CALIFORNIA STATE UNIVERSITY
SACRAMENTO
DEPARTMENT OF ELECTRICAL AND ELECTRONIC ENGINEERING
ENGR 140 – Engineering Economics
Engineering Economic Analysis,Newman, Eschenbach, Lavelle, 14th
Edition, (Oxford Press)
Lecture Set 4
Instructor: Riaz Ahmad
Lecture Set 4 - Topics
➢
Rate of Return
➢
Minimum Attractive Rate of Return (MARR)
➢
Profitably Index (PI)
➢
Internal Rate of Return (IRR)
➢
Practice Problems
Rate of Return
▪ A rate of return is the net gain or loss of an investment over a
specified time period, expressed as a percentage of the
investment’s initial cost.
▪ When calculating the rate of return, you are determining the
percentage change from the beginning of the period until the
end.
▪ The metric of rate of return can be used on a variety of assets,
from stocks to bonds, real estate, and art.
▪ The effects of inflation are not taken into consideration in the
simple rate of return calculation but are in the real rate of return
calculation.
▪ The rate of return works with any asset provided the asset is
purchased at one point in time and produces cash flow at some
point in the future.
▪ Many investors like to pick a required rate of return before
making an investment choice.
▪ The formula to calculate the rate of return is as follows:
Rate of return = [
𝐶𝑢𝑟𝑟𝑒𝑛𝑡 𝑉𝑎𝑙𝑢𝑒−𝐼𝑛𝑖𝑡𝑖𝑎𝑙 𝑣𝑎𝑙𝑢𝑒
𝐼𝑛𝑖𝑡𝑖𝑎𝑙 𝑣𝑎𝑙𝑢𝑒
] x 100
Minimum Attractive Rate of Return (MARR): The Minimum
Attractive Rate of Return (MARR) is a reasonable rate of return
established for the evaluation and selection of alternatives.
▪ A project is not economically viable unless it is expected to
return at least the MARR.
▪ MARR is also referred to as the hurdle rate, cutoff rate,
benchmark rate, and minimum acceptable rate of return.
▪ A minimum acceptable rate of return (MARR) is the minimum
profit an investor expects to make from an investment, taking
into account the risks of the investment and the opportunity
cost of undertaking it instead of other investments.
▪ For most practical situations RoR and MARR can be
considered same.
Example – 01: An engineer invests $5000 annually at year-end for 40 years. For $1
million at retirement, what interest must be earned? Consider interest compounded
annually.
We know for uniformly distributed installment case:
Fv = Pu [
(1+𝑅)𝑇 −1
𝑅
] =======➔ Use iteration
where
Fv = Future value
= $1000000
Pu = uniform installments = $5000
R = Interest rate per period = ?
T= No of periods = 40
1000000 = 5000 [
(1+𝑅)40 −1
𝑅
]
=====➔ By iteration, R = 7.007%
Example – 02: A firm invests $75,000 to save $9000/year in energy costs for 15 years.
What compound interest rate should it get?
Future worth of $75000 can be found as:
Fw = P [ (1+R)T] = 75000*(1+R)15
This future worth will be used in uniformly distributed equation.
We know, for uniformly distributed installment case:
Fv = Pu [
(1+𝑅)𝑇 −1
𝑅
] =======➔ Use iteration
where
Fv = Future value
= 75000*(1+R) 15
Pu = uniform installments = $9000
R = Interest rate per period = ?
T= No of periods = 15
75000 = 9000 [
(1+𝑅)15 −1
𝑅
]
=====➔ By iteration, R = 8.442%
Example – 03: A small restaurant costs $350,000. Expected profits equal $22,000/year
for 6 years when value is $400,000. B. 8.22%
Future worth of $-350000 can be found as:
Fw = P [ (1+R)T] = -350000*(1+R) 6
We know, for uniformly distributed installment case:
Fv = Pu [
(1+𝑅)𝑇 −1
𝑅
] = 22000* [
(1+𝑅)6 −1
𝑅
]
After 6 years restaurant value = $400000
So the interest rate at which
iteration method.
Fw + Fv + 400000 ≈ 0, can be found out by using
R = 8.222%
Profitability Index (PI)
▪ The profitability index (PI) is a measure of the attractiveness of
a project or investment.
▪ The profitability index (PI), alternatively referred to as value
investment ratio (VIR) or profit investment ratio (PIR),
describes an index that represents the relationship between the
costs and benefits of a proposed project.
▪ The profitability index is calculated as the ratio between the
present value of future expected cash flows and the initial
amount invested in the project.
▪ A higher PI means that a project will be considered more
attractive.
▪ Profitability index (PI) =
PV of future cash inflows
𝐼𝑛𝑖𝑡𝑖𝑎𝑙 𝑖𝑛𝑣𝑒𝑠𝑡𝑚𝑒𝑛𝑡
Internal Rate of Return (IRR)
▪ The internal rate of return is a discount rate that makes the net
present value (NPV) of all cash flows from a particular project
or investment equal to zero.
▪ The rate of return using discounted cash flows is also known as
the internal rate of return (IRR).
▪ The internal rate of return (IRR) takes into consideration the
time value of money.
▪ IRR calculations rely on the same formula as NPV does and
utilizes the time value of money (using interest rates). The
formula for IRR is as follows:
▪ Interest rate at which EUAW = 0
▪ On a loan, it is interest rate paid on unpaid balance such that
balance = 0 after final payment
▪ On an investment it is interest rate earned on the un-recovered
investment so un-recovered investment = 0 after last cash flow
▪ Internal rate of return is interest rate where:
A. NPV = 0
B. PV of benefits = PV of costs
C. Annual benefits = Annual costs
Example – 04: Mary invests $40000 in a business project. The business pays her $12000
at the end of each year for the next 5 years at which point she sells her business for
$20000. Prevailing interest rates are currently 5%.
a) What is the net present value for this project.
b) Is it worth investing in?
c) Calculate the profitability index.
d) Determine the internal rate of return for this project.
12000
a)
NPV = -40000 +
b)
Yes! since NPV is positive so do it.
(1.05)1
+
12000
(1.05)2
+
12000
(1.05)3
+
12000
(1.05)4
+
32000
(1.05)5
= $27,624.24
c) Profitability index (PI) =
PV of future cash inflows
𝐼𝑛𝑖𝑡𝑖𝑎𝑙 𝑖𝑛𝑣𝑒𝑠𝑡𝑚𝑒𝑛𝑡
=
67624.24
40000
if PI > 1
===➔ NPV > 0 ===➔ Accept
if PI < 1
===➔ NPV < 0 ===➔ Reject
if PI = 1
===➔ NPV = 0 ===➔ Accept or Reject
= 1.69 >1 Accept
d) Internal rate of return (IRR) = Rate for which NPV=0
-40000 +
12000
(1+𝑅)1
+
12000
(1+𝑅)2
+
12000
(1+𝑅)3
+
12000
(1+𝑅)4
By using iteration method, we found R = 23.8%
+
32000
(1+𝑅)5
=0
=====➔ IRR = 23.8%
Example – 05: Maria borrowed $9000 yearly for 4 years. No interest was charged until
graduation, then rate = 5%.
Maria makes 5 equal annual payments of what amount to pay off the loan?
What is the internal rate of return for Maria’s cash flow?
At point 4 loan amount = 4*9000 = $36000
At point 9 loan amount = 36000*(1.05)5 = $45946.13
Loan payment A =
45946.13∗0.05
[(1.05)5 −1]
NPV of inflow = 9000 +
NPV of outflow =
9000
= $8315.1
1 +
(1+𝑅)
9000
2 +
(1+𝑅)
9000
(1+𝑅)3
8315.1
8315.1
8315.1
8315.1
8315.1
(1+𝑅)
(1+𝑅)
(1+𝑅)
(1+𝑅)
(1+𝑅)9
5 +
6 +
7 +
By using iteration method, we found R = 2.66%
8 +
=====➔ IRR = 2.66%
Practice Problems
P-01: A concrete plant is considering a new piece of equipment with an initial cost of
$75,000 and a salvage value of $15,000 at the end of its 25-year useful life. The annual
maintenance cost for this piece of equipment is projected to be $10,500. The equipment
will produce an annual labor savings of $24,000. What is the approximate projected
before-tax rate of return on the equipment?
17.8%
P-02: A tire manufacturing plant is considering two alternatives for production. A
minimum attractive rate of return (MARR) of 8% is desired. Considering the following,
which alternative should be selected? Choose A since interest rate is greater than the MARR
Year
Alternative A
Alternative B
0
−$350
−$175
1
+$110
+$65
2
+$110
+$65
3
+$110
+$65
4
+$110
+$65
5
+$110
+$65
P-03: The rate of return on an investment of $1500 that doubles in value over a 4-year
period, and produces a $300 annual cash flow, is nearest to which value?
20%
P-04: An investor with a MARR of 20% has been asked to invest in a project that has an
expected rate of return (ROR) of 15%. Should the investor invest in this project? No,
because ROR < MARR
P-05: Here are the data for an asset that is being considered:
Initial cost = $35,000
Salvage value at 5 years = $3000
Rebuild cost at 3 years = $23,000
Annual net cash flow = $22,000 per year
What is the ROR for this asset?
43.9%
P-06: A project has an initial cost of $100,000 and uniform annual benefits of $12,500.
At the end of its 8-year useful life, its salvage value is $30,000. At a 10% interest rate, the
net present worth of the project is approximately.
$19,317.50
P-07: Winners of the Power State Lottery can take $30 million now or payments of $2.5
million per year for the next 15 years. These are equivalent at what annual interest rate?
The answer is closest to what value?
3%
P-08: At what annual compound interest rate is $100 today equivalent to $370,
seventeen years from now?
8%