Chapter 5 Present Worth Analysis
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1 Chapter 5 Present Worth Analysis 1. Net Present Worth (NPW) Analysis NPW is a comparison of alternatives by determining at year 0 (i.e., the present time). At least one of the following three situations will characterize all NPW problems: a) all alternatives have, b) at least one alternative differs from the others in terms of useful life spans, c) there is an analysis period. The above situations determine the analysis period that you will use in each problem (i.e., over what period of time you will compare the alternatives). The cash flows for the analysis period are moved to year zero with an interest rate known as the Minimum Attractive Rate of Return (MARR). The term MARR was discussed earlier in the course. If the NPW equal $0, the alternative earns an interest rate equal to the. A positive NPW indicates that the alternative earns an interest rate that is than the MARR. A negative NPW indicates that the alternative earns an interest rate that is than the MARR. 2. NPW Analysis of Alternatives With Equal Life Spans For each alternative, simply move all associated cash flows (with the MARR) to year zero. 3. Choosing Between Alternatives The option must be considered if it is an option. If donothing is an option, it is chosen if none of the other alternatives have a NPW > 0. If do-nothing is an option and at least one of the remaining alternatives has a positive NPW, simply choose the alternative with the. If do-nothing is not an option, simple choose the alternative with the (whether negative or positive).
2 4. Example of Equal Life Spans Utilize NPW Analysis to decide which, if any, of the following alternatives should be accepted. MARR = 10%. TOPS Corp. Quality, LLC Initial Cost 10,000 $8,000 Benefits $4,000/yr $3,500 in year 1, growing by $500 each year after Costs $1,000/yr $1,500/yr Salvage $1,000 0 Life (yrs) 5 5 CFD for TOPS Corp. NPW = (P/A, 10%, 5) (P/A, 10%, 5) (P/F, 10%, 5) = (3.791) (3.791) (.6209) = $
3 CFD for Quality, LLC 5. NPW Analysis of Alternatives with Unequal Life Spans In order to compare alternatives with unequal lives, must be assumed. Identical replacement assumes that the current option will be replaced at the end of its life with an identical with identical. To compare the options, find the (LCM) of all the alternatives life spans. The LCM determines the number of times that you will need to identically replace each alternative in order to have a proper comparison. After determining the cash flows associated with each alternative (based on identical replacement and the LCM), simply move all cash flows (with the MARR) to year 0.
4 6. Choosing Between Alternatives The same rules discussed for choosing between alternatives with equal life spans also apply for choosing between alternatives with unequal life spans. That is, assuming the procedures described in note #5 are properly followed. 7. Example of Unequal Life Spans Given the following two alternatives and NPW Analysis, determine which, if any, should be chosen. MARR = 15% Alternative #1 Alternative #2 Initial Cost $50,000 $25,000 Benefits $10,000/yr $13,000 in year 1, decreasing by $2,000 each year after Costs $2,000/yr $2,000/yr Salvage $0 $2,000 Life (yrs) 10 5 CFD for Alternative #1 NPW = (P/A, 15%, 10) (P/A, 15%, 10) NPW = (5.019) (5.019) NPW = - $9,848.00
5 CFD for Alternative #2 8. Comparing Alternatives with an Infinite Analysis Period This type of application is often seen in governmental projects. According to page 166, In governmental analyses, a service or condition sometimes must be maintained for an infinite period (roads, dams, pipelines, etc.). Due to these type applications, Civil Engineers are highly likely to see these applications. is the present sum (i.e., the worth) of money that would need to be set aside now, at some interest rate, to yield the funds required to provide the service (or whatever) indefinitely. Formula: (P/A, i%, ) = 1/i thus P = A(1/i) = A/i A = (Equivalent) Uniform Payments that occur an infinite number of times P = (Note: When solving for P, P always lands one period prior to the first (equivalent) uniform payment. i = effective interest rate for the time period between each (equivalent) uniform payment
6 9. An Example of an Infinite Analysis Period Mr. I. B. Rich wants to donate to LaTech a lump sum of money today that will provide ten scholarships worth $4,000. The scholarships are to start one year from today and occur every year, forever. If LaTech can earn 5% per year, compounded annually on this investment, what amount must he donate today? 10. Another Infinite Series Example If the first scholarship in the previous problem did not occur until 5 years from today, how would the capitalized cost change?
7 11. Another Infinite Series Problem Mr. I. B. Rich donates $50,000 to LaTech and desires that the money provide a scholarship once every 4 years, forever. If LaTech can earn 10% on the investment, what amount will be available for the scholarship that occurs every 4 years, forever? 12. Bond Problems When dealing with bond problems, you will notice that two interest rates are given. One will refer to the periodic dividend that will be paid to the bond holder. The other will refer to the interest rate or MARR.
8 13. An Example of a Bond Problem You are interested in a particular bond that has the following characteristics: Face Value = $10,000 Maturity date is in 5 years. Pays dividends at a rate of 6% (paid equally over semi-annual periods) If you desire 8%, compounded semi-annually on any investment you make, what is the most you should pay for the bond?
9 14. More Examples You have been asked to perform an economic analysis on two different investment opportunities. The first alternative considered has an expected lifespan of two (2) years. The second is expected to last for three (3) years. In making your analysis, you have been asked to use Net Present Worth Analysis. The NPW of alternative one has already been calculated for you, and it was found that the NPW = $22,500 (Please note that this NPW has taken into account any necessary information relative to alternative two.). Given this and the following cash flows for alternative two, make the necessary calculations for the second alternative to compare it to the first alternative. Given your calculations, which of the two (if any) should be chosen? Interest is 10%. Alternative #2 Initial Cost $30,000 Benefits $10,000 the first year and growing by $10,000 each year after O&M Costs $2,000 per year Life (in years) 3
10 You are trying to decide whether or not you want to purchase a corporate bond that will mature in 20 years. The face value of the bond is $15,000. The bond will pay equal quarterly payments based on a yearly rate of 6% of the face value. If you want to earn an interest rate of at least 8%, compounded quarterly, what is the most you should pay for the bond?
11 You have been asked to perform an economic analysis on two different possible courses of action (shown below). Using Present Worth Analysis and an interest rate of 7%, determine which course of action should be chosen. Do-nothing is not an option. *Fortunately, the NPW for alternative 1 (over the given life) has already be calculated. It was found that the NPW = $25,333. Alternative #1 Alternative #2 Initial Investment $200,000 $100,000 Benefits $20,000 annually $4,000 the first year and growing by $4,000 each year until it reaches $20,000. After reaching $20,000, it remains the same for the rest of the expected life. O&M Costs $2,000 annually $1,000 annually Salvage $15,000 $0 Expected Life (in years) 30 15
12 Your goal is to have $5,000 available every two (2) years, forever. The first $5,000 is to be available ten years from now (i.e., in year 10). To do this, you are going to invest over the next 8 years. In the first four (4) years, you will invest an increasing amount. The investments are $1,000 for year 1; $2,000 for year 2; $3,000 for year 3; and $4,000 for year 4. Over the remaining four years, you plan on investing a uniform amount per year. Based on your goal of $5,000 every two years, beginning in year 10, what equivalent uniform amount must you invest over years five (5) through eight (8). Interest is 10%.
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