Chapter 5 Present Worth Analysis EGN 3615 ENGINEERING

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

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 2

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 3

Economic Criteria Depending on situation, the economic criterion should be chosen from one of the following 3: Situation Neither input nor output fixed Fixed input Fixed output Criterion Maximize (Output – Input) Maximize output Minimize input Engineering Economics 4

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: 1. All alternatives have the same useful life: Set it as the analysis period. 2. Alternatives have different useful lives: Let the analysis period equal the least common multiple, or some realistic time (based on needs). 3. Infinite analysis period, n=∞ Engineering Economics 5

Net Present Worth (NPW or PW) Here is the basic NPW formula: PW = PW of benefits – PW of cost Engineering Economics 6

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 Minimize present worth of outputs costs or other inputs Engineering Economics 7

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 SAVING USEFUL LIFE DEVICE A $1000 $300 Annually 5 year DEVICE B $1350 $300 The first year and increase $50 annually 5 year Engineering Economics 8

Example Continues A=$300 0 1 2 3 4 5 i=6% P= $1000 Engineering Economics 9

Example Continues $300 0 1 $450 $400 $350 2 3 4 $500 5 i=6% P= $1350 Engineering Economics 10

Example Continues A=$300 0 1 2 3 4 5 i=6% P= $1000 $300 0 1 $450 $400 $350 2 3 $500 4 i=6% P= $1350 5 Work 5 -4 DEVICE B has the larger present worth & is the preferred alternative Engineering Economics 11

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 Investment 1 $2000 $450 Annually Investment 2 $3000 $600 Annually Engineering Economics Useful Salvage Value Life (End of Useful Life) 5 year $100 5 year $700 12

Example Continues Engineering Economics 13

Example Continues Engineering Economics 14

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 INVESTMENT 1 Engineering Economics 15

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 16

Example Continues One method to select an analysis period is the least common multiple of useful lives. EQUIPMENT A 0 1 $200 Original Equipment A Investment 2 3 4 $1500 5 $200 Replacement Equipment A Investment 6 7 8 9 10 $1500 Engineering Economics 17

Question Continues $350 EQUIPMENT B 0 1 2 3 Original Equipment B Investment 4 5 6 7 8 9 10 $1600 Engineering Economics 18

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 19

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 Terminal Value at the end of 10 End Of -year analysis Useful Life period USEFUL LIFE MACHINE A $40, 000 $8, 000 $15, 000 7 year MACHINE B $65, 000 $10, 000 13 year Engineering Economics 20

Example Continues $8, 000 MACHINE A 0 1 2 3 4 5 6 7 $15, 000 8 7 -year life $40, 000 9 10 11 12 13 14 7 -year life $40, 000 Engineering Economics 21

Example Continues MACHINE B 0 1 2 3 4 5 6 7 8 $10, 000 9 10 11 12 13 14 13 -year life $65, 000 Engineering Economics 22

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 23

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 24

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 25

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 Equipment 2 Equipment 3 0 -$2000 -$1500 -$3000 1 +1000 +700 +500 2 +850 +300 +500 3 +700 +300 +550 4 +550 +300 +600 5 +400 +300 +650 6 +400 +700 7 +400 +500 +400 +600 +500 8 Engineering Economics 26

Question Continues $1000 $850 $700 $550 $400 0 $2000 1 2 3 4 5 6 7 8 EQUIPMENT 1 Engineering Economics 27

Question Continues $700 $300 $400 $500 $600 0 $1500 1 2 3 4 5 6 7 8 EQUIPMENT 2 Engineering Economics 28

Question Continues $500 $550 $600 $650 $700 $500 0 $3000 1 2 3 4 5 6 7 8 EQUIPMENT 3 To maximize NPW, choose EQUIPMENT 1 Engineering Economics 29

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 period value range = the cash flow values Engineering Economics 30

Question Continues (MS EXCEL) Year Equipment 1 Equipment 2 Equipment 3 0 ($2, 000) ($1, 500) ($3, 000) 1 1000 700 500 2 850 300 500 3 700 300 550 4 550 300 600 5 400 300 650 6 400 700 7 400 500 8 400 600 500 Interest 10% For Equip 1: NPW=NPV(A 12, B 3: B 10)+B 2 $1, 379. 17 Engineering Economics $763. 15 ($20. 64) 31

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) = – 980000 + 200000(6. 424) + 20000(0. 2292) = $309, 384 As PW > 0, purchase the machine. Or using MS EXCEL PW = -P + pv(0. 12, 13, -200000, -20000) = $309, 293. 17 Terms A(P/A, 0. 12, 13) and F(P/F, 0. 12, 13) are combined! Engineering Economics 32

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 33

Problem 5 -41 Outputs: 2000 lines for years 1~10 4000 lines for years 21~30 i = 10% per year, cables last for at least 30 yrs Option 1: 1 cable with capacity of 4000 lines Cost: $200 k with $15 k annual maintenance cost Option 2: 1 cable with capacity of 2000 lines now 1 cable with capacity of 2000 lines in 10 years Cost: $150 k with $10 k maintenance cost/year/cable (a) (b) Which option to choose? Will answer to (a) change if 2000 additional lines are needed in 5 years, instead of 10 years? Engineering Economics 34

Problem 5 -41 Solution (a) Present worth of cost for option 1 PW 0 f cost = $200 k + $15 k(P/A, 10%, 30) = $341, 400 Present worth of cost for option 2: PW of cost = $150 k + $10 k(P/A, 10%, 30) + $150 k(P/F, 0. 1, 10) + $10 k(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 35

Problem 5 -41 Solution (b) Cost for option 1 will not change. PW 0 f cost = $341, 400 Present worth of cost for option 2: PW of cost = $150 k + $10 k(P/A, 10%, 30) + $150 k(P/F, 0. 1, 5) + $10 k(P/A, 0. 1, 25)(P/F, 0. 1, 5) = $394, 300 Therefore, the answer will change to option 1. Engineering Economics 36

End of Chapter 5 Engineering Economics 37