Study guide
4.3 Nominal vs. Effective Interest Rates & Continuous Compounding
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
4.3 Nominal vs. Effective Interest Rates & Continuous Compounding
Key Takeaways Explain This Topic AI
The nominal interest rate (r) is the stated annual interest rate without intra-year compounding, whereas the effective annual interest rate (i_eff = (1 + r/m)^m - 1) reflects the true annual return accounting for m compounding periods per year.
As compounding frequency approaches infinity (m -> inf), discrete compounding converges to the continuous compounding limit: i_eff = e^r - 1.
Under continuous compounding of single lump sums, future and present worth calculations utilize natural exponential functions: F = P * e^(r n) and P = F * e^(-r n).
When cash flow payment intervals do not match interest compounding intervals, cost engineers must determine the effective rate corresponding precisely to the payment period (i_eff,PP) before applying discrete uniform series factors.
In debt financing and capital project loan covenants, evaluating projects using the nominal Annual Percentage Rate (APR) rather than the effective Annual Percentage Yield (APY) systematically underestimates borrowing costs.
Last updated: August 2026
4.3 Nominal vs. Effective Interest Rates & Continuous Compounding
In capital project financing, interest is rarely compounded on a strictly annual basis. Commercial credit agreements, project finance bonds, equipment lease contracts, and construction bank facilities frequently compound interest on semi-annual, quarterly, monthly, daily, or continuous schedules.
A common source of error on the Certified Cost Professional (CCP) examination involves confusing the nominal annual interest rate ( r r r ) with the effective annual interest rate ( i eff i_{\text{eff}} i eff ) , or failing to harmonize the interest compounding period ( C P CP CP ) with the cash flow payment period ( P P PP PP ). Mastering intra-year compounding and continuous exponential limits is essential for precise engineering economic evaluation.
1. Nominal vs. Effective Rates: Theoretical Distinction +-----------------------------------------------------------------------------+
- | NOMINAL VS. EFFECTIVE INTEREST RATE TAXONOMY |
- | |
- | NOMINAL ANNUAL RATE (r or i_nom / APR): |
- | - The contractual, stated annual interest rate. |
- | - Ignores the earning effect of intra-year compounding. |
- | - Sub-period interest rate: i_sub = r / m |
- | |
- | COMPOUNDING FREQUENCY (m): |
- | - Number of compounding sub-periods per calendar year. |
- | - Semi-Annual (m=2), Quarterly (m=4), Monthly (m=12), Daily (m=365). |
- | |
- | EFFECTIVE ANNUAL RATE (i_eff / APY): |
- | - The TRUE, actual annual interest rate earned or paid over a full year. |
- | - Accounts for compounding on intra-year interest accumulations. |
- | - Governing Formula: i_eff = (1 + r / m)^m - 1 |
- +-----------------------------------------------------------------------------+
- Mathematical Derivation of Effective Annual Rate ( i eff i_{\text{eff}} i eff )
Let r r r be the nominal annual interest rate and m m m be the number of compounding sub-periods per year.
The interest rate charged per sub-period is: i sub = r m i_{\text{sub}} = \frac{r}{m} i sub = m r
If a principal of $1.00 is invested for one full calendar year ( m m m sub-periods), the accumulated future value at year-end is: F 1 = 1.00 × ( 1 + r m ) m F_1 = 1.00 \times \left(1 + \frac{r}{m}\right)^m F 1 = 1.00 × ( 1 + m r ) m
The net effective annual interest rate ( i eff i_{\text{eff}} i eff ) earned on the $1.00 principal is: i eff = F 1 − 1.00 = ( 1 + r m ) m − 1 i_{\text{eff}} = F_1 - 1.00 = \mathbf{\left(1 + \frac{r}{m}\right)^m - 1} i eff = F 1 − 1.00 = ( 1 + m r ) m − 1
2. The Continuous Compounding Limit ( m → ∞ m \to \infty m → ∞ )
In large continuous industrial operations—such as oil refineries, power utilities, and chemical plants—revenues and operational expenses flow continuously throughout the year rather than arriving in discrete quarterly or annual lumps. To model continuous financial velocity, engineering economics extends discrete compounding to its mathematical limit as the compounding frequency approaches infinity ( m → ∞ m \to \infty m → ∞ ). +-----------------------------------------------------------------------------+
- | CONTINUOUS COMPOUNDING MATHEMATICAL DERIVATION |
- | |
- | Effective Annual Rate Limit: |
- | i_eff = lim_{m -> inf} [ (1 + r / m)^m - 1 ] |
- | |
- | Let k = m / r (as m -> inf, k -> inf): |
- | i_eff = lim_{k -> inf} [ (1 + 1 / k)^(k * r) ] - 1 |
- | = [ lim_{k -> inf} (1 + 1 / k)^k ]^r - 1 |
- | |
- | By Euler's limit definition of the base of natural logarithms (e): |
- | lim_{k -> inf} (1 + 1 / k)^k = e = 2.718281828... |
- | |
- | CONTINUOUS EFFECTIVE RATE FORMULA: i_eff = e^r - 1 |
- +-----------------------------------------------------------------------------+
- Single Payment Continuous Compounding Formulas
For single lump sums compounded continuously over n n n years at nominal annual rate r r r :
Future Worth ( F F F ): F = P e r n ⟹ ( F / P , r , n ) ∞ = e r n F = P e^{rn} \implies (F/P, r, n)_{\infty} = e^{rn} F = P e r n ⟹ ( F / P , r , n ) ∞ = e r n
Present Worth ( P P P ): P = F e − r n ⟹ ( P / F , r , n ) ∞ = e − r n P = F e^{-rn} \implies (P/F, r, n)_{\infty} = e^{-rn} P = F e − r n ⟹ ( P / F , r , n ) ∞ = e − r n
Continuous Uniform Cash Flow Compounding
When cash flows at a uniform nominal annual rate of A ˉ \bar{A} A ˉ dollars per year continuously throughout n n n years:
Present Worth of Continuous Cash Flow ( P P P ): P = ∫ 0 n A ˉ e − r t d t = A ˉ [ 1 − e − r n r ] P = \int_0^n \bar{A} e^{-rt} \, dt = \bar{A} \left[ \frac{1 - e^{-rn}}{r} \right] P = ∫ 0 n A ˉ e − r t d t = A ˉ [ r 1 − e − r n ]
Future Worth of Continuous Cash Flow ( F F F ): F = ∫ 0 n A ˉ e r ( n − t ) d t = A ˉ [ e r n − 1 r ] F = \int_0^n \bar{A} e^{r(n-t)} \, dt = \bar{A} \left[ \frac{e^{rn} - 1}{r} \right] F = ∫ 0 n A ˉ e r ( n − t ) d t = A ˉ [ r e r n − 1 ]
3. Comparative Compounding Spectrum Across Frequencies
To visualize how compounding frequency alters financial outcomes, consider an initial project investment of P = $ 100 , 000 P = \text{\textdollar}100,000 P = $ 100 , 000 evaluated at a nominal interest rate of r = 12.0 % r = 12.0\% r = 12.0% per annum over an asset life of n = 5 years n = 5\text{ years} n = 5 years .
Compounding Frequency Sub-Periods per Year ( m m m ) Sub-Period Interest Rate ( i sub = r / m i_{\text{sub}} = r/m i sub = r / m ) Effective Annual Rate ( i eff i_{\text{eff}} i eff ) Total Periods in 5 Years ( N = m ⋅ n N = m \cdot n N = m ⋅ n ) Future Value at Year 5 ( F 5 F_5 F 5 ) Additional Gain vs. Annual
- Annual 1 1 1 12.0000% 12.0000% 5 5 5 $176,234.17 Baseline ($0.00)
- Semi-Annual 2 2 2 6.0000% 12.3600% 10 10 10 $179,084.77 +$2,850.60
- Quarterly 4 4 4 3.0000% 12.5509% 20 20 20 $180,611.12 +$4,376.95
- Monthly 12 12 12 1.0000% 12.6825% 60 60 60 $181,669.67 +$5,435.50
- Weekly 52 52 52 0.2308% 12.7341% 260 260 260 $182,086.04 +$5,851.87
- Daily 365 365 365 0.0329% 12.7475% 1 , 825 1,825 1 , 825 $182,193.91 +$5,959.74
- Continuous ( m → ∞ m \to \infty m → ∞ ) ∞ \infty ∞ → 0 \to 0 → 0 12.7497% ∞ \infty ∞ $182,211.88 +$5,977.71
Key Analytical Observations:
Asymptotic Convergence: As compounding frequency increases from annual to continuous, the effective annual interest rate rises from 12.0000% to 12.7497%. The greatest incremental jump occurs between annual and monthly compounding (+68.25 bps); increasing from daily to continuous yields only an extra +0.22 bps.
Financial Asymmetry: For a borrower, higher compounding frequencies increase the total effective cost of debt. For an investor, higher compounding frequencies maximize terminal asset yield.
4. Resolving Payment Period vs. Compounding Period Mismatches
A paramount requirement on the AACE CCP exam is knowing how to handle scenarios where the Payment Period ( P P PP PP ) does not align with the Compounding Period ( C P CP CP ) . +-----------------------------------------------------------------------------------------+
- | HARMONIZATION OF PAYMENT & COMPOUNDING PERIODS |
- | |
- | THE GOLDEN RULE OF ENGINEERING ECONOMICS: |
- | The interest rate period MUST match the cash flow payment period! |
- | |
- | CASE 1: PP = CP (Standard Case) |
- | - Direct application of discrete formulas using i_sub = r / m and total periods N. |
- | |
- | CASE 2: PP > CP (Payment Period is Longer than Compounding Period) |
- | - Example: Annual deposits (PP = 1 yr) with Monthly compounding (CP = 1 mo). |
- | - SOLUTION: Calculate the Effective Interest Rate for the Payment Period (i_eff,PP): |
- | i_eff,PP = (1 + r / m)^k - 1 [where k = number of CPs per PP] |
- | - Then apply standard discrete formulas using i_eff,PP and n = number of PPs. |
- | |
- | CASE 3: PP < CP (Payment Period is Shorter than Compounding Period) |
- | - Example: Monthly deposits (PP = 1 mo) with Semi-Annual compounding (CP = 6 mo). |
- | - STANDARD CONVENTION: Unless contractually stated otherwise, deposits within an |
- | interest period earn no interest until the start of the next compounding period, |
- | OR calculate the equivalent nominal sub-rate i_PP = (1 + i_CP)^(1/k) - 1. |
- +-----------------------------------------------------------------------------------------+
- 5. Step-by-Step Worked Engineering Calculations
- Example 1: Comparing Commercial Debt Financing Proposals
A project developer is securing a $2,000,000 construction bridge loan for a 1-year development window. Four commercial lending institutions submit competitive credit terms:
Lender A: 8.25% nominal annual rate, compounded annually ( m = 1 m = 1 m = 1 ).
Lender B: 8.05% nominal annual rate, compounded monthly ( m = 12 m = 12 m = 12 ).
Lender C: 8.10% nominal annual rate, compounded quarterly ( m = 4 m = 4 m = 4 ).
Lender D: 8.00% nominal annual rate, compounded continuously ( m → ∞ m \to \infty m → ∞ ).
Determine which lender offers the lowest true cost of capital by calculating the Effective Annual Rate ( i eff i_{\text{eff}} i eff ) for each loan.
Calculations:
Lender A ( m = 1 m = 1 m = 1 ): i eff, A = ( 1 + 0.0825 ) 1 − 1 = 8.2500 % i_{\text{eff, A}} = (1 + 0.0825)^1 - 1 = \mathbf{8.2500\%} i eff, A = ( 1 + 0.0825 ) 1 − 1 = 8.2500%
Lender B ( m = 12 m = 12 m = 12 ): i eff, B = ( 1 + 0.0805 12 ) 12 − 1 = ( 1 + 0.0067083 ) 12 − 1 = 1.083543 − 1 = 8.3543 % i_{\text{eff, B}} = \left(1 + \frac{0.0805}{12}\right)^{12} - 1 = (1 + 0.0067083)^{12} - 1 = 1.083543 - 1 = \mathbf{8.3543\%} i eff, B = ( 1 + 12 0.0805 ) 12 − 1 = ( 1 + 0.0067083 ) 12 − 1 = 1.083543 − 1 = 8.3543%
Lender C ( m = 4 m = 4 m = 4 ): i eff, C = ( 1 + 0.0810 4 ) 4 − 1 = ( 1 + 0.02025 ) 4 − 1 = 1.083494 − 1 = 8.3494 % i_{\text{eff, C}} = \left(1 + \frac{0.0810}{4}\right)^4 - 1 = (1 + 0.02025)^4 - 1 = 1.083494 - 1 = \mathbf{8.3494\%} i eff, C = ( 1 + 4 0.0810 ) 4 − 1 = ( 1 + 0.02025 ) 4 − 1 = 1.083494 − 1 = 8.3494%
Lender D ( m → ∞ m \to \infty m → ∞ ): i eff, D = e 0.0800 − 1 = 2.7182818 0.08 − 1 = 1.083287 − 1 = 8.3287 % i_{\text{eff, D}} = e^{0.0800} - 1 = 2.7182818^{0.08} - 1 = 1.083287 - 1 = \mathbf{8.3287\%} i eff, D = e 0.0800 − 1 = 2.718281 8 0.08 − 1 = 1.083287 − 1 = 8.3287%
Decision Ranking:
1st (Lowest Cost): Lender A ( i eff = 8.2500 % i_{\text{eff}} = 8.2500\% i eff = 8.2500% , Total Year 1 Interest = $165,000.00)
2nd: Lender D ( i eff = 8.3287 % i_{\text{eff}} = 8.3287\% i eff = 8.3287% , Total Year 1 Interest = $166,574.00)
3rd: Lender C ( i eff = 8.3494 % i_{\text{eff}} = 8.3494\% i eff = 8.3494% , Total Year 1 Interest = $166,988.00)
4th (Highest Cost): Lender B ( i eff = 8.3543 % i_{\text{eff}} = 8.3543\% i eff = 8.3543% , Total Year 1 Interest = $167,086.00)
Key Takeaway: Despite Lender B advertising the second-lowest nominal APR (8.05%), monthly compounding causes it to be the most expensive credit facility. Lender A with the highest nominal APR (8.25%) is actually the cheapest loan because it lacks intra-year compounding.
Example 2: Sinking Fund with Mismatched Payment & Compounding Periods
An EPC contractor must accumulate F = $ 500 , 000 F = \text{\textdollar}500,000 F = $ 500 , 000 in a specialized environmental decommissioning reserve account at the end of n = 5 years n = 5\text{ years} n = 5 years . The contractor will make equal quarterly deposits ( A q A_q A q ) ( P P = quarterly PP = \text{quarterly} PP = quarterly , N = 20 quarters N = 20\text{ quarters} N = 20 quarters ). The reserve fund pays a nominal interest rate of r = 8.0 % r = 8.0\% r = 8.0% compounded monthly ( m = 12 m = 12 m = 12 , C P = monthly CP = \text{monthly} CP = monthly ).
Calculate the required quarterly deposit A q A_q A q .
Step 1: Calculate the Effective Interest Rate per Quarter ( i q i_q i q )
There are k = 3 monthly compounding periods k = 3\text{ monthly compounding periods} k = 3 monthly compounding periods per quarter ( 12 / 4 = 3 12 / 4 = 3 12/4 = 3 ). The monthly rate is i m = 0.08 / 12 = 0.0066667 i_m = 0.08 / 12 = 0.0066667 i m = 0.08/12 = 0.0066667 . ∋ q = ( 1 + i m ) 3 − 1 = ( 1 + 0.0066667 ) 3 − 1 = ( 1.0066667 ) 3 − 1 = 1.0201336 − 1 = 0.0201336 ( 2.01336 % ) \ni_q = (1 + i_m)^3 - 1 = (1 + 0.0066667)^3 - 1 = (1.0066667)^3 - 1 = 1.0201336 - 1 = \mathbf{0.0201336 \ (2.01336\%)} ∋ q = ( 1 + i m ) 3 − 1 = ( 1 + 0.0066667 ) 3 − 1 = ( 1.0066667 ) 3 − 1 = 1.0201336 − 1 = 0.0201336 ( 2.01336% )
Step 2: Calculate Required Quarterly Deposit Using Sinking Fund Factor A q = F × ( A / F , i q , 20 ) = F × [ i q ( 1 + i q ) 20 − 1 ] A_q = F \times (A/F, i_q, 20) = F \times \left[ \frac{i_q}{(1 + i_q)^{20} - 1} \right] A q = F × ( A / F , i q , 20 ) = F × [ ( 1 + i q ) 20 − 1 i q ]
Evaluate the compound growth term ( 1 + i q ) 20 (1 + i_q)^{20} ( 1 + i q ) 20 : ( 1 + 0.0201336 ) 20 = ( 1.0066667 ) 60 = 1.489846 (1 + 0.0201336)^{20} = (1.0066667)^{60} = 1.489846 ( 1 + 0.0201336 ) 20 = ( 1.0066667 ) 60 = 1.489846 ( A / F , 2.01336 % , 20 ) = 0.0201336 1.489846 − 1 = 0.0201336 0.489846 = 0.0411019 (A/F, 2.01336\%, 20) = \frac{0.0201336}{1.489846 - 1} = \frac{0.0201336}{0.489846} = \mathbf{0.0411019} ( A / F , 2.01336% , 20 ) = 1.489846 − 1 0.0201336 = 0.489846 0.0201336 = 0.0411019 A q = 500 , 000 × 0.0411019 = $ 20 , 550.95 / quarter A_q = 500,000 \times 0.0411019 = \mathbf{\$20,550.95 / \text{quarter}} A q = 500 , 000 × 0.0411019 = $20 , 550.95/ quarter
6. CCP Exam Alerts & Pitfalls
Warning
Exam Trap: APR vs. APY in Project Finance Commercial lending terms often quote Annual Percentage Rate (APR), which is the nominal rate ( r r r ). However, the true economic cost to the project is always the Annual Percentage Yield (APY), which is the effective rate ( i eff i_{\text{eff}} i eff ). Whenever compounding occurs more frequently than once per year, APY > APR!
Tip
Continuous Compounding Quick Rule: For single-payment continuous compounding questions on the CCP exam, remember that ( F / P , r , n ) = e r n (F/P, r, n) = e^{rn} ( F / P , r , n ) = e r n . On the computer testing software calculator, use the e x e^x e x function with x = r ⋅ n x = r \cdot n x = r ⋅ n .