Matakuliah
Tahun
: D0762 – Ekonomi Teknik
: 2009
Present Worth and Annual Worth
Course Outline 6
Exercises
• Equal live PW Analysis,
A traveling saleswoman expect to purchase a used car this year.
She has collected or estimated the following information : first cost is
$10,000; trade in value will $500 after 4 years; annual maintenance
and insurance cost are $1500; and additional annual income due to
ability to travel is $5000. Will the saleswoman be able to make a rate
of return of 20% per year on her purchase?
Solution :
Compute the PW value of the investment at i= 20%
PW = -10,000 -1,500(P/A,20%,4) + 5000(P/A,20%,4) +
500(P/F,20%,4)
= $698
No, She will not make 20% since PW is less than zero.
(if the PW value had been greater than zero, the rate of return would 2
have exceded 20%)
Exercises
Compare the alternatives shown below on the basis of a present worth analysis. Use an
interest rate of 1% per month
Alternative Y Alternative Z
First Cost $
70.000
90.000
Monthly operating cost $ 1.200
1.400
Savage Value, $
7.000
10.000
Life, years
3
6
Alternative Y
Alternative Z
0
1
2
4
3
10.000
7.000
7.000
5
6
0
1
14.400
70.000
2
3
4
5
6
16.800
70.000
90.000
PY = - 70.000 - 70.000(P/F,12%,3) +
7.000(P/F,12%,3) +7.000(P/F,12%,6) (1.200x12)(P/A,12%,6) = ?
PZ = - 90.000 + 10.000(P/F,12%,6) (1.400x12)(P/A,12%,6) = ?
3
Exercises
• Example 5-5
How much should one set aside to pay $50 per year for
maintenance on a gravesite if interest is assumed to be 4% ? For
perpetual maintenance, the principal sum must remain undiminished
after making the annual disbursement.
Annual disbursement A
Solution
Interest rate i
Capitalized cost P =
= 50/0.04 = $1250
One should set aside $1250
4
Exercises
Example 5.6
• A city plans a pipeline to transport water from a distant watershed
area to the city. The pipeline will cost $8000 million and have an
expected life of seventy years. The city anticipates it will need to
keep the water line in service indefinitely. Compute the capitalized
cost equation .
A
P
Solution
i
We have the capitalized cost equation
$8million
$8million
$8million
…….
70years
70years
$8million
$8million
70years
n=∞
Capitalized Cost P
5
Exercises
•
The $8 million disbursement at the end of 70 years may be resolved into an
equivalent A
$8million
n= 70
A = F(A/F,i,n) = $8 million (A/F,7%,70)
= $8million (0,0062) = $4960
A
Each 70 year period is identical to this one.
Capitalized cost P
= $8million + A/i = 8million +
= $8,071,000
4960
0,07
6
Exercises
•
Example 6.7
Determination of AW
A local pizza shop has just purchased a fleet of five electric-powered mini
vehicles for delivery in an urban area. The initial cost was $6400 per
vehicle, and their expected life and salvage values are 5 years and $300,
respectively. The combined insurance, maintenance, recharge, and
lubrication cost are expected to be $650 the first year and to increase by
$50 per year thereafter. Delivery service will generate an estimated extra
$1200 per year. If a return of 10% per year is required, use the AW method
to determine if the purchase should have been made.
$1,500
$1,200
$650
$23,000
$700
$750
$800
$850
7
Exercises
• Example 6.7 Solution.
Steps 1-3 of the salvage sinking fund method :
A1 = annual cost of fleet purchase
= -5(4600)(A/P,10%,5) + 5(300)(A/F,10%,5)
= $-5822
Steps 4 the annual disbursement and income series can be
combined into an annual net income series that conveniently follows
a decreasing gradient with a base amount of $550($1200-650). The
equivalent annual income A2 is
A2 = 550-50(A/G,10%,5) = $460
The Total AW equals the algebraic sum of the vehicle cost and
income AW values
AW = -5822 + 460 = $-5362
Since AW < 0; a return of less than 10% per year is expected, the 8
Exercises
•
Mutually Exclusive Alternative with Equal Life Project
Size
Cost
Life
Salvage
Efficiency
Energy Cost
Operating Hours
Standard
Motor
Premium
Efficient Motor
25 HP
$13,000
20 Years
$0
89.5%
$0.07/kWh
3,120 hrs/yr.
25 HP
$15,600
20 Years
$0
93%
$0.07/kWh
3,120 hrs/yr.
(a) At i= 13%, determine the operating cost per kWh for each motor.
(b) At what operating hours are they equivalent?
9
Exercises
(a) Operating cost per kWh per unit
•
Determine total input power
•
Conventional motor
input power = 18.650 kW/ 0.895 = 20.838kW
•
PE motor:
input power = 18.650 kW/ 0.930 = 20.054kW
• Determine total kWh per year with 3120 hours of operation
•
Conventional motor:
3120 hrs/yr (20.838 kW) = 65,018 kWh/yr
•
PE motor:
3120 hrs/yr (20.054 kW) = 62,568 kWh/yr
•
Determine annual energy costs at $0.07/kwh:
Conventional motor:
$0.07/kwh  65,018 kwh/yr = $4,551/yr
PE motor:
$0.07/kwh  62,568 kwh/yr = $4,380/yr
10
Solution
(b) break-even
Operating
Hours = 6,742
Exercises
11
Exercises
•
Example 6-5
A firm is considering which of two devises to install to reduce costs in a
particular situation. Both devices cost $1000 and have useful lives of five
years with no salvage value. Device A can be expected to result in $300
savings annually. Device B will provide cost savings of $400 the first year
but will decline $50 annually, making the second year savings $350, the
third year savings $300 and so forth. With interest at 7%, which device
should the firm purchase?
Solution
Device A : AWA
= $300
Device B : AWA
= 400 – 50(A/G,7%,5) = 400-50(1,85)
= $306.75
To maximize equivalent uniform annual benefit (EUAB) select the larger
AW, which is Device B
12
Exercises
• Example 6-9
In the Construction of the aqueduct to expand the water supply of a
city, there a two alternatives for a particular portion of the aqueduct.
Either a tunnel can be constructed through a mountain, or a pipeline
can be laid to go around the mountain. If there is a permanent need
for the aqueduct, should the tunnel or the pipeline be selected for
this particular portion of the aqueduct? Assume a 6% interest rate
Tunnel through mountain
Initial cost
Maintenance
Useful life
Salvage value
Pipeline around mountain
$5,5 million
$5 million
0
0
Permanent
50years
0
0
13
Exercises
• Example 6-9 – Solution
Tunnel : For the tunnel, with its permanent life, we want
(A/P,6%,∞). For an infinite life, the capital recovery is
simply interest on the invested capital. So (A/P,6%,∞) = i
AW = P.i = = $5,5 million (0,06) = $330,000
Pipeline
AW
= $5 million (A/P,6%,50)
= $5 million (0,0634)
= $317,000
For fixed output , minimize Equivalent Uniform Annual
Cost, Select the pipeline
14