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Chapter 6 Annual Equivalence Method
Identifying Cash Inflows and Outflows
6.1
PW(10%) $300 $100( / ,10%,1) $160( / ,10%, 2)
$130( / ,10%,6)
$105.54
AE(10%) $105.44( / ,10%,6)
$24.23
PF PF
PF
AP
=−+ −
++
=
=
=
6.2
6.3
6.4
AE(10%) [ $3, 000 $3, 000( / ,10%, 2)
PA
=−−
6.5
AE(12%) [ $8, 000 $4, 000( / ,12%, 2)
PF
=−−
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6.7
A
AE(15%) $3,500( / ,15%, 4) $600
=−+
AP
6.8
6.9
AE(12%) $4, 400( / ,12%,3) $12,500( / ,12%,3)
AP AF
=−+
6.10 Since the project has the same cash flow cycle during the project life, you just
can consider the first cycle.
6.11
∴ The amount of additional funds should be $226,399.
Capital (Recovery) Cost / Annual Equivalent Cost
6.12
3,880 (25,000 )( / ,10%,10) 0.1S AP S=−+
6.13 Given:
$65,000, $5, 000, 10I SN= = =
years,
9%i=
(a)
1
AE(9%) ($65,000 $5,000)( / ,9%,10)
AP= −
6.14
6.15
CR(12%) ($75, 000 $22,000)( / ,12%,5) (0.12)($22,000)
AP=−+
6.16 Given:
$350,000, $60,000, 5I S N= = =
years,
15%i=
6.17
n
Option 1 (Buy storage tank) Option 2 (Buy KS every year)
0
-$600-$13,760
1
0 -$3,784
2
3
4
6.18 Given
6%i=
compounded annually, N=20 years.
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Amount of water to produce per year:
50,000,000 365 56,007 acre-foot
325,851
×=
Cost per acre-foot:
6.20
(a)
AE(15%) $4,500( / ,15%, 4) $1,000
( $1,000)( / ,15%,2)( / ,15%, 4)
=−+
+−
AP
X PF AP
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(b)
6.21
• Option 1: Purchase-annual installment option:
• Option 2: Cash payment option:
6.22
The total investment consists of the sum of the initial equipment cost and the
installation cost, which is
$145,000.
Let
R
denote the break-even annual
revenue.
R
6.23
• Capital recovery cost:
Unit–Cost Profit Calculation
6.24
• Capital recovery cost
CR(12%) ($30,000 $18, 000)( / ,12%,2) $18,000(0.12)
$9, 260
AP=−+
=
•
6.25
6.26
• Capital cost:
CR(5%) ($10,000,000 $850,000)( / ,5%,10) (0.05)($850,000)
$1, 227, 467
AP=−+
=
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6.27
• Salvage value:
$11,000 −($2,879 +$1,776 +$1,545) =$4,800
• Capital recovery cost:
CR(7%) ($11, 000 $4,800)( / ,7%,3) $4,800(0.07)
$2,698.52
AP=−+
=
6.28
Let
T
denote the total operating hours in full load.
• Motor I (Expensive): Annual power cost:
180 (0.746) (0.05) $8.089
0.83
TT× × ×=
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•Motor II (Less expensive): Annual power cost:
180 (0.746) (0.05) $8.392
0.80
TT× × ×=
6.29
•Option 1: Purchase units from John Holland
Unit cost
=$25 −($35,000 / 20,000) −$3.5 =$19.75
•Option 2: Make units in house
6.30
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(a) Determine the unit profit of air sample test by the TEM (in-house).
•Sub-contract Option:
(b) Let X denote the break-even number of air samples per year.
6.31
•Option 1: Pay employee $0.56 per mile
•Option 2: Provide a car to employee:
6.30
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•Capital costs:
•Annual recharging cost:
•Total annual costs (including annual maintenance cost):
6.33
•Minimum operating hours:
•Annual worth of the generator at full load operation:
•Discounted payback period at full load of operation:
6.32
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n
Investment
Revenue
Maintenance
cost
Net
Cash flow
0
-$32,000
-$32,000
6.34
• Capital recovery cost:
CR(6%) ($170,000 $12,000)( / , 6%,12) (0.06)($12,000)
$19,565.77
AP=−+
=
• Annual operating costs:
6.35
Given: Investment cost
=$10
million, plant capacity
=300,000
lbs/hour, plant
operating hours
=4,000
hours per year, O&M cost
$4=
million per year, useful
life
15=
years, salvage value
=
$800,000, and MARR = 15%.
1
n
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6.36
Given: Investment
=$5
million, plant capacity
0.4=
acre-foot, useful plant life
6.37
•Annual total operating hours:
(0.70)(8, 760) 6,132=
hours per year
•Annual electricity generated:
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Let
X
denote the average number of round-trip passengers per year.
•Capital recovery costs:
•Annual fuel costs for round trips:
($1.10)(3, 280)(2)(3)(52) $1,125,696=
•Annual landing fees:
•Total equivalent annual costs:
AEC(15%) $2, 010,171 $225,000 $1,125,696
$78,000 $403,500 $75
$3, 400
X
X
= ++
++ +
=
Solving for
X
yields
Comparing Mutually Exclusive Alternatives by the AE Method
6.38