1. E CONOMIC A NALYSIS AND E CONOMIC D ECISIONS FOR E NERGY 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. E CONOMIC F ACTORS Capital cost of the technology Long term debt availability Capacity Risks and uncertainties Time Rate of Inflation Competitors Tax and Promotions

5. T HE N EED FOR E CONOMIC A NALYSIS 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. T HE N EED FOR E CONOMIC A NALYSIS ( 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. E CONOMIC E VALUATION M ETHODS Investment profitability analysis Annual Cost Method Present worth method Capitalized Cost Method

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

9. M OST COMMONLY USED PROFITABILITY METHODS ARE : Internal rate of return (IRR) Net present value (NPV) Simple payback period (SPP) Simple rate of return

10. IRR AND NPV METHODS IRR and 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. S IMPLE 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. A NNUAL 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. S IMPLE 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. E XAMPLE A lighting improvement costs $1000. The improvement saves $500 each year. What is the Simple Payback Period?

15. SPP E XAMPLE -- S OLUTION A lighting improvement costs $1000. The improvement saves $500 each year. What is the Simple Payback Period?

16. E XAMPLE 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? YearCash Flow Years from Today : n 0-$1000 1$300 2$500 3$2000

17. T IME V ALUE OF M ONEY A dollar today worth more 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. T IME V ALUE OF M ONEY ( 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. N ET 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 PV FV 12n FV= PV(1+r) n PV= FV/ (1+r) n

21. A NNUITIES CONCEPT YearCash FlowYear to End:nFuture Value 0030 1C2C(1+r) 2 2C1C(1+r) 1 3C0C Cash flow(savings from the retrofit) within the lifetime of the retrofit measure FV=C[(1+r) n ……..+1]

22. E XAMPLE YearsCash flowYears to Discount:n Present value 0-$10000 1$13201 2$14522 NPV 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. A SSIGNMENT : DEA D LINE 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. L IFE C YCLE C OSTING 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. L IFE C YCLE C OSTING ( 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. L IFE C YCLE C OSTING ( CONTINUED ) Need interest tables, a computer, or a calculator to find these PVs

28. L IFE C YCLE C OSTING ( 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. T IME V ALUE OF M ONEY A NALYSIS S = the sum of money at the n th year. i = Annual interest or discount rate n = number of years of life of project The present worth P of S dollars in n th year is

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

31. T HE 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. R EADING THE I NTEREST T ABLES 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 Values of (PWF) and (SPWF) at a compound interest of 10% Year, nPWFSPWF1/PWF 10.9091 1.1000 20.82641.73551.2100 30.75132.48691.3310 40.68303.16991.4641 50.62093.79081.6105 60.56454.35531.7716

33. R EADING THE I NTEREST T ABLES - E XAMPLES 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]= 0.1770 Note 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. E CONOMIC E VALUATION E XAMPLE 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. L IFE C YCLE C OST E XAMPLE 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. S OLUTION ( SKELETON ) LCC Rhino = $30,000 + $15,000 [P/A, 10%, 10] = LCC Elephant = $40,000 + $12,000 [P/A, 10%, 10] =

38. S OLUTION ( 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. T HREE B ASIC E CONOMIC P ROBLEMS Find P given A, i, and n Find A given P, i, and n Find i given P, A, and n

40. E XAMPLE 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. S OLUTION ( SKELETON ) P = A [P/A, i, n ]

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

43. E XAMPLE 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. S OLUTION ( SKELETON ) A = P[A/P, I, N] The A/P factor is called the Capital Recovery Factor.

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

46. E XAMPLE 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. S OLUTION ( 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. S OLUTION ( 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. S OLUTION BY USING A SPREADSHEET PROGRAM ( SUCH AS M ICROSOFT E XCEL ) Investment SAVINGS DURING YEARS 123456789101112 -$50,000 $9,225 IRR=15.0024%

54. A PPENDIX FOR E CONOMIC A NALYSIS This Appendix contains additional economic analysis examples of potential energy projects.

55. A DDITIONAL S OLVED E CONOMIC E XAMPLES Here is a group of additional examples to practice on, and to illustrate more opportunities for energy savings projects. A solution is provided for each of these examples. Each of these examples can also be worked out using the Ten Step Economic Spreadsheet provided.

56. B OILER E CONOMIZER E XAMPLE A boiler economizer will cost $20,000 installed, and will last for five years. How much will it have to save each year to return 12%? Here, P = $20,000, i = 12%, n = 5, A = ? A = P  [A/P, i, n ] = $20,000  [A/P, 12%, 5] = $20,000  [0.2774] = $5548

57. C ASE S TUDY – D ISTRIBUTED G ENERATION A company is investigating the possibility of building a distributed generation (DG) plant with an initial investment cost of $400,000 that will save $60,000 a year in lost production and reduced energy cost. This DG plant has an anticipated life of 20 years and requires an overhaul every 10 years of operation costing $30,000. Conduct a thorough analysis (both Present Value and IRR) to determine whether the investment is a wise one or not. The cost of capital is 15% and salvage value of the plant at the end of the year 20 is $40,000.

58. C ASE S TUDY DG1 - C ONCLUSIONS Is this a wise investment? Explain? At what MARR does this project look attractive or does this project never look attractive?