ParkEndff.qxd
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3:15 PM
Page 2
TABLE 3.4 Summary of Discrete Compounding Formulas with Discrete Payments
Flow Type
Factor
Notation
Formula
Compound
amount
(F/P, i, N)
Present
worth
(P/F, i, N)
Compound
amount
(F/A, i, N)
P
A
Y
M
E
N
T
Sinking
fund
(A/F, i, N)
A = Fc
Present
worth
(P/A, i, N)
P = Ac
G
R
A
D
I
E
N
T
S
E
R
I
E
S
Cash Flow
Diagram
F = P11 + i2N
S
I
N
G
L
E
E
Q
U
A
L
S
E
R
I
E
S
Excel Command
F
= FV1i, N, P,, 02
0
N
P = F11 + i2-N
F = Ac
11 + i2N - 1
i
= PV1i, N, F,, 02
d
F
= PV1i, N, A,, 02
F
0 1 2 3 N⫺1
= PMT1i, N, P, F, 02
i
d
11 + i2N - 1
11 + i2N - 1
i11 + i2N
d
= PV1i, N, A,, 02
N
AAA
AA
AAA
AA
1 2 3 N⫺1N
Capital
recovery
(A/P, i, N)
A = Pc
Linear
gradient
i11 + i2N
11 + i2N - 1
d
Present
worth
(P/G, i, N)
P = Gc
11 + i2N - iN - 1
Conversion factor
(A/G, i, N)
A = Gc
11 + i2N - iN - 1
Geometric
gradient
A1 c
i 11 + i2
2
N
i[11 + i2N - 1]
= PMT1i, N,, P2
d
1 23
P
d
1 - 11 + g2N11 + i2-N
i - g
Present
P = D
N
worth
A1 a
b1if i = g2
1
+ i
1P/A 1, g, i, N2
(N⫺2)G
2G
G
d
N⫺1N
A1(1⫹g)N⫺1
A3
A2
A1
123
P
N
ParkEndff.qxd
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Page 3
Summary of Project Analysis Methods
Analysis
Method
Payback period
PP
Discounted
payback period
Description
Single
Project
Evaluation
Revenue
Projects
PP 6 PP°
Select the
one with
shortest PP
A variation of payback period when
factors in the time value of money.
Management sets the benchmark PP •.
PP1i2 6 PP •
Select the
one with
shortest
PP(i)
An equivalent method which
translates a project’s cash flows into a
net present value
PW1i2 7 0
Select the
one with the
largest PW
Select the
one with the
least negative
PW
An equivalence method variation of
the PW: a project’s cash flows are
translated into a net future value
FW1i2 7 0
Select the
one with the
largest FW
Select the one
with the least
negative FW
An equivalence method variation of
the PW of perpetual or very long-lived
project that generates a constant
annual net cash flow
CE1i2 7 0
Select the
one with
the largest
CE
Select the one
with the least
negative CE
An equivalence method and variation
of the PW: a project’s cash flows are
translated into an annual equivalent sum
AE1i2 7 0
Select the
one with
the largest
AE
Select the one
with the least
negative AE
A relative percentage method which
measures the yield as a percentage of
investment over the life of a project:
The IRR must exceed the minimum
required rate of return (MARR).
IRR 7 MARR
An equivalence method to evaluate
public projects by finding the ratio of
the equivalent benefit over the
equivalent cost
BC1i2 7 1
PW(i)
Future worth
FW(i)
Capitalized
equivalent
CE(i)
Annual
equivalence
AE(i)
Internal rate of
return
IRR
Benefit–cost
ratio
BC(i)
Service
Projects
A method for determining when in a
project’s history it breaks even.
Management sets the benchmark PP°.
PP(i)
Present worth
Mutually Exclusive
Projects
Incremental analysis:
If IRR A2-A1 7 MARR, select the
higher cost investment project, A2.
Incremental analysis:
If BC1i2A2-A1 7 1, select the higher cost investment project, A2.
ParkEndff.qxd
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Page 4
Summary of Useful Excel’s Financial Functions (Part A)
Description
Excel Function
Example
Solution
SinglePayment
Find: F
Given: P
=FV1i%, N, 0, -P2
Find the future worth of
$500 in 5 years at 8%.
=FV18%, 5, 0, -5002
= $734.66
Cash
Flows
Find: P
Given: F
=PV1i%, N, 0, F2
Find the present worth of
$1,300 due in 10 years at
a 16% interest rate.
=PV116%, 10, 0, 13002
=1$294.692
Find: F
Given: A
=FV1i%, N, A2
Find the future worth of
a payment series of $200
per year for 12 years at 6%.
=FV16%, 12, -2002
= $3,373.99
Find: P
Given: A
=PV1i%, N, A2
Find the present worth of
a payment series of $900
per year for 5 years at 8%
interest rate.
=PV18%, 5, 9002
=1$3,593,442
What equal-annual-payment
series is required to repay
$25,000 in 5 years at 9%
interest rate?
=PMT19%, 5, -250002
= $6,427.31
What is the required annual
savings to accumulate
$50,000 in 3 years at 7%
interest rate?
=PMT17%, 3, 0, 500002
=1$15,552.582
EqualPaymentSeries
Find: A
Given: P
Find: A
Given: F
Measures
of
Investment
Worth
=PMT1i%, N, -P2
=PMT1i%, N, 0, F2
Find: NPW
Given: Cash
flow series
=NPV1i%, series2
Find: IRR
Given: Cash
flow series
=IRR1values, guess2
Find: AW
Given: Cash
flow series
=PMT 1i%, N,
-NPW2
Consider a project with the
following cash flow series
at 12% 1n = 0, - $200;
n = 1, $150, n = 2, $300,
n = 3, 2502?
1
A
Period
2
3
4
5
0
1
2
3
B
Cash
Flow
-200
150
300
250
= NPV112%, B3:B52 + B2
= $351.03
=IRR1B2:B5, 10%2
=89%
=PMT112%, 3,
-351.032
= $146.15
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Page 5
Summary of Useful Excel’s Financial Functions (Part B)
Description
Excel Function
=PMT1i%, N, P2
Loan
payment size
Loan
Analysis
Functions
Suppose you borrow
$10,000 at 9% interest
to be paid in 48 equal
monthly payments. Find
the loan payment size.
Solution
=PMT19%/12, 48, 100002
=1$248.452
Interest
payment
=IMPT1i%, n, N, P2
Find the portion of
interest payment for
the 10th payment.
=IPMT19%/12, 10, 48, 100002
=1$62.912
Principal
payment
=PPMT1i%, n, N, P2
Find the portion of
principal payment
for the 10th payment.
=PPMT19%/12, 10, 48, 100002
=1$185.942
Cumulative
interest
payment
=CUMIMPT1i%, N,
P, start_period,
end_period)
Find the total interest
payment over 48
months.
= CUMIMPT19%/12,
48, 10000, 1, 482
= $1,944.82
Interest rate
Depreciation
functions
Example
=RATE1N, A, P, F2
What nominal interest
rate is being paid on
the following financing
arrangement? Loan
amount:$10,000, loan
period: 60 months, and
monthly payment:
$207.58.
=RATE160, 207.58, -100002
= 0.7499%
APR = 0.7499% * 12 = 9%
Find the number of
months required to pay
=NPER112%/12, 200, -100002
off a loan of $10,000
=69.66 months
with 12% APR where
you can afford a monthly
payment of $200.
Number of
payments
=NPER1i%, A, P, F2
Straight-line
=SLN1cost, salvage,
life2
Cost = $100,000,
S = $20,000,
life = 5 years
=SLN1100000, 20000, 52
= $16,000
Declining
balance
=DB1cost, salvage,
life, period, month2
Find the depreciation
amount in period 3.
=DB1100000, 20000, 5, 3, 122
= $14,455
Double
declining
balance
=DDB1cost, salvage,
life, period, factor2
Find the depreciation
amount in period 3
with α = 150%,
=DDB1100000, 20000, 5, 3, 1.52
= $14,700
Declining
balance with
switching to
straight-line
=VDB1cost, salvage,
life, strat_period,
end_period, factor2
Find the depreciation
amount in period 3
with α = 150%, with
switching allowed.
=VDB1100000, 20000, 5, 3, 4, 1.52
= $10,290