SEEM2440
Engineering Economic
Lecture 04 – Evaluating a Single Project
The final test of any system is,
does
it pay?
Frederick W. Taylor (1912)
Copyright (c) 2015. Gabriel Fung. All rights reserved.
Evaluating Alternative
Given so many formulas, when we are given some alternatives,
how to apply those formulas?
(F / P,i%,N) = (1+ i)N
(P / F,i%,N) =
1
(1+ i)N
(1+ i)N − 1
i
i
(A / F,i%,N) =
(1+ i)N − 1
(F / A,i%,N) =
(P / A,i%,N) =
(1+ i) − 1
i(1+ i)N
1
N
[(1+ i)N − 1] −
i
i2
(P / G,i%,N) = (F / G,i%,N)(P / F,i%,N)
Single payment compound amount
(F / G,i%,N) =
Single payment present worth
1 ⎡1 ⎛
1⎞ 1 ⎤
= ⎢ −⎜N + ⎟
⎥
i ⎣i ⎝
i ⎠ (1+ i)N ⎦
(A / G,i%,N) = (F / G,i%,N)(A / F,i%,N)
Unifrom series compound amount
Uniform series present worth
⎡1
⎤
N
=⎢ −
⎥
N
⎣ i (1+ i) − 1 ⎦
N
i(1+ i)N
(A / P,i%,N) =
(1+ i)N − 1
Sinking fund
Capital recovery
(F / P,r%,N) = e rN
(P / F,r%,N) =
(F / A,r%,N) =
⎧ A1[1− (P / F,i%,N)(F / P,i%,N)]
⎪
i− f
P=⎨
⎪
A1N(P / F,i%,1)
⎩
f ≠i
(A / F,r%,N) =
f =i
(P / A,r%,N) =
Copyright (c) 2015. Gabriel Fung. All rights reserved.
(A / P,r%,N) =
1
e rN
e rN − 1
er − 1
er − 1
e rN − 1
e rN − 1
rN
e (e r − 1)
e rN (e r − 1)
e rN − 1
Single payment compound amount
Single payment present worth
Unifrom series compound amount
Uniform series present worth
Sinking fund
Capital recovery
Note
Unless otherwise specified, the end-of-period cash-flow
convention and discrete compounding of interest are used.
Copyright (c) 2015. Gabriel Fung. All rights reserved.
MARR
Minimum Attractive Rate of Return (MARR) is the baseline set
by You
Any project that has a return higher than MARR should be
considered/accepted
If there are several mutually exclusive projects, we should choose
the one with largest MARR.
How to determine MARR?
The easiest way to determine MARR is based on opportunity
cost analysis.
E.g., If you put money in a bank and can earn 3% interest rate,
then this 3% can be considered as MARR.
Copyright (c) 2015. Gabriel Fung. All rights reserved.
MARR (cont’d)
Some simple and effective methods that can apply the concept
of MARR:
Present Worth Method
Future Worth Method
Annual Worth Method
Internal Rate of Return (IRR)
External Rate of Return (ERR)
Copyright (c) 2015. Gabriel Fung. All rights reserved.
Equivalent
MARR – PW / FW / AW
Basic steps of using Present Worth Method / Future Worth
Method / Annual Worth Method (note: step 2 may sometimes
need to revise, which depends on the situation):
1.  Draw the cash flow diagram
2.  Assume the investment return rate i = MARR
3.  If the present value/future value/annuity of the project > 0, the
project should be considered/accepted.
Copyright (c) 2015. Gabriel Fung. All rights reserved.
MARR – IRR
Basic steps of using IRR:
1.  Draw the cash flow diagram
2.  Assume the present value of the project = 0.
3.  Identify an investment return rate i which can solve the cash
flow.
4.  If i > MARR, it means the project should be considered/
accepted.
Copyright (c) 2015. Gabriel Fung. All rights reserved.
MARR – ERR
Basic steps of using ERR:
1.  Draw the cash flow diagram
2.  Assume:
1.  The external reinvestment return rate e = MARR
2.  The profit of the project = The value of investment and all expenses
when they are invested somewhere else at with return rate i
3.  Discount all cash outflow to time zero using e
4.  Compound all cash inflow to period N using e
5.  Identify i so that (3) = (4)
6.  If i > MARR, it means the project should be considered/
accepted.
Copyright (c) 2015. Gabriel Fung. All rights reserved.
More About MARR
Why we need three kinds of methods to do the analysis?
Well… We will know very soon….
Copyright (c) 2015. Gabriel Fung. All rights reserved.
Economic Desirability of a Project
Consider the following investment plan:
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 salvage value of $2,000.
Annual expense 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. Is this a desirable investment?
Copyright (c) 2015. Gabriel Fung. All rights reserved.
Economic Desirability of a Project (cont’d)
Cash flow diagram:
Solution by Present Worth:
…
Copyright (c) 2015. Gabriel Fung. All rights reserved.
Economic Desirability of a Project (cont’d)
Solution by Internal Rate of Return (IRR):
First, we see that the sum of cash flows is positive. Thus, it is
likely that a positive i can be obtained:
PW = 0 = −$10,000 + ($5,311− $3,000)(P / A,i%,5) + $2,000(P / F,i%,5)
We can solve the above equation by linear interpolation:
PW(5%) = $1,572
PW (15%) = −$1,259
15% − 5%
i % − 5%
=
$1,572 − (−$1,259) $1,572 − $0
i % = 10.5%
3
d
$1,572
2
B
A
1
e
0
−1
5%
10%
i '%
15%
Note: the “exact solution” by Excel
is 10%
−2
Copyright (c) 2015. Gabriel Fung. All rights reserved.
Economic Desirability of a Project (cont’d)
Solution using ERR:
Assume the external reinvestment rate, e = MARR = 10%
We have the following relationship to solve for i %:
$10,000(F | P,i%,5) = ($5,311− $3,000)(F | A,e%,5) + $2,000
= ($5,311− $3,000)(F | A,10%,5) + $2,000
i = 10.01%
Because i % > MARR, the project is justified.
Copyright (c) 2015. Gabriel Fung. All rights reserved.
C
−$1, 259
New Equipment Purchase
A piece of new equipment has been proposed by engineers to
increase the productivity:
The investment cost is $25,000 and the equipment will have a
market value of 5,000 at the end of a study period 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 revenue generated by the additional production.
If the firm ‘s MARR is 20% per year, is this proposal a sound
one?
New Equipment Purchase (cont’d)
Solution by PW / FW / AW:
Present Worth = $934.29
Future Worth = $2324.80
Annual Worth = $312.40
New Equipment Purchase (cont’d)
Solution by IRR:
0 = −$25,000 + $8,000(P | A,i%, 5) + $5,000(P | F,i%,5)
i = 21.58%
Since i > MARR, the project is justified.
Copyright (c) 2015. Gabriel Fung. All rights reserved.
New Equipment Purchase (cont’d)
Solution using ERR:
We have the following relationship to solve for i %:
$25,000(F | P,i%,5) = $8,000(F | A,20%,5) + $5,000
i = 20.88%
Because i > MARR, the project is justified.
Copyright (c) 2015. Gabriel Fung. All rights reserved.
Note About PW / FW/ AW
Some implicit Assumptions:
1.  We know the future with certainty, such as the interest rate.
2.  The capital markets are perfect, i.e., borrow and lend money are
at the same interest rate.
Copyright (c) 2015. Gabriel Fung. All rights reserved.
Finding the Current Price of A Bond
Bond is one of the most important financial instruments.
You pay some money to buy a bond.
E.g. You pay $70 now to buy a $100 HSBC 10-year Bond
A bond has a certain “life” (the number of years to “mature”)
E.g. 10 years.
You will be paid for a fixed amount called “coupon” for every
fixed time interval.
E.g. You will receive $10 every 6 months.
You will be paid for a fixed price when the bond “mature”.
E.g. You will receive back $100 after 10 years.
Copyright (c) 2015. Gabriel Fung. All rights reserved.
Bonds
Terminologies:
Face Value (the value of the bond), Z
E.g. Z = $100
Disposal Price (the price when the bond mature), C
Usually, C = Z
Coupon rate, r
E.g. If coupon rate for a bond with Z = $100 is 8%, then the
coupon payment is 8% x $100 = $8.
Yield, I
The internal rate of return of the bond.
Copyright (c) 2015. Gabriel Fung. All rights reserved.
Bonds (cont’d)
Sometimes, a bond does not have any coupon payment
It is know as “Zero-Coupon Bond”
In this situation, the coupon rate, r, is certainly equals to zero.
Copyright (c) 2015. Gabriel Fung. All rights reserved.
Stan Moneymaker Wants to Buy a Bond
Stan Moneymaker has the opportunity to purchase a certain
U.S. Treasury bond that matures in 8 years and has a face
value of $10,000.
The bond stipulates a fixed nominal interest rate of 8% per year
to be made to the bondholder every three months.
Stan would like to earn 10% nominal interest (compounded
quarterly) per year on his investment. How much should Stan be
willing to pay for the bond?
Solution:
V = $10000(P | F,2.5%,32) + $10000 × 0.02(P | A,2.5%,32)
Willing to pay = $8907.55
Copyright (c) 2015. Gabriel Fung. All rights reserved.
A Bond Example
Find the price of a 10-year bond with face value $1,000 paying
6% per year semiannually that is redeemable at par valued if it
is bought by a purchaser to yield 10% per year.
Current price = Present Worth
N = 10 × 2 = 20 periods,
r = 6% / 2 = 3% per period,
i = [(1.10)1 2 − 1] × 100% ! 4.9% per semiannual period,
C = Z = $1,000
Current price = $761.16
Copyright (c) 2015. Gabriel Fung. All rights reserved.
Current Price and Annual Yield of A Bond
A bond with a face value of $5,000 pays coupon 8% per year.
This bond will be redeemed at par value at the end of its 20year life, and the first interest payment is due 1 year from now.
How much should be paid now for this bond in order to receive
a yield of 10% per year on the investment?
If this bond is purchased now for $4,600, what annual yield
would the buyer expect?
Solution
Value = $4,148.44
Annual yield = 8.9%
Copyright (c) 2015. Gabriel Fung. All rights reserved.
Capitalized-Worth (CW) Method
A variation of PW method which determines the PW of all
revenues or expenses over an infinite length of time.
From previous lecture:
A = Pi ⇔ P =
A
i
Hence:
CW = P =
A
i
Copyright (c) 2015. Gabriel Fung. All rights reserved.
Establishing An Endowment
Suppose that a firm wishes to endow a laboratory at CUHK.
If the endowment principal can earn 8% interest per year, it will
be able to cover all expenditures incurred in the establishment
and maintenance of the laboratory for an indefinitely long
period of time.
Cash requirements of the laboratory are estimated to be
$100,000 now (to establish it), $30,000 per year thereafter, and
$20,000 at the end of every fourth year.
What amount of endowment principal is required to establish
the laboratory?
Solution
Answer: Present Worth = –$530,475
Copyright (c) 2015. Gabriel Fung. All rights reserved.
CW of A Bridge
The construction cost of a bridge is $1,900,000 and annual
upkeep is $25,000. Major maintenance work will be conducted
every 8 years at $350,000. Assume the government targets
MARR to be 8%. If the expected life of the bridge is 50 year,
what is the CW of the bridge over a 100-year study period?
Solution
CW (8%) = −1900000 − $1900000(P | F,8%,50)
−[$350000(A | F,8%,8)] / 0.08 − $25000 / 0.08
Copyright (c) 2015. Gabriel Fung. All rights reserved.
“Fly-by-Night” Financing
A company advertises a “bargain 6% plan”:
E.g. If you borrow $10,000 and makes an initial cash payment of
$2,500 and wish to pay the $7,500 balance in 24 monthly:
Loan
$10,000
Initial Payment
2,500
Balance Due
7,500
6% charge in 2 years (0.06 x 2 x 7500)
900
Total reminded
8,400
Monthly Payment (8,400 / 24)
$350
What effective annual rate of interest you are paying actually?
Answer
12%!
Copyright (c) 2015. Gabriel Fung. All rights reserved.
Multiple IRR
For the following project. Is it worthwhile to invest? Assume
that your MARR is 10%.
Hint: Plot the present worth versus interest rate!
Year
Net Cash Flow
0
$500
1
-1,000
2
0
3
250
4
250
5
250
Copyright (c) 2015. Gabriel Fung. All rights reserved.
Multiple IRR (cont’d)
The PW of the net cash flows equals zero at interest rates of
about 30% and 62%
Multiple IRRs exist!
Whenever there are multiple IRRs, it is likely that none of
them are correct!
Copyright (c) 2015. Gabriel Fung. All rights reserved.
Multiple IRR (cont’d)
In this situation, we should use either the Equivalent Worth
Method (i.e., PW / FW / AW) or the ERR Method to solve
the problem.
Assume that the external reinvestment rate is 10% (MARR) per
year, then the ERR is 12.4%.
In addition, PW(15%) = $105
So both the ERR and PW methods indicate that this project is
acceptable when the MARR is 10% per year.
Copyright (c) 2015. Gabriel Fung. All rights reserved.
Which Methods To Use?
For the following project, is it worthwhile to carry out?
Assume that the MARR is 15% and the External Reinvestment rate is
12%
Year
Cash Flow
0
$5,000
1
-7,000
2
2,000
3
2,000
Solution:
IRR does not exist
ERR method = 21.7%
PW method = $1,740.36
Copyright (c) 2015. Gabriel Fung. All rights reserved.
Note When Using ERR
In the pervious slides, we always assume that e = MARR.
However, they may NOT necessarily equals!!!
E.g. e is different from MARR if (for some reasons) part or all of
the funds related to a project are "handled" outside the firm's
normal capital structure.
Copyright (c) 2015. Gabriel Fung. All rights reserved.
Note When Using ERR (cont’d)
Example:
When e = 15% and MARR = 20% per year, determine whether
the following project is acceptable.
$6000
1
2
$6000
4
3
$1000
$6000
$1000
$6000
5
$1000
$6000
6
$1000
$1000
$5000
$10000
Years
Note When Using ERR (cont’d)
Solution
E 0 = $10,000 (k = 0),
E1 = $5,000 (k = 1),
R k = $5,000 for k = 2,3,!6,
[$10,000 + $5,000( P / F , 15%, 1)]( F / P, i′%, 6) = $5,000( F / A, 15%, 5);
i′% = 14.2%
The i'% is less than the MARR = 20%; therefore, this project
would be unacceptable according to the ERR method.
Determination of Monthly Rent
An investment company is considering building a 25-unit
apartment complex in a growing town.
The company think that it could average 90% of full occupancy
for the complex each year. What is the minimum monthly rent
that should be charged if a 12% MARR (per year) is desired?
Land investment cost
$50,000
Building investment cost
$225,000
Study period, N
20 years
Rent per unit per month
?
Upkeep expense per unit per month
$35
Property taxes and insurance per year
10% of total initial investment
Solution:
Rent = $271
Copyright (c) 2015. Gabriel Fung. All rights reserved.