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ME 195A FALL 2012 NICOLE OKAMOTO Overview of Engineering Economics.

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Presentation on theme: "ME 195A FALL 2012 NICOLE OKAMOTO Overview of Engineering Economics."— Presentation transcript:

1 ME 195A FALL 2012 NICOLE OKAMOTO Overview of Engineering Economics

2 Outline Market Analysis – for your ME 195a reports Time Value of Money Present Worth Analysis/Cost-Benefit Analysis Depreciation Taxes

3 Revision of J. Rhee slide -- Engineering Economics and Business 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.

4 Revision of J. Rhee slide -- Engineering Economics and Business Market Research - Example 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

5 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?

6 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 i r : real interest rate; interest rate with inflationary effects removed, showing the real earning power of your money f: annual inflation rate

7 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

8 Compound Interest Interest is compounded at specified periods and is added to the principal. May be compounded yearly, quarterly, daily, continuous, etc.

9 Future Worth/Present Worth Equations If interest is compounded multiple (m) times per year: If interest is compounded once per year:

10 Uniform Payments We can also consider the present value or future worth of a uniform series of payments (annual or other time frame) 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

11 Table Example (from EIT review manual)

12 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)= Multiply your annual payments (A) by P/A *$500=$

13 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%. We can use our equation with n=120 and i= A=P*A/P=$200,000*0.0110=$2200

14 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; r r Nominal escalation rate r n : includes both inflation and escalation rate Additional equations that we won’t get into here can take these effects into account.

15 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)

16 Present Worth Analysis, cont. Step 6: Find the present worth, P, of each period  A n = 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 Or use P/F for each period

17 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.

18 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: Evaluate the economic merit of the acquisition assuming MARR=15%. So we should accept this opportunity! Reference: Chan Park, Fundamentals of Engineering Economics, Prentice-Hall, End of YearNet Cash Flow 0-$75,000 1$24,000 2$27,340 3$55,760

19 Cash Flow Diagram Example Here’s another way to look at the analysis, with a cash flow diagram: 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! $55,760 $27, 340 $24,400 $75,000 Year 0 123

20 Here’s how P changes with MARR Reference: Chan Park, Fundamentals of Engineering Economics, Prentice-Hall, 2004.

21 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 break- even point.

22 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 long- term electricity costs. Other considerations – a more efficient system could help you get LEED certification, which may have benefits.

23 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

24 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  z=year of interest  C 0 =purchase price  S=salvage value at end of lifetime  N=lifetime (years)

25 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

26 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)

27 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; life- cycle economic concepts, costing methodology and applications. Corequisite: MATH 31 and ENGR 10 or equivalent.

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