# Solution Manual for Engineering Economics Financial Decision Making for Engineers, 6th Edition

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Engineering Economics Financial Decision Making for Engineers Canadian 6th Edition Fraser Solutions Manual
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CHAPTER 2
Solutions to Chapter-End Problems
A. Key Concepts
Simple Interest:
2.1
P = 3000
N = 6 months
i = 0.09 per year
= 0.09/12 per month, or 0.09/2 per six months
P + I = P + PiN = P(1 + iN)
= 3000[1 + (0.09/12)(6)] = 3135
or
= 3000[1 + (0.09/2)(1)] = 3135
The total amount due is $3135, which is $3000 for the principal amount
and $135 in interest.
2.2
I = 150
N = 3 months
i = 0.01 per month
P = I/(iN) = 150/[(0.01)(3)] = 5000
A principal amount of $5000 will yield $150 in interest at the end of 3
months when the interest rate is 1% per month.
2.3
P = 2000
N = 5 years
i = 0.12 per year
F = P(1+i)N = 2000(1+0.12)5 = 3524.68
The bank account will have a balance of $3525 at the end of 5 years.
2.4
(a) P = 21 000
i = 0.10 per year
N = 2 years
F = P(1+i)N = 21000(1+0.10)2 = 25 410
The balance at the end of 2 years will be $25 410.
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Chapter 2 – Time Value of Money
(b) P = 2 900
i = 0.12 per year = 0.01 per month
N = 2 years = 24 months
F = P(1+i)N = 2900(1+0.01)24 = 3682.23
The balance at the end of 24 months (2 years) will be $3682.23.
2.5
From: F = P(1 + i)N
P = F/(1 + i)N = 50 000/(1 + 0.01)20 = 40977.22
Greg should invest about $40 977.
2.6
F = P(1 + i)N
50 000 = 20 000(1 + i)20
(1+i)20 = 5/2
i = (5/2)1/20 โ 1 = 0.04688 = 4.688% per quarter = 18.75% per year
The investment in mutual fund would have to pay at least 18.75% nominal
interest, compounded quarterly.
Cash Flow Diagrams:
2.7
Cash flow diagram:
$6000
$2000
0
1
2
3
4
5
6
$900
2.8
Showing cash flow elements separately:
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Chapter 2 – Time Value of Money
$100
0
1
2
3
4
5
6
7
8
9
10
11
12
$300
$500
Showing net cash flow:
$100
0
1
2
3
4
5
6
7
8
9
10
11
12
$200
$500
2.9
Showing cash flow elements separately:
$10000
$10000
0
1
2
3
$15000
4
5
6
$15000
7
8
9
10
$15000
$20000
Showing net cash flow:
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12
$15000
Chapter 2 – Time Value of Money
$10000
$10000
$10000
$10000
$5000
0
1
3
2
4
6
5
7
8
9
$5000
$5000
10
11
12
$5000
$20000
2.10
The calculation of the net cash flow is summarized in the table below.
Time
0
1
2
3
4
5
6
7
8
9
10
11
12
Payment
20
Receipt
Net
โ20
30
33
16.3
39.9
43.9
28.3
53.1
58.5
44.3
70.7
77.8
65.6
30
33
36.3
39.9
43.9
48.3
53.1
58.5
64.3
70.7
77.8
85.6
20
20
20
20
Cash flow diagram:
$70.7
$53.1
$65.5
$58.5
$39.9 $43.9
$30
$77.8
$44.3
$33
$28.3
$16.3
0
1
2
3
4
5
6
7
8
9
10
$20
2.11
(a) functional loss
(b) use-related physical loss
(c) functional loss
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Chapter 2 – Time Value of Money
(d) time-related physical loss
(e) use-related physical loss
(f) use-related physical loss
(g) functional loss
(h) time-related physical loss
2.12
(a) market value
(b) salvage value
(c) scrap value
(d) market value to Liam, salvage value to Jacque
(e) book value
2.13
The book value of the company is $4.5 based on recent financial
statements. The market value is $7 million, assuming that the bid is real
and would actually be paid.
2.14
Since sewing machine technology does not change very quickly nor does
the required functionality, functional loss will probably not be a major
factor in the depreciation of this type of asset. Left unused, but cared for,
the machine will lose some value, and hence time-related loss may be
present to some extent. The greatest source of depreciation on a machine
will likely be use-related and due to wear and tear on the machine as it is
operated.
2.15
A switch will generally not suffer wear and tear due to use, and thus userelated physical loss is not likely to be a big factor. Nor will there likely be
a physical loss due to the passage of time. The primary reason for
depreciation will be functional loss – the price of a similar new unit will
likely have dropped due to development of new technology and
competition in the marketplace.
2.16
The depreciation is certainly not due to use related physical loss, or other
non-physical losses in functionality. The depreciation is a time-related
physical loss because it has not being used and maintained over time.
2.17
(a) BV(1) = 14 000 โ (14 000 โ 3000)/7 = $12 429
(b) BV(4) = 14 000 โ 4ร(14 000 โ 3000)/7 = $7714
(c) BV(7) = 3000
2.18
(a) BV(1) = 14 000(1 โ 0.2) = $11 200
(b) BV(4) = 14 000(1 โ 0.2)4 = $5 734
(c) BV(7) = 14 000(1 โ 0.2)7 = $2936
2.19
(a) d = 1 โ (3000/14 000)1/7 = 19.75%
(b) BV(4) = 14 000(1โ 0.1975)4 = $5806
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Spreadsheet used for chart:
Year
0
1
2
3
4
5
6
7
Straight Line
14000
12429
10857
9286
7714
6143
4571
3000
Declining Balance
14000
11200
8960
7168
5734
4588
3670
2936
14000
12000
Book value ($)
2.20
10000
Straight line
8000
6000
4000
Declining balance
2000
0
0
1
2
3
4
5
Year
10
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Chapter 2 – Time Value of Money
2.21
Spreadsheet used for chart:
Year
0
1
2
3
4
5
6
7
8
9
10
d = 5%
150000
142500
135375
128606
122176
116067
110264
104751
99513
94537
89811
d = 20%
150000
120000
96000
76800
61440
49152
39322
31457
25166
20133
16106
d = 30%
150000
105000
73500
51450
36015
25211
17647
12353
8647
6053
4237
Book value ($)
150000
5%
100000
50000
20%
30%
0
0
5
10
Year
B. Applications
2.22
I = 190.67
P = 550
N = 4 1/3 = 13/3 years
i = I/(PN) = 190.67/[550(13/3)] = 0.08
The simple interest rate is 8% per year.
2.23
F = P(1 + i)N
50 000 = 20 000(1 + 0.01)N
(1.01)N = 5/2
N = ln(5/2)/ln(1.01) = 92.09 quarters = 23.02 years
Greg would have to invest his money for about 23.02 years to reach his
target.
2.24
F = P(1 + i)N
= 20 000(1 + 0.01)20 = 24 403.80
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Greg would have accumulated about $24 404.
2.25
(a) P = 5000
i = 0.05 per six months
F = 8000
From: F = P(1 + i)N
N = ln(F/P)/ln(1 + i) = ln(8000/5000)/ln(1 + 0.05) = 9.633
The answer that we get is 9.633 (six-month) periods. But what does this
mean? It means that after 9 compounding periods, the account will not yet
have reached $8000. (You can verify yourself that the account will contain
$7757). Since compounding is done only every six months, we must, in
fact, wait 10 compounding periods, or 5 years, for the deposit to be worth
more than $8000. At that time, the account will hold $8144.
(b) P = 5000
r = 0.05 (for the full year)
F = 8000
i = r/m = 0.05/2 = 0.025 per six months
From: F = P(1 + i)N
N = ln(F/P)/ln(1 + i) = ln(8000/5000)/ln(1 + 0.025) = 19.03
We must wait 20 compounding periods, or 10 years, for the deposit to be
worth more than $8000.
2.26
P = 500
F = 708.31
i = 0.01 per month
From: F = P(1 + i)N
N = ln(F/P)/ln(1 + i) = ln(708.31/500)/ln(1 + 0.01) = 35.001
The deposit was made 35 months ago.
2.27
(a) P = 1000
i = 0.1
N = 20
F = P(1 + i)N = 1000(1+0.1)20 = 6727.50
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Chapter 2 – Time Value of Money
About $6728 could be withdrawn 20 years from now.
(b) F = PiN = 1000(0.1)(20) = 2000
Without compounding, the investment account would only accumulate
$2000 over 20 years.
2.28
Let P = X and F = 2X.
(a) By substituting F = 2X and P = X into the formula, F = P + I = P + PiN,
we get
2X = X + XiN = X(1 + iN)
2 = 1 + iN
iN = 1
N = 1/i = 1/0.11 = 9.0909
It will take 9.1 years.
(b) From F = P(1 + i)N, we get N = ln(F/P)/ln(1 + i). By substituting F = 2X
and P = X into this expression of N,
N = ln(2X/X)/ln(1 + 0.11) = ln(2)/ln(1.11) = 6.642
Since compounding is done every year, the amount will not double until
the 7th year.
(c) Given r = 0.11 per year, the effective interest rate is i = er โ 1 = 0.1163.
From F = P(1 + i)N, we get N = ln(F/P)/ln(1 + i). By substituting F = 2X and
P = X into this expression of N,
N = ln(2X/X)/ln(1 + 0.1163) = ln(2)/ln(1.1163) = 6.3013
Since interest is compounded continuously, the amount will double after
6.3 years.
2.29
(a) r = 0.25 and m = 2
i e = (1 + r/m)m โ 1 = (1 + 0.25/2)2 โ 1 = 0.26563
The effective rate is approximately 26.6%.
(b) r = 0.25 and m = 4
i e = (1 + r/m)m โ 1 = (1 + 0.25/4)2 โ 1 = 0.27443
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Chapter 2 – Time Value of Money
The effective rate is approximately 27.4%.
(c) i e = er โ 1 = e0.25 โ 1 = 0.28403
The effective rate is approximately 28.4%.
2.30
(a) i e = 0.15 and m = 12
From: i e = (1 + r/m)m โ 1
r = m[(1 + i e )1/m โ 1] = 12[(1 + 0.15)1/12 โ 1] = 0.1406
The nominal rate is 14.06%.
(b) i e = 0.15 and m = 365
From: i e = (1 + r/m)m โ 1
r = m[(1 + i e )1/m โ 1] = 365[(1 + 0.15)1/365 โ 1] = 0.13979
The nominal rate is 13.98%.
(c) For continuous compounding, we must solve for r in i e = er โ 1:
r = ln(1 + i e ) = ln(1 + 0.15) = 0.13976
The nominal rate is 13.98%.
2.31
F = P(1 + i)N
14 800 = 665(1 + i)64
i = 0.04967
The rate of return on this investment was 5%.
2.32
The present value of X is calculated as follows:
F = P(1 + i)N
3500 = X(1 + 0.075)5
X = 2437.96
The value of X in 10 years is then:
F = 2437.96(1 + 0.075/365)3650 = 4909.12
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The present value of X is $2438. In 10 years, it will be $4909.
2.33
r = 0.02 and m = 365
i e = (1 + r/m)m โ 1 = (1 + 0.02/365)365 โ 1 = 0.0202
The effective interest rate is about 2.02%.
2.34
Effective interest for continuous interest account:
i e = er โ 1 = e0.0599 โ 1 = 0.08318 = 6.173%
Effective interest for daily interest account:
i e = (1 + r/m)m โ 1 = (1 + 0.08/365)365 โ 1 = 0.08328 = 8.328%
No, your money will earn less with continuous compounding.
2.35
i e (weekly) = (1 + r/m)m โ 1 = (1 + 0.055/52)52 โ 1 = 0.0565 = 5.65%
i e (monthly) = (1 + r/m)m โ 1 = (1 + 0.07/12)12 โ 1 = 0.0723 = 7.23%
2.36
i e (Victory Visa) = (1 + r/m)m โ 1 = (1 + 0.26/365)365 โ 1 = 0.297 = 29.7%
i e (Magnificent Master Card) = (1 + 0.28/52)52 โ 1 = 0.322 = 32.2%
i e (Amazing Express) = (1 + 0.3/12)12 โ 1 = 0.345 = 34.5%
Victory Visa has the lowest effective interest rate, so based on interest
rate, Victory Visa seems to offer the best deal.
2.37
First, determine the effective interest rate that May used to get $2140.73
from $2000. Then, determine the nominal interest rate associated with the
effective interest:
F = P(1 + i e )N
2140.73 = 2000(1 + i e )1
i e = 0.070365
i e = er โ 1
0.070365 = er โ 1
r = 0.068
The correct effective interest rate is then:
i e = (1 + r/m)m โ 1 = (1 + 0.068/12)12 โ 1 = 0.07016
The correct value of $2000 a year from now is:
F = P(1 + i e )N = 2000(1 + 0.07016)1 = $2140.32
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Chapter 2 – Time Value of Money
2.38
The calculation of the net cash flow is summarized in the table below.
Time
0
1
2
3
4
5
6
7
8
9
10
11
12
Investment A
Payment
Receipt
2400
250
250
250
250
250
250
250
250
250
250
250
200
250
Net
โ2400
250
250
250
250
250
250
250
250
250
250
250
50
Investment B
Payment
Receipt
50
100
150
200
250
300
350
400
450
500
550
600
500
500
500
500
500
500
Cash flow diagram for investment A:
$250
$50
0
1
2
3
4
5
6
7
8
9
10
$2400
Cash flow diagram for investment B:
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12
Net
0
50
โ400
150
โ300
250
โ200
350
โ100
450
0
550
100
Chapter 2 – Time Value of Money
$550
$450
$350
$250
$150
$100
$50
0
1
2
3
4
5
6
7
8
9
10
11
12
$100
$200
$300
$400
Since the cash flow diagrams do not include the time factor (i.e., interest),
it is difficult to say which investment may be better by just looking at the
diagrams. However, one can observe that investment A offers uniform
cash inflows whereas B alternates between positive and negative cash
flows for the first 10 months. On the other hand, investment A requires
$2400 up front, so it may not be a preferred choice for someone who does
not have a lump sum of money now.
2.39
(a) The amount owed at the end of each year on a loan of $100 using 6%
interest rate:
Year
0
1
2
3
4
5
6
7
8
9
10
Simple Interest
Compound Interest
100
100.00
106
106.00
112
112.36
118
119.10
124
126.25
130
133.82
136
141.85
142
150.36
148
159.38
154
168.95
160
179.08
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Chapter 2 – Time Value of Money
180
Amount owed ($)
170
160
Compound interest
150
140
130
Simple interest
120
110
100
0
1
2
3
4
5
6
7
8
9
10
Year
(b) The amount owed at the end of each year on a loan of $100 using
18% interest rate:
Year
0
1
2
3
4
5
6
7
8
9
10
Simple Interest Compound Interest
100
100.00
118
118.00
136
139.24
154
164.30
172
193.88
190
228.78
208
269.96
226
318.55
244
375.89
262
443.55
280
523.38
Amount owed ($)
600
500
400
Compound interest
300
200
Simple interest
100
0
1
2
3
4
5
6
7
8
Year
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Chapter 2 – Time Value of Money
2.40
(a) From F = P(1 + i)N, we get N = ln(F/P)/ln(1 + i).
At i = 12%:
N = ln(1 000 000/0.01)/ln(1 + 0.12) = 162.54 years
At i = 18%:
N = ln(F/P)/ln(1 + i) = ln(1 000 000/0.01)/ln(1 + 0.18) = 111.29 years
(b) The growth in values of a penny as it becomes a million dollars:
Year
0
10
20
30
40
50
60
70
80
90
100
110
120
130
140
150
160
170
180
2.41
At 12%
0.01
0.03
0.10
0.30
0.93
2.89
8.98
27.88
86.58
268.92
835.22
2 594.07
8 056.80
25 023.21
77 718.28
241 381.18
749 693.30
2 328 433.58
7 231 761.26
At 18%
0.01
0.05
0.27
1.43
7.50
39.27
205.55
1 075.82
5 630.68
29 470.04
154 241.32
807 273.70
4 225 137.79
22 113 676.39
115 739 345.70
605 760 702.48
3 170 451 901.72
16 593 623 884.84
86 848 298 654.83
From the table and the charts below, we can see that $100 will double in
(a) 105 months (or 8.75 years) if interest is 8% compounded monthly
(b) 13 six-month periods (6.5 years) if interest is 11% per year,
compounded semi-annually
(c) 5.8 years if interest is 12% per year compounded continuously
Month
0
12
24
36
48
60
72
8%
100.00
108.30
117.29
127.02
137.57
148.98
161.35
11%
100.00
111.30
123.88
137.88
153.47
170.81
190.12
12%
100.00
112.75
127.12
143.33
161.61
182.21
205.44
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Chapter 2 – Time Value of Money
Deposit ($)
84
96
108
174.74
189.25
204.95
211.61
235.53
262.15
231.64
261.17
294.47
12%
300
280
260
240
220
200
180
160
140
120
100
11%
8%
0
20
40
60
80
100
120
Month
2.42
P(1 โ d)n = P โ n(P โ S)/N
245 000(1 โ d)20 = 245 000 โ 20(245 000 โ 10 000)/30
(1 โ d)20 = 88 333.33/245 000 = 0.3605
1 โ d = 0.9503
d = 4.97%
The two will be equal in 20 years with a depreciation rate of 4.97%.
2.43
780 000(1 โ d)20 = 60 000
(1 โ d)20 = 1/13
d = 1 โ (1/13)1/20 = 1 โ 0.8796 = 0.1204
A depreciation rate of about 12% will produce a book value in 20 years
equal to the salvage value of the press.
2.44
(a) BV(4) = 150 000 โ 4[(150 000 โ 25 000)/10]
= 150 000 โ 4(12 500) = 150 000 โ 50 000 = 100 000
DC(5) = (150 000 โ 25 000)/10 = 12 500
(b) BV(n) = 150 000(1 โ 0.2)4 = 150 00(0.8)4 = 61 440
DC(5) = BV(4)ร0.2 = 61 440(0.2) = 12 288
(c) d = 1 โ (25 000/150 000)1/10 = 0.1640 = 16.4%
C. More Challenging Problems
2.45
The present worth of each instalment:
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Instalment
1
2
3
4
5
6
7
8
9
10
F
100000
100000
100000
100000
100000
100000
100000
100000
100000
100000
Total
P
100000
90521
81941
74174
67143
60779
55018
49803
45082
40809
665270
Sample calculation for the third instalment, which is received at the end of
the second year:
P = F/(1 + r/m)N = 100 000/(1 + 0.10/12)24 = 81 941
The total present worth of the prize is $665 270, not $1 000 000.
2.46
The present worth of the lottery is $665 270. If you take $300 000 today,
that leaves a present worth of $365 270. The future worth of $365 270 in 5
years (60 months) is:
F = P(1 + r/m)N = 365 270(1 + 0.10/12)60 = 600 982
The payment in 5 years will be $600 982.
2.47
The first investment has an interest rate of 1% per month (compounded
monthly), the second 6% per 6 month period (compounding semiannually).
(a) Effective semi-annual interest rate for the first investment:
i e = (1 + i s )N โ 1 = (1 + 0.01)6 โ 1 = 0.06152 = 6.152%
Effective semi-annual interest rate for the second investment is 6% as
interest is already stated on that time period.
(b) Effective annual interest rate for the first investment:
i e = (1 + i s )N โ 1 = (1 + 0.01)12 โ 1 = 0.1268 = 12.68%
Effective annual interest rate for the second investment:
i e = (1 + i s )N โ 1 = (1 + 0.06)2 โ 1 = 0.1236 = 12.36%
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(c) The first investment is the preferred choice because it has the higher
effective interest rate, regardless of on what period the effective rate is
computed.
2.48
(a) i = 0.15/12 = 0.0125, or 1.25% per month
The effective annual rate is:
i e = (1 + i)m โ 1 = (1 + 0.0125)12 โ 1 = 0.1608 or 16.08%
(b) P = 50 000
N = 12
i = 0.15/12 = 0.0125, or 1.25% per month
F = P(1 + i)N = 50 000(1 + 0.0125)12 = 58 037.73
You will have $58 038 at the end of one year.
(c) Adam’s Fee = 2% of F = 0.02(58037.73) = 1160.75
Realized F = 58 037.73 โ 1160.75 = 56 876.97
The effective annual interest rate is:
F = P(1 + i)1
56 876.97 = 50 000(1 + i)
i = 56 876.97/50 000 โ 1 = 0.1375 or 13.75%
The effective interest rate of this investment is 13.75%.
2.49
Market equivalence does not apply as the cost of borrowing and lending is
not the same. Mathematical equivalence does not hold as neither 2% nor
4% is the rate of exchange between the $100 and the $110 one year from
now. Decisional equivalence holds as you are indifferent between the
$100 today and the $110 one year from now.
2.50
Decisional equivalence holds since June is indifferent between the two
options. Mathematical equivalence does not hold since neither 8%
compounded monthly (lending) or 8% compounded daily (borrowing) is
the rate of exchange representing the change in the house price ($110
000 now and $120 000 a year later is equivalent to the effective interest
rate of 9.09%). Market equivalence also does not hold since the cost of
borrowing and lending is not the same.
2.51
(a) The amount of the initial deposit, P, can be found from F = P(1 + i)N
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Chapter 2 – Time Value of Money
with F = $3000, N = 36, and i = 0.10/12.
(b) Having determined P = $2225, then we can figure out the size of the
deposit at the end of years 1, 2 and 3.
If you had not invested in the fixed interest rate investment, you would
have obtained interest rates of 8%, 10%, and 14% for each of the three
years. The table below shows how much the initial deposit would have
been worth at the end of each of the three years if you had been able to
reinvest each year at the new rate. Because of the surge in interest rates
in the third year, with 20/20 hindsight, you would have been better off (by
about $60) not to have locked in at 10% for three years.
Deposit amount ($)
Year
0
1
2
3
Fixed Interest Rate
Varying Interest Rate
2225
2225
2458
2410
2715
2662
3000
3060
3200
3000
2800
Fixed interest rate
2600
Varying interest rate
2400
2200
0
1
2
3
Year
2.52
Interest rate i likely has its origins in commonly available interest rates
present in Marleeโs financial activities such as investing or borrowing
money. Interest rate j can only be determined by having Marlee choose
between X and Y to determine at which interest rate Marlee is indifferent
between the choice. Interest rate k probably does not exist for Martlee,
since it is unlikely that she can borrow and lend money at the same
interest rate. If for some reason she could, then k=j.
Also, i could be either greater or less than j.
2.53
BV(0) = 250 000
BV(6) = 250 000ร(1 โ 0.3)6 = 29 412.25
The book values of the conveyor after 7, 8, 9, and 10 years are:
23
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Chapter 2 – Time Value of Money
BV(7) = 29 412.25 โ 29 412.25/4ร1 = 22 059.19
BV(8) = 29 412.25 โ 29 412.25/4ร2 = 14 706.13
BV(9) = 29 412.25 โ 29 412.25/4ร3 = 7353.07
BV(10) = 29 412.25 โ 29 412.25/4ร4 = 0
2.54
d = 1 โ (S/P)1/n = 1 โ (8300/12 500)1/2 = 1โ 0.81486 = 0.18514 = 18.514%
BV db (5) = 12 500 (1 โ 0.18514)5 = 4470.87
Enrique should expect to get about $4471 for his car three years from
now.
24
Copyright ยฉ 2016 Pearson Canada Inc.
Chapter 2 – Time Value of Money
Notes for Case-in-Point 2.1
1)
Close, if the appropriate depreciation method is being used.
2)
It makes sense because it is a new technology.
3)
Because the accounting department is likely using a specific depreciation
method that is not particularly accurate in this case. In particular, they may be
using a depreciation method required for tax purposes.
4)
Bill Fisher is probably not doing anything wrong, but it wouldnโt hurt to
check..
25
Copyright ยฉ 2016 Pearson Canada Inc.
Chapter 2 – Time Value of Money
Notes for Mini-Case 2.1
3)
Money will always be lost over the year. If money could be gained,
everybody would borrow as much money as possible to invest.
Solutions to All Additional Problems
Note: Solutions to odd-numbered problems are provided on the Student CD-ROM.
2S.1
You can assume that one month is the shortest interval of time for which the
quoted rental rates and salaries apply. Assembling the batteries will require 24
person-months, and the associated rental space. To maximize the interest you
receive from your savings, and minimize the interest you pay on your line of
credit, you should defer this expenditure till as late in the year as possible. So
you leave your money in the bank till December 1, then purchase the necessary
materials and rent the industrial space. Assume that salaries will be paid at the
end of the month.
As of December 1, you have $100 000(1.005)11 = $105 640 in the bank.
You need to spend $360 000 on materials and $240 000 to rent space. After
spending all you have in the bank, you therefore need to borrow an additional
$494 360 against your line of credit.
As of December 31, you owe $494 360(1.01) = $499 304 to the bank, and you
owe $240 000 in salaries. So after depositing the government cheque and paying
these debts, you have
1 200 000 โ 499 304 โ 240 000 = $460 696 in the bank.
This example illustrates one of the reasons why Just-in-Time (โJITโ) manufacture
has become popular in recent years: You want to minimize the time that capital is
tied up. An additional motivation for JIT would become evident if you were to
consider the cost of storing the finished batteries before delivery.
Be aware, however, that the JIT approach also carries risks. December is
typically a time when labour, space, and credit are in high demand so there is a
possibility that the resources you need will be unavailable or more expensive
than expected, and there will then be no time to recover. We will look at methods
for managing risk in Chapter 12.
2S.2
26
Copyright ยฉ 2016 Pearson Canada Inc.
Chapter 2 – Time Value of Money
We want to solve the equation
Future worth = Present worth (1+i)N, where the future worth is twice the present
worth.
So we have
2 = (1+i)N
Taking logarithms on both sides, we get
N = log(2) / log(1+i)
For small values of i, log (1+i) is approximately i (this can be deduced from the
Taylor series).
And log(2) = 0.69315. So, expressing i as a percentage rather than a fraction, we
have:
N = 69.3 / i
Since this is only an approximation, we will adjust 69.3 to an easily factored
integer, 72, thus obtaining
N = 72 / i
2S.3
Gita is paying 15% on her loan over a two-week period, so the effective annual
rate is
(1.1526 โ1) ร 100% = 3686%
The Grameen Bank was awarded the Nobel Peace Prize in 2006 for making
loans available to poor investors in Bangladesh at more reasonable rates.
2S.4
Five hundred years takes us beyond the scope of the tables in Appendix A, so
we employ the formula
P = F / (1+i)N
to find the present value of the potential loss.
In this case, we have P = $1 000 000 000 / (1.05)500 = $0.025, or two-and-a-half
cents. This implies that it is not worth going to any trouble to make the waste
repository safe for that length of time.
This is a rather troubling conclusion, because the example is not imaginary; the
U.S., for example, is currently trying to design a nuclear waste repository under
Yucca Mountain in Nevada that will be secure for ten thousand yearsโtwenty
27
Copyright ยฉ 2016 Pearson Canada Inc.
Chapter 2 – Time Value of Money
times as long as in our example. It is not clear how the engineers involved in the
project can rationally plan how to allocate their funds, since the tool we usually
use for that purposeโengineering economicsโgives answers that seem
irrational.
2S.5
There is no โrightโ answer to this question, which is intended for discussion in
class or in a seminar. Some of the arguments that might be advanced are as
follows:
One option is to say, โYou cannot play the numbers game with human lives.
Each life is unique and of inestimable value. Attempting to treat lives on the same
basis as dollars is both cold-blooded and ridiculous.โ
But this really wonโt do. Medical administrators, for example, do have a
responsibility to save lives, and they have limited resources to meet this
responsibility. If they are to apportion their resources rationally, they must be
prepared to compare the results of different strategies.
To support the point of view that future lives saved should be discounted by
some percentage in comparison with present lives, the following arguments
might be offered:
1.
Suppose we make the comparison fifty years in the future. If we spend our
resources on traffic police, we will have saved the lives of those who would have
died in accidents, and, because we spent the money that way, the world of fifty
years hence will also contain the descendants of those who would have died. So
the total number of live humans will be increased by more than fifty.
(This argument assumes that creating a new life is of the same value as
preserving an existing life. We do not usually accept that assumption; for
example, many governments may promote population control by limiting the
number of children born, but it would be unacceptable to control the population
by killing off the old and infirm. To give another example, if I am accused of
murder, it would not be an acceptable defence to argue that I have fathered two
children, and have thus made a greater contribution to society than a law-abiding
bachelor.)
2.
Just as I charge you interest on a loan because of the uncertainty of what
might happen between now and the due dateโyou could go bankrupt, I could
die, etc.โso we should discount future lives saved because we cannot anticipate
how the world will change before the saving is realized. For example, a cure for
cancer could be discovered ten years from now, and then all the money spent on
the anti-smoking campaign will have been wasted, when it could have been used
to save lives lost in highway accidents.
3.
We should not be counting numbers of lives, but years of human life. Thus
it is better to spend the money on preventing accidents, because these kill
people of all ages, while cancer and heart disease are mostly diseases of later
28
Copyright ยฉ 2016 Pearson Canada Inc.
Chapter 2 – Time Value of Money
life; so more years of human life are saved by the first strategy.
4.
The world population is growing. Thus, for humanity as a whole, losing a
fixed number of lives can more easily be born in the future than it can now. (This
is an extension of the argument that it is worse to kill a member of an
endangered species than of a species that is plentiful.)
29
Copyright ยฉ 2016 Pearson Canada Inc.
Engineering Economics
Chapter 2
Time Value of Money
Copyright ยฉ 2017 Pearson Canada Inc.
2-1
Engineering Economics, Sixth Edition
Outline
2.1 Introduction
2.2 Interest and Interest Rates
2.3 Compound and Simple Interest
2.3.1 Compound Interest
2.3.2 Simple Interest
2.4 Effective and Nominal Interest Rates
2.5 Continuous Compounding
2.6 Cash Flow Diagrams
2.7 Depreciation
2.7.1 Reasons for Depreciation
2.7.2 Value of an Asset
2.7.3 Straight-Line Depreciation
2.7.4 Declining Balance Depreciation
2.8 Equivalence
2.8.1 Mathematical Equivalence
2.8.2 Decisional Equivalence
2.8.3 Market Equivalence
Copyright ยฉ 2017 Pearson Canada Inc.
2-2
Engineering Economics, Sixth Edition
2.1 Introduction
โข Engineering decisions frequently involve tradeoffs
among costs and benefits occurring at different times
โ Typically, we invest in project today to gain future benefits
โข Chapter 2 discusses economic methods used to
compare benefits and costs occurring at different times
โข The key to making these comparisons is the use of
interest rates discussed in Sections 2.2 to 2.5
โข Section 2.6 introduces Cash Flow Diagrams
โข Section 2.7 explains Depreciation models
โข Section 2.8 discusses the equivalence of costs and
benefits that occur at different times
Copyright ยฉ 2017 Pearson Canada Inc.
2-3
Engineering Economics, Sixth Edition
2.2 Interest and Interest Rates
โข Interest (I) is compensation for giving up use of money
โ difference between the amount loaned and the amount repaid
โข An amount of money today, P, can be related to a future
amount, F, by the interest amount I, or interest rate i:
F = P + I = P + Pi = P(1 + i)
โข Right to P at beginning is exchanged for right to
F at end, where F = P(1+i)
โข i ๏ interest rate, P ๏ present worth of F
โข F ๏ future worth of P, base period ๏ interest period
Copyright ยฉ 2017 Pearson Canada Inc.
2-4
Engineering Economics, Sixth Edition
2.2 Interest and Interest Rates (contโd)
โข The dimension of an interest rate is (dollars/dollars)/time.
โข i.e. if $1 is lent at a 9% interest rate
โ then $0.09/year would be paid in interest per time period
โข period over which interest calculated is interest period.
CLOSE-UP 2.2 Interest Periods
Copyright ยฉ 2017 Pearson Canada Inc.
2-5
Engineering Economics, Sixth Edition
2.3 Compound and Simple Interest
2.3.1 Compound Interest
โข If amount P is lent for one period at interest rate, i
โ then amount repaid at the end of the period is F = P(1 + i).
โข If more than one period, interest is usually
compounded
โ (i.e. end of each period, interest is added to principal that existed at
the beginning of that period)
โข The interest accumulated is:
F = P(1 + i)N
IC = F โ P = P(1 + i)N โ P
Copyright ยฉ 2017 Pearson Canada Inc.
2-6
Engineering Economics, Sixth Edition
Table 2.1 Compound Interest Computations
Copyright ยฉ 2017 Pearson Canada Inc.
2-7
Engineering Economics, Sixth Edition
Example 2.2
โข With i = 10% per year, how much is owed on
a loan of $100 at the end of 3 years?
โข What is the compound interest amount?
F = P(1 + i)N = 100(1 + 0.10)3
= $133.10
IC = F โ P = $133.10 โ $100.00
= $ 33.10
The amount owed is $133.10 The interest owed
is $33.10 (See Table 2.2 for yearly accrual)
What is the amount owed at each year end?
Copyright ยฉ 2017 Pearson Canada Inc.
2-8
Engineering Economics, Sixth Edition
2.3 Compound and Simple Interest (contโd)
2.3.2 Simple Interest
Simple Interest โ interest without compounding
(interest is not added to principal at end of period)
IS = P i N
โข Compound and simple interest amounts equal if N = 1.
โข As N increases, difference between accumulated interest
amounts for the two methods increases exponentially
โข The conventional approach for computing interest is the
compound interest method
โข Simple interest is rarely used
Copyright ยฉ 2017 Pearson Canada Inc.
2-9
Engineering Economics, Sixth Edition
Figure 2.1 Compound and Simple
Interest at 24% Per Year for 20 Years
Copyright ยฉ 2017 Pearson Canada Inc.
2 – 10
Engineering Economics, Sixth Edition
2.4 Effective and Nominal Interest Rates
โข Interest rates stated for some period, usually a year
โข Computation based on shorter compounding sub-periods
โข In this section we consider the relation between:
โ The nominal interest rate stated for the full period.
โ The effective interest rate that results from the
compounding based on the subperiods.
โข Unless otherwise noted, rates are nominal annual rates
โข Suppose: r is nominal rate stated for a period (1 year)
consisting of m equal compounding periods (sub-periods)
โข If is = r/mโฆ then F = P(1 + iS)m = P(1 + ie)
Copyright ยฉ 2017 Pearson Canada Inc.
2 – 11
Engineering Economics, Sixth Edition
2.4 Effective and Nominal Interest (contโd)
โข Effective interest rate, ie, gives same future amount, F,
over the full period as when sub-period interest rate, iS, is
compounded over m sub-periods F = P(1 + iS)m = P(1 + ie)
EXAMPLE 2.6 Cardex Credit Card Co. charges a nominal 24
percent interest on overdue accounts, compounded daily.
What is the effective interest rate?
Since F = P(1 + iS)m = P(1 + ie), ie = (1 + iS)m โ 1
where is = r/m = 0.24/365 = 0.0006575
then ie = (1 + iS)m โ 1 = (1 + 0.0006575)365 – 1 = 0.271
โข The effective interest rate is 27.1%
Copyright ยฉ 2017 Pearson Canada Inc.
2 – 12
Engineering Economics, Sixth Edition
2.5 Continuous Compounding
โข Suppose that the nominal interest rate is 12% and
interest is compounded semi-annually
โข We compute the effective interest rate as follows:
where r = 0.12, m = 2
is = r/m = 0.12/2 = 0.06
ie = (1 + iS)m โ 1 = (1 + 0.06)2 – 1 = .1236 (12.36%)
What if interest were compounded monthly?
ie = (1 + iS)m โ 1 = (1 + 0.01)12 โ 1 = 0.1268 (12.68%)
โข Daily? ie = 0.127475 or about 12.75%
โข More than daily? โฆContinuous Compounding
Copyright ยฉ 2017 Pearson Canada Inc.
2 – 13
Engineering Economics, Sixth Edition
2.5 Continuous Compounding (contโd)
โข The effective interest rate under continuous
compounding is:
Ie = e r – 1
Copyright ยฉ 2017 Pearson Canada Inc.
2 – 14
Engineering Economics, Sixth Edition
2.5 Continuous Compounding (contโd)
โข To compute effective interest rate for nominal
interest rate of 12% by continuous compounding:
ie = er โ 1 = e0.12 โ 1 = 0.12750 = 12.75%
โข Continuous compounding makes sense in some
situations (i.e large cash flows), but not often used
โข Discrete compounding is the norm
Copyright ยฉ 2017 Pearson Canada Inc.
2 – 15
Engineering Economics, Sixth Edition
2.6 Cash Flow Diagrams
โข Cash flow diagram is a graphical summary of the timing
and magnitude of a set of cash flows
Copyright ยฉ 2017 Pearson Canada Inc.
2 – 16
Engineering Economics, Sixth Edition
Close Up 2.3 Beginning and Ending of Periods
Assumptions:
Cash flows occur at the ends of periods.
End of time period 1 = beginning of time period 2โฆ
Time 0 = โnowโ
Copyright ยฉ 2017 Pearson Canada Inc.
2 – 17
Engineering Economics, Sixth Edition
2.7 Depreciation
โข
โข
โข
โข
โข
Projects involve investment in assets (buildings,
equipmentโฆ) that are put to productive use
Assets lose value, or depreciate, over time
Depreciation taken into account when a firm states the
value of its assets in a Financial Statement (Chapter 6)
Also part of decision as to when to replace an aging
asset as described in Chapter 7
It also impacts taxation as shown in Chapter 8
Copyright ยฉ 2017 Pearson Canada Inc.
2 – 18
Engineering Economics, Sixth Edition
2.7.1 Reasons for Depreciation
โข
Assets depreciate for a variety of reasons:
1. Use related physical loss: usually measured in
units of production, kilometres driven, hours of use
2. Time related physical loss: usually measured in
units of time as an unused car will rust and lose
value over time
3. Functional loss: usually expressed in terms of
function lost including fashion, legislative (i.e.
pollution control, safety devices) and technical
Copyright ยฉ 2017 Pearson Canada Inc.
2 – 19
Engineering Economics, Sixth Edition
2.7.2 Value of an Asset
โข
โข
โข
Depreciation models are used to model
(estimate) value of an asset at any point in time.
Market Value: value of asset in the open market
Book value: value of an asset calculated from a
depreciation model for accounting purposes.
โ This value may be different from the market value.
โ may be several book values given for the same asset
(i.e. different for taxation vs. shareholder reports)
โข Salvage Value: either actual or estimated value
at end of its useful life (when sold)
โข Scrap Value: either actual or estimated value at
end of life (when broken up for material value)
Copyright ยฉ 2017 Pearson Canada Inc.
2 – 20
Engineering Economics, Sixth Edition
2.7.2 Value of an Asset (cont)
To state the book value of an asset a good model of
depreciation is desirable for the following reasons:
1. To make managerial decisions itโs important to know
the value of owned assets (i.e. collateral for a loan)
2. One needs an estimate of the value of assets for
planning purposes (i.e. keep an asset or replace)
3. Tax legislation requires company tax to be paid on
profits. Rules are legislated on how to calculate
income and expenses that includes depreciation
Copyright ยฉ 2017 Pearson Canada Inc.
2 – 21
Engineering Economics, Sixth Edition
2.7.2 Value of an Asset (cont)
Copyright ยฉ 2017 Pearson Canada Inc.
2 – 22
Engineering Economics, Sixth Edition
2.7.3 Straight-Line Depreciation
โข
Straight line depreciation (SLD) assumes rate of
loss of assetโs value is constant over its useful
life.
P = purchase price
S = salvage value at the end of N periods.
N = useful life of asset
โข
โข
โข
Advantage: easy to calculate and understand.
Disadvantage: most assets do not depreciate at
a constant rate.
Hence, market values often differ from book
values when SLD is used.
Copyright ยฉ 2017 Pearson Canada Inc.
2 – 23
Engineering Economics, Sixth Edition
Figure 2.7 Book Value Under Straight-Line Depreciation
Copyright ยฉ 2017 Pearson Canada Inc.
2 – 24
Engineering Economics, Sixth Edition
2.7.3 Straight-Line Depreciation (contโd)
โข
Depreciation in period
n using SLD:
P โS
Dsl (n) =
N
โข
Book Value of the
asset at the end of
period n:
๏ฃซP โS ๏ฃถ
BVsl ( n ) = P โ n ๏ฃฌ
๏ฃท
N
๏ฃญ
๏ฃธ
โข
Accumulated
Depreciation at the
end of period n:
๏ฃซP โS ๏ฃถ
P โ BV sl ( n ) = n ๏ฃฌ
๏ฃท
๏ฃญ N ๏ฃธ
Copyright ยฉ 2017 Pearson Canada Inc.
2 – 25
Engineering Economics, Sixth Edition
2.7.4 Declining-Balance Depreciation
โข
This DBD method models loss in value of an
asset in a period as a constant proportion of
the assetโs current value.
โข
Initial Book Value:
BVdb(0) = P
โข
Book Value at the end
of period n using DBD:
BVdb (n ) = P (1 โ d )n
โข
Depreciation in period n
using DBD:
Copyright ยฉ 2017 Pearson Canada Inc.
Ddb(n) = BVdb(nโ1) d
2 – 26
Engineering Economics, Sixth Edition
Figure 2.8 Book Value Under Declining-Balance
Depreciation
Copyright ยฉ 2017 Pearson Canada Inc.
2 – 27
Engineering Economics, Sixth Edition
Example 2.11
Sherbrooke Data Services has purchased a new mass
storage system for $250 000. It is expected to last six years,
with a $10 000 salvage value. Using both the straight-line
and declining-balance methods, determine the following:
(a) The depreciation charge in year 1
(b) The depreciation charge in year 6
(c) The book value at the end of year 4
(d) The accumulated depreciation at the end of year 4
โข Ideal application for spreadsheet (see Table 2.3)
Copyright ยฉ 2017 Pearson Canada Inc.
2 – 28
Engineering Economics, Sixth Edition
Table 2.3 Spreadsheet for Example 2.11
Copyright ยฉ 2017 Pearson Canada Inc.
2 – 29
Engineering Economics, Sixth Edition
2.8 Equivalence
โข Engineering Economics utilises โtime value of moneyโ
to compare certain values at different points in time.
โข Three concepts of equivalence are distinguished
underlying comparisons of costs/benefits at different
times:
1. Mathematical Equivalence
2. Decisional Equivalence
3. Market Equivalence
Copyright ยฉ 2017 Pearson Canada Inc.
2 – 30
Engineering Economics, Sixth Edition
2.8 Equivalence (contโd)
โข Mathematical Equivalence: Decision-makers exchange
P dollars now for F dollars N periods from now using rate i
and the mathematical relationship:
F = P(1 + i)N
โข Decisional Equivalence: Decision-maker is indifferent as
to P dollars now or F dollars N periods from now.
โ We infer decision-makerโs implied interest rate
โข Market Equivalence: Decision-makers exchange
different cash flows in a market at zero cost.
โ In a financial market, individuals/companies are
lending and borrowing money.
โ i.e. buying a car and owing $15 000; a lender provides
the $15 000 now for $500/month over 36months.
Copyright ยฉ 2017 Pearson Canada Inc.
2 – 31
Engineering Economics, Sixth Edition
2.8 Equivalence (contโd)
โข For the remainder of this text, we assume:
1. market equivalence holds
2. decisional equivalence can be expressed in
monetary terms
โข If these two assumptions are reasonably valid,
then mathematical equivalence can be used
โข Accurate model of costs/benefits relationship
Copyright ยฉ 2017 Pearson Canada Inc.
2 – 32
Engineering Economics Financial Decision Making for Engineers Canadian 6th Edition Fraser Solutions Manual
Full Download: http://testbanklive.com/download/engineering-economics-financial-decision-making-for-engineers-canadian-6th-edition-fraser-solutions-manual/
Engineering Economics, Sixth Edition
Summary
โข Notion of Interest and Interest Rates
โข Compound and Simple Interest
โข Effective and Nominal Interest
โข Continuous Compounding
โข Representing Cash Flows by Diagrams
โข
Depreciation and Depreciation Accounting
โข
โข
โข
โข
Reasons for Depreciation
Value of an Asset
Straight Line Depreciation
Declining Balance Depreciation
โข Mathematical, Decisional, Market Equivalence
Copyright ยฉ 2017 Pearson Canada Inc.
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