1. Matakuliah: D0762 – Ekonomi Teknik Tahun: 2009 Present Worth and Annual Worth Course Outline 6

2. 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) = $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 have exceded 20%)

3. Exercises 3

4. 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 = $1250 One should set aside $1250 4 Annual disbursement A Interest rate i

5. 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 5 Example 5.6 $8million Capitalized Cost P $8million 70years $8million 70years $8million 70years $8million n=∞ …….

6. 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 6 $8million A n= 70 A = F(A/F,i,n) = $8 million (A/F,7%,70) = $8million (0,0062) = $4960 4960 0,07

7. 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. 7 $1,500 $850 $800 $750 $650 $700 $1,200 $23,000

8. 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 purchase in not justified. 8

9. Exercises Mutually Exclusive Alternative with Equal Life Project 9 StandardPremium MotorEfficient Motor25 HP $13,000$15,60020 Years$0 89.5%93%$0.07/kWh3,120 hrs/yr. Size Cost Life Salvage Efficiency Energy Cost Operating Hours (a) At i= 13%, determine the operating cost per kWh for each motor. (b) At what operating hours are they equivalent?

10. 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

11. Exercises 11 Solution (b) break-even Operating Hours = 6,742

12. 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 : AW A = $300 Device B : AW A = 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

13. 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 13 Tunnel through mountainPipeline around mountain Initial cost$5,5 million$5 million Maintenance00 Useful lifePermanent50years Salvage value00

14. 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