final life cycle costing 2020 - ashrae library/professional...the lowest life cycle cost alternative...

Life Cycle Costing AnalysisMark Stetz, P.E., [email protected]

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Course Outline

1. Introduction to Engineering Economics2. Tools and Methods of Life Cycle Costing3. Principles of Life Cycle Costing4. Uncertainty5. Microsoft Excel™ Financial Functions6. Course Summary


Learning Objectives

Explain the principles of engineering economics Describe the tools and methods of life cycle costing Explain the principles of life cycle costing Provide in-class problems for participants to solve Explain the principle of uncertainty Describe the financial functions of Microsoft Excel

spreadsheet


Section 1: Introduction to Engineering Economics

Principles of engineering economics Definitions Cash Flow diagrams Using life cycle costing during the design process Class problem


Money Has Time Value

$1.00 today does not have the same value as $1.00 a year from now Interest that can be earned Purchasing power (inflation) Opportunity cost

Time is Money!


Interest Rate and Discount Rate Definitions

Interest rate or discount rate: the cost of using capitalAmount paid to borrow fundsAmount received when loaning (or saving) fundsAmount of change in value (up or down) of funds over time

Interest rate or discount rate is used to account for:Opportunity cost (how much can be earned on the funds) Inflation (decreased value in the future)Cost of money (cost to borrow)


Rate of Return (ROR) Internal Rate of Return (IRR)

The discount rate at which benefits equal costs for a given time period.

Rate of return is stated in terms of percent, for example: The rate of return on my investment was 15% means

that my investment is worth 15% more than it was previously

After taxes, my rate of return was only 12%


Rate of Return (cont.)

Often used as a measure to determine if a project is economically attractive.

A minimum acceptable rate of return is often established for a particular decision.

Rate of return is calculated after all expenses are paid, for example: If we invest $10,000 on this new machine, our net

rate of return (after expenses, training, etc.) will be 8%.


Study Period (Analysis Period)

Period over which the economic study is performed.

Often equal to the useful life of the facility, expected period of ownership, or loan repayment period.

Increments used (for example, years) must match period for interest rate or discount rate.


Study Period (Analysis Period)

10 CFR 436 defines the appropriate study period. For retrofits, § 436.14(d) says to use the lesser of

the effective equipment life or 40 years. For new construction, § 436.14(d) says to use the

lesser of the effective building life or 40 years.


Cash Flow Diagram

Useful visual tool to capture all income and costs All income is positive (+) All expenses are negative (-) All costs occur at the end of the period Initial costs occur at the end of year 0, which is

also the beginning of year 1 (now)


Cash Flow Diagram Example

0 1 3 42

$5K

$3K

$1K $1K $1K $1K


Economic Equivalence

Two sums are said to be economically equivalent if they have the same time value

Example: $100 invested at an annual return of 10%

becomes $110 in a year $100 today is said to be equivalent to $110

in a year at 10% interest


Economic Equivalence (cont.)

� Economically speaking, there is no reason to prefer $100 today or $110 in one year at a 10% interest rate.


Current Year Dollars vs. Constant Year Dollars

Current Year Dollars Express all costs inflated to expected future value Interest rate used for analysis includes inflation

Constant Year Dollars All costs expressed in today’s dollars Discount rate is adjusted for price inflation


Discount Rate

The discount rate can be used to represent a “net” interest rate that takes into account both the current interest rate and the current rate of inflation


Example: Determining a Discount Rate

Assume 6.6% interest rate (INT) Assume 2.4% inflation rate (INF) Discount rate =

INT-INF = 0.066-0.024 = 0.041 = 4.1%1+INF 1+ 0.024

Discount rate is 4.1% and all dollars should be expressed as constant year dollars


Development of Unique Design Alternatives Alternatives must be independent of each

other Watch for synergistic effects Example:

Building energy consumption can be reduced by installing a more efficient boiler and insulating the roof

Cannot evaluate alternatives individually and add savings together; doing one project changes the baseline energy consumption for evaluation of the other project


Useful Life

Expected life of item Time until technologically obsolete Time until no longer economical to maintain Usually not depreciation life Example:

A personal computer may run for 10 years, but it will be technologically obsolete much earlier and will probably be replaced long before the end of its potential operating life


Relevant Cash Flows

Include initial costs, recurring and non-recurring costs, capital replacement, disposal costs and any residual value.

For analysis purposes to select the best alternative, ignore costs identical in all alternatives.

If projected cash flow is required (for budgeting, for example), all costs must be included.

Sunk cost is an expense already incurred. It is irrelevant to the current analysis as it won’t / can’t change.


Differences in Costs Over the Same Period

Recall economic equivalence. Use life cycle cost techniques to

determine equivalent present costs or equivalent annual costs.


Uncertainty

Estimates of future costs, expected inflation rates, projected interest rates and expected life are just that, estimates

Changing any parameters can alter the outcome

Techniques can be used to account for uncertainty in analysis


Life Cycle Cost

The life cycle cost is the present value (at a given discount rate) of all costs and benefits for a project


Annualized Cost

The annualized cost is the present value or life cycle cost annualized (at a given discount rate) over the life of the study


Interpreting the Results

The lowest life cycle cost alternative is always the best economic alternative of two or more independent alternatives

Alternatively, the lowest annualized cost is always the best economic alternative for two or more independent alternatives

Various measures (payback, rate of return) can be used to further compare alternatives


Using Life Cycle Costing During Design

System selection System optimization Interdependencies When is LCC not an appropriate tool?


Use of Economics for System Selection Determine independent alternatives and

analyze their cost streams Select the alternative with the lowest life

cycle cost Example:

Compare the costs and savings of installing a VAV reheat system with a constant volume reheat system


Use of Economics for System Optimization Examine one system and potential

optimization alternatives Example:

Compare the costs and savings of 1 in. versus 2 in. of pipe insulation


Interdependencies

If two alternatives are interdependent, they must be analyzed together

In the previous example of replacing the boiler and adding roof insulation, there are three potential projects and four scenarios: Do nothing Add insulation to the roof Replace the boiler Add insulation to the roof and replace the boiler

Energy consumption must be modeled for all four alternatives


When is Life Cycle Costing Not an Appropriate Tool?

When the decision will be made for other than economic reasons

Examples: A new law requires replacement of a piece of currently

functional equipment to be in compliance Your maintenance department wants a certain

manufacturer even though they are more expensive When there is not enough information to do a meaningful

economic analysis When comparing two unrelated projects Emergency replacements


Summary for Introduction to Engineering Economics

In this section, we defined basic economic parameters such as: Time value of money Interest rate and discount rate Rate of return Study period and useful life Economic equivalence Current year dollars and constant year dollars

We also learned how to develop a simple cash flow diagram


Summary (cont.)

We discussed how engineering economics can be used to make better engineering decisions, and when the use of economics is not an appropriate decision-making tool


Class Problem 1-1

Draw the cash flow diagram for the following problem: Initial investment $10,000 Annual cost $ 4,000 Income in year 4 $ 6,000 Expense in year 10 $ 7,500 Salvage in year 15 $ 2,000


Solution to Class Problem 1-1

� Draw the cash flow diagram below:

10K7.5K

2K6K

0 5 10 15

4K


Section 2: Tools and Methods

Present, Future, and Annual Worth Factors Discount Factors Evaluation Methods

Rate of Return Simple Payback Savings to Investment Ratio, Benefit / Cost Ratio Net Present Value, Annualized Value

Class problems


Present Worth or Value

Worth or value at time 0 (beginning of analysis period)

P = F (P/F,i,n)

� where: F = future amount� i = interest or discount rate� n = number of years

Used to determine the present value of future single amounts


Present Worth or Value (cont.)

The present worth factor can be computed using the following equation:

P = F * (1 + i )-n = F / (1+i)nwhere: F = future amount

� i = interest or discount rate�n = number of years


Use the present worth factor to convert a sum of money in a future year to an equivalent present sum.

Example: If you require $500,000 in five years, how much must you invest now at an interest rate of 8%?

� P = $500,000 x (1 + 0.08)-5 = $340,291 or� P = $500,000 / (1 + 0.08)5 = $340,291



So, $340,300 invested for five years at 8% interest compounded annually would yield $500,000.



Series present worth factorP = A (P/A,i,n)

� where: A = recurring annual amount� i = interest or discount rate�n = number of years

Used to determine the present value of future annual amounts.

Series Present Worth


The series present worth factor can be computed using the following equation:

�P = A [(1 + i )n – 1]� i (1 + i )n

� where: A = future annual amount� i = interest or discount rate�n = number of years

Series Present Worth (cont.)


Use the series present worth factor to convert future annual sums to an equivalent present amount.

Example: If you require $10,000 a year for the next five years to maintain an HVAC control system, how much must you invest now at an interest rate of 8%?

� P = $10,000 [(1 + 0.08 )5 – 1] = $39,927� 0.08 (1 + 0.08 )5



So, $39,927 invested for five years at 8% compounded annually would yield $10,000 a year for five years.

�



Future Worth or Value

A cost or value at some time in the futureF = P * (F/P,i,n)

� where: F = future amount� i = interest or discount rate�n = number of years

Used to determine the future value of present single amounts


Future Worth or Value (cont.)

The future worth factor can be computed using the following equation:

� F = P * (1 + i )nwhere: P = present amount



Use the future worth factor to convert a present sum of money to an equivalent future sum.

Example: If you invest $50,000 today at 8% interest, how much money will you have available at the end of five years for equipment replacement?

� P = $50,000 (1 + 0.08)5 = $73,466



So, $50,000 invested for five years at 8% interest compounded annually would yield $73,466.



Series future worth factor

F = A * (F/A,i,n)where: A = recurring annual amount

� i = interest or discount raten = number of years

Used to determine the future value of annual amounts.

Series Future Worth


The series future worth factor can be computed using the following equation:

� F = A * [(1 + i )n – 1]� i

where: A = recurring annual amount


Series Future Worth (cont.)


Use the series future worth factor to convert future annual sums to an equivalent future amount.

Example: If you invest $10,000 a year at 8% interest for five years, how much money will you have for HVAC retrofits at that time?

� F = $10,000 [(1 +0.08)5 – 1] = $58,666� 0.08



So, $10,000 invested each year for five years at 8% interest compounded annually would yield $58,666.



Annual Worth or Value

Capital recovery factorA = P * (A/P,i,n)

where: P = present amount

� i = interest or discount rate�n = number of years of equal payments

Used to determine the equivalent annual value of a single present amount.


Annual Worth or Value (cont.)

The annual worth factor for present amounts, called the capital recovery factor, can be computed using the following equation:

� A = P ( i (1 + i )n )� (1 + i )n - 1

where: P = present amount� i = interest or discount rate�n = number of years


Use the capital recovery factor to convert a present amount into a series of regular payments.

Example: If you invest $50,000 today at 8% interest, how much money can you withdraw for maintenance each year for five years?

� A = $50,000 [0.08 (1 + 0.08 )5 ] = $12,522� (1 + 0.08 )5 - 1



So, $50,000 invested for five years at 8% interest compounded annually would yield $12,523 a year for five years.

�



Sinking fund factorA = F (A/F,i,n)

where: F = future amount�i = interest or discount rate�n = number of years

Used to determine the annualized value of a single future amount.



The annual worth factor for future amounts, called the sinking fund factor, can be computed using the following equation:

� A = F ( i )� (1 + i )n - 1

where: F = future amount� i = interest or discount rate�n = number of years



Use the sinking fund factor to convert a future amount into a series of regular payments.

Example: If you require $50,000 in five years, how much must you invest each year at an interest rate of 8% (in equal amounts)?

� A = $50,000 (0.08 ) = $8,522� (1 + 0.08 )5 - 1



So, $50,000 five years from now is equivalent to $8,523 invested a year for five years at 8% interest compounded annually.



Discount Factor Tables

A series of tables has been developed at different interest rates to provide the present worth factors (as well as other factors to be discussed later) for multiple years. These are included in the course handouts. An excerpt from a typical table is shown on the next slide.


Use of Discount Factor Tables

� Tables by interest rate containing computed factors

i=8%n F/P P/F A/F A/P F/A P/A1 1.0800 0.9259 1.0000 1.0800 1.0000 0.92592 1.1664 0.8573 0.4808 0.5608 2.0800 1.78333 1.2597 0.7938 0.3080 0.3880 3.2464 2.57714 1.3605 0.7350 0.2219 0.3019 4.5061 3.31215 1.4693 0.6806 0.1705 0.2505 5.8666 3.9927

Discount Factors


Rate of Return

The interest rate for which expenditures equal benefits. Each of the previous examples has a rate of return of

8%. If the present value of the benefits exceeds the present

value of the expenditures, the rate of return will not be the interest rate used for the analysis.

Use a trial and error solution for i, so benefits equal costs.


Rate of Return Example

Suppose you had a project whose annual benefits equaled $20,550/year for five years and whose total cost was $82,000. If the benefits equal the cost, we can write:

�

� PV (Benefits) = 1� PV (Costs)


Rate of Return Example (cont.)

This equates to:� $20,550 (P/A, i, 5 years) = 1� $82,000� Or (P/A, i, 5 years) = $82,000 = 3.99� $20,550 In the 8% interest table, (P/A, 8%, 5 years) = 3.99,

so the rate of return = 8%.


Simple Payback

The time period to pay back an initial investment while disregarding the time value of money.

Future cash flows are not discounted. If a single investment of $500,000 yields benefits of

$100,000 per year, the simple payback is five years. Note that expected life is only relevant to determine

whether a project pays back during its expected life.


Sample Calculations of Simple Payback

$100,000 Investment

Savings $18,000/yr

Expected life: 10 yrs

SPB = $100,000$18,000

SPB = 5.55 years

Pays Back During Project Life

$200,000 Investment

Savings $18,000/yr

Expected life: 10 yrs

SPB = $200,000$18,000

SPB = 11.1 years >10

Does Not Pay Back During Project Life


Savings-to-Investment Ratio

Defined as present value of savings divided by present value of investments.

Investments include first-cost, major future upgrades/replacements and salvage value.

Savings include energy, maintenance and repair, and operating cost savings.

SIR >1 is cost effective.


Savings-to-Investment Ratio (cont.)

Difficulty comes in determining if repairs involving replacement of parts of a system qualify as “investments”.

Classify as investment if the part replaces a major portion of system and improves it.

Examples would include replacing an air handling unit or a complete control system within a building. Replacement of a fan wheel is a repair, not an investment.


Savings-to-Investment Ratio Example

Initial investment

PV(energy savings)

PV(operations and maint. savings)

PV (capital replacement, yr 5)

PV (repairs, yr 8)

PV (salvage, yr 10)

$100,000

$ 82,000

$142,000

$ 72,000

$ 24,000

$ 15,000


Savings-to-Investment Ratio Example (cont.)

Savings = Energy + O&M - Repairs Savings = $82,000 + $142,000 - $24,000

= $200,000 Investment = Initial + Cap Repl. - Salvage Investment = $100,000 + $72,000 - $15,000

= $157,000 SIR = $200,000 / $157,000

= 1.27


Benefit-to-Cost Ratio

Similar to SIR, except investment defined differently. Present value of all future benefits divided by present value

of all costs. Simple ranking system for projects; B/C >1 is cost effective From previous example:

PV(savings or benefits) = $119,778 PV(costs) = $100,000 B/C = $119,778 / $100,000 = 1.2


Net Savings

Net savings represent the net present value of the total savings of a project.

Computed using present value of cost differences of two projects.

NS >0 is a cost-effective project. An example of the net savings calculations follows.


Net Savings Example

Project A

Initial investment

PV (energy savings)

PV (operations & maint. savings)


PV (repairs, yr 8)

PV (salvage, yr 10)

$100,000

$ 82,000

$142,000

$ 72,000

$ 24,000

$ 15,000


Net Savings Example (cont.)

For Project A: Savings = $82,000 + $142,000

= $224,000

Costs = $100,000 + $72,000 + $24,000 - $15,000 = $181,000

NS = $224,000 - $181,000= $43,000


Project B

Initial investment

PV (energy savings)



PV (repairs, yr 8)

PV (salvage, yr 10)

$ 50,000

$ 40,000

$ 85,000

$ 32,000

$ 10,000

$ 10,000



For Project B:� Savings = $40,000 + $85,000

= $125,000

� Costs = $50,000 + $32,000 + $10,000 - $10,000 = $82,000

� NS = $125,000 - $82,000= $43,000



Although the net savings for both projects equal $43,000, the second project is more cost-efficient as it produces the same net savings for a smaller initial investment



Uniform Annualized Cost Method

The uniform annualized cost method develops an equivalent annual cost for a project. This is done by itemizing all of the costs, taking their present worth, and then converting it to an annualized value using the capital recovery factor (A/P,i,n)


Uniform Annualized Cost Method Example

Initial investment

PV (energy savings)



PV (repairs, yr 8)

PV (salvage, yr 10)

i = 8%, 10 year life

$100,000

$ 82,000

$142,000

$ 72,000

$ 24,000

$ 15,000


Uniform Annualized Cost Method Example (cont.)

Savings = $82,000 + $142,000= $224,000

Costs = $100,000 + $72,000 + $24,000 - $15,000 = $181,000

PV = $224,000 - $181,000= $43,000


At an interest rate of 8%, the annualized net savings is:� $43,000 (A/P, 8%, 10 years) =� $43,000 (0.1490) = $6,407

Therefore, there is a net annual positive cash flow of $6,407 over 10 years for this project



This annualized savings (or cost) can then be compared against the annualized savings (or cost) of another project to select the best alternative



Acceptable Alternatives

An investment is economically attractive if meets any of the following conditions: Has the lowest life-cycle cost Net savings > 0 Savings-to-Investment > 1 IRR > Discount Rate

Note: SPB is not a definitive criteria


Summary for Tools and Methods of Life Cycle Costing

In Section 2, we examined: Present, future and annual worth or value Rate of return Simple payback and discounted payback

We defined other measures, including: Benefit-to-cost ratio Savings-to-investment ratio Net savings Uniform annualized costs


Section 3: Principles of Life Cycle Costing

Developing the cash flow diagram Estimating costs Determining useful life and

replacement interval Life Cycle Cost


Principles of Life Cycle Costing

Environmental issues Analyzing alternatives Other economic factors Selecting among projects when

funds are limited


Developing the Cash Flow Diagram

Initial cost Future replacement cost Recurring costs

Operation and maintenance Energy

Non-recurring costs Disposal costs Salvage value


Initial Cost

Cost of acquisition. Includes all design, construction and start-up costs. If phased in over multiple years:

Good: Lump all costs at year 0 Better: Discount future years


Future Replacement Cost

For any equipment or system with a useful life that is less than the study period, replacement will be necessary.

Estimate cost of replacement in year 0 dollars.

Show replacement in year projected to occur.


Recurring Costs

Any costs that occur repetitively in each period.

Maintenance costs, operating costs and energy costs are typical examples.

Show separately if inflation rates are different (often true for energy costs).


Non-Recurring Costs

Individual costs that occur at discrete times. Major (non-annual) maintenance are

examples. Unscheduled repair costs are

unpredictable. Include an annual budget for unscheduled

repairs.


Disposal Costs

Expenses associated with disposal of property or equipment.

Examples include environmental clean-up, demolition or removal costs.

Although these may be required by code or law, they still need to be included.


Salvage Value

Residual value remaining at the end of the study period.

Important that salvage value reflects real worth, not depreciated value.

If there is no market, there is no salvage value; in fact, there may be a disposal cost.


Estimating Initial Costs

Sources of potential initial costs include: Industry average values R. S. Means construction estimating Experience with similar projects Contractor bids


Estimating Energy & Water Costs

Use 1 -3 years of consumption. Exclude atypical years. Use current energy & water rates.


Estimating Recurring Costs

Sources of O&M costs include: Review of 1 – 3 years of existing O&M costs. Values of O&M contracts. Manufacturer’s O&M requirements.

Note 1: When calculating O&M cost savings, internal labor costs should not be included unless reductions-in-force are anticipated. Use parts costs only.

Note 2: It is not appropriate to assume O&M costs that have not actually been incurred. If you haven’t spent it, you can’t save it.


Determining Useful Life and Replacement Intervals

Recall useful life discussion.Estimated service life tables in 2019

ASHRAE Handbook-HVAC Applications are useful reference.

Historical data indicates normal level of maintenance that can be expected.

If replacement intervals are too optimistic, they will bias the analysis.


Useful Life: ASHRAE HandbookHVAC Applications, Chapter 38 (2019)

Item Median LifetimeA/C, window mount 10A/C, water-cooled package 15Boiler, cast iron 35Fan, centrifugal 25Chiller, reciprocating 20Chiller, centrifugal 23Cooling tower, galvanized steel 20Motor, electric 18Transformers 30


Life Cycle Cost

Once all costs have been determined, the life cycle cost is found by taking the present worth of the cost stream

LCC = PV (Initial Cost + Recurring + Non-recurring + Future Replacement + Disposal -Salvage)

To compare projects, they must have the same study period


Life Cycle Cost (cont.)

Lowest life cycle cost is the best alternative.Can also take present value and convert to an

annual cost (A/P,i,n). Lowest annualized cost is best alternative.Not appropriate to compare two independent

projects on different buildings.


Environmental Issues

Environmental factors can have major impacts on costs

Examples include: Increasing refrigerant costs Clean-up of environmental conditions (asbestos,

PCBs, underground storage tanks) Designing for LEED Certification


Environmental Issues (cont.)

Examples include: Additional monitoring requirements (refrigerant

alarms, underground storage tank leak detection) Special requirements of owners, or government Indoor Air Quality


Analyzing Alternatives

Useful life equals study period Straight present value calculation

Useful life different from study period If shorter than study period, show planned

replacement at end of useful life, estimate “real”salvage value at end of study period

If longer than study period, estimate “real”salvage value at end of study period


Analyzing Alternatives (cont.)

Alternatives with different useful lives Use study period as required by project Use techniques above for useful life different from

study period


Example Using Different Useful Lives

Study period = 25 years Heat pump useful life = 15 years Cash flow diagram:

5 10 15 20 25Salvage

Value

InitialCost

ReplacementCost


Analyzing Alternatives

Interdependent projects Must analyze interdependent alternatives as a combined

project Do not analyze separately and add costs together



For example, projects to add insulation and replace windows, alternatives include: Do nothing Replace windows Add insulation Replace windows and add insulation



Incremental analysis Useful for determining when an additional

increment of funding is economically attractive Use to fund multiple independent projects Use to determine most attractive increment for

a single project


Incremental Analysis Procedure

Rank projects from lowest to highest first-cost. Discard any with individual rates of return that are unacceptable

Determine rate of return of increment of funding from first to second project

If rate of return is higher than target rate for acceptable projects, take the increment

Proceed until all increments have been evaluated


Incremental Analysis Example

First Cost Annual Savings

Project A: Add 1 in. of Pipe Insulation

$20,000 $4,100

Project B: Add 2 in. of Pipe Insulation

$40,000 $6,560

Cost & Savings $20,000 $2,460


Incremental Analysis Example (cont.)

�The individual rate of return for Project A is: �$20,000 = $4,100 (P/A,i,10)�$20,000/$4,100 = 4.878 = (P/A,i,10)

�From interest rate tables, i = 16%



Similarly, the individual rate of return for Project B is: � $40,000 = $6,560 (P/A,i,10)� $40,000/$6,560 = 6.09 = (P/A,i,10)

From interest rate tables, i = 10.5% (approx)



If the minimum rate of return is 6%, then it would appear that the second project should be funded.This may not be the case.



Compute the incremental rate of return between Projects A and B if the study period is 10 years and the minimum rate of return is 6%:

� $20,000 = $2,460 (P/A,i,10)� $20,000/$2,460 = (P/A,i,10)

�From the interest rate tables, i = 4%



In this case, an interest rate of 4% does not exceed the minimum required rate of return and Project B should not be funded.



If the computed rate of return had been acceptable, it would be cost effective to fund Project B in addition to Project A.


Optimizing Project Elements

If multiple, mutually exclusive alternatives are available for selection, present value (or life cycle cost) is the easiest method of ranking

Also can be used to optimize increments (such as insulation thickness)

Lowest life cycle cost increment is the most economical


Optimizing Project Elements Example


Other Economic Factors

Other factors that can affect the economic analysis include: Taxes and depreciation Inflation Cost of financing


Taxes

Taxes affect economic analysis by reducing the effective rate of return on investments

For non-depreciable assets, �

After-tax ROR = � (1 - incremental tax rate) (Before-tax ROR)


Depreciation

Depreciation affects the cost of equipment and facility ownership by affecting taxes

Computing after-tax rate of return requires adjusting the before-tax cash flow to a depreciated cash flow before computing taxes

Rate of return is then computed based on after-tax cash flow


Taxes and Depreciation

Because of the complications introduced by taxes and depreciation, “first cut” analyses often ignore taxes and depreciation

In these cases, the discount rate used to perform the analysis (or the rate of return required for the project) is set high enough to offset the expected tax consequences


Inflation

Inflation affects the costs of goods and services by either raising (inflation) or lowering (deflation) these costs

If analysis is being performed for a study period to start in the future, use the general inflation rate to inflate all costs forward to year 0

Use constant dollar analysis and account for inflation in the discount rate


Inflation (cont.)

� If the discount rate is selected to include inflation, constant year dollars can be used for the analysis. Recall the example from Section 1:

� Assume 6.6% interest rate (INT), 2.4% inflation rate (INF). Discount rate =

INT-INF = 0.066 - 0.024 = 0.041 = 4.1%1+INF 1 + 0.024

� Discount rate is 4.1% and all dollars should be expressed as constant year dollars


Inflation (cont.)

If different types of costs are inflating at different rates, discount factors for each rate of inflation must be computed and used

Inflation does affect after-tax analysis because taxes are paid based on cash flow during the year (and future actual cash flows occur with inflated dollars)


Inflation (cont.)

� Use of multiple inflation rates complicates the analysis. For example, if a different rate is used for electricity, the present value of the annual electricity costs would be computed separately using a discount rate that reflected the appropriate inflation. Continuing with the previous example:� Assume 6.6% interest rate (INT)� Assume 2.4% inflation rate (INF)� Assume 3% inflation rate for electricity (INF’)


Inflation (cont.)

� Discount rate for everything but electricity = � INT-INF = 0.066-0.024 = 0.041 = 4.1%

1+INF 1+0.024� Discount rate for electricity =

INT-INF’ = 0.066-0.03 = 0.035 = 3.5%1+INF’ 1+0.03

� Discount rate is 4.1% for all but electricity, which is 3.5%. All dollars should be expressed as constant year dollars.


Discount Rate for Federal Projects

Discount Rate: For federal projects, use Office of Management & Budget Circular A-94.

For 2020: https://www.whitehouse.gov/wp-content/uploads/2019/12/M-20-07.pdf

Nominal bond yield interest rates are 1.6% - 2.4%. Interest rate is 2.0%. This makes the effective

discount rate ~0%.


Discount Rate for Federal Projects (2020)

3 Year 5 Year 7 Year 10 Year 20 Year 30 YearInterest 1.6 1.7 1.8 2.0 2.3 2.4Discount -0.4 -0.3 -0.2 0.0 0.3 0.4

Use linear interpolation for evaluation horizons in between stated values. For horizons > 30 years, use the 30-year value.


Energy Escalation Rate

For federal projects, use NIST Guidance. Use FEMP’s Energy Escalation Rate Calculator

(EERC) with a 2% inflation rate. https://www.energy.gov/eere/femp/building-life-cycle-

cost-programs Energy costs can be treated separately and vary by

state. NIST Handbook 135 recommends using a single

escalation rate.


Cost of Financing

Cost of financing can be accounted for by treating it as another project cost.

One alternative is to show the total cost of the project as a first-cost and then show only the interest as an annual cost in later years.

Another alternative is to show the first-cost as a series of annual payments that include all financing costs (for example, show as an annual mortgage payment).


Selecting Among Projects With Limited Funds

Projects must be ranked by economic efficiency (use SIR or B/C).

If inadequate funds are available for all projects, select the top projects first.

Combine projects to match funding target and compute combined SIR or B/C to get best combination of projects for dollars available.


5 simple projects available with a total budget of $900,000

Only costs are first-costs All benefits are present value of annual

benefits B/C computed as ratio of present value of

savings divided by first-cost

Example: Projects With Limited Funds


Available projects:

Example: Projects With Limited Funds (cont.)

Project Cost Benefits B/C A $200,000 $343,100 1.72 B $500,000 $629,400 1.26 C $400,000 $541,600 1.35 D $300,000 $400,200 1.33 E $600,000 $904,500 1.51



Projects ordered by B/C:

Project Cost Benefits B/C A $200,000 $343,100 1.72 E $600,000 $904,500 1.51 C $400,000 $541,600 1.35 D $300,000 $400,200 1.33 B $500,000 $629,400 1.26


Possible funding combinations:


Project Cost Benefits B/C E+D $900,000 $1,304,700 1.45

A+C+D $900,000 $1,284,900 1.43 C+B $900,000 $1,171,000 1.30 A+E $800,000 $1,247,600 1.56 B+D $800,000 $1,029,600 1.29


Highest B/C results if projects A and E are funded at a total of $800,000

Highest B/C if all funds are spent involves funding projects E and D



Summary for Principles of Life Cycle Costing

In this section, we discussed: Developing cash flow diagrams Estimating costs, determining useful life and

replacement interval and effects of environmental issues

Analyzing alternatives Other economic factors Selecting among projects when funds are

limited


Section 4: Uncertainty


Dealing with Uncertainty

Definitions of risk, uncertainty and sensitivity Methods for dealing with uncertainty Applying uncertainty techniques to class problem


Risk

Exposure to possible loss, or having a return less than that predicted. Risk can generally be quantified with probabilities of occurrence For example, an investment that is 95% sure

may be considered low risk, while one that is only 5% sure would be very high risk


Uncertainty

Different outcomes are possible, but their probability cannot be easily determined For example, the cost to replace a computer in

four years can be estimated to be less than today due to decreasing costs, or more than today due to increasing features. The probability of either alternative cannot be easily determined.


Sensitivity

The relative magnitude in change of one or more parameters that would cause a change in the decision For example, if a 1% rise in mortgage interest rates

would cause you to forego the purchase of a house, that is the amount of sensitivity your decision has with respect to mortgage rates


Dealing with Uncertainty

Break-even analysis (including graphical techniques)

Sensitivity analysis Reduction of useful life Reducing the uncertainty


Break-Even Analysis

Determines the point in time where two alternatives have the same present value

Useful when examining options that have different cash flows

Refer to the prior example on discounted payback


Break-Even Analysis Example

Recall example on computing discounted payback

Graph net cost by year Break-even point (or payback point) occurs

when the cost curve crosses the 0 axis


Break-Even Analysis Example (cont.)

Multiply each year’s savings by present value factor to get year’s discounted savings

Discounted payback occurs when sum of savings equals investment

i=8%n Savings P/F Discounted Savings Savings by Year0 (100,000)$ 1.0000 (100,000)$ (100,000)$ 1 30,000$ 0.9259 27,777$ (72,223)$ 2 30,000$ 0.8573 25,719$ (46,504)$ 3 30,000$ 0.7938 23,814$ (22,690)$ 4 30,000$ 0.7350 22,050$ (640)$ 5 30,000$ 0.6806 20,418$ 19,778$

Discount Factors


-120000-100000-80000-60000-40000-20000

02000040000

0 1 2 3 4 5Years to Payback

EXAMPLE

Break-Even Analysis Example (cont.)


Sensitivity Analysis

Determine a range of values over which the decision does not change

Identify parameters for which analysis is very sensitive

Concentrate on refining information on those parameters

Recall the previous example on optimizing project elements


Optimizing Project Elements Example

First-Cost PW @ 6%, 10 years

Annual Savings

Project A $20,000 $10,176 $4,100

Project B $40,000 $8,282 $6,560

Cost & Savings

$20,000 -$1,894


Sensitivity Analysis Example

How sensitive is the decision to select Project B to the estimate of annual savings?

To maximize the present value for B: P = x (P/A, 6%, 10 years) - $40,000

where (P/A, 6%, 10) = 7.360 P = $10,176 for benefits B = benefits A For P = $10,176, x = $6,817.

Therefore, Project B’s annual savings needs to be $6,817 before selecting Project B in addition to Project A


Reduction of Useful Life

One technique for ensuring that a decision will be cost effective is to reduce the useful life

If a project still pays back with a reduction of useful life, there is more certainty that it will pay back in normal operation even if unforeseen circumstances occur to reduce the period of use


Reducing the Uncertainty

The techniques for reducing uncertainty include: Reducing useful life Using conservative rather than optimistic

projections for savings Performing detailed analysis to better

predict future costs Using statistical analysis to assess

probabilities of occurrences and adjusting according to your risk tolerance


Summary for Uncertainty

In this section, we covered the following topics: Definitions of risk, uncertainty and sensitivity Methods for dealing with uncertainty


Section 5: Using Excel™ Financial Functions


Functions

Items in [braces] are optional and default to 0.

Type = 0: end of period

Type = 1: beginning of period


Present & Future Worth P/F, F/P

No specific function. Use:

P/F = (1 + i)^-n or 1/(1 + i)^n

F/P = (1 + i)^n


Series Present Worth P/A

Use to calculate present value of future payment stream

= -PV(rate, nper, pymt, [fv], [type]) Function returns negative value; add negative sign

to invert For monthly payments, use:

nper = 12 x year, rate = annual rate/12


Capital Recovery Factor A/P

Used to calculate periodic payments equal to present value

= -PMT(rate, nper, pv, [fv], [type]) Function returns negative value; add negative sign




Principle & Interest Payments

Used to calculate principle and interest payments at a given time

Used to calculate equity = -PPMT(rate, per, nper, pv, [fv], [type]) = -IPMT(rate, per, nper, pv, [fv], [type]) ‘per’ is specific period in ‘nper’


Principle & Interest

Principle & Interest Payments on $100 loan at 7%

$0

$2

$4

$6

$8

$10

$12

$14

$16

1 2 3 4 5 6 7 8 9 10

IPMTPPMT


Series Future Worth F/A

Used to calculate future value of periodic payments

= -FV(rate, nper, pymt, [pv], [type]) Function returns negative value; add negative sign




Sinking Fund Factor, A/F

Used to calculate periodic payments equal to some future amount

= -1/FV(rate, nper, pymt, [pv], [type]) Function returns negative value; add negative sign




Net Present Value

Used to calculate Net Present Value (Present Worth) from irregular amounts at constant intervals

= NPV(rate, array) Array can contain positive or negative values of

any magnitude. For monthly amounts, use:

rate = annual rate/12


Internal Rate of Return

Used to estimate Internal Rate of Return of an irregular cash flow at constant intervals containing positive and negative values

= IRR(array, [guess]) This is an iterative solution and cash flows with

more than one sign change may have more than one answer!


Functions

Items in [braces] are optional and default to 0.

Type = 0: end of period

Type = 1: beginning of period


Summary for Using Excel™ Financial Functions

Functions can be used to develop models Be careful with functions that invert signs Test models with simple data Check for multiple answers when using IRR

function


Section 6: Course Summary


Course Summary

Defined basic economic parameters such as time value of money, interest or discount rate, and study period

Developed cash flow diagrams Demonstrated use of economic factors to

find present, future and annual value Defined useful life and replacement

interval


Course Summary (cont.)

Defined elements of a life cycle cost: Initial cost Future replacement cost Recurring costs Non-recurring costs Disposal costs Salvage value


Demonstrated various analysis techniques, including the use of incremental analysis

Discussed other factors affecting a life cycle cost, including: Taxes Depreciation Inflation Environmental issues



Developed analysis techniques to deal with uncertainty

Provided introduction to financial functions in Microsoft Excel™



When is life cycle costing an appropriate tool for engineering decision-making? System selection System optimization Analyzing interdependent projects Selecting among multiple independent projects for

funding



When should life cycle costing not be used? Whenever the decision will be solely based

on non-economic factors When the information available is

inadequate to do a meaningful economic analysis



� Life cycle costing is one more tool that engineers can use to make effective, efficient decisions in support of their customers



Thank You!

Questions?

Mark Stetz, P.E., [email protected]


ASHRAE values your comments about this course. You will receive your Certificate of Attendance when you finish the online course evaluation form at this URL: www.ashrae.org/GSAincompanyaug2020Be sure to add your appropriate license numbers.

If you have any questions about ASHRAE Certificates, please contact Kelly Arnold, Professional Development Coordinator, at [email protected].

If you have any questions about ASHRAE courses, please contact Tiffany Cox, Course Administrator, at [email protected].

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