ME 443
MEASURING THE WORTH OF
INVESTMENTS
Prof. Dr. Mustafa Gökler
INTRODUCTION
Engineers and systems analysts
solve problems by formulating and
analyzing the problem, generating a
number of feasible solutions
(alternatives) to the problem,
comparing the investment
alternatives, selecting the preferred
solution, and implementing the
solution.
INTRODUCTION
A number of methods and criteria for
measuring the worth of investments
will be discussed in this chapter.
These forms the basis for techniques
used in the next chapter to compare
alternative investments and select
those preferred from the economic
point of view.
INTRODUCTION
Where compound interest
calculations are used, the measures
of investment worth are referred to
as discounted cash flow (DCF)
measures;
the methods used to compute the
values of investment worth are
called discounted cash flow
methods.
INTRODUCTION
The compound interest rate used is
variously referred to as a minimum
attractive rate of return (MARR),
required rate of return, return on
investment, and discount rate.
METHODS OF MEASURING
INVESTMENT WORTH
1. Present worth method (PW).
2. Annual worth method (AW).
3. Future worth method (FW).
4. Internal rate of return method (IRR).
5. External rate of return method (ERR).
6. Savings/investment ratio method (SIR).
7. Payback period method (PBP).
8. Capitalized worth method (CW).
METHODS OF MEASURING
INVESTMENT WORTH
With the exception of the payback
period
and
capitalized
worth
methods all of the measures listed
are equivalent methods of measuring
investment worth. Hence, applying
each of the first six measures of
merit to the same investment
alternative will yield the same
recommendation.
PRESENT WORTH METHOD
Present Worth (PW) method converts
all cash flows to a single sum
equivalent at time zero using i =
MARR.
Ajt : Net cash flow for investment j at the end of
period t
ANNUAL WORTH METHOD
Annual Worth (AW) method converts
all cash flows to an equivalent
uniform annual series of cash flows
over the planning horizon using i =
MARR
ANNUAL WORTH METHOD
FUTURE WORTH METHOD
Future Worth (FW) method converts
all cash flows to a single sum
equivalent at the end of the planning
horizon using
i = MARR.
FUTURE WORTH METHOD
EXAMPLE
A pressure vessel was purchased for
$16,000, kept for 5 years, and sold
for $3000. Annual operating and
maintenance
costs were $4000.
Using a 12% minimum attractive rate
of return, what was the present
worth for the investment?
EXAMPLE (PW)
PW (12%) = - $16,000 - $4000(P|A
12,5)
+
$3000(P|F
12,5)
= -$16,000 - $4000(3.6048)
+ $3000(0.5674)
= -$28,717
EXAMPLE (AW)
AW (12%) = - $16,000(A|P 12,5)-$4000
+ $3000(A|F 12,5)
= -$16,000(.2774) - $4000
+ $3000(.1574)
= -7966.20/year
EXAMPLE (AW)
AW (12%) = PW(12%) (A|P 12,5)
AW (12%) = (-$28,717)(A|P 12,5)
= (-$28,717)(.2774)
= - $7966.10/year
EXAMPLE (FW)
FW (12%) = - $16,000(F|P 12,5)
- $ 4000(F|A 12,5)+$ 3000
= -$16,000(1.7613)
- $4000(6.3528) + $ 3000
= - $ 50608
EXAMPLE (FW)
FW (12%) = PW(12%) (F|P 12,5)
AW (12%) = (-$28,717)(1.7623)
= -$ 50608
INTERNAL RATE OF RETURN
METHOD
Internal Rate of Return (IRR) method
determines the interest rate that
yields a future worth (or present
worth or annual worth) of zero.
INTERNAL RATE OF RETURN
METHOD
Ajt : Net cash flow for investment j
in period t
ij* : Internal rate of return IRR
INTERNAL RATE OF RETURN
METHOD
INTERNAL RATE OF RETURN
METHOD
Rjt : Receipts
Cjt : Cost(Disbursments)
rt : reinvestment rate for positive cash flows
occuring in period t
i’ : the rate of return for negative cash flows
In IRR Method
rt = i’
EXAMPLE
An investment of $ 10 000 will be made in a project
that will produce an uniform annual revenue of
$ 5 310 for 5 years and then has a salvage value of
$ 2 000. Annual disbursment will be $ 3 000 for
operation and maintenance cost. What is IRR ?
EXAMPLE
(5310-3000) (P\A i,5) + 2000 (P\F i,5)- 10 000 = 0
2310 [(1+i)5-1]/[i(1+i)5]+2000 (1+i)-5-10000=0
IRR can be found by trial-and-error.
IRR = i = 10 %
2310(3.7908)+2000(0.6209) – 10000 = 0
EXTERNAL RATE OF RETURN
METHOD
External Rate of Return (ERR)
method determines the interest rate
that yields a future worth of zero,
explicitly assuming reinvestment of
recovered funds at the MARR.
EXTERNAL RATE OF RETURN
METHOD
Rjt : Receipts
Cjt : Cost(Disbursments)
rt : the reinvestment rate for positive cash flows
occuring in period t
rt = MARR
i’ : the external rate of return
(for negative cash flows)
EXAMPLE
An investment of $ 10 000 will be made in a project
that will produce an uniform annual revenue of
$ 5 310 for 5 years and then has a salvage value of
$ 2 000. Annual disbursment will be $ 3 000 for
operation and maintenance cost. MARR = 10%
What is ERR ?
EXAMPLE
(5310-3000) (F\A 10,5) + 2000 - 10 000 (F\P i’,5)= 0
2310 (6.1051)+2000 -10000 (1+i’) 5 =0
16102.781 = 10 000 (1+i’) 5
(1+i’) 5 = 1.6102781
ERR= i’ = 10 %
i’ = 10 %
EXAMPLE (IRR)
Multiple Roots :As an illustration of a cash flow profile
having multiple roots, consider the data given in Table.
The future worth of the cash flow series will be zero
using a 20, 40, or 50 % interest rate.
Cash Flow Profile
EOY
CF
0
- 1000
1
4100
2
- 5580
3
2520
EXAMPLE
FW1(20%)= -1000(1.2)3+ 4100 (1.2)2 -5580(1.2) +2520 = 0
FW1(40%)= -1000(1.4)3+ 4100 (1.4)2 -5580(1.4) +2520 = 0
FW1(50%)= -1000(1.5)3+ 4100 (1.5)2 -5580(1.5) +2520 = 0
By factoring the future worth of third-degree polynomial
FWi(i)= -1000(1.2-x) (1.4-x) (1.5-x) where x= (1+i)
EXAMPE
SAVINGS/INVESTMENT RATIO
METHOD
Savings/Investment Ratio(SIR) (or
Benefit/Cost) method determines the
ratio of the present worth of savings
to the present worth of the
investment.
SAVINGS/INVESTMENT RATIO
METHOD
SAVINGS/INVESTMENT RATIO
METHOD
EXAMPLE
Cash Flow Profile
EOY
CF
0
- 10000
1
2525
2
2525
3
2525
4
3840
5
3840
6
3840
Find SIR
EXAMPLE
=2525 (P\A) 15,3)
+3840(P\A) 15,3) (P\F) 15,3)
= 2525(2.2832)+3840(2.2832)(0.6575)
= 11 529.70
= - 10 000
= -10 000 + 11529.70 = 1529.70
EXAMPLE
Method I : SIR1 = Receipts / Costs
SIR 1 (15%) =
$11,529.70 = 1.15297
$10,000
Method II : SIR1 = Net cash flows / Costs
SIR1(15%) =$1529.70 / $10000 = 0.15297
PAYBACK PERIOD METHOD
Payback Period (PBP) method
determines how long at a zero
interest rate it will take to recover the
initial investment.
PAYBACK PERIOD METHOD
A number of variations of the payback or payout
method have been used by different organizations. The
payback period might be defined as the smallest value
of mj in the following formula,
Also, the payback period may be
noninteger, assuming cash flows are
uniformly distributed throughout the
year.
PAYBACK PERIOD METHOD
In the payback period method, the timing of
cash flows and the duration of the project are
ignored.
The
payback
period
method
is
recommended as a secondary method of
measuring investment worth. In particular, it
is suggested that the payback period method
be used in addition to a method based on the
time value of money.
EXAMPLE
Cash Flow Profile
EOY
CF
0
- 10000
1
2525
2
2525
3
2525
4
3840
5
3840
6
3840
Find Payback Period
PAYBACK PERIOD METHOD
2525+2525+2525=7575 < C10= 10000
2525+2525+2525+3840=11415 > C10= 10000
Consequently, m equals 4, indicating that 4
years are required to pay back the original
investment.
CAPITALIZED WORTH METHOD
Capitalized Worth (CW) method
determines the single sum at time
zero that is equivalent at i = MARR to
a cash flow pattern that continues
indefinitely.
CAPITALIZED WORTH
CAPITALIZED WORTH METHOD
A=Pi
P=A/i
P is a present value that will pay out equal
payments of size A indefinitely, if the
interest rate per period is i.
The present value P is termed the
capitalized worth of A, the size of each of
the perpetual payments.
EXAMPLE (4.12)
What deposit at t = 0 into a fund
paying 9.5 % annually is
required in order to pay out
$5000 each year forever?
P = A / i = 5000/ 0.095 =
52631.58
EXAMPLE (4.13)
Project ABC consists of the following
requirements:
1. A $50,000 first cost at t = 0.
2. A $5000 expense every year.
3. A $25,000 expense every third year
forever, with the first expense occurring at t =
3.
What is the capitalized cost of Project ABC
if i = 15% annually?
EXAMPLE (4.13)
CW1 = 50 000
CW2 = 5 000 / 0.15 = 33 333
i = [ ( 1+0.15)3-1]=0.5209
CW3 = 25000/0.5209 = 47996.16
EXAMPLE (4.13)
The capitalized cost of Project ABC is computed as follows:
1. Capitalized cost of $50,000 first cost
50000
=$
2. Capitalized cost of $5000 every year= $5000/0.15 = $
33333
3. Capitalized cost of $25,000 every third year
47996.16
TOTAL CAPITALIZED COST
=$
=$131329.16
CAPITAL RECOVERY FORMULA
Figure illustrates an investment of $P in an asset
having a life of n years and disposed of for a
salvage value of $F. The capital recovery cost is a
uniform annual amount defined as
CR = P (A\P i,n)- F(A\F i,n)
EXAMPLE
A stand-alone computer is purchased for P
= $82,000, has a service life of n = 7 years,
and a salvage value of F = $5000. At an
interest rate of 15%, the capital recovery
cost is
CR = 82000 (A\P 15,7)- 5000(A\F 15,7)
= 82000(0.2404) - 5000(0.0904)
= 19260.80/year