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Economic Analysis and Economic Decisions for Energy RETROFITTINGS

Economic Analysis and Economic Decisions for Energy RETROFITTINGS.

quincy-mcneil
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Economic Analysis and Economic Decisions for Energy RETROFITTINGS

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  1. Economic AnalysisandEconomic Decisionsfor Energy RETROFITTINGS

  2. This chapter provides an overview of the basic principles of economic analysis that are essential to determine the cost-effectiveness of various energy conservation measures suitable for residential/commercial and/or industrial facilities

  3. In most applications, initial investments are required to implement energy conservation measures. These initial costs must be generally justified in terms of a reduction in the operating costs. For an energy retrofit project to be economically worthwhile, the initial expenses have to be lower than the sum of savings obtained by the reduction in the operating costs over the lifetime of the project.

  4. Economic Factors • Capital cost of the technology • Long term debt availability • Capacity • Risks and uncertainties • Time • Rate of Inflation • Competitors • Tax and Promotions

  5. The Need for Economic Analysis • Economics frequently play a dominant role in the decision whether management/owner will invest in an energy savings/investment project or not • The communication of energy managers with the decision makers is very important in investment decisions. • The energy manager must present projects in economic terms in order to help the decion makers to make their decisions.

  6. The Need for Economic Analysis (continued) • There are various methods for economic evaluation of energy savings/investment projects. • There are many measures of project economic analysis, and many businesses and industries use their own methods or procedures to make their decisions. • The most commonly used economic evaluation methods in energy projects are:

  7. Economic Evaluation Methods • Investment profitability analysis • Annual Cost Method • Present worth method • Capitalized Cost Method

  8. Profitability Analysis Profitability analysis is concerned with the assesing feasibility of a new project from the point of view of its financial results.

  9. Most commonly used profitability methods are: • Internal rate of return (IRR) • Net present value (NPV) • Simple payback period (SPP) • Simple rate of return

  10. IRRand NPV methods IRRand NPV are discounted methods because they take into consideration the entire life of a project and the time factor by discounting the future inflows/savings and outflows to their present values

  11. Simple payback and simple rate of return Simple payback and simple rate of return are usually referred to as simple methods since they do not take into the whole life span of the project

  12. Annual cost method, present worth method and capitilized cost method Annual cost method, present worth method and capitilized cost method are three methods used to compare life time cost of alternative parameters. Life cycle costing (LCC) is important to help the designer/owner see the coupling between the initial cost and the long-term economic performance.

  13. Simple payback period • SPP does not take into the whole life span of the project. • Simple and easy use. • SPP is not an acceptable method for longer time periods.

  14. Example A lighting improvement costs $1000. The improvement saves $500 each year. What is the Simple Payback Period?

  15. SPP Example -- Solution A lighting improvement costs $1000. The improvement saves $500 each year. What is the Simple Payback Period?

  16. Example 2 • A lighting improvement costs $1000. The improvement saves $300 in the first year, $500 in the second year and $2000 in the third year. What is the Simple Payback Period?

  17. Time Value of Money • A dollar today worthmore than a dollar tomorrow because money has earning power. • The dollar today could be invested in a bank and earn interest so that it is worth more than a dollar tomorrow. • This relationship between interest and time is called the time value of money.

  18. Time Value of Money (cont.) • Time value of money should be considered by discounting the future inflows and outflows to their present values. • The fundamental approach to correctly account for cash inflows and outflows at different times is called discounted cash flow analysis.

  19. Net present value (NPV) • PV=Present value • FV=Future value • NPV=Net Present value • r= interest rate • N=Number of period • Impact of time on decision is time value of money

  20. A Time line 0 1 2 n FV PV FV= PV(1+r)n PV= FV/ (1+r)n

  21. Annuities concept Cash flow(savings from the retrofit) within the lifetime of the retrofit measure FV=C[(1+r)n……..+1]

  22. Example r =0.1

  23. Roxanne invested $500,000 in retrofitting measures 6 years ago. The ECMs was expected to pay $8,000 each month for the next 21 years (in excess of all costs). The annual cost of capital (or interest rate) for this type of business was 9%. What is the value of the business today?

  24. Assignment: deaDline 19/oct/2012 Austin needs to purchase a new heating/cooling system for his home. He is thinking about having a geothermal system installed, but he wants to know how long it will take to recoup the additional cost of the system. The geothermal system will cost $20,000. A conventional system will cost $7,000. Austin is eligible for a 30% tax credit to be applied immediately to the purchase. He estimates that he will save $1,500 per year in utility bills with the geothermal system. These cash outflows can be assumed to occur at the end of the year. The cost of capital (or interest rate) for Austin is 7%. How long will Austin have to use the system to justify the additional expense over the conventional model?( i.e, What is the DISCOUNTED payback period in years?. Also at interest rate of 8% what is the NPV of this project

  25. Life Cycle Costing • LCC is required to see the coupling between the initial cost and the long term economic performance. • An energy project life may exceed 20 years. • The value of annual operation expenses is related to the time these expenses occur. • Because of this, the concept present value (PV) must be utilized.

  26. Life Cycle Costing (continued) • Present Value or present worth (PW) is the value of sum of money at the present time that, with compound interest, will have a specified value at a certain time in the future. • Use Present Value (PV) analysis to find lowest life cycle cost (LCC)

  27. Life Cycle Costing (continued) • Need interest tables, a computer, or a calculator to find these PVs

  28. Life Cycle Costing (continued) • A good project has a Net Present Value (NPV) greater than zero NPV = PV (cash inflows/savings) - PV (cash outflows/costs) • The Internal Rate of Return (IRR) is the interest rate (I) at which the PV of the cash inflows/savings equals the PV of the costs (i.e., NPV = 0)

  29. Time Value of Money Analysis • S = the sum of money at the nth year. • i = Annual interestor discount rate • n = number of years of life of project • The present worth P of S dollars in nth year is

  30. The calculation method The term P/S=(1+i)-n is frequently referred to as single payment present worth factor (PWF)

  31. The calculation method (continued) • On many occasions equal amount of equal savings/expenses are required. • Use annual series present worth factor (P/A=SPWF) • Where A = annual savings/payment • P = A  [P/A, i, n] = A  [SPWF, i, n]

  32. Reading the Interest Tables • To find SPWF for i=10% and n=5 years: • Locate the 10% interest table • Locate the column “To find P given A”, (i.e., SPWF) • Locate the row for n=5 • At the intersection of this row and this column, read 3.7908

  33. Reading the Interest Tables - Examples • Find [P/A, 12%, 10] = 1-(1+0.12) -10/0.12=5.5602 • Find [P/A, 15%, 7] = 1-(1+0.15) -7/0.15=4.1604 • Find [A/P, 12%, 10] =1/ [1-(1+0.12) -10/0.12]= 0.1770Note that A/P =1/[P/A] • Find the present value of $1000 per year savings for 8 years at a discount rate of 10%. P = $1000 [P/A, 10% , 8] = $1000 [ 5.3349 ] = $5,334.90

  34. Economic Evaluation Example A combined heat and power DG (distributed generation) system costs $30,000 and saves $10,000 per year. The average life of the project is 7 years. At a discount rate of 10%, what is the NPV of this project? Is this a good project? NPV = PV (savings) – PV (cost) Solution: NPV = A  [P/A, I, N] - Cost NPV = $10,000  [P/A, 10%, 7] - $30,000 = $10,000  - $30,000 =

  35. Solution: NPV = A  [P/A, I, N] - Cost NPV = $10,000  [P/A, 10%, 7] - $30,000 = $10,000  4.8684 - $30,000 = $48,684 - $30,000 = $18,684 NPV > $0 so it is a good project

  36. Life Cycle Cost Example A Rhino air compressor costs $30,000 to buy and costs $15,000/year to operate over its 10-year life. An Elephant air compressor costs $40,000 to buy and costs $12,000/year to operate. Which air compressor has the lowest LCC at a 10% discount rate? LCC = PV(purchase cost) + PV(operating cost)

  37. Solution (skeleton) LCC Rhino = $30,000 + $15,000 [P/A, 10%, 10] = LCC Elephant = $40,000 + $12,000 [P/A, 10%, 10] =

  38. Solution (complete) LCC Rhino = $30,000 + $15,000 [P/A, 10%, 10] = $30,000 + $15,000 (6.1446) = $122,169 LCC Elephant = $40,000 + $12,000 [P/A, 10%, 10] = $40,000 + $12,000 (6.1446) = $113,735 The elephant air compressor has the lowest life cycle cost

  39. Three Basic Economic Problems • Find P given A, i, and n • Find A given P, i, and n • Find i given P, A, and n

  40. Example A facility presently has an old boiler and is considering installing a new boiler in its place. The new boiler will save the facility $5,000/year. How much can the facility pay for the new boiler and make a 12% rate of return if the new system lasts 10 years?

  41. Solution (skeleton) P = A [P/A, i, n]

  42. Solution (complete) P = A [P/A, i, n] = $5,000 [P/A, 12%, 10] = $5,000  (5.6502) = $28,251

  43. Example A facility purchases and installs a new chiller for $100,000. What annual savings is required to return 15% on this investment if the chiller lasts 10 years?

  44. Solution (skeleton) A = P[A/P, I, N] The A/P factor is called the Capital Recovery Factor.

  45. Solution (complete) A = P[A/P, i, n] = $100,000 [A/P, 15%, 10] = $100,000  0.1993 = $19,930/yr

  46. Example An equipment sales company offers your facility a complete “turn-key” installation of a motor retrofit for $50,000, and says it will save you $9,225 per year. If the system lifetime is 12 years, what rate of return (IRR) will your facility make if the estimated savings is correct?

  47. Solution (skeleton) P = A [P/A, i, n] Solve for the P/A factor, and then look through the interest tables – one by one – until you find the page that your P/A factor is on. Then, IRR = the interest rate on that page. $50,000 = $9,225 [P/A, IRR, 12] [P/A, IRR, 12] = IRR from table =

  48. Solution (complete) P = A [P/A, i, n] $50,000 = $9,225 [P/A, IRR, 12]Scan through the tables to find this P/A(SPWF) factor (5.42005) at the intersection of the P/A column, and the n =12 row. The closest number found is 5.4206, and it is on the table for 15% interest rate. Since this is an extremely close number to our desired value of 5.42005, we accept it as close enough; so IRR from table = 15%

  49. Solution by using a spreadsheet program (such as Microsoft Excel)

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