5.2 Evaluating Alternatives: Present Worth, Equivalent Annual Cost, and IRR

Industrial engineers frequently evaluate competing capital investment proposals—such as selecting an automated material handling conveyor, replacing a CNC machine tool, or contracting an outside logistics vendor. To reach an optimal engineering decision, candidate projects must be systematically compared against corporate benchmark criteria using time-tested equivalence metrics.

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1. The Minimum Attractive Rate of Return (MARR)

The Minimum Attractive Rate of Return (MARR), also known as the corporate hurdle rate or cutoff rate, is the minimum nominal rate of return that an engineering project must earn to be considered financially acceptable.

Factors Determining MARR

1. Weighted Average Cost of Capital (WACC): The blended interest rate paid to debt holders (bonds, loans) and equity shareholders.
2. Opportunity Cost: The rate of return foregone by investing capital in the proposed project rather than the next best alternative.
3. Perceived Risk: High-risk R&D endeavors or unproven technologies demand a higher MARR than routine equipment replacements.
4. Capital Rationing Constraints: When available capital budget is tight, MARR is artificially adjusted upward to filter down to only the highest-yield projects.

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2. Present Worth (PW) Analysis and the Unequal Lives Problem

The Present Worth (PW) method discounts all anticipated cash inflows and outflows to the present moment ($t = 0$) using the established MARR:

$PW(MARR) = \sum_{t=0}^n CF_t (1 + MARR)^{-t}$

Decision Criteria

• Independent Projects: Accept any project where $PW(MARR) \ge 0$. Each project is evaluated on its own merits.
• Mutually Exclusive Projects: Only one alternative can be selected. For revenue projects, choose the single alternative that maximizes $PW(MARR)$. For service (cost-only) projects providing identical utility, select the alternative with the least negative $PW$ (lowest present value of costs).

The Unequal Lives Dilemma

A critical error on the FE exam is directly comparing the Present Worth of two alternatives that have different service lives (e.g., comparing Machine A with a 3-year life against Machine B with a 6-year life). This violates the requirement for an identical service comparison window.

Unequal Lives Resolution
 ├── Least Common Multiple (LCM) of Lives ──> Repeat asset cycles until lifetimes align (e.g., 3 yr & 4 yr -> 12 yr LCM)
 ├── Fixed Study Period (Planning Horizon) ──> Evaluate over fixed time window; truncate longer assets with estimated terminal salvage
 └── Annual Worth (AW / EUAC) Method ──> Compare on annual basis; identical repeatability assumption handles life mismatch automatically

1. Least Common Multiple (LCM) of Lives: Evaluate both alternatives over a study period equal to the least common multiple of their lifespans ($n_{\text{LCM}} = \operatorname{lcm}(n_A, n_B)$). This assumes the repeatability assumption: each asset can be repurchased at the end of its life with identical costs and performance.
2. Fixed Study Period (Planning Horizon): When repeatability is unrealistic (e.g., rapid technological obsolescence), an arbitrary horizon (e.g., 5 years) is specified. Assets retired before the horizon are replaced; assets lasting beyond the horizon are credited with their estimated market salvage value at year 5.

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3. Capitalized Cost for Perpetual Infrastructure

Capitalized Cost ($P$) is the present worth of a project whose service life is assumed to be infinite ($n \to \infty$). It is standard practice in evaluating major civil and industrial infrastructure such as dams, railway bridges, municipal drainage systems, and permanent university endowments.

Perpetual Uniform Annual Cash Flow

From the series present worth factor, as $n \to \infty$: $\lim_{n \to \infty} (P/A, i, n) = \lim_{n \to \infty} \frac{(1 + i)^n - 1}{i(1 + i)^n} = \frac{1}{i}$

Therefore, the present worth required to fund a perpetual end-of-year annual disbursement $A$ forever is: $P = \frac{A}{i}$

Periodic Renewal Every $k$ Years

When an asset requires an initial capital outlay $C_0$ and recurring overhauls or replacements costing $C_R$ every $k$ years into perpetuity:

1. Convert the periodic replacement cost $C_R$ into an equivalent uniform annual series $A_R$ over its renewal cycle $k$ using the sinking fund factor: $A_R = C_R (A/F, i, k) = C_R \left[ \frac{i}{(1 + i)^k - 1} \right]$
2. Capitalize $A_R$ over an infinite horizon by dividing by $i$: $P_R = \frac{A_R}{i} = \frac{C_R}{(1 + i)^k - 1}$
3. The total capitalized cost is the sum of initial outlay, capitalized annual maintenance, and capitalized periodic renewals: $P_{\text{total}} = C_0 + \frac{A_{\text{annual}}}{i} + \frac{C_R}{(1 + i)^k - 1}$

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4. Equivalent Uniform Annual Cost / Annual Worth (AW / EUAC)

The Annual Worth (AW)—or Equivalent Uniform Annual Cost (EUAC) for cost-only alternatives—converts all cash flows into an equivalent uniform annual series extending over the asset's life:

$AW = PW \times (A/P, i, n)$

The Major Computational Advantage of AW

Unlike Present Worth, the Annual Worth method does not require an LCM analysis when the repeatability assumption holds. If an asset operates with a 4-year life, its equivalent annual cost over 1 cycle ($n = 4$) is mathematically identical to its equivalent annual cost over 3 repeated cycles ($n = 12$). Therefore, industrial engineers can compare AW directly using each asset's individual lifespan!

Capital Recovery with Salvage Value

The capital recovery ($CR$) component reflects the equivalent annual cost of owning the physical capital asset, factoring in initial cost $P$ and terminal salvage value $S$:

$CR = P(A/P, i, n) - S(A/F, i, n)$

Using the identity $(A/F, i, n) = (A/P, i, n) - i$, this equation rearranges into the convenient Depreciation-Plus-Interest Form:

$CR = (P - S)(A/P, i, n) + S \cdot i$

Total $EUAC$ adds Annual Operating Costs ($AOC$) to Capital Recovery: $EUAC = CR + AOC = (P - S)(A/P, i, n) + S \cdot i + AOC$

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5. Internal Rate of Return (IRR) and Polynomial Roots

The Internal Rate of Return (IRR), denoted $i^*$, is the discount rate that equates the present worth of cash inflows to the present worth of cash outflows—meaning the net present worth is zero:

$PW(i^*) = \sum_{t=0}^n CF_t (1 + i^*)^{-t} = 0$

Solving for IRR via Linear Interpolation

Because the equation is an $n$th-degree polynomial, finding $i^*$ by hand requires bracketing the root between two test rates ($i_1$ and $i_2$) where $PW_1 > 0$ and $PW_2 < 0$, then applying linear interpolation:

$i^* \approx i_1 + (i_2 - i_1) \left[ \frac{PW_1}{PW_1 - PW_2} \right]$

The Multiple IRR Pathology and Descartes' Rule of Signs

A cash flow sequence is conventional if there is exactly one sign reversal in the net cash flows (typically negative initial investment followed by positive operating inflows: $-, +, +, +$). Conventional cash flows guarantee a unique positive real rate of return $i^* > -1$.

A cash flow sequence is non-conventional if the net cash flow signs reverse two or more times (e.g., $-, +, +, -, +$), commonly occurring when major mid-life overhauls or end-of-life environmental site rehabilitations are incurred.

• Descartes' Rule of Signs: The maximum number of positive real roots ($i^* > 0$) for the polynomial $PW(i) = 0$ is equal to the number of sign changes in the net cash flow sequence ${CF_0, CF_1, CF_2, \dots, CF_n}$, or less than that by an even integer.
• Norstrom's Criterion: If the cumulative cash flow series $S_t = \sum_{k=0}^t CF_k$ starts negative and changes sign exactly once, there exists precisely one real positive internal rate of return, even if the individual period cash flows exhibit multiple sign changes.

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6. Step-by-Step Worked Engineering Calculations

Worked Example 5.2.1: Unequal Lives Evaluation via EUAC

Problem: An industrial plant must install a dust filtration system. Two mutually exclusive systems provide identical air cleaning performance at a MARR of 10%:

• System 1: Initial capital cost = $50,000; Service life = 3 years; Salvage value = $5,000; Annual operating cost = $14,000.
• System 2: Initial capital cost = $90,000; Service life = 6 years; Salvage value = $12,000; Annual operating cost = $7,000.

Determine the EUAC of each system and identify which system should be selected.

Solution:

1. Evaluate System 1 ($n = 3, i = 10%$):

   • $(A/P, 10%, 3) = \frac{0.10(1.10)^3}{(1.10)^3 - 1} = \frac{0.1331}{0.331} = 0.40211$
   • Capital Recovery: $CR_1 = (P - S)(A/P, 10%, 3) + S \cdot i$ $CR_1 = (50,000 - 5,000)(0.40211) + 5,000(0.10) = 45,000(0.40211) + 500 = 18,095 + 500 = \$18,595$
   • Total $EUAC_1 = CR_1 + AOC_1 = 18,595 + 14,000 = $32,595$

2. Evaluate System 2 ($n = 6, i = 10%$):

   • $(A/P, 10%, 6) = \frac{0.10(1.10)^6}{(1.10)^6 - 1} = \frac{0.177156}{0.771561} = 0.229607$
   • Capital Recovery: $CR_2 = (P - S)(A/P, 10%, 6) + S \cdot i$ $CR_2 = (90,000 - 12,000)(0.229607) + 12,000(0.10) = 78,000(0.229607) + 1,200 = 17,909 + 1,200 = \$19,109$
   • Total $EUAC_2 = CR_2 + AOC_2 = 19,109 + 7,000 = $26,109$

3. Comparison and Decision: Comparing annual costs: $\Delta EUAC = EUAC_1 - EUAC_2 = 32,595 - 26,109 = \$6,486\text{ annual savings with System 2}$ Select System 2, saving $6,486 per year.

Worked Example 5.2.2: Capitalized Cost of Industrial Infrastructure

Problem: A drainage bypass culvert for a petrochemical refinery requires an initial investment of $800,000. Routine annual inspection and clearing costs $20,000 per year. Every 12 years, the concrete lining must be completely replaced at an estimated cost of $150,000. Assuming an interest rate of 6% and an infinite project lifespan, calculate the capitalized cost of this culvert.

Solution:

1. Initial construction cost: $P_0 = $800,000$.
2. Capitalized cost of annual maintenance ($A = $20,000$): $P_{\text{maint}} = \frac{A}{i} = \frac{20,000}{0.06} = \$333,333$
3. Capitalized cost of recurring relining ($C_R = $150,000, k = 12$ years): $P_{\text{reline}} = \frac{C_R}{(1 + i)^k - 1} = \frac{150,000}{(1.06)^{12} - 1}$ $(1.06)^{12} = 2.012196 \implies (1.06)^{12} - 1 = 1.012196$ $P_{\text{reline}} = \frac{150,000}{1.012196} = \$148,193$
4. Sum all components: $P_{\text{total}} = 800,000 + 333,333 + 148,193 = \$1,281,526$
5. Engineering Conclusion: The capitalized cost to construct and maintain the bypass indefinitely is $1,281,526.

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7. NCEES Reference Handbook Tips & Realistic Exam Traps

• The Direct PW Comparison Trap: If comparing alternatives with lives of 4 and 6 years, never pick the one with the highest 4-year or 6-year Present Worth directly. You must either expand to the 12-year LCM or simply compute $AW$ over their respective single lifespans.
• Salvage Value Sign in Capital Recovery: Using $CR = (P - S)(A/P) + Si$, be certain to subtract salvage value inside the parentheses and add the interest on salvage. If using the formula $CR = P(A/P) - S(A/F)$, remember that salvage carries a minus sign because it is a cash inflow returning capital.
• Descartes' Signs Count: Signs are counted on net periodic cash flows, not cumulative cash flows. Count every time the sequence flips between positive and negative.