compounded monthly, theequivalent uniform monthly cash flow is closest to

a. ",3/0

b. ",203

c. /,017

d. 7,3/0

Solution The negative gradient is 9 ! =*+. Thetotal equivalent monthly cash flow, T, is

olving Typical FE Problems

What to do: 2dentify 3ominal and Effective 2nterest4ates



How to do it:

>ominal and effective interest rates are similar tosimple and compound interest rates, with anominal rate being equivalent to a simple interestrate. ll of the equations e5pressing time value of money are based on compound (i.e., effective)

rates, so if the interest rate that is provided is anominal interest rate, it must be converted intoan effective rate before it can be used in any ofthe formulas. The first step in the process ofinsuring that only effective interest rates are usedis to recogni?e whether an interest rate is

nominal or effective. Table " shows the threeways interest rates may be stated.

Table "5 'arious interest statements andtheir interpretation

(")'nterest @ateStatement

(/)'nterpretation

(3);omment

i ! "/ peryear

i ! effective"/ per yearcompoundedyearly

<hen nocompoundingperiod isgiven, interestrate is aneffective rate,withcompounding

periodassumed to beequal to statedtime period

i ! " permonth

i ! effective" per monthcompoundedmonthly

i ! 3="%/per quarter

i ! effective 3="%/ perquartercompoundedquarterly

i ! - peryear,compoundedmonthly

i ! nominal- per yearcompoundedmonthly

<hencompoundingperiod is givenwithout statingwhether theinterest rate isnominal oreffective, it isassumed to benominal.;ompoundingperiod is asstated.

i ! 7 per

quartercompoundedmonthly

i ! nominal

7 perquartercompoundedmonthly

i ! "7 peryear

compoundedsemiannually

i ! nominal"7 per year

compoundedsemiannually



i ! effective"+ per yearcompoundedmonthly

i ! effective"+ per yearcompoundedmonthly

'f interest rateis stated as aneffective rate,then it is an

effective rate.'fcompoundingperiod is not

given,compoundingperiod isassumed tocoincide withstated timeperiod.

i ! effective2 perquarter

i ! effective2 perquartercompoundedquarterly

i ! effective" per monthcompoundeddaily

i ! effective" per monthcompoundeddaily

The three statements in the top part of the tableshow that an interest rate can be stated oversome designated time period without specifyingthe compounding period. Such interest rates are

assumed to be effective rates withthe compounding period (CP) assumed to be thesame as that of the stated interest rate.

For the interest statements presented in the

middle of Table ", three conditions prevail

". the compounding period is identified,

/. this compounding period is shorter thanthe time period over which the interest isstated, and

3. the interest rate is not designated aseither nominal or effective.

'n such cases, the interest rate is assumed to benominal and the compounding period is equal tothat which is stated. (<e show how to geteffective interest rates from these in the ne5tsection.)

For the third group of interest=rate statements inTable ", the word effective precedes or followsthe specified interest rate and the compoundingperiod is also stated. These interest rates areobviously effective rates over the respective timeperiods stated. Aiewise, the compoundingperiods are equal to those stated. Similarly, if theword nominal had preceded any of the intereststatements, the interest rate would be a nominalrate. Table / contains a listing of several interest



statements (column ") along with theirinterpretations (columns / and 3).

Table #5 Specific eamples of intereststatements and interpretations

(")'nterest

Statement

(/)>ominal orBffective'nterest

(3);ompounding

Period

"* per yearcompoundedmonthly

>ominal Conthly

"* per year Bffective Dearly

Bffective "*per yearcompoundedmonthly

Bffective Conthly

/+ per yearcompounded

quarterly

>ominal Euarterly

>ominal /per monthcompoundedweely

>ominal <eely

/ per month Bffective Conthly

/ per month

compoundedmonthly

Bffective Conthly

Bffective 2per quarter

Bffective Euarterly



Bffective /per monthcompoundeddaily

Bffective aily

" per weecompoundedcontinuously

>ominal ;ontinuously

+." per daycompoundedcontinuously

>ominal ;ontinuously


What to do: Find an Effective 2nterest 4ate from a 3ominal 4ateand 'ice 'ersa

How to do it:

ll of the formulas used in maing time value calculationsare based on effective interest rates. Therefore, wheneverthe interest rate that is provided is a nominal rate, it isnecessary to convert it to an effective interest rate. sshown below, an effective interest rate, i, can be calculatedfor any time period longer than the compounding period.

The most common way that nominal interest rates arestated is in the form G5 per year compounded yG where 5! interest rate and y ! compounding period. n e5ample is"- per year compounded monthly. <hen interest ratesare stated this way, the simplest effective rate to get is theone over the compounding period because all that isrequired is a simple division. For e5ample, from theinterest rate of "- per year compounded monthly, amonthly interest rate of ".* is obtained (i.e., "- peryear%"/ compounding periods per year) and this is aneffective rate because it is the rate per compoundingperiod. To get an effective rate for any period longer thanthe compounding period use the effective interest rateformula.

i ! ("#r%m)m = "

This effective interest rate formula can be solved for r or



r%m as needed to determine a nominal interest rate froman effective rate.

For continuous compounding, the effective rate formula isthe mathematical limit as m increases without bounds, and

the formula reduces to i ! er = ".

Eample !6 For an interest rate of "/ per yearcompounded quarterly, the effective interest rate per year

is closest to

a. 7

b. "/

c. "/.**

d. "/.2-

Solution: n effective interest rate per year is sought.

Therefore, r must be e5pressed per year and m is thenumber of times interest is compounded per year .

Eample !"7 For an interest rate of / per month, theeffective semiannual rate is closest to

a. "".**

b. "/

c. "/.2/

d. /2.-/

Solution: 'n this e5ample, the i on the left=hand side ofthe effective interest rate equation will have units ofsemiannual periods. Therefore, the r must have units ofsemiannual periods (i.e., "/ per si5 months) and m mustbe the number of times interest is compounded persemiannual period, 2 in this e5ample.



The types of calculations used to obtain effective interestrates are summari?ed in Table 3.

Table (5 Summary of .alculations 2nvolved inFinding Effective 4ates

'nterestStatement

To Find i for;ompounding

Period

To Find i for anyPeriod Aonger than

;ompoundingPeriod

i ! " permonth

i is alreadye5pressed overcompoundingperiod

4se effectiveinterest rateequation

i ! "/ per

yearcompoundedquarterly

ivide "/ by 7 4se effective

interestrateequation

i ! nominal "2per yearcompounded

semiannually

ivide "2 by / 4se effectiveinterest rateequation

i ! effective"7 per yearcompoundedmonthly

4se effectiveinterest rateequation andsolve for r%m

For effective ivalues other thanyearly, solve for r ineffective interestrate equation andthen proceed as inprevious twoe5amples


What to do: Find P or F for Single Payments and 3ominal 2nterest4ates

How to do it:

For problems involving single payment amounts, that is, Pand F, there are essentially an infinite number of ways tosolve the problems. This is because any effective interest ratecan be used in the P%F or F%P factors as long as the n has thesame units as the i. That is, if an effective interest rate per



month is used, then n must be the number of monthsbetween the P and F. 'f i is an effective interest rate per year,then n must be the number of years.

Eample !"" 'f you deposit ",+++ now at an interest rateof "/ per year compounded monthly, the amount that willbe in the savings account five years from now is closest to

a. ",2++

b. ",02/

c. ",-"0

d. ",1+*

Solution: The "/ rate is a nominal rate and cannot be useddirectly. The simplest way to wor the problem is to use aninterest rate of " per month (an effective rate) because theF%P factor can be looed up directly in the " table. nyother effective interest rate would involve a fraction. (Fore5ample, the effective yearly rate is "/.2-, which is not ina factor table.)

'f the effective interest rate of "/.2- per year is used in thee5ample above, it is necessary to use the F%P formula tocalculate F, since there is no "/.2- interest table. 'f donecorrectly, however, the answer is the same.


What to do: .alculations 2nvolving $niform Series and 3ominal2nterest 4ates

How to do it:

<hen using the uniform series cash flow equations, there aretwo necessary conditions

". the n must always be equal to the number of payments



involved, and

/. the effective interest rate must have the same timeunits as the payments.

For e5ample, if uniform cash flow values ( or 9) occur overquarterly time periods, the interest rate must be an effectivequarterly rate. Similarly, if the cash flow values occur monthlyor yearly, the interest rate must be an effective monthly oryearly rate, respectively.

Eample !"# n individual deposits "++ per month into anaccount which pays interest at a rate of 2 per yearcompounded monthly. The value of the account after five yearsis closest to

a. *,23+

b. 2,/"+

c. 2,027

d. 2,100

Solution: Since cash flow occurs over monthly time periods,

the interest rate must be an effective monthly rate, which inthis case is +.*.


What to do: Find P, A, or F from 3on5conventional $niform Series .ashFlows

How to do it:<hen a uniform series cash flow begins at a time other than periodone, it is called a non-conventional uniform series. 't is non=conventional since determining the present worth requires at leasttwo different factors. This is because the uniform series presentworth equation is derived with the P one interest period ahead of thefirst value. 'f the first value does not occur in period one, the Pwill not occur at time ?eroH another factor, usually the P%F factor,



must be utili?ed to obtain the P value at time ?ero.

Eample !"( The costs associated with a particular process are

e5pected to be 2,+++ per year for five years, beginning three yearsfrom now. t an interest rate of "+ per year, the present worth ofthese costs is closest to

a. "0,+--

b. "-,012

c. //,07*

d. /1,/"+

Solution: 'f the P% factor is used, the P (call it P/) will be placed inyear two, one period ahead of the first . The value can then bemoved to year ?ero with the P%F factor with n ! /.

0ore to do: Find F for a non5conventional series

To convert a non=conventional uniform series into a future amount,F, the F% factor is used with n equal to the number of periods. The Fvalue is located in the period containing the last payment.

Eample !"): For the cash flow in the previous e5ample, the futureworth in year eight at an interest rate of "+ per year is closest to

a. 32,23"

b. 70,3"+

c. *2,1/3



d. 2-,2"*

Solution: >ow, the n is equal to five and F is located in year eight,as requested.

0ore to do: Find A for a non5conventional series

To convert a non=conventional uniform series cash flow into aconventional uniform series cash flow (i.e., in periods one throughn), the simplest method is to initially find P or F (as described above)

and then use the %P or %F factor.

Eample !"*: For the cash flow in B5ample "3 (2,+++ for *years), the equivalent uniform annual worth, , in years one througheight at i ! "+ per year is closest to

a. /,-0*

b. 3,/+/

c. 3,*//

d. 7,/2/

Solution: Bither the P (B5ample I"3) or the F (B5ample I"7) canbe used to find . 'n either case, n is equal to eight in the %P or %Ffactor. ;alculating P initially (B5ample I"3) to find


What todo:

.ompare Alternatives by PW or AW Analysis



How to doit: present worth (P<) comparison of alternatives involves converting all cash flows to

their present worth and then selecting the one alternative with the lowest cost (orhighest profit). n annual worth (<) analysis, on the other hand, involves convertingall cash flows into equivalent uniform amounts per period (usually years).

Sign .onvention: The sign convention of the FB B5am is used in the analysis below.;osts, such as, first cost and annual operating cost, are given a positive sign,revenues, such as, salvage value, are assigned a negative sign. This is the opposite ofthe te5t material, however it agrees with the FB B5am sign convention.

PW analysis: <hen the alternatives under consideration have different lives, it isnecessary to adopt some procedure, which will yield a comparison for equalservice when using the P< method. The reason for the equal service requirement isobvious, since without it, the alternative with the shortest life is liely to yield thelowest present cost even if it is not the most economical. Jne way to satisfy the equalservice requirement is to compare the alternatives over their least common multiple of years. This will insure that the alternatives under consideration will end at the sametime. @epurchase of each alternative at the same first cost is a common assumption.

AW analysis: 'n the annual worth procedure, it is not necessary to worry about equalservice because the annual worth of one life cycle will be e5actly the same as that fortwo, three, or any number of life cycles. Therefore, compare the < of alternativessimply by calculating each annual worth over the respective life cycle and select theone with the lowest cost (or highest profit).

Eample !"+ company is considering two alternatives for manufacturing a certainpart. Cethod @ will have a first cost of 7+,+++, an annual operating cost of /*,+++,and a "+,+++ salvage value after its five year life. Cethod S will have an initial costof "++,+++, an annual operating cost of "*,+++, and a "/,+++ salvage value afterits "+ year life. t an interest rate of "/ per year, the present worth values of the

two alternatives are closest to

Solution: The least common multiple is "+ years. Cethod @ is repurchased after *years.



Eample !"/ For the alternatives in B5ample "2 above, their annual worth valuesare closest to

Solution: ;alculate < over their respective life cycles of * and "+ years,respectively.


What to do: Find the .apitali8ed .ost of 2nfinite .ash Flow Se9uence

How to do it:

;apitali?ed cost refers to the present worth of cash flows which goon for an infinite period of time. For e5ample, if someone wanted



to now how much money now (a P value) is needed to fund apermanent "+,+++ per year scholarship in their name, thisinvolves a capitali?ed cost calculation. The equation is

P ! %i

Sign .onvention: The sign convention of the FB B5am is used inthe following e5amplesH costs are given a positive sign andrevenues are assigned a negative sign.

Eample !"1 member of congress wants to now thecapitali?ed cost of maintaining a proposed national par. Theannual maintenance cost is e5pected to be /*,+++. t an interestrate of 2 per year, the capitali?ed cost of the maintenance wouldbe closest to

a. ",*++

b. /*,+++

c. "*+,+++

d. 7"2,220

Solution: 'n this problem, ! /*,+++ and i ! +.+2.

0ore to do: 'f the infinite cash flow series occurs in time periodslonger than the stated interest period (for e5ample, every threeyears instead of every year), the easiest way to wor the problemis to convert the recurring cash flow into an value using the %F

factor and then divide by i.

Eample !"6: dam will have a first cost of *,+++,+++, anannual maintenance cost of /*,+++ and minor reconstructioncosts of "++,+++ every five years. t an interest of - per year,the capitali?ed cost of the dam is nearest to



a. /"3,"/*

b. */*,2/*

c. *,3"/,*++

d. *,*/*,2/*

Solution: The "++,+++ which occurs every five years can beconverted to an value using the %F factor. ividing the resulting values by i will yield the capitali?ed cost, Pcap.

'f one want to find the equivalent uniform annual worth (an value) of an infinite cash flow, simply multiply the capitali?ed costby i to obtain .


What to do: Find the .apitali8ed .ost of Finite .ash Flows

How to do it:

s discussed in the previous section, capitali?ed cost refers to thepresent worth of cash flow which goes on for an infinite period of time.'f an asset or alternative has a finite life, its capitali?ed cost isdetermined by first finding the annual worth of the alternative over onelife cycle (which is also its annual worth for infinite service) and thendividing the resulting value by i.

Sign .onvention: The sign convention of the FB B5am is used in thefollowing e5ampleH costs are given a positive sign and revenues areassigned a negative sign.

Eample !#7 n alternative for manufacturing a certain part has afirst cost of *+,+++, an annual cost of "+,+++, and a salvage value of *,+++ after its "+ year life. t an interest rate of "+ per year, thecapitali?ed cost of the alternative is closest to

a. "0,-//



b. "7*,+++

c. "0-,/"*

d. "-1,3**

Solution: Find the equivalent uniform annual worth over one life cycle("+ years) and then divide by i for the capitali?ed cost.


What to do: Find the enefit;.ost 4atio

How to do it:

The benefit%cost ratio (K%;) is an economic analysis technique used commonly,especially by governmental agencies. 'n its purest form, the numerator K consistsof economic consequences to the people (i.e., benefits and disbenefits), while thedenominator ; consists of consequences to the government (i.e., costs andsavings). The units in the calculation can be present worth, annual worth, or

future worth dollars, as long as they are the same in the numerator anddenominator. K%; ratio equal to or greater than " indicates that the pro6ect iseconomically attractive. 'f disbenefits are involved, they are substracted from thebenefitsH if government savings are involved, they are subtracted from the costs.The general K%; is

In B/C analysis, costs are not preceded by a minus sign

Eample !#" federal agency is considering e5panding a national par by



adding recreational facilities. The initial cost of the pro6ect will be ".* million,with an annual upeep cost of *+,+++. Public benefits have been valued at3++,+++ per year, but disbenefits of /++,+++ (initial cost) have also beenrecogni?ed. The par is e5pected to be permanent. t an interest rate of 2 peryear, the K%; ratio is closest to

a. +.0"

b. /.+2

c. /.*+

d. 3.*0

Solution: nnual dollars will be used (arbitrarily chosen over P< or F<) todetermine the K%; ratio. 4se ! Pi to convert present worth estimates to annualworth values.

'nstead of dividing the benefits by the cost to obtain a K%; ratio, the costs couldbe substracted from the benefits (K = ;) to obtain the difference between them. 'f this procedure is followed, a (K = ;) difference of ?ero or greater indicateseconomic attractiveness.


What to do: Find the Present Worth of a ond

How to do it:

bond is a long term note (essentially an 'J4) issued by a corporation orgovernmental entity for the purpose of financing ma6or pro6ects. The borrowerreceives money now in return for a promise to pay later, with interest paid inbetween. The conditions for repayment of the money obtained by the borrowerare specified at the time the bonds are issued. These conditions include the

bond face value, bond interest rate, and bond maturity date.

The bond face value refers to the denomination of the bond (frequently",+++). The face value is important for two reasons (") it represents thelump sum amount the holder will receive on the bond maturity date, and (/) itis used in con6unction with the bond interest rate and bond interest paymentperiod to determine the interest per period the bond holder will receive prior tomaturity. This interest received per period by the bond holder is calculated



according to the following equation

' ! LLLLLLL ( M)(b)LLLLLLLLLLL>o. times interest paid per year

The present worth of a bond represents the amount of money now that isequivalent to the future income or payment stream associated with the bondthe interest, ', received each period and the face value. The bond interestrepresents a uniform series cash flow while the face value, M, represents afuture single payment amount on the bond maturity date. The present worth of a bond can be determined by the following general equation

P<bond ! '(P%,i,n) # M(P%F,i,n)

Eample !## municipal bond with a face value of "+,+++ will mature "*years from now. The bond interest rate is 2 per year, payable quarterly. t aninterest rate of "2 per year compounded quarterly, the present worth of thebond is closest to

a. 7,"03

b. 7,37*

c. *,/00

d. 2,"3*

Solution: The first step is to calculate the bond interest paid per quarter.Then, use this interest as an value and the single amount face value todetermine the present worth. The quarterly interest rate is "2%7 ! 7 for"*(7) ! 2+ quarters.




What to do: Find the Present Worth When 2nflation is .onsidered

How to do it:

There are two ways to tae inflation into account in engineering economicevaluations

". 4se an interest rate that has been corrected for inflation, or

/. ;onvert the cash flows into constant value dollars.

Jnly the first procedure is discussed here. The equation that can be used to ad6ustthe interest rate to account for inflation is the following

if ! i # f # if

'f the inflated interest rate is used in maing present worth calculations, all cashflow amounts are left in :then current: dollars (i.e. inflated or future dollars).

Eample !#( company has the option of building a warehouse now or buildingit three years from now. The cost now would be 7++,+++, but three years fromnow the cost will be *++,+++. 'f the companyGs minimum attractive rate of return(real i) is "/ per year and the inflation rate is "+ per year, the present worthcost of the building in three years when inflation is considered is closest to

a. /2-,0++

b. 3**,1++

c. 30*,2*+

d. 7+/,0++

Solution: First calculate the inflated interest rate, i f . Then, use the inflated interestrate in the P%F formula "%(" # if )

n.




What to do: <etermine Asset <epreciation by S= or 0A.4S 0ethods

How to do it:epreciation is an accounting procedure for systematically reducing thevalue of an asset. epreciation isone of the deductions that reduces ta5ableincome in the general income ta5 equation for corporations.

'ncome ta5 ! (income = deductions) × (ta5 rate)

There are several methods for depreciating an asset but only the twocommonly accepted methods are discussed here Straight Aine (SA) andCodified ccelerated ;ost @ecovery System (C;@S).

The straight=line method is so named because the depreciation charge is thesame each year, resulting in a straight line when the assetGs remaining boo

value (i.e., undepreciated amount, which is discussed in ne5t section) isplotted versus time. The general equation for the annual SA depreciationcharge () is

! K = SMn

Eample !#) machine with a first cost of /*,+++ is e5pected to have a*,+++ salvage value after its five year depreciable life. The depreciationcharge by the straight=line method for year three is closest to

a. 7,+++

b. *,+++

c. "+,+++

d. "*,+++

Solution: ccording to the straight line method, the depreciation charge is

the same each year of the five years.

! /*,+++ = *,+++*

! 7,+++

nswer is (a)



0ore to do: 0A.4S The C;@S method is an accelerated depreciationmethod because more depreciation is charged in early years than lateryears. The annual depreciation rate is tabulated for each acceptabledepreciable life value. The general equation is

! dtKwhere dt ! depreciation rate for year t

K ! first cost or unad6usted base

The dt value is obtained from tables provided by the 4.S. 9overnment. Thedt value is different for each year, decreasing with each year, e5cept betweenyears one and two. The reason for this is that some of the depreciation inyear one is deferred to year (n #"). For e5ample, d t values for a three=yeardepreciable life are 33.33, 77.7*, "7.-", and 0.7" for years t ! ", /,3, and 7, respectively.

>ote that in the C;@S equation for calculating depreciation, the salvagevalue is not subtracted from the first cost as it is in the straight line method.

Eample !#*: machine with a first cost of 7+,+++ is to be depreciatedby the C;@S method. The machine has an estimated "+,+++ salvagevalue after its five year depreciable life. The depreciation charge for year 3 isclosest to

a. *,02+

b. *,1/+

c. 0,2-+

d. "+,+++

Solution: From the C;@S tables, dt for year 3 for a five year recoveryperiod is "1./.


What to do: Find Asset oo> 'alue by S= or 0A.4S

How to do it:

Koo value (KM) represents the remaining, undepreciated amount of an asset after



the depreciation charges to date have been subtracted from the first cost. 'ngeneral equation form, boo value is

For the C;@S method, the depreciation charge is different each year. To find thetotal depreciation, t, the annual depreciation rates must be summed and then

multiplied by K.

Eamples !#+ five=year asset which had a first cost of /+,+++ with a /,+++

salvage value was depreciated by the straight line method. The boo value at theend of year four was closest to

a. 3,2++

b. 7,+++

c. *,2++

d. "2,7++

Solution: ;alculate the annual depreciation charge and use this amount in the boovalue equation.

Eamples !#/ machine with a first cost of -+,+++ is depreciated by theC;@S method. The machine has a depreciable value of "+ years with a "+,+++estimated salvage value. The boo value of the machine after year 2 would be



closest to

a. /3,2++

b. 3+,2*+

c. 71,3*+

d. *2,7++

Solution: From the C;@S depreciation rate table, the rates (in percent) for thefirst si5 years, respectively, are "+, "-, "7.7, "".*/, 1.//, and 0.30, for a total of0+.*". The boo value after si5 years is


What to do: Find the rea>even Point

How to do it:

The most common breaeven analysis problems are composed of two parts a fi5edcost part and a variable cost part. Jften the variable cost is related to the number

of units of something produced or consumed, and in many cases, units is commonto all the alternatives under consideration. Jther times, only one alternative has a

variable cost. 'n either case, the procedure for solving the problem involves settingthe costs of two alternatives (in terms of P, , or F) equal to each other and solvingfor the number of units required for breaeven.

Sign .onvention: The sign convention of the FB B5am is used in the followinge5ampleH costs are given a positive sign and revenues are assigned a negative sign.

Eample !#1 company is considering two methods for obtaining a certain part.Cethod will involve purchasing a machine for *+,+++ with a life of * years, a/,+++ salvage value and a fi5ed annual operating cost of "+,+++. dditionally,each part produced by the method will cost "+.

Cethod K will involve purchasing the part from a subcontractor for /* per part. tan interest rate of "+ per year, the number of parts per year required for the twomethods to brea even is

a. ",333

b. ",*/7



c. ",-*+

d. /,+""

Solution: 'f 5 is the number of parts required per year, the breaeven equation in

terms of annual dollars is

solving typical fe problems

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