enginnering economy L6 PW-AW-FW-GW .pptx

Project management is the application of knowledge, skills, tools, and techniques to project activities to meet project requirements. It’s the practice of planning, organizing, and executing the tasks needed to turn a brilliant idea into a tangible product, service, or deliverable.

In essence, project management involves:

1. **Initiation**: Defining the project at a high level and obtaining authorization.
2. **Planning**: Establishing the scope, objectives, and procedures.
3. **Execution**: Coordinating people and resources to carry out the plan.
4. **Monitoring and Controlling**: Tracking progress and making necessary adjustments.
5. **Closing**: Finalizing all activities and formally closing the project.

Effective project management ensures that a project is completed on time, within budget, and to the required quality standards. It also involves managing risks, communicating with stakeholders, and ensuring that the project aligns with the overall strategic goals of the organization.

Is there a specific aspect of project management you're interested in?

1. Engineering Economy Dr. Abdelhamid Naid Lecture 6

2. EX: University lab is a research contractor to NASA for in-space fuel cell systems that are hydrogen and methanol-based. During lab research, three equal-service machines need to be evaluated economically. Perform the present worth analysis with the costs shown below. The MARR is 10% per year. Electric-Powered Gas-Powered solar-Powered First cost, $ -4500 3500 -6000 Annual operating cost (AOC), $/year -900 -700 -50 Salvage value S, $ 200 350 100 Life, years 8 8 8 QUIZZE 5 G-A

3. PW E=- 4500 - 900( P/A ,10%,8)+ 200( P/F ,10%,8)= $-9208 PW G=- 3500 - 700( P/A ,10%,8) + 350( P/ F ,10%,8) = $-7071 PW S=- 6000 - 50( P/A ,10%,8) + 100( P/F ,10%,8) = $ - 6220

4. Annual Worth Analysis AW is easy to understand by any individual acquainted with annual amounts, for example, dollars per year. The AW value, which has the same interpretation as A used thus far, is the economic equivalent of the PW and FW values at the MARR for n years. All three can be easily determined from each other by the relation AW = PW (A/P, i, n) = FW (A /F, i, n)

5. Annual Worth Analysis Fundamental Decision Rule Single Project Evaluation: •‰If AW(i) > 0, accept the investment. •‰If AW(i) = 0, remain indifferent to the investment •‰If AW(i) < 0, reject the investment Comparing Mutually Exclusive Alternatives: •Service projects: select the alternative with the minimum annual equivalent cost •‰Revenue projects: select the alternative with the maximum AW(i)

6. To compare this against the alternative with a 6-year life, the analysis needs to be extended over 12 years (=LCM) for equal-service comparison. Therefore, we have to extend the cash flow of this alternative to 12 years, i.e. cycle has to be repeated two more times (total of 3 cycles). As also outlined before in capitalized cost calculations, whether we use one cycle, or two cycles, or three or more cycles of this cash flow in calculating the AW value, we get the same value. That is, the AW value obtained by using only one cycle of cash flow applies for every year of the life cycle, and for each additional life cycle. Then, The AW value has to be calculated for only one life cycle. Therefore, it is not necessary to use the LCM of lives, as it is for PW and FW analyses. The procedure in evaluating the alternative(s) is then similar to those for the PW method. The major difference being that AW value will now be calculated by considering the cash flow for one cycle only

7. Example1: Company ABC plans to purchase new equipment to improve productivity. The equipment cost is $25000 and is expected to have a market value of $5000 at the end of its 5-year life. If the expected improvement in productivity will net $8000 per year and Company’s MARR is 20% per year, should the Company purchase this equipment? AW = -25000(A/P,20%,5) + 8000 + 5000(A/F,20%,5) = 312.40 Since AW > 0, then the equipment should be purchased .

8. We only need to consider one cycle of cash flow to calculate AW values: AWA = -62000(A/P,15%,4) – 15000 + 8000(A/F,15%,4) = - 35114.58 AWB = -77000(A/P,15%,6) – 21000 + 10000(A/F,15%,6) = - 40204.08 We select A because its AW value is numerically larger

9. Considering only one cycle of cash flow for each alternative: AWA = -200000.(A/P,20%,8) + (160000 – 100000) + 50000.(A/F,20%,8) = -200000x0.2606 + 60000 + 50000x0.0606 = 10910 AWB = -300000.(A/P,20%,11) + (190000 – 110000) + 100000.(A/F,20%,11) = 13780 and AWC = -150000.(A/P,20%,7) + (200000 – 150000) + 50000.(A/F,20%,7) = 12258 Select B since its AW is numerically largest and also, since it is better than DN. (b) Select all three since AW > 0 for each one

10. AW = -100,000(0.20) – 30,000 – 50,000(A/F,20%,5) = -20,000 – 30,000 – 50,000(0.13438) = $-56,719 per yea

11. Calculation of Capital Recovery and AW Values The annual worth (AW) value for an alternative is comprised of two components: capital recovery for the initial investment P at a stated interest rate (usually the MARR) and the equivalent annual amount A. The symbol CR is used for the capital recovery component. In equation form, AW = CR + A Accordingly, CR is calculated as CR = -P (A/P, i , n) + S (A/F, i , n)

12. Annual Worth Analysis Calculation of Capital Recovery and AW Values An alternative should have the following cash flow estimates: Initial investment P. This is the total first cost of all assets and services required to initiate the alternative. When portions of these investments take place over several years, their present worth is an equivalent initial investment. Use this amount as P . Salvage value S. This is the terminal estimated value of assets at the end of their useful life. The S is zero if no salvage is anticipated; S is negative when it will cost money to dispose of the assets. For study periods shorter than the useful life, S is the estimated market value or trade-in value at the end of the study period. Annual amount A. This is the equivalent annual amount (costs only for cost alternatives; costs and receipts for revenue alternatives). Often this is the annual operating cost (AOC) or M&O cost, so the estimate is already an equivalent A value. The annual worth (AW) value for an alternative is comprised of two components: capital recovery for the initial investment P at a stated interest rate (usually the MARR) and the equivalent annual amount A. The symbol CR is used for the capital recovery component. In equation form, AW = CR + A Capital recovery (CR) is the equivalent annual amount that the asset, process, or system must earn (new revenue) each year to just recover the initial investment plus a stated rate of return over its expected life. Any expected salvage value is considered in the computation of CR. CR= - P ( A/P,i,n ) + S ( A/F,i,n ) (**)

13. Annual Worth Analysis EXA: Lockheed Martin is increasing its booster thrust power in order to win more satellite launch contracts from European companies interested in opening up new global communications markets. A piece of earth-based tracking equipment is expected to require an investment of $13 million, with $8 million committed now and the remaining $5 million expended at the end of year 1 of the project. Annual operating costs for the system are expected to start the first year and continue at $0.9 million per year. The useful life of the tracker is 8 years with a salvage value of $0.5 million. Calculate the CR and AW values for the system, if the corporate MARR is 12% per year.

14. SOL: Capital recovery: Determine P in year 0 of the two initial investment amounts, followed by the use of Equation [**] to calculate the capital recovery. In $1 million units, P= 8 + 5( P/F ,12%,1) = $12.46 1- CR = - 12.46( A/P ,12%,8) +0.5( A/F ,12%,8) = $ -2.47 2- AW= - 2.47 - 0.9 = $ -3.37 million per year The correct interpretation of this result is very important to Lockheed Martin. It means that each and every year for 8 years, the equivalent total net revenue from the tracker must be at least $2,470,000 just to recover the initial present worth investment plus the required return of 12% per year. This does not include the AOC of $0.9 million each year. Annual worth: To determine AW, the cash flows in Figure *a must be converted to an equivalent AW series over 8 years ( * b). Since CR = $2.47 million is an equivalent annual cost, as indicated by the minus sign, total AW is determined.

15. Annual Worth Analysis Equivalent uniform annual cost EUAC = P (A/P,7,10) = 1000 (Table Factor ) = 142.4 $ Question: How much did your furniture cost? Answer A: It cost me $1000, but it will last for 10 years. Answer B: It’s costing me $142.40 every year for 10 years. Both answers are “equivalent”

16. Annual Worth Analysis EXA: If the student now believes the furniture can sold at the end of 10 years for $ 200 under these circumstances, what is the equivalent uniform annual cost? Question: How much did your furniture cost? Answer A : My furniture cost me $1000, but it will last for 10 years, and at that time I can sell it for $200; the Present Worth of its cost is $898. Answer B : It’s costing me $127.92 every year for 10 years. “Both answers are “equivalent” Which is easier to understand?

17. Annual Worth Analysis EXA : Consider the following alternatives, Based on an 8% interest rate, which alternative should be selected? A B First Cost $ 5,000 $ 10,000 Annual Maintenance $ 500 $ 200 Salvage Value $ 600 $ 1000 Useful life 5 15 {

18. {EUAC}A = - 5,000(A/P,8%,5) – 500 + 600(A/F,8%,5) = - 5,000(0.2505) - 500 + 600(0.1705) = - $1,650 EUAC}B = - 10,000(A/P, 8%, 15) - 200 +1000(A/F, 8%, 15) = - 10,000(0.1168) - 200 + 1,000(0.0368) = - $1,331

19. Annual Worth Analysis Example : The following costs are estimated for two equal-service machines to be evaluated by a canning plant manager. If the minimum required rate of return is 15% per year, help the manager decide which machine to select A B First Cost,$ 26,000 36,000 Annual maintenance cost,$ 800 300 Annual labor cost,$ 11,000 7000 Extra annual income taxes ---- 2,600 Salvage value,$ 2,000 3000 Life, years 6 10 Solution: Machine A: AWA= -26,000(A/P,15%,6) + 2,000 (A/F,15%,6) -11,800 = $-18,442 AWA over 6 years = AWA over LCM 30 yrs = -18,442 Machine B: : AWB= -36,000(A/P,15%,10) + 3,000 (A/F,15%,10) -9,900 = $-16,925 AWB over 10 years = AWB over LCM 30 yrs = -16,925 So, Select machine B

20. Annual Worth Analysis Example: Assume the company in previous example is planning to exit the business in 4 years. At that time, the company expects to sell machine A for $12,000 or machine B for $15,000. All other costs are expected to remain the same. Which machine should the company purchase under these conditions? Solution : Machine A: AWA= -26,000(A/P,15%,4) +12,000 (A/F,15%,4) –11,800 = $-18,504 Machine B: AWB= -36,000(A/P,15%,4) +15,000 (A/F,15%,4) –9,900= $-19.506 So, Select machine A

21. Annual Worth Analysis Analysis Period (n) & Alternative Lives : Analysis period is infinite, i.e., n = ∞ •If an investment has infinite life, it is called a perpetual (permanent) investment •Reminder: –Capitalized Worth (CW) for a uniform series A of end-of-period cash flows: CW = A/i –If P is the present worth of that investment, then: AW= P*I Example : What is the difference in annual worth between an investment of $100,000 per year for 100 years and an investment of $100,000 per year forever at an interest rate of 10% per year? Solution : = 100,000(A/P,10%,100) = $10,001 = 100,000(0.10) = $10,000 Difference is $1.

22. Annual Worth Analysis Analysis Period (n) & Alternative Lives Analysis period is infinite, i.e., n = ∞ Example : In the construction of an aqueduct to expand the water supply of a city, there are 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.

23. Annual Worth Analysis Advantages of Annual Worth •Popular Analysis Technique •Easily understood results are reported in $ / time period •Eliminates the LCM problem associated with the present worth method –Only have to evaluate one life cycle of a project •Applicable to a variety of engineering economy studies : Asset Replacement Breakeven Analysis Make or Buy Decisions Studies dealing with manufacturing Costs Economic Value Added analysis (EVA)

24. 5.9 Two methods can be used for producing expansion anchors. Method A costs $80,000 initially and will have a $15,000 salvage value after 3 years. The operating cost with this method will be $30,000 per year. Method B will have a first cost of $120,000, an operating cost of $8000 per year, and a $40,000 salvage value after its 3-year life. At an interest rate of 12% per year, which method should be used on the basis of a present worth analysis? PWA = -80,000 – 30,000(P/A,12%,3) + 15,000(P/F,12%,3) = -80,000 – 30,000(2.4018) + 15,000(0.7118) = $-141,377 PWB = -120,000 – 8,000(P/A,12%,3) + 40,000(P/F,12%,3) = -120,000 – 8,000(2.4018) + 40,000(0.7118) = $-110,742 Select Method B

25. 5.10 Sales of bottled water in the United States totaled 16.3 gallons per person in 2004. Evian Natural Spring Water costs 40¢ per bottle. A municipal water utility provides tap water for $2.10 per 1000 gallons. If the average person drinks 2 bottles of water per day or uses 5 gallons per day in getting that amount of water from the tap, what are the present worth values of drinking bottled water or tap water per person for 1 year? Use an interest rate of 6% per year, compounded monthly, and 30 days per month. Bottled water: Cost/ mo = -(2)(0.40)(30) = $24.00 PW = -24.00(P/A,0.5%,12) = -24.00(11.6189) = $-278.85 Municipal water: Cost/ mo = -5(30)(2.10)/1000= $0.315 PW = -0.315(P/A,0.5%,12) = -0.315(11.6189) = $-3.66