Study guide
4.2 Discrete Compounding Formulas & Uniform Series Factors
AACE Certified Cost Professional CCP
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1 Chapter 1: Introduction & Total Cost Management (TCM) Framework
1.1 AACE CCP Exam Structure, Eligibility & the In-Exam Memo 1.2 The Total Cost Management (TCM) Framework & PDCA Cycle 1.3 Strategic Asset Management vs. Project Control Processes
2 Chapter 2: Cost Estimating Fundamentals & Classification
2.1 AACE Recommended Practice 18R-97 Estimate Classification System 2.2 Direct vs. Indirect Costs and Cost Element Breakdown 2.3 Quantity Takeoff, Standard Cost Databases & Pricing Sources
3 Chapter 3: Conceptual & Parametric Estimating Techniques
3.1 Capacity Factoring & the Six-Tenths Rule 3.2 Parametric Estimating, Lang Factors & Hand Factors 3.3 Cost Indices, Location Adjustments & Escalation Modeling
4 Chapter 4: Cost Engineering Economics & Time Value of Money
4.1 Simple & Compound Interest, Equivalence & Cash Flow Diagrams 4.2 Discrete Compounding Formulas & Uniform Series Factors 4.3 Nominal vs. Effective Interest Rates & Continuous Compounding
5 Chapter 5: Project Profitability & Capital Investment Decision-Making
5.1 Net Present Value (NPV) & Present Worth Analysis 5.2 Internal Rate of Return (IRR) & Incremental IRR Analysis 5.3 Payback Period, Benefit-Cost Ratio (BCR) & Breakeven Analysis
6 Chapter 6: Depreciation Methods, Taxes & Asset Valuation
6.1 Classical Depreciation Methods (Straight-Line, DDB, Sum-of-Years-Digits) 6.2 Modified Accelerated Cost Recovery System (MACRS) & Tax Shield 6.3 After-Tax Cash Flow Analysis & Asset Replacement Economics
7 Chapter 7: Project Scope, Work Breakdown Structure (WBS) & Baselines
7.1 Work Breakdown Structure (WBS) Principles & the 100% Rule 7.2 Organizational Breakdown Structure (OBS) & Control Accounts 7.3 Establishing the Performance Measurement Baseline (PMB)
8 Chapter 8: Critical Path Method (CPM) Planning & Scheduling Logic
8.1 Precedence Diagramming Method (PDM) & Dependency Relationships 8.2 Forward Pass, Backward Pass & Float Calculations (Total vs. Free Float) 8.3 Schedule Baseline Development, Calendar Constraints & Milestones
9 Chapter 9: Schedule Optimization, Compression & Probabilistic PERT
9.1 Schedule Crashing & Least-Cost Slope Optimization 9.2 Fast-Tracking Risks, Overlap Analysis & Execution Strategies 9.3 PERT Three-Point Estimating & Probabilistic Duration Variance
10 Chapter 10: Resource Management, Productivity & Learning Curves
10.1 Resource Loading, Leveling vs. Resource Smoothing 10.2 Labor Productivity Factors, Work Sampling & Efficiency Loss 10.3 Learning Curve Theory (Wright Cumulative vs. Crawford Unit Models)
11 Chapter 11: Earned Value Management (EVM) Core Principles & Metrics
11.1 ANSI/EIA-748 EVM Framework & Core Metrics (PV, EV, AC) 11.2 EVM Variances & Performance Indices (CV, SV, CPI, SPI) 11.3 Work Progress Measurement Techniques (0/100, 50/50, Milestone Weighting)
12 Chapter 12: EVM Forecasting, Indices & Baseline Change Control
12.1 Estimate at Completion (EAC) & Estimate to Complete (ETC) Formulas 12.2 To-Complete Performance Index (TCPI) & Variance at Completion (VAC) 12.3 Integrated Change Control, Baseline Revisions & Trend Management
13 Chapter 13: Risk Management Planning & Qualitative Risk Analysis
13.1 Risk Management Framework & Risk Breakdown Structure (RBS) 13.2 Qualitative Risk Assessment & Probability-Impact (P-I) Matrix 13.3 Risk Response Strategies & Threat/Opportunity Management
14 Chapter 14: Quantitative Risk Analysis, Monte Carlo & Contingency Estimating
14.1 Expected Monetary Value (EMV) & Decision Tree Analysis 14.2 Monte Carlo Simulation & Probability Distributions in Cost/Schedule 14.3 Contingency Estimating & AACE Recommended Practice 40R-08
15 Chapter 15: Contract Types, Procurement Strategy & Commercial Terms
15.1 Contract Types & Risk Allocation (Firm Fixed Price, Cost-Plus, T&M) 15.2 Project Delivery Systems & Procurement Lifecycle (DBB, DB, EPC, CMAR) 15.3 Commercial Terms, Bonds, Retainage & Claims Management
16 Chapter 16: Quality Management, Value Engineering & Technical Communication
16.1 Quality Management & Cost of Quality (CoQ) 16.2 Value Engineering, Function Analysis System Technique (FAST) & Life Cycle Costing 16.3 The CCP Memo, Cost Report Writing, Data Presentation & AACE Ethics 16.4 AACE CCP Capstone Review & High-Yield Exam Strategies
17 Chapter 17: Cost Accounting, Budget Administration & Cost System Governance
17.1 Historical Cost Databases, Data Organization & Benchmarking 17.2 Committed Costs, Accruals, Work-in-Place & Invoice Validation 17.3 Budget Creation, Operating Budgets & Funding Requirement Forecasting 17.4 Cost System Governance, the Code of Accounts & Preserving Project Controls Data
18 Chapter 18: Estimate Preparation, Review & Cost Driver Analysis
18.1 ROM, Summary-Level and Detailed Estimates: Preparing and Reviewing 18.2 Verifying the Basis of Estimate & Analyzing Estimates 18.3 Cost Drivers, Change Order Impact & Resource Utilization Analysis
19 Chapter 19: Interfacing with Other Disciplines & Stakeholders
19.1 Coordinating with Internal Disciplines & Supporting Project Leadership 19.2 Communicating Cost Estimates and Cost/Schedule Key Performance Indicators 19.3 Supporting Tenders, Procurement, Audits, Field Verification & Recovery
20 Chapter 20: Cost Reporting, Dashboards & Profitability Analysis
20.1 Cost Reports, Performance Dashboards & the Reporting Cycle 20.2 Variance Reports & Explaining Project Financial Health 20.3 Profit Margin Analysis & Profitability Enhancement
21 Chapter 21: Cost Support to Scheduling & the CCP Exam Memo
21.1 Cost and Resource Loading Summaries from the Project Schedule 21.2 Translating Schedule Change into Cost Impact 21.3 Writing the CCP Exam Memo: A Worked Domain 7 Walkthrough
4.2 Discrete Compounding Formulas & Uniform Series Factors
Key Takeaways Explain This Topic AI
Standard discrete compounding functional notation (X/Y, i, n) defines the factor multiplier used to find unknown quantity X given known quantity Y at interest rate i over n periods.
The six fundamental discrete compounding factors are interrelated through reciprocal pairs: (P/F) = 1/(F/P), (P/A) = 1/(A/P), and (A/F) = 1/(F/A).
The Capital Recovery Factor (A/P, i, n) converts an initial capital expenditure into an equivalent uniform annual series and equals the Sinking Fund Factor plus the interest rate: (A/P, i, n) = (A/F, i, n) + i.
Arithmetic gradient series model cash flows changing by a constant dollar increment G per period starting at t=2, converted to present worth via (P/G, i, n) or uniform series via (A/G, i, n).
Geometric gradient series model cash flows changing at a constant compound percentage rate g per period, requiring the piecewise formula P = A1[1 - (1+g)^n(1+i)^(-n)] / (i - g) when i != g, and P = n*A1/(1+i) when i = g.
Last updated: August 2026
4.2 Discrete Compounding Formulas & Uniform Series Factors
In cost engineering economics, cash flow streams frequently follow structured patterns rather than isolated single payments. Project lifecycle models routinely feature recurring annual operating expenses, level annualized debt service, periodic capital equipment replacement sinking funds, and escalating maintenance gradients.
To rapidly manipulate these cash flow streams without performing tedious period-by-period discounting, cost engineers utilize discrete compounding factors . AACE International expects Certified Cost Professional candidates to possess complete fluency in the derivation, functional notation, algebraic identities, and numerical execution of all six primary interest factors, as well as arithmetic and geometric gradient series.
1. Standard Functional Notation Architecture
To standardize engineering economics calculations globally, professional societies (including AACE, IISE, and IEEE) adopted a standardized functional notation format: ( X / Y , i , n ) \mathbf{(X/Y, i, n)} ( X/Y , i , n ) +-----------------------------------------------------------------------------+
- | STANDARD FUNCTIONAL NOTATION BREAKDOWN |
- | |
- | ( X / Y , i , n ) |
- | ^ ^ ^ ^ |
- | | | | +---- Number of Compounding Periods (n) |
- | | | +---------- Effective Interest Rate per Period (i) |
- | | +---------------- Known / Given Cash Flow Parameter (Y) |
- | +---------------------- Unknown Cash Flow Parameter to Find (X) |
- | |
- | GOVERNING EQUATION: Target Parameter (X) = Given Parameter (Y) * Factor |
- +-----------------------------------------------------------------------------+
- Fundamental Reciprocal Identities
Every standard compounding factor possesses an exact algebraic reciprocal: ( P / F , i , n ) = 1 ( F / P , i , n ) (P/F, i, n) = \frac{1}{(F/P, i, n)} ( P / F , i , n ) = ( F / P , i , n ) 1 ( A / P , i , n ) = 1 ( P / A , i , n ) (A/P, i, n) = \frac{1}{(P/A, i, n)} ( A / P , i , n ) = ( P / A , i , n ) 1 ( A / F , i , n ) = 1 ( F / A , i , n ) (A/F, i, n) = \frac{1}{(F/A, i, n)} ( A / F , i , n ) = ( F / A , i , n ) 1
2. Mathematical Derivations of the Six Core Discrete Compounding Factors +-----------------------------------------------------------------------------------------+
- | THE SIX CORE DISCRETE COMPOUNDING FACTORS |
- | |
- | SINGLE PAYMENT FACTORS: |
- | 1. Compound Amount (F/P, i, n) = (1 + i)^n [Finds F given P] |
- | 2. Present Worth (P/F, i, n) = (1 + i)^(-n) [Finds P given F] |
- | |
- | UNIFORM SERIES TO FUTURE WORTH: |
- | 3. Series Compound Amount (F/A, i, n) = [(1+i)^n - 1] / i [Finds F given A] |
- | 4. Sinking Fund (A/F, i, n) = i / [(1+i)^n - 1] [Finds A given F] |
- | |
- | UNIFORM SERIES TO PRESENT WORTH: |
- | 5. Series Present Worth (P/A, i, n) = [(1+i)^n - 1] / [i(1+i)^n] [Finds P given A] |
- | 6. Capital Recovery (A/P, i, n) = [i(1+i)^n] / [(1+i)^n - 1] [Finds A given P] |
- +-----------------------------------------------------------------------------------------+
- Factor 1: Single Payment Compound Amount Factor (SPCAF) — ( F / P , i , n ) (F/P, i, n) ( F / P , i , n )
Objective: Compute the future worth F F F accumulated after n n n periods from an initial present investment P P P .
Formula: F = P ( 1 + i ) n ⟹ ( F / P , i , n ) = ( 1 + i ) n F = P(1 + i)^n \implies (F/P, i, n) = (1 + i)^n F = P ( 1 + i ) n ⟹ ( F / P , i , n ) = ( 1 + i ) n
Factor 2: Single Payment Present Worth Factor (SPPWF) — ( P / F , i , n ) (P/F, i, n) ( P / F , i , n )
Objective: Compute the present worth P P P required today to yield a specific future amount F F F at period n n n .
Formula: P = F ( 1 + i ) − n = F ( 1 + i ) n ⟹ ( P / F , i , n ) = ( 1 + i ) − n P = F(1 + i)^{-n} = \frac{F}{(1 + i)^n} \implies (P/F, i, n) = (1 + i)^{-n} P = F ( 1 + i ) − n = ( 1 + i ) n F ⟹ ( P / F , i , n ) = ( 1 + i ) − n
Factor 3: Uniform Series Compound Amount Factor (USCAF) — ( F / A , i , n ) (F/A, i, n) ( F / A , i , n )
Objective: Determine the future lump-sum F F F accumulated at the end of period n n n resulting from a series of n n n equal end-of-period payments A A A .
Mathematical Derivation: Each payment A A A compounds for a different duration. The payment at t = 1 t=1 t = 1 compounds for ( n − 1 ) (n-1) ( n − 1 ) periods, the payment at t = 2 t=2 t = 2 for ( n − 2 ) (n-2) ( n − 2 ) periods, and the final payment at t = n t=n t = n compounds for 0 0 0 periods: F = A ( 1 + i ) n − 1 + A ( 1 + i ) n − 2 + ⋯ + A ( 1 + i ) 1 + A ( 1 + i ) 0 F = A(1+i)^{n-1} + A(1+i)^{n-2} + \dots + A(1+i)^1 + A(1+i)^0 F = A ( 1 + i ) n − 1 + A ( 1 + i ) n − 2 + ⋯ + A ( 1 + i ) 1 + A ( 1 + i ) 0 F = A ∑ t = 0 n − 1 ( 1 + i ) t F = A \sum_{t=0}^{n-1} (1+i)^t F = A t = 0 ∑ n − 1 ( 1 + i ) t Applying the finite geometric series summation formula ( S n = a ( r n − 1 ) r − 1 S_n = \frac{a(r^n - 1)}{r - 1} S n = r − 1 a ( r n − 1 ) where a = 1 a=1 a = 1 and r = ( 1 + i ) r=(1+i) r = ( 1 + i ) ): F = A [ ( 1 + i ) n − 1 ( 1 + i ) − 1 ] = A [ ( 1 + i ) n − 1 i ] F = A \left[ \frac{(1+i)^n - 1}{(1+i) - 1} \right] = A \left[ \frac{(1+i)^n - 1}{i} \right] F = A [ ( 1 + i ) − 1 ( 1 + i ) n − 1 ] = A [ i ( 1 + i ) n − 1 ] ( F / A , i , n ) = ( 1 + i ) n − 1 i (F/A, i, n) = \frac{(1+i)^n - 1}{i} ( F / A , i , n ) = i ( 1 + i ) n − 1
Factor 4: Sinking Fund Factor (SFF) — ( A / F , i , n ) (A/F, i, n) ( A / F , i , n )
Objective: Determine the uniform periodic payment A A A that must be deposited at the end of each of n n n periods to accumulate a targeted future sum F F F .
Formula: A = F [ i ( 1 + i ) n − 1 ] ⟹ ( A / F , i , n ) = i ( 1 + i ) n − 1 A = F \left[ \frac{i}{(1+i)^n - 1} \right] \implies (A/F, i, n) = \frac{i}{(1+i)^n - 1} A = F [ ( 1 + i ) n − 1 i ] ⟹ ( A / F , i , n ) = ( 1 + i ) n − 1 i
Factor 5: Uniform Series Present Worth Factor (USPWF) — ( P / A , i , n ) (P/A, i, n) ( P / A , i , n )
Objective: Compute the present lump-sum P P P equivalent to a series of n n n uniform end-of-period cash flows A A A .
Mathematical Derivation: Substitute the future worth F = P ( 1 + i ) n F = P(1+i)^n F = P ( 1 + i ) n into the ( F / A ) (F/A) ( F / A ) equation: P ( 1 + i ) n = A [ ( 1 + i ) n − 1 i ] ⟹ P = A [ ( 1 + i ) n − 1 i ( 1 + i ) n ] P(1+i)^n = A \left[ \frac{(1+i)^n - 1}{i} \right] \implies P = A \left[ \frac{(1+i)^n - 1}{i(1+i)^n} \right] P ( 1 + i ) n = A [ i ( 1 + i ) n − 1 ] ⟹ P = A [ i ( 1 + i ) n ( 1 + i ) n − 1 ] ( P / A , i , n ) = ( 1 + i ) n − 1 i ( 1 + i ) n = 1 − ( 1 + i ) − n i (P/A, i, n) = \frac{(1+i)^n - 1}{i(1+i)^n} = \frac{1 - (1+i)^{-n}}{i} ( P / A , i , n ) = i ( 1 + i ) n ( 1 + i ) n − 1 = i 1 − ( 1 + i ) − n
Factor 6: Capital Recovery Factor (CRF) — ( A / P , i , n ) (A/P, i, n) ( A / P , i , n )
Objective: Determine the uniform annual revenue or savings A A A required over n n n periods to fully recover an initial capital investment P P P and provide a return on investment at rate i i i .
Formula: A = P [ i ( 1 + i ) n ( 1 + i ) n − 1 ] = P [ i 1 − ( 1 + i ) − n ] ⟹ ( A / P , i , n ) = i ( 1 + i ) n ( 1 + i ) n − 1 A = P \left[ \frac{i(1+i)^n}{(1+i)^n - 1} \right] = P \left[ \frac{i}{1 - (1+i)^{-n}} \right] \implies (A/P, i, n) = \frac{i(1+i)^n}{(1+i)^n - 1} A = P [ ( 1 + i ) n − 1 i ( 1 + i ) n ] = P [ 1 − ( 1 + i ) − n i ] ⟹ ( A / P , i , n ) = ( 1 + i ) n − 1 i ( 1 + i ) n
The Critical Capital Recovery Identity
A celebrated mathematical relationship tested frequently on the CCP exam connects Capital Recovery to the Sinking Fund Factor: ( A / P , i , n ) = ( A / F , i , n ) + i (A/P, i, n) = (A/F, i, n) + i ( A / P , i , n ) = ( A / F , i , n ) + i
Proof: ( A / F , i , n ) + i = i ( 1 + i ) n − 1 + i [ ( 1 + i ) n − 1 ] ( 1 + i ) n − 1 = i + i ( 1 + i ) n − i ( 1 + i ) n − 1 = i ( 1 + i ) n ( 1 + i ) n − 1 = ( A / P , i , n ) (A/F, i, n) + i = \frac{i}{(1+i)^n - 1} + \frac{i[(1+i)^n - 1]}{(1+i)^n - 1} = \frac{i + i(1+i)^n - i}{(1+i)^n - 1} = \frac{i(1+i)^n}{(1+i)^n - 1} = (A/P, i, n) ( A / F , i , n ) + i = ( 1 + i ) n − 1 i + ( 1 + i ) n − 1 i [( 1 + i ) n − 1 ] = ( 1 + i ) n − 1 i + i ( 1 + i ) n − i = ( 1 + i ) n − 1 i ( 1 + i ) n = ( A / P , i , n )
Economic Meaning: The annual amount required to recover capital ( C R F CRF CRF ) equals the amount needed to repay the principal into a sinking fund ( S F F SFF SFF ) plus the annual interest on the outstanding principal ( i i i ).
3. Arithmetic Gradient Series Factors
In real industrial operations, operating and maintenance (O&M) expenditures rarely remain perfectly uniform. Equipment wear, component degradation, and rising maintenance labor cause costs to escalate by a constant monetary amount each year. +-----------------------------------------------------------------------------------------+
- | ARITHMETIC GRADIENT CASH FLOW STRUCTURE |
- | |
- | Base Uniform Amount (A1) PLUS Constant Periodic Increment (G): |
- | - End of Year 1: CF1 = A1 |
- | - End of Year 2: CF2 = A1 + 1G |
- | - End of Year 3: CF3 = A1 + 2G |
- | - End of Year n: CFn = A1 + (n - 1)G |
- | |
- | NOTICE: At t = 0, Gradient = 0. At t = 1, Gradient = 0*G. At t = 2, Gradient = 1*G. |
- +-----------------------------------------------------------------------------------------+
- 1. Arithmetic Gradient Present Worth Factor — ( P / G , i , n ) (P/G, i, n) ( P / G , i , n )
To find the present worth P G P_G P G of the gradient component alone ( 0 0 0 at t = 1 t=1 t = 1 , 1 G 1G 1 G at t = 2 t=2 t = 2 , … \dots … , ( n − 1 ) G (n-1)G ( n − 1 ) G at t = n t=n t = n ): ( P / G , i , n ) = 1 i [ ( 1 + i ) n − 1 i ( 1 + i ) n − n ( 1 + i ) n ] = ( 1 + i ) n − 1 − i ⋅ n i 2 ( 1 + i ) n (P/G, i, n) = \frac{1}{i} \left[ \frac{(1+i)^n - 1}{i(1+i)^n} - \frac{n}{(1+i)^n} \right] = \frac{(1+i)^n - 1 - i \cdot n}{i^2(1+i)^n} ( P / G , i , n ) = i 1 [ i ( 1 + i ) n ( 1 + i ) n − 1 − ( 1 + i ) n n ] = i 2 ( 1 + i ) n ( 1 + i ) n − 1 − i ⋅ n
2. Arithmetic Gradient Uniform Series Factor — ( A / G , i , n ) (A/G, i, n) ( A / G , i , n )
To convert an arithmetic gradient into an equivalent uniform annual series A G A_G A G : ( A / G , i , n ) = ( P / G , i , n ) ( P / A , i , n ) = 1 i − n ( 1 + i ) n − 1 (A/G, i, n) = \frac{(P/G, i, n)}{(P/A, i, n)} = \frac{1}{i} - \frac{n}{(1+i)^n - 1} ( A / G , i , n ) = ( P / A , i , n ) ( P / G , i , n ) = i 1 − ( 1 + i ) n − 1 n
Total Equivalent Series Formulations
To evaluate an entire cash flow stream comprising a base series A 1 A_1 A 1 and an arithmetic gradient G G G : P total = A 1 ( P / A , i , n ) ± G ( P / G , i , n ) \mathbf{P_{\text{total}} = A_1(P/A, i, n) \pm G(P/G, i, n)} P total = A 1 ( P/A , i , n ) ± G ( P/G , i , n ) A total = A 1 ± G ( A / G , i , n ) \mathbf{A_{\text{total}} = A_1 \pm G(A/G, i, n)} A total = A 1 ± G ( A/G , i , n )
(Use + + + for escalating costs/revenues and − - − for declining series).
4. Geometric Gradient Series Factors
A Geometric Gradient models cash flows that change by a constant compound percentage rate ( g g g ) each period (such as energy inflation, union labor rate escalations, or product demand decay): C F t = A 1 ( 1 + g ) t − 1 for t = 1 , 2 , … , n CF_t = A_1(1 + g)^{t-1} \quad \text{for } t = 1, 2, \dots, n C F t = A 1 ( 1 + g ) t − 1 for t = 1 , 2 , … , n +-----------------------------------------------------------------------------------------+
- | GEOMETRIC GRADIENT PRESENT WORTH EQUATIONS |
- | |
- | CASE 1: Discount Rate (i) != Escalation Rate (g) |
- | P = A1 * [ 1 - (1 + g)^n * (1 + i)^(-n) ] / (i - g) |
- | = A1 * [ 1 - ((1 + g) / (1 + i))^n ] / (i - g) |
- | |
- | CASE 2: Discount Rate (i) == Escalation Rate (g) |
- | P = A1 * [ n / (1 + i) ] = n * A1 * (1 + i)^(-1) |
- +-----------------------------------------------------------------------------------------+
Important
Special Case ( i = g i = g i = g ): When the discount rate equals the escalation rate ( i = g i = g i = g ), attempting to apply Case 1 results in division by zero. Cost engineers must recognize this condition immediately and apply the simplified formula: P = n ⋅ A 1 1 + i P = \frac{n \cdot A_1}{1 + i} P = 1 + i n ⋅ A 1 .
5. Master Engineering Economics Factor Reference Matrix
Factor Name Standard Functional Notation Mathematical Closed-Form Formula Solves For (Given) Common Cost Engineering Application
Single Payment Compound Amount ( F / P , i , n ) (F/P, i, n) ( F / P , i , n ) ( 1 + i ) n (1 + i)^n ( 1 + i ) n F = P × ( F / P ) F = P \times (F/P) F = P × ( F / P ) Future value of an upfront capital outlay.
Single Payment Present Worth ( P / F , i , n ) (P/F, i, n) ( P / F , i , n ) ( 1 + i ) − n (1 + i)^{-n} ( 1 + i ) − n P = F × ( P / F ) P = F \times (P/F) P = F × ( P / F ) Present value of single salvage value or future cost.
Uniform Series Compound Amount ( F / A , i , n ) (F/A, i, n) ( F / A , i , n ) ( 1 + i ) n − 1 i \frac{(1+i)^n - 1}{i} i ( 1 + i ) n − 1 F = A × ( F / A ) F = A \times (F/A) F = A × ( F / A ) Accumulated total in an escrow/reserve fund.
Sinking Fund ( A / F , i , n ) (A/F, i, n) ( A / F , i , n ) i ( 1 + i ) n − 1 \frac{i}{(1+i)^n - 1} ( 1 + i ) n − 1 i A = F × ( A / F ) A = F \times (A/F) A = F × ( A / F ) Required annual deposit for future decommissioning.
Uniform Series Present Worth ( P / A , i , n ) (P/A, i, n) ( P / A , i , n ) ( 1 + i ) n − 1 i ( 1 + i ) n \frac{(1+i)^n - 1}{i(1+i)^n} i ( 1 + i ) n ( 1 + i ) n − 1 P = A × ( P / A ) P = A \times (P/A) P = A × ( P / A ) Present worth of annual operational savings/revenues.
Capital Recovery ( A / P , i , n ) (A/P, i, n) ( A / P , i , n ) i ( 1 + i ) n ( 1 + i ) n − 1 \frac{i(1+i)^n}{(1+i)^n - 1} ( 1 + i ) n − 1 i ( 1 + i ) n A = P × ( A / P ) A = P \times (A/P) A = P × ( A / P ) Equivalent Uniform Annual Cost (EUAC) of CAPEX.
Arithmetic Gradient Present Worth ( P / G , i , n ) (P/G, i, n) ( P / G , i , n ) ( 1 + i ) n − 1 − i n i 2 ( 1 + i ) n \frac{(1+i)^n - 1 - i n}{i^2(1+i)^n} i 2 ( 1 + i ) n ( 1 + i ) n − 1 − in P G = G × ( P / G ) P_G = G \times (P/G) P G = G × ( P / G ) Present worth of linearly escalating O&M costs.
Arithmetic Gradient Uniform Series ( A / G , i , n ) (A/G, i, n) ( A / G , i , n ) 1 i − n ( 1 + i ) n − 1 \frac{1}{i} - \frac{n}{(1+i)^n - 1} i 1 − ( 1 + i ) n − 1 n A G = G × ( A / G ) A_G = G \times (A/G) A G = G × ( A / G ) Annualized equivalent of a step-wise gradient.
Geometric Gradient Present Worth ( i ≠ g i \neq g i = g ) ( P / A 1 , g , i , n ) (P/A_1, g, i, n) ( P / A 1 , g , i , n ) 1 − ( 1 + g ) n ( 1 + i ) − n i − g \frac{1 - (1+g)^n(1+i)^{-n}}{i - g} i − g 1 − ( 1 + g ) n ( 1 + i ) − n P = A 1 × ( P / A 1 ) P = A_1 \times (P/A_1) P = A 1 × ( P / A 1 ) Present worth of percentage-escalating revenues.
- 6. Step-by-Step Worked Engineering Calculations
- Example 1: Capital Recovery with Salvage Value (EUAC Analysis)
An industrial gas production facility is installing a high-pressure cryogenic compressor system. Capital cost is P = $ 600 , 000 P = \text{\textdollar}600,000 P = $ 600 , 000 . The asset has an anticipated operational lifespan of n = 8 years n = 8\text{ years} n = 8 years and a projected market salvage value of S = $ 80 , 000 S = \text{\textdollar}80,000 S = $ 80 , 000 at the end of Year 8. The corporate MARR is i = 8.0 % i = 8.0\% i = 8.0% per annum.
Calculate the Equivalent Uniform Annual Cost (EUAC) representing the capital recovery of this asset.
Method 1: Standard Capital Recovery Minus Sinking Fund Salvage E U A C = P ( A / P , 8 % , 8 ) − S ( A / F , 8 % , 8 ) EUAC = P(A/P, 8\%, 8) - S(A/F, 8\%, 8) E U A C = P ( A / P , 8% , 8 ) − S ( A / F , 8% , 8 )
Calculate ( A / P , 8 % , 8 ) (A/P, 8\%, 8) ( A / P , 8% , 8 ) : ( A / P , 8 % , 8 ) = 0.08 ( 1.08 ) 8 ( 1.08 ) 8 − 1 = 0.08 ( 1.850930 ) 1.850930 − 1 = 0.148074 0.850930 = 0.174015 (A/P, 8\%, 8) = \frac{0.08(1.08)^8}{(1.08)^8 - 1} = \frac{0.08(1.850930)}{1.850930 - 1} = \frac{0.148074}{0.850930} = \mathbf{0.174015} ( A / P , 8% , 8 ) = ( 1.08 ) 8 − 1 0.08 ( 1.08 ) 8 = 1.850930 − 1 0.08 ( 1.850930 ) = 0.850930 0.148074 = 0.174015
Calculate ( A / F , 8 % , 8 ) (A/F, 8\%, 8) ( A / F , 8% , 8 ) : ( A / F , 8 % , 8 ) = 0.08 ( 1.08 ) 8 − 1 = 0.08 0.850930 = 0.094015 (A/F, 8\%, 8) = \frac{0.08}{(1.08)^8 - 1} = \frac{0.08}{0.850930} = \mathbf{0.094015} ( A / F , 8% , 8 ) = ( 1.08 ) 8 − 1 0.08 = 0.850930 0.08 = 0.094015 (Notice that ( A / P ) − ( A / F ) = 0.174015 − 0.094015 = 0.080000 = i (A/P) - (A/F) = 0.174015 - 0.094015 = 0.080000 = i ( A / P ) − ( A / F ) = 0.174015 − 0.094015 = 0.080000 = i , confirming the fundamental identity!)
Calculate Annualized Capital Recovery: E U A C = [ 600 , 000 × 0.174015 ] − [ 80 , 000 × 0.094015 ] EUAC = [600,000 \times 0.174015] - [80,000 \times 0.094015] E U A C = [ 600 , 000 × 0.174015 ] − [ 80 , 000 × 0.094015 ] E U A C = 104 , 409.00 − 7 , 521.20 = $ 96 , 887.80 / year EUAC = 104,409.00 - 7,521.20 = \mathbf{\$96,887.80 / \text{year}} E U A C = 104 , 409.00 − 7 , 521.20 = $96 , 887.80/ year
Method 2: Depreciation Plus Interest on Salvage (Check) E U A C = ( P − S ) ( A / P , i , n ) + S ⋅ i EUAC = (P - S)(A/P, i, n) + S \cdot i E U A C = ( P − S ) ( A / P , i , n ) + S ⋅ i E U A C = ( 600 , 000 − 80 , 000 ) ( 0.174015 ) + ( 80 , 000 × 0.08 ) EUAC = (600,000 - 80,000)(0.174015) + (80,000 \times 0.08) E U A C = ( 600 , 000 − 80 , 000 ) ( 0.174015 ) + ( 80 , 000 × 0.08 ) E U A C = ( 520 , 000 × 0.174015 ) + 6 , 400.00 = 90 , 487.80 + 6 , 400.00 = $ 96 , 887.80 / year EUAC = (520,000 \times 0.174015) + 6,400.00 = 90,487.80 + 6,400.00 = \mathbf{\$96,887.80 / \text{year}} E U A C = ( 520 , 000 × 0.174015 ) + 6 , 400.00 = 90 , 487.80 + 6 , 400.00 = $96 , 887.80/ year
(Both methods yield identical results, providing an indispensable exam check mechanism).
Example 2: Maintenance Escalation via Arithmetic Gradient
A water treatment plant pump station requires A 1 A_1 A 1 = $20,000 in maintenance at the end of Year 1. Due to hydraulic component wear, maintenance costs increase by G G G = $4,000 each year through Year 6 ( n = 6 n = 6 n = 6 ). The corporate discount rate is i = 10.0 % i = 10.0\% i = 10.0% .
Calculate the Total Present Worth ( P total P_{\text{total}} P total ) and the Equivalent Uniform Annual Cost ( A total A_{\text{total}} A total ) of maintenance.
Calculate ( P / A , 10 % , 6 ) (P/A, 10\%, 6) ( P / A , 10% , 6 ) : ( P / A , 10 % , 6 ) = ( 1.10 ) 6 − 1 0.10 ( 1.10 ) 6 = 1.771561 − 1 0.177156 = 4.355261 (P/A, 10\%, 6) = \frac{(1.10)^6 - 1}{0.10(1.10)^6} = \frac{1.771561 - 1}{0.177156} = \mathbf{4.355261} ( P / A , 10% , 6 ) = 0.10 ( 1.10 ) 6 ( 1.10 ) 6 − 1 = 0.177156 1.771561 − 1 = 4.355261
Calculate ( P / G , 10 % , 6 ) (P/G, 10\%, 6) ( P / G , 10% , 6 ) : ( P / G , 10 % , 6 ) = ( 1.10 ) 6 − 1 − ( 0.10 × 6 ) ( 0.10 ) 2 ( 1.10 ) 6 = 1.771561 − 1 − 0.60 0.01 × 1.771561 = 0.171561 0.017716 = 9.684170 (P/G, 10\%, 6) = \frac{(1.10)^6 - 1 - (0.10 \times 6)}{(0.10)^2(1.10)^6} = \frac{1.771561 - 1 - 0.60}{0.01 \times 1.771561} = \frac{0.171561}{0.017716} = \mathbf{9.684170} ( P / G , 10% , 6 ) = ( 0.10 ) 2 ( 1.10 ) 6 ( 1.10 ) 6 − 1 − ( 0.10 × 6 ) = 0.01 × 1.771561 1.771561 − 1 − 0.60 = 0.017716 0.171561 = 9.684170
Calculate Total Present Worth ( P total P_{\text{total}} P total ): P total = A 1 ( P / A , 10 % , 6 ) + G ( P / G , 10 % , 6 ) P_{\text{total}} = A_1(P/A, 10\%, 6) + G(P/G, 10\%, 6) P total = A 1 ( P / A , 10% , 6 ) + G ( P / G , 10% , 6 ) P total = ( 20 , 000 × 4.355261 ) + ( 4 , 000 × 9.684170 ) P_{\text{total}} = (20,000 \times 4.355261) + (4,000 \times 9.684170) P total = ( 20 , 000 × 4.355261 ) + ( 4 , 000 × 9.684170 ) P total = 87 , 105.22 + 38 , 736.68 = $ 125 , 841.90 P_{\text{total}} = 87,105.22 + 38,736.68 = \mathbf{\$125,841.90} P total = 87 , 105.22 + 38 , 736.68 = $125 , 841.90
Calculate Equivalent Uniform Annual Series ( A total A_{\text{total}} A total ): A total = P total × ( A / P , 10 % , 6 ) = 125 , 841.90 4.355261 = $ 28 , 894.23 / year A_{\text{total}} = P_{\text{total}} \times (A/P, 10\%, 6) = \frac{125,841.90}{4.355261} = \mathbf{\$28,894.23 / \text{year}} A total = P total × ( A / P , 10% , 6 ) = 4.355261 125 , 841.90 = $28 , 894.23/ year (Alternatively: A total = A 1 + G ( A / G , 10 % , 6 ) = 20 , 000 + 4 , 000 × ( 9.684170 / 4.355261 ) = 20 , 000 + 4 , 000 ( 2.223558 ) = $ 28 , 894.23 A_{\text{total}} = A_1 + G(A/G, 10\%, 6) = 20,000 + 4,000 \times (9.684170 / 4.355261) = 20,000 + 4,000(2.223558) = \text{\textdollar}28,894.23 A total = A 1 + G ( A / G , 10% , 6 ) = 20 , 000 + 4 , 000 × ( 9.684170/4.355261 ) = 20 , 000 + 4 , 000 ( 2.223558 ) = $ 28 , 894.23 ).