2.3 Percent Applications, Financial Models & Depreciation

2.3 Percent Applications, Financial Models & Depreciation

Percentage Foundations and the Base-Value Principle

A percent is a dimensionless ratio expressing parts per hundred ($r\% = \frac{r}{100}$). While percentage computations are algebraically straightforward, applied percentage modeling is prone to frequent errors regarding the reference quantity, known as the base value. The formal definition of relative percentage change is:

$$\text{Percent Change} = \frac{\text{Final Value} - \text{Base Value}}{\text{Base Value}} \times 100\% = \left(\frac{V_{\text{final}} - V_{\text{base}}}{V_{\text{base}}}\right) \times 100\%$$

In every percentage calculation, the denominator is strictly the initial value before change occurs. When comparing two distinct quantities $A$ and $B$, saying "$A$ is $25\%$ greater than $B$" means:

$$A = B + 0.25B = 1.25B \implies \frac{A}{B} = 1.25$$

Conversely, expressing $B$ as a percentage decrease relative to $A$ changes the base to $A$:

$$\frac{B - A}{A} = \frac{0.80A - A}{A} = -0.20 = -20\%$$

Thus, $A$ being $25\%$ greater than $B$ means $B$ is $20\%$ less than $A$.

---

Successive Percentage Changes and the Asymmetry of Loss and Recovery

A pervasive student misconception is the belief that consecutive percentage increases and decreases are additive. When an initial quantity $V_0$ increases by $r\%$ (where $r$ is expressed in decimal form) and subsequently decreases by the same $r\%$, the changes do not cancel each other out.

The Asymmetry Proof

Let $V_0$ be the initial value:

1. After an $r\%$ increase, the intermediate value is: $$V_1 = V_0(1 + r)$$
2. After an $r\%$ decrease applied to $V_1$, the final value is: $$V_2 = V_1(1 - r) = V_0(1 + r)(1 - r)$$
3. Expanding the binomial product reveals a difference of squares: $$V_2 = V_0(1 - r^2)$$

Because $r^2 > 0$ for any non-zero rate $r$, $1 - r^2 < 1$. Consequently, an $r\%$ increase followed by an $r\%$ decrease always results in a net decrease equal to:

$$\text{Net Loss} = r^2 \times 100\%$$

For example, if a stock rises by $20\%$ ($r = 0.20$) and then falls by $20\%$, the final value is:

$$V_2 = V_0(1 - 0.20^2) = V_0(1 - 0.04) = 0.96 V_0$$

This represents an absolute net loss of $4\%$, not a return to break-even.

The Compensatory Gain Formula

If an asset experiences a fractional loss $d$ (where $0 < d < 1$), the percentage gain $g$ required to restore the original value $V_0$ must satisfy:

$$V_0(1 - d)(1 + g) = V_0 \implies (1 - d)(1 + g) = 1 \implies 1 + g = \frac{1}{1 - d} \implies g = \frac{d}{1 - d}$$

• A $20\%$ loss ($d = 0.20$) requires a gain of $g = \frac{0.20}{0.80} = 0.25 = 25\%$ to break even.
• A $50\%$ loss ($d = 0.50$) requires a gain of $g = \frac{0.50}{0.50} = 1.00 = 100\%$ to break even.
• An $80\%$ loss ($d = 0.80$) requires a gain of $g = \frac{0.80}{0.20} = 4.00 = 400\%$ to break even.

---

Commercial Mathematics: Multipliers, Taxes, Discounts, and Commissions

Commercial financial transactions are modeled most efficiently through composite multiplying factors rather than piecemeal addition or subtraction.

Successive Discounts

Retailers often market chained promotions, such as "Take 40% off original price, plus an additional 25% off at the register." Adding these percentages ($40\% + 25\% = 65\%$) is mathematically invalid because the second discount applies exclusively to the already-discounted price. The correct composite multiplier is:

$$\text{Multiplier} = (1 - 0.40)(1 - 0.25) = (0.60)(0.75) = 0.45$$

The final price is $45\%$ of the original price, representing an effective total discount of $1 - 0.45 = 0.55 = 55\%$.

Taxes, Tips, and Retail Markups

If an item priced at $P_0$ is marked up by $m\%$, discounted by $d\%$, and subject to sales tax $t\%$, the final price is:

$$P_{\text{final}} = P_0(1 + m)(1 - d)(1 + t)$$

Because multiplication in $\mathbb{R}$ is commutative, the sequence of discounts and taxes produces the same final numerical total regardless of order:

$$P_0(1 - d)(1 + t) = P_0(1 + t)(1 - d)$$

However, legal statutes generally dictate that sales tax must be assessed on the actual transaction price (the discounted subtotal).

Tiered Commission Models

Sales commissions often implement piecewise-linear functions to incentivize high performance:

$$\text{Commission}(S) = \begin{cases} r_1 S, & 0 \le S \le S_1 \\ r_1 S_1 + r_2(S - S_1), & S > S_1 \end{cases}$$

A sales professional earning a base salary $B$, with $5\%$ on the first $\text{\textdollar}30,000$ and $8\%$ on sales exceeding $\text{\textdollar}30,000$, earns for $\text{\textdollar}75,000$ in total sales:

$$\text{Income} = B + 0.05(30,000) + 0.08(75,000 - 30,000) = B + 1,500 + 3,600 = B + \$5,100$$

---

Financial Growth: Simple, Discrete Compound, and Continuous Interest

Financial growth models represent transitions across linear, discrete exponential, and continuous exponential functions.

1. Simple Interest (Linear Growth)

Simple interest is calculated exclusively on the original principal $P$:

$$I = Prt \implies A(t) = P + I = P(1 + rt)$$

where $P$ is principal, $r$ is annual nominal interest rate, and $t$ is time in years. The growth rate is constant: $\frac{dA}{dt} = Pr$, creating a straight line with slope $Pr$ and $y$-intercept $P$.

2. Discrete Compound Interest (Geometric Growth)

When earned interest is added to the principal at discrete intervals, future interest accrues on prior interest. If interest compounds $n$ times per year across $t$ years:

$$A(t) = P\left(1 + \frac{r}{n}\right)^{nt}$$

Common compounding frequencies include annual ($n=1$), semiannual ($n=2$), quarterly ($n=4$), monthly ($n=12$), and daily ($n=365$).

3. Continuous Compounding (Euler's Limit)

As the compounding frequency approaches infinity ($n \to \infty$):

$$\lim_{n \to \infty} P\left(1 + \frac{r}{n}\right)^{nt} = P \left[ \lim_{n \to \infty} \left(1 + \frac{1}{n/r}\right)^{n/r} \right]^{rt}$$

Letting $m = \frac{n}{r}$, as $n \to \infty$, $m \to \infty$. By the definition of Euler's constant $e = \lim_{m \to \infty} \left(1 + \frac{1}{m}\right)^m \approx 2.71828$:

$$A(t) = P e^{rt}$$

4. Effective Annual Rate (EAR / APY)

Because compounding generates interest on interest, the actual annual return exceeds the stated nominal rate $r$:

$$\text{EAR}_{\text{discrete}} = \left(1 + \frac{r}{n}\right)^n - 1, \qquad \text{EAR}_{\text{continuous}} = e^r - 1$$

For a nominal rate of $6\%$ compounded monthly:

$$\text{EAR} = \left(1 + \frac{0.06}{12}\right)^{12} - 1 = (1.005)^{12} - 1 \approx 1.061678 - 1 = 6.168\%$$

---

Asset Depreciation: Linear vs. Exponential Models

Depreciation quantifies the decline in monetary value of physical assets (machinery, vehicles, electronics) over time due to wear, obsolescence, and age.

Linear (Straight-Line) Depreciation

In straight-line depreciation, the asset loses a constant absolute dollar amount each period:

$$V(t) = V_0 - dt, \quad \text{where } d = \frac{V_0 - V_{\text{salvage}}}{T_{\text{useful}}}$$

• $V_0$: initial purchase price
• $V_{\text{salvage}}$: anticipated residual salvage value
• $T_{\text{useful}}$: useful lifespan in years
• $d$: constant annual depreciation rate in dollars ($\frac{dV}{dt} = -d$)

Linear depreciation reaches the salvage value at precisely $t = T_{\text{useful}}$.

Exponential (Declining-Balance) Depreciation

In declining-balance depreciation, the asset loses a constant proportion $r$ of its remaining book value each year:

$$V(t) = V_0(1 - r)^t = V_0 e^{-kt}, \quad \text{where } k = -\ln(1 - r)$$

• Dollar loss in year 1 is $r V_0$.
• Dollar loss in year 2 is $r V_1 = r V_0 (1 - r) < r V_0$.
• Because absolute losses decrease over time, the function exhibits asymptotic decay toward zero without ever reaching a negative balance.

Comparative Analytical Example

A business purchases industrial hardware for $V_0 = \text{\textdollar}50,000$.

• Linear Depreciation: Fixed loss of $\text{\textdollar}6,000/\text{year}$ with a $\text{\textdollar}2,000$ salvage value after 8 years: $$V(t) = 50,000 - 6,000t \implies V(3) = 50,000 - 18,000 = \$32,000$$
• Exponential Depreciation: Constant $15\%$ annual loss rate ($r = 0.15$): $$V(t) = 50,000(0.85)^t \implies V(3) = 50,000(0.85)^3 = 50,000(0.614125) = \$30,706.25$$

---

Financial and Depreciation Modeling Matrix