S3 · Supportive Knowledge (HVAC and Refrigeration)

Foundation lesson · 18 minute study session

Compare costs at the same point in time.

Use incremental cash flows to distinguish simple payback from discounted value.

By the end of this lesson

Define the base alternative and the incremental investment.
Discount uniform end-of-year savings on a stated time basis.
Keep simple payback, lifecycle cost and net present value distinct.

Draw the cash-flow boundary.

An economic comparison uses differences between alternatives over a common study period. Include first cost, energy, maintenance, replacement and residual value when supplied. Shared costs cancel. State whether cash flows occur now or at year end before choosing a factor.

Discounting weights timing.

A future saving has a present value determined by the stated interest rate and timing. A uniform annual series can use a present-worth factor. Keep nominal dollars with a nominal rate, or constant dollars with an appropriate real rate. Do not add an inflation correction when the problem already incorporates it.

Payback answers a narrower question.

Simple payback divides an initial incremental cost by a constant annual saving. It ignores discounting and cash flows after payback. Net present value evaluates the whole stated period. A short payback does not by itself prove the best lifecycle choice.

A decision map for this lesson

Put competing cash flows on one time basis

Use the supplied study period, interest convention and incremental costs; simple payback answers a narrower question than NPV.

Choose the governing relationship and catch the wrong route
Condition or question	Governing decision	Error to avoid
Simple payback is requested	Divide incremental first cost by the stated constant annual net saving.	Using gross energy saving while omitting additional annual maintenance.
NPV or present worth is requested	Discount each supplied future cash flow to the same time origin.	Adding dollars from different years without a time-value adjustment.
Real or nominal flows are supplied	Use a consistent discount-rate basis and apply inflation only once.	Discounting inflated nominal flows with an unadjusted real rate.
Replacement or residual value is material	Include it at the specified time within the study-period comparison.	Using short payback as permission to omit later costs.

On a narrow screen, scroll across the table to compare the decision and the error.

Relationships and their limits

Present value of uniform end-year savings

P=A\frac{1-(1+i)^{-n}}{i};\qquad NPV=P-C_0

Constant end-of-year net savings, n years, interest fraction i per year. Include other supplied cash flows separately; at i = 0, P = nA.

Single future cash flow

P=F(1+i)^{-n}

Future amount F occurs at the stated end of year n; use its sign consistently.

Simple payback

t_{pb}=C_0/A

Constant positive annual savings; ignores discounting, residual value and later cash flows.

Symbols and unit basis
Symbol	Meaning	Units or basis
$C_0$	Incremental investment now	dollars
$A$	Constant annual net saving	dollars/year
$i$	Stated annual discount rate	dimensionless
$n$	Study duration	years
$P$	Present value	dollars
$NPV$	Net present value relative to base	dollars
$F$	Single future cash flow at year n	dollars
$t_{pb}$	Simple payback period	years

One decision, fully worked

Apply the method.

An efficiency upgrade costs $10,000 more now than the base option and saves $3,000 at the end of each of five years. The annual discount rate is 8%. No other cost differences or residual value apply. Calculate simple payback and incremental NPV.

Use incremental cost and net savings; base costs cancel.
Discount the five end-year savings using the uniform-series factor.
Subtract the immediate extra investment to find NPV.

t_{pb}=10{,}000/3000=3.33\ \mathrm{years}

P=3000\frac{1-(1.08)^{-5}}{0.08}=11{,}978.13\ \mathrm{dollars}

NPV=11{,}978.13-10{,}000=1978.13\ \mathrm{dollars}

Decision

Simple payback is 3.33 years and incremental NPV is about $1,978. On the supplied costs, rate and five-year period, the upgrade has a positive present-value benefit.

Independent check

Undiscounted savings total $15,000; discounted savings of $11,978 are smaller but remain above the $10,000 investment.

Specialist extension

Compare lifecycle costs on one time and money basis

Choose a common study period and timing convention. Discount annual operating costs, replacements and residual values on the same real or nominal basis as the discount rate. Simple payback can be a useful early screen, but omits later costs and timing. At zero interest the present value of equal annual payments is their undiscounted sum, so a formula with division by i needs its zero-rate limit.

A new applied case

Option B costs $4,000 more initially and saves $1,000 at each year-end for five years. Use supplied P/A = 4.3295 at 5%. Other costs and residual values are equal.

P_{savings}=1000(4.3295)=4329.50

P_{net\ benefit}=4329.50-4000=329.50\ \mathrm{dollars}

Decision. B has a $329.50 present-value advantage under the supplied period and assumptions.

Cross-check. Five undiscounted savings total $5,000, which overstates present value at a positive discount rate.

Unit basis. Cash flows are dollars; P/A is the supplied dimensionless five-year uniform-series factor; payments occur at year-end.

NCEES PE Mechanical Reference Handbook v2.1; see the module reference path.

Open targeted practice in the study studio →

Transfer the method

Supportive Knowledge (HVAC and Refrigeration) · HV-LC-S3

Try a new decision.

Select one

An upgrade costs $8,000 more now and saves $2,000 at each year end for five years. Use 10% annual discounting, no residual value and no other cost differences. The supplied five-year present-worth factor P/A is 3.79079. Which conclusion follows?

ASimple payback is 4 years and NPV is about −$418.BSimple payback is 4 years and NPV is +$2,000.CSimple payback is 3.79079 years and NPV is zero.DNPV is +$7,582, so the extra investment is justified.

A lesson check supports practice, not a readiness score. Your response stays in this page session and clears on reload.

Catch the common errors

Discounting a cost paid now as though it occurs in year one.
Using gross energy savings while omitting a supplied maintenance penalty.
Calling simple payback a lifecycle-cost comparison.

Handbook navigation drill

Find it, then explain it.

Search: Present Worth; Uniform Series

Find the uniform-series present-worth factor. Identify the factor direction and the first payment date before substituting.

Check your navigation

Section 1.6 distinguishes economic factors. A/P and P/A are reciprocals with different requested outputs; select by the cash-flow diagram.

Reference path: §1.6, pp. 48–58. These are printed handbook page numbers; section names and search terms remain the primary navigation method. See technical review status.

Theory sources

NCEES PE Mechanical Reference Handbook v2.1: §1.6, pp. 48–58
Mechanical and Electrical Equipment for Buildings, 12th ed.: mechanical systems and energy principles. Supporting theory; not an examination reference.

Worked examples and lesson prose are original ReasonSpan^™ material. Obtain authorized publications separately; this preview does not reproduce the supplied books or handbook.