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Chapter 7,8,9 WHITE Spring 2010

# Chapter 7,8,9 WHITE Spring 2010 - CHAPTER 7 Present Worth...

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CHAPTER 7 Present Worth Method of Comparing Alternatives

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HW # Due Tuesday 02/16/10 Ch.7 A.16 A.20 A.22 B.5 Ch.8 C.7 D.4 E.3
NPW = Net Present Worth 2 Types of Comparing Alternatives 1. Same Life Span (n) and n H Find the Present worth of all \$ Compare 2. Different Life Spans Can not compare directly Must use multiple lives or shorten the “n” of the longer life alternative

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Different Life Spans Assume duplicate lives Assume shorter lived alternative is replaced at the end of its life Assume the purchase price of the alternative does not increase (no inflation) Replacements are made as many times as are needed until you have equivalent “n”’s for both alternatives Shorten life Assume some salvage value for the longer life alternative at the end of the shortest “n”
Important Information Capitalized Cost Meaning perpetual life, n goes to infinity Trade Ins Often promotion of new sales by inflating the trade in allowance For example, you have a crappy 1970 Ford Pinto and the dealer offers you a \$7,000 trade in Often misinterpretation of trade in as not being equal to cash – the trade in is also an asset. Lost Potential Revenue We deal with trade ins as LOST POTENTIAL REVENUE This is to keep all costs associated with a given alternative with that alternative.

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Lost Potential Revenue We deal with trade ins as LOST POTENTIAL REVENUE This is to keep all costs associated with a given alternative with that alternative. For example, you want to buy a car for \$15,000 – your current car is worth \$3,000 – the dealer offers you 2 choices, either pay the \$15,000 or pay \$12,000 and give them your current car. Your car although not \$3,000 in cash, is still a \$3,000 asset. So instead of the new car costing \$12,000 If you get the new car, you are loosing out on \$3,000 of potential revenue from selling your old car.
Cash Flow Diagrams Purchase New Car \$15,000 \$3,000 – trade in Maintain Old Car \$3,000 – lost potential revenue By not selling the used car, you are loosing out on the equivalent of \$3,000 – or have lost potential revenue associated with retaining the older vehicle This keeps all \$ associated with a given alternative together

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Example 7.1 Unequal Period Comparision 7.1 Given i = 6% Type A Type B Cost New \$ 20,000 \$ 5,000 Annual Maintenance \$ 1,000 \$ 2,000 Estimated Life 10 yrs 5 yrs Salvage Value 2,500 0 F = \$2,500 A = \$1,000 P = \$20,000 10 A = \$2,000 P =\$5,000 5
7.1 F = \$2,500 A = \$1,000 P = \$20,000 10 A = \$2,000 P =\$5,000 5 NPW = -\$20,000 – \$1000(P/A, i, 10) + \$2500 (P/F, i, 10) NPW = -\$5,000 – \$2000(P/A, i, 10) – \$5000(P/F, i, 5) 10 P =\$5,000 A = \$2,000

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Example 7.2 7.2 Given i = 15% Type A Type B Cost New \$ 150,000 \$100,000 Annual Maintenance\$ 3,000 \$2,000 Estimated Life 8 yrs 4 yrs Select an alternative A = \$3,000 P =\$150,000 8 NPW = -\$150,000 – \$3000(P/A, 15, 8) = -\$163,462 A = 2,000 P =100,000 4 NPW = -100,000 – 2000(P/A, i, 8) – 100,000(P/F, i, 4) = -166,150 8 P =100,000
Example 7.3 7.3 Given – Need to earn 10% Cost = \$100,000 Income = \$12,000 /year Expenses = \$2,700 / year Salvage = \$50,000 at the end of 10 years Assume i = 10% assuming i = 10% .. If Income > Expenses .. we earned more than 10% assuming i = 10% .. If Income < Expenses .. we earned less than 10% assuming i = 10% .. If Income = Expenses .. we earned exactly 10%

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7.3
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Chapter 7,8,9 WHITE Spring 2010 - CHAPTER 7 Present Worth...

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