Engineering Investment Evaluation: Present, Future & Annual Worth

Posted on Jan 3, 2026 in Mathematics

Present, Future & Annual Equivalent — Equal & Unequal Lives

1. Purpose of Investment Evaluation

Used to compare different engineering projects based on their economic worth.

Projects may have:

  • Different lifetimes
  • Different cash-flow patterns
  • Different initial investments

Goal → Select the economically best alternative.

2. Interest Factors (Essential)

(These appear multiple times, so memorize.)

Let i = interest rate per period, n = number of periods.

(P/F, i, n)

Present worth of single future sum:

P = F(1 + i)-n

(F/P, i, n)

Future worth of present sum:

F = P(1 + i)n

(P/A, i, n)

Present worth of equal annual series (annuity):

P = A × [(1 + i)n – 1] / [i(1 + i)n]

(A/P, i, n)

To convert present worth → annual equivalent:

A = P × [i(1 + i)n] / [(1 + i)n – 1]

A. Present Worth (PW) Method

Definition

All cash inflows and outflows are converted to their present values (time = 0) using interest rate i.

1. Revenue-Dominated Cash Flow

(Inflow > Outflow)

Steps

  1. Assign + to revenues and − to costs.
  2. Convert every cash flow to present value using interest factors.
  3. Compute:

PW = Sum of Rj(P/F, i, j) − P

4. Choose the alternative with maximum PW.

Example (PW, Equal Lives)

Technology comparison:

  • Tech 1: P = ₹1,200,000; A = ₹400,000; n = 10
  • Tech 2: P = ₹2,000,000; A = ₹600,000; n = 10
  • i = 20%

PW = −P + A(P/A, 20%, 10)

Compute factor from tables: (P/A) ≈ 4.193.

Tech 1: PW = −1,200,000 + 400,000 × 4.193 = 477,200

Tech 2: PW = −2,000,000 + 600,000 × 4.193 = 515,800

Answer: Choose Technology 2

2. Cost-Dominated Cash Flow

(Cost minimization problem)

Signs reversed:

  • Costs → +
  • Revenues/salvage → −

Steps

  1. Compute PW of cost stream:

PW = P + Sum of Cj(P/F, i, j) − S(P/F, i, n)

2. Select the alternative with minimum PW cost.

Example: Elevator bid problem → whichever has smaller present worth cost wins.

B. Future Worth (FW) Method

Definition

All cash flows are converted to their future worth at year n.

Revenue-Dominated

FW = −P(1 + i)n + Sum of Rj(1 + i)n−j + S

Choose maximum FW.

Cost-Dominated

Same but all costs are +, and revenues and salvage are −.

Choose minimum FW.

Example Idea

Alternative A, B with inflows from years 1–4 → compute FW at year 4.

C. Annual Equivalent Worth (AE / AW) Method

Definition

Convert each project’s entire cash flow into a uniform annual equivalent.

Useful when:

  • Project lives differ
  • Vehicle comparison
  • Loan repayments
  • Salary/operations comparisons

Formula

AE = PW × (A/P, i, n)

Rule

  • Revenue-dominated → choose maximum AE
  • Cost-dominated → choose minimum AE

Example (Similar to PDF Example 2)

Annual fuel cost increasing yearly → compute AE using:

  • Gradient-series factor (A/G, i, n)
  • Uniform-series factor (A/P, i, n)

D. Unequal Lives Comparison

When alternatives have different lives, three valid approaches:

  1. Repeatability Assumption — repeat each alternative until reaching a common life (LCM of lives). Then compare PW or AE.
  2. Common Study Period — choose a fixed period (e.g., 10 years). Truncate or extend cash flows.
  3. Annual Equivalent (Best Method) — AE method automatically handles unequal lives. AE = PW(A/P, i, n)

Example:

Machine A life = 4 years, Machine B life = 6 years.

Use AE:

  • Compute PW of each
  • Convert to AE
  • Choose smaller cost (cost-dominated)

Evaluation of Investment Proposals — Full, Detailed Notes

Why evaluate investment proposals?

Engineers often must choose between alternative projects (buy machine A or B, build bridge or not, accept contract or reject). Money today and money later are not the same — time value of money. Evaluation methods translate all future cash flows into a single comparable number so you can decide which alternative is better economically.

Key idea: money has interest. If you get ₹100 now, you could invest it and have more later. Conversely, ₹100 later is worth less today.

Fundamental Building Blocks — Interest Factors

You must be fluent with these because they are used repeatedly.

Let i = interest rate per period (decimal), n = number of periods (years usually).

1) Future value of a present single sum (F = P(1 + i)n)

If you invest P now, after n periods: F = P(1 + i)n

So the factor (F/P, i, n) = (1 + i)n.

2) Present value of a future single sum (P = F(1 + i)-n)

If you will receive F after n years, its value today is: P = F(1 + i)-n

Factor (P/F, i, n) = (1 + i)-n.

3) Present value of an equal annual series A (annuity)

If you receive (or pay) A at the end of each year for n years, present worth P is:

P = A × [1 − (1 + i)-n] / i

This factor is called (P/A, i, n). Equivalently: A = P × i / [1 − (1 + i)-n] (A/P factor)

4) Future value of an equal annual series

If you get A at end of each year, future worth at year n:

F = A × [(1 + i)n − 1] / i

Factor (F/A, i, n).

5) Gradient (arithmetic) series

If benefits or costs increase by a constant G each year (0 in year 1, G in year 2, 2G in year 3, …), there are standard formulas. Usually used via tables or combined as PG = G(P/G, i, n).

A. Present Worth (PW) Method — Concept & Steps

Intuition

Convert every cash flow (positive = benefit/inflow, negative = cost/outflow) to its value at time 0 using interest rate i and add them up. The project is acceptable if PW ≥ 0 (for single project vs do-nothing) or choose the alternative with the largest PW (for mutually exclusive projects).

Step-by-step Algorithm

  1. Draw a simple timeline showing all cash flows (year 0 is today).
  2. Mark signs: inflows (+), outflows (−). For revenue-dominated problems we usually treat incomes as + and costs as − (opposite for cost-dominated comparisons).
  3. For each cash flow at year t, compute present value: PV = casht × (1 + i)-t.
  4. Sum all PVs → PW.
  5. Decision:
  • If comparing to doing nothing: accept if PW ≥ 0.
  • If choosing among alternatives: choose the one with max PW.

Worked Example — Full Arithmetic

Problem: Project needs initial investment P = ₹300,000 at t = 0. It produces net revenues at year ends: R1 = ₹70,000, R2 = ₹90,000, R3 = ₹120,000. Salvage at end of year 3 = ₹20,000. Evaluate at i = 10%. Accept or reject?

Solution:

Compute PV of each:

  • PV of initial = −₹300,000 (already at time 0).
  • PV of R1 = 70,000 / 1.1 = ₹63,636.36
  • PV of R2 = 90,000 / 1.21 = ₹74,380.17
  • PV of R3 = 120,000 / 1.331 = ₹90,225.56
  • PV of salvage S = 20,000 / 1.331 = ₹15,037.59

Sum PW = (−300,000) + 63,636.36 + 74,380.17 + 90,225.56 + 15,037.59 = −56,720.32

Decision: PW < 0 → reject the project at 10%.

Cost-dominated vs revenue-dominated

Revenue-dominated: positive flows (sales) are +, costs are −. Choose max PW.

Cost-dominated: you’re minimizing cost; treat costs as + and benefits (savings) as −, then choose minimum PW.

B. Future Worth (FW) Method — Concept & Steps

Intuition

Instead of bringing everything to present, push everything to a common future date (often the end of the project life) and compare the future amounts. Same idea as PW but at future time.

Steps

  1. Choose the future date (commonly the final year of each project).
  2. For each cash flow at year t, compute its future value at year n: FV = casht × (1 + i)n−t.
  3. Sum all FVs → FW.
  4. Decision: pick project with larger FW (if revenue-dominated) or smaller FW (if cost-dominated).

C. Annual Equivalent (AE) Method — Concept & Steps

Convert the entire project cash flow into an equivalent uniform annual amount (A) that would be economically equivalent to that project. Great when comparing projects with different life spans.

Steps

  1. Compute PW of project (or FW then convert).
  2. Convert PW to AE: A = PW × (A/P, i, n), where (A/P) is the capital recovery factor.
  3. If revenue-dominated → choose max AE. If cost-dominated → choose min AE.

Example (Full)

Two machines:

  • Machine A: life 4 yr, initial cost ₹200,000, annual net revenue ₹70,000, salvage 0. i = 10%
  • Machine B: life 6 yr, initial cost ₹300,000, annual net revenue ₹110,000, salvage 0.

Compute AE for both and compare.

Machine A:

PWA = −200,000 + 70,000 × (P/A, 10%, 4).

(P/A,10%,4) = 3.16986.

PWA = −200,000 + 70,000 × 3.16986 = 21,890.2.

AEA = PWA × (A/P,10%,4). (A/P) = 0.315486.

AEA = 21,890.2 × 0.315486 = 6,906.5 (net annual benefit).

Machine B:

(P/A,10%,6) = 4.35526

PWB = −300,000 + 110,000 × 4.35526 = 179,078.6

(A/P,10%,6) = 0.22817

AEB = 179,078.6 × 0.22817 = 40,868.6

Comparison: AEB ≫ AEA → Choose Machine B.

D. Handling Salvage & Modernization Costs

  • Salvage value at end year n: treat as inflow at year n and discount to P (or convert to AE).
  • Modernization (mid-life) cost: treat as additional outflow at that year; include in cash-flow timeline and discount.

E. Unequal Lives — Three Standard Approaches

  1. Annual Equivalent (recommended) — compute AE for each and choose the best AE.
  2. Common study period (repeatability method) — find a common comparison horizon (usually LCM of lives).
  3. Present worth with equivalent annualization — convert each to AE then compare.

F. Gradient Cases

Total benefit series = uniform part + gradient part.

If B1 = base at year 1 and growth G each year:

  1. A uniform series = B1 for all years → discount using (P/A).
  2. A gradient series = 0, G, 2G, … → use gradient PV formula (P/G) or convert gradient to AE using (A/G) factor.

G. Decisions Based on B/C Ratio (Benefit–Cost Ratio)

Used for public projects where you compare total PV of benefits to PV of costs.

B/C = PW(Benefits) / PW(Costs)

Rules:

  • If B/C > 1 → project acceptable (benefits exceed costs)
  • If B/C < 1 → reject

Economic Appraisal Techniques

(Payback Period, NPV, IRR, MARR, Cost–Benefit Analysis)

1. Payback Period Method

Concept: Payback period = How long a project takes to recover its initial investment from its cash inflows.

Types of Payback:

  1. Simple/Traditional Payback: Does NOT consider time value of money. Just cumulative inflows until they equal initial cost.

Formula (if inflows are equal each year): Payback Period = Initial Investment / Annual Inflow

2. Discounted Payback Period: Considers time value → inflows are discounted. Then cumulate PVs until they equal initial cost.

Worked Example (Full)

A machine costs ₹100,000 and gives cash inflow:

  • Year 1: 30,000
  • Year 2: 40,000
  • Year 3: 50,000
  • Year 4: 30,000

Cumulative inflow:

Yr 1 = 30,000; Yr 2 = 70,000; Yr 3 = 120,000 → exceeds cost.

Cost reached between Year 2 and 3.

Remaining amount after year 2 = 100,000 − 70,000 = 30,000

Fraction of year = 30,000 / 50,000 = 0.6

Payback = 2.6 years

2. Net Present Value (NPV)

NPV = Present value of inflows − Present value of outflows.

This is the most accurate profitability method.

Formula: NPV = Sum [Rt / (1 + i)t] − P

Decision Rule:

  • NPV > 0 → accept
  • NPV < 0 → reject

Worked Example (Detailed)

Investment = ₹500,000. i = 12%

Returns:

  • Yr 1: 150,000 → PV = 133,928.57
  • Yr 2: 180,000 → PV = 143,400
  • Yr 3: 200,000 → PV = 142,130
  • Yr 4: 220,000 → PV = 139,616

Sum PV inflows = 559,075

NPV = 559,075 − 500,000 = ₹59,075 (positive → accept)

3. Internal Rate of Return (IRR)

Concept: IRR is the interest rate at which NPV = 0.

Decision Rule:

  • IRR ≥ MARR → accept
  • IRR < MARR → reject

Interpolation Method (Most Important in Exams)

  1. Find NPV using a low rate i1 → NPV positive.
  2. Find NPV using a high rate i2 → NPV negative.
  3. Interpolate:

IRR = i1 + [NPV1 / (NPV1 − NPV2)] × (i2 − i1)

Worked Example (Full)

Initial = 100,000. Returns = 30,000 each for 5 years.

Step 1: Compute NPV at 10%. (P/A,10%,5) = 3.7908. PV = 113,724. NPV1 = +13,724.

Step 2: Try 20%. (P/A,20%,5) = 2.9906. PV = 89,718. NPV2 = −10,282.

Step 3: Interpolate:

IRR = 10 + [13,724 / (13,724 − (−10,282))] × 10

IRR = 10 + 5.72 = 15.72%

4. MARR (Minimum Attractive Rate of Return)

MARR = lowest acceptable return for the firm.

Sources: Cost of capital, Opportunity cost, Risk premium.

5. Cost–Benefit Analysis (B/C Ratio)

Used for public projects.

B/C = PW(Benefits) / PW(Costs)

If B/C > 1 → project justified; if B/C < 1 → reject.

Depreciation Calculation

Key Terms

  1. First Cost (P): Initial cost/purchase price.
  2. Salvage Value (S): Estimated value at the end of asset life.
  3. Useful Life (n): Total number of years expected to be used.
  4. Book Value (Bt): Value of asset remaining in books at end of year t.

A. Straight Line Method of Depreciation (SLM)

Concept: Asset loses equal amount of value every year.

Formula: Dt = (P − S) / n

Book Value: Bt = P − t × Dt

B. Declining Balance Method (DBM)

Concept: Depreciation each year = constant % of previous book value.

Formula: Dt = K × Bt−1

Bt = (1 − K)t × P

Where K = depreciation rate. Double declining balance means K = 2 / n.

C. Sum of the Years’ Digits Method (SYD)

Concept: More depreciation in early years.

  1. Compute SYD = n(n + 1)/2.
  2. Yearly Depreciation Formula: Dt = [(n − t + 1) / SYD] × (P − S).

D. Sinking Fund Method (SFM)

Concept: Assumes you put aside a fixed amount A every year into an account earning interest i. At the end of n years, this fund becomes exactly P − S.

  1. Compute Annual Deposit (A): A = (P − S)(A/F, i, n)
  2. Depreciation in year t: Dt = A(1 + i)t−1

E. Service Output Method

Concept: Depreciation based on usage, not time.

Depreciation per unit = (P − S) / Total Expected Service

Dt = (Depreciation per unit) × Actual usage in year t

Cost Analysis & Break-Even Analysis

Types of Costs

  1. Fixed Costs (FC): Do NOT change with output (e.g., rent).
  2. Variable Costs (VC): Change with output (e.g., raw materials).
  3. Total Cost (TC): TC = FC + VC.

Marginal Cost

Marginal Cost (MC): Extra cost of producing one more unit. MC = change in TC / change in Q.

Relationship Between Cost Curves

  • When MC < AC → AC is falling
  • When MC > AC → AC is rising
  • MC intersects AC at its minimum point

Break-Even Analysis

At Break-Even Point (BEP): Total Revenue (TR) = Total Cost (TC). Profit = 0.

Formulas:

Let s = selling price, v = variable cost.

Contribution per unit (C) = s − v

P/V Ratio = C / s

BEP (Quantity) = FC / (s − v)

BEP (Sales Value) = FC / (P/V Ratio)

Margin of Safety (MOS) = Actual Sales − BEP Sales

Solved Break-Even Example

Selling price (s) = ₹72, Variable cost (v) = ₹42, Fixed cost (FC) = ₹300,000.

C = 72 − 42 = 30.

BEP Quantity = 300,000 / 30 = 10,000 units.

If actual sales = 15,000 units: MOS = 15,000 − 10,000 = 5,000 units.

Example 2: Sales & Profit Method

2022: Sales = 250,000, Profit = 50,000

2023: Sales = 290,000, Profit = 70,000

Step 1: P/V Ratio = Change in Profit / Change in Sales = 20,000 / 40,000 = 50%

Step 2: Fixed Cost

Profit = (P/V × Sales) − FC

50,000 = (0.5 × 250,000) − FC

FC = 125,000 − 50,000 = 75,000

Step 3: BEP Sales = FC / (P/V) = 75,000 / 0.5 = 150,000

Law of Variable Proportion & Returns to Scale

Law of Variable Proportion (Short Run)

Applies when at least one factor is fixed.

1. Total Product (TP): Total output.

2. Average Product (AP): TP / Labor.

3. Marginal Product (MP): Extra output from one additional worker.

Three Stages:

  1. Stage 1 (Increasing Returns): TP increases at increasing rate. MP rising.
  2. Stage 2 (Diminishing Returns): TP increases at decreasing rate. MP falling but positive. Rational firms operate here.
  3. Stage 3 (Negative Returns): TP decreases. MP is negative.

Returns to Scale (Long Run)

Applies when all inputs are variable.

  • Increasing Returns to Scale (IRS): Inputs double → output more than doubles.
  • Constant Returns to Scale (CRS): Inputs double → output exactly doubles.
  • Diminishing Returns to Scale (DRS): Inputs double → output less than doubles.

Types of Markets & Price-Output Decisions

1. Perfect Competition

  • Large number of buyers/sellers.
  • Homogeneous product.
  • Price taker.
  • Demand curve horizontal (P = MR = AR).
  • Equilibrium: MC = MR.

2. Monopoly

  • Single seller.
  • No close substitutes.
  • Price maker.
  • Downward sloping demand curve (MR < AR).
  • Equilibrium: MC = MR (but price is read from demand curve).

3. Monopolistic Competition

  • Many sellers.
  • Product differentiation (branding).
  • Free entry/exit.
  • In long run: Normal profit but excess capacity.

4. Oligopoly

  • Few large firms.
  • High entry barriers.
  • Interdependence.
  • Kinked demand curve model (price rigidity).

Price–Output Decision Summary

  • Perfect Competition: P = MC.
  • Monopoly/Monopolistic: P > MC.
  • All markets maximize profit where MC = MR.