Matakuliah : D0762 – Ekonomi Teknik Tahun : 2009

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

2. 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 have exceded 20%) 2

3. Exercises 3

4. Exercises Annual disbursement A Interest rate i • 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

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

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

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

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 Standard Premium Motor Efficient Motor 25 HP 25 HP $13,000 $15,600 20 Years 20 Years $0 $0 89.5% 93% $0.07/kWh $0.07/kWh 3,120 hrs/yr. 3,120 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? Mutually Exclusive Alternative with Equal Life Project 9

10. Exercises • 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 (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 10

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

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 : 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

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

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