Chapter 3: Interest Rate and Economic Equivalence
Types of Interest
3.1
• Simple interest:
3.2
• Simple interest:
3.3
• Option 1: Compound interest with 8%:
3.4
End of Year
Principal
Repayment
Interest
payment
Remaining
Balance
0
$15,000.00
Equivalence Concept
3.5
$22,000( / ,5%,5) $22,000(0.7835) $17,237.58P PF= = =
Single Payments (Use of F/P or P/F Factors)
3.9
$250,000( / ,6%,10) $447,712F FP= =
3.10
(a)
$5,000( / ,7%,5) $7,013F FP= =
1
$10,379.50
2
3
3.11
$300,000( / ,8%,10) $138,958P P F= =
3.12
(a)
$25,500( / ,12%,8) $10,299P PF= =
3.13
(a)
3.14
3.15
2 (1 0.06)
N
F PP
= = +
3.16
3.17
(a)
$18( / , ,49) $190,500
F Pi
=
Uneven Payment Series
3.18
$1,000 +$1,000
1.1 +$1,500
1.13=$1,210
1.12+X
1.14
X=$2,981
3.19
3.20
$2,000( / ,6%,10) $2,500( / ,6%,8) $3,000( / ,6%,6)
$11,822
F FP FP FP= ++
=
3.21
3.22
$9,000( / ,8%,2) $6,000( / ,8%,5) $3,000( / ,8%,7) $13,550P PF PF PF= + +=
Equal Payment Series
3.23
(a) With deposits made at the end of each year
3.24
$10,000( / ,6%,20) $367,856F F A= =
3.25
(a)
$8,000( / ,11.75%,5) $50,571F FA= =
3.26
(a)
$45,000( / ,8%,11) $2,703A AF= =
3.27
3.28
3.29
$15,000 ( / ,11%,5)
$2,408.57
AF A
A
=
=
3.30
3.31
(a)
$15,000( / ,3.5%,6) $2,815.02A AP= =
3.32
(a) The capital recovery factor
),,/( NiPA
for
N
6%
7%
(b) The equal payment series present–worth factor
)85,,/( iAP
for
3.33
• Equal annual payment:
$50,000( / ,12%,3) $20,817.45A A P= =
3.34
$15,000( / ,9%,10) $2,337.30A AP= =
3.35
(a)
$1,000( / ,7.2%,8) $5,925.29P PA= =
3.36
3.37
(a)
$9,600,000 $15,100,000( / ,6%,1)
$17,100,000( / ,6%,2)… $19,250,000( / ,6%,5)
P PF
PF PF
= +
++
Linear Gradient Series
3.38
$20,000( / ,6%,5) $5,000( / ,6%, 5)
$20,000( / ,6%,5) $5,000( / ,6%,5)( / , 6%, 5)
$165,833
F FA FG
FA A
G FA
= +
= +
=
3.39
3.40
3.41
$15, 000 $1, 000( / , 8%, 12)
$10,404.25
A AG= −
=
3.42
3.43 Using the geometric gradient series present worth factor, we can establish
the equivalence between the loan amount $150,000 and the balloon payment
series as
3.44
3.45
(a)
076,372,21$)7%,12%,10,/(000,000,6$ 1=−= APP
(b) Note that the oil price increases at the annual rate of 5% while the oil
production decreases at the annual rate of 10%. Therefore, the annual
revenue can be expressed as follows:
3.46
20
1
20
1
1
20
1
(1 )
(2,000,000) (1.06) (1.06)
1.06
(2,00
0,000/1.06) ( )
1.06
$396,226,415
n
n
n
n n
n
n
n
P A i
n
n
−
=
− −
=
=
= +
=
=
=
∑
∑
∑
3.47
(a) The withdrawal series would be
Period
Withdrawal
Amount
11
$15,000
$15,000
12
$15,000(1.08)
$16,200
13
$15,000(1.08)(1.08)
$17,496
14
$15,000(1.08)(1.08)(1.08)
$18,896
15
$15,000(1.08)(1.08)(1.08)(1.08)
$20,407
(b)
10 1
$15,000( / ,8%,6%,5) $73,476P PA= =
$73,476 ( / ,6%,10)
$5,574.47
AF A
A
=
=
Various Interest Factor Relationships
3.48
(a)
(P/F,8%,67) =(P/F,8%,50)(P/F,8%,17) =0.0058
(P/F,8%,67) =(1+0.08)
−67
=0.0058
3.49
(a)
N
N
N
N
i
i
i
i
ii
NiAFiNiPF
)1(
11)1(
1
1)1(
)1(
1),,/(),,/(
+=
+−+=
+
−+
=+
+=
(d)
1)1(
)1(
)1(
1
)1(
)1(1)1(
)1(
)],,/(1[
),,/(
−+
+
=
+
−
+
+
=
−+
+
−
=
N
N
NN
NN
N
i
ii
ii
i
i
i
ii
NiFP
i
NiPA
Equivalence Calculations
3.50
3.51
(1.08) $200 $200( / ,8%,1) $120( / ,8%,2) $120( / ,8%,3)
$300( / ,8%,4)
$373.92
P PF PF PF
PF
P
+= + +
+
=
3.52
3.53
3.54
96.0$
)12%,10,/(20$)5%,10,/(20$
=
−= APGPP
$20
0
1
2
3
4
5
6
7
8
9
10
11
12
$20
Establish economic equivalent at
N=8
:
3.56
The original cash flow series is
$40
$60
$80
N A
N
N A
N
0 0 6 $900
1 $800 7 $920
3.57
13.297$
)4872.9()1436.2(277.092,4$
)7%,10,/()8%,10,/(2)3%,10,/(200$)8%,10,/(300$
=
+=
+=+
C
CC
AFCPFCAFAF
3.58
3.59
Computing equivalence at
N=5
$3,000( / ,9%,5) $3,000( / ,9%,5) $29,623.08X FA PA=+=
3.63 (a)
3.64 (b)
Solving for an Unknown Interest Rate of Unknown Interest Periods
3.66
5
1/5
2 (1 )
21
14.87%
PP i
i
i
= +
= +
=
$1,000,000 $2,000( / ,6%, )
(1 0.06) 1
500 0.06
31 (1 0.06)
log31 log1.06
58.93 59years
N
N
FA N
N
N
=
+−
=
= +
=
= ≈
3.70
3.71
Assuming that annual renewal fees are paid at the beginning of each year,
(a)
3.72
Assuming that premiums paid at the end of each year, the maximum amount
to invest in the prevention program is
Short Case Studies
ST 3.1
(a)
ST 3.2
Establish the following equivalence equation:
ST 3.3
(a)
Contract
$5,600,000 $7,178,000( / ,6%,1)
P P F
= +