Matakuliah : D0762 – Ekonomi TeknikTahun : 2009

Present Worth and Annual Worth Course Outline 6

Exercises

2

• 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)

= $698No, 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 have exceded 20%)

Exercises

3

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 3 4 5 6

70.000

14.400

7.000 7.000

70.000

0 2 4 5 6

90.000

10.000

1

16.800

3

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) = ?

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.Solution Capitalized cost P =

= 50/0.04 = $1250One should set aside $1250

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Annual disbursement A

Interest rate i

Exercises

• 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 .Solution We have the capitalized cost equation

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Example 5.6

i

A P

$8million

Capitalized Cost P

$8million

70years

$8million

70years

$8million

70years

$8million

n=∞…….

Exercises• The $8 million disbursement at the end of 70 years may be

resolved into an equivalent A

Each 70 year period is identical to this one. Capitalized cost P = $8million + A/i = 8million +

= $8,071,000

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$8million

A

n= 70A = F(A/F,i,n) = $8 million (A/F,7%,70)

= $8million (0,0062) = $4960

49600,07

Exercises

• Example 6.7 Determination of AWA 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.

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$1,500

$850$800

$750$650

$700

$1,200

$23,000

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) = $460The Total AW equals the algebraic sum of the vehicle cost and income AW values AW = -5822 + 460 = $-5362Since AW < 0; a return of less than 10% per year is expected, the purchase in not justified. 8

Exercises

• Mutually Exclusive Alternative with Equal Life Project

9

Standard Premium Motor Efficient Motor

25 HP 25 HP$13,000 $15,60020 Years 20 Years$0 $089.5% 93%$0.07/kWh $0.07/kWh3,120 hrs/yr. 3,120 hrs/yr.

SizeCostLifeSalvageEfficiencyEnergy CostOperating Hours

(a) At i= 13%, determine the operating cost per kWh for each motor.(b) At what operating hours are they equivalent?

Exercises

10

(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

Exercises

11

Solution

(b) break-evenOperating Hours = 6,742

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)

= $306.75To maximize equivalent uniform annual benefit (EUAB) select the larger AW, which is Device B

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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

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Tunnel through mountain Pipeline around mountain

Initial cost $5,5 million $5 million

Maintenance 0 0

Useful life Permanent 50years

Salvage value 0 0

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%,∞) = iAW = P.i = = $5,5 million (0,06) = $330,000Pipeline 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