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Chapter 5 Present Worth Analysis
EGN 3615 ENGINEERING ECONOMICS WITH SOCIAL AND GLOBAL IMPLICATIONS 1

Three Economic Analysis Methods
There are three major economic analysis techniques: Present Worth Analysis Annual Cash Flow Analysis Rate of Return Analysis This chapter discusses the first techniques

Chapter Contents Economic Criteria Considering Project Life
Net Present Worth Applying Present Worth Techniques Useful Lives Equal the Analysis Period Useful Lives Different from the Analysis Period Infinite Analysis Period: Capitalized Cost Multiple Alternatives Spreadsheet Solution

Economic Criteria Situation Criterion
Depending on situation, the economic criterion should be chosen from one of the following 3: Situation Criterion Neither input nor output fixed Maximize (Output – Input) Fixed input Maximize output Fixed output Minimize input Engineering Economics

Analysis Period Specific time period, same for each alternative, called the analysis period, planning horizon, or project life Three different analysis-period situations may be considered: All alternatives have the same useful life: Set it as the analysis period. Alternatives have different useful lives: Let the analysis period equal the least common multiple, or some realistic time (based on needs). Infinite analysis period, n=∞ Engineering Economics

Net Present Worth (NPW or PW)
Here is the basic NPW formula: PW = PW of benefits – PW of cost Engineering Economics

Present Worth Techniques
Mutually exclusive alternatives: Resolve their consequences to the present time. Situation Criterion Neither input nor output fixed Maximize net present worth Amount of money or other input resources are fixed Maximize present worth of benefits or other outputs Fixed task, benefit, or other outputs Minimize present worth of costs or other inputs Engineering Economics

Present Worth—Equal Useful Lives
Example: Consider two mechanical devices to install to reduce cost. Expected costs and benefits of machines are shown in the following table for each device. If interest rate is 6%, which device should be purchased? DEVICE COST COST SAVING USEFUL LIFE DEVICE A \$1000 \$300 Annually 5 year DEVICE B \$1350 \$300 The first year and increase \$50 annually Engineering Economics

Example Continues 1 2 3 4 5 A=\$300 P= \$1000 i=6% Engineering Economics

Example Continues \$500 \$450 \$400 \$350 \$300 i=6% P= \$1350 1 2 3 4 5
1 2 3 4 \$300 P= \$1350 i=6% \$350 \$400 \$450 \$500 5 Engineering Economics

Example Continues A=\$300 i=6% P= \$1000 \$500 \$450 \$400 \$350 \$300 i=6%
1 2 3 4 5 A=\$300 P= \$1000 i=6% 1 2 3 4 \$300 P= \$1350 i=6% \$350 \$400 \$450 \$500 5 Work 5-4 DEVICE B has the larger present worth & is the preferred alternative Engineering Economics

Present Worth—Equal Useful Lives
Example: Consider two investments with expected costs and benefits shown below for each investment. If investments have lives equal to the 5-year analysis period, which one should be selected at 10% interest rate? Investment Cost Benefit Useful Life Salvage Value (End of Useful Life) Investment 1 \$2000 \$450 Annually 5 year \$100 Investment 2 \$3000 \$600 Annually \$700 Engineering Economics

Example Continues Engineering Economics

Example Continues Engineering Economics

Example Continues Salvage value is considered as another positive cash flow. Since criterion is to maximize PW (= present worth of benefits – present worth of costs), the preferred alterative is INVESTMENT1 Engineering Economics

Alternatives with different Useful Lives
Example: Consider two new equipments to perform desired level of (fixed) output. expected costs and benefits of machines are shown in the below table for each equipment. If interest rate is 6%, which equipment should be purchased? EQUIPMENT COST SALVAGE VALUE USEFUL LIFE EQUIPMENT A \$1500 \$200 5 year EQUIPMENT B \$1600 \$350 10 year Engineering Economics

Replacement Equipment A Investment
Example Continues One method to select an analysis period is the least common multiple of useful lives. 1 2 3 4 5 \$200 \$1500 6 7 8 9 10 Replacement Equipment A Investment Original Equipment A Investment EQUIPMENT A Engineering Economics

Question Continues \$350 \$1600 EQUIPMENT B 1 2 3 4 5 6 7 8 9 10
1 2 3 4 5 \$1600 6 7 8 9 10 Original Equipment B Investment EQUIPMENT B Engineering Economics

Question Continues EQUIPMENT A EQUIPMENT B
For fixed output of 10 years of service of equipments, Equipment B is preferred because it has a smaller cost. Engineering Economics

Present Worth-Useful Lives are Different from the Analysis Period
Example: Consider two alternative production machines with expected initial costs & salvage values shown below. If interest rate is 10%, compare these alternatives over a (suitable) 10-year analysis period (by using the present worth method)? MACHINE INITIAL COST Salvage Value at the End Of Useful Life Terminal Value at the end of 10-year analysis period USEFUL LIFE MACHINE A \$40,000 \$8,000 \$15,000 7 year MACHINE B \$65,000 \$10,000 13 year Engineering Economics

Example Continues \$40,000 \$8,000 \$15,000 MACHINE A 1 2 3 4 5 6 7 8 9
1 2 3 4 5 \$8,000 6 7 8 9 10 \$15,000 7-year life MACHINE A 11 12 13 14 Engineering Economics

Example Continues \$10,000 \$65,000 MACHINE B 1 2 3 4 5 6 7 8 9 10 11 12
1 2 3 4 5 6 7 8 9 10 \$10,000 \$65,000 MACHINE B 11 12 13 14 13-year life Engineering Economics

Example Continues MACHINE A MACHINE B
For fixed output of 10 years of service of equipments, Machine A is preferred because it has a smaller cost. Engineering Economics

Infinite Analysis Period (Capitalized Cost)
Capitalized cost is the present sum of money that is set aside now at a given interest rate to yield the funds (future interest earned) required to provide the service indefinitely. (5-2) Engineering Economics

Infinite Analysis Period (Capitalized Cost)
Example: How much should one set aside to pay \$1000 per year for maintenance on an equipment if interest rate is 2.5% per year and the equipment is kept in service indefinitely (perpetual maintenance)? Engineering Economics

Multiple (3+) Alternatives
Question: Cash flows (costs and incomes) for three pieces of construction equipments are shown below. For 10% interest rate, which alternative should be selected? Year Equipment 1 Equipment2 Equipment3 -\$2000 -\$1500 -\$3000 1 +1000 +700 +500 2 +850 +300 3 +550 4 +600 5 +400 +650 6 7 8 Engineering Economics

Question Continues \$2000 \$400 \$1000 \$850 \$700 \$550 EQUIPMENT 1 1 2 3 4
1 2 3 4 5 \$400 6 7 8 EQUIPMENT 1 \$1000 \$850 \$700 \$550 Engineering Economics

Question Continues \$300 \$1500 \$600 \$700 \$400 \$500 EQUIPMENT 2 1 2 3 4
1 2 3 4 5 \$300 \$1500 6 7 8 \$600 EQUIPMENT 2 \$700 \$400 \$500 Engineering Economics

Question Continues \$650 \$3000 \$500 \$550 \$600 \$700 EQUIPMENT 3
1 2 3 4 5 \$650 \$3000 6 7 8 \$500 EQUIPMENT 3 \$550 \$600 \$700 To maximize NPW, choose EQUIPMENT 1 Engineering Economics

Question Continues (MS EXCEL)
Use function: npv(rate, value range) - Return the net present value of a series of future cash flows “value range” at interest “rate”/period. rate = interest rate per period value range = the cash flow values Engineering Economics

Question Continues (MS EXCEL)
Year Equipment 1 Equipment2 Equipment3 (\$2,000) (\$1,500) (\$3,000) 1 1000 700 500 2 850 300 3 550 4 600 5 400 650 6 7 8 Interest For Equip 1: NPW=NPV(A12,B3:B10)+B2 10% \$1,379.17 \$763.15 (\$20.64) Engineering Economics

Problem 5-15 Solution i = 12% P = \$980,000 purchase cost
F = \$20,000 salvage value after 13 years A = \$200,000 annual benefit for 13 years PW = –P + A(P/A, 0.12, 13) + F(P/F, 0.12, 13) = – (6.424) (0.2292) = \$309,384 As PW > 0, purchase the machine. Or using MS EXCEL PW = -P + pv(0.12, 13, , ) = \$309,293.17 Terms A(P/A, 0.12, 13) and F(P/F, 0.12, 13) are combined! Engineering Economics

Problem 5-23 Solution i = 18%/12 = 1.5% per month
A = \$500 payment/month n = 36 payments P = ? price of a car she can afford P = A(P/A, 0.015, 36) = 500(27.661) = \$13,831 What is P, if r = 6%? i = 6%/12 = 0.5% P = pv(0.005, 36, -500) = \$16,435.51 Do Problems 5-24, 5-25, 5-26! Engineering Economics

Problem 5-41 Outputs: 2000 lines for years 1~10
i = 10% per year, cables last for at least 30 yrs Option 1: 1 cable with capacity of 4000 lines Cost: \$200k with \$15k annual maintenance cost Option 2: 1 cable with capacity of 2000 lines now 1 cable with capacity of 2000 lines in 10 years Cost: \$150k with \$10k maintenance cost/year/cable (a) Which option to choose? (b) Will answer to (a) change if 2000 additional lines are needed in 5 years, instead of 10 years? Engineering Economics

Problem 5-41 Solution (a) Present worth of cost for option 1
PW 0f cost = \$200k + \$15k(P/A, 10%, 30) = \$341,400 Present worth of cost for option 2: PW of cost = \$150k + \$10k(P/A, 10%, 30) + \$150k(P/F, 0.1, 10) + \$10k(P/A, 0.1, 20)(P/F, 0.1, 10) = \$334,900 Select option 2, as it has a smaller PW of cost. Engineering Economics

Problem 5-41 Solution (b) Cost for option 1 will not change.
PW 0f cost = \$341,400 Present worth of cost for option 2: PW of cost = \$150k + \$10k(P/A, 10%, 30) + \$150k(P/F, 0.1, 5) + \$10k(P/A, 0.1, 25)(P/F, 0.1, 5) = \$394,300 Therefore, the answer will change to option 1. Engineering Economics

End of Chapter 5 Engineering Economics

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