7.2 Capital Budgeting: Present Worth, Annual Worth, IRR, and B/C Ratio

7.2 Capital Budgeting: Present Worth, Annual Worth, IRR, and B/C Ratio

Methodology Overview: Capital budgeting evaluates and ranks alternative engineering investments to ensure capital is deployed efficiently. Choosing the correct decision criterion—Present Worth, Annual Worth, Internal Rate of Return, or Benefit-Cost ratio—depends on project ownership (private vs. public), service life constraints, and capital availability.

Present Worth (PW) and Annual Worth (AW) Analysis

Present Worth (PW) Method

Present Worth (PW) converts all cash inflows and outflows throughout a project's service life into an equivalent single dollar sum at $t=0$ using the Minimum Attractive Rate of Return (MARR):

$PW(i) = -P_0 + \sum_{t=1}^n CF_t (1+i)^{-t} + S_n (1+i)^{-n}$

Where $P_0$ is initial capital investment, $CF_t$ is net annual cash flow at year $t$, $S_n$ is net salvage value at year $n$, and $i = \text{MARR}$.

• Decision Standard for Independent Projects: Accept any project with $PW(\text{MARR}) \ge 0$.
• Decision Standard for Mutually Exclusive Projects: Select the single alternative with the largest positive Present Worth (for revenue options) or lowest negative Present Worth (for service/cost-only options).

The Equal Service Life Requirement & Least Common Multiple (LCM) Rule

Present Worth comparison is valid only if all alternatives are evaluated over equal study periods. If Alternative A has a 3-year life and Alternative B has a 6-year life:

• A direct PW comparison over 3 years versus 6 years is mathematically invalid because Option B provides twice as many years of service.
• Least Common Multiple (LCM) Rule: Repeat each alternative over a study period equal to the LCM of their individual service lives (e.g., $\text{LCM}(3, 6) = 6$ years). Option A is assumed to be replaced at $t=3$ with identical costs.

Annual Worth (AW) Method

Annual Worth (AW) converts all cash flows into an equivalent uniform annual series $A$ over the asset's service life $n$:

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

$\text{Or explicitly: } AW(i) = -P_0(A/P, i, n) + A_{\text{net}} + S_n(A/F, i, n)$

Why AW is Preferred for Unequal Service Lives

Key Exam Rule: The Annual Worth of an alternative over one life cycle is identical to its Annual Worth over two, three, or $k$ repeated life cycles (under the assumption of identical replacement costs). Therefore, Annual Worth analysis does NOT require the LCM rule and can compare options with different service lives directly over their respective single life spans.

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Capital Recovery Cost and Capitalized Cost

Capital Recovery Cost (CR)

Capital Recovery Cost (CR) represents the equivalent uniform annual cost of purchasing and owning an asset over $n$ years, taking into account initial cost $P$, salvage value $S$, and interest rate $i$.

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

By substituting $(A/F, i, n) = (A/P, i, n) - i$, the Capital Recovery formula simplifies to:

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

Where $(P - S)(A/P, i, n)$ is the annual recovery of depreciated capital, and $S \cdot i$ is the annual interest earned on unrecovered salvage value.

Summary of Capital Budgeting Decision Criteria

Capitalized Cost ($n \to \infty$)

Capitalized Cost is the present worth of an asset or facility designed to provide service in perpetuity ($n = \infty$), such as dams, bridges, canals, or municipal water mains.

For a perpetual annual expenditure $A$ occurring every year indefinitely at interest rate $i$:

$P_{\text{cap}} = \frac{A}{i}$

If an asset has an initial cost $P_0$, perpetual annual O&M cost $A$, and a periodic replacement cost $A_{\text{repl}}$ occurring every $k$ years indefinitely:

$P_{\text{cap}} = P_0 + \frac{A}{i} + \frac{A_{\text{repl}}}{(1+i)^k - 1}$

Where $\frac{A_{\text{repl}}}{(1+i)^k - 1}$ converts the $k$-year periodic expenditure into a capitalized present worth.

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Internal Rate of Return (IRR) Analysis

The Internal Rate of Return (IRR) is the break-even discount rate $i^$ at which the present worth of cash inflows equals the present worth of cash outflows ($PW(i^) = 0$).

$PW(i^*) = -P_0 + \sum_{t=1}^n \frac{CF_t}{(1+i^*)^t} + \frac{S_n}{(1+i^*)^n} = 0$

Linear Interpolation Formula for IRR

When solving for $i^*$ manually without a financial calculator, evaluate $PW(i)$ at two trial discount rates $i_1$ (yielding $PW_1 > 0$) and $i_2$ (yielding $PW_2 < 0$):

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

Incremental IRR Analysis ($\Delta \text{IRR}$) for Mutually Exclusive Options

Exam Trap Warning: Choosing the alternative with the highest individual IRR among mutually exclusive options can produce wrong decisions because individual IRR ignores the scale of initial investment.

To compare mutually exclusive alternatives using IRR, perform Incremental Analysis:

1. Rank options in order of increasing initial capital investment ($P_0$).
2. Verify each option individually against do-nothing ($IRR \ge \text{MARR}$). drop unviable options.
3. Compute the incremental cash flow between the higher-cost option ($B$) and lower-cost option ($A$): $\Delta CF_t = CF_{t,B} - CF_{t,A}$.
4. Calculate the incremental rate of return $\Delta i^$ such that $PW_{\Delta}(\Delta i^) = 0$.
5. Decision Rule: If $\Delta i^* \ge \text{MARR}$, select the higher-cost alternative ($B$). If $\Delta i^* < \text{MARR}$, select the lower-cost alternative ($A$).

Benefit-Cost (B/C) Ratio Analysis

Public sector infrastructure projects (such as highways, flood barriers, public parks, and transit systems) are evaluated using Benefit-Cost (B/C) ratio analysis rather than commercial profit metrics.

Definitions of Terms

• Benefits ($B$): Quantifiable advantage realized by the general public.
• Disbenefits ($D$): Quantifiable disadvantages or costs incurred by the public (e.g., loss of property value, environmental damage).
• Costs ($C$): Expenditures incurred by the government entity (initial capital cost $I$ and net annual maintenance/operating costs $O&M$).

Conventional B/C Ratio Formula

All terms are expressed as equivalent Present Worths ($PW$) or Equivalent Uniform Annual Worths ($AW$) evaluated at the public discount rate:

$\text{Conventional } B/C = \frac{PW(B) - PW(D)}{PW(I) + PW(O\&M) - PW(S)}$

Modified B/C Ratio Formula

In the modified B/C formulation, recurring maintenance costs ($O&M$) are subtracted from public benefits in the numerator rather than added to government costs in the denominator:

$\text{Modified } B/C = \frac{PW(B) - PW(D) - PW(O\&M)}{PW(I) - PW(S)}$

• Decision Rule (Single Option): If $B/C \ge 1.0$, the project is economically justified.
• Incremental B/C Rule (Mutually Exclusive Options): Calculate incremental benefits and costs between higher-cost alternative $B$ and lower-cost alternative $A$: $\Delta B / \Delta C = \frac{PW(B_B) - PW(B_A)}{PW(C_B) - PW(C_A)}$. Select Option B if $\Delta B / \Delta C \ge 1.0$.

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Payback Period Analysis

Simple Payback Period

The Simple Payback Period ($n_p$) is the number of years required for cumulative un-discounted net cash inflows to equal the initial capital investment $P_0$:

$P_0 = \sum_{t=1}^{n_p} CF_t$

If annual net cash inflows are constant ($A$): $n_p = \frac{P_0}{A}$

Limitation: Simple payback ignores the time value of money and all cash flows occurring after $n_p$.

Discounted Payback Period

The Discounted Payback Period accounts for the time value of money by finding $n_p$ such that:

$P_0 = \sum_{t=1}^{n_p} CF_t (1+i)^{-t} = A (P/A, i, n_p)$

Worked Capital Budgeting Examples

Worked Example 1: Annual Worth Comparison (Unequal Service Lives)

Problem: An engineering firm compares two industrial pumps at MARR = 10%:

• Pump A: Initial cost $P = $30,000$, annual O&M $= $5,000$, life $n = 3$ years, salvage value $S = $4,000$.
• Pump B: Initial cost $P = $55,000$, annual O&M $= $3,000$, life $n = 6$ years, salvage value $S = $7,000$.

Determine which pump should be selected based on Equivalent Uniform Annual Cost (EUAC).

Solution:

1. Calculate EUAC for Pump A ($n=3$): $(A/P, 10\%, 3) = \frac{0.10(1.10)^3}{(1.10)^3 - 1} = \frac{0.1331}{0.331} \approx 0.40211$ $(A/F, 10\%, 3) = \frac{0.10}{(1.10)^3 - 1} = \frac{0.10}{0.331} \approx 0.30211$ $\text{EUAC}_A = P_A(A/P, 10\%, 3) + \text{O\&M}_A - S_A(A/F, 10\%, 3)$ $\text{EUAC}_A = 30,000(0.40211) + 5,000 - 4,000(0.30211) = 12,063.30 + 5,000 - 1,208.44 = 15,854.86 \text{ dollars}$

2. Calculate EUAC for Pump B ($n=6$): $(A/P, 10\%, 6) = \frac{0.10(1.10)^6}{(1.10)^6 - 1} = \frac{0.177156}{0.771561} \approx 0.22961$ $(A/F, 10\%, 6) = \frac{0.10}{(1.10)^6 - 1} = \frac{0.10}{0.771561} \approx 0.12961$ $\text{EUAC}_B = 55,000(0.22961) + 3,000 - 7,000(0.12961) = 12,628.55 + 3,000 - 907.27 = 14,721.28 \text{ dollars}$

3. Decision: Compare $\text{EUAC}_A = 15,854.86 \text{ dollars}$ vs $\text{EUAC}_B = 14,721.28 \text{ dollars}$. Select Pump B because its annual equivalent cost is lower by $1,133.58 per year.

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Worked Example 2: Capitalized Cost Calculation

Problem: A bridge project costs $5,000,000 initially. Annual maintenance costs $60,000 per year. In addition, major resurfacing costing $400,000 is required every 12 years in perpetuity. At an interest rate of 5% per year, calculate the total capitalized cost of the bridge.

Solution:

1. Formulate components: $P_{\text{cap}} = P_0 + P_{\text{annual O\&M}} + P_{\text{periodic resurfacing}}$ $P_0 = 5,000,000 \text{ dollars}$ $P_{\text{annual O\&M}} = \frac{A}{i} = \frac{60,000}{0.05} = 1,200,000 \text{ dollars}$

2. Capitalized cost of 12-year periodic resurfacing: $P_{\text{periodic}} = \frac{A_{\text{repl}}}{(1+i)^k - 1} = \frac{400,000}{(1.05)^{12} - 1}$ $(1.05)^{12} = 1.795856$ $P_{\text{periodic}} = \frac{400,000}{1.795856 - 1} = \frac{400,000}{0.795856} = 502,603.66 \text{ dollars}$

3. Total Capitalized Cost: $P_{\text{cap}} = 5,000,000 + 1,200,000 + 502,603.66 = 6,702,603.66 \text{ dollars}$