t#2 solution
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School
University of Windsor *
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Course
3130
Subject
Industrial Engineering
Date
Jan 9, 2024
Type
Pages
2
Uploaded by UltraSeaLionMaster403
NAME.____
=
SID:
GENG-3130
(01
&
02)
Engineering
Economics
Tutorial
Assignment
#2
Due:
12PM,
Sept.
22,
2023
(submission
through
Brightspace
only,
no
e-mail
submission)
Problem
#1:
How
much
should
you
invest
today
at
6%
interest
(compounded
monthly)
to
accumulate
$1,000,000
in
30
years?
Perform
calculations
using
the
discounting
formula.
Discounting
Formula:
P =
F/(1+i)V
Solution
F
=
1,000,000
nominal
rate
per
year
i
=
6%
monthly
rate
=
0.06/12
=
0.0051
=
0.5%
N
=
30
years
=
360
months
Using
the
formula:
P
=F/(1
+i)M
=
1,000,000/(1
+
0.005)3%
=
166,041.93
You
can
also
calculate
effective
annual
rate
and
use
‘year’
as
calculation
period.
Effective
annual
rate
=
(1+0.005)'2
-1
=
6.1678%
P
=F/(1
+i)M
=
1,000,000/(1
+
0.061678)%®
=
166,041.05
Problem
#2
(3
pts):
(a)
Using
straight-line
depreciation,
what
is
the
book
value
after
8
years
for
an
asset
costing
$500,000
that
has
a
salvage
value
of
$100,000
after 10
years?
What
is
the
depreciation
charge
in
the
8th
year?
(b)
Using
decline-balance
depreciation
with
d=8%,
what
is
the
book
value
after
5
years
for
an
asset
costing
$500,000?
What
is
the
depreciation
charge
in
the
6™
year?
(c)
What
is
the
depreciation
rate
using
declining-balance
for
an
asset
costing
$250,000
and
having
a
salvage
value
of
$50,000
after 10
years?
Solution
(a)
BV(8)
=
500,000
—
8[(500,000
—
100,000)/10]
=
500,000
-
8(40,000)
=
500,000
—
320,000
=
180,000
DC(8)
=
(500,000
—
100,000)/10
=
40,000.
(b)
BV(5)
=
500,000(1-0.08)5
=
500,000(0.92)5
=
329,540.76
DC(6)
=
BV(5)x0.08
=
329,540.76(0.08)=26,363.26
(c)
d =
1
—
(50,000/250,000)""°
=
0.14866
=
14.866%
TAs
on
duty
(marking):
Liu,
Yinting;
Tian,
Jingnan
NAME:
SID:
Problem
#3
(2
pts):
A
machine
costs
$500,000
and
has
a
planned
life
of
20
years,
with
a
salvage
value
of
$100,000
(at
the
end
of
20
years)
using
straight-line
depreciation.
What
depreciation
rate
will
result
in
the
same
book
values,
for
both
the
declining-balance
and
straight-line
methods
at
the
end
of
year
10?
P(1-d)"=P-n(P-S)N
500,000(1
—
d)'°
=
500,000
—
10(500,000
—
100,000)/20
(1
-
d)"°
=
300,000/500,000
=
0.6
1-d=0.95020
d=4.98%
Problem
#4
(3
pts):
Based
on
the
following
cash
flow
diagram,
compute
the
equivalent
value
of
these
cash
flows
at
n=5
(yr),
using
i=10%
(annually).
30000
F=?
20000
10000
T
|
|
1
|
o
1
2
3
5
6
L
1
1
1
1
!
o
1
2
3
4
5
6
-20000
Fs
=
10000*(1+0.10)°
+
20000*(1+0.10)°
-20000*(1+0.1)"
+(30000)/(1+0.1)
=16105.1
+
26620.0
-22000.0
+
27272.73
=47,997.83
TAs
on
duty
(marking):
Liu,
Yinting;
Tian,
Jingnan
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