1. Applications of Money – Time
Relationships
Chapter 5
Topics
• Evaluating the economic profitability of a single proposed
problem solution (alternative)
– Multiple alternatives in Chapter 6
• Minimum attractive rate of return (MARR) concept
• Five methods for evaluation
– Present worth
– Future worth
– Annual worth
– Internal rate of return
– External rate of return
• Payback period method as a measure of the speed with which
an investment is recovered by the cash inflows it produces
1
2. Project Evalution Example:
Federal Express
Nature of Project:
• Equip 40,000 couriers
with PowerPads
• Save 10 seconds per
pickup stop
• Investment cost: $150
million
• Expected savings: $20
Federal Express
million per year
3
Ultimate Questions
• Is it worth investing $150 million to save
$20 million per year, say over 10 years?
• How long does it take to recover the initial
investment?
• What kind of interest rate should be used in
evaluating business investment
opportunities?
4
2
3. Project Evalution Example:
Mr. Bracewell’s Investment Problem
• Built a hydroelectric plant using his personal savings of $800,000
• Power generating capacity of 6 million kWh
• Estimated annual power sales after taxes: $120,000
• Expected service life of 50 years
Bracewell's $800,000 investment a wise one?
Was
long does he have to wait to recover his initial investment,
How
and will he ever make a profit?
5
Mr. Bracewell’s Hydro Project
6
3
4. Bank Loan vs. Investment Project
Bank Loan: Loan cash flow
Loan
Bank Customer
Repayment
Investment Project: Project cash flow
Investment
Company Project
Return
7
Describing Project Cash Flows
Year Cash Inflows Cash Outflows Net
(n) (Benefits) (Costs) Cash Flows
0 0 $650,000 -$650,000
1 215,500 53,000 162,500
2 215,500 53,000 162,500
… … … …
8 215,500 53,000 162,500
8
4
5. The Present Worth Method
• The Present Worth (PW) method is based
on the concept of equivalent worth of all
cash flows relative to some beginning point
in time called the present.
• All cash inflows and outflows are
discounted to the present point in time at an
interest rate that is generally the MARR.
9
MARR: Minimum Attractive Rate of Return
• The rate at which the firm can always invest
the money in its investment pool.
• ―hurdle rate‖
• Possibly change over the life of project
10
5
6. Determining the Minimum Rate of Return
(MARR)
Example 5.1: Estimation of the MARR,
Using an Opportunity Cost Viewpoint
• Consider the following schedule,
Expected Investment Cumulative
which shows prospective annual Annual Rate Requirements Investment
rates of profit for a company’s of Profit (Thousands of
portfolio of capital investment Dollars)
projects (this is the demand for 40% and over $2,200 $2,200
capital).
30-39.9% 3,400 5,600
• If the supply of capital obtained
20.29.9% 6,800 12,400
from internal and external
sources has a cost of 15% per 10-19.9% 14,200 26,600
year for the first $5,000,000 Below 10% 22,800 49,400
invested and then increases 1% Note: All projects with a rate of profit of 10%
for every $5,000,000 thereafter, or greater are acceptable.
what is the company’s MARR
when using an opportunity cost
viewpoint?
6
7. Example 5.1: Estimation of the MARR,
Using an Opportunity Cost Viewpoint
The Present Worth Method
• The PW of an investment alternative is a
measure of how much money an individual
or a firm could afford to pay for the
investment in excess of its cost.
• A positive PW for an investment project is a
dollar amount of profit over the minimum
amount required by investors.
• It is assumed that cash generated by the
alternative is available for other uses that
earn interest at a rate equal to the MARR.
14
7
8. The Present Worth Method
(Net Present Worth Measure)
Principle: Compute the equivalent net surplus at n = 0 for a
given interest rate of i.
Decision Rule: Accept the project if the net surplus is
positive.
Inflow
0 1
2 3 4 5
Outflow Net surplus
PW(i)inflow
PW(i) > 0
0
PW(i)outflow
15
Example: Tiger Machine Tool Company
inflow
$55,760
$24,400 $27,340
0
1 2 3
outflow
$75,000
PW (15%) inflow $24,400( P / F ,15%,1) $27,340( P / F ,15%,2)
$55,760( P / F ,15%,3)
$78,553
PW (15%) outflow $75,000
PW (15%) $78,553 $75,000
$3,553 0, Accept
16
8
10. The Present
Worth Method
The higher the
interest rate and
the farther into
the future a cash
flow occurs, the
lower its PW is.
Meaning of Present Worth
Two possible perspectives
• Investment pool: All funds in the firm’s treasury
can be placed in investments that yield a return
equal to the MARR.
• Borrowed funds (project balance): If no funds
are available for investment, the firm can borrow
them at MARR from the capital market.
20
10
11. Investment Pool Concept
• The company has $75,000 available for
investment. Two choices:
– Do not invest, keep the money in the
investment pool and earn at MARR.
– Invest in the project
21
Meaning of Net Present Worth
N=3
How much would you have if the
Investment is made?
Investment pool $24,400(F/P,15%,2) = $32,269
$27,340(F/P,15%,1) = $31,441
$55,760(F/P,15%,0) = $55,760
$75,000 $119,470
How much would you have if the
investment was not made?
$55,760
$27,340 $75,000(F/P,15%,3) = $114,066
$24,400
What is the net gain from the
investment?
Project
0 1 2 3
$119,470 - $114,066 = $5,404
PW(15%) = $5,404(P/F,15%,3) = $3,553 Net future worth
of the project 22
11
13. Example 5.2: Economic Desirability of a
Project, Using Present Worth
• An investment of $10,000 can be made in a project
that will produce a uniform annual revenue of
$5,311 for five years and then have a market
(salvage) value of $2,000. Annual expenses will be
$3,000 each year. The company is willing to accept
any project that will earn 10% per year or more on
all invested capital. Show whether this is a desirable
investment by using the PW method.
Example 5.3: Evaluation of New
Equipment Purchase, Using Present Worth
• A piece of new equipment has been proposed by
engineers to increase 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 the additional production. If the firm’s
MARR is 20% per year, is this proposal a sound
one? Use the PW method.
13
14. Example 5.3: Evaluation of New
Equipment Purchase, Using Present Worth
Example 5.3: Evaluation of New
Equipment Purchase, Using Present Worth
14
15. Example 5.3: Evaluation of New
Equipment Purchase, Using Present Worth
Bond Value
• Excellent example of commercial value as being the
PW of the future net cash flows that are expected to
be received through ownership of an interest-bearing
certificate.
– The value of a bond, at any time, is the PW of future cash
receipts
15
16. Bond Value
Z: face, or par, value
C: redemption or disposal price (usually equal to Z)
r: bond rate (nominal interest) per interest period
N: number of periods before redemption
i: bond yield rate per period
VN: value (price) of the bond N interest periods prior
to redemption – this is a PW measure of merit
Bond Value
• The owner of a bond is paid two types of payments
by the borrower:
– Series of periodic interest payments (each equal to rZ)
until bond is retired – these constitute an annuity
– When retired, a single payment of C
• The present worth of the bond is:
VN C P | F , i%, N rZ P | A, i%, N
16
17. Example 5.4: Finding the Current Price
(PW) of a Bond
• Find the current price (PW) of a 10-year
bond paying 6% per year (payable
semiannually) that is redeemable at par
value, if bought by a purchaser to yield 10%
per year. The face value of the bond is
$1,000.
Example 5.5: Current Price and Annual
Yield of Bond Calculations
• A bond with a face value of $5,000 pays interest
of 8% per year. This bond will be redeemed at par
value at the end of its 20-year life, and the first
interest payment is due 1 year from now.
a) How much should be paid now for this bond in
order to receive a yield of 10% per year on the
investment?
b) If this bond is purchased now for $4,600, what
annual yield would the buyer receive?
17
18. Example 5.6: Stan Moneymaker Wants to
Buy a Bond
• Stan Moneymaker has the opportunity to purchase a
certain U.S. Treasury bond that matures in eight years and
has a face value of $10,000. This means that Stan will
receive $10,000 cash when the bond’s maturity date is
reached. The bond stipulates a fixed nominal interest rate
of 8% per year, but interest payments are made to the
bondholder every three months; therefore, each payment
amounts to 2% of the face value.
• Stan would like to earn 10% nominal interest
(compounded quarterly) per year on his investment,
because interest rates in the economy have risen since the
bond was issued. How much should Stan be willing to pay
for the bond?
The Capitalized Worth (CW) Method
(The Capitalized Equivalent (CE) Method)
• Special variation of the PW method involving the
determination of the PW of all revenues or
expenses over an infinite length of time
• If only expenses are considered, sometimes referred
to as capitalized cost
18
19. Capitalized Equivalent Worth
Principle: PW for a project with an annual
receipt of A over infinite (or extremely long)
service life
Equation:
CW(i) or CE(i) = A(P/A, i, ) = A/i
A
0
P = CE(i) N ∞
Computing PW: capitalization of project cost
Cost is called capitalized cost (the amount of money that must be invested
today in order to yield a certain return A at the end of each and every
37
period forever, assuming an interest rate of i
Given: i = 10%, N =
Find: P or CE (10%)
$2,000
$1,000
0
10
$1,000 $1,000
CE(10%) ( P / F,10%,10)
0.10 0.10
P = CE (10%) = ? $10,000(1 0.3855)
$13,855
38
19
20. Comparion of Present Worth for
Long Life and Infinite Life
• Built a hydroelectric plant using his personal savings of $800,000
• Power generating capacity of 6 million kwhs
• Estimated annual power sales after taxes: $120,000
• Expected service life of 50 years
Bracewell's $800,000 investment a wise one?
Was
long does he have to wait to recover his initial investment,
How
and will he ever make a profit?
39
Mr. Bracewell’s Hydro Project
40
20
21. Long Service Life: 50 years
• Equivalent lump sum investment
V1 = $50K(F/P, 8%, 9) + $50K(F/P, 8%, 8) +
. . . + $100K(F/P, 8%, 1) + $60K
= $1,101K
• Equivalent lump sum benefits
V2 = $120K(P/A, 8%, 50)
= $1,468K
• Equivalent net worth
V2 - V1 = $367K > 0, Good Investment
41
Infinite Project Life
• Equivalent lump sum investment
V1 = $50K(F/P, 8%, 9) + $50K(F/P, 8%, 8) +
. . . + $100K(F/P, 8%, 1) + $60K
= $1,101K
• Equivalent lump sum benefits assuming N =
V2 = $120(P/A, 8%, )
= $120/0.08
= $1,500K
• Capitalized equivalent worth
CE(8%) = V2 - V1
= $399K > 0
Difference (50 years vs infinite) = $32,000
42
21
22. Long life vs infinite life at 12%
Long life: -302K
Infinite life: -299K
The difference is $3,000 (smaller than for 8%)
1. Selection of i very critical
2. We can approximate PW of long cash flows by capitalized
worth (approximation improves as i increases)
3. 12% not profitable
8% profitable
43
Example: Bridge Construction
Construction cost = $2,000,000
Annual Maintenance cost = $50,000
Renovation cost = $500,000 every 15 years
Planning horizon = infinite period
Interest rate = 5%
44
22
24. Alternate way to calculate P3
• Concept: Find the effective interest rate per payment period
0 15 30 45 60
$500,000 $500,000 $500,000 $500,000
• Effective interest rate for a 15-year cycle
i = (1 + 0.05)15 - 1 = 107.893%
• Capitalized equivalent worth
P3 = $500,000/1.07893
= $463,423
47
Example 5.7: Determining the Capitalized
Worth (CW) of an Endowment
• Suppose that a firm wishes to endow an advanced biological processes
laboratory at a university. The endowment principal will earn interest
that averages 8% per year, which will be sufficient to cover all
expenditures incurred in the establishment and maintenance of the
laboratory for an indefinitely long period of time (forever). Cash
requirements of the laboratory are estimated to be $100,000 now (to
establish it), $30,000 per year indefinitely, and $20,000 at the end of
every fourth year (forever) for equipment replacement.
a) For this type of problem, what analysis period is, practically speaking,
defined to be ―forever‖?
b) What amount of endowment principal is required to establish the
laboratory and then earn enough interest to support the remaining cash
requirements of this laboratory forever?
24
25. The Future Worth Method
(Future Worth Criterion)
• Given: Cash flows
and MARR (i)
• Find: The net
equivalent worth at $55,760
$27,340
a time period other $24,400
than 0 (typically at 0
3
1 2
the end of project
life) $75,000
Project life
49
Future Worth Criterion
FW (15%)inflow $24,400( F / P,15%,2) $27,340( F / P,15%,1)
$55,760( F / P,15%,0)
$119,470
FW (15%)outflow $75,000( F / P,15%,3)
$114,066
FW (15%) $119,470 $114,066
$5,404 0, Accept
50
25
26. Example 5.8: The Relationship between
Future Worth and Present Worth
• (From Example 5.3) A piece of new equipment has been
proposed by engineers to increase 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 the additional production. The
firm’s MARR is 20% per year. Evaluate the FW of the
potential improvement project. Show the relationship
between FW and PW for this example.
The Annual Worth (AW) Method
(Annual Equivalence (AE) Analysis)
• The Annual Worth
(AW) of a project 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
52
26
27. Annual Worth Analysis
Principle: Measure investment worth on annual basis
The AW of a project is annual equivalent revenues (R)
or savings minus annual equivalent expenses (E)
less its annual equivalent Capital Recovery (CR)
amount
Benefit:
• Annual reports, yearly budgets
• Seek consistency of report format
• Determine unit cost (or unit profit)
• Facilitate unequal project life comparison
53
Annual Equivalent Worth
AE(i)=PW(i)*(A|P,i,N)
If AE(i) > 0, accept
AE(i) < 0, reject
AE(i)=0, remain indifferent
Since (A|P,i,N) > 0 for –1<i<
AE(i) > 0 if and only if PW(i) > 0
54
27
29. • First Cycle:
PW(10%) = -$1,000 + $500 (P/F, 10%, 1)
+ . . . + $400 (P/F, 10%, 5)
= $1,155.68
AE(10%) = $1,155.68 (A/P, 10%, 5) = $304.87
• Both Cycles:
PW(10%) = $1,155.68 + $1,155.68 (P/F, 10%, 5)
= $1,873.27
AE(10%) = $1,873.27 (A/P, 10%,10) = $304.87
57
Annual Equivalent Cost
• When only costs are
Annual Equivalent Costs
involved, the AE Capital
method is called the costs
annual equivalent cost. +
• Revenues must cover
Operating
two kinds of costs: costs
Operating costs and
capital costs.
58
29
30. Capital and Operating Costs
• Capital costs are incurred by purchasing assets to
be used in production and service. Normally, they
are nonrecurring (one-time) costs.
• Operating costs are incurred by the operation of
physical plant or equipment needed to provide
service (e.g. labor and raw materials). Normally,
they recur for as long as an asset is owned.
• Operating costs are on annual basis anyway.
Annual equivalent of a capital cost is called
capital recovery cost, CR(i).
Remember: (A|P, i, N) is called the capital recovery factor.
59
Capital (Ownership) Costs
Def: The cost of owning an equipment S
is associated with two 0
transactions—(1) its initial cost (I) N
and (2) its salvage value (S).
• Capital costs: Taking into these I
sums, we calculate the capital costs
as:
CR(i) I ( A / P, i, N ) S( A / F, i, N ) 0 1 2 3 N
( I S)( A / P, i, N ) iS
CR(i)
since (A|F,i,N)=(A|P,i,N)-i
60
30
31. Example - Capital Cost Calculation
$50,000
• Given:
I = $200,000 0
N = 5 years 5
S = $50,000
i = 20%
• Find: CR(20%) $200,000
CR (i ) = ( I - S) ( A / P, i, N ) + iS
CR (20%) = ($200,000 - $50,000) ( A / P, 20%, 5)
+ (0.20)$50,000
= $60,157
Annual cost of owning the asset at 20% 61
Calculation of Equivalent Annual CR Amount
• Consider a device that will cost $10,000, last five
years, and have a salvage (market) value of
$2,000. Thus, the loss in value of this asset over
five years is $8,000. The MARR is 10% per year.
31
32. Justifying an investment based on AE
Method
Given: I = $20,000, S =
$4,000, N = 5 years, i =
10%
Find: see if an annual
revenue of $4,400 is
enough to cover the
capital costs.
Solution:
CR(10%) = $4,620.76
Conclusion: Need an
additional annual revenue
in the amount of $220.76.
63
Applying Annual Worth Analysis
•Unit Cost (Profit) Calculation
• Unequal Service Life Comparison
• Minimum Cost Analysis
64
32
33. Equivalent Worth per Unit of Time
$55,760
$24,400 $27,340
0
1 2 3
$75,000
Operating Hours per Year
2,000 hrs. 2,000 hrs. 2,000 hrs.
• PW (15%) = $3553
• AE (15%) = $3,553 (A/P, 15%, 3)
= $1,556
• Savings per Machine Hour Note: 3553/6000=0.59/hour:
= $1,556/2,000 instant savings in present
= $0.78/hr. worth for each hourly use;
does not consider the time
over which the savings occur 65
Equivalent Worth per Unit of Time
(cont’d)
$55,760
$24,400 $27,340
0
1 2 3
$75,000
Operating Hours per Year
1,500 hrs. 2,500 hrs. 2,000 hrs.
•Let C denote the equivalent annual savings per machine hour
• $1,556=[(C)(1500)(P|F,15%,1)
+(C)(2500)(P|F,15%,2)
+(C)(2000) (P|F,15%,3)] (A|P,15%,3)
C=$0.79/hr
66
33
34. Breakeven Analysis
Problem: Year Miles Total costs
(n) Driven (Ownership &
Operating)
At i = 6%, what
1 14,500 $4,680
should be the
2 13,000 $3,624
reimbursement rate
3 11,500 $3,421
per mile so that Sam
can break even?
Total 39,000 $11,725
67
First Year Second Year Third Year
Depreciation $2,879 $1,776 $1,545
Scheduled maintenance 100 153 220
Insurance 635 635 635
Registration and taxes 78 57 50
Total ownership cost $3,693 $2,621 $2,450
Nonscheduled repairs 35 85 200
Replacement tires 35 30 27
Accessories 15 13 12
Gasoline and taxes 688 650 522
Oil 80 100 100
Parking and tolls 135 125 110
Total operating cost $988 $1,003 $971
Total of all costs $4,680 $3,624 $3,421
Expected miles driven 14,500 miles 13,000 miles 11,500 miles
68
34
35. • Equivalent annual cost of owning and operating the car
[$4,680 (P/F, 6%, 1) + $3,624 (P/F, 6%, 2) +
$3,421 (P/F, 6%, 3)] (A/P, 6%,3)
= $3,933 per year
• Equivalent annual reimbursement
Let X = reimbursement rate per mile
[14,500X(P/F, 6%, 1) + 13,000X(P/F, 6%, 2)
+ 11,500 X (P/F, 6%, 3)] (A/P, 6%,3)
= 13.058X
• Break-even value
13.058X = 3,933
X = 30.12 cents per mile 69
Annual equivalent reimbursement as a
function of cost per mile
Annual equivalent
cost of owning and
operating ($3,933)
Gain
Annual equivalent ($)
4000
3000 Loss Minimum
reimbursement
2000 requirement ($0.3012)
Annual reimbursement
1000 amount
0 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40
Reimbursement rate ($) per mile (X)
70
35
36. Example 5.9: Using Annual Worth to Evaluate
the Purchase of New Equipment
• (From Example 5.3) A piece of new equipment has been
proposed by engineers to increase 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 the additional production. The
firm’s MARR is 20% per year. By using the AW method,
determine whether the equipment should be recommended.
Example 5.10: Determination of Monthly
Rent by Using the Annual Worth Method
• An investment company is
considering building a 25-unit
Land investment $50,000
apartment complex in a growing cost
town. Because of the long-term
Building $225,000
growth potential of the town, it is
investment cost
felt that the company could
Study period 20 years
average 90% of full occupancy for
the complex each year. If the Rent per unit per ?
following items are reasonably month
accurate estimates, what is the Upkeep expense $35
minimun monthly rent that should per unit per month
be charged if a 12% MARR (per Property taxes and 10% of total initial
year) is desired. Use the AW insurance per year investment
method.
36
37. Rate of Return
• Definition: A relative percentage method which
measures the yield as a percentage of investment
over the life of a project
• Example: Vincent Van Gogh’s painting ―Irises‖
• John Whitney Payson bought the art at $80,000.
• John sold the art at $53.9 million in 40 years.
• What is the rate of return on John’s investment?
Rate of Return
$53.9M
• Given: P =$80,000, F
= $53.9M, and N = 40
years
• Find: i 0
• Solution: 40
$53.9 M $80,000(1 i ) 40 $80,000
i 17.68%
37
38. Why is the ROR measure so popular?
• This project will bring in a 15% rate of return on
investment.
• This project will result in a net surplus of $10,000
in NPW.
Which statement is easier to understand?
Meaning of Rate of Return
In 1970, when Wal-Mart Stores, Inc. went
public, an investment of 100 shares cost
$1,650. That investment would have
been worth $13,312,000 on January 31,
2000.
What is the rate of return on that
investment?
38
39. Solution: $13,312,000
0
30
$1,650
Given: P = $1,650
F = $13,312,000
N = 30
Find i:
F P(1 i ) N
$13,312,000 = $1,650 (1 + i )30
i = 34.97% Rate of Return
Suppose that you invested that amount ($1,650) in
a savings account at 6% per year. Then, you could
have only $9,477 on January, 2000.
What is the meaning of this 6% interest here?
This is your opportunity cost if putting money in
savings account was the best you can do at that
time!
39
40. So, in 1970, as long as you earn more than 6%
interest in another investment, you will take that
investment.
Therefore, that 6% is viewed as a minimum
attractive rate of return (or required rate of return).
So, you can apply the following decision rule, to see
if the proposed investment is a good one.
ROR > MARR
Return on Investment
(ROR Defn 1)
• Definition 1: Rate of return (ROR) is defined
as the interest rate earned on the unpaid balance
of an installment loan.
• Example: A bank lends $10,000 and receives
annual payment of $4,021 over 3 years. The
bank is said to earn a return of 10% on its loan
of $10,000.
40
41. Loan Balance Calculation
A = $10,000 (A/P, 10%, 3)
= $4,021
Unpaid Return on Unpaid
balance unpaid balance
at beg. balance Payment at the end
Year of year (10%) received of year
0 -$10,000 -$10,000
1 -$10,000 -$1,000 +$4,021 -$6,979
2 -$6,979 -$698 +$4,021 -$3,656
3 -$366 -$366 +$4,021 0
A return of 10% on the amount still outstanding at the
beginning of each year
Return on Investment
(ROR Defn 2)
• Definition 2: Rate of return (ROR) is the
break-even interest rate, i*, which equates the
present worth of a project’s cash outflows to the
present worth of its cash inflows.
• Mathematical Relation:
PW (i* ) PW (i* )cash inflows PW (i* )cash outflows
0
41
42. Return on Invested Capital
(ROR Defn 3)
• Definition 3: Return on invested capital is
defined as the interest rate earned on the
unrecovered project balance of an investment
project such that, when the project terminates,
the uncovered project balance is zero. It is
commonly known as internal rate of return
(IRR).
• Example: A company invests $10,000 in a
computer and results in equivalent annual labor
savings of $4,021 over 3 years. The company is
said to earn a return of 10% on its investment of
$10,000.
Project Balance Calculation
0 1 2 3
Beginning
-$10,000 -$6,979 -$3,656
project balance
Return on
-$1,000 -$697 -$365
invested capital
Cash generated
-$10,000 +$4,021 +$4,021 +$4,021
from project
Ending project
-$10,000 -$6,979 -$3,656 0
balance
The firm earns a 10% rate of return on funds that remain internally invested
in the project. Since the return is internal to the project, we call it internal
rate of return. Investment firm lender, project borrower
42
43. The Internal Rate of Return Method
• The Internal Rate of Return method 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).
– Equivalent worth may be computed with any of the three
methods (PW, FW, AW).
• The resultant interest rate is termed the Internal Rate
of Return (IRR).
The Internal Rate of Return Method
• For a single alternative, from the lender’s
viewpoint, the IRR is not positive unless
1) Both receipts and expenses are present in the cash
flow pattern
2) The sum of receipts exceeds the sum of all cash
outflows
43
44. The Internal Rate of Return Method
General form of PW
versus interest rate
for an alternative
with a single
investment cost at
the present time
followed by a series
of positive cash
inflows over N
Plot of PW versus Interest Rate
Investment Balance Diagram
Investment Balance Diagram Showing IRR
44
45. Investment Balance Diagram
• Investment balance diagram: How much of the original
investment in an alternative is still to be recovered as a
function of time.
• Downward arrows represent annual returns, Rk-Ek
against the unrecovered investment
• Dashed lines indicate the opportunity cost of interest, or
profit, on the beginning-of-year investment balance
• The IRR is the value of i that causes the unrecovered
investment balance to exactly equal 0 at the end of the
study period and thus represents the internal earning
rate of a project
Methods for Finding Rate of Return
• Investment Classification
– Simple Investment
– Nonsimple Investment
• Computational Methods
– Direct Solution Method
– Trial-and-Error Method
– Computer Solution Method
45
46. Investment Classification
Simple Investment Nonsimple Investment
• Def: Initial cash flows • Def: Initial cash flows
are negative, and only are negative, but more
one sign change than one sign changes
in the remaining cash
occurs in the net cash flow series.
flows series.
• Example: -$100, $300,
• Example: -$100, $250, -$120 (-, +, -)
$300 (-, +, +) • ROR: A possibility of
• ROR: A unique ROR multiple RORs
Period Project A Project B Project C
(N)
0 -$1,000 -$1,000 +$1,000
1 -500 3,900 -450
2 800 -5,030 -450
3 1,500 2,145 -450
4 2,000
Project A is a simple investment.
Project B is a nonsimple investment.
Project C is a simple borrowing.
46
47. Computational Methods
Direct Direct Trial & Computer
Solution Solution Error Solution
Method Method
Log Quadratic
n Project A Project B Project C Project D
0 -$1,000 -$2,000 -$75,000 -$10,000
1 0 1,300 24,400 20,000
2 0 1,500 27,340 20,000
3 0 55,760 25,000
4 1,500
Direct Solution Methods
• Project A • Project B
$1,300 $1,500
$1,000 $1,500( P / F , i ,4) PW (i ) $2,000 0
(1 i ) (1 i ) 2
$1,000 $1,500(1 i ) 4 1
Let x , then
0.6667 (1 i ) 4 1 i
PW (i ) 2,000 1,300 x 1,500 x 2
ln 0.6667
ln(1 i ) Solve for x:
4
x 0.8 or -1.667
0.101365 ln(1 i )
Solving for i yields
e 0.101365 1 i
1 1
0.8 i 25%, 1667
. i 160%
i e 0.101365 1 1 i 1 i
10.67% Since 100% i , the project's i * 25%.
47
48. Trial and Error Method – Project C
• Step 1: Guess an interest • Step 4: If you bracket the
rate, say, i = 15% solution, you use a linear
• Step 2: Compute PW(i) interpolation to approximate
at the guessed i value. the solution
PW (15%) = $3,553 3,553
0
• Step 3: If PW(i) > 0, then -749
increase i. If PW(i) < 0,
then decrease i. 15% i 18%
PW(18%) = -$749 3,553
i 15% 3%
3,553 749
17.45%
Graphical Solution – Project D
• Step 1: Create a NPW plot
using Excel.
• Step 2: Identify the point
at which the curve crosses
the horizontal axis closely
approximates the i*.
• Note: This method is
particularly useful for
projects with multiple
rates of return, as most
financial softwares would
fail to find all the multiple
i*s.
48
49. Basic Decision Rule
(single project evaluation):
If ROR > MARR, Accept
This rule does not work for a situation where
an investment has multiple rates of return
Multiple Rates of Return Problem
$2,300
$1,000 $1,320
• Find the rate(s) of return:
$2,300 $1,320
PW (i) $1,000
1 i (1 i)2
0
49
50. 1
Let x . Then,
1 i
$2,300 $1,320
PW (i ) $1,000
(1 i ) (1 i ) 2
$1,000 $2,300 x $1,320 x 2
0
Solving for x yields,
x 10 / 11 or x 10 / 12
Solving for i yields
i 10% or 20%
NPW Plot for a Nonsimple Investment
with Multiple Rates of Return
50
51. Project Balance Calculation
i* =20%
n=0 n=1 n=2
Beg. Balance -$1,000 +$1,100
Interest -$200 +$220
Payment -$1,000 +$2,300 -$1,320
Ending Balance -$1,000 +$1,100 $0
Cash borrowed (released) from the project is assumed to
earn the same interest rate through external investment
as money that remains internally invested.
Critical Issue: Can the company be able to invest the money
released from the project at 20% externally in
Period 1?
If your MARR is exactly 20%, the answer is ―yes‖, because it
represents the rate at which the firm can always invest
the money in its investment pool. Then, the 20% is also true
IRR for the project.
Suppose your MARR is 15% instead of 20%. The assumption
used in calculating i* is no longer valid.
Therefore, neither 10% nor 20% is a true IRR.
51
52. How to Proceed: If you encounter multiple rates of
return, abandon the IRR analysis and use the NPW
criterion (or use alternative procedures such as the
external rate of return method).
• If NPW criterion is used at MARR = 15%
PW(15%) = -$1,000
+ $2,300 (P/F, 15%, 1)
- $1,320 (P/F, 15%, 2 )
= $1.89 > 0
Accept the investment
Decision Rules for Nonsimple Investment
• A possibility of multiple RORs.
• If PW (i) plot looks like this,
then, IRR = ROR.
i*
If IRR > MARR, Accept i
• If PW(i) plot looks like this,
PW (i)
Then, IRR ROR (i*).
• Find the true IRR by using
other procedures or, i*
• Abandon the IRR method
and use the PW method. i* i
52
53. Example 5.11
• A capital investment of $10,000 can be
made in a project that will produce a
uniform annual revenue of $5,311 for five
years and then have a salvage value of
$2,000. Annual expenses will be $3,000.
The company is willing to accept any
project that will earn at least 10% per year
on all invested capital. Determine whether it
is acceptable, by using the IRR method.
53
54. Example 5.11
Example 5.12
• A piece of new equipment has been proposed by
engineers to increase 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 the
additional production. Evaluate the IRR of the
proposed equipment. Is the investment a good one?
The MARR is 20% per year.
54
56. Example 5.12
Investment (or Project) Balance Diagram
Example 5.13
• In 1915, Albert Epstein borrowed $7,000 from a large New
York bank on the condition that he would repay 7% of the
loan every three months, until a total of 50 payments had
been made. At the time of the 50th payment, the $7,000
loan would be completely repaid.
• Albert computed his annual interest rate to be
[0.07(7000)*4]/7000=0.28 (28%).
1. What true effective annual interest rate did Albert pay?
2. What, if anything, was wrong with his calculation?
56
57. Example 5.13
Example 5.13:
What was wrong with his calculation?
His answer is the same (28% ) whether he
makes 20, 50, 70 payments (that is, insensitive
to how long the payments are made).
20 quarterly 50 quarterly 70 quarterly
payments of payments of payments of
$490 $490 $490
True annual 14.5% 30% 31%
effective rate
57
58. Example 5.14
• The Fly-by-Night finance company advertises a ―bargain 6%
plan‖ for financing the purchase of automobiles.
• To the amount of loan being financed, 6% is added for each
year money is owed. This total is then divided by the number
of months over which the payments are to be made, and the
result is the amount of the monthly payments. For example, a
woman purchases a $10,000 automobile under this plan and
makes an initial cash payment of $2,500. She wishes to pay
the $7,500 balance in 24 monthly payments.
• What effective annual rate of interest does she actually pay?
Example 5.14
58
59. Example 5.15
• A small airline executive charter company needs to borrow
$160,000 to purchase a prototype synthetic vision system
(SVS) for one of its business jets. The SVS is intended to
improve the pilots’ situational awareness when visibility is
impaired. The local (and only) banker makes this statement:
• ―We can loan you $160,000 at a very favorable rate of 12%
per year for a five-year loan. However, to secure this loan, you
must agree to establish a checking account (with no interest) in
which the minimum average balance is $32,000. In addition,
your interest payments are due at the end of each year, and the
principal will be repaid in a lump-sum amount at the end of
year five. ‖
• What is the true effective annual interest rate being charged?
Example 5.15
59
60. Difficulties with the IRR Method
• The reinvestment assumption of the IRR method
may not be valid.
– Suppose MARR=20% and, IRR for a project is 42.4%: it
may not be possible to reinvest net cash proceeds from the
project at much more than 20%.
– PW, AW, FW method assume that positive recovered
funds are reinvested at the MARR
• Computational difficulty
• Possibility of multiple rates of return in some
problems
The External Rate of Return
(ERR) Method
• Directly takes into account the interest rate
() external to a project at which net cash
flow 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.
60
61. The External Rate of Return
(ERR) Method
Step 1: Discount all net cash outflows to 0.
Step 2: Compound all net cash inflows to 0.
Step 3: ERR is the rate that establishes equivalence
between the two quantities.
Project acceptable when the ERR is greater than or
equal to the firm’s MARR.
The External Rate of Return
(ERR) Method
• Two advantages over IRR:
– Easier to obtain
– Not subject to the possibility of multiple rates of return
61
62. Example 5.16
• Referring to Example 5.12, suppose that
= MARR = 20% per year.
• What is the project’s ERR, and is the
project acceptable?
Example 5.16
62
63. Example 5.17
• When =15% and MARR = 20% per year,
determine whether the project (with the
cash flow given next) is acceptable. Notice
in this example that the use of an %
different from the MARR is illustrated. This
might occur if, for some reason, part or all
of the funds related to a project are
―handled‖ outside the firm’s normal capital
structure.
Example 5.17
63
64. Appendix 5-A The Multiple Rate of Return
Problem with the IRR Method:
Example 5-A-1
• Plot the present worth versus interest rate for the
following cash flows. Are there multiple IRRs? If
so, what do they mean?
Year k Net Cash Flow i% PW(i%)
0 500 0 250
1 -1000 10 150
2 0 20 32
3 250 30 0
4 250 40 -11
5 250 62 0
80 24
Example 5-A-1
64
65. Example 5-A-2
• Use the ERR method to
analyze the cash flow Year Cash Flows
pattern in the accompanying 0 5,000
table. 1 -7,000
• The IRR is indeterminant 2 2,000
(none exists), so the IRR is 3 2,000
not a workable procedure.
The external reinvestment
rate is 12% per year, and the
MARR equals 15%.
Example 5-A-2
65
66. Summary of Decision Rules
Payback (Payout) Period Method
How fast can I recover my initial investment?
• Conventional payback method (time value of money
ignored)
• Discounted payback method
• Payback screening - If the payback period is within
acceptable range, formal project evaluation may start (a
high-tech firm may have a short time limit since products
rapidly become obsolete).
132
66
67. Conventional-Payback Method
Principle:
How fast can I recover my initial investment?
Method:
Based on cumulative cash flow
Screening Guideline:
If the payback period is less than or equal to
some specified payback period, the project
would be considered for further analysis.
Weakness:
Does not consider the time value of money
133
Example: Conventional Payback Period
N Cash Flow Cum. Flow
0 -$105,000+$20,000 -$85,000
1 $15,000 -$70,000
2 $25,000 -$45,000
3 $35,000 -$10,000
4 $45,000 $35,000
5 $45,000 $80,000
6 $35,000 $115,000
Payback period should occurs somewhere
between N = 3 and N = 4.
134
67
68. $45,000 $45,000
$35,000 $35,000
Annual cash flow
$25,000
$15,000
0
1 2 3 4 5 6
Years
$85,000
150,000
Cumulative cash flow ($)
100,000 3.2 years
Payback period
50,000
0
-50,000
-100,000
0 1 2 3 4 5 6
135
Years (n)
Payback Method Pitfall
n Project 1 Project 2
Payback:3 years Payback:3.6 years
0 -90000 -90000
1 30000 25000
2 30000 25000
3 30000 25000
4 1000 25000
5 1000 25000
6 1000 25000
• Fails to measure profitability
• Ignores timing of cash flows 136
68
69. Discounted Payback Method
Principle:
How fast can I recover my initial investment
plus interest?
Method:
Based on cumulative discounted cash flow
Screening Guideline:
If the discounted payback period (DPP) is less
than or equal to some specified payback period,
the project would be considered for further
analysis.
Weakness:
Cash flows occurring after DPP are ignored
137
Discounted Payback Period Calculation
Period Cash Flow Cost of Funds Cumulative
(15%) Cash Flow
0 -$85,000 0 -$85,000
1 15,000 -$85,000(0.15)= -$12,750 -82,750
2 25,000 -$82,750(0.15)= -12,413 -70,163
3 35,000 -$70,163(0.15)= -10,524 -45,687
4 45,000 -$45,687(0.15)=-6,853 -7,540
5 45,000 -$7,540(0.15)= -1,131 36,329
6 35,000 $36,329(0.15)= 5,449 76,778
-85000(1.15)+15000= -82750 138
69
