Solving Typical FE Problems What to do:
Find P or F from Single Amounts
How to do it: The simplest problems to solve in engineering economic analysis are those which involve finding the value of a single amount of of money at an earlier or later date than that which is given. Such problems involve finding the future worth (F) of a specified present amount (P), or vice versa. These problems involve using the equations F ! P(" # i)n or P or P ! F$"% (" # i)n& 'n terms of standard factor notation, the equation on 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 a money maret account which pays interest at a rate of - per year. The amount that would be in the account at the end of ten years is most nearly a.
/ , 01/
b.
1 , +++
c.
"+,01*
d.
"/,"2*
Solution: The Solution: The *,+++ represents a present amount, P. The future amount, F, is
Eample !#: !#: small company wants to deposit a single amount of money now so that it will have enough to purchase a new truc costing *+,+++ five years from now. 'f the money can be deposited into an account which earns interest at "+ per year, the amount that must be deposited is most
nearly a.
"+,+++
b.
3",+*+
c.
33,/*+
d.
3"1,"2+
Solution: The Solution: The *+,+++ is a future amount in year five that must be moved to the present.
Solving Typical FE Problems What to do:
Find P from a $niform Series %A& and 'ice 'ersa
How to do it: 4niform series cash flows are represented by the symbol . uniform series refers to cash flows which (") occur in consecutive interest periods, and (/) are the same amount each time. To solve for P for these types of problems, the following equation is used
P! 'n standard factor notation, the equation is P ! (P%,i,n). 't is important to note in using this equation that the present worth, P, is located one of the first . 't is also interest period period ahead of important to remember that n must be equal to the number of values and the interest rate, i, must be e5pressed in the same time units as n. For e5ample, if n is in months, i must be an effective interest rate per month. This standard equation can be used in reverse to convert a present worth into a uniform series amount using the f orm ! P(%P,i,n). P(%P,i,n). This, for e5ample, is used to determine the monthly p ayment
nearly a.
"+,+++
b.
3",+*+
c.
33,/*+
d.
3"1,"2+
Solution: The Solution: The *+,+++ is a future amount in year five that must be moved to the present.
Solving Typical FE Problems What to do:
Find P from a $niform Series %A& and 'ice 'ersa
How to do it: 4niform series cash flows are represented by the symbol . uniform series refers to cash flows which (") occur in consecutive interest periods, and (/) are the same amount each time. To solve for P for these types of problems, the following equation is used
P! 'n standard factor notation, the equation is P ! (P%,i,n). 't is important to note in using this equation that the present worth, P, is located one of the first . 't is also interest period period ahead of important to remember that n must be equal to the number of values and the interest rate, i, must be e5pressed in the same time units as n. For e5ample, if n is in months, i must be an effective interest rate per month. This standard equation can be used in reverse to convert a present worth into a uniform series amount using the f orm ! P(%P,i,n). P(%P,i,n). This, for e5ample, is used to determine the monthly p ayment
associated with a car purchase or house loan for a compound interest rate of i.
Eample !(: !(: company e5pects the material cost of a certain manufacturing operation to be /+,+++ per year. t an interest rate of - per year, the present worth of this cost over a five year pro6ect period 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 its investment 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 Solution: The value is per month.
Solving Typical FE Problems What to do:
Find F from a $niform Series %A& and 'ice 'ersa
How to do it: 'n the previous problem type, the procedure for converting a uniform series into an equivalent present amount was discussed. 8ere a uniform series is converted into a future amount instead of a present one. The equation for doing so is
F! The standard notation form is F! (F%,i,n). 't is important to remember that the F occurs in the sameperiod as the last . s before, the n is equal to the number of values and the i used in the calculation must be e5pressed over the same time units as n.
Eample !*: 'f a person deposits "++ per month into an account which pays interest at a rate of 2 per year compounded monthly, the amount in the account at the end of f ive years would be nearest to a.
*27
b.
3,21
c.
2,100
d.
0,11/
Solution Since the cash flow (i.e., values) occurs over monthly interest periods, the n and i must have monthly time units.
s in the previous problem type, the standard equation can be set up and solved in reverse to find an value from a given future worth, F, using ! F(%F,i,n). Eample !+ small company wants to have enough money saved to purchase a new /++,+++ warehouse in five years. 'f the company can invest money at "- per year, the amount that must be invested each year is closest to a.
/0,12+
b.
32,1/+
c.
71,2*+
d.
23,12+
Solution
Solving Typical FE Problems What to do:
Find P, A, or F from $niform -radient .ash Flows
How to do it: uniform gradient cash flow is one wherein the cash flow changes (increases or decreases) by the same amount in each payment period. For e5ample, if the cash flow in period " is -++, and in period two it is 1++, with amounts increasing by "++ in each subsequent period, this is a uniform gradient cash flow series with the gradient, 9, equal to "++. The standard factor equation to find the present worth of the gradient is represented as P ! 9(P%9,i,n). This equation finds the value of only the gradient , not the amount of money that the gradient was :built on: (i.e., the base amount) in period one. The base amount in period one must be handled separately as a uniform series cash flow. Thus, the general equation to find the present worth of a uniform gradient cash f low series is P ! (P%,i,n) # 9(P%9,i,n)
'f the gradient is negative, the total cash flow decreases from one period to the ne5t. The only difference is that the plus sign becomes a minus sign in the equation above.
Eample !/: company e5pects the cost of equipment maintenance to be *,+++ in year one, *,*++ in year two, and amounts increasing by *++ per year through year "+. t an interest rate of "+ per year, the present worth of the maintenance 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 .ash Flows ;onvert a gradient cash flow into an equivalent uniform series, , in one of two ways ".
Find the present worth, P, of the cash flows as discussed above and then convert the P value to an value using the !P(%P,i,n) factor, or
/. 4se the uniform gradient annual worth factor, (%9,i,n), directly in ! 9(%9,i,n).
value, 6ust lie the P%9 factor affects only the gradient. The base amount in year one is simply added 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 found by either first finding P and then using the F%P factor, or first finding and then using the F% factor.
Eample !1 The cash flow associated with a stripper oil well is e5pected to be 3,+++ in month one, /,1*+ in month two, and amounts decreasing by *+ each month through year five. t an interest rate of "/ per year compounded monthly, the equivalent uniform monthly cash flow is closest to a.
",3/0
b.
",203
c.
/,017
d.
7,3/0
Solution The negative gradient is 9 ! =*+. The total equivalent monthly cash flow, T, is
olving Typical FE Problems What to do:
2dentify 3ominal and Effective 2nterest 4ates
How to do it: >ominal and effective interest rates are similar to simple and compound interest rates, with a nominal rate being equivalent to a simple interest rate. 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 a nominal interest rate, it must be converted into an effective rate before it can be used in any of the formulas. The first step in the process of insuring that only effective interest rates are used is to recogni?e whether an interest rate is nominal or effective. Table " shows the three ways interest rates may be stated. Table "5 'arious interest statements and their interpretation
(") 'nterest @ate Statement
i ! "/ per year
i ! " per month
i ! 3="%/ per quarter
(/) 'nterpretation
(3) ;omment
i ! effective "/ per year compounded yearly
i ! - per year, compounded monthly
i ! nominal - per year compounded monthly
i ! 7 per quarter compounded monthly
i ! nominal 7 per quarter compounded monthly
i ! "7 per year compounded semiannually
i ! nominal "7 per year compounded semiannually
i ! effective "+ per year compounded monthly
i ! effective "+ per year compounded monthly
i ! effective 2 per quarter
i ! effective 2 per quarter compounded quarterly
i ! effective " per month compounded daily
i ! effective " per month compounded daily
'f interest rate is stated as an effective rate, then it is an effective rate. 'f compounding period is not given, compounding period is assumed to coincide with stated time period.
The three statements in the top part of the table show that an interest rate can be stated over some designated time period without specifying the compounding period. Such interest rates are assumed to be effective rates with the compounding period (CP) assumed to be the same 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 than the time period over which the interest is stated, and
3.
the interest rate is not designated as either nominal or effective.
'n such cases, the interest rate is assumed to be nominal and the compounding period is equal to that which is stated. (
statements (column ") along with their interpretations (columns / and 3). Table #5 Specific eamples of interest statements and interpretations
(") 'nterest Statement
(/) >ominal or Bffective 'nterest
(3) ;ompounding Period
"* per year compounded monthly
>ominal
Conthly
"* per year
Bffective
Dearly
Bffective "* per year compounded monthly
Bffective
Conthly
/+ per year compounded quarterly
>ominal
Euarterly
>ominal / per month compounded weely
>ominal
/ per month
Bffective
Conthly
/ per month compounded monthly
Bffective
Conthly
Bffective 2 per quarter
Bffective
Euarterly
Bffective / per month compounded daily
Bffective
aily
" per wee compounded continuously
>ominal
;ontinuously
+." per day compounded continuously
>ominal
;ontinuously
Solving Typical FE Problems What to do:
Find an Effective 2nterest 4ate from a 3ominal 4ate and 'ice 'ersa
How to do it: ll of the formulas used in maing time value calculations are based on effective interest rates. Therefore, whenever the interest rate that is provided is a nominal rate, it is necessary to convert it to an effective interest rate. s shown below, an effective interest rate, i, can be calculated for any time period longer than the compounding period. The most common way that nominal interest rates are stated is in the form G5 per year compounded yG where 5 ! interest rate and y ! compounding period. n e5ample is "- per year compounded monthly.
Eample !6 For an interest rate of "/ per year compounded 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 the number of times interest is compounded per year .
Eample !"7 For an interest rate of / per month, the effective semiannual rate is closest to a.
"".**
b.
"/
c.
"/.2/
d.
/2.-/
Solution: 'n this e5ample, the i on the left=hand side of the effective interest rate equation will have units of semiannual periods. Therefore, the r must have units of semiannual periods (i.e., "/ per si5 months) and m must be the number of times interest is compounded p er semiannual period, 2 in this e5ample.
The types of calculations used to obtain effective interest rates are summari?ed in Table 3. Table (5 Summary of .alculations 2nvolved in Finding Effective 4ates
'nterest Statement
To Find i for ;ompounding Period
To Find i for any Period Aonger than ;ompounding Period
i ! " per month
i is already e5pressed over compounding period
4se effective interest rate equation
i ! "/ per year compounded quarterly
ivide "/ by 7
4se effective interestrate equation
i ! nominal "2 per year compounded semiannually
ivide "2 by /
4se effective interest rate equation
i ! effective "7 per year compounded monthly
4se effective interest rate equation and solve for r%m
For effective i values other than yearly, solve for r in effective interest rate equation and then proceed as in previous two e5amples
Solving Typical FE Problems What to do:
Find P or F for Single Payments and 3ominal 2nterest 4ates
How to do it: For problems involving single payment amounts, that is, P and F, there are essentially an infinite number of ways to solve the problems. This is because any effective interest rate can be used in the P%F or F%P factors as long as the n has the same units as the i. That is, if an effective interest rate per
month is used, then n must be the number of months between 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 rate of "/ per year compounded monthly, the amount that will be 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 used directly. The simplest way to wor the problem is to use an interest rate of " per month (an effective rate) because the F%P factor can be looed up directly in the " table. ny other effective interest rate would involve a fraction. (For e5ample, the effective yearly rate is "/.2-, which is not in a factor table.)
'f the effective interest rate of "/.2- per year is used in the e5ample above, it is necessary to use the F%P f ormula to calculate F, since there is no "/.2- interest table. 'f done correctly, however, the answer is the same.
Solving Typical FE Problems What to do:
.alculations 2nvolving $niform Series and 3ominal 2nterest 4ates
How to do it:
the n must always be equal to the number of payments
involved, and /.
the effective interest rate must have the same time units as the payments.
For e5ample, if uniform cash flow values ( or 9) occur over quarterly time periods, the interest rate must be an effective quarterly rate. Similarly, if the cash flow values occur monthly or yearly, the interest rate must be an effective monthly or yearly rate, respectively.
Eample !"# n individual deposits "++ per month into an account which pays interest at a rate of 2 per year compounded monthly. The value of the account af ter five years is 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 in this case is +.*.
Solving Typical FE Problems What to do:
Find P, A, or F from 3on5conventional $niform Series .ash Flows
How to do it:
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 years from now. t an interest rate of "+ per year, the present worth of these 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 in year two, one period ahead of the first . The value can then be moved 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 F value is located in the period containing the last payment.
Eample !"): For the cash flow in the previous e5ample, the future worth 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 a conventional uniform series cash flow (i.e., in periods one through n), 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 through eight 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) can be used to find . 'n either case, n is equal to eight in the %P or %F factor. ;alculating P initially (B5ample I"3) to find
Solving Typical FE Problems What to do:
.ompare Alternatives by PW or AW Analysis
How to do it:
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 (or highest profit). n annual worth (<) analysis, on the other hand, involves converting all 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 oppo site of the te5t material, however it agrees with the FB B5am sign convention. PW analysis:
Eample !"+ company is considering two alternatives fo r manufacturing a certain part. 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 cost of "++,+++, an annual operating cost of "*,+++, and a "/,+++ salvage value after its "+ year life. t an in terest 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 ab ove, their annual worth values are closest to
Solution: ;alculate < over their respective life cycles of * and "+ years, respectively.
Solving Typical FE Problems 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 go on for an infinite per iod of time. For e5ample, if someone wanted
to now how much money now (a P value) is needed to fund a permanent "+,+++ per year scholarship in their name, this involves a capitali?ed cost calculation. The equation is P ! %i
Sign .onvention: The sign convention of the FB B5am is used in the following e5amplesH costs are given a positive sign and revenues are assigned a negative sign.
Eample !"1 member of congress wants to now the capitali?ed cost of maintaining a proposed national par. The annual maintenance cost is e5pected to be /*,+++. t an interest rate of 2 per year, the capitali?ed cost of the maintenance would be 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 periods longer than the stated interest period (for e5ample, every three years instead of every year), the easiest way to wor the problem is 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 *,+++,+++, an annual maintenance cost of /*,+++ and minor reconstruction costs 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 f ive years can be converted to an value using the %F f actor. 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 cost by i to obtain .
Solving Typical FE Problems 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 the present 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 is determined by first finding the annual worth of the alternative over one life cycle (which is also its annual wo rth for infinite service) and then dividing the resulting value by i. Sign .onvention: The sign convention of the FB B5am is used in the following e5ampleH costs are given a positive sign and revenues are assigned a negative sign.
Eample !#7 n alternative for manufacturing a certain part has a first cost of *+,+++, an annual cost of "+,+++, and a salvage value of *,+++ after its "+ year life. t an interest rate of "+ per year, the capitali?ed cost of the alternative is closest to a.
"0,-//
b.
"7*,+++
c.
"0-,/"*
d.
"-1,3**
Solution: Find the equivalent uniform annual wo rth over one life cycle ("+ years) and then divide by i for the capitali?ed cost.
Solving Typical FE Problems 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 consists of economic consequences to the people (i.e., benefits and disbenefits), while the denominator ; consists of consequences to the government (i.e., costs and savings). The units in the calculation can be p resent worth, annual worth, or future worth dollars, as long as they are the same in the numerator and denominator. K%; ratio equal to or greater than " indicates that the pro6ect is economically attractive. 'f disbenefits are involved, they are substracted from the benefitsH 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 at 3++,+++ per year, but disbenefits of /++,+++ (initial cost) have also been recogni?ed. The par is e5pected to be permanent. t an interest rate of 2 per year, 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<) to determine the K%; ratio. 4se ! Pi to convert present worth estimates to annual worth values.
'nstead of dividing the benefits by the cost to obtain a K%; ratio, the costs could be substracted from the benefits (K = ;) to obtain the difference between them. 'f this procedure is followed, a (K = ;) difference of ?ero or greater indicates economic attractiveness.
Solving Typical FE Problems 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 or governmental entity for the purpose of financing ma6or pro6ects. The borrower receives money now in return for a promise to pay later, with interest paid in between. The conditions for repayment of the money obtained by the borrower are specified at the time the bonds are issued. These conditions include the bond face value, bond interest rate, and bo nd maturity date. The bond face value refers to the denomination of the bond (frequently ",+++). The face value is important for two reasons (") it represents the lump sum amount the holder will receive on the bond maturity date, and (/) it is used in con6unction with the bond interest rate and bond interest payment period to determine the interest per period the bond holder will receive prior to maturity. 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 is equivalent to the future income or payment stream associated with the bond the interest, ', received each period and the face value. The bond interest represents a uniform series cash flow while the face value, M, represents a future single payment amount on the bond maturity date. The present worth of a bond can be determined by the following general equation P
Eample !## municipal bond with a face value of "+,+++ will mature "* years from now. The bond interest rate is 2 per year, payable quarterly. t an interest rate of "2 per year compounded q uarterly, the present worth of the bond 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 to determine the present worth. The quarterly interest rate is "2%7 ! 7 for "*(7) ! 2+ quarters.
Solving Typical FE Problems
What to do:
Find the Present Worth When 2nflation is .onsidered
How to do it: There are two ways to tae infla tion into account in engineering economic evaluations ".
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 ad6ust the 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 cash flow amounts are left in :then current: dollars (i.e. inflated or future dollars).
Eample !#( company has the option of building a warehouse now or building it three years from now. The cost now would be 7++,+++, but three years from now 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 worth cost 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 interest rate in the P%F formula "%(" # if )n.
Solving Typical FE Problems What to do:
How to do it: epreciation is an accounting procedure for systematically reducing the value of an asset. epreciation isone of the deductions that reduces ta5able income 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 two commonly accepted methods are discussed here Straight Aine (SA) and Codified ccelerated ;ost @ecovery System (C;@S). The straight=line method is so named because the depreciation charge is the same each year, resulting in a straight line when the assetGs remaining boo value (i.e., undepreciated amount, which is discussed in ne5t section) is plotted versus time. The general equation for the annual SA depreciation charge () is ! K = SM n
Eample !#) machine with a first cost of /*,+++ is e5pected to have a *,+++ salvage value after its five year depreciable life. The depreciation charge 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 depreciation method because more depreciation is charged in early years than later years. The annual depreciation rate is tabulated for each acceptable depreciable life value. The general equation is
dt
! dtK ! depreciation rate for year t
K
! first cost or unad6usted base
where
The dt value is obtained from tables p rovided by the 4.S. 9overnment. The dt value is different for each year, decreasing with each year, e5cept between years one and two. The reason for this is that some of the depreciation in year one is deferred to year (n #"). For e5ample, d t values for a three=year depreciable 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 salvage value 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 depreciated by the C;@S method. The machine has an estimated "+,+++ salvage value after its five year depreciable life. The depreciation charge for year 3 is closest to a.
*,02+
b.
*,1/+
c.
0,2-+
d.
"+,+++
Solution: From the C;@S tables, d t for year 3 for a five year recovery period is "1./.
Solving Typical FE Problems 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. 'n general equation form, boo value is
For the C;@S method, the depreciation charge is different each year. To find the total 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 the end 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 boo value equation.
Eamples !#/ machine with a first cost of -+,+++ is depreciated by the C;@S method. The machine has a depreciable value of "+ years w ith 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 the first si5 years, respectively, are "+, "-, "7.7, "".*/, 1.//, and 0.30, for a total of 0+.*". The boo value after si5 years is
Solving Typical FE Problems What to do:
Find the rea>even Point
How to do it: The most common breaeven analysis problems are composed of two parts a fi5ed cost 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 common to all the alternatives under consideration. Jther times, only one alternative has a variable cost. 'n either case, the procedure for solv ing the problem involves setting the costs of two alternatives (in terms of P, , or F) equal to each other and solving for the number of units requi red for breaeven. Sign .onvention: The sign convention of the FB B5am is used in the following e5ampleH 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. t an interest rate of "+ per year, the number of parts per year required for the two methods to brea even is a.
",333
b.
",*/7
