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COMPUTER APPLICATIONS IN HVAC SYSTEM LIFE CYCLE COSTING

Fundamental Principals and Methods For Analyzing and Justifying System Installation and Upgrade Costs

Communication White Paper by: Craig J. Gann, P.E. Carrier Corporation, Syracuse, NY

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Computer Applications in HVAC System Life Cycle Costing

by Craig J. Gann, P.E.

SUMMARY

This paper first explores the fundamentals of Life Cycle Cost Analysis, familiarizes the reader with commonly-used terminology and concepts, then demonstrates computerized applications to perform an actual Life Cycle Cost Analysis example. The methodology used encompasses the concepts of Net Present Value and Internal Rate of Return. Emphasis is placed on HVAC equipment, however the fundamental concepts discussed herein are applicable to any building component under consideration.

INTRODUCTION

With the strong trend in the HVAC industry emphasizing energy savings, there is an equally powerful force, primarily owner-driven, which dictates that installed costs of new and replacement systems be as economical as possible. There is a delicate balance between these two influences since reducing energy consumption by installing higher-efficiency equipment often must be achieved at the cost of sacrificing installed first-cost savings. The owner or owner's agent must carefully analyze various systems and installations to determine those with the most economical overall cost, which involves simultaneously analyzing future operating and maintenance costs against the initial first-cost purchase price of the equipment. There are many types of software programs available to analyze both total building energy consumption and to predict annual operating costs. Using a simple baseline system, an analyst may compare various system alternatives, that is chilled water verses direct expansion, air-cooled verses water-cooled or other parameters to determine the most energy efficient system for a given installation. Once the annual operating costs have been determined, the next logical step is to obtain first-cost data for the proposed equipment and installation and maintenance cost data. All data is then input into a software program and a Life Cycle Cost Analysis is performed to determine the most cost effective system.

FUNDAMENTAL TERMINOLOGY & CONCEPTS

Life Cycle Cost Analysis: An economic technique used to compare various design alternatives by projecting (discounting or compounding) associated costs over the economic life of the project, (also called the "Life Cycle Period"), to a common period of time. All costs which may occur at various times are considered, such as first cost, installation costs, maintenance costs and any additional expenditures such as replacement of components. Bringing all of these costs, which

occur at different times, to a common point in time allows the analyst to compare various design alternatives and select the alternative with the lowest total cost. Life Cycle Period: The time of consideration over which the economic analysis will be performed, usually corresponds to the estimated useful life of the equipment. Minimum Attractive Rate of Return (MARR): A building owner typically makes capital budgeting decisions based on a fixed amount of available capital. This fixed amount of available capital is competed for by various alternative investments. For example, should the owner invest in higher-efficient HVAC equipment or upgrade the building lighting system to higher-efficient ballasts and light bulbs, or should the building owner take this same amount of capital and invest it into marketable securities such as stocks or bonds? The building owner must decide the Minimum Attractive Rate of Return for any proposed capital expenditure. Proposals which meet or exceed this minimum required return are deemed acceptable. Discount Rate (d): Interest rate used to determine the net present worth of future cash flows. Net Present Worth (NPW): A method used to evaluate investment decisions whereby all future cash flows are discounted back to the beginning of the analysis period. The discount rate (factor) used is the minimum attractive rate of return. If the net present worth is positive, the proposal's forecasted return exceeds the minimum attractive rate of return and the proposal is acceptable. The proposal with the highest net present worth is the most attractive alternative. Internal Rate of Return (IRR): This is the discount rate at which the present value of a future stream of cash flows of an investment are equal to the first cost of the investment. This value is determined by a trial-and-error method by setting the net present value of cash outflows (cost of the investment) and cash inflows (returns on the investment) equal to zero. This discount rate is the Internal Rate of Return. Capital Investment Costs: Expenditures made for assets such as first-cost of equipment, replacement cost of equipment, buildings or other items held by a company. These costs may be either financed or paid for with cash. Non-capital Investment Costs: These are costs associated with non-capital expenditures such as labor, replacement parts, maintenance supplies and utility company rebates. Operating Costs: Costs incurred to operate equipment, which include energy, maintenance labor, insurance and taxes.

TIME VALUE OF MONEY

The concept of money having a time value is fundamental to understanding any economic or financial analysis of alternative investments. Money has a time value since it can be invested at some particular interest rate and will grow in value over time.

FUTURE VALUE (FV) If an investor places $100 in a bank account which pays 7% interest, compounded yearly, at the end of the first year the investor will have accumulated $107, that is $7.00 interest plus the original principal of $100. If the investor leaves his or her money in the account for another year, the

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account balance will grow to $114.49. An additional $0.49 has accrued over and above the first year's interest because the account has accumulated (compounded) interest on interest. To compute the future value (FV) of any investment amount (P) at an interest rate (i) over (N) number of years, the following formula may be used:

FV = P (1 + i ) N

For our above example using (Eq. 1) above:

(Eq. 1)

FV = 100(1 + .07)2 = 100(1.1449) = $114.49

It is helpful to draw a time line (Fig. 1) when analyzing time value of money problems. This gives you a visual tool which makes it easier to see the expenditures as they occur.

PV Periods FV

0

1

2

$107 $114.49

where:

PV = Present Value; FV = Future Value

Figure 1. Future Value of a Single Payment

The user must be careful when using (Eq. 1) to ensure that the interest rate (i) used corresponds to the number of times the interest is compounded per year. The annual interest rate is adjusted by dividing it by the number of compounding periods per year. The exponent (N) is adjusted by multiplying it by the same number of compounding periods per year. For our previous example, suppose the interest rate quoted was 7%, compounded monthly rather than compounded annually as before. The interest rate used would be (7% /12), and the exponent (N) would be (N x 12). Plugging these values into (Eq. 1) yields:

FV = 100(1 + .07/12)2 x 12 = 100(1 + .0058)24 = 100(1.1498) = $114.98

In many instances, investments are made more often than just at the beginning of the analysis period. Sometimes investors deposit periodic payments into an account which greatly adds to the effect of compounding. When the same amount of money is paid, or received, periodically it is referred to as an annuity. When the money is invested or received at the end of the period it is referred to as an ordinary annuity. To calculate the Future Value of an ordinary annuity (A), the following formula may be used:

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(1 + i ) N - 1 FV = A i

(Eq. 2)

For example, lets assume you deposit $100 at the end of each year into a savings account paying 7%, compounded annually. How much money would you accumulate at the end of 5 years? Using our ordinary annuity formula:

(1+.07) 5 - 1 .4026 FV = 100 = 100 .07 = $575.07 .07

As before, it is easier to visualize these expenditures with a time line as shown in Figure 2:

PV

P e r io d s

FV

0

1 100

2 100

3 100

4 100

5 1 0 0 ( 1 + .0 7 ) 0 = $ 1 0 0 .0 0 1 0 0 ( 1 + .0 7 ) 1 = $ 1 0 7 .0 0 1 0 0 ( 1 + .0 7 ) 2 = $ 1 1 4 .4 9 1 0 0 ( 1 + .0 7 ) 3 = $ 1 2 2 .5 0 1 0 0 ( 1 + .0 7 ) 4 = $ 1 3 1 .0 8 $ 5 7 5 .0 7

Figure 2. Future Value of an Annuity

PRESENT VALUE (PV) There is another method of calculating the time value of money, similar to the future value method outlined previously, which is called the Present Value. This is especially useful for calculating the amount of money which must be set aside today in order to accumulate a certain desired amount in the future (FV). For example, let's assume you wanted to accumulate $20,000 to help pay for your child's college tuition. Let's also assume that your child is 10 years away from attending college. The bank is offering you a CD (Certificate of Deposit) which pays 6.5%, compounded annually. How much money would you need to deposit today at 6.5% interest to accumulate $20,000 in 10 years? Substituting (PV) for (P) in (Eq. 1) and solving for the Present Value (PV):

1 PV = FV N (1 + i )

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(Eq. 3)

Substituting the information from the example above into (Eq. 3):

1 PV = $20,000 10 = $20,000[.5327] = $10,654.52 (1+.065)

In other words you would need to deposit slightly over $10,600 today if you want it to grow to $20,000 in ten years. Let's look at another example. Suppose there are a future series of payments which will be made at a regular time interval. For example, let's assume a building maintenance contract has been sold and costs the building owner $500/month. The owner wishes to determine the present value (PV) of all of these future payments for the next 5 years, assuming an interest rate of 8%, compounded annually. A formula similar to the future value of an ordinary annuity formula is the Present Value of an ordinary annuity (A):

1 1 - (1 + i ) N PV = A i

(Eq. 4)

Since there are 12 months per year and 5 years, (N = 12 x 5 = 60). Also, since the payments are monthly, the interest rate must be expressed as a monthly rate, i.e. .08/12. Substituting the data from the above example into (Eq. 4):

1 - PV = 500

.08 (1 + 12 ) .3288 = 500 .0067 = $24,659.22 .08 12

60

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INTERNAL RATE OF RETURN (IRR) The actual return on any investment is calculated by computing the equivalent interest rate that will make the present value of all cash flows (inflows & outflows) from the investment equal to the investment cost. This equivalent interest rate is called the Internal Rate of Return (IRR). It is calculated manually by trial-and-error by setting the present value of all future cash flows equal to the investment amount as the following formula illustrates:

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p=

C1 C2 C3 CN + + +..... + 1 2 3 (1 + i ) (1 + i ) (1 + i ) (1 + i ) N

CN = cash flow amount in year N N = number of years p = investment amount i = internal rate of return

(Eq. 5)

where:

For this example, let's assume a building owner is considering adding an outdoor air economizer to a rooftop HVAC unit to save energy. From a previously conducted energy analysis estimate, the following cost data has been collected: Annual Energy Cost without economizer $926 $973 $1021 $1072 $1125 $5117 Annual Energy Cost with economizer $648 $681 $714 $750 $787 $3580 Annual Energy Savings $278 $292 $307 $322 $338 $1537

Year 1 2 3 4 5 Total

Table 1. Annual Energy Cost Data For Example In addition, we will assume that the total cost to purchase and install the economizer is $1200.00, and that the building owner can borrow money from the bank at 7.5%. Since we now know all of the variables in (Eq. 5) except for i, we must perform a trial-and-error calculation assuming a value of i that will make the cost of the economizer and the present value of all of the annual energy cost savings equal. Let's first assume i = 10%:

p = $1200 =

$278 $292 $307 $322 $338 + + + + 1 2 3 4 (1+.10) (1+.10) (1+.10) (1+.10) (1+.10) 5

= 252. 73 + 241. 32 + 230. 65 + 219. 93 + 209. 87 = $1154. 50

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Since the calculated value of p is not equal to $1,200, we must assume a new value for i. Let's assume i = 8%:

p = $1200 =

$278 $292 $307 $322 $338 + + + + 1 2 3 4 (1+. 08) (1+. 08) (1+. 08) (1+. 08) (1+. 08) 5

= 257. 41 + 250. 34 + 243. 71 + 236. 68 + 230. 04 = $1218.18

Now, the calculated value of p is close to $1,200, but not exactly equal. We now know that the actual value of i, which makes the annual savings equal to the investment amount, is somewhere between 8 and 10%. By interpolation, the correct value of i may be determined as follows:

8% i 10%

------> $1,218.18 ------> $1,200.00 ----> ------> $1,154.50 i = 8.57%

Once the actual internal rate of return (IRR) is known, it should be compared to the minimum attractive rate of return (MARR). If the IRR is greater than the MARR, the proposal is feasible. In this case since the bank lending rate is 7.5%, the economizer is a good investment.

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COMPUTER APPLICATIONS

Now that the reader has a basic understanding of the fundamental concepts and terminology used to calculate various cash flow alternatives, the procedure may be greatly simplified by the use of readily available software programs. In addition, computer software allows the analyst to perform much more complex analyses with consideration of variables such as cost escalation rates, depreciation, taxes and insurance. The following example will illustrate the use of a software program to analyze a more complex example. Chiller Comparison Example A consulting engineer requires a new 200-ton (703 kW) chiller for a new building being constructed. A decision has been made to utilize a chilled-water system with central station air handling units. The engineer wishes to analyze three possible types of chillers to determine the machine which yields the best ROI (return on investment) for the owner. The engineer is considering three alternatives: an air-cooled reciprocating chiller, a water-cooled reciprocating chiller and a water-cooled centrifugal chiller. The building owner has established a MARR (minimum attractive rate of return) of 10% for the analysis. To simplify the analysis, an assumption has been made to consider only the purchase cost of the chiller, installation cost, operating and maintenance costs for the chiller, cooling tower, pumps and chilled-water piping. The remainder of the system components will be assumed to be the same for each chiller, i.e. air handling units, ductwork, air terminals, heating equipment and electrical service. In addition, for simplicity, salvage value, insurance and property taxes will be neglected. The engineer has compiled all of the cost data for all three alternatives in the following table:

Cost Component

Purchase Costs: 1. Chiller 2. Cooling Tower 3. CW Pump & Piping 4. Cond. Water Pump & Piping Installation Costs: 5. Chiller 6. Cooling Tower 7. CW Pump & Piping 8. Cond. Water Pump & Piping Total Initial Investment Energy Costs: 9. Electrical Energy Maintenance Costs: 10. Chiller 11. Cooling Tower 12. Pump & Water Treatment Total Operating Cost / Yr.

W/C Centrif.

$67,950 $12,320 $ 4,410 $ 3,675 $23,730 $ 1,340 $ 2,520 $ 2,100 $118,045 $12,310 $ 3,000 $ 1,200 $ 2,005 $18,515

W/C Recip.

$58,410 $12,320 $ 4,410 $ 3,675 $11,870 $ 1,340 $ 2,520 $ 2,100 $96,645 $13,920 $ 3,000 $ 1,200 $ 2,005 $20,125

A/C Recip.

$81,620 $0 $ 4,410 $0 $14,300 $0 $2,520 $0 $102,850 $15,770 $ 2,500 $0 $ 775 $19,045

Table 2. Cost Data for Chiller Comparison Example

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Explanation of Miscellaneous Data 1. Purchase costs for all equipment (items 1-4) and installation labor (items 5-8) were obtained from Means Mechanical Cost Data, 19th Edition, 1996, R.S. Means Company, Inc., Construction Consultants and Publishers, Kingston, MA, USA. The Means handbook contains useful cost estimating data derived from surveys of the construction industry in the United States. This cost data may not be reflective of your local conditions, therefore all cost data for any actual job analysis should be obtained locally. Energy cost data for item 9 was obtained from energy simulations run with the Carrier HAP (Hourly Analysis Program) for a sample office building in Atlanta, Georgia, USA. In these simulations, the following full-load chiller efficiencies were assumed: Water-cooled centrifugal: Water-cooled reciprocating: Air-cooled reciprocating: 3. 4. 0.60 kW/ton (0.17 kW/kW) 1.12 kW/ton (0.32 kW/kW) 1.20 kW/ton (0.34 kW/kW)

2.

The purchase costs (items 1-4) will be classified as capital investment costs because these are the costs of assets owned by the firm. These assets will be subject to depreciation. The installation labor costs (items 5-8) will be classified as non-capital investment costs. The installation cost is part of the cost of investment of equipment. However, these costs are not for the acquisition of assets, nor are they subject to depreciation, therefore, they will be non-capital investments. The remaining costs (items 9-12) will be classified as operating costs. These costs are required each year to keep the equipment operating. For entry into the computer program, items 10-12 will be added together to form a single maintenance cost item. All equipment purchase and installation costs will be cash investments, i.e. non-financed, and will occur at the start of the analysis period. They will be entered as occurring at the end of year zero which is equivalent to the beginning of year 1. Operation of the equipment begins in year 1, therefore the first energy and maintenance costs will occur at the end of year 1. Capital investments will be depreciated using the straight-line-method with a depreciation life of 31 years which is allowable under 1991 U.S. business tax regulations. Maintenance costs are assumed to increase at an average rate of 5% per year, and electrical energy costs are assumed to increase at an average rate of 6% per year. The incremental income tax rate for this company will be assumed to be 40%. For simplicity, salvage value, insurance costs and property taxes will be neglected.

5.

6.

7. 8. 9. 10. 11.

The next four pages contain computer printout data of the design case inputs as well as the Life Cycle Summary for all three design alternatives.

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LIFE CYCLE DESIGN CASE INPUTS Design Case : W/C Centrifugal Chiller 09-26-96 Prepared By : Carrier Corporation 6013192200 Engineering Economic Analysis v2.00 Page 1 of 1 ************************************************************************* 1. DESIGN CASE DATA ------------------------------------------------------------------------Design Case Name...............: W/C Centrifugal Chiller Currency Symbol................: $ Length of Analysis.............: 25 years First Year of Operation........: 1 ************************************************************************* 2. CAPITAL INVESTMENT COSTS ------------------------------------------------------------------------Cost Salvage L DNP P D T I E D DL Investment Name $ $ Yr % P Yr Yr % % M Yr ------------------------------------------------------------------------Chiller purchase 67950 0 25 100 1 0 1 0.00 0.00 2 31 Cooling Tower Purchase 12320 0 25 100 1 0 1 0.00 0.00 2 31 Chilled Water Pump & Pipe 4410 0 25 100 1 0 1 0.00 0.00 2 31 Cond. Water Pump & Piping 3675 0 25 100 1 0 1 0.00 0.00 2 31 ------------------------------------------------------------------------L = Useful Life E = Investment Escalation Rate DNP=% Downpayment DM= Depreciation Method PP= Payment Plan 1-No Depreciation 1-Equal Payments 2-Straight Line 2-Interest Only Payments 3-Sum-of-Years-Digits 3-Equal Principal Payments 4-Double Declining Balance D = Date 5-Dbl Declining Balance to Straight Line T = Term of Loan 6-Modified Accelerated Cost Recovery System I = Loan Interest Rate DL= Depreciation Life ************************************************************************* 3. NON-CAPITAL INVESTMENT COSTS ------------------------------------------------------------------------Benefit Esc. Tax or Cost Value Rate Exempt Name (B/C) $ Date Number Period (%) Status ------------------------------------------------------------------------Chiller installation cost C 23730 0 1 1 0.00 N Clg. Tower Installation C 1340 0 1 1 0.00 N CW Pump & Piping Install C 2520 0 1 1 0.00 N Cond Water Pump/Pipe Inst C 2100 0 1 1 0.00 N ************************************************************************* 4. OPERATING COSTS ------------------------------------------------------------------------Cost Escalation Cost Component $ Rate (%) -------------------------------------------------Maintenance...............: 6205 5.00 Energy: Electric..........: 12310 6.00 Energy: Natural Gas.......: 0 0.00 Energy: Fuel Oil..........: 0 0.00 Energy: Propane...........: 0 0.00 -------------------------------------------------Rate Escalation Cost Component (%) Rate (%) -------------------------------------------------Insurance.................: 0.00 0.00 Property Tax..............: 0.00 0.00 ************************************************************************* LIFE CYCLE DESIGN CASE INPUTS

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Design Case : W/C Reciprocating Chiller 09-26-96 Prepared By : Carrier Corporation 6013192200 Engineering Economic Analysis v2.00 Page 1 of 1 ************************************************************************* 1. DESIGN CASE DATA ------------------------------------------------------------------------Design Case Name...............: W/C Reciprocating Chiller Currency Symbol................: $ Length of Analysis.............: 25 years First Year of Operation........: 1 ************************************************************************* 2. CAPITAL INVESTMENT COSTS ------------------------------------------------------------------------Cost Salvage L DNP P D T I E D DL Investment Name $ $ Yr % P Yr Yr % % M Yr ------------------------------------------------------------------------Chiller purchase 58410 0 25 100 1 0 1 0.00 0.00 2 31 Cooling Tower Purchase 12320 0 25 100 1 0 1 0.00 0.00 2 31 Chilled Water Pump & Pipe 4410 0 25 100 1 0 1 0.00 0.00 2 31 Cond. Water Pump & Piping 3675 0 25 100 1 0 1 0.00 0.00 2 31 ------------------------------------------------------------------------L = Useful Life E = Investment Escalation Rate DNP=% Downpayment DM= Depreciation Method PP= Payment Plan 1-No Depreciation 1-Equal Payments 2-Straight Line 2-Interest Only Payments 3-Sum-of-Years-Digits 3-Equal Principal Payments 4-Double Declining Balance D = Date 5-Dbl Declining Balance to Straight Line T = Term of Loan 6-Modified Accelerated Cost Recovery System I = Loan Interest Rate DL= Depreciation Life ************************************************************************* 3. NON-CAPITAL INVESTMENT COSTS ------------------------------------------------------------------------Benefit Esc. Tax or Cost Value Rate Exempt Name (B/C) $ Date Number Period (%) Status ------------------------------------------------------------------------Chiller installation cost C 11870 0 1 1 0.00 N Clg. Tower Installation C 1340 0 1 1 0.00 N CW Pump & Piping Install C 2520 0 1 1 0.00 N Cond Water Pump/Pipe Inst C 2100 0 1 1 0.00 N ************************************************************************* 4. OPERATING COSTS ------------------------------------------------------------------------Cost Escalation Cost Component $ Rate (%) -------------------------------------------------Maintenance...............: 6205 5.00 Energy: Electric..........: 13920 6.00 Energy: Natural Gas.......: 0 0.00 Energy: Fuel Oil..........: 0 0.00 Energy: Propane...........: 0 0.00 -------------------------------------------------Rate Escalation Cost Component (%) Rate (%) -------------------------------------------------Insurance.................: 0.00 0.00 Property Tax..............: 0.00 0.00 *************************************************************************

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LIFE CYCLE DESIGN CASE INPUTS Design Case : A/C Reciprocating Chiller 09-26-96 Prepared By : Carrier Corporation 6013192200 Engineering Economic Analysis v2.00 Page 1 of 1 ************************************************************************* 1. DESIGN CASE DATA ------------------------------------------------------------------------Design Case Name...............: A/C Reciprocating Chiller Currency Symbol................: $ Length of Analysis.............: 25 years First Year of Operation........: 1 ************************************************************************* 2. CAPITAL INVESTMENT COSTS ------------------------------------------------------------------------Cost Salvage L DNP P D T I E D DL Investment Name $ $ Yr % P Yr Yr % % M Yr ------------------------------------------------------------------------Chiller purchase 81620 0 25 100 1 0 1 0.00 0.00 2 31 Chilled Water Pump & Pipe 4410 0 25 100 1 0 1 0.00 0.00 2 31 ------------------------------------------------------------------------L = Useful Life E = Investment Escalation Rate DNP=% Downpayment DM= Depreciation Method PP= Payment Plan 1-No Depreciation 1-Equal Payments 2-Straight Line 2-Interest Only Payments 3-Sum-of-Years-Digits 3-Equal Principal Payments 4-Double Declining Balance D = Date 5-Dbl Declining Balance to Straight Line T = Term of Loan 6-Modified Accelerated Cost Recovery System I = Loan Interest Rate DL= Depreciation Life ************************************************************************* 3. NON-CAPITAL INVESTMENT COSTS ------------------------------------------------------------------------Benefit Esc. Tax or Cost Value Rate Exempt Name (B/C) $ Date Number Period (%) Status ------------------------------------------------------------------------Chiller installation cost C 14300 0 1 1 0.00 N CW Pump/Piping Install C 2520 0 1 1 0.00 N ************************************************************************* 4. OPERATING COSTS ------------------------------------------------------------------------Cost Escalation Cost Component $ Rate (%) -------------------------------------------------Maintenance...............: 3275 5.00 Energy: Electric..........: 15770 6.00 Energy: Natural Gas.......: 0 0.00 Energy: Fuel Oil..........: 0 0.00 Energy: Propane...........: 0 0.00 -------------------------------------------------Rate Escalation Cost Component (%) Rate (%) -------------------------------------------------Insurance.................: 0.00 0.00 Property Tax..............: 0.00 0.00 *************************************************************************

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LIFE CYCLE SUMMARY Job Name: Chiller Life Cycle Study 09-26-96 Prepared By: Carrier Corporation 6013192200 Engineering Economic Analysis v2.00 Page 1 ************************************************************************* Life Cycle Period.......................................: 25 yrs Minimum Attractive Rate of Return.......................: 10.0 % Income Tax Rate.........................................: 40.0 % ************************************************************************* TABLE 1. EXECUTIVE SUMMARY: ------------------------------------------------------------------------Economic Criteria Most Favorable Design Alternative ------------------------------------------------------------------------1. Internal Rate of Return and NPW..( 3) A/C Reciprocating Chiller 2. Lowest total present worth.......( 3) A/C Reciprocating Chiller 3. Lowest first cost................( 2) W/C Reciprocating Chiller ------------------------------------------------------------------------TABLE 2. DESIGN CASE DATA ------------------------------------------------------------------------After Tax Before Tax First Cost Total PW Total PW Design Case ($) ($) ($) ------------------------------------------------------------------------( 2) W/C Reciprocating Chiller 96,645 267,684 408,982 ( 3) A/C Reciprocating Chiller 102,850 265,832 402,493 ( 1) W/C Centrifugal Chiller 118,045 267,764 404,617 TABLE 3. COMPARATIVE ANALYSIS DATA: ------------------------------------------------------------------------First Net Pres. Before After Cost Worth Tax Tax Payback Comparison of Difference Savings IRR IRR MARR Period Case vs Case ($) ($) (%) (%) (%) (years) ------------------------------------------------------------------------3 vs 2 6,205 1,852 20.6 13.1 10.0 8 1 vs 3 15,195 -1,932 8.7 8.2 10.0 N/A ------------------------------------------------------------------------N/A = Not available because value could not be calculated. TABLE 4. COMPARATIVE DECISION DATA: ------------------------------------------------------------------------Comparison of Case vs Case Winner Reason ------------------------------------------------------------------------3 vs 2 3 (IRR>MARR) After tax internal rate of return is greater than MARR. 1 vs 3 3 (IRR=<MARR) After tax internal rate of return is equal to or less than MARR. -------------------------------------------------------------------------

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Life Cycle Summary Discussion In Table 1 of the Life Cycle Summary, we see that the Air-Cooled Reciprocating case was the overall winner of the incremental rate of return analysis. This means the Air-Cooled Reciprocating case had the largest investment cost that could be justified. In subsequent tables the reason for this conclusion will be documented. The Air-Cooled Reciprocating case also had the lowest total present worth (TPW), while the Water-Cooled Reciprocating case had the lowest first cost. Table 2 provides key information for each design case. Design cases in this table are listed in order of increasing first cost. "First Cost" is the sum of capital and non-capital investment costs occurring in year zero. The Water-Cooled Reciprocating case has the lowest first cost. Examining Table 2, we can see this is because the Water-Cooled Reciprocating case has the lowest chiller purchase and installation costs. Even though the Air-Cooled Reciprocating case does not require a cooling tower or condenser water pumps or piping, its first cost is higher because of the large chiller purchase and installation costs. The Water-Cooled Centrifugal case has the highest overall first cost. In spite of the fact that the Water-Cooled Reciprocating case has the lowest first cost, it does not have the lowest TPW. This is due to the fact that its annual operating costs are larger than the other two cases, and over the 25-year analysis period this offsets the initial investment cost advantage. Note that in this analysis present worth values are computed using a discount rate equal to the minimum attractive rate of return (MARR), which is 10% for this example. Table 3 lists the key final results from the case-by-case economic comparisons. The incremental internal rate of return (IRR) analysis used systematically compares successive pairs of design cases. For the first comparison, the baseline case is the design case with the lowest first cost, which in this analysis is the Water-Cooled Reciprocating case. The alternative case is the design case ranked next in order of first cost, which is the Air-Cooled Reciprocating case. Comparing these cases, we see the Air-Cooled Reciprocating case requires an additional $6,205 in first cost investment. The goal of the rate of return analysis will be to determine if a sufficiently high rate of return can be achieved to justify this investment. Next, the net present worth (NPW) savings is $1,852. A positive NPW value indicates a savings at the discount rate of 10% for this study. This usually means that the internal rate of return (IRR) for the comparison will be greater than the MARR (10%). In fact it is, at 13.1%, after taxes. Therefore, the additional first cost difference of $6,205 is justified and the Air-Cooled Reciprocating case is the "winner" between the two cases. A payback period for this comparison is eight years. Payback is calculated using present worth cash flows calculated in this analysis using a discount rate of 10% (the MARR). The eight year payback period means that during the 8th year of the analysis, cumulative operating cost savings surpass the additional investment cost required by the alternative case. As the winner of comparison #1, the Air-Cooled Reciprocating case now becomes the baseline case for comparison #2. The next case in the first cost rankings will be the alternative case, which is the Water-Cooled Centrifugal case. For this comparison the Water-Cooled Centrifugal case requires an additional $15,195 initial investment over the baseline case. The comparison will evaluate whether this additional investment can be justified by a high enough rate of return. The NPW for this comparison is -$1,932. A negative value indicates a loss rather than a savings at the discount rate of 10% for this study. This is an indication that the return on the additional investment will be lower than 10%. In fact, the after tax IRR is 8.2% which is less than the 10%

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MARR. Therefore, the extra investment ($15,195) required for the Water-Cooled Centrifugal case cannot be justified and the Air-Cooled Reciprocating case is the overall winner. Note that the Centrifugal case did yield a positive IRR of 8.2%, however this was below the MARR for the firm which was 10%. Finally, a payback period for comparison #2 cannot be computed because of the fact that over the 25-year study period, the operating cost savings are not sufficient to offset the additional investment cost, therefore, no payback occurs for this alternative. To conclude, Table 4 on the printout provides supporting documentation for the decisions made in each life cycle comparison. This table lists the alternative and baseline cases, the winner and the reason for choosing the winner for each comparison.

WHERE TO GO FOR MORE INFORMATION The software programs utilized to perform this computer analysis, (Engineering Economic Analysis and Hourly Analysis Program), are available by contacting your local Carrier sales office. You may also call the software systems hotline in Syracuse, NY at 1-800-253-1794 or log-on to Carrier's Commercial Systems & Services Web site at http://www.Carrier-Commercial.com for more information.

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BIBLIOGRAPHY The Handbook of Fixed Income Securities, Third Edition, Frank J. Fabozzi; Business One Irwin, Homewood, IL, USA, 1991; pp. 60-80 Financial Decision Making, Fourth Edition, John J. Hampton; Prentice Hall, Englewood Cliffs, NJ, USA; 1989; pp. 325-29 Barron's Dictionary of Finance and Investment Terms, Third Edition, John Downes & Jordan Elliot Goodman; Barron's Educational Series, Inc., Hauppauge, NY, USA, 1991 Engineering Economic Analysis User's Manual, Carrier Corporation, Syracuse, NY, USA; 1992, pp. 5-1 to 5-10 Hourly Analysis Program User's Manual, Carrier Corporation, Syracuse, NY, USA; 1995 Mean's Mechanical Cost Data - 19th Annual Edition, Copyright 1995, R.S. Means Co., Inc., Kingston, MA

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