Solution manual engineering economic analysis 9th edition ch04

Chapter 4: More Interest Formulas
4-1
(a)
100 100 100 100
100100100 100

0

1

2

3

4
R

R = $100(F/A, 10%, 4) = $100(4.641)
= $464.10

(b)
$15

$50
$0

0

1

$50

2

$10 0
0

3

4

S

S = 50 (P/G, 10%, 4)
= 218.90

= 50 (4.378)

(c)
90
30

T

T

T

60

$90

T

T

= 30 (A/G, 10%, 5)
= 54.30

120

T

= 30 (1.810)


4-2
(a)
$100
$100

$100
$0

$0

B

B

= $100 (P/F, 10%, 1) + $100 (P/F, 10%, 3) + $100 (P/F, 10%, 5)

= $100 (0.9091 + 0.7513 + 0.6209)
= $228.13

(b)
$200 $200 $200
$200

i=?
$63
4

$634

= $200 (P/A, i%, 4)

(P/A, i%, 4)

= $634/$200 = 3.17

From compound interest tables, i = 10%.
(c)
$10

$10

$10

$10

V

V

= $10 (F/A, 10%, 5) - $10
= $10 (6.105) - $10
= $51.05


(d)

2x

x

3x

4x

$50
0

$500
$500

= x (P/A, 10%, 4) + x (P/G, 10%, 4)
= x (3.170 + 4.378)

x

= $500/7.548

= $66.24

4-3
(a)

$25

$50

$75

$0

$50
0

C

= $25 (P/G, 10%, 4)
= $25 (4.378)
= $109.45

(b)
A = $140

i=?
$50
0

$500

= $140 (P/A, i%, 6)

(P/A, i%, 6)

= $500/$140 = 3.571


Performing linear interpolation:
(P/A, i%, 6) i
3.784
15%
3.498
18%
i

= 15% + (18% - 15%) ((3.487 – 3.571)/(3.784 – 3.498)
= 17.24%

(c)
$50
$0

$75

$10
0

$25

F

P

F

= $25 (P/G, 10%, 5) (F/P, 10%, 5)
= $25 (6.862) (1.611)
= $276.37

(d)

$0

P

A

4-4
(a)

$40

A

$12
$80 0

A

A

A

= $40 (P/G, 10%, 4) (F/P, 10%, 1) (A/P, 10%, 4)
= $40 (4.378) (1.10) (0.3155)
= $60.78


$10
0

$75

$50
$25

W

W

= $25 (P/A, 10%, 4) + $25 (P/G, 10%, 4)
= $25 (3.170 + 4.378)
= $188.70

(b)
$15

$10 0
0

$0

$0 $50

x

= $100 (P/G, 10%, 4) (P/F, 10%, 1)
= $100 (4.378) (0.9091)
= $398.00

x

(c)
$30
0

$20
0

$10
0

Y

Y

(d)

= $300 (P/A, 10%, 3) - $100 (P/G, 10%, 3)

= $300 (2.487 – 2.329)
= $513.20


$10
0

$50

$10
0

Z

Z

= $100 (P/A, 10%, 3) - $50 (P/F, 10%, 2)
= $100 (2.487) - $50 (0.8264)
= $207.38

4-5

$10
0

$15
0

$20
0

$25
0

P

P = $100 + $150 (P/A, 10%, 3) + $50 (P/G, 10%, 3)
= $100 + $150 (2.487) + $50 (2.329)
= $589.50

4-6

$30
0

$40
0

$50
0

$40
0

$30
0

x

x

= $300 (P/A, 10%, 5) + $100 (P/G, 10%, 3) + $100 (P/F, 10%, 4)
= $300 (3.791) + $100 (2.329) + $100 (0.6830)
= $1,438.50


4-7
$80
$60 $70

$50
$0

P

P = $10 (P/G, 15%, 5) + $40 (P/A, 15%, 4)(P/F, 15%, 1)
= $10 (5.775) + $40 (2.855) (0.8696)
= $157.06

4-8

B

$800
$800

$800

O
B

B

1.5B

Receipts (upward) at time O:
PW

= B + $800 (P/A, 12%, 3)

= B + $1,921.6

Expenditures (downward) at time O:
PW= B (P/A, 12%, 2) + 1.5B (P/F, 12%, 3)
Equating:
B + $1,921.6

= 2.757B

B = $1,921.6/2.757
= $1,093.70

4-9
F
= A (F/A, 10%, n)
$35.95
= 1 (F/A, 10%, n)
(F/A, 10%, n)
= 35.95

= 2.757B


From the 10% interest table, n = 16.
4-10
P
$1,000

= A (P/A, 3.5%, n)
= $50 (P/A, 3.5%, n)

(P/A, 3.5%, n)

= 20

From the 3.5% interest table, n = 35.

4-11
$100 $100 $100

F

F

P’

J

J

J

= $100 (F/A, 10%, 3) = $100 (3.310)= $331

P’ = $331 (F/P, 10%, 2) = $331 (1.210)= $400.51
J

= $400.51 (A/P, 10%, 3)

= $400.51 (0.4021)

= $161.05

Alternate Solution:
One may observe that J is equivalent to the future worth of $100 after five interest periods,
or:
J

= $100 (F/P, 10%, 5) = $100 (1.611) = $161.10


4-12
$20
0

$10
0

$30
0

P

P’

C

C

C

P = $100 (P/G, 10%, 4) = $100 (4.378)= $437.80
P’ = $437.80 (F/P, 10%, 5)

= $437.80 (1.611)

= $705.30

C = $705.30 (A/P, 10%, 3)

= $705.30 (0.4021)

= $283.60

4-13

G

2G

3G

4G

5G

6G

0

$500 $500
P

Present Worth P of the two $500 amounts:
P = $500 (P/F, 12%, 2) + $500 (P/F, 12%, 1)
= $500 (0.7972) + $500 (0.7118)
= $754.50
Also:
P
$754.50

= G (P/G, 12%, 7)
= G (P/G, 12%, 7)
= G (11.644)

G

= $754.50/11.644
= $64.80

4-14

$10
0

0

D
D

D

$20
0

$30
0

D
P

Present Worth of gradient series:
P = $100 (P/G, 10%, 4) = $100 (4.378)= $437.80
D = $437.80 (A/F, 10%, 4)

= $4.7.80 (0.2155)

= $94.35

4-15
$30
0

$20
0

$10
0

$20
0

$10
0

E

$20
0

$30
0

E

P

P = $200 + $100 (P/A, 10%, 3) + $100 (P/G, 10%, 3) + $300 (F/P, 10%, 3) + $200 (F/P,
10%, 2) + $100 (F/P, 10%, 1)

= $200 + $100 (2.487) + $100 (2.329) + $300 (1.331) + $200 (1.210) + $100 (1.100)
= $1,432.90
E = $1,432.90 (A/P, 10%, 2)

4-16

= $1,432.90 (0.5762) = $825.64


$10
0

$20
0

$30
0

$40
0

P

4B

3B

2B

B

P = $100 (P/A, 10%, 4) + $100 (P/G, 10%, 4)
= $100 (3.170 + 4.378)
= $754.80
Also:
P = 4B (P/A, 10%, 4) – B (P/G, 10%, 4)
Thus,

4B (3.170) – B (4.378) = $754.80
B = $754.80/8.30 = $90.94

4-17
P = $1,250 (P/A, 10%, 8) - $250 (P/G, 10%, 8) + $3,000 - $250 (P/F, 10%, 8)
= $1,250 (5.335) - $250 (16.029) + $3,000 - $250 (0.4665)
= $5,545

4-18
Cash flow number 1:
P01 = A (P/A, 12%, 4)
Cash flow number 2:
P02 = $150 (P/A, 12%, 5) + $150 (P/G, 12%, 5)
Since P01 = P02,
A (3.037) = $150 (3.605) + $150 (6.397)


A

= (540.75 + 959.55)/3.037
= $494

4-19
F=?
F

n = 180 months

i = 0.50% /month

A = $20.00

= A (F/A, 0.50%, 180)

Since the ½% interest table does not contain n = 180, the problem must be split into
workable components. On way would be:

A = $20

n = 90

n = 90
F’

F

F

= $20 (F/A, ½%, 90) + $20 (F/A, ½%, 90)(F/P, ½%, 90)
= $5,817

Alternate Solution

Perform linear interpolation between n = 120 and n = 240:
F

= $20 ((F/A, ½%, 120) – (F/A, ½%, 240))/2
= $6,259

Note the inaccuracy of this solution.


4-20

Dec. 1

Nov. 1

March 1

A = $30

F
F’

Amount on Nov 1:
F’ = $30 (F/A, ½%, 9)

= $30 (9.812) = $275.46

Amount on Dec 1:
F

= $275.46 (F/P, ½%, 1)

= $275.46 (1.005)

= 276.84

4-21
B
B

B

B

B

B

F

The solution may follow the general approach of the end-of-year derivation in the book.
(1) F

= B (1 + i)n + …. + B (1 + i)1

Divide equation (1) by (1 + i):
(2) F (1 + i)-1

= B (1 + i)n-1 + B (1 + i)n-2 + … + B

Subtract equation (2) from equation (1):
(1) – (2)

F – F (1 + i)-1 = B [(1 + i)n – 1]

Multiply both sides by (1 + i):
F (1 + i) – F

= B [(1 + i)n+1 – (1 + i)]


So the equation is:
= B[(1 + i)n+1 – (1 + i)]/i

F

Applied to the numerical values:
F

= 100/0.08 [(1 + 0.08)7 – (1.08)]
= $792.28

B = $200
n = 15

i = 7%

4-22

…………..

F
F’

F

= $200 (F/A, i%, n)
= $5,025.80

F’ = F (F/P, i%, n)
= $5,377.61

= $200 (F/A, 7%, 15) = $200 (25.129)
= $5,025.80 (F/P, 7%, 1)

= $5,025.80 (1.07)

4-23
F

= $2,000 (F/A, 8%, 10) (F/P, 8%, 5)
= $2,000 (14.487) (1.469)
= $42,560

4-24
A = $300

i = 5.25%

P = A (P/A, 5.25%, 10)

= A [(1 + i)n – 1]/[i(1 + i)n]
= $300 [(1.0525)10 – 1]/[0.0525 (1.0525)10]
= $300 (7.62884)
= $2,289

P=?

n = 10 years


4-25
P = $10,000

i = 12%

F = $30,000

n=4

$10,000 (F/P, 12%, 4) + A (F/A, 12%, 4) = $30,000
$10,000 (1.574) + A (4.779)
A = $2,984

= $30,000

4-26
Let X = toll per vehicle. Then:
A = 20,000,000 X

i = 10%

20,000,000 X (F/A, 10%, 3)
20,000,000 X (3.31)
X

F = $25,000,000

n=3

= $25,000,000
= $25,000,000
= $0.38 per vehicle

4-27
From compound interest tables, using linear interpolation:
(P/A, i%,
i
10)
7.360
6%
7.024
7%
(P/A, 6.5%, 10)

= ½ (7.360 – 7.024) + 7.024
= 7.192

Exact computed value:
(P/A, 6.5%, 10)

= 7.189

Why do the values differ? Since the compound interest factor is non-linear, linear
interpolation will not produce an exact solution.


4-28
A = $4,000

A = $600

A=?
x

To have sufficient money to pay the four $4,000 disbursements,
x

= $4,000 (P/A, 5%, 4) = $4,000 (3.546)
= $14,184

This $14,184 must be accumulated by the two series of deposits.
The four $600 deposits will accumulate by x (17th birthday):
F

= $600 (F/A, 5%, 4) (F/P, 5%, 10)
= $600 (4.310) (1.629)
= $4,212.59

Thus, the annual deposits between 8 and 17 must accumulate a future sum:
= $14,184 - $4,212.59

= $9,971.41
The series of ten deposits must be:
A

= $9,971.11 (A/F, 5%, 10)
= $792.73

= $9,971.11 (0.0745)

4-29
P = A (P/A, 1.5%, n)
$525
= $15 (P/A, 1.5%, n)
(P/A, 1.5%, n)

= 35

From the 1.5% interest table, n = 50 months.


4-30
P=1

F=2

i = 1%

n=?

$2 = $1 (F/P, 1%, n)

(F/P, 1%, n)
=2
From the 1%, table:
n

= 70 months

4-31

A = $10

………….

$156

P = $156
$156

n=?

i = 1.5%

A = $10

= $10 (P/A, 1.5%, n)

(P/A, 1.5%, n)

= $156/$10
= 15.6

From the 1.5% interest table, n is between 17 and 18. Therefore, it takes 18 months to
repay the loan.

4-32
A = $500 (A/P, 1%, 16) = $500 (0.0679)
= $33.95
4-33
This problem may be solved in several ways. Below are two of them:
Alternative 1:
$5000
= $1,000 (P/A, 8%, 4) + x (P/F, 8%, 5)
= $1,000 (3.312) + x (0.6806)
= $3,312 + x (0.6806)
x

= ($5,000 - $3,312)/0.6806


= $2,480.16
Alternative 2:
P = $1,000 (P/A, 8%, 4)
= $1,000 (3.312)
= $3,312
($5,000 - $3,312) (F/P, 8%, 5)

= $2,479.67

4-34
A = P (A/P, 8%, 6)

= $3,000 (0.2163)
= $648.90
The first three payments were $648.90 each.

A = $648.90

P = $3,000

A’ = ?

P’ = Balance
Due after
3rd payment

Balance due after 3rd payment equals the Present Worth of the originally planned last three
payments of $648.90.
P’ = $648.90 (P/A, 8%, 3)
= $1,672.22

= $648.90 (2.577)

Last three payments:
A’ = $1,672.22 (A/P, 7%, 3)
= $637.28

= $1,672.22 (0.3811)


4-35

$15

A = $10

…….

n=?
$15
0

($150 - $15)

= $10 (P/A, 1.5%, n)

(P/A, 1.5%, n)

= $135/$10

= 13.5

From the 1.5% interest table we see that n is between 15 and 16. This indicates that there
will be 15 payments of $10 plus a last payment of a sum less than $10.
Compute how much of the purchase price will be paid by the fifteen $10 payments:
P = $10 (P/A, 1.5%, 15) = $10 (13.343)
= $133.43
Remaining unpaid portion of the purchase price:
= $150 - $15 - $133.43
= $1.57
16th payment

= $1.57 (F/P, 1.5%, 16)
= $1.99

4-36
A A A

$12,000

A A

Final
Payment

A = $12,000 (A/P, 4%, 5)
= $12,000 (0.2246)
= $2,695.20
The final payment is the present worth of the three unpaid payments.
Final Payment

= $2,695.20 + $2,695.20 (P/A, 4%, 2)
= $2,695.20 + $2,695.20 (1.886)
= $7,778.35


4-37
A=?

$3,00
0

Pay off loan

Compute monthly payment:
$3,000
A

= A + A (P/A, 1%, 11)
= A + A (10.368)
= 11.368 A
= $3,000/11.368
= $263.90

Car will cost new buyer:
= $1,000 + 263.90 + 263.90 (P/A, 1%, 5)
= $1263.90 + 263.90 (4.853)
= $2,544.61

4-38
(a)
A=?

i = 8%

P = $120,000

P = $150,000 - $30,000 = $120,000
A

= P (A/P, i%, n)
= $120,000 (A/P, 8%, 15)

= $120,000 (0.11683)
= $14,019.55

RY
RY

= Remaining Balance in any year, Y
= A (P/A, i%, n – Y)

R7

= $14,019.55 (P/A, 8%, 8)
= $14,019.55 (5.747)
= $80,570.35

(b) The quantities in Table 4-38 below are computed as follows:

n = 15 years


Column 1 shows the number of interest periods.
Column 2 shows the equal annual amount as computed in part (a) above.
The amount $14,019.55 is the total payment which includes the principal and interest
portions for each of the 15 years. To compute the interest portion for year one, we must
first multiply the interest rate in decimal by the remaining balance:
Interest Portion
TABLE 4-38:
YEAR
0
1

2
3
4
5
6
7*
8
9
10
11
12
13
14
15

= (0.08) ($120,000)

= $9,600

SEPARATION OF INTEREST AND PRINCIPAL
ANNUAL
PAYMENT

INTEREST
PORTION

PRINCIPAL
PORTION

$14,019.55

$14,019.55
$14,019.55
$14,019.55
$14,019.55
$14,019.55
$14,019.55
$14,019.55
$14,019.55
$14,019.55
$14,019.55
$14,019.55
$14,019.55
$14,019.55
$14,019.55

$9,600
$9,246.44
$8,864.59
$8,452.19
$8,006.80
$7,525.78
$7,006.28
$6,445.22
$5,839.27
$5,184.85
$4,478.07
$3,714.76
$2,890.37
$2,000.04
$1,038.48

$4,419.55
$4,773.11
$5,154.96
$5,567.36
$6,012.75
$6,493.77
$7,013.27
$7,574.33
$8,180.28
$8,834.70
$9,541.48
$10,304.79
$11,129.18
$12,019.51
$12,981.00

REMAINING
BALANCE
$120,000.00
$115,580.45
$110,807.34
$105,652.38
$100,085.02
$94,072.27
$87,578.50
$80,565.23
$72,990.90
$64,810.62
$55,975.92

$46,434.44
$36,129.65
$25,000.47
$12,981.00
0

Subtracting the interest portion of $9,600 from the total payment of $14,019.55 gives the
principal portion to be $4,419.55, and subtracting it from the principal balance of the loan
at the end of the previous year (y) results in the remaining balance after the first
payment is made in year 1 (y1), of $115,580.45. This completes the year 1 row. The
other row quantities are computed in the same fashion. The interest portion for row two,
year 2 is:
(0.08) ($115,580.45) = $9,246.44
*NOTE: Interest is computed on the remaining balance at the end of the preceding year
and not on the original principal of the loan amount. The rest of the calculations proceed
as before. Also, note that in year 7, the remaining balance as shown on Table 4-38 is
approximately equal to the value calculated in (a) using a formula except for round off
error.


4-39
Determine the required present worth of the escrow account on January 1, 1998:
A = $8,000

i = 5.75%

PW = ?

n = 3 years

PW= A (P/A, i%, n)
= $8,000 + $8,000 (P/A, 5.75%, 3)
= $8,000 + $8,000 [(1 + i)n – 1]/[i(1 + i)n]
= $8,000 + $8,000 [(1.0575)3 – 1]/[0.0575(1.0575)3]
= $29,483.00
It is necessary to have $29,483 at the end of 1997 in order to provide $8,000 at the end of
1998, 1999, 2000, and 2001. It is now necessary to determine what yearly deposits should
have been over the period 1981–1997 to build a fund of $29,483.
A=?

i = 5.75%

A = F (A/F, i%, n)

F = $29,483

n = 18 years

= $29,483 (A/F, 5.75%, 18)

= $29,483 (i)/[(1 + i)n – 1]
= $29,483 (0.575)/[(1.0575)18 – 1]
= $29,483 (0.03313)
= $977
4-40
Amortization schedule for a $4,500 loan at 6%
Paid monthly for 24 months
P = $4,500
Pmt. #
1

2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18

i = 6%/12 mo = 1/2% per month
Amt. Owed
BOP
4,500.00
4,323.06
4,145.24
3,966.52
3,786.91
3,606.41
3,425.00
3,242.69
3,059.46

2,875.32
2,690.25
2,504.26
2,317.35
2,129.49
1,940.70
1,750.96
1,560.28
1,368.64

Int. Owed
(this pmt.)
22.50
21.62
20.73
19.83
18.93
18.03
17.13
16.21
15.30
14.38
13.45
12.52
11.59
10.65
9.70
8.75
7.80
6.84

Total Owed
(EOP)
4,522.50
4,344.68
4,165.97
3,986.35
3,805.84
3,624.44
3,442.13
3,258.90
3,074.76
2,889.69
2,703.70
2,516.79
2,328.93
2,140.14
1,950.40
1,759.72
1,568.08
1,375.48

Principal
(This pmt)
176.94
177.82
178.71
179.61
180.51
181.41

182.32
183.23
184.14
185.06
185.99
186.92
187.85
188.79
189.74
190.69
191.64
192.60

Monthly
Pmt.
199.44
199.44
199.44
199.44
199.44
199.44
199.44
199.44
199.44
199.44
199.44
199.44
199.44
199.44
199.44

199.44
199.44
199.44


19
20
21
22
23
24

1,176.04
982.48
787.96
592.46
395.98
198.52

TOTALS

5.88
4.91
3.94
2.96
1.98
0.99

1,181.92
987.40

791.90
595.42
397.96
199.51

286.63

193.56
194.53
195.50
196.48
197.46
198.45

199.44
199.44
199.44
199.44
199.44
199.44

4499.93

B12 = $4,500.00 (principal amount)
B13 = B12 - E12 (amount owed BOP- principal in this payment)
Column C = amount owed BOP * 0.005
Column D = Column B + Column C (principal + interest)
Column E = Column F - Column C (payment - interest owed)
Column F = Uniform Monthly Payment (from formula for A/P)

4-41
Amortization schedule for a $4,500 loan at 6%
Paid monthly for 24 months
P = $4,500
Pmt. #
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22

i = 6%/12 mo = 1/2% per month
Amt. Owed

BOP
4,500.00
4,323.06
4,145.24
3,966.52
3,786.91
3,606.41
3,425.00
3,242.69
2,758.90
2,573.25
2,306.12
2,118.21
1,929.36
1,739.57
1,548.83
1,357.13
1,164.48
970.86
776.27
580.71
384.18
186.66

Int. Owed
(this pmt.)
22.50
21.62
20.73
19.83

18.93
18.03
17.13
16.21
13.79
12.87
11.53
10.59
9.65
8.70
7.74
6.79
5.82
4.85
3.88
2.90
1.92
0.93

Total Owed
(EOP)
4,522.50
4,344.68
4,165.97
3,986.35
3,805.84
3,624.44
3,442.13
3,258.90
2,772.69

2,586.12
2,317.65
2,128.80
1,939.01
1,748.27
1,556.57
1,363.92
1,170.30
975.71
780.15
583.61
386.10
187.59

Principal
(This pmt)
176.94
177.82
178.71
179.61
180.51
181.41
182.32
483.79
185.65
267.13
187.91
188.85
189.79
190.74

191.70
192.65
193.62
194.59
195.56
196.54
197.52
186.66

Monthly
Pmt.
199.44
199.44
199.44
199.44
199.44
199.44
199.44
500.00
199.44
280.00
199.44
199.44
199.44
199.44
199.44
199.44
199.44
199.44
199.44

199.44
199.44
187.59


23
24

0.00
0.00

TOTALS

0.00
0.00

0.00
0.00

256.95

0.00
0.00

0.00
0.00

4500.00

B12 = $4,500.00 (principal amount)

B13 = B12 - E12 (amount owed BOP- principal in this payment)
Column C = amount owed BOP * 0.005
Column D = Column B + Column C (principal + interest)
Column E = Column F - Column C (payment - interest owed)
Column F = Uniform Monthly Payment (from formula for A/P)
Payment 22 is the final payment. Payment amount = $187.59

4-42
Interest Rate per Month = 0.07/12

= 0.00583/month

Interest Rate per Day

= 0.000192/day

= 0.07/365

Payment = P[i(1 + i)n]/[(1 + i)n – 1]
= $80,000 [0.00583 (1.00583)12]/[(1.00583)12 – 1]
= $532.03
Principal in 1st payment = $532.03 – $80,000 (0.00583)
= $65.63
Loan Principal at beginning of month 2 = $80,000 - $65.63
= $79.934.37
Interest for 33 days

= Pin = $79,934 (33) (0.000192)

Principal in 2nd payment = $532.03 – 506.46

4-43
(a) F16
F10

= $10,000 (1 + 0.055/4)16
= $12,442.11
= $12,442.11 (1 + 0.065/4)24
= $18,319.24

(b) $18,319.24

= (1 + i)10 ($10,000)

(1 + i)10 = $18,319.24/$10,000= 1.8319
10 ln (1 + i)

= ln (1.8319)

= $25.57

= $506.46


ln (1 + i)

= (ln (1.8319))/10
= 0.0605

(1 + i)

= 1.0624
i

= 0.0624

= 6.24%

Alternative Solution
$18,319.24

= $10,000 (F/P, i, 10)

(F/P, i, 10)
= 1.832
Performing interpolation:
(F/P, i%, 10)
1.791
1.967
i

i
6%
7%

= 6% + [(1.832 – 1.791)/(1.967 – 1.791)]

= 6.24%

4-44

Correct equation is (2).
$50 (P/A, i%, 5) + $10 (P/G, i%, 5) + $50 (P/G, i%, 5)
100

=1