13.1 Time Value of Money and Project Selection

13.1 Time Value of Money and Project Selection

Quick Answer: The Time Value of Money (TVM) is the core principle of engineering economics, stating that a dollar today is worth more than a dollar in the future due to its earning capacity (interest). The NCEES FE Reference Handbook provides standard formulas and interest tables for discrete cash flows. Project selection requires converting cash flows to a common basis using Present Worth (PW), Future Worth (FW), Annual Worth (AW), or the Rate of Return (ROR). For comparing projects with unequal lives, the Annual Worth (AW) method is the most efficient because it does not require repeating cycles to a Least Common Multiple (LCM) of lives.

Fundamental Principles of Time Value of Money

Engineering decisions almost always involve cash flows occurring at different times—such as capital investments today, annual operating expenses, periodic equipment replacements, and salvage values at the end of a project's life. To make valid comparisons, all cash flows must be moved to the same point in time. This process is called equivalence.

The relationship between present worth $P$ and future worth $F$ after $n$ periods at interest rate $i$ per period is:

$$F = P(1 + i)^n$$

This is the single-payment compound-amount factor, denoted as $(F/P, i, n)$. Conversely, the present value of a future sum is:

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

This is the single-payment present-worth factor, denoted as $(P/F, i, n)$.

Nominal and Effective Interest Rates

When interest compounding frequency does not match the annual payment period, we distinguish between the nominal interest rate $r$ and the effective annual interest rate $i_e$. The relationship is given by:

$$i_e = \left(1 + \frac{r}{m}\right)^m - 1$$

where $m$ is the number of compounding periods per year. If compounding is continuous, the effective interest rate is:

$$i_e = e^r - 1$$

On the FE exam, always check the compounding period. If a problem states "12% compounded monthly," the rate per month is $12\% / 12 = 1\%$, and the number of periods in 3 years is $3 \times 12 = 36$.

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Cash Flow Equivalence Factors

The FE Reference Handbook summarizes these standard discrete compounding factors. Below is a summary of the most critical factors:

Arithmetic and Geometric Gradients

An arithmetic gradient is a series where the cash flow increases by a constant amount $G$ each period (e.g., $100, $150, $200). The present worth of an arithmetic gradient is:

$$P = G(P/G, i, n) = G \left[ \frac{(1+i)^n - in - 1}{i^2(1+i)^n} \right]$$

A geometric gradient occurs when cash flows increase by a constant percentage $g$ each period (e.g., fuel cost increasing by 5% annually). The present worth is calculated as:

$$P = A_1 \left[ \frac{1 - (1+g)^n(1+i)^{-n}}{i - g} \right] \quad (\text{for } i \neq g)$$ $$P = A_1 \cdot n(1 + i)^{-1} \quad (\text{for } i = g)$$

where $A_1$ is the cash flow at the end of Year 1.

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Project Selection Metrics

To choose among competing alternatives, engineers evaluate:

1. Present Worth (PW): Converts all cash flows to the present time ($t = 0$). A project is viable if $PW(i) \ge 0$ at the Minimum Attractive Rate of Return (MARR).
2. Annual Worth (AW): Converts all cash flows to an equivalent uniform annual series. It is useful for comparing projects because $AW = PW(A/P, i, n)$.
3. Future Worth (FW): Evaluates cash flows at the end of the project life ($t = n$).
4. Rate of Return (ROR): The interest rate at which the present worth of net cash flows equals zero (i.e., $PW(ROR) = 0$). A project is acceptable if $ROR \ge MARR$.

Capitalized Cost

Capitalized cost is the present worth of an asset or project that is assumed to have an infinite life. This concept is typically applied to public works (e.g., dams, bridges) or permanent endowments. For a series of annual expenses $A$ stretching to infinity, the capitalized cost is:

$$Capitalized\ Cost = \frac{A}{i}$$

If the project involves a recurring replacement cost $C$ every $k$ periods, the capitalized cost is:

$$Capitalized\ Cost = P_{initial} + \frac{A}{i} + \frac{C}{(1+i)^k - 1}$$

where $P_{initial}$ is the first-time cost and $C$ is the replacement cost (less any salvage value) that occurs every $k$ periods.

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Comparison of Alternatives with Unequal Lives

When comparing mutually exclusive alternatives, the comparison must cover the same timeframe to be valid.

Equal Lives

For alternatives with the same lifespan, we directly compare their $PW$, $AW$, or $FW$ at the MARR. The option with the highest algebraically worth (least negative for cost-only projects, or highest positive for revenue-generating projects) is selected.

Unequal Lives

If the lifespans differ, comparing $PW$ over their individual lives is a fatal error. We must use one of the following methods:

• Least Common Multiple (LCM) of Lives: Repeat each project's cycle until they reach a common endpoint (e.g., for a 3-year and a 4-year project, the study period is 12 years). This assumes the projects can be renewed at identical costs.
• Annual Worth (AW) Method: Calculate the equivalent annual worth over each project's individual life. Because the AW represents the equivalent annual cost/benefit of a single cycle, it is automatically valid for infinite cycles of renewal. This is the preferred and fastest method on the FE exam.

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Lease vs. Buy vs. Make Decisions

In process plant design, engineers must choose between leasing equipment, purchasing it outright, or manufacturing a component in-house:

• Lease: Reduces initial capital expenditure, keeps debt off the balance sheet, and provides flexibility. However, it usually results in higher operating cash outflows over time and offers no salvage value or depreciation tax shield.
• Buy: Requires significant initial capital expenditure but results in lower operating costs, ownership of the asset (including salvage value), and tax benefits through depreciation.
• Make (In-house Production): Feasible when a chemical plant has excess utility capacity or raw material streams. The decision hinges on comparing the total cost of purchasing a chemical or service vs. the fixed capital investment and variable costs of producing it in-house. The break-even quantity is found where:

$$Total\ Cost_{Buy} = Total\ Cost_{Make} \Rightarrow Price \times Q = Fixed\ Cost + Variable\ Cost \times Q$$

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Worked Examples

Example 1: Unequal Lives Comparison

Compare two heat exchangers at a MARR of 10%:

• Exchanger A: Capital cost = $15,000, annual maintenance = $1,000, life = 3 years, salvage value = $3,000.
• Exchanger B: Capital cost = $25,000, annual maintenance = $500, life = 6 years, salvage value = $5,000.

Solution using the Annual Worth (AW) method: For Exchanger A:

$$AW_A = -15,000(A/P, 10\%, 3) - 1,000 + 3,000(A/F, 10\%, 3)$$

From interest tables or formulas: $(A/P, 10\%, 3) = \frac{0.1(1.1)^3}{(1.1)^3 - 1} = 0.4021$ $(A/F, 10\%, 3) = \frac{0.1}{(1.1)^3 - 1} = 0.3021$

$$AW_A = -15,000(0.4021) - 1,000 + 3,000(0.3021) = -6,031.50 - 1,000 + 906.30 = -\text{\textdollar}6,125.20$$

For Exchanger B:

$$AW_B = -25,000(A/P, 10\%, 6) - 500 + 5,000(A/F, 10\%, 6)$$

$(A/P, 10\%, 6) = \frac{0.1(1.1)^6}{(1.1)^6 - 1} = 0.2296$ $(A/F, 10\%, 6) = \frac{0.1}{(1.1)^6 - 1} = 0.1296$

$$AW_B = -25,000(0.2296) - 500 + 5,000(0.1296) = -5,740 - 500 + 648 = -\text{\textdollar}5,592$$

Conclusion: Exchanger B has a lower equivalent annual cost ($-\text{\textdollar}5,592$ vs. $-\text{\textdollar}6,125.20$) and is the preferred option.

Example 2: Capitalized Cost Calculation

A water treatment system has an initial installation cost of $50,000. Every 5 years, the filter membranes must be replaced at a cost of $12,000. The annual operating cost is $2,000. Calculate the capitalized cost of this system at a 6% annual interest rate.

Solution: The capitalized cost is the present worth of an infinite series of these expenses:

$$Capitalized\ Cost = Initial\ Installation\ Cost + PW_{operating} + PW_{replacements}$$ $$PW_{operating} = \frac{A}{i} = \frac{2,000}{0.06} = \text{\textdollar}33,333.33$$ $$PW_{replacements} = \frac{C}{(1+i)^k - 1} = \frac{12,000}{(1.06)^5 - 1} = \frac{12,000}{1.3382 - 1} = \frac{12,000}{0.3382} = \text{\textdollar}35,481.96$$ $$Capitalized\ Cost = 50,000 + 33,333.33 + 35,481.96 = \text{\textdollar}118,815.29$$