Overview of Engineering Economics ME 195A FALL 2012 NICOLE OKAMOTO Outline  Market Analysis – for your ME 195a reports  Time Value of Money 

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Transcript Overview of Engineering Economics ME 195A FALL 2012 NICOLE OKAMOTO Outline  Market Analysis – for your ME 195a reports  Time Value of Money 

Overview of Engineering
Economics
ME 195A FALL 2012
NICOLE OKAMOTO
Outline
 Market Analysis – for your ME 195a reports
 Time Value of Money
 Present Worth Analysis/Cost-Benefit Analysis
 Depreciation
 Taxes
Market Research
Market research backs up the project motivation.
A typical analysis includes:
Target Market Size (size in $$, large > $50M)
 Growth (aggressive growth > 20%)
 Current Major Players in the Market
 Trends
 Your Competitive Advantages
 Customers

All claims should be backed up by reliable and
reputable information sources. This kind of information
will bolster your sections on “Motivation” and
“Significance of your Project” for your Chapter 1’s.
Revision of J. Rhee slide -- Engineering Economics and Business
Market Research - Example
Solar Icemaker – Dr. Rhee, Fall 2010
Target Market/Segment: Corporate data centers (spent $2.25B
on electricity for AC in 2006)
Growth: Projected to double from 2006 to 2011
Major Players: Commercial HVAC (Trane, Honeywell, etc…)
Trends: RE<C (Google, 2007); Sustainable IT Lab (HP, 2008);
Ice Energy (2008)
Competitive Advantage: very little electricity required; very few
moving parts, no emissions/consumables during operation
Customers: Data center owners
Revision of J. Rhee slide -- Engineering Economics and Business
Time Value of Money
 Funds placed in a secure investment will increase in
value, depending on the time elapsed and interest
rate.
 $1000 placed in a bank account at the beginning of
the year with a 5% interest rate, compounded at the
end of the year, will give you $1050 at the end of the
year.
 This understanding is important to choose the best
design, from an economic perspective.

For example, product x may be more expensive but also more
efficient than product y. Which is better in the long run?
Important Definitions/Terms
 F: future worth; the amount in the future a certain




cash value today will be worth
P: present value; the value today of a certain cash
amount at some point
i: market interest rate; percent added per year to a
transaction
ir: real interest rate; interest rate with inflationary
effects removed, showing the real earning power of
your money
f: annual inflation rate
Important Definitions/Terms, cont.
 m: number of compounding periods per year
 n: number of years
 MARR: minimum acceptable (or attractive) rate of
return; minimum expected rate of return that a
company will accept before beginning a project;
often is the rate at which they can invest money
Compound Interest
 Interest is compounded at specified periods and is
added to the principal.
 May be compounded yearly, quarterly, daily,
continuous, etc.
Future Worth/Present Worth Equations
If interest is compounded once per year:
F  P1  i 
n
P  F 1  i 
n
If interest is compounded multiple (m) times per
year:
mn
i 

F  P1  
 m
Uniform Payments
 We can also consider the present value or
future worth of a uniform series of payments
(annual or other time frame)
F 1  i   1

A
i
n
P 1  i   1

n
A
i1  i 
n
 Terminology is often in the form of (P/A, i, n)
 If interest is compounded multiple times per
year, replace i with i/m and n with mn
Table Example (from EIT review manual)
PW Example
 What is the present worth of payments of $500/year
for 10 years with an interest rate of 6%?
 from previous slide
(P/A, 6%, 10)=7.3601
 Multiply your annual payments (A) by P/A
7.3601*$500=$3680.50
Monthly Compounding Example.
 Consider a 10-year mortgage where the principal
amount P is $200,000 and the annual interest rate is
6%, compounded monthly. Find the monthly
payment.
 The number of monthly payments is 120 and the
monthly interest rate is 6/12=0.5%.
n


P 1  i 1
 We can use our equation

n
A
with n=120 and i=0.005.
i1  i 
 A=P*A/P=$200,000*0.0110=$2200
What if prices escalate at an annual rate?
 Real interest rate must take this into account
 Inflation: general rise in price levels associated with
increase in available currency and credit without
increase in goods and services; f
 Escalation: changes in cost due to increase in
demand, resource depletion, and technology
advances; rr
 Nominal escalation rate rn: includes both inflation
and escalation rate
 Additional equations that we won’t get into here can
take these effects into account.
Present Worth Analysis
 This is one example of a cost-benefit analysis to compare





alternatives and see which investment choice is best.
 Estimate and total the “Equivalent Money Value” of the
BENEFITS and COSTS to the planned projects or
investments to establish whether they are worthwhile
Step 1: Determine interest rate firm wishes to earn on
investments; MARR – minimum attractive (or acceptable)
rate of return
Step 2: Estimate service life of project
Step 3: Estimate cash inflow for each period over service
life
Step 4: Estimate cash outflow for each period over service
life
Step 5: Determine net cash flows for each period (inflow
minus outflow)
Present Worth Analysis, cont.
 Step 6: Find the present worth, P, of each period
N
An
P
n


1

i
n 0




Or use P/F for each period
An= net cash flow at the end of period n
N=lifetime
n=year
i=MARR (or interest rate)
 Step 7: Decide
 P>0 accept
 P=0 indifferent
 P<0 reject
Choosing Between Alternatives
 If you have multiple alternatives
 Select the one with the highest P if the service lives are the
same length, all else being equal.
 If revenues are the same, choose the one with the smallest P of
costs.
Example
 Tiger Machine Tool Company is considering the acquisition of a new
metal-cutting machine. The required initial investment of $75000 and
projected cash benefits over a three-year project life are as follows:
End of Year
0
1
2
3
Net Cash Flow
-$75,000
$24,000
$27,340
$55,760
 Evaluate the economic merit of the acquisition assuming MARR=15%.

P(15%)  75000 $24400P

 55760P
F

F


,15%,1  $27340P
F
,15%,2

,15%,3  $3553
 So we should accept this opportunity!
Reference: Chan Park, Fundamentals of Engineering Economics, Prentice-Hall, 2004.
Cash Flow Diagram Example
 Here’s another way to look at the analysis, with a cash
flow diagram:
$55,760
$24,400
$27, 340
0
1
2
3
Year
$75,000
 P (15%)inflow=$24,400(P/F, 15%, 1)+$ 27,340(P/F,
15%,2)+ $55,760(P/F, 15%,3)=$78553
 P (15%)outflow=$75000
 P (15%)=$78553-$75000>0 accept!
Here’s
how P
changes
with
MARR
Reference: Chan Park, Fundamentals of Engineering Economics, Prentice-Hall, 2004.
Another Example
 By replacing a condenser on a building’s AC system
with a more efficient one, you can save 10,000
kWh/year over its 15 year life. It needs replacing
anyway. How much more can you afford to pay for
the more efficient one? Assume that the company
can get a 15-year loan at i=6%, and electricity costs
5¢/kWh.
 Solution: Set the present worth of condenser A equal
to the present work of condenser B to find the breakeven point.
Solution
 Cost A + present worth of electricity cost=Cost B + present
worth of electricity cost
 Cost B-Cost A=difference in present worth of electricity
costs=10,000 kWh*$0.05/kWh*(P/A, 6%,
15)=$500*9.7122=$4856
 If you spend less than $4856 on the more efficient
condenser, you will come out ahead.
 The big uncertainty in this analysis is the stability of longterm electricity costs.
 Other considerations – a more efficient system could help
you get LEED certification, which may have benefits.
Cash Outflow Sources
Cash outflow (costs) sources:
• Initial planning cost
• Capital investment
• Operating (both direct and indirect costs)
• Maintenance
• Marketing and promotion
• Taxes
• Quality assurance and warranty
• Depreciation
• Others
Depreciation
 Value of equipment decreases with time
 Depreciation can be deducted from income as a
business expense for tax purposes.
 Book value = original cost – depreciation
 Three common ways to calculate depreciation
 Straight line
C0  S
DPz 
N
z=year of interest
 C0=purchase price
 S=salvage value at end of lifetime
 N=lifetime (years)

Depreciation, cont.
 Other ways to calculate depreciation:Sum-of-Years Digits,
Declining Balance, Modified Accelerated Cost Recovery
System (MACRS)
Burmeister, Elements of Thermal-Fluid System Design, Prentice Hall
Income Taxes
 Percentage can vary widely
 CA is a very high tax state
 Many companies incorporate in NV instead, and a lot of
manufacturing has been moving to surrounding states
 Local taxes may be negotiable
 Property taxes 1-4% of assessed value
 Tax deductible expenditures include
 Fuel costs
 O&M
 Interest on debt
 Depreciation
 State income taxes (for federal)
 Federal income taxes (for state)
For more information:
 Take ISE 102 Engineering Economic Systems as an
elective
 Get approval from your advisor first
 Course Description: Systems analysis applied to
economic decisions in engineering; comparison of
alternatives based on cost breakdown structure and
time value of money; system life-cycle process; lifecycle economic concepts, costing methodology and
applications. Corequisite: MATH 31 and ENGR 10 or
equivalent.