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© 2012 by McGraw-Hill, New York, N.Y All Rights Reserved 1-1
Lecture slides to accompany
Engineering Economy 7th edition
Leland Blank
Anthony Tarquin
Chapter 1 Foundations Of
Engineering Economy
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© 2012 by McGraw-Hill, New York, N.Y All Rights Reserved 1-2
LEARNING OUTCOMES
1. Role in decision making
2. Study approach
3. Ethics and economics
4. Interest rate
5. Terms and symbols
6. Cash flows
7. Economic equivalence
8. Simple and compound interest
9. Minimum attractive rate of return
10. Spreadsheet functions
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© 2012 by McGraw-Hill, New York, N.Y All Rights Reserved 1-3
Why Engineering Economy is Important to Engineers
Engineers design and create Designing involves economic decisions Engineers must be able to incorporate economic
analysis into their creative efforts Often engineers must select and implement from
multiple alternatives Understanding and applying time value of money,
economic equivalence, and cost estimation are vital for engineers
A proper economic analysis for selection and execution is a fundamental task of engineering
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Time Value of Money (TVM)
Description: TVM explains the change in the amount of money over time for funds owed by or owned by a corporation (or individual)
Corporate investments are expected to earn a return Investment involves money Money has a ‘time value’
The time value of money is the most
important concept in engineering economy
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Engineering Economy Engineering Economy involves Formulating Estimating, and Evaluating expected economic outcomes of alternatives
designed to accomplish a defined purpose Easy-to-use math techniques simplify the
evaluation Estimates of economic outcomes can be
deterministic or stochastic in nature
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© 2012 by McGraw-Hill, New York, N.Y All Rights Reserved 1-6
General Steps for Decision Making Processes
1. Understand the problem – define objectives 2. Collect relevant information 3. Define the set of feasible alternatives 4. Identify the criteria for decision making 5. Evaluate the alternatives and apply
sensitivity analysis 6. Select the “best” alternative 7. Implement the alternative and monitor
results
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Steps in an Engineering Economy Study
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Ethics – Different Levels Universal morals or ethics – Fundamental
beliefs: stealing, lying, harming or murdering another are wrong Personal morals or ethics – Beliefs that an
individual has and maintains over time; how a universal moral is interpreted and used by each person Professional or engineering ethics – Formal
standard or code that guides a person in work activities and decision making
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Code of Ethics for Engineers All disciplines have a formal code of ethics. National Society of
Professional Engineers (NSPE) maintains a code specifically for engineers; many engineering professional societies have their own code
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Interest and Interest Rate Interest – the manifestation of the time value of money
• Fee that one pays to use someone else’s money • Difference between an ending amount of money and
a beginning amount of money
Interest = amount owed now – principal
Interest rate – Interest paid over a time period expressed as a percentage of principal
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Rate of Return
Interest earned over a period of time is expressed as a percentage of the original amount (principal)
interest accrued per time unitRate of return (%) = x 100%original amount
Borrower’s perspective – interest rate paid
Lender’s or investor’s perspective – rate of return earned
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Interest paid Interest earned
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Interest rate Rate of return
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Commonly used Symbols t = time, usually in periods such as years or months P = value or amount of money at a time t
designated as present or time 0 F = value or amount of money at some future
time, such as at t = n periods in the future A = series of consecutive, equal, end-of-period
amounts of money n = number of interest periods; years, months i = interest rate or rate of return per time period;
percent per year or month
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Cash Flows: Terms Cash Inflows – Revenues (R), receipts,
incomes, savings generated by projects and activities that flow in. Plus sign used Cash Outflows – Disbursements (D), costs,
expenses, taxes caused by projects and activities that flow out. Minus sign used
Net Cash Flow (NCF) for each time period: NCF = cash inflows – cash outflows = R – D
End-of-period assumption: Funds flow at the end of a given interest period
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Cash Flows: Estimating Point estimate – A single-value estimate of a
cash flow element of an alternative Cash inflow: Income = $150,000 per month
Range estimate – Min and max values that estimate the cash flow
Cash outflow: Cost is between $2.5 M and $3.2 M
Point estimates are commonly used; however, range estimates with probabilities attached provide a better understanding of variability of economic parameters used to make decisions
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Cash Flow Diagrams What a typical cash flow diagram might look like
0 1 2 … … … n - 1 n
Draw a time line
One time period
0 1 2 … … … n-1 n
Show the cash flows (to approximate scale)
Cash flows are shown as directed arrows: + (up) for inflow
- (down) for outflow
Always assume end-of-period cash flows
Time
F = $100
P = $-80
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Cash Flow Diagram Example
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Plot observed cash flows over last 8 years and estimated sale next year for $150. Show present worth (P) arrow at present time, t = 0
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Economic Equivalence Definition: Combination of interest rate (rate of
return) and time value of money to determine different amounts of money at different points in time that are economically equivalent
How it works: Use rate i and time t in upcoming
relations to move money (values of P, F and A) between time points t = 0, 1, …, n to make them equivalent (not equal) at the rate i
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Example of Equivalence Different sums of money at different times may
be equal in economic value at a given rate
0 1
$100 now
$110
Rate of return = 10% per year
$100 now is economically equivalent to $110 one year from now, if the $100 is invested at a rate of 10% per year.
Year
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Simple and Compound Interest
Simple Interest Interest is calculated using principal only
Interest = (principal)(number of periods)(interest rate) I = Pni
Example: $100,000 lent for 3 years at simple i = 10% per year. What is repayment after 3 years?
Interest = 100,000(3)(0.10) = $30,000
Total due = 100,000 + 30,000 = $130,000
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Simple and Compound Interest Compound Interest
Interest is based on principal plus all accrued interest That is, interest compounds over time
Interest = (principal + all accrued interest) (interest rate)
Interest for time period t is
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Compound Interest Example Example: $100,000 lent for 3 years at i = 10% per
year compounded. What is repayment after 3 years?
Interest, year 1: I1 = 100,000(0.10) = $10,000 Total due, year 1: T1 = 100,000 + 10,000 = $110,000
Interest, year 2: I2 = 110,000(0.10) = $11,000 Total due, year 2: T2 = 110,000 + 11,000 = $121,000
Interest, year 3: I3 = 121,000(0.10) = $12,100 Total due, year 3: T3 = 121,000 + 12,100 = $133,100
Compounded: $133,100 Simple: $130,000
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Minimum Attractive Rate of Return MARR is a reasonable rate
of return (percent) established for evaluating and selecting alternatives
An investment is justified economically if it is expected to return at least the MARR
Also termed hurdle rate, benchmark rate and cutoff rate
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MARR Characteristics MARR is established by the financial
managers of the firm MARR is fundamentally connected to the cost
of capital Both types of capital financing are used to
determine the weighted average cost of capital (WACC) and the MARR MARR usually considers the risk inherent to a
project
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Types of Financing
Equity Financing –Funds either from retained earnings, new stock issues, or owner’s infusion of money. Debt Financing –Borrowed funds from outside
sources – loans, bonds, mortgages, venture capital pools, etc. Interest is paid to the lender on these funds
For an economically justified project
ROR ≥ MARR > WACC
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Opportunity Cost Definition: Largest rate of return of all projects not
accepted (forgone) due to a lack of capital funds If no MARR is set, the ROR of the first project not undertaken
establishes the opportunity cost
Example: Assume MARR = 10%. Project A, not funded due to lack of funds, is projected to have RORA = 13%. Project B has RORB = 15% and is funded because it costs less than A
Opportunity cost is 13%, i.e., the opportunity to make an additional 13% is forgone by not funding project A
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Introduction to Spreadsheet Functions Excel financial functions
Present Value, P: = PV(i%,n,A,F) Future Value, F: = FV(i%,n,A,P) Equal, periodic value, A: = PMT(i%,n,P,F) Number of periods, n: = NPER((i%,A,P,F) Compound interest rate, i: = RATE(n,A,P,F) Compound interest rate, i: = IRR(first_cell:last_cell) Present value, any series, P: = NPV(i%,second_cell:last_cell) + first_cell
Example: Estimates are P = $5000 n = 5 years i = 5% per year Find A in $ per year Function and display: = PMT(5%, 5, 5000) displays A = $1154.87
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Chapter Summary Engineering Economy fundamentals Time value of money Economic equivalence Introduction to capital funding and MARR Spreadsheet functions
Interest rate and rate of return Simple and compound interest
Cash flow estimation Cash flow diagrams End-of-period assumption Net cash flow Perspectives taken for cash flow estimation
Ethics Universal morals and personal morals Professional and engineering ethics (Code of Ethics)
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© 2012 by McGraw-Hill, New York, N.Y All Rights Reserved 2-1
Lecture slides to accompany
Engineering Economy 7th edition
Leland Blank
Anthony Tarquin
Chapter 2 Factors: How Time and Interest Affect
Money
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© 2012 by McGraw-Hill, New York, N.Y All Rights Reserved 2-2
LEARNING OUTCOMES
1. F/P and P/F Factors
2. P/A and A/P Factors
3. F/A and A/F Factors
4. Factor Values
5. Arithmetic Gradient
6. Geometric Gradient
7. Find i or n
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Single Payment Factors (F/P and P/F) Single payment factors involve only P and F. Cash flow diagrams are as follows:
F = P(1 + i ) n P = F[1 / (1 + i ) n] Formulas are as follows:
Terms in parentheses or brackets are called factors. Values are in tables for i and n values
Factors are represented in standard factor notation such as (F/P,i,n),
where letter to left of slash is what is sought; letter to right represents what is given
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F/P and P/F for Spreadsheets
Future value F is calculated using FV function:
= FV(i%,n,,P)
Present value P is calculated using PV function:
= PV(i%,n,,F)
Note the use of double commas in each function
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Example: Finding Future Value A person deposits $5000 into an account which pays interest at a rate of 8% per year. The amount in the account after 10 years is closest to:
(A) $2,792 (B) $9,000 (C) $10,795 (D) $12,165
The cash flow diagram is: Solution: F = P(F/P,i,n )
= 5000(F/P,8%,10 )
= $10,794.50
Answer is (C)
= 5000(2.1589)
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Example: Finding Present Value A small company wants to make a single deposit now so it will have enough money to purchase a backhoe costing $50,000 five years from now. If the account will earn interest of 10% per year, the amount that must be deposited now is nearest to:
(A) $10,000 (B) $ 31,050 (C) $ 33,250 (D) $319,160
The cash flow diagram is: Solution: P = F(P/F,i,n )
= 50,000(P/F,10%,5 ) = 50,000(0.6209)
= $31,045
Answer is (B)
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Uniform Series Involving P/A and A/P
0 1 2 3 4 5
A = ?
P = Given
The cash flow diagrams are:
Standard Factor Notation P = A(P/A,i,n) A = P(A/P,i,n)
Note: P is one period Ahead of first A value
(1) Cash flow occurs in consecutive interest periods The uniform series factors that involve P and A are derived as follows:
(2) Cash flow amount is same in each interest period
0 1 2 3 4 5
A = Given
P = ?
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Example: Uniform Series Involving P/A
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A chemical engineer believes that by modifying the structure of a certain water treatment polymer, his company would earn an extra $5000 per year. At an interest rate of 10% per year, how much could the company afford to spend now to just break even over a 5 year project period? (A) $11,170 (B) 13,640 (C) $15,300 (D) $18,950
The cash flow diagram is as follows:
P = 5000(P/A,10%,5) = 5000(3.7908) = $18,954
Answer is (D) 0 1 2 3 4 5
A = $5000
P = ? i =10%
Solution:
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Uniform Series Involving F/A and A/F
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(1) Cash flow occurs in consecutive interest periods The uniform series factors that involve F and A are derived as follows:
(2) Last cash flow occurs in same period as F
0 1 2 3 4 5
F = ?
A = Given
0 1 2 3 4 5
F = Given
A = ?
Note: F takes place in the same period as last A
Cash flow diagrams are:
Standard Factor Notation F = A(F/A,i,n) A = F(A/F,i,n)
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Example: Uniform Series Involving F/A An industrial engineer made a modification to a chip manufacturing process that will save her company $10,000 per year. At an interest rate of 8% per year, how much will the savings amount to in 7 years?
(A) $45,300 (B) $68,500 (C) $89,228 (D) $151,500
The cash flow diagram is:
A = $10,000
F = ?
i = 8%
0 1 2 3 4 5 6 7
Solution: F = 10,000(F/A,8%,7)
= 10,000(8.9228) = $89,228
Answer is (C)
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Factor Values for Untabulated i or n
3 ways to find factor values for untabulated i or n values
Use formula Use spreadsheet function with corresponding P, F, or A value set to 1 Linearly interpolate in interest tables
Formula or spreadsheet function is fast and accurate Interpolation is only approximate
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Example: Untabulated i Determine the value for (F/P, 8.3%,10)
Formula: F = (1 + 0.083)10 = 2.2197
Spreadsheet: = FV(8.3%,10,,1) = 2.2197
Interpolation: 8% ------ 2.1589 8.3% ------ x 9% ------ 2.3674 x = 2.1589 + [(8.3 - 8.0)/(9.0 - 8.0)][2.3674 – 2.1589] = 2.2215
Absolute Error = 2.2215 – 2.2197 = 0.0018
OK
OK
(Too high)
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Arithmetic Gradients Arithmetic gradients change by the same amount each period
The cash flow diagram for the PG of an arithmetic gradient is:
0
1 2 3 n
G 2G
4
3G (n-1)G
PG = ? G starts between periods 1 and 2
(not between 0 and 1)
This is because cash flow in year 1 is usually not equal to G and is handled
separately as a base amount (shown on next slide)
Note that PG is located Two Periods Ahead of the first change that is equal
to G Standard factor notation is
PG = G(P/G,i,n)
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Typical Arithmetic Gradient Cash Flow PT = ?
i = 10% 0 1 2 3 4 5
400 450
500 550
600
PA = ? i = 10%
0 1 2 3 4 5
400 400 400 400 400
PG = ? i = 10%
0 1 2 3 4 5
50 100
150 200
+
This diagram = this base amount plus this gradient
PA = 400(P/A,10%,5) PG = 50(P/G,10%,5)
PT = PA + PG = 400(P/A,10%,5) + 50(P/G,10%,5)
Amount in year 1 is base amount
Amount in year 1 is base amount
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Converting Arithmetic Gradient to A
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i = 10% 0 1 2 3 4 5
G 2G
3G 4G
i = 10% 0 1 2 3 4 5
A = ?
Arithmetic gradient can be converted into equivalent A value using G(A/G,i,n)
General equation when base amount is involved is
A = base amount + G(A/G,i,n)
0 1 2 3 4 5
G 2G
3G 4G
For decreasing gradients, change plus sign to minus
A = base amount - G(A/G,i,n)
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Example: Arithmetic Gradient The present worth of $400 in year 1 and amounts increasing by $30 per year through year 5 at an interest rate of 12% per year is closest to:
(A) $1532 (B) $1,634 (C) $1,744 (D) $1,829
0
1 2 3 Year
430 460
4
490 520
PT = ?
5
400
i = 12%
G = $30
= 400(3.6048) + 30(6.3970) = $1,633.83
Answer is (B)
PT = 400(P/A,12%,5) + 30(P/G,12%,5)
The cash flow could also be converted into an A value as follows:
A = 400 + 30(A/G,12%,5) = 400 + 30(1.7746) = $453.24
Solution:
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Geometric Gradients
© 2012 by McGraw-Hill, New York, N.Y All Rights Reserved 2-17
Geometric gradients change by the same percentage each period
0
1 2 3 n
A1 A 1(1+g)1
4
A 1(1+g)2
A 1(1+g)n-1
Pg = ? There are no tables for geometric factors Use following equation for g ≠ i:
Pg = A1{1- [(1+g)/(1+i)]n}/(i-g) where: A1 = cash flow in period 1 g = rate of increase If g = i, Pg = A1n/(1+i)
Note: If g is negative, change signs in front of both g values
Cash flow diagram for present worth of geometric gradient
Note: g starts between periods 1 and 2
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Example: Geometric Gradient
© 2012 by McGraw-Hill, New York, N.Y All Rights Reserved 2-18
Find the present worth of $1,000 in year 1 and amounts increasing by 7% per year through year 10. Use an interest rate of 12% per year.
(a) $5,670 (b) $7,333 (c) $12,670 (d) $13,550
0
1 2 3 10
1000 1070
4
1145
1838
Pg = ? Solution:
Pg = 1000[1-(1+0.07/1+0.12)10]/(0.12-0.07) = $7,333
Answer is (b) g = 7%
i = 12%
To find A, multiply Pg by (A/P,12%,10)
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2-19
Unknown Interest Rate i Unknown interest rate problems involve solving for i,
given n and 2 other values (P, F, or A) (Usually requires a trial and error solution or interpolation in interest tables)
A contractor purchased equipment for $60,000 which provided income of $16,000 per year for 10 years. The annual rate of return of the investment was closest to:
(a) 15% (b) 18% (c) 20% (d) 23%
Can use either the P/A or A/P factor. Using A/P: Solution:
60,000(A/P,i%,10) = 16,000 (A/P,i%,10) = 0.26667
From A/P column at n = 10 in the interest tables, i is between 22% and 24% Answer is (d)
Procedure: Set up equation with all symbols involved and solve for i
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Unknown Recovery Period n Unknown recovery period problems involve solving for n,
given i and 2 other values (P, F, or A) (Like interest rate problems, they usually require a trial & error solution or interpolation in interest tables)
Procedure: Set up equation with all symbols involved and solve for n
A contractor purchased equipment for $60,000 that provided income of $8,000 per year. At an interest rate of 10% per year, the length of time required to recover the investment was closest to:
(a) 10 years (b) 12 years (c) 15 years (d) 18 years
Can use either the P/A or A/P factor. Using A/P: Solution: 60,000(A/P,10%,n) = 8,000
(A/P,10%,n) = 0.13333
From A/P column in i = 10% interest tables, n is between 14 and 15 years Answer is (c)
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Summary of Important Points
© 2012 by McGraw-Hill, New York, N.Y All Rights Reserved 2-21
In P/A and A/P factors, P is one period ahead of first A
In F/A and A/F factors, F is in same period as last A
To find untabulated factor values, best way is to use formula or spreadsheet
For arithmetic gradients, gradient G starts between periods 1 and 2
Arithmetic gradients have 2 parts, base amount (year 1) and gradient amount
For geometric gradients, gradient g starts been periods 1 and 2
In geometric gradient formula, A1 is amount in period 1
To find unknown i or n, set up equation involving all terms and solve for i or n
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© 2012 by McGraw-Hill, New York, N.Y All Rights Reserved 3-1
Lecture slides to accompany
Engineering Economy 7th edition
Leland Blank
Anthony Tarquin
Chapter 3 Combining Factors and Spreadsheet
Functions
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© 2012 by McGraw-Hill, New York, N.Y All Rights Reserved 3-2
LEARNING OUTCOMES
1. Shifted uniform series
2. Shifted series and single cash flows
3. Shifted gradients
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Shifted Uniform Series A shifted uniform series starts at a time other than period 1
0 1 2 3 4 5
A = Given
PA = ?
The cash flow diagram below is an example of a shifted series Series starts in period 2, not period 1
Shifted series usually
require the use of
multiple factors
Remember: When using P/A or A/P factor, PA is always one year ahead of first A
When using F/A or A/F factor, FA is in same year as last A
FA = ?
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Example Using P/A Factor: Shifted Uniform Series
P0 = ?
A = $10,000
0 1 2 3 4 5 6
i = 10%
The present worth of the cash flow shown below at i = 10% is: (a) $25,304 (b) $29,562 (c) $34,462 (d) $37,908
Solution: (1) Use P/A factor with n = 5 (for 5 arrows) to get P1 in year 1 (2) Use P/F factor with n = 1 to move P1 back for P0 in year 0
P0 = P1(P/F,10%,1) = A(P/A,10%,5)(P/F,10%,1) = 10,000(3.7908)(0.9091) = $34,462 Answer is (c)
0 1 2 3 4 5
P1 = ?
Actual year Series year
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How much money would be available in year 10 if $8000 is deposited each year in years 3 through 10 at an interest rate of 10% per year?
0 1 2 3 4 5 6 7 8 9 10
FA = ?
A = $8000
i = 10%
Solution: Re-number diagram to determine n = 8 (number of arrows)
0 1 2 3 4 5 6 7 8
Cash flow diagram is:
FA = 8000(F/A,10%,8) = 8000(11.4359) = $91,487
Example Using F/A Factor: Shifted Uniform Series
Actual year Series year
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Shifted Series and Random Single Amounts For cash flows that include uniform series and randomly placed single amounts:
Uniform series procedures are applied to the series amounts
Single amount formulas are applied to the one-time cash flows
The resulting values are then combined per the problem statement
The following slides illustrate the procedure
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Example: Series and Random Single Amounts Find the present worth in year 0 for the cash flows
shown using an interest rate of 10% per year.
0 1 2 3 4 5 6 7 8 9 10
PT = ?
A = $5000
i = 10%
First, re-number cash flow diagram to get n for uniform series: n = 8
$2000
0 1 2 3 4 5 6 7 8 9 10
PT = ?
A = $5000
i = 10%
$2000
0 1 2 3 4 5 6 7 8
Solution:
Actual year
Series year
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© 2012 by McGraw-Hill, New York, N.Y All Rights Reserved 1-8
A = $5000
i = 10%
0 1 2 3 4 5 6 7 8
Actual year
Series year
0 1 2 3 4 5 6 7 8 9 10
$2000
Use P/A to get PA in year 2: PA = 5000(P/A,10%,8) = 5000(5.3349) = $26,675
Move PA back to year 0 using P/F: P0 = 26,675(P/F,10%,2) = 26,675(0.8264) = $22,044
Move $2000 single amount back to year 0: P2000 = 2000(P/F,10%,8) = 2000(0.4665) = $933
Now, add P0 and P2000 to get PT: PT = 22,044 + 933 = $22,977
Example: Series and Random Single Amounts
PT = ? PA
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© 2012 by McGraw-Hill, New York, N.Y All Rights Reserved 3-9
Example Worked a Different Way (Using F/A instead of P/A for uniform series)
0 1 2 3 4 5 6 7 8 9 10
PT = ?
A = $5000
i = 10%
$2000
0 1 2 3 4 5 6 7 8
The same re-numbered diagram from the previous slide is used
Solution: Use F/A to get FA in actual year 10: FA = 5000(F/A,10%,8) = 5000(11.4359) = $57,180
Move $2000 single amount back to year 0: P2000 = 2000(P/F,10%,8) = 2000(0.4665) = $933 Now, add two P values to get PT: PT = 22,043 + 933 = $22,976 Same as before
Move FA back to year 0 using P/F: P0 = 57,180(P/F,10%,10) = 57,180(0.3855) = $22,043
As shown, there are usually multiple ways to work equivalency problems
FA = ?
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Example: Series and Random Amounts
© 2012 by McGraw-Hill, New York, N.Y All Rights Reserved 3-10
Convert the cash flows shown below (black arrows) into an equivalent annual worth A in years 1 through 8 (red arrows)
at i = 10% per year.
0 1 2 3 4 5 6 7 8
A = $3000
i = 10%
$1000
0 1 2 3 4 5
A = ?
Solution:
1. Convert all cash flows into P in year 0 and use A/P with n = 8 2. Find F in year 8 and use A/F with n = 8
Solve for F: F = 3000(F/A,10%,5) + 1000(F/P,10%,1) = 3000(6.1051) + 1000(1.1000) = $19,415
Find A: A = 19,415(A/F,10%,8) = 19,415(0.08744) = $1698
Approaches:
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Shifted Arithmetic Gradients
© 2012 by McGraw-Hill, New York, N.Y All Rights Reserved 3-11
Shifted gradient begins at a time other than between periods 1 and 2
Present worth PG is located 2 periods before gradient starts
Must use multiple factors to find PT in actual year 0
To find equivalent A series, find PT at actual time 0 and apply (A/P,i,n)
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© 2012 by McGraw-Hill, New York, N.Y All Rights Reserved 3-12
Example: Shifted Arithmetic Gradient
Solution:
John Deere expects the cost of a tractor part to increase by $5 per year beginning 4 years from now. If the cost in years 1-3 is $60, determine the present worth in year 0 of the cost through year 10 at an interest rate of 12% per year.
0
1 2 3 10 4 5
60 60 60 65
70 95
PT = ? i = 12%
First find P2 for G = $5 and base amount ($60) in actual year 2
P2 = 60(P/A,12%,8) + 5(P/G,12%,8) = $370.41
Next, move P2 back to year 0 P0 = P2(P/F,12%,2) = $295.29
Next, find PA for the $60 amounts of years 1 and 2 PA = 60(P/A,12%,2) = $101.41
Finally, add P0 and PA to get PT in year 0 PT = P0 + PA = $396.70
G = 5
0 1 2 3 8 Gradient years Actual years
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Shifted Geometric Gradients
Shifted gradient begins at a time other than between periods 1 and 2
Equation yields Pg for all cash flows (base amount A1 is included)
For negative gradient, change signs on both g values
Pg = A 1{1 - [(1+g)/(1+i)]n/(i-g)} Equation (i ≠ g):
There are no tables for geometric gradient factors
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Example: Shifted Geometric Gradient Weirton Steel signed a 5-year contract to purchase water treatment chemicals from a local distributor for $7000 per year. When the contract ends, the cost of the chemicals is expected to increase by 12% per year for the next 8 years. If an initial investment in storage tanks is $35,000, determine the equivalent present worth in year 0 of all of the cash flows at i = 15% per year.
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Example: Shifted Geometric Gradient
Gradient starts between actual years 5 and 6; these are gradient years 1 and 2. Pg is located in gradient year 0, which is actual year 4
Move Pg and other cash flows to year 0 to calculate PT
PT = 35,000 + 7000(P/A,15%,4) + 49,401(P/F,15%,4) = $83,232
Pg = 7000{1-[(1+0.12)/(1+0.15)]9/(0.15-0.12)} = $49,401
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© 2012 by McGraw-Hill, New York, N.Y All Rights Reserved 3-16
Negative Shifted Gradients For negative arithmetic gradients, change sign on G term from + to -
For negative geometric gradients, change signs on both g values
All other procedures are the same as for positive gradients
General equation for determining P: P = present worth of base amount - PG
Pg = A1{1-[(1-g)/(1+i)]n/(i+g)}
Changed from + to -
Changed from + to -
Changed from - to +
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Example: Negative Shifted Arithmetic Gradient
For the cash flows shown, find the future worth in year 7 at i = 10% per year F = ?
0 1 2 3 4 5 6 7
700 650
500 450 550 600
G = $-50 Gradient G first occurs between actual years 2 and 3; these are gradient years 1 and 2 Solution:
0 1 2 3 4 5 6
Actual years
Gradient years
PG is located in gradient year 0 (actual year 1); base amount of $700 is in gradient years 1-6
PG = 700(P/A,10%,6) – 50(P/G,10%,6) = 700(4.3553) – 50(9.6842) = $2565 F = PG(F/P,10%,6) = 2565(1.7716) = $4544
i = 10% PG = ?
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© 2012 by McGraw-Hill, New York, N.Y All Rights Reserved 3-18
Summary of Important Points
P for shifted uniform series is one period ahead of first A; n is equal to number of A values
F for shifted uniform series is in same period as last A; n is equal to number of A values
For gradients, first change equal to G or g occurs between gradient years 1 and 2
For negative arithmetic gradients, change sign on G from + to -
For negative geometric gradients, change sign on g from + to -
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4-1
Lecture slides to accompany
Engineering Economy 7th edition
Leland Blank
Anthony Tarquin
Chapter 4 Nominal and
Effective Interest Rates
© 2012 by McGraw-Hill All Rights Reserved
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4-2
LEARNING OUTCOMES 1. Understand interest rate statements
2. Use formula for effective interest rates
3. Determine interest rate for any time period
4. Determine payment period (PP) and compounding period (CP) for equivalence calculations
5. Make calculations for single cash flows
6. Make calculations for series and gradient cash flows with PP ≥ CP
7. Perform equivalence calculations when PP < CP
8. Use interest rate formula for continuous compounding
9. Make calculations for varying interest rates
© 2012 by McGraw-Hill All Rights Reserved
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4-3
Interest Rate Statements The terms ‘nominal’ and ‘effective’ enter into consideration when the interest period is less than one year.
Interest period (t) – period of time over which interest is expressed. For example, 1% per month.
New time-based definitions to understand and remember
Compounding period (CP) – Shortest time unit over which interest is charged or earned. For example,10% per year compounded monthly.
Compounding frequency (m) – Number of times compounding occurs within the interest period t. For example, at i = 10% per year, compounded monthly, interest would be compounded 12 times during the one year interest period.
© 2012 by McGraw-Hill All Rights Reserved
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4-4
Understanding Interest Rate Terminology A nominal interest rate (r) is obtained by multiplying an interest rate that is expressed over a short time period by the number of compounding periods in a longer time period: That is:
r = interest rate per period x number of compounding periods
Example: If i = 1% per month, nominal rate per year is r = (1)(12) = 12% per year)
IMPORTANT: Nominal interest rates are essentially simple interest rates. Therefore, they can never be used in interest formulas.
Effective rates must always be used hereafter in all interest formulas.
Effective interest rates (i) take compounding into account (effective rates can be obtained from nominal rates via a formula to be discussed later).
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© 2012 by McGraw-Hill All Rights Reserved 4-5
More About Interest Rate Terminology There are 3 general ways to express interest rates as shown below
When no compounding period is given, rate is effective
Sample Interest Rate Statements Comment i = 2% per month i = 12% per year
(1)
When compounding period is given and it is not the same as interest period, it is nominal
i = 10% per year, comp’d semiannually i = 3% per quarter, comp’d monthly
(2)
i = effective 9.4%/year, comp’d semiannually i = effective 4% per quarter, comp’d monthly (3)
When compounding period is given and rate is specified as effective, rate is effective over stated period
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Effective Annual Interest Rates
4-6 © 2012 by McGraw-Hill All Rights Reserved
Nominal rates are converted into effective annual rates via the equation:
ia = (1 + i)m – 1
where ia = effective annual interest rate i = effective rate for one compounding period
m = number times interest is compounded per year
Example: For a nominal interest rate of 12% per year, determine the nominal and effective rates per year for (a) quarterly, and (b) monthly compounding
Solution: (a) Nominal r / year = 12% per year
Effective i / year = (1 + 0.03)4 – 1 = 12.55% per year (b) Nominal r /month = 12/12 = 1.0% per year
Effective i / year = (1 + 0.01)12 – 1 = 12.68% per year
Nominal r / quarter = 12/4 = 3.0% per quarter
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4-7
Effective Interest Rates Nominal rates can be converted into effective rates
for any time period via the following equation:
i = (1 + r / m)m – 1 where i = effective interest rate for any time period
r = nominal rate for same time period as i m = no. times interest is comp’d in period specified for i
Example: For an interest rate of 1.2% per month, determine the nominal and effective rates (a) per quarter, and (b) per year
(a) Nominal r / quarter = (1.2)(3) = 3.6% per quarter Effective i / quarter = (1 + 0.036/3)3 – 1 = 3.64% per quarter
(b) Nominal i /year = (1.2)(12) = 14.4% per year Effective i / year = (1 + 0.144 / 12)12 – 1 = 15.39% per year
Solution:
Spreadsheet function is = EFFECT(r%,m) where r = nominal rate per period specified for i
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4-8
Equivalence Relations: PP and CP New definition: Payment Period (PP) – Length of time between cash flows
In the diagram below, the compounding period (CP) is semiannual and the payment period (PP) is monthly
$1000
0 1 2 3 4 5
F = ?
A = $8000
i = 10% per year, compounded quarterly
0 1 2 3 4 5 6 7 8 Years
Semi-annual periods
Similarly, for the diagram below, the CP is quarterly and the payment period (PP) is semiannual
Semi-annual PP © 2012 by McGraw-Hill All Rights Reserved
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4-9
Single Amounts with PP > CP
There are two equally correct ways to determine i and n
For problems involving single amounts, the payment period (PP) is usually longer than the compounding period (CP). For these problems, there are an infinite number of i and n combinations that can be used, with only two restrictions:
(1) The i must be an effective interest rate, and (2) The time units on n must be the same as those of i (i.e., if i is a rate per quarter, then n is the number of quarters between P and F)
Method 1: Determine effective interest rate over the compounding period CP, and set n equal to the number of compounding periods between P and F
Method 2: Determine the effective interest rate for any time period t, and set n equal to the total number of those same time periods.
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4-10
Example: Single Amounts with PP ≥ CP How much money will be in an account in 5 years if $10,000 is deposited now at an interest rate of 1% per month? Use three different interest rates: (a) monthly, (b) quarterly , and (c) yearly.
(a) For monthly rate, 1% is effective [n = (5 years)×(12 CP per year = 60] F = 10,000(F/P,1%,60) = $18,167
(b) For a quarterly rate, effective i/quarter = (1 + 0.03/3)3 –1 = 3.03% F = 10,000(F/P,3.03%,20) = $18,167
(c) For an annual rate, effective i/year = (1 + 0.12/12)12 –1 = 12.683% F = 10,000(F/P,12.683%,5) = $18,167
effective i per month months
effective i per quarter quarters
effective i per year years
i and n must always have same time units
i and n must always have same time units
i and n must always have same time units
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When PP ≥ CP, the only procedure (2 steps) that can be used is as follows:
Series with PP ≥ CP
4-11
For series cash flows, first step is to determine relationship between PP and CP
(1) First, find effective i per PP Example: if PP is in quarters, must find effective i/quarter (2) Second, determine n, the number of A values involved Example: quarterly payments for 6 years yields n = 4×6 = 24
Determine if PP ≥ CP, or if PP < CP
© 2012 by McGraw-Hill All Rights Reserved
Note: Procedure when PP < CP is discussed later
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4-12
Example: Series with PP ≥ CP
Solution: First, find relationship between PP and CP PP = six months, CP = one month; Therefore, PP > CP
Since PP > CP, find effective i per PP of six months
Step 1. i /6 months = (1 + 0.06/6)6 – 1 = 6.15%
Next, determine n (number of 6-month periods)
Step 2: n = 10(2) = 20 six month periods
Finally, set up equation and solve for F
F = 500(F/A,6.15%,20) = $18,692 (by factor or spreadsheet)
How much money will be accumulated in 10 years from a deposit of $500 every 6 months if the interest rate is 1% per month?
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4-13
Series with PP < CP Two policies: (1) interperiod cash flows earn no interest (most common) (2) interperiod cash flows earn compound interest
For policy (1), positive cash flows are moved to beginning of the interest period in which they occur and negative cash flows are moved to the end of the interest period
Note: The condition of PP < CP with no interperiod interest is the only situation in which the actual cash flow diagram is changed
For policy (2), cash flows are not moved and equivalent P, F, and A values are determined using the effective interest rate per payment period
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4-14
Example: Series with PP < CP
Solution:
A person deposits $100 per month into a savings account for 2 years. If $75 is withdrawn in months 5, 7 and 8 (in addition to the deposits), construct the cash flow diagram to determine how much will be in the account after 2 years at i = 6% per year, compounded quarterly. Assume there is no interperiod interest.
Since PP < CP with no interperiod interest, the cash flow diagram must be changed using quarters as the time periods
0 1 2 3 4 5 6 7 8 9 10 21 24
300 300 300 300 300
Months Quarters 1 2 3 7 8
75 150 F = ?
to
this 0 1 2 3 4 5 6 7 8 9 10 23 24
100
75
from
F = ?
this
75 75
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4-15
Continuous Compounding When the interest period is infinitely small, interest is
compounded continuously. Therefore, PP > CP and m increases.
Take limit as m → ∞ to find the effective interest rate equation
i = er – 1 Example: If a person deposits $500 into an account every 3 months at an interest rate of 6% per year, compounded continuously, how much will be in the account at the end of 5 years?
Solution: Payment Period: PP = 3 months Nominal rate per three months: r = 6%/4 = 1.50%
Effective rate per 3 months: i = e0.015 – 1 = 1.51%
F = 500(F/A,1.51%,20) = $11,573 © 2012 by McGraw-Hill All Rights Reserved
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4-16
Varying Rates
For the cash flow shown below, find the future worth (in year 7) at i = 10% per year.
Solution:
When interest rates vary over time, use the interest rates associated with their respective time periods to find P
Example: Find the present worth of $2500 deposits in years 1 through 8 if the interest rate is 7% per year for the first five years and 10% per year thereafter.
P = 2,500(P/A,7%,5) + 2,500(P/A,10%,3)(P/F,7%,5) = $14,683
An equivalent annual worth value can be obtained by replacing each cash flow amount with ‘A’ and setting the equation equal to the calculated P value
14,683 = A(P/A,7%,5) + A(P/A,10%,3)(P/F,7%,5) A = $2500 per year
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i = (1 + r / m)m – 1
4-17
Summary of Important Points
Must understand: interest period, compounding period, compounding frequency, and payment period
Always use effective rates in interest formulas
Interest rates are stated different ways; must know how to get effective rates
For single amounts, make sure units on i and n are the same
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4-18
Important Points (cont’d)
For uniform series with PP < CP and no interperiod interest, move cash flows to match compounding period
For continuous compounding, use i = er – 1 to get effective rate
For the cash flow shown below, find the future worth (in year 7) at i = 10% per year.
For varying rates, use stated i values for respective time periods
© 2012 by McGraw-Hill All Rights Reserved
For uniform series with PP ≥ CP, find effective i over PP
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© 2012 by McGraw-Hill, New York, N.Y All Rights Reserved 5-1
Lecture slides to accompany
Engineering Economy 7th edition
Leland Blank
Anthony Tarquin
Chapter 5 Present Worth
Analysis
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5-2
LEARNING OUTCOMES
1. Formulate Alternatives
2. PW of equal-life alternatives
3. PW of different-life alternatives
4. Future Worth analysis
5. Capitalized Cost analysis
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5-3
Formulating Alternatives
Two types of economic proposals
Mutually Exclusive (ME) Alternatives: Only one can be selected; Compete against each other
Independent Projects: More than one can be selected; Compete only against DN
© 2012 by McGraw-Hill All Rights Reserved
Do Nothing (DN) – An ME alternative or independent project to maintain the current approach; no new costs, revenues or savings
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5-4
Revenue: Alternatives include estimates of costs (cash outflows) and revenues (cash inflows)
Two types of cash flow estimates
Cost: Alternatives include only costs; revenues and savings assumed equal for all alternatives; also called service alternatives
Formulating Alternatives
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5-5
Convert all cash flows to PW using MARR Precede costs by minus sign; receipts by plus sign
For mutually exclusive alternatives, select one with numerically largest PW
For independent projects, select all with PW > 0
PW Analysis of Alternatives
© 2012 by McGraw-Hill All Rights Reserved
For one project, if PW > 0, it is justified EVALUATION
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5-6
For the alternatives shown below, which should be selected selected selected if they are (a) mutually exclusive; (b) independent?
Project ID Present Worth
A $30,000 B $12,500
C $-4,000 D $ 2,000
Solution: (a) Select numerically largest PW; alternative A (b) Select all with PW > 0; projects A, B & D
Selection of Alternatives by PW
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5-7
Example: PW Evaluation of Equal-Life ME Alts.
Alternative X has a first cost of $20,000, an operating cost of $9,000 per year, and a $5,000 salvage value after 5 years. Alternative Y will cost $35,000 with an operating cost of $4,000 per year and a salvage value of $7,000 after 5 years. At an MARR of 12% per year, which should be selected?
Solution: Find PW at MARR and select numerically larger PW value
PWX = -20,000 - 9000(P/A,12%,5) + 5000(P/F,12%,5) = -$49,606
PWY = -35,000 - 4000(P/A,12%,5) + 7000(P/F,12%,5) = -$45,447
Select alternative Y
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5-8
PW of Different-Life Alternatives
Must compare alternatives for equal service (i.e., alternatives must end at the same time)
Two ways to compare equal service:
(The LCM procedure is used unless otherwise specified)
Least common multiple (LCM) of lives
Specified study period
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Assumptions of LCM approach Service provided is needed over the LCM or
more years
Selected alternative can be repeated over each life cycle of LCM in exactly the same manner
Cash flow estimates are the same for each life cycle (i.e., change in exact accord with the inflation or deflation rate)
1-9 © 2012 by McGraw-Hill All Rights Reserved
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5-10
Example: Different-Life Alternatives Compare the machines below using present worth analysis at i = 10% per year
Machine A Machine B First cost, $ Annual cost, $/year Salvage value, $ Life, years
20,000 30,000 9000 7000 4000 6000
3 6
Solution: PWA = -20,000 – 9000(P/A,10%,6) – 16,000(P/F,10%,3) + 4000(P/F,10%,6)
= $-68,961 PWB = -30,000 – 7000(P/A,10%,6) + 6000(P/F,10%,6)
= $-57,100
LCM = 6 years; repurchase A after 3 years
Select alternative B © 2012 by McGraw-Hill All Rights Reserved
20,000 – 4,000 in year 3
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PW Evaluation Using a Study Period Once a study period is specified, all cash flows after
this time are ignored
Salvage value is the estimated market value at the end of study period
Short study periods are often defined by management
when business goals are short-term
Study periods are commonly used in equipment replacement analysis
1-11 © 2012 by McGraw-Hill All Rights Reserved
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5-12
Example: Study Period PW Evaluation Compare the alternatives below using present worth analysis at i = 10% per year and a 3-year study period
Machine A Machine B First cost, $ Annual cost, $/year Salvage/market value, $
Life, years
-20,000 -30,000 -9,000 -7,000 4,000 6,000 (after 6 years)
10,000 (after 3 years) 3 6
Solution:
PWA = -20,000 – 9000(P/A,10%,3) + 4000(P/F,10%,3) = $-39,376
PWB = -30,000 – 7000(P/A,10%,3) + 10,000(P/F,10%,3) = $-39,895
Study period = 3 years; disregard all estimates after 3 years
Marginally, select A; different selection than for LCM = 6 years © 2012 by McGraw-Hill All Rights Reserved
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5-13
Future Worth Analysis
Must compare alternatives for equal service (i.e. alternatives must end at the same time)
Two ways to compare equal service:
(The LCM procedure is used unless otherwise specified)
Least common multiple (LCM) of lives
Specified study period
FW exactly like PW analysis, except calculate FW
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5-14
FW of Different-Life Alternatives Compare the machines below using future worth analysis at i = 10% per year
Machine A Machine B First cost, $ Annual cost, $/year Salvage value, $ Life, years
-20,000 -30,000 -9000 -7000 4000 6000
3 6
Solution: FWA = -20,000(F/P,10%,6) – 9000(F/A,10%,6) – 16,000(F/P,10%,3) + 4000
= $-122,168 FWB = -30,000(F/P,10%.6) – 7000(F/A,10%,6) + 6000
= $-101,157
LCM = 6 years; repurchase A after 3 years
Select B (Note: PW and FW methods will always result in same selection) © 2012 by McGraw-Hill All Rights Reserved
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Capitalized Cost (CC) Analysis
5-15
CC refers to the present worth of a project with a very long life, that is, PW as n becomes infinite
Basic equation is: CC = P = A i
“A” essentially represents the interest on a perpetual investment
For example, in order to be able to withdraw $50,000 per year forever at i = 10% per year, the amount of capital required is 50,000/0.10 = $500,000
For finite life alternatives, convert all cash flows into an A value over one life cycle and then divide by i
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5-16
Example: Capitalized Cost
Solution:
Compare the machines shown below on the basis of their capitalized cost. Use i = 10% per year
Machine 1 Machine 2 First cost,$ Annual cost,$/year Salvage value, $ Life, years
-20,000 -100,000 -9000 -7000 4000 -----
3 ∞
Convert machine 1 cash flows into A and then divide by i A1 = -20,000(A/P,10%,3) – 9000 + 4000(A/F,10%,3) = $-15,834
CC1 = -15,834 / 0.10 = $-158,340
CC2 = -100,000 – 7000/ 0.10 = $-170,000
Select machine 1 © 2012 by McGraw-Hill All Rights Reserved
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5-17
Summary of Important Points PW method converts all cash flows to present value at MARR
PW comparison must always be made for equal service
Alternatives can be mutually exclusive or independent Cash flow estimates can be for revenue or cost alternatives
Equal service is achieved by using LCM or study period
Capitalized cost is PW of project with infinite life; CC = P = A/i
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© 2012 by McGraw-Hill, New York, N.Y All Rights Reserved 6-1
Lecture slides to accompany
Engineering Economy 7th edition
Leland Blank
Anthony Tarquin
Chapter 6 Annual Worth
Analysis
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6-2
LEARNING OUTCOMES
1. Advantages of AW
2. Capital Recovery and AW values
3. AW analysis
4. Perpetual life
5. Life-Cycle Cost analysis
© 2012 by McGraw-Hill All Rights Reserved
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6-3
Advantages of AW Analysis
AW calculated for only one life cycle
Assumptions: Services needed for at least the LCM of lives of alternatives
Selected alternative will be repeated in succeeding life cycles in same manner as for the first life cycle All cash flows will be same in every life cycle (i.e., will change by only inflation or deflation rate)
© 2012 by McGraw-Hill All Rights Reserved
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6-4
Initial investment, P – First cost of an asset Salvage value, S – Estimated value of asset at end of useful life
Annual amount, A – Cash flows associated with asset, such as annual operating cost (AOC), etc.
© 2012 by McGraw-Hill All Rights Reserved
Alternatives usually have the following cash flow estimates
Relationship between AW, PW and FW
AW = PW(A/P,i%,n) = FW(A/F,i%,n)
n is years for equal-service comparison (value of LCM or specified study period)
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6-5
Calculation of Annual Worth
An asset has a first cost of $20,000, an annual operating cost of $8000 and a salvage value of $5000 after 3 years. Calculate the AW for one and two life cycles at i = 10%
AWone = - 20,000(A/P,10%,3) – 8000 + 5000(A/F,10%,3) = $-14,532
AWtwo = - 20,000(A/P,10%,6) – 8000 – 15,000(P/F,10%,3)(A/P,10%,6) + 5000(A/F,10%,6)
= $-14,532
AW for one life cycle is the same for all life cycles!!
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Capital Recovery and AW
Capital recovery (CR) is the equivalent annual amount that an asset, process, or system must earn each year to just recover the first cost and a stated rate of return over its expected life. Salvage value is considered when calculating CR.
CR = -P(A/P,i%,n) + S(A/F,i%,n)
Use previous example: (note: AOC not included in CR ) CR = -20,000(A/P,10%,3) + 5000(A/F,10%,3) = $ – 6532 per year
Now AW = CR + A AW = – 6532 – 8000 = $ – 14,532
6-6 © 2012 by McGraw-Hill All Rights Reserved
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Selection Guidelines for AW Analysis
6-7 © 2012 by McGraw-Hill All Rights Reserved
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6-8
Solution:
ME Alternative Evaluation by AW Not necessary to use LCM for different life alternatives
A company is considering two machines. Machine X has a first cost of $30,000, AOC of $18,000, and S of $7000 after 4 years. Machine Y will cost $50,000 with an AOC of $16,000 and S of $9000 after 6 years. Which machine should the company select at an interest rate of 12% per year?
AWX = -30,000(A/P,12%,4) –18,000 +7,000(A/F,12%,4) = $-26,412 AWY = -50,000(A/P,12%,6) –16,000 + 9,000(A/F,12%,6) = $-27,052
Select Machine X; it has the numerically larger AW value © 2012 by McGraw-Hill All Rights Reserved
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6-9
AW of Permanent Investment
Solution: Find AW of C over 5 years and AW of D using relation A = Pi
Select alternative C
Use A = Pi for AW of infinite life alternatives Find AW over one life cycle for finite life alternatives
Compare the alternatives below using AW and i = 10% per year C D First Cost, $ -50,000 -250,000 Annual operating cost, $/year -20,000 -9,000 Salvage value, $ 5,000 75,000 Life, years 5 ∞
AWC = -50,000(A/P,10%,5) – 20,000 + 5,000(A/F,10%,5) = $-32,371 AWD = Pi + AOC = -250,000(0.10) – 9,000 = $-34,000
© 2012 by McGraw-Hill All Rights Reserved
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6-10
Typical Life-Cycle Cost Distribution by Phase
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6-11
Life-Cycle Cost Analysis LCC analysis includes all costs for entire life span,
from concept to disposal
Best when large percentage of costs are M&O
Includes phases of acquisition, operation, & phaseout
© 2012 by McGraw-Hill All Rights Reserved
Apply the AW method for LCC analysis of 1 or more cost alternatives Use PW analysis if there are revenues and other benefits considered
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6-12
Summary of Important Points AW method converts all cash flows to annual value at MARR
AW comparison is only one life cycle of each alternative
Alternatives can be mutually exclusive, independent, revenue, or cost
For infinite life alternatives, annualize initial cost as A = P(i)
Life-cycle cost analysis includes all costs over a project’s life span
© 2012 by McGraw-Hill All Rights Reserved
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7-1
Lecture slides to accompany
Engineering Economy 7th edition
Leland Blank
Anthony Tarquin
Chapter 7 Rate of Return
One Project
© 2012 by McGraw-Hill All Rights Reserved
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7-2
LEARNING OUTCOMES
1. Understand meaning of ROR
2. Calculate ROR for cash flow series
3. Understand difficulties of ROR
4. Determine multiple ROR values
5. Calculate External ROR (EROR)
6. Calculate r and i for bonds © 2012 by McGraw-Hill All Rights Reserved
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7-3
Interpretation of ROR Rate paid on unrecovered balance of borrowed money
such that final payment brings balance to exactly zero with interest considered
ROR equation can be written in terms of PW, AW, or FW
Numerical value can range from -100% to infinity
Use trial and error solution by factor or spreadsheet
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ROR Calculation and Project Evaluation To determine ROR, find the i* value in the relation
PW = 0 or AW = 0 or FW = 0
Alternatively, a relation like the following finds i*
PWoutflow = PWinflow
For evaluation, a project is economically viable if
i* ≥ MARR
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Finding ROR by Spreadsheet Function Using the RATE function
= RATE(n,A,P,F)
P = $-200,000 A = $-15,000 n = 12 F = $435,000
Function is = RATE(12,-15000,-200000,450000)
Display is i* = 1.9%
Using the IRR function
= IRR(first_cell, last_cell)
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= IRR(B2:B14)
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7-6
ROR is the unique i* rate at which a PW, FW, or AW relation equals exactly 0
ROR Calculation Using PW, FW or AW Relation
Since i* > MARR = 15%, the company should buy the machine
Example: An investment of $20,000 in new equipment will generate income of $7000 per year for 3 years, at which time the machine can be sold for an estimated $8000. If the company’s MARR is 15% per year, should it buy the machine?
Solution:: The ROR equation, based on a PW relation, is:
Solve for i* by trial and error or spreadsheet: i* = 18.2% per year
0 = -20,000 + 7000(P/A,i*,3) + 8000(P/F,i*,3)
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7-7
Incremental analysis necessary for multiple alternative evaluations (discussed later)
Special Considerations for ROR May get multiple i* values (discussed later)
i* assumes reinvestment of positive cash flows earn at i* rate (may be unrealistic)
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7-8
Multiple ROR Values Multiple i* values may exist when there is more than one sign
change in net cash flow (CF) series. Such CF series are called non-conventional
Two tests for multiple i* values:
Descarte’s rule of signs: total number of real i* values is ≤ the number of sign changes in net cash flow series.
Norstrom’s criterion: if the cumulative cash flow starts off negatively and has only one sign change, there is only one positive root . © 2012 by McGraw-Hill All Rights Reserved
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7-9 © 2012 by McGraw-Hill All Rights Reserved
Plot of PW for CF Series with Multiple ROR Values
i* values at ~8% and ~41%
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7-10
Example: Multiple i* Values
Solution:
Determine the maximum number of i* values for the cash flow shown below Year Expense Income
0 -12,000 - 1 -5,000 + 3,000 2 -6,000 +9,000 3 -7,000 +15,000 4 -8,000 +16,000 5 -9,000 +8,000
Therefore, there is only one i* value ( i* = 8.7%)
Net cash flow
-12,000 -2,000 +3,000 +8,000
-1,000 +8,000
Cumulative CF
-12,000 -14,000 -11,000 -3,000 +5,000 +4,000
The cumulative cash flow begins negatively with one sign change
The sign on the net cash flow changes twice, indicating two possible i* values
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7-11
Removing Multiple i* Values
Two approaches to determine External ROR (EROR) • (1) Modified ROR (MIRR) • (2) Return on Invested Capital (ROIC)
Two new interest rates to consider: Investment rate ii – rate at which extra funds are invested external to the project
Borrowing rate ib – rate at which funds are borrowed from an external source to provide funds to the project
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7-12
Modified ROR Approach (MIRR)
Four step Procedure: Determine PW in year 0 of all negative CF at ib
Determine FW in year n of all positive CF at ii
Calculate EROR = i’ by FW = PW(F/P,i’,n)
If i’ ≥ MARR, project is economically justified
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7-13
Example: EROR Using MIRR Method For the NCF shown below, find the EROR by the MIRR method if MARR = 9%, ib = 8.5%, and ii = 12%
Year 0 1 2 3 NCF +2000 -500 -8100 +6800
Solution: PW0 = -500(P/F,8.5%,1) - 8100(P/F,8.5%,2) = $-7342
FW3 = 2000(F/P,12%,3) + 6800 = $9610
PW0(F/P,i’,3) + FW3 = 0 -7342(1 + i’)3 + 9610 = 0
i’ = 0.939 (9.39%)
Since i’ > MARR of 9%, project is justified © 2012 by McGraw-Hill All Rights Reserved
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7-14
Return on Invested Capital Approach Measure of how effectively project uses funds that remain internal to project
ROIC rate, i’’, is determined using net-investment procedure
Three step Procedure (1) Develop series of FW relations for each year t using: Ft = Ft-1(1 + k) + NCFt where: k = ii if Ft-1 > 0 and k = i’’ if Ft-1 < 0 (2) Set future worth relation for last year n equal to 0 (i.e., Fn= 0); solve for i’’ (3) If i’’ ≥ MARR, project is justified; otherwise, reject
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ROIC Example
7-15
For the NCF shown below, find the EROR by the ROIC method if MARR = 9% and ii = 12%
Year 0 1 2 3 NCF +2000 -500 -8100 +6800
Solution: Year 0: F0 = $+2000 F0 > 0; invest in year 1 at ii = 12% Year 1: F1 = 2000(1.12) - 500 = $+1740 F1 > 0; invest in year 2 at ii = 12% Year 2: F2 = 1740(1.12) - 8100 = $-6151 F2 < 0; use i’’ for year 3 Year 3: F3 = -6151(1 + i’’) + 6800 Set F3 = 0 and solve for i’’ -6151(1 + i’’) + 6800 = 0 i’’= 10.55%
Since i’’ > MARR of 9%, project is justified © 2012 by McGraw-Hill All Rights Reserved
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Important Points to Remember About the computation of an EROR value
EROR values are dependent upon the selected investment and/or borrowing rates Commonly, multiple i* rates, i’ from MIRR and i’’ from ROIC have different values
About the method used to decide For a definitive economic decision, set the MARR value and use the PW or AW method to determine economic viability of the project
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7-17
ROR of Bond Investment Bond is IOU with face value (V), coupon rate (b), no. of payment periods/year (c), dividend (I), and maturity date (n). Amount paid for the bond is P.
I = Vb/c General equation for i*: 0 = - P + I(P/A,i*,nxc) + V(P/F,i*,nxc)
Solution: (a) I = 10,000(0.06)/4 = $150 per quarter ROR equation is: 0 = -8000 + 150(P/A,i*,20) + 10,000(P/F,i*,20)
By trial and error or spreadsheet: i* = 2.8% per quarter
(b) Nominal i* per year = 2.8(4) = 11.2% per year Effective i* per year = (1 + 0.028)4 – 1 = 11.7% per year
A $10,000 bond with 6% interest payable quarterly is purchased for $8000. If the bond matures in 5 years, what is the ROR (a) per quarter, (b) per year?
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7-18
Summary of Important Points
More than 1 sign change in NCF may cause multiple i* values
Descarte’s rule of signs and Norstrom’s criterion useful when multiple i* values are suspected
ROR equations can be written in terms of PW, FW, or AW and usually require trial and error solution
i* assumes reinvestment of positive cash flows at i* rate
EROR can be calculated using MIRR or ROIC approach. Assumptions about investment and borrowing rates is required.
General ROR equation for bonds is 0 = - P + I(P/A,i*,nxc) + V(P/F,i*,nxc)
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8-1
Lecture slides to accompany
Engineering Economy 7th edition
Leland Blank
Anthony Tarquin
Chapter 8 Rate of Return
Multiple Alternatives
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8-2
LEARNING OUTCOMES 1. Why incremental analysis is required in ROR
2. Incremental cash flow (CF) calculation
3. Interpretation of ROR on incremental CF
4. Select alternative by ROR based on PW relation
5. Select alternative by ROR based on AW relation
6. Select best from several alternatives using ROR method
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8-3
Why Incremental Analysis is Necessary Selecting the alternative with highest ROR may not yield highest return on available capital
Must consider weighted average of total capital available
Capital not invested in a project is assumed to earn at MARR Example: Assume $90,000 is available for investment and MARR = 16% per year. If alternative A would earn 35% per year on investment of $50,000, and B would earn 29% per year on investment of $85,000, the weighted averages are:
Overall RORA = [50,000(0.35) + 40,000(0.16)]/90,000 = 26.6% Overall RORB = [85,000(0.29) + 5,000(0.16)]/90,000 = 28.3% Which investment is better, economically?
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Why Incremental Analysis is Necessary
© 2012 by McGraw-Hill All Rights Reserved 8-4
If selection basis is higher ROR: Select alternative A (wrong answer)
If selection basis is higher overall ROR: Select alternative B
Conclusion: Must use an incremental ROR analysis to make a consistently correct selection
Unlike PW, AW, and FW values, if not analyzed correctly, ROR
values can lead to an incorrect alternative selection. This is called the ranking inconsistency problem (discussed later)
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8-5
Incremental cash flow = cash flowB – cash flowA where larger initial investment is Alternative B
Calculation of Incremental CF
Example: Either of the cost alternatives shown below can be used in a grinding process. Tabulate the incremental cash flows.
B A First cost, $ -40,000 - 60,000
Annual cost, $/year -25,000 -19,000
Salvage value, $ 8,000 10,000
The ROR on the extra $20,000 investment in B determines which alternative to select (as discussed later)
The incremental CF is shown in the (B-A) column
-20,000
+6000
+2000
B - A
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8-6
Interpretation of ROR on Extra Investment
For independent projects, select all that have ROR ≥ MARR (no incremental analysis is necessary)
Based on concept that any avoidable investment that does not yield at least the MARR should not be made.
Once a lower-cost alternative has been economically justified, the ROR on the extra investment (i.e., additional amount of money associated with a higher first-cost alternative) must also yield a ROR ≥ MARR (because the extra investment is avoidable by selecting the economically-justified lower-cost alternative).
This incremental ROR is identified as ∆i*
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8-7
ROR Evaluation for Two ME Alternatives (1) Order alternatives by increasing initial investment cost (2) Develop incremental CF series using LCM of years (3) Draw incremental cash flow diagram, if needed (4) Count sign changes to see if multiple ∆i* values exist (5) Set up PW, AW, or FW = 0 relation and find ∆i*B-A Note: Incremental ROR analysis requires equal-service comparison.
The LCM of lives must be used in the relation (6) If ∆i*B-A < MARR, select A; otherwise, select B
If multiple ∆i* values exist, find EROR using either MIRR or ROIC approach.
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8-8
Example: Incremental ROR Evaluation
Either of the cost alternatives shown below can be used in a chemical refining process. If the company’s MARR is 15% per year, determine which should be selected on the basis of ROR analysis?
B A First cost ,$ -40,000 -60,000 Annual cost, $/year -25,000 -19,000 Salvage value, $ 8,000 10,000 Life, years 5 5
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Initial observations: ME, cost alternatives with equal life estimates and no multiple ROR values indicated
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8-9
Example: ROR Evaluation of Two Alternatives
B A First cost , $ -40,000 -60,000 Annual cost, $/year -25,000 -19,000 Salvage value, $ 8,000 10,000 Life, years 5 5
Solution, using procedure:
Order by first cost and find incremental cash flow B - A
-20,000 +6000 +2000
B - A
0 = -20,000 + 6000(P/A,∆i*,5) + 2000(P/F,∆i*,5)
∆i*B-A = 17.2% > MARR of 15% ROR on $20,000 extra investment is acceptable: Select B
Write ROR equation (in terms of PW, AW, or FW) on incremental CF
Solve for ∆i* and compare to MARR
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Breakeven ROR Value An ROR at which the PW, AW or FW values:
Of cash flows for two alternatives are exactly
equal. This is the i* value
Of incremental cash flows between two alternatives are exactly equal.
This is the ∆i* value
© 2012 by McGraw-Hill All Rights Reserved 8-10
If MARR > breakeven ROR, select lower-investment
alternative
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8-11
ROR Analysis – Multiple Alternatives Six-Step Procedure for Mutually Exclusive Alternatives
(1) Order alternatives from smallest to largest initial investment (2) For revenue alts, calculate i* (vs. DN) and eliminate all with i* < MARR; remaining alternative with lowest cost is defender. For cost alternatives, go to step (3) (3) Determine incremental CF between defender and next lowest-cost alternative (known as the challenger). Set up ROR relation (4) Calculate ∆i* on incremental CF between two alternatives from step (3) (5) If ∆i* ≥ MARR, eliminate defender and challenger becomes new defender against next alternative on list (6) Repeat steps (3) through (5) until only one alternative remains. Select it.
For Independent Projects Compare each alternative vs. DN and select all with ROR ≥ MARR
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8-12
Example: ROR for Multiple Alternatives The five mutually exclusive alternatives shown below are under consideration for improving visitor safety and access to additional areas of a national park. If all alternatives are considered to last indefinitely, determine which should be selected on the basis of a rate of return analysis using an interest rate of 10%. A B C D E_ First cost, $ millions -20 -40 -35 -90 -70 Annual M&O cost, $ millions -2 -1.5 -1.9 -1.1 -1.3
Solution: Rank on the basis of initial cost: A,C,B,E,D; calculate CC values C vs. A: 0 = -15 + 0.1/0.1 ∆i* = 6.7% (eliminate C) B vs. A: 0 = -20 + 0.5/0.1 ∆i* = 25% (eliminate A) E vs. B: 0 = -30 + 0.2/0.1 ∆i* = 6.7% (eliminate E) D vs. B: 0 = -50 + 0.4/0.1 ∆i* = 8% (eliminate D)
Select alternative B © 2012 by McGraw-Hill All Rights Reserved
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8-13
Must consider incremental cash flows for mutually exclusive alternatives
Summary of Important Points
Incremental cash flow = cash flowB – cash flowA where alternative with larger initial investment is Alternative B
Eliminate B if incremental ROR ∆i* < MARR; otherwise, eliminate A
For multiple mutually exclusive alternatives, compare two at a time and eliminate alternatives until only one remains
For independent alternatives, compare each against DN and select all that have ROR ≥ MARR
© 2012 by McGraw-Hill All Rights Reserved
Breakeven ROR is i* between project cash flows of two alternatives, or ∆i* between incremental cash flows of two alternatives
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© 2012 by McGraw-Hill All Rights Reserved 9-1
Lecture slides to accompany
Engineering Economy 7th edition
Leland Blank
Anthony Tarquin
Chapter 9 Benefit/Cost
Analysis
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9-2
LEARNING OUTCOMES
1. Explain difference in public vs. private sector projects
2. Calculate B/C ratio for single project
3. Select better of two alternatives using B/C method
4. Select best of multiple alternatives using B/C method
5. Use cost-effectiveness analysis (CEA) to evaluate service sector projects
6. Describe how ethical compromises may enter public sector projects
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© 2012 by McGraw-Hill All Rights Reserved 9-3
Differences: Public vs. Private Projects Characteristic Public Private Size of Investment Large Small, medium, large Life Longer (30 – 50+ years) Shorter (2 – 25 years) Annual CF No profit Profit-driven Funding Taxes, fees, bonds, etc. Stocks, bonds, loans, etc. Interest rate Lower Higher Selection criteria Multiple criteria Primarily ROR Environment of evaluation Politically inclined Economic
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Types of Contracts Contractors does not share project risk
Fixed price - lump-sum payment Cost reimbursable - Cost plus, as negotiated
Contractor shares in project risk Public-private partnerships (PPP), such as: Design-build projects - Contractor responsible from
design stage to operations stage Design-build-operate-maintain-finance (DBOMF)
projects - Turnkey project with contractor managing financing (manage cash flow); government obtains funding for project
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9-5
Cash Flow Classifications and B/C Relations
Must identify each cash flow as either benefit, disbenefit, or cost
Benefit (B) -- Advantages to the public Disbenefit (D) -- Disadvantages to the public Cost (C) -- Expenditures by the government
Note: Savings to government are subtracted from costs
Conventional B/C ratio = (B–D) / C Modified B/C ratio = [(B–D) – C] / Initial Investment Profitability Index = NCF / Initial Investment
Note 1: All terms must be expressed in same units, i.e., PW, AW, or FW Note 2: Do not use minus sign ahead of costs
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Decision Guidelines for B/C and PI Benefit/cost analysis
If B/C ≥ 1.0, project is economically justified at discount rate applied
If B/C < 1.0, project is not economically acceptable
Profitability index analysis of revenue projects
If PI ≥ 1.0, project is economically justified at discount rate applied
If PI < 1.0, project is not economically acceptable
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9-7
B/C Analysis – Single Project
Conventional B/C ratio = B - D C
Modified B/C ratio = B – D – M&O C
If B/C ≥ 1.0, accept project; otherwise, reject
PI = PW of initial investment
PW of NCFt Denominator is initial investment
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If PI ≥ 1.0, accept project; otherwise, reject
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9-8
B = $175,000
Example: B/C Analysis – Single Project
A flood control project will have a first cost of $1.4 million with an annual maintenance cost of $40,000 and a 10 year life. Reduced flood damage is expected to amount to $175,000 per year. Lost income to farmers is estimated to be $25,000 per year. At an interest rate of 6% per year, should the project be undertaken?
Solution: Express all values in AW terms and find B/C ratio
D = $25,000 C = 1,400,000(A/P,6%,10) + $40,000 = $230,218
B/C = (175,000 – 25,000)/230,218 = 0.65 < 1.0
Do not build project © 2012 by McGraw-Hill All Rights Reserved
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9-9
Defender, Challenger and Do Nothing Alternatives
General approach for incremental B/C analysis of two ME alternatives:
Lower total cost alternative is first compared to Do-nothing (DN) If B/C for the lower cost alternative is < 1.0, the DN option is compared to ∆B/C of the higher-cost alternative If both alternatives lose out to DN option, DN prevails, unless overriding needs requires selection of one of the alternatives
When selecting from two or more ME alternatives, there is a: Defender – in-place system or currently selected alternative Challenger – Alternative challenging the defender Do-nothing option – Status quo system
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9-10
Alternative Selection Using Incremental B/C Analysis – Two or More ME Alternatives
(1) Determine equivalent total cost for each alternative (2) Order alternatives by increasing total cost (3) Identify B and D for each alternative, if given, or go to step 5 (4) Calculate B/C for each alternative and eliminate all with B/C < 1.0 (5) Determine incremental costs and benefits for first two alternatives (6) Calculate ∆B/C; if >1.0, higher cost alternative becomes defender (7) Repeat steps 5 and 6 until only one alternative remains
Procedure similar to ROR analysis for multiple alternatives
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Example: Incremental B/C Analysis Compare two alternatives using i = 10% and B/C ratio
Alternative X Y First cost, $ 320,000 540,000 M&O costs, $/year 45,000 35,000 Benefits, $/year 110,000 150,000 Disbenefits, $/year 20,000 45,000 Life, years 10 20
Solution: First, calculate equivalent total cost
AW of costsX = 320,000(A/P,10%,10) + 45,000 = $97,080 AW of costsY = 540,000(A/P,10%,20) + 35,000 = $98,428
Order of analysis is X, then Y X vs. DN: (B-D)/C = (110,000 – 20,000) / 97,080 = 0.93 Eliminate X Y vs. DN: (150,000 – 45,000) / 98,428 = 1.07 Eliminate DN
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Example: ∆B/C Analysis; Selection Required Must select one of two alternatives using i = 10% and ∆B/C ratio Alternative X Y
First cost, $ 320,000 540,000 M&O costs, $/year 45,000 35,000 Benefits, $/year 110,000 150,000 Disbenefits, $/year 20,000 45,000 Life, years 10 20
Solution: Must select X or Y; DN not an option, compare Y to X AW of costsX = $97,080 AW of costsY = $98,428
Incremental values: ∆B = 150,000 – 110,000 = $40,000 ∆D = 45,000 – 20,000 = $25,000 ∆C = 98,428 – 97,080 = $1,348 Y vs. X: (∆B - ∆D) / ∆C = (40,000 – 25,000) / 1,348 = 11.1 Eliminate X
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B/C Analysis of Independent Projects
Independent projects comparison does not require incremental analysis Compare each alternative’s overall B/C with DN option
+ No budget limit: Accept all alternatives with B/C ≥ 1.0
+ Budget limit specified: capital budgeting problem; selection follows different procedure (discussed in chapter 12)
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Cost Effectiveness Analysis Service sector projects primarily involve intangibles, not physical facilities; examples include health care, security programs, credit card services, etc.
Cost-effectiveness analysis (CEA) combines monetary cost estimates with non-monetary benefit estimates to calculate the
Cost-effectiveness ratio (CER)
Equivalent total costs Total effectiveness measure = C/E
CER =
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9-15
CER Analysis for Independent Projects Procedure is as follows: (1) Determine equivalent total cost C, total effectiveness measure E and CER (2) Order projects by smallest to largest CER (3) Determine cumulative cost of projects and compare to budget limit b (4) Fund all projects such that b is not exceeded
Example: The effectiveness measure E is the number of graduates from adult training programs. For the CERs shown, determine which independent programs should be selected; b = $500,000. Program CER, $/graduate Program Cost, $ A 1203 305,000 B 752 98,000 C 2010 126,000 D 1830 365,000
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9-16
Example: CER for Independent Projects First, rank programs according to increasing CER: Cumulative Program CER, $/graduate Program Cost, $ Cost, $ B 752 98,000 98,000 A 1203 305,000 403,000 D 1830 365,000 768,000 C 2010 126,000 894,000
Next, select programs until budget is not exceeded
Select programs B and A at total cost of $403,000
Note: To expend the entire $500,000, accept as many additional individuals as possible from D at the per-student rate
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9-17
CER Analysis for Mutually Exclusive Projects Procedure is as follows
(1) Order alternatives smallest to largest by effectiveness measure E (2) Calculate CER for first alternative (defender) and compare to DN option (3) Calculate incremental cost (∆C), effectiveness (∆E), and incremental measure ∆C/E for challenger (next higher E measure) (4) If ∆C/Echallenger < C/Edefender challenger becomes defender (dominance); otherwise, no dominance is present and both alternatives are retained (5) Dominance present: Eliminate defender and compare next alternative to new defender per steps (3) and (4). Dominance not present: Current challenger becomes new defender against next challenger, but old defender remains viable (6) Continue steps (3) through (5) until only 1 alternative remains or only non-dominated alternatives remain (7) Apply budget limit or other criteria to determine which of remaining non-dominated alternatives can be funded
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9-18
Example: CER for ME Service Projects The effectiveness measure E is wins per person. From the cost and effectiveness values shown, determine which alternative to select. Cost (C) Effectiveness (E) CER Program $/person wins/person $/win A 2200 4 550 B 1400 2 700 C 6860 7 980
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9-19
Example: CER for ME Service Projects
Order programs according to increasing effectiveness measure E Cost (C) Effectiveness (E) CER Program $/person wins/person $/win B 1,400 2 700 A 2,200 4 550
C 6,860 7 980
Solution:
B vs. DN: C/EB = 1400/2 = 700 A vs. B: ∆C/E = (2200 – 1400)/(4 – 2) = 400 Dominance; eliminate B C vs. A: ∆C/E = (6860 – 2200)/(7 – 4) = 1553 No dominance; retain C
Must use other criteria to select either A or C
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9-20
Ethical Considerations
Engineers are routinely involved in two areas where ethics may be compromised:
Public policy making – Development of strategy, e.g., water system management (supply/demand strategy; ground vs. surface sources)
Public planning - Development of projects, e.g., water operations (distribution, rates, sales to outlying areas)
Engineers must maintain integrity and impartiality and always adhere to Code of Ethics
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9-21
B/C method used in public sector project evaluation
Summary of Important Points
Can use PW, AW, or FW for incremental B/C analysis, but must be consistent with units for B,C, and D estimates
For multiple mutually exclusive alternatives, compare two at a time and eliminate alternatives until only one remains
For independent alternatives with no budget limit, compare each against DN and select all alternatives that have B/C ≥ 1.0
CEA analysis for service sector projects combines cost and nonmonetary measures
Ethical dilemmas are especially prevalent in public sector projects © 2012 by McGraw-Hill All Rights Reserved
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Slide Sets to accompany Blank & Tarquin, Engineering Economy, 6th Edition, 2005
© 2005 by McGraw-Hill, New York, N.Y All Rights Reserved 10-1
Developed By:
Dr. Don Smith, P.E.
Department of Industrial Engineering
Texas A&M University
College Station, Texas
Executive Summary Version
Chapter 10 Making Choices: The Method, MARR, and Multiple Attributes
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Slide Sets to accompany Blank & Tarquin, Engineering Economy, 6th Edition, 2005
© 2005 by McGraw-Hill, New York, N.Y All Rights Reserved 10-2
LEARNING OBJECTIVES
1. Choose a Method
2. Cost of capital and MARR
3. WACC – Weighted Average Cost of Capital
4. Cost of debt capital
5. Cost of equity capital
6. High D-E mixes
7. Multiple attributes
8. Weighted attribute methods
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Slide Sets to accompany Blank & Tarquin, Engineering Economy, 6th Edition, 2005
© 2005 by McGraw-Hill, New York, N.Y All Rights Reserved 10-3
Sct 10.1 Comparing Mutually Exclusive Alternatives by Different Evaluation Methods
Different problem types lend themselves to different engineering economy methods Different information is available from
different evaluation methods Primary criteria for what method to apply Speed Ease of performing the analysis
See Tables 10-1 & 10-2 for a concise summary
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Slide Sets to accompany Blank & Tarquin, Engineering Economy, 6th Edition, 2005
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Evaluation Times
Equal lives of the alternatives PW, AW, FW
LCM of lives PW approach
Specified study period Normally exercised in industry
Infinity (capitalized cost)
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Slide Sets to accompany Blank & Tarquin, Engineering Economy, 6th Edition, 2005
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Decision Guidelines
Select the alternative with: Numerically largest PW, FW, or AW value
For ROR and B/C Apply the incremental analysis approach
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Slide Sets to accompany Blank & Tarquin, Engineering Economy, 6th Edition, 2005
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Sct 10.2 MARR Relative to the Cost of Capital
Establishing the MARR within the enterprise Requires: Cost of equity capital (cost of corporate funds) Cost of retained earnings included here
Cost of debt capital (cost of borrowed funds) Debt Capital $$ acquired from borrowing outside of the firm
Equity Capital $$ acquired from the owners and retained earnings
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Slide Sets to accompany Blank & Tarquin, Engineering Economy, 6th Edition, 2005
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Cost of Capital and the MARR
Established MARR is the sum of: (expressed as a % cost) Cost of capital + Expected return + Risk factor
MARR will vary from firm to firm and from project to project Cost of capital
(%) CC Min. MARR
Expected return (%) ER
CC + x% = ER
Established MARR Risk factor added (R%)
ER + R%
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Slide Sets to accompany Blank & Tarquin, Engineering Economy, 6th Edition, 2005
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Factors Impacting the MARR Perceived project risk
Higher the risk – higher the MARR for that project Investment opportunity
Expansion opportunity – may set a lower MARR Maintain flexibility
Tax structure Higher tax rate – higher MARR Federal reserve monetary policy – interest rates
Limited capital Tighter constraints on capital – higher MARR
Market rates of other firms Competitors alter their MARR - the firm could follow suit
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Slide Sets to accompany Blank & Tarquin, Engineering Economy, 6th Edition, 2005
© 2005 by McGraw-Hill, New York, N.Y All Rights Reserved 10-9
Sct 10.3 Debt-Equity Mix and WACC
D/E ratio (Debt to Equity mix) Ex.: 40-60 DE = {40% from debt, 60% from equity}
Weighted Average Cost of Capital (WACC) WACC = (equity fraction)(cost of equity capital) + (debt fraction)(cost of debt capital) Both ‘costs’ are expressed as a percentage cost Example: WACC = 0.6(4%) + 0.6(9%) = 7.8%
A variety of “models” exist that will approximate the
WACC for a given firm
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Slide Sets to accompany Blank & Tarquin, Engineering Economy, 6th Edition, 2005
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WACC: Example 10.3 Source of Capital Amount ($) Cost (%) Common Stock $5 million 13.7%
Retained Earnings
$2 million 8.9%
Debt from bonds $3 million 7.5%
CS = 50%; RE = 20%; Bonds = 30%
WACC = (0.50)(13.7) + (0.20)(8.9) + (0.30)(7.5) = 10.88%
This firm’s MARR must be > 10.88%
Sum: $10 million
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Slide Sets to accompany Blank & Tarquin, Engineering Economy, 6th Edition, 2005
© 2005 by McGraw-Hill, New York, N.Y All Rights Reserved 10-11
Tax Implications (detailed in Chapter 17)
WACC values are computed: Before-tax basis After-tax basis After-tax WACC = (Before Tax WACC)(1- Te) Where Te represents the effective tax rate
composed of: o Federal rate o State rate o Local rate(s)
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Slide Sets to accompany Blank & Tarquin, Engineering Economy, 6th Edition, 2005
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Sct 10.4 Determining Cost of Debt Capital
Debt financing Loans (borrowing)
$ borrowed from banks $ borrowed from Insurance companies, etc
Issuance of bonds (borrowing) Interest on loans and bonds are tax deductible in the US Bonds are sold (floated) within a bond market by investment
bankers on behalf of the firm Subject to extensive state and federal regulations
Interest payments from the firm to the lenders is tax deductible – important cost consideration
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Slide Sets to accompany Blank & Tarquin, Engineering Economy, 6th Edition, 2005
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Tax Savings from Debt Financing
The cost of financing by debt is lower than the actual interest rate charged because of the tax deductibility of the interest payments
Assume Te = the effective tax rate (%) Tax Savings = ($ expenses)(Te) Net Cash Flow = {$ expenses - $ tax savings}
NCF = expenses (1 – Te)
See Example 10.4
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Slide Sets to accompany Blank & Tarquin, Engineering Economy, 6th Edition, 2005
© 2005 by McGraw-Hill, New York, N.Y All Rights Reserved 10-14
Example of Tax Deductibility Impact on Cost of Debt Capital
Assume a loan has a 10% interest rate charged to the borrower The effective tax rate is 30% The after-tax cost of borrowing at 10% is (0.10)(1 – 0.30) = (0.10)(0.70) = 0.07 or 7%
Observations Due to tax deductibility the effective cost is 7% after tax Higher tax rates result in lower after-tax borrowing rates
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Slide Sets to accompany Blank & Tarquin, Engineering Economy, 6th Edition, 2005
© 2005 by McGraw-Hill, New York, N.Y All Rights Reserved 10-15
Sct 10.5 Determination of the Cost of Equity Capital and the MARR
Sources of equity capital 1. Sale of preferred stock (PS) 2. Sale of common stock (CS) 3. Use of retained earnings (RE) RE = past profits retained within the firm This money belongs to the owners of the firm
Sale of new stock is handled by investment bankers and brokerage firms – highly regulated – charge the firm for these sales
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Slide Sets to accompany Blank & Tarquin, Engineering Economy, 6th Edition, 2005
© 2005 by McGraw-Hill, New York, N.Y All Rights Reserved 10-16
Types of Stock Preferred Stock A form of ownership Pays a stated dividend per share periodically Generally a conservative type of stock
Common Stock A form of ownership Carries more risk than preferred No guarantee of dividends to be paid
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Slide Sets to accompany Blank & Tarquin, Engineering Economy, 6th Edition, 2005
© 2005 by McGraw-Hill, New York, N.Y All Rights Reserved 10-17
Cost of Equity Capital
Cost of equity capital generally applies some form of a dividend growth model or valuation model Basic model
“g” is the estimated annual increase in
returns to the shareholders
e
1e
first-year dividendR expected growth rateprice of the stockDVR =
Pg
= +
+
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Slide Sets to accompany Blank & Tarquin, Engineering Economy, 6th Edition, 2005
© 2005 by McGraw-Hill, New York, N.Y All Rights Reserved 10-18
Capital Asset Pricing Model -- CAPM Re for equity capital is specified by Re = risk-free return + premium above risk-free return Re = Rf + β(Rm – Rf) β = volatility of firm’s common stock relative to other stocks
o β = 1 is the norm Rm = return on stocks is a defined market portfolio as measured
by a prescribed index Rf = quoted US Treasury Bill rate (considered a safe investment) (Rm – Rf) = premium paid above the safe or “risk-free” rate
See Example 10.6
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Slide Sets to accompany Blank & Tarquin, Engineering Economy, 6th Edition, 2005
© 2005 by McGraw-Hill, New York, N.Y All Rights Reserved 10-19
Sct 10.6 Effect of Debt-Equity Mix on Investment Risk
D-E mix (Review Section 10.3) As the proportion of debt increasesDue to t the
calculated cost of capital tends to decrease Tax advantage of deducting interest
But…..leverage offered by larger percentage of debt capital increases the risks of funding future projects within the company
Too much debt is a “bad thing” Objective – strive for a balance between debt and
equity funding
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Too Much Debt….. Use of larger percentages of debt capital
increases the risk that is assumed by Investors (owners) and Lenders
Over time, investor confidence in the firm may diminish and the value of the stock could well decline Difficult to attract new investment funds Lenders will charge higher and higher interest
rates to hedge the risk
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Slide Sets to accompany Blank & Tarquin, Engineering Economy, 6th Edition, 2005
© 2005 by McGraw-Hill, New York, N.Y All Rights Reserved 10-21
Sct 10.7 Multiple Attribute Analysis: Identification and Importance of Each Attribute Refer back to Chapter 1 and 7-steps in Figure 10-5
Up to now we have focused on one attribute of a decision making problem Economic attribute!
Complex problems possess more than one attribute Multiple attribute analysis is often required Quantitative attributes Subjective attributes
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Slide Sets to accompany Blank & Tarquin, Engineering Economy, 6th Edition, 2005
© 2005 by McGraw-Hill, New York, N.Y All Rights Reserved 10-22
Identification of Key Attributes
Must ID the key attributes Comparison Input from experts Surveys Group discussion Delphi methods
Tabulate and then agree on the critical mix of subjective and objective attributes
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Slide Sets to accompany Blank & Tarquin, Engineering Economy, 6th Edition, 2005
© 2005 by McGraw-Hill, New York, N.Y All Rights Reserved 10-23
Importance of Each Attribute Determine the extent of importance of each
attribute Implies some form of weighting – wi Given m attributes we want:
11.0
m
ii
W=
=∑
Value ratings Vi j
Tabular format of attributes vs. alternatives
Weights for each
attribute
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Slide Sets to accompany Blank & Tarquin, Engineering Economy, 6th Edition, 2005
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Weighting Methodologies Equal Weighting All defined attributes are assigned equal weights Default model May or may not be appropriate
Rank Order m attributes are ranked in order of increasing importance (1 =
least important; 2, 3, ….) Weighted Rank Order m attributes ranked in order of importance and apply:
1
ii m
ii
sWs
=
=
∑
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Slide Sets to accompany Blank & Tarquin, Engineering Economy, 6th Edition, 2005
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Value Rating of Attributes Each alternative is assigned a value rating –
Vij for each attribute i Can apply a scale of 0-100 Can apply a Likert Scale 4-5 graduations (prefer an even number of choices)e.g.
o Very Poor o Poor o Good o Very good
See Table 10.4
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Slide Sets to accompany Blank & Tarquin, Engineering Economy, 6th Edition, 2005
© 2005 by McGraw-Hill, New York, N.Y All Rights Reserved 10-26
Sct 10.8 Evaluation Measure for Multiple Attributes
Weighted Attribute Method
Selection guideline Choose the alternative with the largest Rj value Assumes increasing weights mean more important
attributes Increasing Vij mean better performance for a given
alternative See Example 10.10
1
n
j i ijj
R WV=
=∑
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Chapter Summary Best methods for economic evaluation PW and AW at the stated MARR
Public projects Use the B/C ratio
The interest rate used is based upon the cost of capital, mix between equity and debt, and risk levels Multiple attributes incorporate more than
objective measures and permit the incorporation of criteria that is not totally economic based
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Chapter 10 End of Set
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© 2012 by McGraw-Hill All Rights Reserved 11-1
Lecture slides to accompany
Engineering Economy 7th edition
Leland Blank
Anthony Tarquin
Chapter 11 Replacement & Retention
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11-2
LEARNING OUTCOMES 1. Explain replacement terminology and basics
2. Determine economic service life (ESL)
3. Perform replacement/retention study
4. Understand special situations in replacement
5. Perform replacement study over specified time
6. Calculate trade-in value of defender
© 2012 by McGraw-Hill All Rights Reserved
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11-3
Replacement Study Basics
1. Reduced performance 2. Altered requirements 3. Obsolescence
Terminology Defender – Currently installed asset Challenger – Potential replacement for defender Market value (MV) – Value of defender if sold in open market Economic service life – No. of years at which lowest AW of cost occurs Defender first cost – MV of defender; used as its first cost (P) in analysis Challenger first cost – Capital to recover for challenger (usually its P value) Sunk cost – Prior expenditure not recoverable from challenger cost Nonowner’s viewpoint – Outsider’s (consultant’s) viewpoint for objectivity
Reasons for replacement study
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11-4
Example: Replacement Basics An asset purchased 2 years ago for $40,000 is harder to maintain than expected. It can be sold now for $12,000 or kept for a maximum of 2 more years, in which case its operating cost will be $20,000 each year, with a salvage value of $9,000 two years from now. A suitable challenger will have a first cost of $60,000 with an annual operating cost of $4,100 per year and a salvage value of $15,000 after 5 years. Determine the values of P, A, n, and S for the defender and challenger for an AW analysis.
Solution:
Defender: P = $-12,000; A = $-20,000; n = 2; S = $9,000 Challenger: P = $-60,000; A = $-4,100; n = 5; S = $15,000
© 2012 by McGraw-Hill All Rights Reserved
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Overview of a Replacement Study Replacement studies are applications of the AW method Study periods (planning horizons) are either specified or unlimited Assumptions for unlimited study period:
1. Services provided for indefinite future 2. Challenger is best available now and for future, and will be
repeated in future life cycles 3. Cost estimates for each life cycle for defender and challenger
remain the same If study period is specified, assumptions do not hold Replacement study procedures differ for the two cases
© 2012 by McGraw-Hill All Rights Reserved 11-5
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11-6
Economic Service Life
Economic service life (ESL) refers to the asset retention time (n) that yields its lowest equivalent AW
Determined by calculating AW for 1, 2, 3,…n years
General equation is: Total AW = capital recovery – AW of annual operating costs = CR – AW of AOC
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11-7
Example: Economic Service Life Determine the ESL of an asset which has the costs shown below. Let i = 10%
Year Cost,$/year Salvage value,$ 0 - 20,000 - 1 -5,000 10,000 2 -6,500 8,000 3 - 9,000 5,000 4 -11,000 5,000 5 -15,000 3,000
AW1 = - 20,000(A/P,10%,1) – 5000(P/F,10%,1)(A/P,10%,1) + 10,000(A/F,10%,1) = $ -17,000
AW2 = - 20,000(A/P,10%,2) – [5000(P/F,10%,1) + 6500(P/F,10%,2)](A/P,10%,2) + 8000(A/F,10%,2) = $ -13,429 Similarly, AW3 = $ -13,239 AW4 = $ -12,864 AW5 = $ -13,623
Economic service life is 4 years
Solution:
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11-8
Performing a Replacement Study
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11-9
Performing a Replacement Study -- Unlimited Study Period
1. Calculate AWD and AWC based on their ESL; select lower AW 2. If AWC was selected in step (1), keep for nC years (i.e.,
economic service life of challenger); if AWD was selected, keep defender one more year and then repeat analysis (i.e., one-year-later analysis)
3. As long as all estimates remain current in succeeding years, keep defender until nD is reached, and then replace defender with best challenger
4. If any estimates change before nD is reached, repeat steps (1) through (4)
Note: If study period is specified, perform steps (1) through (4) only through end of study period (discussed later)
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11-10
Example: Replacement Analysis
An asset purchased 2 years ago for $40,000 is harder to maintain than expected. It can be sold now for $12,000 or kept for a maximum of 2 more years, in which case its operating cost will be $20,000 each year, with a salvage value of $10,000 after 1 year or $9000 after two years. A suitable challenger will have an annual worth of $-24,000 per year. At an interest rate of 10% per year, should the defender be replaced now, one year from now, or two years from now?
Solution: First, determine ESL for defender AWD1 = -12,000(A/P,10%,1) – 20,000 + 10,000(A/F,10%,1) = $-23,200
AWD2 = -12,000(A/P,10%,2) – 20,000 + 9,000(A/F,10%,2) = $-22,629
ESL is n = 2 years; AWD = $-22,629
Lower AW = $-22,629 Replace defender in 2 years AWC = $-24,000
Note: conduct one-year-later analysis next year © 2012 by McGraw-Hill All Rights Reserved
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11-11
Additional Considerations Opportunity cost approach is the procedure that was previously presented for obtaining P for the defender. The opportunity cost is the money foregone by keeping the defender (i.e., not selling it). This approach is always correct
Cash flow approach subtracts income received from sale of defender from first cost of challenger.
Potential problems with cash flow approach: Provides falsely low value for capital recovery of challenger Can’t be used if remaining life of defender is not same as that of challenger
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11-12
Replacement Analysis Over Specified Study Period
Same procedure as before, except calculate AW values over study period instead of over ESL years of nD and nC
It is necessary to develop all viable defender-challenger combinations and calculate AW or PW for each one over study period
Select option with lowest cost or highest income
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11-13
Example: Replacement Analysis; Specified Period
An asset purchased 2 years ago for $40,000 is harder to maintain than expected. It can be sold now for $12,000 or kept for a maximum of 2 more years, in which case its operating cost will be $20,000 each year, with a salvage value of $10,000 after 1 year or $9000 after two. A suitable challenger will have an annual worth of $-24,000 per year. At an interest rate of 10% per year and over a study period of exactly 2 years, determine when the defender should be replaced.
Solution: From previous analysis, AWD for 1 and 2 years, and AWC are: AWD2 = $-22,629
AWC = $-24,000
© 2012 by McGraw-Hill All Rights Reserved
AWD1 = $-23,200
Option Year 1, $ Year 2, $ Year 3, $ AW, $ 1 (C, C, C) -24,000 -24,000 -24,000 -24,000 2 (D, C, C) -23,200 -24,000 -24,000 -23,708 3 (D, D, C) -22,629 -22,629 -24,000 -23,042
Decision: Option 3; Keep D for 2 years, then replace
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11-14
Replacement Value Replacement value (RV) is market/trade-in value of
defender that renders AWD and AWC equal to each other
If defender can be sold for amount > RV, challenger is the better option, because it will have a lower AW value
Set up equation AWD = AWC except use RV in place of P for the defender; then solve for RV
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11-15
Example: Replacement Value An asset purchased 2 years ago for $40,000 is harder to maintain than expected. It can be sold now for $12,000 or kept for a maximum of 2 more years, in which case its operating cost will be $20,000 each year, with a salvage value of $10,000 at the end of year two. A suitable challenger will have an initial cost of $65,000, an annual cost of $15,000, and a salvage value of $18,000 after its 5 year life. Determine the RV of the defender that will render its AW equal to that of the challenger, using an interest rate of 10% per year. Recommend a course of action.
- RV(A/P,10%,2) - 20,000 + 10,000(A/F,10%,2) = - 65,000(A/P,10%,5) -15,000 +18,000(A/F,10%,5)
RV = $24,228
Solution: Set AWD = AWC
Thus, if market value of defender > $24,228, select challenger © 2012 by McGraw-Hill All Rights Reserved
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11-16
In replacement study, P for presently-owned asset is its market value
Summary of Important Points
Economic service life is the n value that yields lowest AW
In replacement study, if no study period is specified, calculate AW over the respective life of each alternative
When study period is specified, must consider all viable defender-challenger combinations in analysis
Replacement value (RV) is P value for defender that renders its AW equal to that of challenger
© 2012 by McGraw-Hill All Rights Reserved
Opportunity cost approach is correct, it recognizes money foregone by keeping the defender, not by reducing challenger’s first cost
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12-1
Lecture slides to accompany
Engineering Economy 7th edition
Leland Blank
Anthony Tarquin
Chapter 12 Independent Projects with
Budget Limitation
© 2012 by McGraw-Hill All Rights Reserved
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LEARNING OBJECTIVES
12-2 © 2012 by McGraw-Hill All Rights Reserved
1. Capital rationing basics
2. Projects with equal lives
3. Projects with unequal lives
4. Linear program model
5. Ranking options
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Overview of Capital Rationing
• Capital is a scarce resource; never enough to fund all projects
• Each project is independent of others; select one, two, or more projects; don’t exceed budget limit b
• ‘Bundle’ is a collection of independent projects that are mutually exclusive (ME)
• For 3 projects, there are 23 = 8 ME bundles, e.g., A, B,
C, AB, AC, BC, ABC, Do nothing (DN)
12-3 © 2012 by McGraw-Hill All Rights Reserved
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Capital Budgeting Problem
Each project selected entirely or not selected at all Budget limit restricts total investment allowed Projects usually quite different from each other and have
different lives Reinvestment assumption: Positive annual cash flows
reinvested at MARR until end of life of longest-lived project
12-4 © 2012 by McGraw-Hill All Rights Reserved
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Capital Budgeting for Equal-Life Projects
Procedure
Develop ≤ 2m ME bundles that do not exceed budget b Determine NCF for projects in each viable bundle Calculate PW of each bundle j at MARR (i)
12-5 © 2012 by McGraw-Hill All Rights Reserved
Note: Discard any bundle with PW < 0; it does not return at least MARR
Select bundle with maximum PW (numerically largest)
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Bundle, j Projects NCFj0 , $ NCFjt , $ SV, $ PWj , $
1 A -25,000 +6,000 +4,000 -5,583
2 B -20,000 +9,000 0 +5,695
3 C -50,000 +15,000 +20,000 +4,261
4 A, B 45,000 +15,000 +4,000 +112
5 B, C 70,000 +24,000 +20,000 +9,956
6 DN 0 0 0 0
Example: Capital Budgeting for Equal Lives
Select projects to maximize PW at i = 15% and b = $70,000
12-6 © 2012 by McGraw-Hill All Rights Reserved
Project
Initial investment, $
Annual NCF, $
Life, years
Salvage value, $
A -25,000 +6,000 4 +4,000
B -20,000 +9,000 4 0
C -50,000 +15,000 4 +20,000
Solution: Five bundles meet budget restriction. Calculate NCF and PW values
Conclusion: Select projects
B and C with max PW
value
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Capital Budgeting for Unequal-Life Projects
LCM is not necessary in capital budgeting; use PW over respective lives to select independent projects
Same procedure as that for equal lives
Example: If MARR is 15% and b = $20,000 select projects
12-7 © 2012 by McGraw-Hill All Rights Reserved
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Example: Capital Budgeting for Unequal Lives Solution: Of 24 = 16 bundles, 8 are feasible. By spreadsheet:
12-8 © 2012 by McGraw-Hill All Rights Reserved
Reject with PW < 0 Conclusion: Select projects A and C
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Capital Budgeting Using LP Formulation
Why use linear programming (LP) approach? -- Manual approach not good for large number of projects as 2m ME bundles
grows too rapidly
Apply 0-1 integer LP (ILP) model to: Objective: Maximize Sum of PW of NCF at MARR for projects Constraints: Sum of investments ≤ investment capital limit
Each project selected (xk = 1) or not selected (xk = 0)
LP formulation strives to maximize Z
12-9 © 2012 by McGraw-Hill All Rights Reserved
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Example: LP Solution of Capital Budgeting Problem
MARR is 15%; limit is $20,000; select projects using LP
LP formulation for projects A, B, C, D labeled k = 1, 2, 3, 4 and b = $20,000 is:
Maximize: 6646x1 - 1019x2 + 984x3 - 748x4
Constraints: 8000x1 + 15,000x2 +8000x3 + 8000x4 ≤ 20,000 x1, x2, x3, and x4 = 0 or 1
12-10 © 2012 by McGraw-Hill All Rights Reserved
PW @ 15%, $
6646 -1019 984 -748
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Example: LP Solution of Capital Budgeting Problem
Use spreadsheet and Solver tool to solve LP problem
Select projects A (x1 = 1) and C (x3 = 1) for max PW = $7630
12-11 © 2012 by McGraw-Hill All Rights Reserved
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Different Project Ranking Measures Possible measures to rank and select projects:
‡ PW (present worth) – previously used to solve capital budgeting problem; maximizes PW value) ‡ IROR (internal ROR) – maximizes overall ROR; reinvestment
assumed at IROR value ‡ PWI (present worth index) – same as PI (profitability index); provides most money for the investment amount over life of the project, i.e.,
maximizes ‘bang for the buck’. PWI measure is:
12-12 © 2012 by McGraw-Hill All Rights Reserved
Projects selected by each measure can be different since each measure maximizes a different parameter
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Summary of Important Points
12-13 © 2012 by McGraw-Hill All Rights Reserved
Capital budgeting requires selection from independent projects. The PW measure is commonly maximized with a limited investment budget
Manual solution requires development of ME bundles of projects
Solution using linear programming formulation and the Solver tool on a spreadsheet is used to maximize PW at a stated MARR
Projects ranked using different measures, e.g., PW, IROR and PWI, may select different projects since different measures are maximized
Equal service assumption is not necessary for a capital budgeting solution
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13-1
Lecture slides to accompany
Engineering Economy 7th edition
Leland Blank
Anthony Tarquin
Chapter 13 Breakeven and
Payback Analysis
© 2012 by McGraw-Hill All Rights Reserved
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LEARNING OUTCOMES
© 2012 by McGraw-Hill All Rights Reserved 13-2
1. Breakeven point – one parameter
2. Breakeven point – two alternatives
3. Payback period analysis
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Breakeven Point
© 2012 by McGraw-Hill All Rights Reserved 13-3
The parameter (or variable) can be an amount of revenue, cost, supply, demand, etc. for one project or between two alternatives
One project - Breakeven point is identified as QBE. Determined using linear or non-linear math relations for revenue and cost
Between two alternatives - Determine one of the parameters P, A, F, i, or n with others constant
Solution is by one of three methods: Direct solution of relations Trial and error Spreadsheet functions or tools (Goal Seek or Solver)
Value of a parameter that makes two elements equal
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Cost-Revenue Model ― One Project
© 2012 by McGraw-Hill All Rights Reserved 13-4
Quantity, Q — An amount of the variable in question, e.g., units/year, hours/month Breakeven value is QBE
Fixed cost, FC — Costs not directly dependent on the variable, e.g., buildings, fixed overhead, insurance, minimum workforce cost
Variable cost, VC — Costs that change with parameters such as production level and workforce size. These are labor, material and marketing costs. Variable cost per unit is v Total cost, TC — Sum of fixed and variable costs, TC = FC + VC
Revenue, R — Amount is dependent on quantity sold
Revenue per unit is r
Profit, P — Amount of revenue remaining after costs
P = R – TC = R – (FC+VC)
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Breakeven for linear R and TC
© 2012 by McGraw-Hill All Rights Reserved 13-5
Set R = TC and solve for Q = QBE
R = TC rQ = FC + vQ FC r – v
When variable cost, v, is lowered, QBE decreases
(moves to left)
QBE =
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Example: One Project Breakeven Point
© 2012 by McGraw-Hill All Rights Reserved
Solution: Find QBE and compare to 15,000; calculate Profit
QBE = 75,000 / (8.00-2.50) = 13,636 units/month
Production level is above breakeven Profit
Profit = R – (FC + VC) = rQ – (FC + vQ) = (r-v)Q – FC = (8.00 – 2.50)(15,000) – 75,000 = $ 7500/month
13-6
A plant produces 15,000 units/month. Find breakeven level if FC = $75,000 /month, revenue is $8/unit and variable cost is $2.50/unit. Determine expected monthly profit or loss.
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Breakeven Between Two Alternatives
© 2012 by McGraw-Hill All Rights Reserved 13-7
To determine value of common variable between 2 alternatives, do the following:
1. Define the common variable 2. Develop equivalence PW, AW or FW relations as function of common
variable for each alternative 3. Equate the relations; solve for variable. This is breakeven value
Selection of alternative is based on anticipated value of common variable: Value BELOW breakeven; select higher variable cost Value ABOVE breakeven; select lower variable cost
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Example: Two Alternative Breakeven Analysis
© 2012 by McGraw-Hill All Rights Reserved 13-8
Perform a make/buy analysis where the common variable is X, the number of units produced each year. AW relations are:
AWmake = -18,000(A/P,15%,6)
+2,000(A/F,15%,6) – 0.4X
AWbuy = -1.5X
Solution: Equate AW relations, solve for X
-1.5X = -4528 - 0.4X X = 4116 per year
X, 1000 units per year
Breakeven value of X
1 2 3 4 5
AWbuy
AWmake
If anticipated production > 4116, select make alternative (lower variable cost)
AW, 1000 $/year 8 7 6 5 4 3 2 1 0
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Breakeven Analysis Using Goal Seek Tool
© 2012 by McGraw-Hill All Rights Reserved 13-9
Spreadsheet tool Goal Seek finds breakeven value for the common variable between two alternatives
Problem: Two machines (1 and 2) have following estimates. a) Use spreadsheet and AW analysis to select one at MARR = 10%. b) Use Goal Seek to find the breakeven first cost.
Machine 1 2 P, $ -80,000 -110,000 NCF, $/year 25,000 22,000 S, $ 2,000 3,000 n, years 4 6
Solution: a) Select machine A with AWA = $193
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Breakeven Analysis Using Goal Seek Tool
© 2012 by McGraw-Hill All Rights Reserved 13-10
Solution: b) Goal Seek finds a first-cost breakeven of $96,669 to make machine B economically equivalent to A
Spreadsheet after Goal Seek is applied
Changing cell
Target cell
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Payback Period Analysis
© 2012 by McGraw-Hill All Rights Reserved 13-11
Caution: Payback period analysis is a good initial screening tool, rather than the primary method to justify a project or select an alternative (Discussed later)
Payback period: Estimated amount of time (np) for cash inflows to recover an initial investment (P) plus a stated return of return (i%)
Types of payback analysis: No-return and discounted payback
1. No-return payback means rate of return is ZERO (i = 0%) 2. Discounted payback considers time value of money (i > 0%)
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Payback Period Computation
© 2012 by McGraw-Hill All Rights Reserved 13-12
Formula to determine payback period (np) varies with type of analysis.
NCF = Net Cash Flow per period t
Eqn. 1
Eqn. 2
Eqn. 3
Eqn. 4
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Points to Remember About Payback Analysis
© 2012 by McGraw-Hill All Rights Reserved 13-13
• No-return payback neglects time value of money, so no return is expected for the investment made
• No cash flows after the payback period are considered in the analysis. Return may be higher if these cash flows are expected to be positive.
• Approach of payback analysis is different from PW, AW, ROR and B/C analysis. A different alternative may be selected using payback.
• Rely on payback as a supplemental tool; use PW or AW at the MARR for a reliable decision
• Discounted payback (i > 0%) gives a good sense of the risk involved
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Example: Payback Analysis
© 2012 by McGraw-Hill All Rights Reserved 13-14
System 1 System 2 First cost, $ 12,000 8,000 NCF, $ per year 3,000 1,000 (year 1-5) 3,000 (year 6-14) Maximum life, years 7 14
Problem: Use (a) no-return payback, (b) discounted payback at 15%, and (c) PW analysis at 15% to select a system. Comment on the results. Solution: (a) Use Eqns. 1 and 2 np1 = 12,000 / 3,000 = 4 years np2 = -8,000 + 5(1,000) + 1(3,000) = 6 years
Select system 1
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Example: Payback Analysis (continued)
© 2012 by McGraw-Hill All Rights Reserved 13-15
System 1 System 2
First cost, $ 12,000 8,000 NCF, $ per year 3,000 1,000 (year 1-5) 3,000 (year 6-14) Maximum life, years 7 14
Solution: (b) Use Eqns. 3 and 4 System 1: 0 = -12,000 + 3,000(P/A,15%,np1) np1 = 6.6 years
System 2: 0 = -8,000 + 1,000(P/A,15%,5) + 3,000(P/A,15%,np2 - 5)(P/F,15%,5) np1 = 9.5 years
Select system 1
(c) Find PW over LCM of 14 years PW1 = $663 PW2 = $2470
Select system 2
Comment: PW method considers cash flows after payback period. Selection changes from system 1 to 2
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Summary of Important Points
© 2012 by McGraw-Hill All Rights Reserved 13-16
Breakeven amount is a point of indifference to accept or reject a project
One project breakeven: accept if quantity is > QBE
Two alternative breakeven: if level > breakeven, select lower variable cost alternative (smaller slope)
Payback estimates time to recover investment. Return can be i = 0% or i > 0%
Use payback as supplemental to PW or other analyses, because np neglects cash flows after payback, and if i = 0%, it neglects time value of money
Payback is useful to sense the economic risk in a project
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14-1
Lecture slides to accompany
Engineering Economy 7th edition
Leland Blank
Anthony Tarquin
Chapter 14 Effects of Inflation
© 2012 by McGraw-Hill All Rights Reserved
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14-2
LEARNING OUTCOMES
1. Understand inflation/deflation
2. Calculate PW of cash flows with inflation
3. Calculate FW with inflation considered
4. Calculate AW with inflation considered
© 2012 by McGraw-Hill All Rights Reserved
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© 2012 by McGraw-Hill All Rights Reserved 14-3
Understanding Inflation
Inflation: Increase in amount of money needed to purchase same amount of goods or services. Inflation results in a decrease in purchasing power, i.e., one unit of money buys less goods or services
Constant-value dollars = future dollars = then-current dollars (1+ f)n (1+ f)n
(1) Convert to constant value (CV) dollars, then use real rate i. If f = inflation rate (% per year), the equation is:
(2) Leave money amounts as is and use interest rate adjusted for inflation, if
if = i + f + (i)(f)
Two ways to work problems when considering inflation:
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14-4
Example: Constant Value Dollars
Solution: Solve for future dollars
How much would be required today to purchase an item that increased in cost by exactly the inflation rate? The cost 30 years ago was $1000 and inflation has consistently averaged 4% per year.
© 2012 by McGraw-Hill All Rights Reserved
Future dollars = constant value dollars(1 + f)n
= 1000(1 + 0.04)30 = $3243
Note: This calculation only accounts for the decreased purchasing power of the currency. It does not take into account the time value of money (to be discussed)
Deflation: Opposite of inflation; purchasing power of money is greater in future than at present; however, money, credit, jobs are ‘tighter’
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Three Different Rates
14-5
► Real or inflation rate i – Rate at which interest is earned when effects of inflation are removed; i represents the real increase in purchasing power
► Market or inflation-adjusted rate if – Rate that takes inflation into
account. Commonly stated rate everyday ► Inflation rate f – Rate of change in value of currency
Relation between three rates is derived using the relation
Market rate is: if = i + f + (i)(f) © 2012 by McGraw-Hill All Rights Reserved
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Example: Market vs. Real Rate
14-6
Money in a medium-risk investment makes a guaranteed 8% per year. Inflation rate has averaged 5.5% per year. What is the real rate of return on the investment?
Solution: Solve for the real rate i in relation for if
if = i + f + (i)(f) if – f 1 + f
0.08 – 0.055 1 + 0.055 0.024
Investment pays only 2.4% per year in real terms vs. the stated 8%
i =
=
=
© 2012 by McGraw-Hill All Rights Reserved
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14-7
PW Calculations with Inflation
© 2012 by McGraw-Hill All Rights Reserved
Two ways to account for inflation in PW calculations
(2) Express cash flow in future (then-current ) dollars and use inflated interest rate where if = i + f + (i)(f)
( Note: Inflated interest rate is the market interest rate)
(1) Convert cash flow into constant-value (CV) dollars and use regular i where: CV = future dollars/(1 + f)n = then-current dollars/(1 + f)n
f = inflation rate
(Note: Calculations up to now have assumed constant-value dollars)
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14-8
Example: PW with Inflation
© 2012 by McGraw-Hill All Rights Reserved
A honing machine will have a cost of $25,000 (future cost) six years from now. Find the PW of the machine, if the real interest rate is 10% per year and the inflation rate is 5% per year using (a) constant-value dollars, and (b) future dollars.
CV = 25,000 / (1 + 0.05)6 = $18,655
PW = 18,655(P/F,10%,6) = $10,530
Solution: (a Determine constant-value dollars and use i in PW equation
(b) Leave as future dollars and use if in PW equation
if = 0.10 + 0.05 + (0.10)(0.05) = 15.5%
PW = 25,000(P/F,15.5%,6) = $10,530
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14-9
FW Calculations with Inflation
© 2012 by McGraw-Hill All Rights Reserved
FW values can have four different interpretations
(2) The purchasing power in terms of CV dollars of the future amount Use if in FW equation and divide by (1+f)n or use real i where real i = (if – f)/(1 + f) FW = PW(F/P,i,n)
(3) The number of future dollars required to have the same purchasing power as a dollar today with no time value of money considered Use f instead of i in F/P factor FW = PW(F/P,f,n)
(1) The actual amount accumulated Use if in FW equation FW = PW(F/P, if, n)
(4) The amount required to maintain the purchasing power of the present sum and earn a stated real rate of return
Use if in FW equation FW = PW(F/P, if, n)
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14-10
Example: FW with Inflation
© 2012 by McGraw-Hill All Rights Reserved
An engineer invests $15,000 in a savings account that pays interest at a real 8% per year. If the inflation rate is 5% per year, determine (a) the amount of money that will be accumulated in 10 years, (b) the purchasing power of the accumulated amount (in terms of today’s dollars), (c) the number of future dollars that will have the same purchasing power as the $15,000 today, and (d) the amount to maintain purchasing power and earn a real 8% per year return.
(a) The amount accumulated is a function of the market interest rate, if if = 0.08 + 0.05 + (0.08)(0.05) = 13.4%
Amount Accumulated = 15,000(F/P,13.4%,10) = $52,750
Solution:
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14-11
Example: FW with Inflation (cont’d)
© 2012 by McGraw-Hill All Rights Reserved
(c) The number of future dollars required to purchase goods that cost $15,000 now is the inflated cost of the goods Number of future dollars = 15,000(F/P,5%,10) = $24,434
(b) To find the purchasing power of the accumulated amount deflate the inflated dollars Purchasing power = 15,000(F/P,13.4%,10) / (1 + 0.05)10 = $32,384
(d) In order to maintain purchasing power and earn a real return, money must grow by the inflation rate and the interest rate, or if = 13.4%, as in part (a) FW = 15,000(F/P,13.4%,10) = $52,750
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14-12
Capital Recovery with Inflation
© 2012 by McGraw-Hill All Rights Reserved
The A/P and A/F factors require the use of if when inflation is considered
If a small company invests $150,000 in a new production line machine, how much must it receive each year to recover the investment in 5 years? The real interest rate is 10% and the inflation rate is 4% per year.
Solution: Capital recovery (CR) is the AW value if = 0.10 + 0.04 + (0.10)(0.04) = 14.4%
CR = AW = 150,000(A/P,14.4%,5) = $44,115 per year
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14-13
Summary of Important Points
During deflation, purchasing power of money is greater in future than at present
Two ways to account for inflation in economic analyses: (1) Convert all cash flows into constant-value dollars and use i (2) Leave cash flows as inflated dollars and use if
Inflation occurs because value of currency has changed
© 2012 by McGraw-Hill All Rights Reserved
Future worth values can have four different interpretations, requiring different interest rates to find FW
Use if in calculations involving A/P or A/F when inflation is considered
Inflation reduces purchasing power; one unit buys less goods ort services
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15-1
Lecture slides to accompany
Engineering Economy 7th edition
Leland Blank
Anthony Tarquin
Chapter 15 Cost Estimation
and Indirect Costs
© 2012 by McGraw-Hill All Rights Reserved
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LEARNING OUTCOMES
© 2012 by McGraw-Hill All Rights Reserved 15-2
1. Approaches to estimation
2. Unit method
3. Cost indexes
4. Cost-capacity equations
5. Factor method
6. Indirect cost rates and allocation
7. ABC allocation
8. Ethical considerations
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Direct and Indirect Cost Estimates
© 2012 by McGraw-Hill All Rights Reserved 15-3
Direct cost examples • Physical assets • Maintenance and operating
costs (M&O) • Materials • Direct human labor (costs
and benefits) • Scrapped and reworked
product • Direct supervision of
personnel
Indirect cost examples • Utilities • IT systems and networks • Purchasing • Management • Taxes • Legal functions • Warranty and guarantees • Quality assurance • Accounting functions • Marketing and publicity
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What Direct Cost Estimation Includes
© 2012 by McGraw-Hill All Rights Reserved 15-4
Direct costs are more commonly estimated than revenue in an engineering environment. Preliminary decisions required are:
What cost components should be estimated? What approach to estimation is best to apply? How accurate should the estimates be? What technique(s) will be applied to estimate costs?
Sample direct cost components: first costs and its elements (P); annual costs (AOC or M&O); salvage/market value (S) Approaches: bottom-up; design-to-cost (top down) Accuracy: feasibility stage through detailed design estimates require more exacting estimates Some techniques: unit; factor; cost estimating relations (CER)
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Different Approaches to Cost Estimation
© 2012 by McGraw-Hill All Rights Reserved 15-5
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Accuracy of Cost Estimates
© 2012 by McGraw-Hill All Rights Reserved 15-6
Characteristic curve of accuracy vs. time to make estimates
General guidelines for accuracy Conceptual/Feasibility stage – order-of-magnitude estimates are in range of ±20% of actual costs
Detailed design stage - Detailed estimates are in range of ±5% of actual costs
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Unit Method
© 2012 by McGraw-Hill All Rights Reserved 15-7
• Commonly used technique for preliminary design stage estimates
• Total cost estimate CT is per unit cost (u) times number of units (N)
CT = u × N • Example uses:
Cost to operate a car at 60¢/mile for 500 miles: CT = 0.60 × 500 = $300 Cost to build a 250 m2 house at $2250/m2: CT = 2250 × 250 = $562,500 • Cost factors must be updated periodically to remain timely
When several components are involved, estimate cost of each component and add to determine total cost estimate CT
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Cost Indexes
© 2012 by McGraw-Hill All Rights Reserved 15-8
Definition: Cost Index is ratio of cost today to cost in the past • Indicates change in cost over time; therefore, they account for
the impact of inflation • Index is dimensionless • CPI (Consumer Price Index) is a good example
Formula for total cost is
Formula for total cost is
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Example: Cost Index Method
© 2012 by McGraw-Hill All Rights Reserved 15-9
Problem: Estimate the total cost of labor today in US dollars for a maritime construction project using data from a similar project in Europe completed in 1998.
Labor index, 1998: 789.6 Cost in 1998: €3.9 million Labor index, current: 1165.8 Currently, 1 € = 1.5 US$ Solution: Let t = today and 0 = 1998 base Ct = 3.9 million × (1165.8/789.6) = €5.76 million = €5.76 × 1.5 = $8.64 million
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Finding Cost Indexes
© 2012 by McGraw-Hill All Rights Reserved 15-10
Cost indexes are maintained in areas such as construction, chemical and mechanical industries • Updated monthly and annually; many include regionalized and international project indexes • Indexes in these areas are often subdivided into smaller components and can be used in preliminary, as well as detailed design stages Examples are: Chemical Engineering Plant Cost Index (CEPCI) www.che.com/pci McGraw-Hill Construction Index www.construction.com US Department of Labor, Bureau of Labor Statistics www.bls.gov
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Cost-Estimating Relationships (CER)
© 2012 by McGraw-Hill All Rights Reserved 15-11
CER equations are used in early design stages to estimate plant, equipment and construction costs CERs are generically different from index relations, because they estimate based on design variables (weight, thrust, force, pressure, speed, etc.)
Two commonly used CERs
Cost-capacity equation (relates cost to capacity) Factor method (total plant cost estimator, including indirect costs)
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Cost-Capacity Equation
© 2012 by McGraw-Hill All Rights Reserved 15-12
Also called power law and sizing model
Exponent defines relation between capacities
x = 1, relationship is linear x < 1, economies of scale (larger capacity is less costly than linear) x > 1, diseconomies of scale
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Cost-Capacity Combined with Cost Index
© 2012 by McGraw-Hill All Rights Reserved 15-13
Example: A 100 hp air compressor costs $3000 five years ago when the cost index was 130. Estimate the cost of a 300 hp compressor today when the cost index is 255. The exponent for a 300 hp air compressor is 0.9. Solution: Let C300 represent the cost estimate today
C300 = 3000(300/100)0.9(255/130) = $15,817
Multiply the cost-capacity equation by a cost index (It/I0) to adjust for time differences and obtain estimates of current cost (in constant-value dollars)
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Factor Method
© 2012 by McGraw-Hill All Rights Reserved 15-14
Factor method is especially useful in estimating total plant cost in processing industries Both direct and indirect costs can be included
Total plant cost estimate CT is overall cost factor (h) times total cost of major equipment items (CE)
CT = h × CE
Overall cost factor h is determined using one of two bases:
Delivered-equipment cost (purchase cost of major equipment)
Installed-equipment cost (equipment cost plus all make-ready costs)
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Cost Factor h
© 2012 by McGraw-Hill All Rights Reserved 15-15
The cost factor is commonly the sum of a direct cost component and an indirect cost component, that is,
h = 1 + Σfi
for i = 1, 2, …, n components, including indirect costs Example: Equipment is expected to cost $20 million delivered to a
new facility. A cost factor for direct costs of 1.61 will make the plant ready to operate. An indirect cost factor of 0.25 is used. What will the plant cost?
Solution: h = 1 + 1.61 + 0.25 = 2.86
CT = 20 million (2.86) = $57.2 million
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Cost Factor h If indirect costs are charged separately against all direct costs,
the indirect cost component is added separately, that is,
h = 1 + Σfi (direct costs components)
and CT = hCE(1 + findirect) Example: Conveyor delivered-equipment cost is $1.2 million. Factors for installation costs (0.4) and training (0.2) are determined. An indirect cost factor of 0.3 is applied to all direct costs. Estimate total cost.
Solution: h = 1 + 0.4 + 0.2 = 1.6
CT = hCE(1 + findirect)
= 1.6(1.2 million)(1 + 0.3) = $2.5 million © 2012 by McGraw-Hill All Rights Reserved 15-16
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Indirect Costs
© 2012 by McGraw-Hill All Rights Reserved 15-17
Indirect costs (IDC) are incurred in production, processes and service delivery that are not easily tracked and assignable to a specific function.
Indirect costs (IDC) are shared by many functions because they are necessary to perform the overall objective of the company
Indirect costs make up a
significant percentage of the overall costs in many organizations – 25 to 50%
Sample indirect costs IT services Quality assurance Human resources Management Safety and security Purchasing; contracting Accounting; finance; legal
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Indirect Cost Allocation - Traditional Method
Cost center -- Department, function, or process used by the cost accounting system to collect both direct and indirect costs
Indirect-cost rate – Traditionally, a predetermined rate is used to allocate indirect costs to a cost center using a specified basis. General relation is:
© 2012 by McGraw-Hill All Rights Reserved 15-18
Estimated total indirect costs Estimated basis level
Example:
Allocation rates for $50,000 to each machine
Indirect-cost rate =
Machine 1: Rate = $50,000/100,000 = $0.50 per DL $ Machine 2: Rate = $50,000/2,000 = $25 per DL hour Machine 3: Rate = $50,000/250,000 = $0.20 per DM $
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Example: AW Analysis - Traditional IDC Allocation
© 2012 by McGraw-Hill All Rights Reserved 15-19
MAKE/BUY DECISION
Buy: AW = $-2.2 million per year
Make: P = $-2 million S = $50,000 n = 10 years MARR = 15% • Direct costs of $800,000 per year are detailed below • Indirect cost rates are established by department
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© 2012 by McGraw-Hill All Rights Reserved 15-20
Example: Indirect Cost Analysis - Traditional Method
INDIRECT COST ALLOCATION FOR MAKE ALTERNATIVE
Dept A: Basis is -- Direct labor hours 25,000(10) = $250,000 Dept B: Basis is -- Machine hours 25,000(5) = $125,000 Dept C: Basis is -- Direct labor hours 10,000(15) = $150,000
ECONOMIC COMPARISON AT MARR = 15%
AOCmake = direct labor + direct materials + indirect allocation = 500,000 + 300,000 + 525,000 = $1.325 M AWmake = - 2 M(A/P,15%,10) + 50,000(A/F,15%,10) - 1.325 M = $-1.72 M AWbuy = $-2.2 M
$525,000
Conclusion: Cheaper to make
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ABC Allocation Activity-Based Costing ─ Provides excellent allocation strategy and analysis of
costs for more advanced, high overhead, technologically-based systems
Cost Centers (cost pools) ─ Final products/services that receive allocations
Activities ─ Support departments that generate indirect costs for distribution to cost centers (maintenance, engineering, management)
Cost drivers ─ These are the volumes that drive consumption of shared resources (# of POs, # of machine setups, # of safety violations, # of scrapped items)
© 2012 by McGraw-Hill All Rights Reserved 15-21
Steps to implement ABC: 1. Identify each activity and its total cost (e.g., maintenance at $5 million/year) 2. Identify cost drivers and expected volume (e.g., 3,500 requested repairs and 500 scheduled maintenances per year) 3. Calculate cost rate for each activity using the relation:
ABC rate = total activity cost/volume of cost driver
4. Use ABC rate to allocate IDC to cost centers for each activity
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Example: ABC Allocation
Use ABC to allocate safety program costs to plants in US and Europe
Cost centers: US and European plants Activity and cost: Safety program costs $200,200 per year Cost driver: # of accidents Volume: 560 accidents; 425 in US plants and 135 in European plants
© 2012 by McGraw-Hill All Rights Reserved 15-22
Solution: ABC rate for accident basis = 200,200/560 = $357.50/accident
US allocation: 357.50(425) = $151,938 Europe allocation: 357.50(135) = $48,262
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Example: Traditional Allocation Comparison
© 2012 by McGraw-Hill All Rights Reserved 15-23
Solution: Rate for employee basis = 200,200/1400 = $143/employee
US allocation: 143(900) = $128,700 Europe allocation: 143(500) = $71,500
Use traditional rates to allocate safety costs to US and EU plants
Cost centers: US and European plants Activity and cost: Safety program costs $200,200 per year Basis: # of employees Volume: 1400 employees; 900 in US plants and 500 in European plants
Comparison: US allocation went down; European allocation increased
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Traditional vs. ABC Allocation
© 2012 by McGraw-Hill All Rights Reserved 15-24
o Traditional method is easier to set up and use o Traditional method is usually better when making
cost estimates o ABC is more accurate when process is in operation o ABC is more costly, but provides more information
for cost analysis and decision making o Traditional and ABC methods complement each
other: Traditional is good for cost estimation and allocation ABC is better for cost tracking and cost control
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Ethics and Cost Estimating Unethical practices in estimation may be the result of:
Personal gain motivation
Bias
Deception
Favoritism toward an individual or organization
Intentional poor accuracy
Pre-arranged financial favors (bribes, kickbacks)
When making any type of estimates, always comply with the
Code of Ethics for Engineers
© 2012 by McGraw-Hill All Rights Reserved 15-25
Avoid deceptive acts
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Summary of Important Points
© 2012 by McGraw-Hill All Rights Reserved 15-26
Required accuracy of cost estimates depends on the stage of a system design; accuracy varies from ±20% to ±5% of actual cost
Costs can be updated using the unit method and cost indexes, where time differences are considered (inflation over time)
The factor method estimates total plant costs, including indirect costs
Traditional indirect cost allocation use bases such as direct labor hours, costs, and direct materials
Indirect costs comprise a large percentage of product and service costs
The ABC method of indirect cost allocation uses cost drivers to allocate to cost centers; it is better for understanding and analyzing cost accumulation
Unethical practices in cost estimation result from personal financial motives, deception, financial pre-arrangements. Avoid deceptive acts
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16-1
Lecture slides to accompany
Engineering Economy 7th edition
Leland Blank
Anthony Tarquin
Chapter 16 Depreciation
Methods
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16-2
LEARNING OUTCOMES
1. Understand basic terms of asset depreciation
2. Apply straight line method of depreciation
3. Apply DB and DDB methods of depreciation; switch between DDB and SL methods
4. Apply MACRS method of depreciation
5. Select asset recovery period for MACRS
6. Explain depletion and apply cost depletion & percentage depletion methods
© 2012 by McGraw-Hill All Rights Reserved
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© 2012 by McGraw-Hill All Rights Reserved 16-3
Depreciation Terminology
Definition: Book (noncash) method to represent decrease in value of a tangible asset over time
Two types: book depreciation and tax depreciation
Book depreciation: used for internal accounting to track value of assets
Tax depreciation: used to determine taxes due based on tax laws
In USA only, tax depreciation must be calculated using MACRS; book depreciation can be calculated using any method
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16-4
Common Depreciation Terms
© 2012 by McGraw-Hill All Rights Reserved
First cost P or unadjusted basis B: Total installed cost of asset
Book value BVt: Remaining undepreciated capital investment in year t
Recovery period n: Depreciable life of asset in years
Market value MV: Amount realizable if asset were sold on open market
Salvage value S: Estimated trade-in or MV at end of asset’s useful life
Depreciation rate dt: Fraction of first cost or basis removed each year t
Personal property: Possessions of company used to conduct business
Real property: Real estate and all improvements (land is not depreciable)
Half-year convention: Assumes assets are placed in service in midyear
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16-5
Straight Line Depreciation
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Book value decreases linearly with time
Dt = B - S n
Where: Dt = annual depreciation charge t = year B = first cost or unadjusted basis S = salvage value n = recovery period
BVt = B - tDt Where: BVt = book value after t years
SL depreciation rate is constant for each year: d = dt = 1/n
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16-6
Example: SL Depreciation
© 2012 by McGraw-Hill All Rights Reserved
Solution: (a ) D3 = (B – S)/n = (20,000 – 5,000)/5 = $3,000
(b) BV3 = B – tDt
= 20,000 – 3(3,000) = $11,000
An argon gas processor has a first cost of $20,000 with a $5,000 salvage value after 5 years. Find (a) D3 and (b) BV3 for year three. (c) Plot book value vs. time.
(c) Plot BV vs. time
20,000
11,000
3 5
5,000
0 Year, t
BVt
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16-7
Declining Balance (DB) and Double Declining Balance (DDB) Depreciation
© 2012 by McGraw-Hill All Rights Reserved
Determined by multiplying BV at beginning of year by fixed percentage d
Max rate for d is twice straight line rate, i.e., d ≤ 2/n Cannot depreciate below salvage value
Depreciation for year t is obtained by either relation: Dt = dB(1 – d)t-1 = dBVt-1 Where: Dt = depreciation for year t d = uniform depreciation rate (2/n for DDB)
B = first cost or unadjusted basis BVt -1 = book value at end of previous year
Book value for year t is given by:
BVt = B(1 – d)t
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16-8
Example: Double Declining Balance
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(b) BV3 = B(1 – d)t
= 20,000(1 – 0.4)3 = $4320
A depreciable construction truck has a first cost of $20,000 with a $4,000 salvage value after 5 years. Find the (a) depreciation, and (b) book value after 3 years using DDB depreciation.
Solution: (a) d = 2/n = 2/5 = 0.4 D3 = dB(1 – d)t-1 = 0.4(20,000)(1 – 0.40)3-1 = $2880
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16-9
Spreadsheet Functions for Depreciation
© 2012 by McGraw-Hill All Rights Reserved
Straight line function: SLN(B,S,n)
Declining balance function: DB(B,S,n,t)
Double declining balance function: DDB(B,S,n,t,d)
Note: It is better to use the DDB function for DB and DDB depreciation. DDB function checks for BV < S and is more accurate than the DB function.
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Switching Between Depreciation Methods Switch between methods to maximize PW of depreciation
PWD = ∑ Dt (P/F,i%,t)
A switch from DDB to SL in latter part of life is usually better
Can switch only one time during recovery period
16-10 © 2012 by McGraw-Hill All Rights Reserved
t = 1
t = n
Procedure to switch from DDB to SL: 1) Each year t compute DDB and SL depreciation using the relations
DDDB = d(BVt-1) and DSL = BVt-1 / (n-t+1)
2) Select larger depreciation amount, i.e., Dt = max[DDDB, DSL]
3) If required, calculate PWD
Alternatively, use spreadsheet function VDB(B,S,n,start_t,end_t) to determine Dt
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16-11
MACRS Depreciation
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Required method to use for tax depreciation in USA only
Originally developed to offer accelerated depreciation for economic growth
Dt = dtB Where: Dt = depreciation charge for year t B = first cost or unadjusted basis dt = depreciation rate for year t (decimal)
BVt = B - ∑Dj Where: Dj = depreciation in year j ∑ Dj = all depreciation through year t j = 1
j = t
Get value for dt from IRS table for MACRS rates
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16-12
MACRS Depreciation
© 2012 by McGraw-Hill All Rights Reserved
Always depreciates to zero; no salvage value considered
Incorporates switching from DDB to SL depreciation
MACRS recovery time is always n+1 years; half-year convention assumes purchase in midyear
Standardized recovery periods (n) are tabulated
No special spreadsheet function; can arrange VDB function to display MACRS depreciation each year
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16-13
Example: MACRS Depreciation
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Solution:
A finishing machine has a first cost of $20,000 with a $5,000 salvage value after 5 years. Using MACRS, find (a) D and (b) BV for year 3.
(a) From table, d3 = 19.20 D3 = 20,000(0.1920) = $3,840
(b) BV3 = 20,000 - 20,000(0.20 + 0.32 + 0.1920) = $5,760
Note: Salvage value S = $5,000 is not used by MACRS and BV6 = 0
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© 2012 by McGraw-Hill All Rights Reserved 16-14
MACRS Recovery Period Recovery period (n) is function of property class
Two systems for determining recovery period:
general depreciation system (GDS) – fastest write-off allowed
alternative depreciation system (ADS) – longer recovery; uses SL
IRS publication 946 gives n values for an asset. For example: MACRS n value Asset description GDS ADS range Special manufacturing devices, racehorses, tractors 3 3 - 5 Computers, oil drilling equipment, autos, trucks, buses 5 6 - 9.5 Office furniture, railroad car, property not in another class 7 10 – 15 Nonresidential real property (not land itself) 39 40
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Unit-of-Production (UOP) Depreciation Depreciation based on usage of equipment, not time Depreciation for year t obtained by relation Dt = (B – S) Example: A new mixer is expected to process 4 million yd3 of concrete over 10-year life time. Determine depreciation for year 1 when 400,000 yd3 is
processed. Cost of mixer was $175,000 with no salvage expected. Solution: D1 = (175,000 – 0) = $17,500
© 2012 by McGraw-Hill All Rights Reserved 16-15
actual usage for year t expected total lifetime usage
400,000 4,000,000
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16-16 © 2012 by McGraw-Hill All Rights Reserved
Depletion Methods Depletion: book (noncash) method to represent decreasing value of natural resources
Two methods: cost depletion (CD) and percentage depletion (PD)
Cost depletion: Based on level of activity to remove a natural resource Calculation: Multiply factor CDt by amount of resource removed Where: CDt = first cost / resource capacity Total depletion can not exceed first cost of the resource
Percentage depletion: Based on gross income (GI) from resource Calculation: Multiply GI by standardized rate (%) from table Annual depletion can not exceed 50% of company’s taxable income (TI)
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© 2012 by McGraw-Hill All Rights Reserved 16-17
Example: Cost and Percentage Depletion A mine purchased for $3.5 million has a total expected yield of one million ounces of silver. Determine the depletion charge in year 4 when 300,000 ounces are mined and sold for $30 per ounce using (a) cost depletion, and (b) percentage depletion. (c) Which is larger for year 4?
Solution: Let depletion amounts equal CDA4 and PDA4
(a) Factor, CD4 = 3,500,000/ 1,000,000 = $3.50 per ounce CDA4 = 3.50(300,000) = $1,050,000
(b) Percentage depletion rate for silver mines is 0.15 PDA4 = (0.15)(300,000)(30) = $1,350,000
(c) Claim percentage depletion amount, provided it is ≤ 50% of TI
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16-18
Summary of Important Points
© 2012 by McGraw-Hill All Rights Reserved
Two methods of depletion: cost (amount resource removed × CDt factor) and percentage (gross income × tabulated %)
Two types for depreciation: tax and book
In USA only, MACRS method is required for tax depreciation
Determine MACRS recovery period using either GDS or ADS
Depletion (instead of depreciation) used for natural resources
Classical methods are straight line and declining balance
Switching between methods is allowed; MACRS switches automatically from DDB to SL to maximize write-off