Economic Analysis

1
Economic Analysis Life Cycle Costing for Energy
Efficiency Alternatives
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Introduction

• Engineering economics is a tool to help investors
  choose among different investment options,
  equipment, retrofits, etc.
• Many factors cannot be translated into dollars.
  Economic analysis is not and should not be the
  only criteria in accepting or rejecting a design
  or investment option

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Simple Payback Period (SPP)

• SPP or Payback Period does not include the time
  value of money
• It is simple and easy to use, and that is why
  many organizations use it
• For periods of 1 to 2 years, this is mostly OK
• SPP is not an acceptable method for longer time
  periods

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SPP Example

• A Lighting system improvement costs 1,000
• The improvement saves 500 each year
• SPP
• SPP 1,000 / 500/yr 2 years

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Rate of Return

• The Rate of Return is the reciprocal of the
  simple payback
• Rate of Return 1 / SP
• 50 / yr

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Time Value of Money

• Because of the potential for economic growth, it
  is generally better to have a given amount of
  money now, than a year from now
• For example, most people would rather have 100
  today than one year from now
• By investing the 100 in the bank at 5 interest
  would yield 105 at the end of the year
• Money is said to be a time-value component

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Example Time Value of Money

• Option 1 A traditional furnace costs 100 and
  consumes 40 per year in fuel over its 10 year
  lifetime.
• Option 2 A high-efficiency furnace costs 200
  and consumes 20 per year in fuel over its 10
  year lifetime.
• By comparing the cash flow diagram for each
  investment we can see which is more favorable

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Example Cash Flow Diagram
Cash Flow Out (-)
Cash Flow In ()
We do not want to compare these options by
calculating 100 (10 40) 500 vs. 200
(10 20) 400 because the Time Value of money
isnt considered
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Cash Flows

• To compare investment options that have cash
  flows occurring over time, it is necessary to
  covert the values of all cash flows to a common
  time
• Calculating the present value of multiple
  investments, then compare present values to see
  which one is better
• Most types of cash flows involving future amounts
  of money can be converted to a present value
  using one of three equations

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Equation 1 The Present Value of a Future Amount

• Consider a present amount P of 100 invested at
  an interest rate i of 5 per year. The future
  value of the investment after n years Fn would
  be
• F0 100
• F1 100 100(.05) P Pi P (1i)
• F2 100 100(.05) 100 100(.05)(.05)
• P(1i) P(1i)i P(1i)2
• Fn P (1i)n
• Or P F(1i)-n present value of a future amount

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Example - Present Value

• Someone promises to pay you 1,000 in 5 years.
  If the interest rate is 10 per year, what amount
  would you take today that is equivalent to 1,000
  in 5 years (i.e. what is the present value of
  1,000 5 years from now) ?
• P F(1i)-n 1,000 (1.10)-5 621
• The factor (1i)-n is called the present worth
  factor, PWF(i,n). Thus,
• P F(1i)-n F PWF(i,n)

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Exponential Growth

• The equation, Fn P (1i)n, shows how the future
  amount of a present value invested at interest
  rate i grows over time
• This type of growth is called exponential growth
  
• If the exponent is less than one, it is called
  exponential decay
• With exponential growth, the rate of growth
  increases over time thus small initial growth
  rates may become surprisingly large over time

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Example Exponential Growth

• The legend goes that a clever engineer was once
  able to avert a potential famine during a severe
  drought by designing and building a dam and
  irrigation system
• The modest engineer declined the many treasures
  he was offered as payment saying that he would
  instead accept only
• the amount of rice equal to placing one grain of
  rice on the first square of a chess board,
• two grains on the second square,
• four grains on the third square, etc.
• until all 64 squares on the board were filled up
  
• How much grain did the engineer ask for?
• Answer F P(1i)n 1(1 1)64 1.84 x 1019
  grains total annual rice harvest of India

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Equation 2 Annual Payments

• Consider a series of n annual payments of amount
  A such as in the following cash flow diagram

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Equation 2 Present Value of An Annual Amount
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Now Doing a Little Algebra
The factor is called the series present worth
factor, SWPF(i,n). Thus Pn A SWPF (i,n)
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Example Annual Payments

• Returning to our furnace comparison
• If interest rates are 10 per year, which is the
  better investment?

Ptrad costs 100 40 346
Phigh-eff costs 200 20 323
Cost Comparison Ptrad Phigh-eff 346 -
323 The high efficiency furnace is cheaper by 23
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Example (cont)

• Or, we could have looked at the problem in terms
  of savings for the high-efficiency furnace
• Phigh-eff -100 20 23

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Equation 3 Present Value of an Escalating
Series

• Sometimes an annual payment is expected to
  increase over time at some rate e.
• For example, as equipment gets older it often
  requires more and more maintenance.
• The present value of an escalation series is

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Present Value of an Escalating Series
The factor or (depending on whether i
e) is called the escalating series present
worth factor, ESPWF(i,e,n). Thus P A
ESPWF(i,e,n) Note that when e 0, ESPWF(i,0,n)
SPWF(i,n)
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Example PV of an Escalating Series

• Assume it will cost 50 this year to maintain an
  aging piece of equipment
• And that the equipment will require 5 more
  maintenance every year for the next 10 years
• New replacement equipment would cost 400
• And would require only 10 of maintenance per
  year with no expected escalation over the next 10
  years
• Which is a better investment if interest rates
  are 5?

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Example (cont)
0 1 2 3 10

• Current Equipment
• Pcurr costs A ESPWF (0.05, 0.05, 10)
• 476
• New Equipment
• Pnew costs 400 A SPWF(0.05,10)
• 400 10 477 Conclusion the costs
  are the same

0 1 2 3 10
10
400
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The Discount Rate

• So far, we have referred to i as the rate of
  interest. More formally, i is the discount rate.
• i discount rate
• expected real rate of return from an
  alternative investment
• The alternative investment could be interest from
  a bank, stock market appreciation or expected
  profits from ones own company
• If we use a minimum return, say 25, as the value
  for i, we can evaluate whether new equipment
  investments should be undertaken

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Converting to Net Present Value

• We can make a direct comparison between
  investments by considering life-cycle costs
• State/Federal governments require Life Cycle
  Costing using time value of money
• Use Net Present Value Analysis to find lowest
  life cycle cost
• Net present value and net present worth are the
  same thing

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Net Present Value

• A good project has a Net Present Value (NPV)
  greater than zero
• The Internal Rate of Return (IRR) is the interest
  rate at which the NPV of the savings equals the
  NPV of the costs
• Need interest tables, a computer, or a calculator
  to find these NPVs

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Using Interest and Discount Rates

• To Compute the Present Value (P)
• P AP/A,I,N
• A annual amount (usually yr savings)
• I interest of discount rate
• N number of years in life of project

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The Calculation Method

• P AP/A,I,N or A PA/P,I,N
• P/A Present Worth Factor
• P/A is found for a particular interest rate and
  project life from an interest table
• From the interest tables, if I 10, and N
  5yrs, then P/A 3.791
• If there were no discounting, P/A 5

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Economic Evaluation Example

• A lighting system improvement costs 30,000
  and saves 10,000/yr. The average life of the
  project is 7 yrs. At a discount rate of 10, what
  is the NPV of this project ? Is this a good
  project ?
• Solution NPV AP/A, I, N cost
• NPV 10,000P/A, 10, 7 - 30,000
• 10,000 4.8684 - 30,000
• 18,684
• GOOD PROJECT !

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Life Cycle Cost Example

• A Rhino motor costs 3,000 to buy and costs
  15,000/yr to operate over its 10yr life. An
  Elephant motor costs 4,000 to buy and costs
  14,500/yr to operate. Which motor has the lowest
  LCC at a 10 discount rate ?
• Rhino 3,000 15,000/yrP/A,10,10
• 95,160/yr
• Elephant 4,00014,500/yrP/A,10,10
• 93,088/yr

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Four Basic Economic Problems

• Find P given A, I, and N
• Find A given P, I, and N
• Find I given P, A, and N
• Find N given P, A, and I

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Find P given A, I, and N

• Consider
• A 15,000 , I 10 , N 10
• Solution
• P AP/A,I,N
• 15,000 6.144
• 92,160

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Find A given P, I, and N

• Consider
• P 15,000 , I 10 , N 7
• Solution
• A PP/A,I,N
• 15,000 0.2054
• 3,081

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Find I given P, A, and N

• Consider
• P 25,000 , A 15,000 , N 8
• Solution
• P AP/A,I,N
• 25,000 15,000 1.67
• i gt50 60

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Find N given P, A, and I

• Consider
• P 25,000 , A 15,000 , I 10
• Solution
• P AP/A,I,N
• 25,000 15,000 1.67
• N 2 years

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SIR Decision Criterion

• SIR is the Savings to Investment Ratio
• The Savings is the Present Value
• The Investment or Cost is in Present Value (now)
• If SIR gt 1, the project is cost effective
• The bigger the SIR, the better the project