Solving Typical Fe Problems 1p3764
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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 engineering economic analysis are those which involve finding the value of a single amount 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(1 + i)n or P = F[1/ (1 + i)n] In 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).
Example: #1: A person deposits $5,000 into a money market which pays interest at a rate of 8% per year. The amount that would be in the at the end of ten years is most nearly: a.
$2,792
b.
$9,000
c.
$10,795
d.
$12,165
Solution: The $5,000 represents a present amount, P. The future amount, F, is
Example #2: A small company wants to deposit a single amount of money now so that it will have enough to purchase a new truck costing $50,000 five years from now. If the money can be deposited into an which earns interest at 10% per year, the amount that must be deposited is most
nearly a.
$10,000
b.
$31,050
c.
$33,250
d.
$319,160
Solution: The $50,000 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 Uniform Series (A) and Vice Versa
How to do it: Uniform series cash flows are represented by the symbol A. A uniform series refers to cash flows which: (1) occur in consecutive interest periods, and (2) are the same amount each time. To solve for P for these types of problems, the following equation is used:
P=A In standard factor notation, the equation is P = A(P/A,i,n). It is important to note in using this equation that the present worth, P, is located one interest period ahead of the first A. It is also important to that n must be equal to the number of A values and the interest rate, i, must be expressed in the same time units as n. For example, 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 form A = P(A/P,i,n). This, for example, is used to determine the monthly payment
associated with a car purchase or house loan for a compound interest rate of i%.
Example #3: A company expects the material cost of a certain manufacturing operation to be $20,000 per year. At an interest rate of 8% per year, the present worth of this cost over a five year project period is closest to: a.
$29,386
b.
$56,220
c.
$79,854
d.
$117,332
Solution:
Example #4: A piece of machinery has a first cost of $31,000 with a monthly operating cost of $10,000. If the company wants to recover its investment in five years at an interest rate of 1% per month, the monthly income must be closest to: a.
$5,498
b.
$6,386
c.
$8,295
d.
$10,688
Solution: The A value is per month.
Solving Typical FE Problems What to do:
Find F from a Uniform Series (A) and Vice Versa
How to do it: In the previous problem type, the procedure for converting a uniform series into an equivalent presentamount was discussed. Here a uniform series is converted into a future amount instead of a present one. The equation for doing so is:
F=A The standard notation form is F= A(F/A,i,n). It is important to that the F occurs in the sameperiod as the last A. As before, the n is equal to the number of A values and the i used in the calculation must be expressed over the same time units as n.
Example #5: If a person deposits $100 per month into an which pays interest at a rate of 6% per year compounded monthly, the amount in the at the end of five years would be nearest to: a.
$564
b.
$3,69
c.
$6,977
d.
$7,992
Solution: Since the cash flow (i.e., A values) occurs over monthly interest periods, the n and i must have monthly time units.
As in the previous problem type, the standard equation can be set up and solved in reverse to find an A value from a given future worth, F, using A = F(A/F,i,n). Example #6: A small company wants to have enough money saved to purchase a new $200,000 warehouse in five years. If the company can invest money at 18% per year, the amount that must be invested each year is closest to: a.
$27,960
b.
$36,920
c.
$49,650
d.
$63,960
Solution:
Solving Typical FE Problems What to do:
Find P, A, or F from Uniform Gradient Cash Flows
How to do it: A uniform gradient cash flow is one wherein the cash flow changes (increases or decreases) by the same amount in each payment period. For example, if the cash flow in period 1 is $800, and in period two it is $900, with amounts increasing by $100 in each subsequent period, this is a uniform gradient cash flow series with the gradient, G, equal to $100. The standard factor equation to find the present worth of the gradient is represented as P = G(P/G,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 flow series is: P = A(P/A,i,n) + G(P/G,i,n)
If the gradient is negative, the total cash flow decreases from one period to the next. The only difference is that the plus sign becomes a minus sign in the equation above.
Example #7: A company expects the cost of equipment maintenance to be $5,000 in year one, $5,500 in year two, and amounts increasing by $500 per year through year 10. At an interest rate of 10% per year, the present worth of the maintenance cost is nearest to a.
$38,220
b.
$42,170
c.
$46,660
d.
$51,790
Solution: This is an increasing gradient (use + sign) with G = $500 and base amount A = $5,000:
More to do: Find A from Uniform Gradient Cash Flows Convert a gradient cash flow into an equivalent uniform series, A, in one of two ways: 1.
Find the present worth, P, of the cash flows as discussed above and then convert the P value to an A value using the A =P(A/P,i,n) factor, or
2.
Use the uniform gradient annual worth factor, (A/G,i,n), directly in A= G(A/G,i,n).
When using (A/G,i,n), it is important to that this factor converts only the gradient into an A
value, just like the P/G factor affects only the gradient. The base amount in year one is simply added to the A value obtained from A= G(A/G,i,n). The general equation for the total A value, AT, is AT = A1 + AG
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 A and then using the F/A factor.
Example #8: The cash flow associated with a stripper oil well is expected to be $3,000 in month one, $2,950 in month two, and amounts decreasing by $50 each month through year five. At an interest rate of 12% per year compounded monthly, the equivalent uniform monthly cash flow is closest to: a.
$1,327
b.
$1,673
c.
$2,794
d.
$4,327
Solution: The negative gradient is G = -$50. The total equivalent monthly cash flow, AT, is:
olving Typical FE Problems What to do:
Identify Nominal and Effective Interest Rates
How to do it: Nominal and effective interest rates are similar to simple and compound interest rates, with a nominal rate being equivalent to a simple interest rate. All of the equations expressing 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 recognize whether an interest rate is nominal or effective. Table 1 shows the three ways interest rates may be stated. Table 1- Various interest statements and their interpretation
(1) Interest Rate Statement
i = 12% per year
i = 1% per month
i = 3-1/2% per quarter
(2) Interpretation
(3) Comment
i = effective 12% per year compounded yearly
When no compounding period is given, interest rate is an i = effective effective rate, 1% per month with compounded compounding monthly period i = effective 3- assumed to be equal to stated 1/2% per time period quarter compounded quarterly
i = 8% per year, compounded monthly
i = nominal 8% per year compounded monthly
i = 4% per quarter compounded monthly
i = nominal 4% per quarter compounded monthly
i = 14% per year compounded semiannually
i = nominal 14% per year compounded semiannually
When compounding period is given without stating whether the interest rate is nominal or effective, it is assumed to be nominal. Compounding period is as stated.
i = effective 10% per year compounded monthly
i = effective 10% per year compounded monthly
i = effective 6% per quarter
i = effective 6% per quarter compounded quarterly
i = effective 1% per month compounded daily
i = effective 1% per month compounded daily
If interest rate is stated as an effective rate, then it is an effective rate. If 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 () assumed to be the same as that of the stated interest rate. For the interest statements presented in the middle of Table 1, three conditions prevail: 1.
the compounding period is identified,
2.
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.
In such cases, the interest rate is assumed to be nominal and the compounding period is equal to that which is stated. (We show how to get effective interest rates from these in the next section.) For the third group of interest-rate statements in Table 1, the word effective precedes or follows the specified interest rate and the compounding period is also stated. These interest rates are obviously effective rates over the respective time periods stated. Likewise, the compounding periods are equal to those stated. Similarly, if the word nominal had preceded any of the interest statements, the interest rate would be a nominal rate. Table 2 contains a listing of several interest
statements (column 1) along with their interpretations (columns 2 and 3). Table 2- Specific examples of interest statements and interpretations
(1) Interest Statement
(2) Nominal or Effective Interest
(3) Compounding Period
15% per year compounded monthly
Nominal
Monthly
15% per year
Effective
Yearly
Effective 15% per year compounded monthly
Effective
Monthly
20% per year compounded quarterly
Nominal
Quarterly
Nominal 2% per month compounded weekly
Nominal
Weekly
2% per month
Effective
Monthly
2% per month compounded monthly
Effective
Monthly
Effective 6% per quarter
Effective
Quarterly
Effective 2% per month compounded daily
Effective
Daily
1% per week compounded continuously
Nominal
Continuously
0.1% per day compounded continuously
Nominal
Continuously
Solving Typical FE Problems What to do:
Find an Effective Interest Rate from a Nominal Rate and Vice Versa
How to do it: All of the formulas used in making 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. As 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 'x% per year compounded y' where x = interest rate and y = compounding period. An example is 18% per year compounded monthly. When interest rates are stated this way, the simplest effective rate to get is the one over the compounding period because all that is required is a simple division. For example, from the interest rate of 18% per year compounded monthly, a monthly interest rate of 1.5% is obtained (i.e., 18% per year/12 compounding periods per year) and this is an effective rate because it is the rate per compounding period. To get an effective rate for any period longer than the compounding period use the effective interest rate formula. i = (1+r/m)m - 1
This effective interest rate formula can be solved for r or
r/m as needed to determine a nominal interest rate from an effective rate. For continuous compounding, the effective rate formula is the mathematical limit as m increases without bounds, and the formula reduces to i = er - 1.
Example #9: For an interest rate of 12% per year compounded quarterly, the effective interest rate per year is closest to: a.
4%
b.
12%
c.
12.55%
d.
12.68%
Solution: An effective interest rate per year is sought. Therefore, r must be expressed per year and m is the number of times interest is compounded per year.
Example #10: For an interest rate of 2% per month, the effective semiannual rate is closest to: a.
11.55%
b.
12%
c.
12.62%
d.
26.82%
Solution: In this example, 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., 12% per six months) and m must be the number of times interest is compounded per semiannual period, 6 in this example.
The types of calculations used to obtain effective interest rates are summarized in Table 3. Table 3- Summary of Calculations Involved in Finding Effective Rates
Interest Statement
To Find i for Compounding Period
To Find i for any Period Longer than Compounding Period
i = 1% per month
i is already expressed over compounding period
Use effective interest rate equation
i = 12% per year compounded quarterly
Divide 12% by 4
Use effective interestrate equation
i = nominal 16% per year compounded semiannually
Divide 16% by 2
Use effective interest rate equation
i = effective 14% per year compounded monthly
Use 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 examples
Solving Typical FE Problems What to do:
Find P or F for Single Payments and Nominal Interest Rates
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. If i is an effective interest rate per year, then n must be the number of years.
Example #11: If you deposit $1,000 now at an interest rate of 12% per year compounded monthly, the amount that will be in the savings five years from now is closest to: a.
$1,600
b.
$1,762
c.
$1,817
d.
$1,905
Solution: The 12% rate is a nominal rate and cannot be used directly. The simplest way to work the problem is to use an interest rate of 1% per month (an effective rate) because the F/P factor can be looked up directly in the 1% table. Any other effective interest rate would involve a fraction. (For example, the effective yearly rate is 12.68%, which is not in a factor table.)
If the effective interest rate of 12.68% per year is used in the example above, it is necessary to use the F/P formula to calculate F, since there is no 12.68% interest table. If done correctly, however, the answer is the same.
Solving Typical FE Problems What to do:
Calculations Involving Uniform Series and Nominal Interest Rates
How to do it: When using the uniform series cash flow equations, there are two necessary conditions: 1.
the n must always be equal to the number of payments
involved, and 2.
the effective interest rate must have the same time units as the payments.
For example, if uniform cash flow values (A or G) 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.
Example #12: An individual deposits $100 per month into an which pays interest at a rate of 6% per year compounded monthly. The value of the after five years is closest to: a.
$5,630
b.
$6,210
c.
$6,764
d.
$6,977
Solution: Since cash flow occurs over monthly time periods, the interest rate must be an effective monthly rate, which in this case is 0.5%.
Solving Typical FE Problems What to do:
Find P, A, or F from Non-conventional Uniform Series Cash Flows
How to do it: When a uniform series cash flow begins at a time other than period one, it is called a non-conventionaluniform series. It is nonconventional since determining the present worth requires at least two different factors. This is because the uniform series present worth equation is derived with the P one interest period ahead of the first A value. If the first A value does not occur in period one, the P will not occur at time zero; another factor, usually the P/F factor,
must be utilized to obtain the P value at time zero.
Example #13: The costs associated with a particular process are expected to be $6,000 per year for five years, beginning three years from now. At an interest rate of 10% per year, the present worth of these costs is closest to: a.
$17,088
b.
$18,796
c.
$22,745
d.
$29,210
Solution: If the P/A factor is used, the P (call it P2) will be placed in year two, one period ahead of the first A. The value can then be moved to year zero with the P/F factor with n = 2.
More to do: Find F for a non-conventional series To convert a non-conventional uniform series into a future amount, F, the F/A factor is used with n equal to the number of periods. The F value is located in the period containing the last payment.
Example #14: For the cash flow in the previous example, the future worth in year eight at an interest rate of 10% per year is closest to: a.
$36,631
b.
$47,310
c.
$56,923
d.
$68,615
Solution: Now, the n is equal to five and F is located in year eight, as requested.
More to do: Find A for a non-conventional series To convert a non-conventional uniform series cash flow into a conventional uniform series cash flow (i.e., A in periods one through n), the simplest method is to initially find P or F (as described above) and then use the A/P or A/F factor.
Example #15: For the cash flow in Example 13 ($6,000 for 5 years), the equivalent uniform annual worth, A, in years one through eight at i = 10% per year is closest to: a.
$2,875
b.
$3,202
c.
$3,522
d.
$4,262
Solution: Either the P (Example #13) or the F (Example #14) can be used to find A. In either case, n is equal to eight in the A/P or A/F factor. Calculating P initially (Example #13) to find A:
Solving Typical FE Problems What to do:
Compare Alternatives by PW or AW Analysis
How to do it:
A present worth (PW) 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). An annual worth (AW) analysis, on the other hand, involves converting all cash flows into equivalent uniform amounts per period (usually years). Sign Convention: The sign convention of the FE Exam is used in the analysis below. Costs, 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 of the text material, however it agrees with the FE Exam sign convention. PW analysis: When the alternatives under consideration have different lives, it is necessary to adopt some procedure, which will yield a comparison for equal service when using the PW method. The reason for the equal service requirement is obvious, since without it, the alternative with the shortest life is likely to yield the lowest present cost even if it is not the most economical. One way to satisfy the equal service 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 same time. Repurchase of each alternative at the same first cost is a common assumption. AW analysis: In the annual worth procedure, it is not necessary to worry about equal service because the annual worth of one life cycle will be exactly the same as that for two, three, or any number of life cycles. Therefore, compare the AW of alternatives simply by calculating each annual worth over the respective life cycle and select the one with the lowest cost (or highest profit).
Example #16: A company is considering two alternatives for manufacturing a certain part. Method R will have a first cost of $40,000, an annual operating cost of $25,000, and a $10,000 salvage value after its five year life. Method S will have an initial cost of $100,000, an annual operating cost of $15,000, and a $12,000 salvage value after its 10 year life. At an interest rate of 12% per year, the present worth values of the two alternatives are closest to:
Solution: The least common multiple is 10 years. Method R is repurchased after 5 years.
Example #17: For the alternatives in Example 16 above, their annual worth values are closest to:
Solution: Calculate AW over their respective life cycles of 5 and 10 years, respectively.
Solving Typical FE Problems What to do:
Find the Capitalized Cost of Infinite Cash Flow Sequence
How to do it: Capitalized cost refers to the present worth of cash flows which go on for an infinite period of time. For example, if someone wanted
to know how much money now (a P value) is needed to fund a permanent $10,000 per year scholarship in their name, this involves a capitalized cost calculation. The equation is P = A/i
Sign Convention: The sign convention of the FE Exam is used in the following examples; costs are given a positive sign and revenues are assigned a negative sign.
Example #18: A member of congress wants to know the capitalized cost of maintaining a proposed national park. The annual maintenance cost is expected to be $25,000. At an interest rate of 6% per year, the capitalized cost of the maintenance would be closest to: a.
$1,500
b.
$25,000
c.
$150,000
d.
$416,667
Solution: In this problem, A= $25,000 and i = 0.06.
More to do: If the infinite cash flow series occurs in time periods longer than the stated interest period (for example, every three years instead of every year), the easiest way to work the problem is to convert the recurring cash flow into an A value using the A/F factor and then divide by i.
Example #19: A dam will have a first cost of $5,000,000, an annual maintenance cost of $25,000 and minor reconstruction costs of $100,000 every five years. At an interest of 8% per year, the capitalized cost of the dam is nearest to:
a.
$213,125
b.
$525,625
c.
$5,312,500
d.
$5,525,625
Solution: The $100,000 which occurs every five years can be converted to an A value using the A/F factor. Dividing the resulting A values by i will yield the capitalized cost, Pcap.
If one want to find the equivalent uniform annual worth (an A value) of an infinite cash flow, simply multiply the capitalized cost by i to obtain A.
Solving Typical FE Problems What to do:
Find the Capitalized Cost of Finite Cash Flows
How to do it: As discussed in the previous section, capitalized cost refers to the present worth of cash flow which goes on for an infinite period of time. If an asset or alternative has a finite life, its capitalized cost is determined by first finding the annual worth of the alternative over one life cycle (which is also its annual worth for infinite service) and then dividing the resulting A value by i. Sign Convention: The sign convention of the FE Exam is used in the following example; costs are given a positive sign and revenues are assigned a negative sign.
Example #20: An alternative for manufacturing a certain part has a first cost of $50,000, an annual cost of $10,000, and a salvage value of $5,000 after its 10 year life. At an interest rate of 10% per year, the capitalized cost of the alternative is closest to: a.
$17,822
b.
$145,000
c.
$178,215
d.
$189,355
Solution: Find the equivalent uniform annual worth over one life cycle (10 years) and then divide by i for the capitalized cost.
Solving Typical FE Problems What to do:
Find the Benefit/Cost Ratio
How to do it: The benefit/cost ratio (B/C) is an economic analysis technique used commonly, especially by governmental agencies. In its purest form, the numerator B consists of economic consequences to the people (i.e., benefits and disbenefits), while the denominator C consists of consequences to the government (i.e., costs and savings). 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 and denominator. A B/C ratio equal to or greater than 1 indicates that the project is economically attractive. If disbenefits are involved, they are substracted from the benefits; if government savings are involved, they are subtracted from the costs. The general B/C is:
In B/C analysis, costs are not preceded by a minus sign.
Example #21: A federal agency is considering expanding a national park by
adding recreational facilities. The initial cost of the project will be $1.5 million, with an annual upkeep cost of $50,000. Public benefits have been valued at $300,000 per year, but disbenefits of $200,000 (initial cost) have also been recognized. The park is expected to be permanent. At an interest rate of 6% per year, the B/C ratio is closest to: a.
0.71
b.
2.06
c.
2.50
d.
3.57
Solution: Annual dollars will be used (arbitrarily chosen over PW or FW) to determine the B/C ratio. Use A = Pi to convert present worth estimates to annual worth values.
Instead of dividing the benefits by the cost to obtain a B/C ratio, the costs could be substracted from the benefits (B - C) to obtain the difference between them. If this procedure is followed, a (B - C) difference of zero or greater indicates economic attractiveness.
Solving Typical FE Problems What to do:
Find the Present Worth of a Bond
How to do it: A bond is a long term note (essentially an IOU) issued by a corporation or governmental entity for the purpose of financing major projects. 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 bond maturity date. The bond face value refers to the denomination of the bond (frequently $1,000). The face value is important for two reasons: (1) it represents the lump sum amount the holder will receive on the bond maturity date, and (2) it is used in conjunction 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 I = _______ ( V)(b)___________ No. 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, I, received each period and the face value. The bond interest represents a uniform series cash flow while the face value, V, 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: PWbond = I(P/A,i,n) + V(P/F,i,n) Example #22: A municipal bond with a face value of $10,000 will mature 15 years from now. The bond interest rate is 6% per year, payable quarterly. At an interest rate of 16% per year compounded quarterly, the present worth of the bond is closest to: a.
$4,173
b.
$4,345
c.
$5,277
d.
$6,135
Solution: The first step is to calculate the bond interest paid per quarter. Then, use this interest as an A value and the single amount face value to determine the present worth. The quarterly interest rate is 16%/4 = 4% for 15(4) = 60 quarters.
Solving Typical FE Problems
What to do:
Find the Present Worth When Inflation is Considered
How to do it: There are two ways to take inflation into in engineering economic evaluations: 1.
Use an interest rate that has been corrected for inflation, or
2.
Convert the cash flows into constant value dollars.
Only the first procedure is discussed here. The equation that can be used to adjust the interest rate to for inflation is the following: if = i + f + if
If the inflated interest rate is used in making present worth calculations, all cash flow amounts are left in "then current" dollars (i.e. inflated or future dollars).
Example #23: A company has the option of building a warehouse now or building it three years from now. The cost now would be $400,000, but three years from now the cost will be $500,000. If the company's minimum attractive rate of return (real i) is 12% per year and the inflation rate is 10% per year, the present worth cost of the building in three years when inflation is considered is closest to: a.
$268,700
b.
$355,900
c.
$375,650
d.
$402,700
Solution: First calculate the inflated interest rate, if. Then, use the inflated interest rate in the P/F formula 1/(1 + if)n.
Solving Typical FE Problems What to do:
Determine Asset Depreciation by SL or MACRS Methods
How to do it: Depreciation is an ing procedure for systematically reducing the value of an asset. Depreciation isone of the deductions that reduces taxable income in the general income tax equation for corporations. Income tax = (income - deductions) (tax rate) There are several methods for depreciating an asset but only the two commonly accepted methods are discussed here: Straight Line (SL) and Modified Accelerated Cost Recovery System (MACRS). The straight-line method is so named because the depreciation charge is the same each year, resulting in a straight line when the asset's remaining book value (i.e., undepreciated amount, which is discussed in next section) is plotted versus time. The general equation for the annual SL depreciation charge (D) is: D = B - SV n
Example #24: A machine with a first cost of $25,000 is expected to have a $5,000 salvage value after its five year depreciable life. The depreciation charge by the straight-line method for year three is closest to: a.
$4,000
b.
$5,000
c.
$10,000
d.
$15,000
Solution: According to the straight line method, the depreciation charge is the same each year of the five years. D = 25,000 - 5,000 5 = $4,000 Answer is (a)
More to do: MACRS The MACRS 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
where:
dt
D = dt B = depreciation rate for year t
B
= first cost or unadjusted base
The dt value is obtained from tables provided by the U.S. Government. The dt value is different for each year, decreasing with each year, except between years one and two. The reason for this is that some of the depreciation in year one is deferred to year (n +1). For example, dt values for a three-year depreciable life are 33.33%, 44.45%, 14.81%, and 7.41% for years t = 1, 2, 3, and 4, respectively. Note that in the MACRS equation for calculating depreciation, the salvage value is not subtracted from the first cost as it is in the straight line method.
Example #25: A machine with a first cost of $40,000 is to be depreciated by the MACRS method. The machine has an estimated $10,000 salvage value after its five year depreciable life. The depreciation charge for year 3 is closest to: a.
$5,760
b.
$5,920
c.
$7,680
d.
$10,000
Solution: From the MACRS tables, dt for year 3 for a five year recovery period is 19.2%.
Solving Typical FE Problems What to do:
Find Asset Book Value by SL or MACRS
How to do it: Book value (BV) represents the remaining, undepreciated amount of an asset after
the depreciation charges to date have been subtracted from the first cost. In general equation form, book value is
For the MACRS method, the depreciation charge is different each year. To find the total depreciation, Dt, the annual depreciation rates must be summed and then multiplied by B.
Examples #26: A five-year asset which had a first cost of $20,000 with a $2,000 salvage value was depreciated by the straight line method. The book value at the end of year four was closest to: a.
$3,600
b.
$4,000
c.
$5,600
d.
$16,400
Solution: Calculate the annual depreciation charge and use this amount in the book value equation.
Examples #27: A machine with a first cost of $80,000 is depreciated by the MACRS method. The machine has a depreciable value of 10 years with a $10,000 estimated salvage value. The book value of the machine after year 6 would be
closest to: a.
$23,600
b.
$30,650
c.
$49,350
d.
$56,400
Solution: From the MACRS depreciation rate table, the rates (in percent) for the first six years, respectively, are 10, 18, 14.4, 11.52, 9.22, and 7.37, for a total of 70.51%. The book value after six years is:
Solving Typical FE Problems What to do:
Find the Breakeven Point
How to do it: The most common breakeven analysis problems are composed of two parts: a fixed cost part and a variable cost part. Often 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. Other times, only one alternative has a variable cost. In either case, the procedure for solving the problem involves setting the costs of two alternatives (in of P, A, or F) equal to each other and solving for the number of units required for breakeven. Sign Convention: The sign convention of the FE Exam is used in the following example; costs are given a positive sign and revenues are assigned a negative sign.
Example #28: A company is considering two methods for obtaining a certain part. Method A will involve purchasing a machine for $50,000 with a life of 5 years, a $2,000 salvage value and a fixed annual operating cost of $10,000. Additionally, each part produced by the method will cost $10. Method B will involve purchasing the part from a subcontractor for $25 per part. At an interest rate of 10% per year, the number of parts per year required for the two methods to break even is a.
1,333
b.
1,524
c.
1,850
d.
2,011
Solution: If x is the number of parts required per year, the breakeven equation in of annual dollars is:
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 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(1 + i)n or P = F[1/ (1 + i)n] In 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).
Example: #1: A person deposits $5,000 into a money market which pays interest at a rate of 8% per year. The amount that would be in the at the end of ten years is most nearly: a.
$2,792
b.
$9,000
c.
$10,795
d.
$12,165
Solution: The $5,000 represents a present amount, P. The future amount, F, is
Example #2: A small company wants to deposit a single amount of money now so that it will have enough to purchase a new truck costing $50,000 five years from now. If the money can be deposited into an which earns interest at 10% per year, the amount that must be deposited is most
nearly a.
$10,000
b.
$31,050
c.
$33,250
d.
$319,160
Solution: The $50,000 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 Uniform Series (A) and Vice Versa
How to do it: Uniform series cash flows are represented by the symbol A. A uniform series refers to cash flows which: (1) occur in consecutive interest periods, and (2) are the same amount each time. To solve for P for these types of problems, the following equation is used:
P=A In standard factor notation, the equation is P = A(P/A,i,n). It is important to note in using this equation that the present worth, P, is located one interest period ahead of the first A. It is also important to that n must be equal to the number of A values and the interest rate, i, must be expressed in the same time units as n. For example, 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 form A = P(A/P,i,n). This, for example, is used to determine the monthly payment
associated with a car purchase or house loan for a compound interest rate of i%.
Example #3: A company expects the material cost of a certain manufacturing operation to be $20,000 per year. At an interest rate of 8% per year, the present worth of this cost over a five year project period is closest to: a.
$29,386
b.
$56,220
c.
$79,854
d.
$117,332
Solution:
Example #4: A piece of machinery has a first cost of $31,000 with a monthly operating cost of $10,000. If the company wants to recover its investment in five years at an interest rate of 1% per month, the monthly income must be closest to: a.
$5,498
b.
$6,386
c.
$8,295
d.
$10,688
Solution: The A value is per month.
Solving Typical FE Problems What to do:
Find F from a Uniform Series (A) and Vice Versa
How to do it: In the previous problem type, the procedure for converting a uniform series into an equivalent presentamount was discussed. Here a uniform series is converted into a future amount instead of a present one. The equation for doing so is:
F=A The standard notation form is F= A(F/A,i,n). It is important to that the F occurs in the sameperiod as the last A. As before, the n is equal to the number of A values and the i used in the calculation must be expressed over the same time units as n.
Example #5: If a person deposits $100 per month into an which pays interest at a rate of 6% per year compounded monthly, the amount in the at the end of five years would be nearest to: a.
$564
b.
$3,69
c.
$6,977
d.
$7,992
Solution: Since the cash flow (i.e., A values) occurs over monthly interest periods, the n and i must have monthly time units.
As in the previous problem type, the standard equation can be set up and solved in reverse to find an A value from a given future worth, F, using A = F(A/F,i,n). Example #6: A small company wants to have enough money saved to purchase a new $200,000 warehouse in five years. If the company can invest money at 18% per year, the amount that must be invested each year is closest to: a.
$27,960
b.
$36,920
c.
$49,650
d.
$63,960
Solution:
Solving Typical FE Problems What to do:
Find P, A, or F from Uniform Gradient Cash Flows
How to do it: A uniform gradient cash flow is one wherein the cash flow changes (increases or decreases) by the same amount in each payment period. For example, if the cash flow in period 1 is $800, and in period two it is $900, with amounts increasing by $100 in each subsequent period, this is a uniform gradient cash flow series with the gradient, G, equal to $100. The standard factor equation to find the present worth of the gradient is represented as P = G(P/G,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 flow series is: P = A(P/A,i,n) + G(P/G,i,n)
If the gradient is negative, the total cash flow decreases from one period to the next. The only difference is that the plus sign becomes a minus sign in the equation above.
Example #7: A company expects the cost of equipment maintenance to be $5,000 in year one, $5,500 in year two, and amounts increasing by $500 per year through year 10. At an interest rate of 10% per year, the present worth of the maintenance cost is nearest to a.
$38,220
b.
$42,170
c.
$46,660
d.
$51,790
Solution: This is an increasing gradient (use + sign) with G = $500 and base amount A = $5,000:
More to do: Find A from Uniform Gradient Cash Flows Convert a gradient cash flow into an equivalent uniform series, A, in one of two ways: 1.
Find the present worth, P, of the cash flows as discussed above and then convert the P value to an A value using the A =P(A/P,i,n) factor, or
2.
Use the uniform gradient annual worth factor, (A/G,i,n), directly in A= G(A/G,i,n).
When using (A/G,i,n), it is important to that this factor converts only the gradient into an A
value, just like the P/G factor affects only the gradient. The base amount in year one is simply added to the A value obtained from A= G(A/G,i,n). The general equation for the total A value, AT, is AT = A1 + AG
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 A and then using the F/A factor.
Example #8: The cash flow associated with a stripper oil well is expected to be $3,000 in month one, $2,950 in month two, and amounts decreasing by $50 each month through year five. At an interest rate of 12% per year compounded monthly, the equivalent uniform monthly cash flow is closest to: a.
$1,327
b.
$1,673
c.
$2,794
d.
$4,327
Solution: The negative gradient is G = -$50. The total equivalent monthly cash flow, AT, is:
olving Typical FE Problems What to do:
Identify Nominal and Effective Interest Rates
How to do it: Nominal and effective interest rates are similar to simple and compound interest rates, with a nominal rate being equivalent to a simple interest rate. All of the equations expressing 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 recognize whether an interest rate is nominal or effective. Table 1 shows the three ways interest rates may be stated. Table 1- Various interest statements and their interpretation
(1) Interest Rate Statement
i = 12% per year
i = 1% per month
i = 3-1/2% per quarter
(2) Interpretation
(3) Comment
i = effective 12% per year compounded yearly
When no compounding period is given, interest rate is an i = effective effective rate, 1% per month with compounded compounding monthly period i = effective 3- assumed to be equal to stated 1/2% per time period quarter compounded quarterly
i = 8% per year, compounded monthly
i = nominal 8% per year compounded monthly
i = 4% per quarter compounded monthly
i = nominal 4% per quarter compounded monthly
i = 14% per year compounded semiannually
i = nominal 14% per year compounded semiannually
When compounding period is given without stating whether the interest rate is nominal or effective, it is assumed to be nominal. Compounding period is as stated.
i = effective 10% per year compounded monthly
i = effective 10% per year compounded monthly
i = effective 6% per quarter
i = effective 6% per quarter compounded quarterly
i = effective 1% per month compounded daily
i = effective 1% per month compounded daily
If interest rate is stated as an effective rate, then it is an effective rate. If 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 () assumed to be the same as that of the stated interest rate. For the interest statements presented in the middle of Table 1, three conditions prevail: 1.
the compounding period is identified,
2.
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.
In such cases, the interest rate is assumed to be nominal and the compounding period is equal to that which is stated. (We show how to get effective interest rates from these in the next section.) For the third group of interest-rate statements in Table 1, the word effective precedes or follows the specified interest rate and the compounding period is also stated. These interest rates are obviously effective rates over the respective time periods stated. Likewise, the compounding periods are equal to those stated. Similarly, if the word nominal had preceded any of the interest statements, the interest rate would be a nominal rate. Table 2 contains a listing of several interest
statements (column 1) along with their interpretations (columns 2 and 3). Table 2- Specific examples of interest statements and interpretations
(1) Interest Statement
(2) Nominal or Effective Interest
(3) Compounding Period
15% per year compounded monthly
Nominal
Monthly
15% per year
Effective
Yearly
Effective 15% per year compounded monthly
Effective
Monthly
20% per year compounded quarterly
Nominal
Quarterly
Nominal 2% per month compounded weekly
Nominal
Weekly
2% per month
Effective
Monthly
2% per month compounded monthly
Effective
Monthly
Effective 6% per quarter
Effective
Quarterly
Effective 2% per month compounded daily
Effective
Daily
1% per week compounded continuously
Nominal
Continuously
0.1% per day compounded continuously
Nominal
Continuously
Solving Typical FE Problems What to do:
Find an Effective Interest Rate from a Nominal Rate and Vice Versa
How to do it: All of the formulas used in making 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. As 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 'x% per year compounded y' where x = interest rate and y = compounding period. An example is 18% per year compounded monthly. When interest rates are stated this way, the simplest effective rate to get is the one over the compounding period because all that is required is a simple division. For example, from the interest rate of 18% per year compounded monthly, a monthly interest rate of 1.5% is obtained (i.e., 18% per year/12 compounding periods per year) and this is an effective rate because it is the rate per compounding period. To get an effective rate for any period longer than the compounding period use the effective interest rate formula. i = (1+r/m)m - 1
This effective interest rate formula can be solved for r or
r/m as needed to determine a nominal interest rate from an effective rate. For continuous compounding, the effective rate formula is the mathematical limit as m increases without bounds, and the formula reduces to i = er - 1.
Example #9: For an interest rate of 12% per year compounded quarterly, the effective interest rate per year is closest to: a.
4%
b.
12%
c.
12.55%
d.
12.68%
Solution: An effective interest rate per year is sought. Therefore, r must be expressed per year and m is the number of times interest is compounded per year.
Example #10: For an interest rate of 2% per month, the effective semiannual rate is closest to: a.
11.55%
b.
12%
c.
12.62%
d.
26.82%
Solution: In this example, 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., 12% per six months) and m must be the number of times interest is compounded per semiannual period, 6 in this example.
The types of calculations used to obtain effective interest rates are summarized in Table 3. Table 3- Summary of Calculations Involved in Finding Effective Rates
Interest Statement
To Find i for Compounding Period
To Find i for any Period Longer than Compounding Period
i = 1% per month
i is already expressed over compounding period
Use effective interest rate equation
i = 12% per year compounded quarterly
Divide 12% by 4
Use effective interestrate equation
i = nominal 16% per year compounded semiannually
Divide 16% by 2
Use effective interest rate equation
i = effective 14% per year compounded monthly
Use 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 examples
Solving Typical FE Problems What to do:
Find P or F for Single Payments and Nominal Interest Rates
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. If i is an effective interest rate per year, then n must be the number of years.
Example #11: If you deposit $1,000 now at an interest rate of 12% per year compounded monthly, the amount that will be in the savings five years from now is closest to: a.
$1,600
b.
$1,762
c.
$1,817
d.
$1,905
Solution: The 12% rate is a nominal rate and cannot be used directly. The simplest way to work the problem is to use an interest rate of 1% per month (an effective rate) because the F/P factor can be looked up directly in the 1% table. Any other effective interest rate would involve a fraction. (For example, the effective yearly rate is 12.68%, which is not in a factor table.)
If the effective interest rate of 12.68% per year is used in the example above, it is necessary to use the F/P formula to calculate F, since there is no 12.68% interest table. If done correctly, however, the answer is the same.
Solving Typical FE Problems What to do:
Calculations Involving Uniform Series and Nominal Interest Rates
How to do it: When using the uniform series cash flow equations, there are two necessary conditions: 1.
the n must always be equal to the number of payments
involved, and 2.
the effective interest rate must have the same time units as the payments.
For example, if uniform cash flow values (A or G) 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.
Example #12: An individual deposits $100 per month into an which pays interest at a rate of 6% per year compounded monthly. The value of the after five years is closest to: a.
$5,630
b.
$6,210
c.
$6,764
d.
$6,977
Solution: Since cash flow occurs over monthly time periods, the interest rate must be an effective monthly rate, which in this case is 0.5%.
Solving Typical FE Problems What to do:
Find P, A, or F from Non-conventional Uniform Series Cash Flows
How to do it: When a uniform series cash flow begins at a time other than period one, it is called a non-conventionaluniform series. It is nonconventional since determining the present worth requires at least two different factors. This is because the uniform series present worth equation is derived with the P one interest period ahead of the first A value. If the first A value does not occur in period one, the P will not occur at time zero; another factor, usually the P/F factor,
must be utilized to obtain the P value at time zero.
Example #13: The costs associated with a particular process are expected to be $6,000 per year for five years, beginning three years from now. At an interest rate of 10% per year, the present worth of these costs is closest to: a.
$17,088
b.
$18,796
c.
$22,745
d.
$29,210
Solution: If the P/A factor is used, the P (call it P2) will be placed in year two, one period ahead of the first A. The value can then be moved to year zero with the P/F factor with n = 2.
More to do: Find F for a non-conventional series To convert a non-conventional uniform series into a future amount, F, the F/A factor is used with n equal to the number of periods. The F value is located in the period containing the last payment.
Example #14: For the cash flow in the previous example, the future worth in year eight at an interest rate of 10% per year is closest to: a.
$36,631
b.
$47,310
c.
$56,923
d.
$68,615
Solution: Now, the n is equal to five and F is located in year eight, as requested.
More to do: Find A for a non-conventional series To convert a non-conventional uniform series cash flow into a conventional uniform series cash flow (i.e., A in periods one through n), the simplest method is to initially find P or F (as described above) and then use the A/P or A/F factor.
Example #15: For the cash flow in Example 13 ($6,000 for 5 years), the equivalent uniform annual worth, A, in years one through eight at i = 10% per year is closest to: a.
$2,875
b.
$3,202
c.
$3,522
d.
$4,262
Solution: Either the P (Example #13) or the F (Example #14) can be used to find A. In either case, n is equal to eight in the A/P or A/F factor. Calculating P initially (Example #13) to find A:
Solving Typical FE Problems What to do:
Compare Alternatives by PW or AW Analysis
How to do it:
A present worth (PW) 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). An annual worth (AW) analysis, on the other hand, involves converting all cash flows into equivalent uniform amounts per period (usually years). Sign Convention: The sign convention of the FE Exam is used in the analysis below. Costs, 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 of the text material, however it agrees with the FE Exam sign convention. PW analysis: When the alternatives under consideration have different lives, it is necessary to adopt some procedure, which will yield a comparison for equal service when using the PW method. The reason for the equal service requirement is obvious, since without it, the alternative with the shortest life is likely to yield the lowest present cost even if it is not the most economical. One way to satisfy the equal service 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 same time. Repurchase of each alternative at the same first cost is a common assumption. AW analysis: In the annual worth procedure, it is not necessary to worry about equal service because the annual worth of one life cycle will be exactly the same as that for two, three, or any number of life cycles. Therefore, compare the AW of alternatives simply by calculating each annual worth over the respective life cycle and select the one with the lowest cost (or highest profit).
Example #16: A company is considering two alternatives for manufacturing a certain part. Method R will have a first cost of $40,000, an annual operating cost of $25,000, and a $10,000 salvage value after its five year life. Method S will have an initial cost of $100,000, an annual operating cost of $15,000, and a $12,000 salvage value after its 10 year life. At an interest rate of 12% per year, the present worth values of the two alternatives are closest to:
Solution: The least common multiple is 10 years. Method R is repurchased after 5 years.
Example #17: For the alternatives in Example 16 above, their annual worth values are closest to:
Solution: Calculate AW over their respective life cycles of 5 and 10 years, respectively.
Solving Typical FE Problems What to do:
Find the Capitalized Cost of Infinite Cash Flow Sequence
How to do it: Capitalized cost refers to the present worth of cash flows which go on for an infinite period of time. For example, if someone wanted
to know how much money now (a P value) is needed to fund a permanent $10,000 per year scholarship in their name, this involves a capitalized cost calculation. The equation is P = A/i
Sign Convention: The sign convention of the FE Exam is used in the following examples; costs are given a positive sign and revenues are assigned a negative sign.
Example #18: A member of congress wants to know the capitalized cost of maintaining a proposed national park. The annual maintenance cost is expected to be $25,000. At an interest rate of 6% per year, the capitalized cost of the maintenance would be closest to: a.
$1,500
b.
$25,000
c.
$150,000
d.
$416,667
Solution: In this problem, A= $25,000 and i = 0.06.
More to do: If the infinite cash flow series occurs in time periods longer than the stated interest period (for example, every three years instead of every year), the easiest way to work the problem is to convert the recurring cash flow into an A value using the A/F factor and then divide by i.
Example #19: A dam will have a first cost of $5,000,000, an annual maintenance cost of $25,000 and minor reconstruction costs of $100,000 every five years. At an interest of 8% per year, the capitalized cost of the dam is nearest to:
a.
$213,125
b.
$525,625
c.
$5,312,500
d.
$5,525,625
Solution: The $100,000 which occurs every five years can be converted to an A value using the A/F factor. Dividing the resulting A values by i will yield the capitalized cost, Pcap.
If one want to find the equivalent uniform annual worth (an A value) of an infinite cash flow, simply multiply the capitalized cost by i to obtain A.
Solving Typical FE Problems What to do:
Find the Capitalized Cost of Finite Cash Flows
How to do it: As discussed in the previous section, capitalized cost refers to the present worth of cash flow which goes on for an infinite period of time. If an asset or alternative has a finite life, its capitalized cost is determined by first finding the annual worth of the alternative over one life cycle (which is also its annual worth for infinite service) and then dividing the resulting A value by i. Sign Convention: The sign convention of the FE Exam is used in the following example; costs are given a positive sign and revenues are assigned a negative sign.
Example #20: An alternative for manufacturing a certain part has a first cost of $50,000, an annual cost of $10,000, and a salvage value of $5,000 after its 10 year life. At an interest rate of 10% per year, the capitalized cost of the alternative is closest to: a.
$17,822
b.
$145,000
c.
$178,215
d.
$189,355
Solution: Find the equivalent uniform annual worth over one life cycle (10 years) and then divide by i for the capitalized cost.
Solving Typical FE Problems What to do:
Find the Benefit/Cost Ratio
How to do it: The benefit/cost ratio (B/C) is an economic analysis technique used commonly, especially by governmental agencies. In its purest form, the numerator B consists of economic consequences to the people (i.e., benefits and disbenefits), while the denominator C consists of consequences to the government (i.e., costs and savings). 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 and denominator. A B/C ratio equal to or greater than 1 indicates that the project is economically attractive. If disbenefits are involved, they are substracted from the benefits; if government savings are involved, they are subtracted from the costs. The general B/C is:
In B/C analysis, costs are not preceded by a minus sign.
Example #21: A federal agency is considering expanding a national park by
adding recreational facilities. The initial cost of the project will be $1.5 million, with an annual upkeep cost of $50,000. Public benefits have been valued at $300,000 per year, but disbenefits of $200,000 (initial cost) have also been recognized. The park is expected to be permanent. At an interest rate of 6% per year, the B/C ratio is closest to: a.
0.71
b.
2.06
c.
2.50
d.
3.57
Solution: Annual dollars will be used (arbitrarily chosen over PW or FW) to determine the B/C ratio. Use A = Pi to convert present worth estimates to annual worth values.
Instead of dividing the benefits by the cost to obtain a B/C ratio, the costs could be substracted from the benefits (B - C) to obtain the difference between them. If this procedure is followed, a (B - C) difference of zero or greater indicates economic attractiveness.
Solving Typical FE Problems What to do:
Find the Present Worth of a Bond
How to do it: A bond is a long term note (essentially an IOU) issued by a corporation or governmental entity for the purpose of financing major projects. 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 bond maturity date. The bond face value refers to the denomination of the bond (frequently $1,000). The face value is important for two reasons: (1) it represents the lump sum amount the holder will receive on the bond maturity date, and (2) it is used in conjunction 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 I = _______ ( V)(b)___________ No. 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, I, received each period and the face value. The bond interest represents a uniform series cash flow while the face value, V, 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: PWbond = I(P/A,i,n) + V(P/F,i,n) Example #22: A municipal bond with a face value of $10,000 will mature 15 years from now. The bond interest rate is 6% per year, payable quarterly. At an interest rate of 16% per year compounded quarterly, the present worth of the bond is closest to: a.
$4,173
b.
$4,345
c.
$5,277
d.
$6,135
Solution: The first step is to calculate the bond interest paid per quarter. Then, use this interest as an A value and the single amount face value to determine the present worth. The quarterly interest rate is 16%/4 = 4% for 15(4) = 60 quarters.
Solving Typical FE Problems
What to do:
Find the Present Worth When Inflation is Considered
How to do it: There are two ways to take inflation into in engineering economic evaluations: 1.
Use an interest rate that has been corrected for inflation, or
2.
Convert the cash flows into constant value dollars.
Only the first procedure is discussed here. The equation that can be used to adjust the interest rate to for inflation is the following: if = i + f + if
If the inflated interest rate is used in making present worth calculations, all cash flow amounts are left in "then current" dollars (i.e. inflated or future dollars).
Example #23: A company has the option of building a warehouse now or building it three years from now. The cost now would be $400,000, but three years from now the cost will be $500,000. If the company's minimum attractive rate of return (real i) is 12% per year and the inflation rate is 10% per year, the present worth cost of the building in three years when inflation is considered is closest to: a.
$268,700
b.
$355,900
c.
$375,650
d.
$402,700
Solution: First calculate the inflated interest rate, if. Then, use the inflated interest rate in the P/F formula 1/(1 + if)n.
Solving Typical FE Problems What to do:
Determine Asset Depreciation by SL or MACRS Methods
How to do it: Depreciation is an ing procedure for systematically reducing the value of an asset. Depreciation isone of the deductions that reduces taxable income in the general income tax equation for corporations. Income tax = (income - deductions) (tax rate) There are several methods for depreciating an asset but only the two commonly accepted methods are discussed here: Straight Line (SL) and Modified Accelerated Cost Recovery System (MACRS). The straight-line method is so named because the depreciation charge is the same each year, resulting in a straight line when the asset's remaining book value (i.e., undepreciated amount, which is discussed in next section) is plotted versus time. The general equation for the annual SL depreciation charge (D) is: D = B - SV n
Example #24: A machine with a first cost of $25,000 is expected to have a $5,000 salvage value after its five year depreciable life. The depreciation charge by the straight-line method for year three is closest to: a.
$4,000
b.
$5,000
c.
$10,000
d.
$15,000
Solution: According to the straight line method, the depreciation charge is the same each year of the five years. D = 25,000 - 5,000 5 = $4,000 Answer is (a)
More to do: MACRS The MACRS 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
where:
dt
D = dt B = depreciation rate for year t
B
= first cost or unadjusted base
The dt value is obtained from tables provided by the U.S. Government. The dt value is different for each year, decreasing with each year, except between years one and two. The reason for this is that some of the depreciation in year one is deferred to year (n +1). For example, dt values for a three-year depreciable life are 33.33%, 44.45%, 14.81%, and 7.41% for years t = 1, 2, 3, and 4, respectively. Note that in the MACRS equation for calculating depreciation, the salvage value is not subtracted from the first cost as it is in the straight line method.
Example #25: A machine with a first cost of $40,000 is to be depreciated by the MACRS method. The machine has an estimated $10,000 salvage value after its five year depreciable life. The depreciation charge for year 3 is closest to: a.
$5,760
b.
$5,920
c.
$7,680
d.
$10,000
Solution: From the MACRS tables, dt for year 3 for a five year recovery period is 19.2%.
Solving Typical FE Problems What to do:
Find Asset Book Value by SL or MACRS
How to do it: Book value (BV) represents the remaining, undepreciated amount of an asset after
the depreciation charges to date have been subtracted from the first cost. In general equation form, book value is
For the MACRS method, the depreciation charge is different each year. To find the total depreciation, Dt, the annual depreciation rates must be summed and then multiplied by B.
Examples #26: A five-year asset which had a first cost of $20,000 with a $2,000 salvage value was depreciated by the straight line method. The book value at the end of year four was closest to: a.
$3,600
b.
$4,000
c.
$5,600
d.
$16,400
Solution: Calculate the annual depreciation charge and use this amount in the book value equation.
Examples #27: A machine with a first cost of $80,000 is depreciated by the MACRS method. The machine has a depreciable value of 10 years with a $10,000 estimated salvage value. The book value of the machine after year 6 would be
closest to: a.
$23,600
b.
$30,650
c.
$49,350
d.
$56,400
Solution: From the MACRS depreciation rate table, the rates (in percent) for the first six years, respectively, are 10, 18, 14.4, 11.52, 9.22, and 7.37, for a total of 70.51%. The book value after six years is:
Solving Typical FE Problems What to do:
Find the Breakeven Point
How to do it: The most common breakeven analysis problems are composed of two parts: a fixed cost part and a variable cost part. Often 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. Other times, only one alternative has a variable cost. In either case, the procedure for solving the problem involves setting the costs of two alternatives (in of P, A, or F) equal to each other and solving for the number of units required for breakeven. Sign Convention: The sign convention of the FE Exam is used in the following example; costs are given a positive sign and revenues are assigned a negative sign.
Example #28: A company is considering two methods for obtaining a certain part. Method A will involve purchasing a machine for $50,000 with a life of 5 years, a $2,000 salvage value and a fixed annual operating cost of $10,000. Additionally, each part produced by the method will cost $10. Method B will involve purchasing the part from a subcontractor for $25 per part. At an interest rate of 10% per year, the number of parts per year required for the two methods to break even is a.
1,333
b.
1,524
c.
1,850
d.
2,011
Solution: If x is the number of parts required per year, the breakeven equation in of annual dollars is:
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