Lecture 6: Annual Cash Flow Analysis

Introduction to Annual Cash Flow Analysis

Annual cash flow analysis is the second of the three major economic analysis techniques. While present worth (PW) analysis converts consequences into a single equivalent sum at Time 0, annual cash flow analysis resolves alternatives into their Equivalent Uniform Annual Cash Flows.

The comparison of alternatives is based on three related measures:

Equivalent Uniform Annual Cost (EUAC)
Equivalent Uniform Annual Benefit (EUAB)
Equivalent Uniform Annual Worth (EUAW), which is defined as the difference: $\text{EUAW} = \text{EUAB} - \text{EUAC}$.

This method is highly versatile, especially when dealing with projects that have unequal lives or a continuing requirement.

Annual Cash Flow Calculations

The goal of annual cash flow analysis is to convert all costs, benefits, and salvage values into an equivalent uniform annual cost ($A$) or benefit, typically at the end of each period.

Resolving a Present Cost ($P$)

The simplest case converts a present sum ($P$) into an equivalent uniform end-of-period cash flow ($A$) over $n$ periods, using the capital recovery factor:

\[\text{EUAC} = P(A/P, i, n)\]

Capital Recovery Costs (Including Salvage Value)

Capital recovery cost is the equivalent uniform annual cost of owning an asset, considering its initial cost ($P$) and its final salvage value ($S$). When salvage value exists, it reduces the equivalent annual cost.

The capital recovery cost can be calculated in three algebraically equivalent ways:

Cash Flow Perspective: Annualized salvage value is subtracted from the annualized first cost: \[\text{EUAC} = P(A/P, i, n) - S(A/F, i, n)\]
Unrecovered Investment Perspective: Annual cost due to the decline in value $(P - S)$ plus the annual interest cost on the initial investment $P$: \[\text{EUAC} = (P - S)(A/F, i, n) + Pi\]
Capital Decline and Salvage Interest Perspective: Annual cost of the decline in value $(P - S)$ plus the annual interest cost on the salvage value $S$: \[\text{EUAC} = (P - S)(A/P, i, n) + Si\]

This quantity is also referred to as the capital recovery cost of the project.

Example 1: Capital Recovery Cost Calculation

Problem: A machine costs $\$30,000$ to purchase. The salvage value ($S$) after 5 years will be $\$7000$. Calculate the machine’s capital recovery cost (EUAC) for an interest rate of $10\%$.

Solution (Using the most common method - Equation 1):

\[\text{EUAC} = P(A/P, i, n) - S(A/F, i, n)\] \[\text{EUAC} = \$30,000(A/P, 10\%, 5) - \$7,000(A/F, 10\%, 5)\]

If $(A/P, 10\%, 5) \approx 0.2638$ and $(A/F, 10\%, 5) \approx 0.1638$

\[\text{EUAC} = \$30,000(0.2638) - \$7,000(0.1638)\] \[\text{EUAC} = \$7,914 - \$1,146.60\] \[\text{EUAC} = \$6,767.40\]

Irregular Cash Flows and Gradients

When dealing with irregular cash disbursements (such as mid-life overhaul costs) or arithmetic gradients, a two-step process is often convenient:

Convert the irregular cash flow series into an equivalent Present Worth of Cost (PW of cost).
Convert the total PW of Cost into an EUAC using the capital recovery factor:

\[\text{EUAC} = (\text{PW of cost})(A/P, i, n)\]

For cash flows that change by a constant amount $G$ (arithmetic gradient), the EUAC can be computed quickly by converting the gradient portion into an equivalent uniform series using the arithmetic gradient uniform series factor, $(A/G, i, n)$.

Example 2: EUAC for Irregular Costs (Gradient and Overhaul)

Problem: Compute the equivalent uniform annual cost (EUAC) over 5 years for a generator with $7\%$ interest, given the maintenance costs start at $\$545$ in Year 1 and increase by $\$45$ annually (gradient), and an overhaul cost of $\$750$ occurs in Year 3.

Solution: We resolve the costs into three components: the initial uniform amount, the arithmetic gradient, and the overhaul.

Uniform Component ($A_{base} = \$545$): \[\text{EUAC}_{A} = \$545\]
Arithmetic Gradient Component ($G = \$45$): \[\text{EUAC}_{G} = G(A/G, 7\%, 5) = \$45(1.865) \approx \$83.93\]
Overhaul Component ($F_3 = \$750$): We convert the future sum $F_3$ to a PW, then convert that PW to an EUAC over $n=5$ years: \[\text{PW}_{Overhaul} = \$750(P/F, 7\%, 3) \approx \$750(0.8163) = \$612.23\] \[\text{EUAC}_{Overhaul} = \text{PW}_{Overhaul}(A/P, 7\%, 5) \approx \$612.23(0.2439) \approx \$149.33\]

\[\text{Total EUAC} = \$545 + \$83.93 + \$149.33 = \$778.26\]

Annual Cash Flow Analysis Criteria

The selection criterion for annual cash flow analysis mirrors the present worth criteria, ensuring maximized economic efficiency based on the problem type.

Input/Output Situation	Criterion
Neither input nor output is fixed (Typical, general situation)	Maximize EUAW ($\text{EUAW} = \text{EUAB} - \text{EUAC}$)
Fixed Input (Amount of resources is limited)	Maximize EUAB
Fixed Output (A fixed task must be accomplished)	Minimize EUAC

Since EUAW is directly proportional to Net Present Worth ($\text{NPW} = \text{EUAW} \times (P/A, i, n)$), maximizing EUAW is equivalent to maximizing NPW.

Example 3: Maximizing EUAW

Problem: Device A costs $\$10,000$ and yields $\$3000$ in annual savings. Device B costs $\$13,500$ and yields savings starting at $\$3000$ and increasing by $\$500$ annually (arithmetic gradient). Both have a 5-year life and no salvage value. If interest is $7\%$, which device is preferred?

Solution: The criterion is to maximize EUAW ($\text{EUAW} = \text{EUAB} - \text{EUAC}$).

Device A: \[\text{EUAC}_{A} = P(A/P, 7\%, 5) = \$10,000(0.2439) \approx \$2,439\] \[\text{EUAB}_{A} = \$3,000 \text{ (Uniform annual savings)}\] \[\text{EUAW}_{A} = \$3,000 - \$2,439 = \$561\]

Device B: Requires converting the cost ($P$) to an EUAC, and the gradient benefit series to an EUAB. \[\text{EUAC}_{B} = P(A/P, 7\%, 5) = \$13,500(0.2439) \approx \$3,293\] \[\text{EUAB}_{B} = A_1 + G(A/G, 7\%, 5)\] \[\text{EUAB}_{B} = \$3,000 + \$500(1.865) \approx \$3,932.50\] \[\text{EUAW}_{B} = \$3,932.50 - \$3,293 = \$639.50\]

Since $\text{EUAW}_{B} (\$639.50)$ is greater than $\text{EUAW}_{A} (\$561)$, Device B is the preferred alternative.

Analysis Period Considerations

For annual cash flow analysis, comparison requires careful consideration of the analysis period, especially when alternatives have unequal lives.

Useful Lives Equal the Analysis Period: The study is based directly on that life.
Analysis Period is a Common Multiple of Alternative Lives: The customary assumption is identical replacement—when an alternative reaches the end of its life, it is replaced by an identical item with the same costs and performance. Under this assumption, the EUAC calculated over the individual asset’s useful life is equivalent to the EUAC over the entire longer analysis period.
Continuing Requirement (Indefinite Life): When the requirement for the asset is continuous or perpetual, the analysis period is assumed to be long or undefined. In this case, unequal-lived alternatives can be compared directly based on the EUAC calculated over each alternative’s own useful life. This simplified comparison relies on the assumption of identical replacement occurring indefinitely.
Infinite Analysis Period (Perpetual): This is a special case of the continuing requirement. When $n = \infty$, the capital recovery factor $(A/P, i, n)$ approaches the interest rate $i$. The EUAC computed for the limited life of a renewable asset is equal to the EUAC for the entire infinite analysis period.

Spreadsheets in Annual Cash Flow Analysis

Spreadsheets are invaluable for annual cash flow analysis, particularly for complex problems or loan analysis.

Amortization Schedules

An amortization schedule details, for each payment period, the breakdown of the uniform loan payment ($A$) into interest paid, principal paid, and the remaining balance.

Interest Paid: Calculated as $i \times (\text{Balance remaining from previous period})$.
Principal Paid: Calculated as $\text{Payment} - \text{Interest Paid}$.
New Balance Due: Calculated as $\text{Previous Balance} - \text{Principal Paid}$.

Spreadsheet annuity functions like PMT, IPMT (Interest Payment), and PPMT (Principal Payment) simplify these calculations.

Finding the Balance Due

The balance due on a loan at any point in time is equivalent to the present worth (PV) of the remaining payments. The spreadsheet function $\text{PV}(i, n_{remaining}, A, F)$ can compute this readily, where $n_{remaining}$ is the number of payments left.

Annuity Due

Most engineering economic problems assume cash flows occur at the end of the period (ordinary annuity). However, payments like rent, leases, insurance, and tuition often occur at the beginning of the period (annuity due).

To handle annuity due payments in spreadsheets, the optional Type argument in financial functions (like PV or PMT) is set to 1.

Example 4: Loan Amortization (Monthly Payment)

Problem: A student loan totals $\$18,000$ at graduation. The nominal interest rate is $6\%$. Payments are made monthly for 60 months (5 years) starting one month after graduation. What is the monthly payment?

Given: $P=\$18,000$, $n=60$ months, Nominal $r=6\%$.

Solution: First, calculate the monthly interest rate $i_{monthly} = 6\% / 12 = 0.5\%$.

We use the PMT function to find the uniform monthly payment ($A$):

\[A = \text{PMT}(0.5\%, 60, -\$18,000, 0) \approx \$347.99\]

The monthly payment is $$347.99.