Table of Contents
Chapter
Quiz 1
Introduction to Engineering Economy
Analysis for Compounding Periods Not Equal to Payments
Arithmetic Gradient Series
Bond Valuation
Bond Valuation Problem Set
Derivation of Linear Gradient Formulas
Financing Decisions
Interest and Money – Time Relationships Part I
Interest and Money – Time Relationships Part II
Selections in Present Economy
Selections in Present Economy Problem Set
Quiz 2
Depreciation
Applications of Money-Time Relationships
Applications of Money-Time Relationships for Multi-Business Project Analysis
Interest and Money-Time Relationship Problem Set
Evaluating Single Projects Problem Set
Money-Time Relationships Problem Set
Quiz 3
After Tax Cash Flow Analysis, Capital Gains (Loss) and Replacement Analysis
Inflation
Inflation Problem Set
Risk in Capital Budgeting
Risk in Capital Budgeting Problem Set
Dealing with Uncertainty
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INTRODUCTION TO ENGINEERING ECONOMY
Engineering economy
–
The science that deals with the study of how economic theories and laws
are used to solve engineering problems.
–
Discipline concerned with the economic aspects of engineering and
involves the systematic evaluation of the costs and benefits of proposed
technical and business projects and ventures.
–
To be economically acceptable (affordable), solutions to engineering
problems must demonstrate a positive balance of long term benefits over
long term costs and they must also:
 Promote the well being and survival of the organization
 Embody creative and innovative technology and ideas
 Permit identification and scrutiny of their estimated outcomes
 Translate profitability to the “bottom line” through a valid and
acceptable measure of merit.
Applications of Engineering Economy
1. Set additional objectives or uses for a particular product or multi-product
2. Discover key factors for the success/failure of the company
3. Comparison of alternatives
4. Analysis of investment of capital
Principles of Engineering Economy
1. Develop the alternatives.
The choice is among alternatives. The alternatives need to be identified and
then defined for subsequent analysis.
2. Focus on the differences
Only the differences in expected future outcomes among the alternatives are
relevant to their comparison and should be considered in the decision.
3. Use a consistent point of view
The prospective outcomes of the alternatives economic and other, should be
consistently developed from a defined viewpoint (perspective).
4. Use a common unit of measure
Using a common unit of measurement to enumerate as many of the
prospective outcomes as possible will make easier the analysis and
comparison of the alternatives.
5. Consider all relevant criteria
1
Selection of a preferred alternative (decision making) requires the use of a
criterion (or several criteria). The decision process should consider both the
outcomes enumerated in the monetary unit and those expressed in some
other unit of measurement or made explicit in a descriptive manner.
6. Make uncertainty explicit
Uncertainty is inherent in projecting (or estimating) the future outcomes of
the alternatives and should be recognized in these analysis and comparison.
7. Revisit your decisions
Improved decision making results from an adaptive process; to the extent
practicable, the initial projected outcomes of the selected alternative should
be subsequently compared with actual results achieved.
Procedure of Engineering Economic Analysis
1. Problem recognition, definition and evaluation
2. Development of feasible alternatives
3. Development of the cash flows for each alternative
4. Selection of a criterion
5. Analysis and comparison of alternatives
6. Selection of preferred alternative
7. Performance monitoring and post evaluation
Accounting and Engineering Economy Studies
General Accounting / Financial Accounting / Cost Accounting
–
Procedures that provide financial events relating to the investment to
determine financial performance.
–
These accounting data are primarily concerned with past and current
financial events even though such data are often used to make projections
about the future.
–
Controls are established and utilized to aid in guiding the operation toward
desired financial goals
General Accounting – source of much of the past financial data needed in making
estimates of future financial conditions. Accounting is also a source of data for
analysis that might be made regarding how well the results of a capital investment
turned out compared to the results that were predicted in the engineering economic
analysis.
Cost Accounting / Management Accounting – is a phase of accounting that is of
particular importance because it is concerned principally with decision making and
control in a firm
2
 It is also the source of some of the cost data that are needed in engineering
economic studies.
 Modern cost accounting may satisfy any or all of the following objectives:
1. Determination of the cost of products or services
2. Provision of a rational basis for pricing goods or services
3. Provision of a means for controlling expenditures
4. Provision of information on which operating decisions may be based
and the results evaluated.
ANALYSIS FOR COMPOUNDING PERIODS NOT EQUAL TO PAYMENTS
Most Common Solution:
Convert interest rate to a period equal to that of payment by using effective interest rate concept
A. Compounding More Frequent Than Payments
Suppose you make equal quarterly deposits of $1,000 into a fund that pays interest at a rate of
12% compounded monthly. Find the balance at the end of 2 years.
First solution: convert interest period to payment period
Second solution: convert payment period into interest period
B. Compounding Less Frequent Than Payments
Suppose you make P500 monthly deposits to a tax deferred retirement plan that pays interest at a
rate of 10% compounded quarterly. Compute the balance at the end of 10 years.
First solution: convert interest period to payment period
Second solution: convert payment period into interest period
Predictions and
projections
Data for projections in
the future
Engineering Economy
- gives information on
which decisions pertaining
to the FUTURE operation
of an organization can be
based
Accounting
3
- is concerned with the
past and current financial
events so that the financial
performance can be
determined.
ARITHMETIC GRADIENT SERIES (Applicable to price changes due to inflation)
2
3
n-1
C(1+g)
4
C(1+g)
2
C(1+g)
C(1+g)
C
1
n-2
Derivation of Formulas:
n-1
n
P
F
P  C 1  i   C 1  g 1  i   ...  C 1  g 
1
2
C 1  g 
C C 1  g 


 ... 
n
1 i
1 i
1  i 
2
n 1
n 1
 1
1 g  

1 g
C

 ... 

2
n
1  i 1  i 
1  i  
Multiply
1
by
n 1
1
1 g
1 g
4
1  i 
n
g = rate of increase


2
n 

1  g  
C 1  g 1  g 
P

 ... 
2
n
1  g  1  i 1  i 
1 i 

 1 4 4 4 4 4 2 4 4 4 4 43 
Sum of Geometric Sequence


a
1 g
1 i
n
1 g
1 i
  1  g n 
 1

 C  1  g    1  i 




1  g   1  i    1  g   1 
  1 i 






n
n


 C   1  g  1  i   1 


1  i   1  g   1  i  


1  i 


 1  g n 1  i  n  1 
C
 if g  i
g i


But if g  i ;
Cn
1 g
For Future Worth Value:
P
if i  g
 1  g n 1  i  n  1 
n
F C
 1  i  

g i

 
 1  g n  1  i n 
C

g i


if i  g
F
Cn 1  i 
n
1 g
1
P
2
5
3
4
P1762.84
P1573.52
P1404.93
P1254.40
P1120
P1000
Sample Problems:
1. For the cash flow shown below, find the values of P and F if i  15% per year:
5
6
g  12% ,
i  15%
2. Eleanor makes year-end deposits of P500 the first year, P550 the second year, P605 the third
year, and so on increasing the net year’s deposit by 105 of the deposit in the preceding year
until the end of the tenth year. Ronald makes equal year-end deposits of P700 each year for
10 years. If interest on both funds is 12% compounded annually, who will be able to save
more at the end of 10 yrs. And by what amount?
Bond Valuation
Value of a Bond – simply the present value of the cash flows the asset (bond)is expected to produce.
M
INT INT INT INT
Kd%
Bond’s
Value
1
N
N
Bond’s Value (VB) =
2
INT
 (1  k
t 1
d
)
t
3

INT
4
M
(1  k d ) N
if coupon interest payment is annual
Annuity
Where: kd = bond’s market interest rate
N = no. of years (or periods) before the bond matures
INT= interest paid each year (or period)
= coupon rate x par value
M = the par or maturity value of the bond
6
If coupon interest payment is:
a. Semi-annual
N
INT
M
Bond’s Value (VB) = 
; INT is interest received semi-annually

kd t
kd N
t 1
(1  )
(1  )
2
2
b. Quarterly
N
INT
M
Bond’s Value (VB) = 
; INT is interest received quarterly

kd t
kd N
t 1
(1  )
(1  )
4
4
c. Monthly
N
INT
M
Bond’s Value (VB) = 
; INT is interest received monthly

kd t
kd N
t 1
(1  )
(1  )
12
12
Example:
Allied Foods issued $1,000 par value 15 years life bonds with 10% coupon rate to be paid at the end
of each year. Compute for the value of this bond is market interest rate at the time of issue is:
a. 10%
b. 5%
c. 15%
Notes:
1. Whenever the going rate of interest, kd is equal to the coupon rate, a fixed-rate bond will sell at
its par value. Normally, the coupon rate is equal to the going rate when a bond is issued, causing
it to sell at par initially.
2. Interest rates do change over time, but the coupon rate remains fixed after the bond has been
issued. Whenever the going rate of interest rises above the coupon rate, a fixed rate bond’s price
will fall below its par value. Such a bond is called a discount bond.
3. Whenever the going rate of interest falls below the coupon rate, a fixed rate bond’s price will rise
above its par value. Such a bond is called a premium bond.
4. Increase in kd will cause the prices of outstanding bonds to fall, whereas a decrease in rates will
cause bond’s prices to rise.
5. The market value of a bond will always approach its par value as its maturity date approaches,
provided the firm will not go bankcrupt.
Bond Yields
1. YTM – Yield to Maturity (rate of interest one would earn if a bond will be held to maturity)
N
INT
M
Bond’s Value (VB) = 
; if kd is unknown, it becomes the YTM.

t
(1  k d ) N
t 1 (1  k d )
(Formula assumes annual coupon payment.)
2. YTC-Yield to Call (rate of interest one would earn if a bond will be called)
N
INT
Call Pr ice
Bond’s Value (VB) = 
; if kd is unknown, it becomes the YTC.

t
(1  k d ) N
t 1 (1  k d )
(Formula assumes annual coupon payment.)
7
3. Current Yield = annual interest/bond’s current price
- does not represent the return that investors should expect to receive from holding the bond.
Example:
Suppose Allied’s bonds had a provision that permitted the company if it is desired, to call the bonds
10 years after the issue date at a price of $1,100. I f you bought the bonds one year after issuance at a
price of $1,494.932, what is the YTC of the bond?
Zero Coupon Bond
Example:
Vandenburg Corporation, a shopping center developer is in need of $50M. The company does not
anticipate major cash flows from the project for about 5 years. Therefore, it decided to issue a 5-year
zero coupon bond with maturity value of $1,000. Bond-yield at issue time is 6%.
a. What should be the issue price of the bond?
b. What should be the total face value of the bonds the company must offer to raise $50M?
Problems on Bond Valuation
1. The Pennington Corporation issued a new series of bonds on January 1, 1974. The bonds
were sold at par ($1,000), have a 12 percent coupon, and mature in 30 years, on December 31,
2003. Coupon payments are made semiannually (on June 30 and December31).
a. What was the YTM of Pennington’s bonds on January 1,1974
b. What was the price of the bond on January 1, 1979, 5 years later, assuming that the level of
interest rates had fallen to 10 percent?
c. On July 1, 1997, Pennington’s bonds sold for $916.42. What was the YTM at that date?
d. Now, assume that you purchased and outstanding Pennington bond on March 1, 1997, when
the going rate of interest was 15.5 percent. How large a check must you have written to
complete the transaction?
2. Callaghan Motors’ bonds have 10 years remaining to maturity. Interest is paid annually, the
bonds have a $1,000 par value, and the coupon interest rate is 8 percent. The bonds have a yield
to maturity of 9 percent. What is the current market price of these bonds?
3. The Heymann Company’s bonds have 4 years remaining to maturity. Interest is paid
annually; the bonds have a $1,000 par value; and the coupon interest rate is 9 percent.
a. What is the yield to maturity at a current market price of (1) $829 or (2) $1,104?
b. Would you pay $829 for one of these bonds if you thought that the appropriate rate of
interest was 12 percent- that is, if kd = 12% Explain your answer.
8
4. Six years ago, The Singleton Company sold a 20-year bond issue with a 14 percent annual
coupon rate and a 9 percent call premium. Today, Singleton called the bonds. The bonds
originally were sold at their face value of $1,000. Compute the realized rate or return for
investors who purchased the bonds when they were issued and who surrender them today in
exchange for the call price.
5. A 10-year, 12 percent semiannual coupon band, with a par value of $1,000, may
be called
in 4 years at a call price of $1,060. The sells for $1,100 (Assume that the bond has been
issued.)
a. What is the bond’s yield to maturity?
b. What is the bond’s current yield?
c. What is the bond’s yield to call?
6. You just purchased a bond which matures in 5 years. The bond has a face value of $1,000,
and has an 8 percent annual coupon. The bond has a current yield of 8.21 percent. What is the
bond’s yield to maturity?
7. A bond which matures in 7 years sells for $1,020 . The bond has a face value of $ 1,000 and a
yield to maturity of 10.5883 percent. The bond pays coupons semiannually. What is the bond’s
current yield?
8. Suppose Ford Motor Company sold an issue of bonds with a 10-year maturity, a $1,000 par
value, a 10 percent coupon rate, and semiannual interest payments.
a. Two years after the bonds were issued, the going rate of interest on bonds such as
these fell to 6 percent. At what price would the bonds sell?
b. Suppose that, 2 years after the initial offering, the going interest rate had risen to12
percent. At what price would the bonds sell?
c. Suppose that the conditions in Part a existed- that is, interest rates fell to6 percent 2
years after the issue date. Suppose further that the interest rate remained at six
percent for the next 8 years. What would happen to the price of the Ford Motor
Company bond over time?
9
DERIVATION OF LINEAR GRADIENT FORMULAS
1
G
2G
3G
(n-2)G (n-1)G
2
3
4
n-1
n
P
F
P  G1  i   2G1  i   ...  n  2G1  i 
2
Multiply1
3
 n 1
 n  1G1  i 
1
1
by 1  i  :
 1
n  2  n  1 
2
P1  i   G 

 ... 
1
2
1  i n  2 1  i n 1 
 1  i  1  i 
Subtract
n
from
2



 n  1 
1
1
1
Pi  G 


...


n 1
n 1
 1  i 1 1  i 2
1 i
1 i 


1 4 4 4 4 4 2 4 4 4 4 4 3

Geometric Sequence to n-1


10
2
Sum of Geometric Sequence -
a
1
1
,r 
1  i 
1  i 
 rn 1

sum  a
 r 1 
1
1  i n
a  ar  ar 2  ar 3  ...  ar n


  1 n




1



1
n 1 
 1   1  i 

Pi  G 
  1  i n  1  i n 
1

1  i 

1




1  i 








G  1  i   1  1  n  1



i  1  1  i 
1  i n 
n
P
n
G  1  i   1
n 
P 


i 
i
1  i n 
G 1  1  i 
P 
i 
i
n
F  P1  i 
n
G 1  1  i 

i 
i
n


n

G  1  i 
F 
 n
i  i



n
 1  i 
1  i  
n
n




1  i n 
n
 i1  i n 
A  P

n
 1  i   1
n
n
G 1  1  i 
n   i1  i  
 



i 
i
1  i n   1  i n  1
n

G  1  i   1
in
 


i  1  i n  1 1  i n  1
1

n
A  G 

n
 i 1  i   1
11
Financing Decision
- determine the optimal capital structures of the firm
Capital Structures
- a mix of long-term sources of funds of the co.
Liabilities
Equity
(creditors)
(owners)
Objective: determine the mix that will maximize the market value of the firm
Types of Financing
1. Debt Financing
2. Equity Financing
Preferred Stock Financing
Common Stock Financing
Debt financing
- the firm borrows from an external source(s) to raise funds.
- Principal amt. loaned and interests to be paid are stipulated in a debt contract
(including when and how much will be paid)
- Company is obligated to pay principal and interests regardless of its earnings
12
-
-
Lenders (creditors) do not/cannot exercise direct control over the company but can
place certain restrictions on the company w/c when violated gives them the right to
exert some influences on the company’s direction
Claim of lenders (creditors) are prioritized over that of equity holders in case of
liquidation
Interest paid to creditors are tax-deductible (as seen in income statement of the
borrowing company)
Bonds – a contract or agreement that a borrower has to pay interest and principal at specific
dates. (actually, a certificate of indebtedness)
4 Main types of Bonds:
1. Treasury Bonds – referred to as government bonds, these are issued by the government.
2. Corporate Bonds – bonds issued by corporations and are exposed to default risk (inability
to make promised principal and interest payments)
3. Municipal Bonds – or “munis”, these are issued by state and local governments and are
exposed to default risk. (main advantage: interest earned by holders are exempted from
taxes)
4. Foreign Bonds – bonds issued by foreign governments or corporations. Main risk:
currency exchange movement. (happen if bonds are denominated in a currency other than
of that of the investor’s home currency)
Key Characteristics of bonds:
- all bonds have some common characteristics, they usually differ only on some
contractual features.
Bond Terminology:
1. Par Value – face value of a stock or bond
2. Coupon Payment – specified amt. of dollars/pesos that are to be paid each period on a
per par value basis
3. Coupon Interest Rate – stated “annual” rate of interest on a bond.
 coupon payment
par value
4. Floating Rate Bond – bond whose interest rate fluctuates with shifted in the general level
at interest rates (maybe tied to T-bonds’ rate)
- coupon rate is set for initial period after which it can be adjusted
from time to time
- can include upper and lower limits (“cups” and “flours”) or how
high or low the yield (interest rate) can go.
- Can cause market value of bonds to stabilize in times of varying
market rates.
5. Zero Coupon Bond – bond paying no periodic interest but is sold at a discount below par,
thus providing corporation Co investors on the form of capital appreciation
6. Maturity date – a specified date on which the par value of a bond must be repaid.
13
Continued in PowerPoint slides
INTEREST AND MONEY-TIME RELATIONSHIPS ( I )
Capital:
- refers to wealth in the form of money or property that can be used to produce more wealth.
Two Types of Capital:
a. Equity Capital – those owned by individuals who have invested their money or property in a
business project or venture in the hope of receiving a profit.
b. Debt Capital / Borrowed Capital – obtained from lenders for investment. Lenders in turn
receive interest payment from the borrowers.
The General Concepts of Interest:
“ What may be unfamiliar to us is the idea that, in the financial world, money itself is a
commodity, and like other goods that are bought and sold, money costs money. “
- C.S. Park -
Interest - the cost of money often expressed as a percentage that is periodically applied
and added to an amount ( or varying amounts ) of money over a specified length
of time.
- the return obtainable from the productive investment and efficient use of
money resources during a specific time period.
- the compensation ( return ) for the administrative expenses of making the loan,
for the risk that the loan will not be repaid, and for the earnings forgone had the
money been placed in other investments.
- it is the profit from lending money, or the earning power of money.
14
- the amount paid for the use of borrowed money, the cost of borrowing money.
The Time Value of Money:
- the economic value of a sum of money depends on when it is received, because money has
earning power over time ( a dollar received today has a greater value than a dollar received
at some future time ).
- the changing worth of money through time can be because of its earning potential over time,
or to its decrease in value due to inflation over time , or to both.
Ways of Calculating Interest:
a. Simple Interest
- when total interest earned or charged is linearly proportional to the initial
amount of the loan ( principal, or the amount of money borrowed ).
- General Formula :
I=(P)(N)(i)
Where : P = principal amount lent or borrowed
N = number of interest period ( e.g. years )
i = interest rate per interest period
- interest period is that time period for which the interest is calculated; usually, it is
annually, semi-annually, quarterly or monthly but it can be any desired time interval.
Illustrative Problems:
1. Php 400 is loaned for 5 quarters at a simple interest rate of 3% per quarter.
a. What is the total amount of interest to be paid?
b. What is the total amount owed after 5 quarters ( that is if principal and interest amount
are to be paid only at the end of 5 quarters ) ?
2. Find the total interest earned by a principal loan of Php 1000 given a simple interest rate of
12% per month if the loan is for 3 weeks, if the interest is 3.5% per quarter for 4 months, and
if the interest rate is 12.55% per annum for 27 months.
b. Compound Interest
- it is based on the total amount owed at the end of the previous period which consists of
the original principal loaned plus the accumulated interests which had not been paid when due.
- when the interest charge for any interest period is based on the remaining principal
amount plus any accumulated interest charges up to the beginning of that period ( compounding
).
Illustrative Problem:
1. If Php 400 is loaned for 5 quarters at a compound interest of 3% per quarter compounded
quarterly, find the total amount owed at the end of the last quarter.
Notations and Cash Flow Diagrams:
The following notations are commonly utililized in formulas for compound interest calculations
and other engineering economy problems :
i = effective interest rate per period
15
N = number of compounding periods
P = present sum of money; the equivalent value of one or more cash flows at a
reference point in time called the present
F = future sum of money; the equivalent value of one or more cash flows at a
reference point in time called the future
A = end-of- period cash flows ( or any other equivalent end-of- period values ) in
A uniform series continuing for a specified number of periods, starting at the
end of the first period and continuing through the last period
With the use of cash flow diagrams, the timing occurrence of flows become more apparent and
therefore reducing careless errors.
Some concepts in constructing a cash flow diagram ( c.f.d ):
1. Define first the time frame over which the cash flows occur. This establishes the horizontal
scale divided into time periods. Time period progresses as one move from left to right of the
horizontal scale.
2. Cash receipts ( incoming ) and disbursements ( outgoing ) are then located on the time scale
according to data projected by the investment. Cash receipts are represented by upward
arrows, and cash disbursements are represented by downward arrows.
3. Direction of arrows depend on whose point of view is the cash flow diagram drawn for.
Illustrative Example :
Mrs. Green has just purchased a new car for $12,000. She makes a down payment of 30% of the
negotiated price and then makes payments for $303.68 per month thereafter for 36 months.
Furthermore, she believes the car can be sold for $3,500 at the end of three years. Draw a cash
flow diagram of this situation from Mrs. Green’s viewpoint.
Nominal Interest Versus Effective Interest Rate
Nominal Rate of Interest ( r ) :
- for compounded interest, the rate of interest usually quoted is the nominal rate of
interest which is the specific rate of interest and the number of interest period per year. This is
because it has become customary to quote interest rates on an annual basis, followed by the
compounding period if different from one year in length.
Ex: a nominal rate of 8% compounded quarterly
i = 8% / 4 = 2%
Effective Rate of Interest :
- the actual or exact rate of interest earned on the principal during one year .
- this is usually expressed on annual basis unless specifically stated otherwise.
Converting Nominal Interest to Effective Interest Rate:
ieff = ( 1+ r/N )N – 1
where : ieff is the effective interest rate
r is the nominal rate of interest
N is the number of compounding period per year
16
Note: effective interest rate is only equal to nominal rate of interest when compounding is on
annual basis. When N>1, i>r.
Sample Problem:
1. Find the nominal rate which if converted quarterly could be used instead of 12%
compounded monthly. What is the corresponding effective rate?
Interest Formulas Relating Present and Future Equivalent Values of Single Cash Flows
Future Worth:
If any amount P is invested at a point in time and i% is the interest rateper period, the amount
will grow to a future amount of P+Pi = P(1+i) by the end of one period; by the end of two
periods, the amount will grow to P(1+i)(1+i)=P(1+i)2; by the end of three periods, the amount
will grow to P(1+i)2(1+i)=P(1+i)3; and by the end of N periods, the amount will grow to:
F = P(1+i)N
Where : F = future worth of money
N = number of interest period
The quantity (1+i)N is commonly called the single payment compound amount factor. It is also
symbolize by ( F/P, i%, N ).
Present Worth:
To find the present worth of a certain amount of money given it’s future value.
P = F(1+i)-N
The quantity ( 1+i)-N is called the single payment present worth factor. It is symbolize by ( P/F,
i%, N ).
Illustrative Problem:
1. A chemical engineer wished to accumulate a total of Php 10,000 in a savings account at the
end of 10 years. If the bank pays only 4% compounded quarterly, what should be the initial
deposit?
2. What amount can be withdrawn two years from now from a bank offering a nominal rate of
40% compounded quarterly if Php 1,000 is deposited now?
3. Solve problem 2 by using the effective rate of interest.
4. By the conditions of a will, the sum of Php 25,000 is left to a girl to be held in trust by her
guardian until it amounts to Php 45,000. When will the girl receive the money if the fund is
invested at 8% compounded quarterly?
5. If Php 1,000 becomes Php 5,743 after 15 years, when invested at an unknown rate of interest
compounded semi-annually, determine the unknown nominal rate and the corresponding
effective rate.
17
INTEREST AND MONEY-TIME RELATIONSHIPS (II)
The Five Types of Cash Flows:
1. Single Cash Flow
- the simplest case, involves equivalence of a single present amount and its future worth.
2. Equal ( Uniform ) Series
- probably the most familiar category which includes transactions arranged as a series of
equal cash flows at regular intervals.
3. Linear Gradient Series
- cash flow when common pattern of variation occurs in terms of cash flow in a series
increases ( or decrease ) by a fixed amount.
4. Geometric Gradient Series
- another kind of gradient series that is formed when the cash flow is determined not by
some fixed amount but by some fixed rate, expressed as percentage.
5. Irregular Series
- a series of cash flow in that it does not exhibit a regular overall pattern.
Annuities
- is a series of equal payments occurring at equal periods of time.
P
1
2
3
n
0
18
N
A
Formulas Relating Annuity , Present Worth and Future Worth Amounts:
Finding P given A:
Finding F given A:
n
(1  i )  1
(1  i) n  1
PA
FA
[i]
i(1  i) n






Illustrative Problems:
1. What are the present worth and the accumulated amount of a 10 year annuity paying Php
10,000 at the end of each year, with interest at 15% compounded annually?
2. A plastic manufacturing company is intending to expand its production facilities starting
1988. The program requires the following estimated expenditures:
Php 1,000,000 at the end of 1988
Php 1,200,000 at the end of 1990
Php 1,500,000 at the end of 1993
To accumulate the required funds, it established a sinking fund consisting of 15 uniform
annual deposits, the first deposit having been made at the end of 1979. The interest rate of the
fund is 12% per annum. Calculate the annual deposit. What will be the balance in the fund on
January 1, 1990?
3. The officers and board of directors of the institute of Integrated Electrical Engineers desire to
award a Php 3,600 scholarship annually to deserving electrical engineering students for as
long as its scholarship fund shall last. The fund was started July 1, 1987 by a donor in the
amount of Php 18,000. The IIEE invested this amount at that time at 8% per annum and plans
on adding Php 600 each year to the fund from its dues starting July 1, 1988 for as long as
awards are made.
a. For how many years starting July 1, 1988 can scholarship be awarded?
b. What will be the balance in the fund after the last award is made?
4. A chemical engineer wishes to set up a special fund by making uniform semi-annual end-ofperiod deposits for 20 years. The fund is to provide Php 100,000 at the end of each of the last
five years of the 20 year period. If interest is 8% compounded semi-annually, what is the
required semi-annual deposit to be made?
Deferred Annuity:
Ordinary Annuity - involves the first cash flow being made at the end of the first period.
Deferred Annuity - if cash flow does not begin until some later date; one where the first payment
is made several periods after the beginning of the annuity.
P
m periods
1
2
3
0
n periods
0
m
1
2
A
A
A ( P/A,i%,n)(P/F,i%,m)
19
3
4
A
A
A ( P/A, i%, n )
n-1 n
A
A
P = A (P/A,i%,n)(P/F,i%,m)
P = A { [ 1-(1+i)-n ] / i } (1+i)-m
Sample Problems:
1. On the day his grandson was born, a man deposited to a trust company a sufficient amount of
money so that the boy could receive five annual payments of Php 10,000 each for his college
fees, starting with his 18th birthday. Interest at the rate of 12% per annum was to be paid on all
amounts on deposit. There was also a provision that the grandson could elect to withdraw no
annual payments and receive a single lump amount on his 25th birthday. The grandson chose this
option.
a. How much did the boy receive as single payment?
b. How much did the grandfather deposit?
2. If Php 10,000 is deposited each year for 9 years, how much annuity can a person get annually
from the bank every 8 years starting 1 year after the 9th deposit is made. Cost of money is 14%.
Perpetuity:
- an annuity in which payments continue indefinitely.
P
1
2
3
n to 
4
0
P = A { [ 1- (1+i)-n]/i} = A { [1-(1+i)-]/i}
P = A/i
Sample Problem:
1. What amount of money invested today at 15% interest can provide the following scholarships:
Php 30,000 at the end of each yearfor 6 years; Php 40,000 for the next 6 years and Php 50,000
thereafter?
Linear Gradient Series:
- involves receipts and expenses that are projected to increase or decrease by a uniform
amount each period, thus constituting an arithmetic sequence of cash flows.
- ex: maintenance and repair expenses on specific equipment may increase by a relatively
constant amount each period; suppose that the maintenance expense on a certain machine is Php
2,00020
1,500
1,000
2,500
3,000
1000 at the end of the first year and increasing at a constant rate of
four years :
Php 500 each for the next
This cash flow may be resolved into two components:
2G
3G
(n-1)G
G
A
A
A
A
A
0
+
1
2
3
4
A = Php 1,000 n = 5
0
5
G PisA known as the uniform gradient amount:
Finding P when given G:
P = PA + P G
PA = A {( [ 1+i]n - 1 ) / i(1+i)n }
PG = G/i { [ (1+i)n - 1]/i - (n) } { 1/(1+i)n }
1
PG
2
3
4
5
G = Php 500 n = 5
Finding F when given G:
F = G/i { [(1+i)n - 1]/i } - nG/i
Finding A when given G:
A = G { 1/i - n/[(1+i)n - 1] }
Sample Problem:
1. A loan was to be amortized by a group of four end of year payments forming an ascending
arithmetic progression. The initial payment was to Php 5,000 and the difference between
successive payments was to be Php 400. But the loan was renegotiated to provide for the
payment of equal rather than uniformly varying sums. If the interest rate of the loan was 15%
what was the annual payment?
2. Find the equivalent annual payment of the following obligations at 20% interest rate.
End of year
Payment
1
Php 8,000
2
7,000
3
6,000
4
5,000
Additional Problem:
21
1. A man bought a lot worth Php 1,000,000 if paid in cash. On the installment basis, he paid a
down payment of Php 200,000; Php 300,00 at the end of one year, Php 400,000 at the end of
three years and a final payment at the end of five years. What was the final payment if interest
was 20%.
Selections in Present Economy : Problem Set
Method:
1. In a mining site in Mindoro, the ore contains, on the average, one ounce of gold per ton. One
method of processing “A” costs Php 1,500 per ton and recovers 90% of the gold. Another
method “B” costs only Php 1,200 per ton and recovers 80% of the gold. If the gold can be
sold at Php 4,000 per ounce, which method is better and by how much?
2. The making of rivet holes in structural steel members can be done by two methods. The first
method consists of laying out the position of the holes in the members and using a drill press
costing Php 30,000. The machinist is paid Php 20 per hour and he can drill 30 holes per hour.
The second method makes use of a multiple-punch machine costing Php 27,500. The punch
operator is paid Php 18 an hour and he can punch out 4 holes every minute. This method also
requires an expense of Php 0.70 per hole to set the machine.
a. If all other costs are assumed equal, what is the total cost for each machine for 6,000
holes, assuming that the total cost of each machine is charged to these holes?
b. For how many holes will the costs be equal?
3. The monthly demand for ice cans being manufactured by Mr. Cool is 3,200 pieces. With a
manually operated guillotine, the unit cutting cost is Php 25. An electrically operated
hydraulic guillotine was offered to Mr. Cool at a price of Php 275,000 and which will cut by
30% the unit cutting cost. Disregarding the cost of money, how many months will Mr. Cool
be able to recover the cost of the machine if he decides to buy now?
22
Material:
1. GMA will be visiting an automobile plant. It was decided by the plant management that they
must paint the plant.
Boysen paint costs Php 70 per gallon and covers 350 sq. ft. per gallon. The manufacturer
claims that it will last 5 years and can be applied at a rate of 100 sq. ft. per hour.
General paint costs Php 100 per gallon and covers 500 sq. ft. per gallon. It will last for 4
years and can be applied at a rate of 125 sq. ft. per hour. If the painter is Paid Php 15 per
hour, which paint should be used?
2. The volume of raw material required for a certain product is 2.02 cu. cm. The finished
product volume is 1.05 cu. cm. The time for machining each piece of the product is 45
seconds for steel and 30 seconds for brass. The cost of steel is Php 26.50 per kg. and the
value of steel scrap is negligible. The cost of brass is Php 51.25 per kg. and the value of brass
scrap is Php 16 per kg. The wage of the operator is Php 25 per hour and the overhead cost of
the machine is Php 15 per hour. The weight of steel and brass are 0.0081 and 0.0088 kg. per
cu cm., respectively. Which material will you recommend?
3. High carbon steel or alloy steel can be used for the set of tools on a lathe. The tools must be
sharpened periodically. Data for each are as follows:
High Carbon Steel
Alloy Steel
Output per hour
60 pcs.
70 pcs.
Time between tool grinds
4 hours
6 hours
Time required to change tools
1 hour
1 hour
The wage of the lathe operator is Php 24 per hour, based on the actual working hours. The tool
changer costs Php 30 per hour. Overhead costs for the lathe are Php 14 per hour, including toolchange time. A set of unsharpened high carbon steel costs Php 500 and can be ground ten times;
a set of unsharpened alloy steel costs Php 650 and can be ground five times. Which type of steel
should be used?
Design:
1. A company manufactures 1,000,000 units of a product yearly. A new design of the product
will reduce materials cost by 12%, but will increase processing cost by 2%. If materials cost
is Php 1.20 per unit and processing will cost Php 0.40 per unit, how much can the company
afford to pay for the preparation of making the new design and making changes in
equipment?
Site Selection:
1. A certain masonry dam requires 200,000 cu. M. Of gravel for its construction. The contractor
found two possible sources for the gravel with the following data:
Source A
Source B
Average distance, gravel pit
3.0 km.
1.2 km.
to dam site
Gravel cost / cu. m. At pit
NA
PHP 10.00
23
Purchase price of pit
Road construction
necessary
Overburdened to be
removed
@ Php 4.20 / cu. m.
Hauling cost per cu. m. per
km.
Php 800,000
Php 450,000
NA
NA
NA
90,000 cu. m.
Php 4.00
Php 4.00
Which of the two sites will give lesser cost?
Equipment Maintenance:
1. A machine used for cutting materials in a factory has the following outputs per hour at
various speeds and requires periodic tool regrinding at the intervals cited.
Speed
Output per Hour
Tool Regrinding
A
200 pieces
Every 8 hours
B
280 pieces
Every 5 hours
A set of tools costs Php 1,260 and can be ground twenty times. Each regrinding costs Php 54
and the time needed to regrind and change tool is 1 hour. The machine operator is paid Php 19.20
per hour, including the time the tool is changed. The tool grinder who also sets the tools to the
machine is paid Php 21 per hour. The hourly rate chargeable against the machine (based on
machine operating time) is Php 38, regardless of machine speed. Which speed is the most
economical?
Economy and Proficiency of Labor / Workers:
1. An electrical contractor has a job which should be completed in 100 days. At present, he has
80 men on the job and it is estimated that they will finish the work in 130 days. If of the 80
men, 50 are paid Php 62 a day, 25 at Php 68 a day, and 5 at Php 75 a day and if for each day
beyond the original 100 days, the contractor has to pay Php 250 liquidated damages:
a. How many more men should the contractor add so he can complete the work on time?
b. Of the additional men, 5 are paid Php 68 a day and the rest at Php 62 a day. Would the
contractor save money by employing more men and not paying the fine?
2. A certain product is being made by hand in a small factory. The workers were paid Php 0.20
per acceptable piece produced. It was found that if a worker produced 80 pieces per day, 5%
would be rejected. If 90 pieces were produced per day, 10% would be rejected, and at the rate
of 100 pieces, 20% would be rejected. The cost of materials was Php 0.50 per piece, and the
24
materials in any rejected piece had to be thrown away. There was a fixed overhead expense
of Php 10.00 per day per worker, regardless of considerable change in output.
a. At which of the three outputs did the worker make the highest wage?
b. At what output did the factory achieve the lowest unit cost?
3. An executive receives an annual salary of Php 240,000 and his secretary a salary of Php
60,000 a year. A certain task can be performed by the executive working alone in 4 hours. If
he delegates the task to his secretary it will require him 30 minutes to explain the work and
another 45 minutes to check the finished work. Due to the unfamiliarity of the secretary to do
task, it takes her an additional time of 6 hours after being instructed. Considering salary cost
only, determine the cost of performing the task by each method if secretary works 2,400
hours a year and the executive 3,000 hours a year.
IEMECON
Selections in the Present Economy
Problem Set
1. A building contractor can source door frames from either a nearby shop or a far-off forest area.
The cost details are summarized in the table below. The total requirement of wood for the
construction work is 75 tons. Find the best alternative for buying the wooden frames. Also find
the economic advantage of the best alternative.
Items
Nearby Shop Far-off Forest Area
Distance to site
Negligible
900 km
Transportation cost per ton per km Negligible
$ 100
Material cost per ton
$ 2,000
$ 1,250
2. A company is considering the possibility of buying part No.010 from an outside supplier instead
of manufacturing the part as it is now. The annual requirement for part No. 010 is 50,000 units.
The cost to manufacture this part is:
Direct Material
Php 4
Direct Labor
Php 2
Manufacturing Overhead
Variable
Php 2
25
Fixed
Total Unit Cost
Php 4
Php 15
An offer was received from a supplier to supply the part at Php 13 per unit. Should the company
make or buy the part?
3. In the design of a jet engine part, the designer has a choice of specifying either an aluminum alloy
casting or steel casting. Either material will provide equal service, but the aluminum casting will
weigh 1.2 kg compared with 1.35 kg for the steel casting.
The aluminum can be cast for $ 80.00 per kg and the steel one for $ 35.00 per kg. The cost
of machining per unit is $ 150.00 for aluminum and $ 170.00 for steel. Every kilogram of excess
weight is associated with a penalty of $ 1,300 due to increased fuel consumption. Which material
should be specified and what is the economic advantage of the selection per unit?
4.
The Pampanga Corporation manufactures a single product called “Mangyanan” Under normal
operating conditions; the company manufactures and sells 90,000 units of this product in a 6
month period. The contribution to fixed costs and profits of each unit of product is P 8. The fixed
overhead costs for six months amount to P 320,000.
Other companies buying this product are currently encountering labor difficulties and this resulted
to a reduction in sales to only 4,000 units per month. A sales volume of only 4,000 units monthly
will definitely result to a loss. Therefore, the management of the firm plans to close the plant for
six months, anticipating that market will be back to normal after six months.
Studies indicated that the 6-month fixed overhead costs of P 320,000 can be cut down to
P225,000 if the plant is closed. However, additional costs to protect the facilities and start up
costs have been estimated at P 31,000.
Should the company close the plant for six months or continue its operation?
5. A cement kiln with production capacity of 130 tons per day (24 hours) of clinker has at its
burning zone about 45 tons of magnesite chrome bricks being replaced periodically, depending on
some operational factors and the life of the bricks.
If locally produced bricks cost P 25,000 per ton and have a life of 4 months, while certain
imported bricks costing P 30,000 per ton and have a life of 6 months, determine the more
economical bricks and by how much. (ME Board Problem- Oct.1987)
6.
An equipment installation job in the completion stage can be completed in 40 days of 8 hour day
work, with 40 men working. With the contract expiring in 30 days, the mechanical engineer
contractor decided to add 10 men on the job, overtime not being permitted.
26
If the liquidated damage is P 2,000 per day of delay, and the men are paid P 80 per day, will the
engineer be able to complete the jobe on time? Would he save money with the addition of
workers?(ME Board Problem- April 1988)
7. The chief engineer of refinery operations is not satisfied with the preliminary design for storage
tanks to be used as part of a plant expansion programme. The engineer who submitted the design
was called in and asked to reconsidfer the overall dimensions in the light of an article in the
Chemical Engineer, entitled “How to size future process vessels?”
The original design submitted called for 4 tanks 5.2 m in diameter and 7m in height. From a graph
of the article, the engineer found that the present ratio of height to diameter of 1.35 is 111% of the
minimum cost and that the minimum cost for a tank was when the ratio of height to diameter was
4:1. The cost for the tank design as originally submitted was estimated to be $ 9,000,000. What
are the optimum task dimensions if the volume remains the same as for the original design? What
total savings may be expected through the redesign?
8. Suarez Corporation manufactures a certain product XYZ. This product can be sold at the end of a
particular stage of production or can be further processed and sold as a completely processed
product.
A partially processed product sells for P 45 per unit and the manufacturing costs amount to P 30
per unit. If the product is further processed, variable cost of P 15 will be spent and the product can
be sold at P 55 per unit. The estimate annual sales for this product is 30,000 units.
9. Two alternatives are under consideration for a tapered fastening pin. Either design will serve the
purpose and will involve the same material and manufacturing cost except for the lathe and
grinder operations.
Design A will require 16 hours of lathe time and 4.5 hours of grinder time per 1,000 units. Design
B will require 7 hours of lathe time and 12 hours of grinder time per 1,000 units. The operating
cost of the lathe including labour is $200 per hour. The operating cost of the grinder including
labour is 150 per hour. Which design should be adopted if 1,000 units are required per year and
what is the economic advantage of the best alternative?
10. A company manufactures dining tables which mainly consist of a wooden frame and a table top.
The different materials used to manufactire the tables and their costs are given in Table 2.1
Table 2.1
Description of Item
Quantity
Wood for frame and legs
0.1 m3
Table top with sunmica finish 1
Leg bushes
4
Nails
100 g
Total labour
15 hr
27
Cost ($)
12,000/m3
3,000
10/bush
300/kg
50/hr
In view of the growing awareness towards deforestation and environmental conservation, the
company feels that the use of wood should be minimal. The wooden top therefore could be
replaced with a granite top. This would require additional wood for the frame and legs to take the
extra weight of the granite top. The materials and labour requirements along with cost details to
manufacture a table with granite top are given in Table 2.2
Table 2.2
Description of Item Quantity
Wood for frame and legs 0.15 m3
Granite Table top
1.62 m2
Leg bushes
4
Nails
50 g
Total labour
8 hr
Cost ($)
12,000/m3
800/m2
25/bush
300/kg
50/hr
If the cost of the dining table with a granite top works out to be lesser than that of the table with
wooden top, the company is willing to manufacture dining tales with granite tops. Compute the
cost of manufacture of the table under each of the alternatives described above and suggest the
best alernative. Also, find the economic advantage of the best alternative.
11. In the design of buidlings to be constructed in Alpha State, the designer is considering the type of
window frame to satisfy. Either steel or aluminum windrow frames will satisfy the design criteria.
Because of the remote location of the building site and lack of building materials in Alpha state,
the window frames will be purchased in Beta State and transported for a distance of 2,500 km to
the site. The price of window frames of the type required is $ 1,000 each for steel frames and $
1,500 each for aluminum frames. The weight of steel window frames is 75 kg each and that of
aluminim window frame is 28 kg each. The shipping rate is $ 1 per kg per 100 km. Which design
should be specified and what is the economic advantage of the selection?
12. The process planning engineer of a firm listed the sequences of operations as shown in the table
2.3 to produce a component.
Table 2.3
Sequence
Process Sequence
1
Turning-Milling-Shaping-Drilling
2
Turning-Milling-Drilling
3
All operations are performed with CNC machines
The details of processing times of the component for various operations and their machine hour
rates are summarized in Table 2.4.
Table 2.4
Machine Hour Rates and Processing Times (minutes)
Machine
Operation
Process Sequence
Hour Rate
1
2
Turning
200
5
5
Milling
400
8
14
28
3
-
Shaping
Drilling
CNC
350
300
1,000
10
3
-
3
-
8
Find the most economical sequence of operations to manufacture the component.
13. A paint manufacturing company uses a sand mill for fine grinding of paint with an output of 100
liters per hour using glass beads as grinding media. Media load in the mill is 25 kg, costing P 200
per kg, and is fully replenished in 2 months time, at 8 hours per day operation, 25 days per month.
A ceramics grinding media is offered to this paint company costing P400 per kg, and needs 30 kg
in the sand mill, but guarantees an output of 120 liters per hour and full replenishment of media in
3 months.
If profits on paint production is P 15 liter, would you recommend the change in media? Show by
calculations comparative cost to justify your answer. (ME Board Problem- Oct.1987).
DEPRECIATION
Depreciation---is the decrease in value of physical properties with the passage
of time and use. More specifically, depreciation is an accounting concept that
establishes an annual deduction against a before-tax income such that the
effect of time and use on an asset’s value can be reflected in a firm’s financial
statements. Annual depreciation deductions are intended to “match” yearly
fraction of value used by an asset in the production of income over the asset’s
actual economic life.
The actual amount of depreciation can never be
established until the asset is retired from service.
Types of Depreciation
1. Normal Depreciation
(a) Physical depreciation---due to the lessening of a property’s physical ability
to produce results.
(b) Functional depreciation or obsolescence---due to the lessening in the
demand for the function that the property was designed to render.
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2. Depreciation due to change in price levels, when price levels rise during
inflationary periods. Even if original purchase price was recovered through
proper depreciation procedure. Identical replacement can not be attained,
capital depreciated.
3. Depletion---used to indicate the value of the resource base when natural
resources are being consumed in producing products or services.
Symbols Used
P-original installed cost of the depreciable property
N-depreciable life of the asset
F-salvage value of the asset after its useful life of N years
dn-depreciation charge for year n and is dependent on the method used
Dn-accumulated depreciation charges from the acquisition time to the end of
year n
BVn-book value of the asset at the end of year n
Depreciation Methods
1. Straight Line Method (SL)---is the simplest and most widely used
depreciation method. It assumes that a constant amount is depreciated
each year over the depreciable life of the asset. To solve using the SLM, the
following formula applies:
PF
N
D n  nd n
dn 
BVn  P  D n
2. Declining Balance Method (DB)---sometimes called the Matheson Formula,
assumes that the annual cost of depreciation is a fixed percentage of the BV
at the beginning of the year. The ratio of the depreciation in any one year to
the BV at the beginning of the year is constant throughout the life of the
asset and is designated by R (0 to 1). In this method, R=2/N when 200%
declining balance is used and R=1.5/N when 150% declining balance is
used.
F
P
BVn  P(1  k ) n
k  1 N
d n  kBVn 1  kP(1  k ) n 1
D n  P  BVn
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3. Sum-of-the-Years Digit Method (SYD)---to compute the depreciation
deduction by the SYD method, the digits corresponding to the number of
each permissible year of life are first listed in reversed order. The sum of
these digits is then determined. The depreciation factor for any year is the
number from the reverse-ordered listing for that year divided by the sum of
the digits.
 2( N  n  1) 
d n  ( P  f )

 N( N  1) 
 ( 2N  n  1)n 
D n  ( P  F )

 N( N  1) 
BVn  P  D n
4. Sinking Fund Formula---it is assumed that a sinking fund is established in
which funds will accumulate for replacement purposes and will bear
interest.


i
d n  ( P  F )

N
 (1  i )  1 
 (1  i ) n  1 
Dn  d n 

i


BVn  I  D n
5. Service Output or Units-of-Production Method---the decrease in value for
(P  F)
d
N
BVn  I  D n
this type of depreciation method is due to the number of hours that the
asset can be used before its end life. In this case, the life of the asset is in
terms of time that it can produce before replacement.
6. Double Declining Balance --- same as Declining Balance only:
k
2
N
Illustrative Problems:
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(1) A company purchased a machine for $15,000. It paid sales tax and
shipping costs of $1,000 and nonrecurring installation costs amounting to
$1,200. At the end of five years, the company had no further use for the
machine, it was sold for $1,000.
Required:
(a) What is the depreciation each year using SL, DB, DDB, and SYD?
(b)What is the book value at the end of the second year using the different
methods mentioned above.
(2) An asset for drilling was purchased and placed in service by a petroleum
production company. Its cost basis is $60,000 and it has an estimated
salvage value of $12,000 at the end of 14 years. The interest for the
sinking fund is 10%.
Required:
(a) What is the depreciation each year using the sinking fund formula?
(b) What is the book value at the end of 10 years?
(3) A piece of equipment used in a business has a cost basis of $50,000 and is
expected to have a $10,000 salvage value when replaced after 20,000 hours
of use. Find its depreciation rate per hour of use, and find its book value
after 10,000 hours of operation.
(4) In 1992, Silicon Chemical Company purchased a special purpose machine
that had a fair price of $120,000. This new machine is expected to produce
100,000 units throughout its estimated useful life of 10 years. Up to the
end of the third year it had produced 45,000 units and during the fourth
year it produced 15,000 units. If the estimated market value is $12,000 at
the end of 10 years, compute for the depreciation amount in the third year
and the book value at the end of the fourth year by each of these methods:
(a) straight-line method
(b) sum-of-the-years digit method
(c) 200% declining balance method
(d) sinking fund method (i=10%)
(e) service output or units-of-production method
[5] Show in tabular form [depreciation schedule] the computation for the
annual depreciation expenses for a machine worth P1,000,000 with a salvage
value of 10% the original cost and a depreciable life of 5 years, using.
[a] straight line method
[b] sum of year’s digits
[c] declining balance
[d] double declining balance
[6] A contractor imported a bulldozer for his job, paying P250,000 to the
manufacturer. Freight and insurance charges amounted to P18,000; custom’s
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broker’s fees, P8,500; taxes, permits, and other initial expenses, P25,000. If
the contractor estimates life of the bulldozer to be 10 years, with a salvage
value of P20,000, determine the book value at the end of six years, using the:
[a] straight line formula
[b] sinking fund formula at 8%
[c] declining balance method
[d] SYD method
[7] A machine that cost P68,000 was depreciated on a straight line basis for
five years under the assumption that it would have an eight year life and a
P4,000 trade-in value. At the point, it was recognized that the machine had
five years of remaining useful lie, after which it would have an estimated
P3,000 scrap value. Using the straight line method.
[a] Determine the machine’s book value at the end of its fifth year.
[b] Determine the amount of depreciation to be charged against the machine
during each of the remaining years of its life.
APPLICATIONS OF MONEY-TIME RELATIONSHIPS
All engineering economy studies of capital projects should consider the return that a given project will or
should produce. By using several methods of making engineering economy studies to evaluate the economic
profitability of a proposed problem solution.
The Present Worth (PW), Annual Worth (AW) and the Future Worth (FW) methods convert cash flows
resulting from a proposed solution into their equivalent worth at some point (or points) in time by using an interest
rate known as the Minimum Attractive Rate of Return (MARR). The Internal Rate of Return (IRR), Explicit
Reinvestment Rate of Return (ERRR) and the External Rate of Return (ERR) methods produce annual rates of
profit, or returns, resulting from an investment, and are then compared against the MARR. The last method, the
payback period, is a measure of speed with which an investment is recovered by the inflows it produces.
DETERMINING THE MINIMUM ATTRACTIVE RATE OF RETURN
The MARR is usually a policy issue resolved by the top management of an organization in view of
numerous considerations. Among these are:
33
1.
2.
3.
4.
The amount of money available for investment, and the source and cost of this funds (i.e., equity funds or
borrowed funds).
The number of good projects available for investment and their purpose.
The amount of perceived risk associated with investment opportunities available to the firm and the estimated
cost of administering projects over short planning horizons versus long planning horizons.
The type of organization involved (i.e., government, public utility, or competitive industry)
Illustrative Problem.
1. A firm has the following capital structure:
Source of Fund
Common Stock
Preferred Stock
Debt
Retained Earnings
Determine the MARR.
Capital
20M
6M
45M
19M
Cost of Capital
15%
8%
9%
15%
BASIC METHODS IN MAKING ENGINEERING ECONOMY STUDIES
Present Worth (PW) – is based on the concept of equivalent worth of all cash flows relative to some base or
beginning point in time called the present. The PW is a measure of how much money an individual or firm
could afford to pay for the investment in excess of its cost. Or, stated differently, a positive PW for an
investment project is a dollar amount of profit over the minimum amount required by investors. Therefore, a
positive PW would accept the proposed solution.
To find PW, it is necessary to discount future amounts to the present by using the formula:
1.
N
PW   Fk 1  i 
k
k 0
Where: i = effective interest rate, or MARR, per compounding period
k = index for each compounding period (0  k  N)
Fk = future cash flow at the end of period k
N = number of compounding periods in the planning horizon (i.e., study
period)
I = investment
S = salvage value
Illustrative Problem.
1. A special purpose machine is to be acquired by paying a P15,000 initial cash payment plus a debt
assumption of P135,000. The machine will generate additional net annual cash inflow of P40,000 for the
firm throughout the 10 years of useful life of the asset. At the end of its life, a salvage value of 10% of its
initial cost will be realized. Assuming that the MARR is 13.2% per annum, is the project justified using
the PW method?
2.
Annual Worth (AW) – is an equal annual series of dollar amounts, for a stated study period, that is equivalent to
the cash inflows and outflows at an interest rate that is generally the MARR. As long as the AW is greater than
or equal to zero, the project is economically attractive; otherwise, it is not. An AW of zero means that an
annual return exactly equal to the MARR has been earned.
N
N

 k  i 1  i 
AW   Fk 1  i  

N
k 0
 1  i  1
Where: i = effective interest rate, or MARR, per compounding period
k = index for each compounding period (0  k  N)
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Fk = future cash flow at the end of period k
N = number of compounding periods in the planning horizon (i.e., study
period)
I = investment
S = salvage value
Illustrative Problem.
1. A manufacturing process can be carried out by using a machine which can be acquired for P20,000. The
annual receipts generated by the investment is P150,000 and the annual disbursement is P138,000. The
asset will have a life of 10 years in which it will have a salvage value of P2,000 at the end of its life.
MARR = 20% Is the project acceptable using the AW method?
3.
Future Worth (FW) – is based on the equivalent worth of all cash inflows and outflows at the end of the
planning horizon at an interest rate that is generally the MARR. A positive FW would result to acceptance of
the problem solution
N
FW   Fk 1  i 
N k
k 0
Where: i = effective interest rate, or MARR, per compounding period
k = index for each compounding period (0  k  N)
Fk = future cash flow at the end of period k
N = number of compounding periods in the planning horizon (i.e., study
period)
I = investment
S = salvage value
Illustrative Problem.
1. Given: I = 150,000; N = 5 years; S = 15,000; MARR = 25%
Year
Cash flow
1
-25,000
2
-10,000
3
50,000
4
70,000
5
90,000
Is the project justified using the FW method?
Internal Rate of Return (IRR) – is the most widely used rate of return method for performing engineering
economic analyses.
This method solves for the interest rate that equates the equivalent worth of an
alternative’s cash inflows (receipts or savings) to the equivalent worth of cash outflows (expenditures, including
investment costs). The IRR must consequently be compared with the MARR, an IRR that is greater than the
MARR would result to acceptance of the problem solution.
To compute for the IRR, equate the PW to zero, then solve for the i by either “trial and error” or “interpolation”.
4.
N

PW   Fk 1  i*

k
0
k 0
Where: i* = internal rate of return
k = index for each compounding period (0  k  N)
Fk = future cash flow at the end of period k
N = number of compounding periods in the planning horizon (i.e., study
period)
Illustrative Problem.
1. Given: MARR = 15%
35
EOY
Cash flow
0
-100,000
1
30,000
2
40,000
3
70,000
4
50,000
5
20,000
Determine the IRR?
5.
Explicit Reinvestment Rate of Return (ERRR) – solves for the ratio of earning power. It takes into account
the interest rate that equates the annual receipts and disbursements to the amount of investment. An ERRR that
is greater than the MARR would result to acceptance of the problem solution.

R  D   I  S 
i
 1  i 
ERRR 
N
I


1 Net Income

Investment
Where: i = effective interest rate, or MARR, per compounding period
R = annual receipts
D = annual disbursements
I = investment
S = salvage value
Illustrative Problem.
1. On land worth P800,000, an investor constructs a building worth P3,000,000, containing a theater, a bank,
stores and offices. The owner estimates that the annual receipts from rental will be P720,000 and the
annual disbursements to cover taxes, insurance and maintenance of the building will be P80,000. He also
estimates that the land can be sold for P1,200,000, the building for P2,000,000 at the end of 20 years. If
his money is now earning 15% before taxes, will this property earn enough for the investment to be
justified?
External Rate of Return (ERR) – it directly takes into account the interest rate external to a project at which net
cash flows generated (or required) by the project over its life can be reinvested (or borrowed). If this external
reinvestment rate, which is usually the firm’s MARR, happens to equal the project’s IRR, then the ERR method
produces results identical to those of the IRR method.
To compute, convert cash inflows and outflows to their future equivalent using the reinvestment rate which is
usually the MARR, equate this to the future equivalent of the investment, and then solve for the ERR.
6.
N
I 1  ERR    Rk  Dk 1  i 
N
N k
S
k 0
Where: i = effective interest rate, or MARR, per compounding period
R = annual receipts
D = annual disbursements
I = investment
Illustrative Problem.
1. Given: I = 150,000; N = 5 years; S = 15,000; MARR = 25%
Year
Cash flow
Is the project justified?
1
-25,000
2
-10,000
36
3
50,000
4
70,000
5
90,000
7.
Payback (Payout) Period – used as a measure of a project’s riskiness, since liquidity deals with how fast an
investment can be recovered. Quite simply, the payback method calculates the number of years required for
cash inflows to just equal cash outflows.
Simple Payback Period – the length of time required to recover the first cost of an investment from the net cash
flow produced by that investment for an interest rate of zero. This method fails to consider the time value of
money. It fails to consider the consequences of the investment following the payback period, including the
magnitude and timing of the cash flows and expected life of the investment. It is more desirable to have a shorter
payback period, this means that the investment provides revenues early in its life, enough to cover initial outlay.
This quick return in capital also shortens the time span over which the investment is susceptible to losses.

 R
k 1
k
 Dk   I  0
Where: R = annual receipts
D = annual disbursements
I = investment
 = payback period
Discounted Payback Period –considers the time value of money. This method determines the length of time
required until the investments equivalent receipt exceeds the equivalent capital outlay.

 R
k 1
 Dk 1  i   I  0
k
k
Where: i = effective interest rate, or MARR, per compounding period
R = annual receipts
D = annual disbursements
I = investment
Illustrative Problem.
1. A piece of new equipment has been proposed by engineers to increase the productivity of a certain manual
welding operation. The initial investment is P25,000 and the equipment will have a salvage value of
P5,000. Increased productivity attributable to the equipment will amount to P8,000 per year after extra
operating costs have been subtracted from the value of the additional production. MARR = 20%. In view
of the minimum acceptable payback period of 5 years, is this alternative attractive?
Exercise.
1.
A distillation column can be purchased and installed for P 2,000,000. It will incur P35,000 in operations
and maintenance expenses and yet can start generating revenues of P125,000 each month as soon as it is installed.
The distillation column could be sold at the end of its useful life of 4 years at an estimated price of P150,000.
The cost of capital is 15% per annum. The reinvestment rate is 12 %. Evaluate the project using:
a. Present worth method
e. ERR method
b. Annual worth method
f. ERRR method
c. Future worth method
g. Simple payback
d. IRR method
h. Discounted payback
For methods (a) to (f) recommend whether the column is acceptable or not.
For methods (g) and (h), Should the column be accepted if the acceptable payback period is 2.5 years?
37
APPLICATIONS OF MONEY-TIME RELATIONSHIPS FOR
MULTI-BUSINESS PROJECT ANALYSIS
Definition of Terms :
Mutually Exclusive Projects - engineering and business projects that can be accomplished by
more than one feasible alternative wherein the selection of one of these alternatives excludes the
choice and implementation of the others.
“ Do-Nothing Alternative “ - means no alternative is selected because none is economically
justifiable therefore leading to the point where no project is implemented or to continued
reliance in a current system.
Analysis Period / Study Period / Planning Horizon - the time span over which the economic
effects of an investment or project will be evaluated.
Types of Projects / Alternatives:
a. Service Projects / Cost Alternative - those whose revenue do not depend on the choice of the
project or those with all negative cashflows except for a possible positive cashflow element from
disposal of asstes at the end of the project’s useful life. This situation occurs when the
organization must take some action and the decision involves the most economical way of doing
it.
b. Revenue Projects / Investment Alternative - those whose revenue depend on the choice of the
alternative or those with initial capital investment(s) that produce positive cashflows from
increased revenue, savings through reduced costs, or both.
Main Principle in Multi-Business Project Analysis :
- Business or engineering projects must be compared over equal time span or over the same
study period; therefore projects must have equal lives before analysis is initiated. Two Cases for Multi-Project Analysis :
1. Useful lives are the same for all alternatives and equal to study period.
2. Useful lives are different among the alternatives and at least one does not match the
study period.
CASE 1 : USEFUL LIVES ARE EQUAL TO STUDY PERIOD
- USE OF EQUIVALENT WORTH METHODS:
1. For investment alternatives or revenue projects, choose the project that gives the highest
equivalent worth value.
2. For cost alternative or service projects, choose the project that gives the least or lowest
negative equivalent worth value ( meaning the project that requires the lowest cost ).
- USE OF RATE OF RETURN METHODS:
38
In this category, the incremental investment analysis technique is utilized.
4 basic steps for incremental analysis:
1. Arrange the feasible alternatives based on increasing capital investment.
2. Determine the base alternative. For cost alternatives, the base alternative is the one that
requires the least capital investment. For investment alternative, the base alternative is the project
which requires the lowest possible investment and which justifies itself ( i.e. IRR  MARR ).
3. Determine the incremental cashflows from the lower to the next higher-investment project and
compare the incremental rate of return value with the MARR. A higher investment alternative
will be chosen if and only if , the additional funds required will have a rate of return that satisfies
the minimum amount.
Incremental Cash Flow = Cash Flow of B - Cash Flow of A
where : A is the base alternative
B is the next higher investment alternative
4. Repeat step 3 until all alternatives have been exhausted in the analysis.
- If no alternative among the given qualifies to become the base alternative then the “ do-nothing
“ alternative is taken.
Sample Problem:
1. Suppose your given the cash flow data of 3 project alternatives in the table below with 3 years
of useful lives each and a study period of 3 years under consideration.
Alpha
Beta
Gamma
Investment Cost
$ 70,000
$ 50,000
$ 120,000
1st Year- Revenue
$ 40,000
$ 30,000
$ 60,000
2nd Year- Revenue
$ 50,000
$ 40,000
$ 60,000
3rd Year- Revenue
$ 60,000
$ 50,000
$ 60,000
Salvage Value
$ 22,750
$ 5,000
$ 46,000
What project would you choose if MARR is 9.762% compounded semi-annually, using
a. Present Worth Method ?
b. Internal Rate of Return Method ?
CASE 2 : USEFUL LIVES NOT EQUAL TO STUDY PERIOD
AND DIFFERENT AMONG ALTERNATIVES
In this case, the equivalent worth and rate of return methods are equally applicable except that
adjustments to project or alternative cash flows have to be made before they can be used.
“ Unequal lives among alternatives somewhat complicate their analysis and comparison. To
make engineering economy studies in such cases, we adopt the rule of comparing mutually
exclusive alternatives over the same time period. The repeatability assumption and the
coterminated assumption are the two types of assumptions used for these comparisons. “
39
Repeatability Assumption:
1. Study period over which the alternatives are being compared is either indefinitely long or
equal to a common multiple of the lives of the alternatives.
2. The economic consequences that are estimated to happen in an alternative’s initial life span
will also happen in all succeeding life spans.
Coterminated Assumption:
1. Uses a finite and identical study period for all alternatives.
2. If an alternative has a useful life shorter than the study period, the estimated annual cost of
contracting for the activities involved might be used during the remaining years. Similarly, if
useful life of an alternative is longer than the study period, a re-estimated market value is
normally used as terminal cash flow at the end of a project’s coterminated life.
For Infinite Study Period:
- either the repeatability or coterminated assumption can be used depending on which is easier to
apply.
For Finite Study Period :
- if useful life is longer than study period , apply coterminated assumption.
- if useful life is shorter than study period, apply repeatability assumption.
( Exception might apply to investment alternatives with different lives which require only a
one-time investment because the task or need within the firm is a one time task or need.)
Sample Problems:
1. Three machines are being considered by LOG Inc. to be used in their production of a special
product which the company plans to offer to the consumer market for only five years. The capital
investment requirement, annual benefits and disbursements required by each machine as well as
their useful lives and salvage values are given in the table below:
Machines
A
B
C
Investment Cost
$ 6,000
$ 7,600
$ 12,400
Useful Life ( years )
5
7
12
Revenues
$ 19,000
$ 18,000
$ 18,800
Disbursements
$ 7,800
$ 7,282
$ 6,298
Dep’n Method
Straight Line
SYD
Declining Balance
Salvage Value
$ 500
$ 600
$ 1,500
If MARR = 20% per year, which machine would you select based on:
a. Future Worth Method ?
b. ERR Method ?
2. The Smith Novelty Company, a mail-order firm, wants to install an automatic mailing system
to handle product announcement and invoices for the next 15 years. The firm has a choice
between 3 different types of systems. The monetary information about the systems is given in
the table below:
40
System
A
B
C
Initial Investment
$ 50,000
$ 100,000
$ 80,000
Yearly
$ 15,000
$ 3,000
$ 9,000
Maintainance
Expense Required
Useful Life ( years )
10
15
12
Salvage Value
$ 10,000
$ 16,000
$ 20,000
Dep’n Method
SYD
Straight Line
Declining Balance
If yearly minimum rate of return established by the company’s upper management is 20%, which
system is economically the best to implement ? Use the IRR method.
3. WWW Corp. is a private water company. In line with it’s vision of expanding and widening
it’s service area coverage; it plans to establish a water piping system connection to WalangTubig
Subdivision so as to be able to provide the latter with clean water in a fast and efficient way for a
long-term period. The company is now considering two piping system and additional
information are given as follows:
Year
Pipe System A
Pipe System B
0
- $ 12,500
- $ 15,000
1
- $ 5,000
- $ 4,000
2
- $ 5,000
- $ 4,000
3
- $ 5,000
- $ 4,000
4
None
- $ 4,000
Salvage Value
$ 2,000 ( on 3rd year )
$ 1,500 ( on 4th year )
Which piping system would you recommend to the company if MARR is 15% using the
repeatability assumption with the Present Worth Method?
41
Interest and Money Time Relationship
Problem Set
1.
A loan of P 2,000 is made for a period of 13 months, from January 1 to January 31 the following
year, at a simple interest rate of 20%. What future amount is due at the end of the loan period?
2. Find the present worth of a future payment of P100,000 to be made in 5 years with an interest of
10% compounded annually (ECE Board problem- Sept 21, 1985).
3. A man wishes his son to receive a P200,000 ten years from now. What amount should he invest
now if it will earn interest of 10 % compounded annually during the first 5 years and 12 %
compounded quarterly during the next 5 years?
4. J purchased a small lot in a subdivision, paying P20,000 down and promising to pay P1,500 every
3 months for the next 10 years. The seller figured interest is 12 % compounded quarterly. What
was the cash price of the lot?
5. A new company developed a program in which the employees will be allowed to purchase shares
of stocks of the company at the end of its fifth (5th) year of operation, when the company is
thought to have gained stability already, at par value of P100.00 per share.
Believing in the good potential of the company, an employee decided to save in a bank the amount of
P8,000 at the end of every year which will earn for him 9% interest, compounded yearly.
How much shared of stocks will he be able to purchase at the end of the fifth (5th) year of his yearly
deposits? (ME Board Problem-April 26, 1990).
6. A young woman, 22 years old, has just graduated from college. She accepts a good job and
desires to establish her own retirement fund. At the end of each year thereafter she plans to
deposit P2,000 in a fund at 15% annual interest. How old will she be when the fund has an
accumulated value of P1,000,000?
7. Mr. Reyes borrows P60,000 at 12% compounded annually, agreeing to repay the loan in 15 equal
annual payments. How much of the original principal is still unpaid after he made the 8th
payment?
8. On January 1 of a certain year, ABC Corporation borrowed P1,450,000 for 12 years at 16%
interest. The terms of the loan obliged the firm to establish a sinking fund in which the following
deposits was to be made: P300,000 at the end of the 2nd to the 6th year; P450,000 at the end of the
7th to the 11th year, and one for the balance of the loan at the end of the 12th year. The interest rate
earned by the sinking fund was 12%. What was the amount of the final deposit?
9. An asphalt road requires no upkeep until the end of 2 years when P60,000 will be needed for
repairs. After this P90,000 will be needed for repairs at the end of each year for the next 5 years,
then P120,000 at the end of each year for the next 5 years.
If money is worth 14% compounded annually, what was the equivalent uniform annual cost for the 12
year period?
42
Problem Set: Evaluating Single Projects
1.
2.
3.
4.
It is estimated that insulation of steam pipes in a factory will reduce the fuel bill by as much as 20%. The
cost of insulation is Php9,000 and the annual cost of insurance and taxes is 5% of the initial cost. With the
insulation, the annual fuel is Php18,000. If the insulation is worthless after 6 years of use and a minimum
return of 12% is desired, would it be worthwhile to invest in the insulation? Use ERR method.
A small company XYZ Corp. is in the business of cleaning bottles for large soft-drink makers. It presently
owns a bottle cleaning machine that operates for 5,000 hours per year at the rate of 50 bottles being cleaned
per hour. XYZ charges Php0.14 per cleaned bottle. The machine may be overhauled to increase its output
by an additional 20 cleaned bottles per hour, with no increase in its current annual operating cost of
Php10,000. The overhaul will cost Php50,000 (Php10,000 financed through the issuance and sale of bond
while the remaining Php40,000 from the company’s retained earnings). This will extend the machine’s life
to 5 years and will result to zero salvage value at the end of that time. Explicit cost for bonds is 15% while
for retained earnings is 20%. Is the renovation acceptable using IRR method?
An existing warehouse is worth Php500,000 and the average value of the goods stored in it is Php800,000.
Total insurance cost on the warehouse and inventory is Php10,000 annually. If a proposed sprinkling
system is installed in the warehouse, the insurance cost will be reduced to three fourths of the original cost.
The cost of the sprinkler system is Php20,000 and the additional annual cost for maintenance and
inspection would be Php500. The sprinkler system has a useful life of 20 years, at the end of which a
disposal cost of Php2,500 for dismantling the system will be incurred. The operation of the warehouse is
now providing the firm a 12% return on the original investment. Is the sprinkler system justified using the
ERR method?
A company has an opportunity to increase its production level of Mr. Kleen detergent bar. The price of the
detergent will remain at Php5 per unit for the next 10 years. The additional annual operating expenses
include a fixed amount of Php10,000 for electricity plus Php2 per unit of detergent produced in selling
expenses. The project will last for 10 years. To execute this project, the company must purchase a new
machine for Php200,000 that will have no value at the end of its life. How many additional units of Mr.
Kleen must be sold each year so that the project will have an IRR of 15%?
43
Problem Set
Evaluating Single Projects
1.
A single project is available for investment. It requires an initial investment of $25,000. The life is expected
to be 7 years and the salvage value of the investment is $ 2,500. The net revenues are expected to be $
5,500/year and MARR is 6%.
a. Determine the PW for this investment. Should the investment be made?
b. Find the annual worth for this investment. Based on AW, what is your recommendation regarding
the investment in this project?
2.
Determine the PW, FW, and AW of the following engineering project when the MARR is 15% per year. Is
the project acceptable?
Proposal A
Investment Cost
$ 10,000
Expected Life
5 years
Market (salvage) value
- $ 1000
Annual Receipts
$ 8,000
Annual Expenses
$ 4,000
3.
The G&K Corporation is planning to expand one of its warehouse. Total first costs is: the land costs
$250,000; the building costs $ 650,000, the equipment costs $ 200.000 and a $ 150,000 start-up cost.
Expected net revenues are $6000,000/year for 10 years at which time the land and building can be sold for
$475,000. and the equipment for $50,000. Annual cost are $350,000. What is the IRR for this investment?
If the company requires a MARR of 15% determine if they should expand the warehouse or not?
4.
Solve for the IRR and then decide whether the investment is justified or not.
Cash Inflows:
P 25,000
P 100,000
P 90,000
P 95,000
P 75,000
0
1
2
3
4
P 60,000
5
.
Cash Outflows
44
0
1
2
P 50,000
P 60,000
3
4
P 75,000
5
P 47,500
P77,500
P 80,000
P 20,000
5.
An existing warehouse is worth Php500,000 and the average value of the goods stored in it is Php800,000.
Total insurance cost on the warehouse and inventory is Php10,000 annually. If a proposed sprinkling
system is installed in the warehouse, the insurance cost will be reduced to three fourths of the original cost.
The cost of the sprinkler system is Php20,000 and the additional annual cost for maintenance and
inspection would be Php500. The sprinkler system has a useful life of 20 years, at the end of which a
disposal cost of Php2,500 for dismantling the system will be incurred. The operation of the warehouse is
now providing the firm a 12% return on the original investment. Is the sprinkler system justified using the
ERR method?
6.
A piece of new equipment has been proposed by engineers to increase the productivity of a certain manual
welding operation. The investment cost is $ 25,000, and the equipment will have a market (salvage) value
of $5,000 at the end of its expected life of five years. Increased productivity attributable to the equipment
will amount to $8,000 per year after extra operating costs have been subtracted from the value of
production. Suppose the MARR is 20% per year. What is the project’s ERR and is the project acceptable?
45
PROBLEM SET: Money-Time Relationships
Evaluating Single Project
1. A machine can be overhauled today for P5,000, and will save P500 in operating expenses
every year for 5 years. If the MARR is 10% per year, is the overhaul desirable?
2. A printing firm is to acquire P10,000 worth of printing equipment with an expected life of 15
years. Auxiliary equipment that can be coupled with the printing equipment can reduce
operating costs by P10,000 annually. If the firm has a 25% cost of capital, what must the
maximum cost be for the auxiliary equipment?
3. A project is estimated to require an initial investment of P80,000 in physical assets, which
would have an economic life of 10 years and be worth P10,000 in salvage value at the end of
that time. The estimated gross annual savings will be P17,000 and the out of pocket
expenses will be P1,400 per year. The opportunity cost of capital before income taxes, is
10%. Determine the advisability of the investment using: a) AW, b) PW, c) FW, and d) IRR.
4. It is estimated that insulation of steam pipes in a factory will reduce the fuel bill by as much
as 20%. The cost of insulation is P9,000 and the annual cost of taxes and insurance is 5% of
the initial cost. With the insulation, the annual fuel bill is P18,000. If the insulation is
worthless after 6 years of use, and a minimum return of 12% is desired, would it be
worthwhile to invest in the insulation? Use ERRR method.
5. A small company XYZ Corporation is in the business of cleaning bottles for large soft drink
makers. It presently owns a bottle-cleaning machine that operates for 5,000 hours per year at
the rate of 50 bottles being cleaned per hour. XYZ charges P0.10 per cleaned bottle. The
machine may be renovated to increase its output by an additional 20 cleaned bottles per hour,
with no increase in its current operating cost which amounts to P10,000 annually. The
renovation will cost P50,000 (as financed by bonds, P10,000; and retained earnings,
P440,000) will extend the machine’s life to 5 years, and will result to zero salvage value at
the end of that time. Is the renovation acceptable? Explicit cost for bonds is 15% while for
retained earnings is 20%.
6. An existing warehouse is worth P500,000, and the average value of the goods stored is
P800,000. Total insurance cost on the warehouse and inventory is P10,000 annually. If a
proposed sprinkling system is installed in the warehouse, the insurance cost will be reduced
to three fourths of the original cost. The cost of the sprinkler system is P20,000 and the
additional annual cost for maintenance and inspection would be P500. The required write-off
period for the entire investment in the sprinkler system is 20 years. The operation of the
warehouse is now providing the firm a 12% return on the original investment.
7. A company has an opportunity to increase its production level of Mr. Kleen detergent bar.
The price of the detergent will remain at P5 per unit for the next 10 years. The additional
annual operating expenses include a fixed amount of P10,000 for electricity plus P2 per unit
of detergent produced in selling expenses. The project will last for ten years. To execute this
project, the company must purchase a new machine for P200,000 that will have no value at
the end of its life. How many units of Mr. Kleen must be sold each year so that the project
will have an IRR of 10%?
8. A computer facility is requested to serve the data processing needs of a manufacturing firm.
In addition, after office hours, computer time may be sold to outside users at a rate of P50 per
hour. The computer will cost the company P200,000 and the expected useful life is 5 years
46
at the end of which, the facility can be sold for P10,000. Annual revenues from this endeavor
are estimated at a fixed P200,000. Maintenance for the asset will amount to P150,000 yearly
for the tapes, diskettes and air-conditioning. Assume MARR = 25% per annum. How many
after-office hours should be used to make the proposal to be barely acceptable?
Evaluating Different Projects (Same Lives)
1. A company is planning to install a new automated plastic molding press. Four different
presses are available. The essential differences in initial investment and operating costs for
these mutually exclusive cost alternatives are as follows:
Alternative
Investment
Useful Life (yr.)
Annual Operation & Maintenance
Power
Labor
Maintenance
Property taxes & insurance
Total annual Costs
A
P6,000
5
B
P7,600
5
C
P12,400
5
D
P13,000
5
680
6,600
400
120
P7,800
680
6,000
450
152
P7,282
1,200
4,200
650
248
P6,298
1,260
3,700
500
260
P5,720
Each press will produce the same number of units. However, because of different degrees of mechanization, some
require different amounts and grades of labor and have different operation and maintenance costs. None is expected
to have a salvage value, and the selected study period is 5 years, the same as the common useful life. Any capital
invested is expected to earn at least 10% before taxes. Which press should be chosen? Use PW, AW & FW
methods.
2. In an automotive parts plant, a study team is analyzing an improvement project to increase
the productivity of a flexible manufacturing center. The estimated cash flows for the three
feasible alternatives being compared are shown in the table. The analysis period is 6 years,
and the MARR for the capital investments at the plant is 20%. Using ERR method, which
alternative should be selected?
EOY
A
B
C
0
-640,000 -680,000 -755,000
1
262,000
-40,000
205,000
2
290,000
392,000
406,000
3
302,000
380,000
400,000
4
310,000
380,000
390,000
5
310,000
380,000
390,000
6
260,000
380,000
324,000
3. The estimated initial investment cost, and the annual operating and maintenance costs (based
on 1,500 hours of operation per year), for four alternative designs of a diesel-powered air
compressor are shown, as well as the estimated salvage value for each design at the end of
the common 5-year useful life. The perspective of these cost estimates is that of the typical
user (construction company, plant facilities department, government highway department,
47
and so on). The study period is 5 years, and the MARR is 20%. Based on this information,
determine the preferred alternative design using the IRR method.
A
B
C
D
Investment
P100,000 P140600 P148,200 P122,000
Annual cost
29,000
16,900
14,800
22,100
Useful Life (yr.)
5
5
5
5
Salvage Value
10,000
14,000
25,600
14,000
Evaluating Different Projects (Different Lives)
1. Bulaklak Oil Company must install anti-pollution equipment in a new refinery in Bataan to
meet clean air legislation. It is considering five types of equipment, and the following data
are available:
Kampupot
Kalasutsi
Katuray
Investment
P1,500,000 P1,000,000 P1,400,000
Revenue/yr.
594,210
480,000
590,433
Disbursement/yr.
Power
24,000
16,000
18,000
Labor
63,000
53,328
56,000
Maintenance
20,000
15,000
16,000
Estimated Life, yr.
8
5
6
Salvage Value (% I)
10
12
15
Method of Depreciation
SL
SYD
DDB
If MARR is 20%, use IRR method to determine which proposal to accept.
2. MARR = 20%. Use PW method.
A
Initial Cost
P190,000
Life (yr.)
4
Salvage Value
Net Cash Flows
1
75,000
2
75,000
3
80,000
4
80,000
5
6
7
Depreciation Method
DDB
B
C
P130,000
P100,000
6
7
10% of initial cost
50,000
60,000
70,000
80,000
70,000
70,000
SL
48
40,000
50,000
60,000
50,000
60,000
60,000
60,000
SYD
D
P50,000
4
40,000
40,000
40,000
50,000
40,000
DB
After Tax Cash Flow Analysis, Capital Gains (Loss) and
Replacement Analysis
After Tax Cash Flow Analysis:
Important Equations:
MARR After Tax = MARR Before Tax ( 1- Income Tax Rate )
NIBT = BTCF – Depreciation
Tax Paid = -( Income Tax Rate ) * NIBT
ATCF = BTCF + Tax Paid
Where : NIBT = Net Income Before Tax
BTCF = Before Tax Cash Flow
ATCF = After Tax Cash Flow
Problems:
1. Project Orpheus has the following cash flows:
Investment Cost
- Php 100,000
st
1 Year - Revenue
Php 30,000
2nd Year - Revenue
Php 50,000
rd
3 Year - Revenue
Php 70,000
4th Year - Revenue
Php 90,000
Salvage Value
Php 10,000
If the depreciation method used for this particular project is SYD and the company has an
MARR before tax of 25% per annum. What should be the action taken if the company is in the
40% income tax bracket? Use the PW method.
2. A certain productivity project can be undertaken by two alternatives. Information about these
alternatives are given as follows:
Alt. Euriphides
Alt. Antigone
Php 150,000
Php 75,000
Initial Cost
3
5
Life in Years
Php 15,000
Php 7,500
Salvage Value
Net Cash Flow ( Php )
Php 60,000
Php 30,000
Year 1
Php 70,000
Php 30,000
Year 2
Php 80,000
Php 30,000
Year 3
Php 30,000
Year 4
Php 30,000
Year 5
Declining Balance
Straight Line
Depreciation Method
If study period is 3 years, income tax rate is 50% and MARRBT is 25%, which of the two
alternatives is best? Use the IRR method.
49
Capital Gains / Loss:
Capital Gain ( Loss ) = Market Value – Book Value
Capital Gain when Market Value > Book Value
Capital Loss when Market Value < Book Value
Problem:
1. Using Project Orpheus given in the previous problem with the same cash flows and a capital
gains tax of 30% :
a. If the machines used in the project was sold for Php 30,000 at the end of 4 years, construct
the ATCF table. What is the new PW value ?
b. If the machines were in turn sold for Php 5,000 at the end of 4 years, construct the ATCF
table. What is the new PW value?
Replacement Analysis:
Problems:
1. Machine A has been used for 10 years and currently has a book value of Php 2,000. A
decision must be made concerning the most economic action to take : Keep A, replace A w/
B , or replace A w/ C.
Machine A, if continued in service, can be used for 6 more years and scrapped at zero
salvage value. Annual operating costs are at Php 10,000 ( depreciation cost included, straight
line depreciation ).
If Machine A is replaced with Machine B, A can be sold for Php 2,500. B will cost Php
12,000, will have a 6 year life and a salvage value of Php 3,000 at the end of that time.
Annual operating costs are estimated at Php 8,000 ( depreciation cost included, straight line
depreciation ).
On the other hand, if A is replaced with Machine C, C will cost Php 17,500, will have a 6
year life and a salvage value of Php 1,750. Annual operating costs are estimated at Php 6,000
( depreciation cost included, straight line depreciation ).
If the MARR before tax is 15%, income tax = 40% and capital gains tax = 25%, what course
of action should be taken?
2. A firm is considering replacing a generator which was purchased 4 years ago for
Php
50,000. Currently, the generator has a book value of Php 30,000 using Straight Line
depreciation. If the generator is retained, it will be used for 4 more years, at which time it
will have a salvage value of Php 10,000. Operating cost for the old generator is Php 7,000
per year.
50
A new generator can be purchased for Php 60,000. It is estimated to have a life of 4 years
with a salvage value of Php 30,000 at the end of that time. Operating costs are Php 5,000 per
year. If the old generator is replaced , it can be sold for Php 20,000.
Using an after tax analysis and a MARR = 10% after tax, should the generator be replaced or
not? Assume that the firm is in the 55% income tax bracket.
51
INFLATION
Inflation - exists when the general price levels of goods and services in an economy are
increasing. This is caused by several complex forces at work within and without a country’s
economy. As prices, rise, (the value of money) its purchasing power, consequently decreases.
Inflation Rate - is a percentage of increase of price levels based on the previous year’s price
levels.
Definition of Terms
f = inflation rate
Actual Pesos, AP : the actual number of pesos at the point in time that they occur; this is the
actual paper and coin denominations of money.
Real Pesos, RP :
the number of pesos that equals the purchasing power of the money; usually
it is the amount of goods and services that can be bought by Actual Pesos.
Conversion Formula:
RP = AP(1 + f )-k
Where: f = inflation rate during the intervening period between the base year and
the year in which AP occurs
k = number of intervening periods between the base year and the year in
which AP occurs
The real interest rate, i, is the interest rate that will make the Real Pesos inflows
equivalent to Real Pesos outflows in a cash flow diagram.
If one undertakes an analysis involving cash flow diagrams of Real (uninflated) Pesos,
then Real (uninflated) interest rate i should be used. If the cashflow diagrams of Actual
(inflated) Pesos are given, then the interest rate should be the combined interest rate, ic:
ic = ir + f + ir(f)
One can either use AP or RP provided that the appropriate interest rate is used. Usually,
the MARR given in the project analysis is a combined interest rate.
Illustrative Problem
1.
EOY
Cash Flow, AP
0
-15000
1
3000
2
4000
3
5000
4
6000
Assume inflation rate is 5% per year and the real monetary interest rate is 12% per year.
Is the project desirable? Use PW.
52
Additional Problems on Inflation:
1. Annual expenses for two alternatives have been estimated on different bases as follows:
End of Year
1
2
3
4
Alternative A
Annual Expenses Estimated in
Actual Pesos
-120,000
-132,000
-148,000
-160,000
Alternative B
Annual Expenses Estimated in Real
Pesos with base period = year 0
-100,000
-110,000
-120,000
-130,000
If the average general price inflation rate is expected to be 6% per year and the real rate of
interest is 9% per year, answer the following questions:
a. What is the combined interest rate?
b. What is the PW of Alternative A?
c. What is the PW of Alternative B?
d. Convert the year 1 and 2 cash flows of
Alternative A to their equivalent real peso
values.
Year 1:
e. Convert the year 3 and 4 cash flows of
Alternative B to their equivalent actual
peso values.
Year 3:
Year 2:
Year 4:
2. Suppose that your salary is $35,000 in year one, will increase at 6% per year through
year 4, and is expressed in actual dollars as follows:
End of Year
Salary (A$)
1
$35,000
2
37,100
3
39,326
4
41,685
53
If the general price inflation is expected to average 8% per year for the first two years and 7% per year
for the last two years, what is the
Final Answer
Question
What is the real dollar equivalent of
these actual dollar salary amounts?
Assume that the base period is year
one.
End of Year
1
2
3
4
If your personal MARR is 10% per year,
calculate the real interest rate.
54
Salary (R$)
Risk In Capital Investment and Budgeting
Objectives:
1. To consider uncertainty of cash flow in each period of a capital investment’s life.
2. To measure the overall riskiness of an investment proposal when the probability
distributions of cash flow outcomes for different periods are not necessarily the same
(both expected value of cash flow and the dispersion of probability distributions can
change over time).
1st Case: Assumption of Independence
-Outcome in period t does not depend on what happened in period t-1; no causative
relationship between cash flows from period to period.
- Mean of probability distribution of possible NPV values for the proposal:

n
NPV  
t 0
At
(1  R )
t
f

where:
At - expected cash flow in period t
Rf – risk-free rate
(the risk premium will come in when we consider the probability
distributions of NPV to judge the risk of the proposal)
n – no. of periods over which cash flows are expected
- Standard deviation of probability distribution of NPV:

 (1 
n
t 0
2
t
R)
2t
f
where : t = standard deviation of the probability distribution of possible net cash
flows in period t.

t 
 ( A  At ) p xt
2
xt
where: Pxt is the probability of xth cash flow in period t.
(Note: Main difficulty in capital budgeting is obtaining cash flow estimates from people
for risky investments. Be careful of biases in obtaining cash-flow information.)
2nd Case: Dependence of Cash Flows Over Time
- Cash flow in one future period depends in part or in whole on the cash flows in the
previous periods.
- Consequence: correlated cash flows gives higher standard deviation of the probability
distributions of possible NPVs. Higher degree of correlation, higher dispersion of
probability distribution.
55
- The mean NPV stays the same regardless of the degree of correlation of cash flows
over time.
A. Perfect Correlation
- If cash flows deviate in exactly the same relative manner, that is, if actual cash flows
for all periods show the same relative deviation from the means of their respective
probability distributions of expected cash flows.
- Standard deviation of probability distribution of NPV with perfectly correlated stream of
cash flows over time:

n
 
t o
t
(1  R f ) t
B. Moderate Correlation
- Standard deviation of probability distribution of NPV will be somewhere between that
of assumption of mutual independence and perfect correlation.
- Can be beautifully captured with a series of conditional probability distributions which
enables us to take account of the correlation of cash flows over time.
- Standard deviation of probability distribution of NPV values:

_____
l
 ( NPV  NPV ) 2 Px
x
x 1
where: NPVx is the net present value for series x of net cash flows covering all
periods
_______
NPV is the mean NPV of the proposal (computed in the same manner
as that of mutual independence)
l = no. of possible series of cash flows
Other Ways of Assessing Project Risk (aside from ):
1. Sensitivity Analysis
- involves changing one of the key variables at a time and observing how sensitive
cash flows (or NPVs) are to changes
- can highlight factors that have the greatest effect on cash flows
2. Scenario Analysis
- involves setting optimistic, pessimistic, and most likely values for the key variables
affecting the proposed investment, the NPV is then computed under each of these
three conditions to have a picture of the project’s riskiness
3. Monte Carlo Simulation
- risk analysis technique in which probable future events are simulated on a
computer, generating estimated rates of return and risk indexes
56
- values of variables affecting cash flows are generated randomly and for which NPV
is computed, this process is done repeatedly thus generating the probability
distribution for the NPV and allows for the location of its mean and a view to its
dispersion.
Total Risk of Multiple Investments (Portfolio)
- Diversification can be done with capital investments just as it can be practiced with
securities investments. However, diversification with respect to capital assets is
more “lumpy” or “difficult” than with that of securities.
- Mean NPV of Multiple Capital Investments: Simply the sum of mean NPVs of the
investment projects making up the portfolio.
- Standard deviation, risk, of Portfolio of Risky Capital Investments: Depends to a
large extent on the degree of correlation between the investments and the standard
deviation of possible NPVs for each project.

m
m
 r  
j 1 k 1
jk
j
k
where: m – total no. of capital investments in the portfolio
rjk – expected correlation between NPV for investment j and k
j – standard deviation of investment j
k – standard deviation of investment k
(Projects in the same general line of business tend to be highly correlated with each
other, while projects in essentially unrelated lines of business tend to have low degrees
of correlation.)
Feasible Combinations and Dominance
Feasible Combinations: Portfolios having Mean NPVs greater than equal zero.
(Corresponds to opportunity set of security portfolios.)
Certain portfolios will dominate others because:
1. They have higher NPV and the same standard deviation.
2. They have a lower standard deviation and the same NPV.
3. They have both a higher NPV and a lower standard deviation.
The portfolios that dominate others represent the efficient frontier of the opportunity set
curve for capital investment portfolios.
57
Sample Problems:
1. The Dewitt Corporation has determined the following discrete probability
distributions for net cash flows generated by a contemplated project:
Period 1
Prob.
Cash Flow
0.10
$1,000
0.20
2,000
0.30
3,000
0.40
4,000
Period 2
Prob.
Cash Flow
0.20
$1,000
0.30
2,000
0.40
3,000
0.10
4,000
Period 3
Prob.
Cash Flow
0.30
$1,000
0.40
2,000
0.20
3,000
0.10
4,000
a. Assume that probability distributions of cash flows for future periods are
independent. Also, assume that the risk-free cost of capital is 7%. If the proposal will
require an initial outlay of $5,000, determine the mean net present value.
b. Determine the standard deviation about the mean.
c. If the total distribution is approximately normal and assumed continuous, what is the
probability of the net present value being zero or less?
d. What is the probability that the net present value will be greater than zero?
e. What is the probability that the profitability index will be 1.00 or less?
f. What is the probability that the profitability index will be greater than 2.00?
g. If perfect correlation exists between cash flows from one period to another, what are
the mean NPV and standard deviation about this mean?
2. Ponape Lumber Company is evaluating a new saw with a life of 2 years. The saw
costs $3,000, and future after-tax cash flows depend on demand for the company’s
products. The probability tree of possible future cash flows associated with the hew
saw is:
Year 1
Year 2
Initial Probability
Cash Flow
Conditional
Cash Flow
Probability
0.30
$1,000
0.40
$1,500
0.40
$1,500
0.30
$2,000
0.40
$2,000
0.60
$2,500
0.40
$2,500
0.20
$3,000
a. What are the joint probabilities of occurrence of the various outcomes in cash flow?
b. If the risk-free WACC is 10%, what are the mean and standard deviation of the
probability distribution of possible net present values?
c. Assuming a normal distribution, what is the probability the actual net present value
will be less than zero?
3. Xonics Graphics is evaluating a new technology for its reproduction equipment. The
technology will have a 3-year life and cost $1,000. Its impact on cash flows is subject
to risk. Management estimates that there is a 50-50 chance that the technology will
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either save the company $1,000 in the first year or save it nothing at all. If nothing at
all, savings in the least 2 years would be zero. Even worse, in the second year an
additional outlay of $300 may be required to convert back to the original process, for
the new technology may result in less efficiency. Management attaches a 40%
probability to this occurrence, given the fact that the new technology “bombs out” in
the first year. If the technology proves itself, second year cash flows maybe either
$1,800, $1,400 or $1,000, with probabilities 0.20, 0.60, and 0.20, respectively. In the
third year, cash inflows are expected to be $200 greater or less than the cash flow in
period 2, with an equal chance of occurrence. (Again, these cash flows depend on
the cash flow in period 1 being $1,000.) All the cash flows are after taxes.
a. Set up a probability tree to depict the foregoing cash-flow possibilities.
b. Calculate a net present value for each three-year possibility, using a risk-free capital
cost of 5%.
c. What is the risk of the project.
4. The Windrop Company will invest in two of three possible proposals, the cash flows
of which are normally distributed. The expected net present value (discounted at the
WACC) and the standard deviation for each proposal are given as follows:
Proposal
Expected NPV
Standard Deviation
1
$10,000
4,000
2
$8,000
3,000
3
$6,000
4,000
Assuming the following correlation coefficients for each possible combination, which two
proposals dominate?
Proposals
Correlation Coefficient
1
1.00
2
1.00
3
1.00
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1&2
0.60
1&3
0.40
2&3
0.70
I.
ABC Corporation is considering a special purpose piece of machinery costing $9,000 with a life of 2
years, after which there is no expected value. The possible incremental cash flows are:
Year 1
Year 2
Probability
Cash flow (in $)
Probability
Cash flow (in $)
0.3
6,000
0.3
2,000
0.5
3,000
0.2
4,000
0.4
7,000
0.3
4,000
0.4
5,000
0.3
6,000
0.3
8,000
0.2
6,000
0.5
7,000
0.3
8,000
The company’s required rate of return for this investment is 8%:
1. What are the joint probabilities of occurrence of the various branches?
2. Calculate the mean net present value and standard deviation of the probability distribution of
possible net present values?
3. Assuming a normal distribution, what is the probability the actual net present value will be less
than zero? And greater than zero?
II.
Expected cash flow of the company’s project in 3 years, with an initial investment of $10,000 in
illustrated below:
Year 1
Year 2
Year 3
Probability
Cash flow (in $) Probability
Cash flow (in $) Probability
Cash flow (in $)
0.10
3,000
0.10
2,000
0.10
2,000
0.25
4,000
0.25
3,000
0.25
3,000
0.30
5,000
0.30
4,000
0.30
4,000
0.25
6,000
0.25
5,000
0.25
5,000
0.10
7,000
0.10
6,000
0.10
6,000
The company’s required rate of return for this investment is 8%:
1. What is the mean net present value of the project?
2. What is the standard deviation about the mean?
3. If the total distribution is approximately normal and assumed continuous, what is the probability
of the net present value being above $1,000?
4. What is the probability that the net present value will be greater than zero?
5. Assuming that there is perfect correlation among the cash flows, what is the standard deviation
of the project and net present value?
III.
Suppose a firm has a single investment project, 1 and it is considering an additional project, 2. The
projects have the following net present values, standard deviations, and correlation coefficients:
Project
Expected NPV ($)
Standard Deviation ($) Correlation Coefficient
1
12,000
14,000
1.00
2
8,000
6,000
1.00
1 and 2
0.40
What is the NPV of the portfolio?
What is the standard deviation of the portfolio?
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Dealing with Uncertainty
Decisions under Risk – decisions in which the analyst models the decision problem in terms of
assumed possible future outcomes, or scenarios, whose probabilities of occurrence can be
estimated.
Decisions under Uncertainty – a decision problem characterized by several unknown futures
for which probabilities of occurrence cannot be estimated.
Sources of Uncertainty
1.
2.
3.
4.
Possible inaccuracy of the cash flow estimates used in the study
Type of business involved in relation to the future health of the economy
Type of physical plant and equipment involved
Length of the study period used in the analysis
Common Non-probabilistic Methods for Dealing with Uncertainty
1. Breakeven analysis – commonly used when the selection among alternatives is heavily
dependent on a single uncertain factor such as capacity utilization. A breakeven point for
the factor is determined such that two alternatives are equally desirable from an economic
standpoint. It is then possible to choose between the alternatives by estimating the most
likely value of the uncertain factor and comparing this estimate to the breakeven value.
2. Sensitivity Analysis – a basic technique often employed when one or more factors are
subject to uncertainty. Sensitivity, in general, means the relative magnitude of change in
the measure of merit (such as PW or IRR) caused by one or more changes in estimated
study factor values. Sometimes it is more specifically defined to mean the relative
magnitude of the change in one or more factors that will reverse the decision among
project alternatives or a decision about the economic acceptability of a project.
The questions that sensitivity analysis attempts to resolve are these: (a) What is
the behavior of the measure of merit (e.g., PW) to x% changes in each individual
factor? (b) What is the amount of change in a particular factor that will cause a
reversal in preference for an alternative? (c) What is the change in the measure of
merit to selected combinations of changes in two or more factors?
A sensitivity graph or a spiderplot allows the decision maker to visualize the effects of
changes in the measure of merit.
3. Risk-adjusted MARRs – sometimes used to deal with estimation uncertainties. This
method involves the use of higher MARRs for alternatives that are classified as highly
uncertain and lower MARRs for projects for which there appear to be fewer uncertainties.
4. Reduction of the useful life of an alternative – the estimated project life of each
alternative is reduced by a fixed percentage and each alternative is evaluated regarding its
acceptability over only this reduced life span.
Sample Problems
1. Suppose that there are two alternative electric motors that provide 100-hp output. An Alpha
motor can be purchased for $12,500 and has an efficiency of 74%, an estimated useful life of 10
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years, and estimated maintenance expenses of $500 per year. A Beta motor will cost $16,000
and has an efficiency of 92%, a useful life of 10 years, and annual maintenance expenses of
$250. Annual taxes and insurance expenses on either motor will be 1 1/2 % of the investment. If
the MARR is 15% per year, how many hours per year would the motors have to be operated at
full load for the annual costs to be equal? Specify the range of operating hours for which each
alternative would be beneficial. Assume that market values at the end of 10 years for both are
negligible and that electricity costs $0.05 per kilowatt-hour.
Ans. X = 508 hours per year; X<508, select Alpha; X>508, select Beta (based on Equivalent
Uniform Annual Cost or EUAC metric) where X = number of hours of operation per year
2.The Universal Postal Service is considering the possibility of putting wind deflectors on the
tops of 500 of their long-haul tractors. Three types of deflectors, with the following
characteristics, are being considered (MARR = 10% per year):
Capital Investment
Drag Reduction
Maintenance/year
Useful Life
Type A
$1,000
20%
$10
10 years
Type B
$400
10%
$5
10 years
Type C
$1,200
25%
$5
5 years
If 5% in drag reduction means 2% in fuel savings per mile, how many miles do the tractors have
to be driven per year before Type A deflector is favored over the other deflectors? Over what
range of miles driven per year is Type C deflector the best choice? (Note: Fuel cost is expected
to be $1.00 per gallon and average fuel consumption is five miles per gallon without the
deflectors.)
Ans. X≦12,831, select Type B; 12,831<X≦37,203, select Type A; X>37,203, select Type C
where X = number of miles driven per year
3. The best (most likely) cash flow estimates for a new piece of equipment being considered for
immediate installation are as follows:
Capital Investment, I
Revenues per year
Expenses per year
Market Value, MV
Useful Life, N
$11,500
5,000
2,000
1,000
6 years
Because of the new technology built into this new machine, it is desired to investigate its PW
over a range of ±40% changes in the estimates for (a) capital investment, (b) annual net cash
flow, (c) market value, and (d) useful life. Based on these best estimates, plot a diagram that
summarizes the sensitivity of present worth to percent deviation changes in each separate factor
estimate when the MARR = 10% per year. (Note: In the Spiderplot, the PW value serves as the
y-axis whereas the % deviation changes in factor estimate serves as the x-axis. Further note that
the relative degree of sensitivity of the present worth to each factor is indicated by the slope of
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the curves. The steeper the slope of the curve, the more sensitive the present worth is to the
factor.)
4. The Atlas Corporation is considering two alternatives, both affected by uncertainty to
different degrees, for increasing the recovery of a precious metal from its smelting process. The
following data concern capital investment requirements and estimated annual savings of both
alternatives:
End of Year
0
1
2
3
4
Alternative P
- $160,000
120,000
60,000
0
60,000
Alternative Q
-$160,000
20,827
60,000
120,000
60,000
The firm’s MARR for its risk-free investments is 10% per year. Because of the technical
considerations involved, Alternative P is thought to be more uncertain than Alternative Q.
Therefore according to the company’s engineering economy handbook, the risk-adjusted MARR
applied to P will be 20% per year and 17% per year for Alternative Q. Which alternative should
be recommended?
Ans. PW of Alt. P = $10,602; PW of Alt. Q = $8,575; Select Alt. P
5. Suppose that Atlas Corporation decided not to use risk-adjusted interest rates as a means of
recognizing uncertainty in their engineering economy studies. Instead, they have decided to
truncate the study period at 75% of the most likely estimate of useful life. Hence, all cash flows
past the third year would be ignored in the analysis of alternatives. By using this method, should
Alternative P or Q be selected when MARR = 10% per year?
Ans. PW of Alt. P = -$1,324; PW of Alt. Q = -$1,324; None of the two should be selected
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