6.39
(a)
A
AE(12%) $40,000( / ,12%, 4) [$19,120 $1, 280( / ,12%, 4)]
AP AG
=− +−
6.40
1
AEC(10%) ($95, 000 $12, 000)( / ,10%,3) (0.1)$12,000
AP
=−+
6.41
• Capital recovery cost for both motors:
• Annual operating cost for both motors:
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(a) Savings per kWh:
(b)
CV
18.650kW
AE 0.75 $0.08 / kWh 1.2503
0.895
TT
= × × ×=
6.42
• New lighting system cost:
• Old lighting system cost:
6.43
1
AEC(15%) ($100,000 $15, 000)( / ,15%,7) (0.15)$15,000
$47,000
$69,680.63
AP
=−+
+
=
$69,680.63 $66,054.05 0.1195X
= +
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6.44
Given:
6%i=
interest compounded monthly, the effective annual interest
12
(1.005) 1 6.17%= − =
per year, effective semiannual interest
6
(1.005) 1 3.04%= − =
per semiannual
• Option 1: Buying a bond
• Option 2: Buying and holding a growth stock for 3 years
• Option 3: Receiving $150 interest per year for 3 years
6.45
• Equivalent annual cost:
• Processing cost per ton:
18 | Page
6.46
Let
X
be the number of machines per year
1
AEC(10%) $40,000
X
=
Life-Cycle Cost Analysis
6.47
• Capital recovery cost for both systems:
• Annual operating cost for both systems:
• Annual equivalent compressor replacement cost:
• Total annual cost for both systems:
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6.48
Assumption: jet fuel cost
=$2.10
/gallon
• System A : Equivalent annual fuel cost: A1 = ($2.10/gal)(40,000 gals/1,000
• System B : Equivalent annual fuel cost: A1 = ($2.10/gal)(32,000 gals/1,000
hours)(2,000 hours)
=$134,400
• Equivalent operating cost (including capital cost) per hour:
6.49
Since the required service period is 12 years and the future replacement cost
for each truck remains unchanged, we can easily find the equivalent annual
cost over a 12-year period by simply finding the annual equivalent cost of the
first replacement cycle for each truck.
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• Truck A: Four replacements are required
• Truck B: Three replacements are required
6.50
(a) Number of decision alternatives (required service period = 5 years):
Alternative Description
A1 Buy Machine A and use it for 4 years.
Then lease a machine for one year.
(b) With lease, the O&M costs will be paid by the leasing company:
• For A1:
21 | Page
• For A2:
2
2
PW(10%) $8,500 $1, 000( / ,10%,5)
$520( / ,10%,5) $280( / ,10%, 4)
$10,042
AEC(10%) $10, 042( / ,10%,5)
$2,649
PF
PA PF
AP
=−+
−−
= −
=
=
• For A3:
6.51
• Option 1:
• Option 2:
2
AEC(18%) ($0.05 $0.215)(180,000,000)
= +
6.52
Given: Required service period
=
indefinite, analysis period
=
indefinite
22 | Page
Plan A: Incremental investment strategy:
• Capital investment :
• Supporting equipment:
• Total equivalent annual worth:
A
AEC(10%) $49,576 $456 $40,056 $90,088= ++ =
Plan B: One time investment strategy:
• Capital investment:
• Total equivalent annual worth:
Minimum Cost Analysis
6.53
(a)
• Energy loss in kilowatt-hour:
• Material weight in pounds:
• Capital recovery cost:
24 | Page
(b) Minimum annual equivalent total cost:
(c)
6.54
We assume the friction factor is 0.02.
25 | Page
Short Case Studies
ST 6.1
We assume
10%i=
.
1
PW(10%) $52, 740( / ,10%,10)
PA=
ST 6.2
This case problem appears to be a trivial decision problem as one alternative
(laser blanking method) dominates the other (conventional method). A problem
of this nature (from an engineer’s point of view) involves more strategic
planning issues than comparing the accounting data. We will first calculate the
unit cost under each production method. Since all operating costs are already
given in dollars per part, we need to convert the capital expenditure into the
required capital recovery cost per unit.
• Conventional method:
26 | Page
Blanking Method
Description
Conventional
Laser
Steel cost/part
$14.98
$8.19
It appears that the window frame production by the laser blanking technique
would save about $8.76 for each part produced. If Ford decides to make the
ST 6.3
Given: annual energy requirement
145, 000, 000,000=
BTUs, 1 ton=2,204.6 lbs,
net proceeds from demolishing the old boiler unit
$1, 000=
(a) Annual fuel costs for each alternative:
• Alternative 1:
• Alternative 2:
$0.42
$0.40
$5.72
$23.49
27 | Page
(b) Unit cost per steam pound:
• Alternative 1: Assuming a zero salvage value of the investment
• Alternative 2:
(c) Select alternative 1.
ST 6.4
Assuming that the cost of your drainage pipe has experienced a 4% annual
inflation rate, I could estimate the cost of the pipe 20 years ago as follows.
You can view this number as the annual amount he expects to recover from his
investment considering the cost of money. With only a 20-year’s usage, he still
has 30 more years to go. So, the unrecovered investment at the current point is
28 | Page
Assumed interest rate
Claim cost
0%
$1,152.24
3%
$1,462.98
4%
$1,545.89
5%
$1,617.15