Difference Equations & Dynamic Systems: Models, Solving & Business Apps

Core Concepts

1. What is a Difference Equation?

A difference equation relates the value of a function at one time step to its values at previous steps. Unlike differential equations (which model continuous change), difference equations model discrete change—ideal for business scenarios where data is recorded at fixed intervals (e.g., monthly sales, quarterly profits).

Key Idea: For a first-order difference equation like $y_t=a{y}_{t-1}+b$, each term depends only on the immediately preceding term. This mirrors real-world systems where today’s output depends on yesterday’s inputs (e.g., a bank’s loan portfolio growing based on last month’s interest).

2. Types of Difference Equations

Visual Trace:

flowchart TD
    A["y_t = 0.6y_{t-1} + 21"] -->|"t=0"| B["y_0 = 50"]
    B -->|"t=1"| C["y_1 = 0.6*50 + 21 = 51"]
    C -->|"t=2"| D["y_2 = 0.6*51 + 21 = 51.6"]
    D -->|"t=3"| E["y_3 = 0.6*51.6 + 21 ≈ 52.96"]
    E -->|"..."| F["y_t → 105 (equilibrium)"]

Solving Difference Equations

Method 1: Iterative (Step-by-Step)

For $y_t=0.6y_{t-1}+21$, $y_0=50$:

1. Compute $y_1=0.6\times 50+21=51$.
2. Compute $y_2=0.6\times 51+21=51.6$.
3. Continue until $y_t$ stabilizes (converges to equilibrium).

Worked Example: Pathao’s Rider Earnings Pathao’s daily rider earnings $E_t$ depend on yesterday’s earnings plus a fixed bonus: $$E_t=0.7E_{t-1}+1500$$ If a rider earns NPR 20,000 on Day 0, what are earnings on Day 5? Solution: Answer: On Day 5, earnings ≈ NPR 11,369.55.

Method 2: Closed-Form Solution

For $y_t=a{y}_{t-1}+b$, the general solution is: $$y_t=(y_0-\bar{y})a^t+\bar{y}$$ where $\bar{y}=\frac{b}{1-a}$ is the equilibrium value.

Worked Example: NTC’s Monthly Subscriber Growth NTC’s subscribers grow monthly by 5% of last month’s count, plus 5,000 new sign-ups: $$S_t=1.05S_{t-1}+5000$$ If $S_0=1,000,000$, find $S_5$ and the long-term equilibrium. Solution:

1. Equilibrium: $\bar{S}=\frac{5000}{1-1.05}=-50,000$ (invalid! This means the system diverges—subscribers grow without bound).
2. Closed-form: $$S_t=(1,000,000-(-50,000))\times 1.05^t-50,000$$ $$S_5=1,050,000\times 1.05^5-50,000\approx 1,286,432$$

Graph of Divergence:

Method 3: Homogeneous Equations

For $y_t+c{y}_{t-1}=0$, the solution is: $$y_t=A(-c{)}^t$$ where $A$ is a constant determined by initial conditions.

Worked Example: Daraz’s Order Backlog Daraz’s daily order backlog $B_t$ decreases by 20% each day (no new orders): $$B_t=0.8B_{t-1}$$ If $B_0=10,000$, find $B_5$. Solution: $$B_t=10,000\times 0.8^t$$ $$B_5=10,000\times 0.32768=3,276.8$$ Graph:

In the Real World

1. Khalti’s Loan Repayment Model

   • Idea Used: Non-homogeneous first-order difference equation.
   • How: Khalti’s "Easy EMI" loans use $L_t=L_{t-1}-P$ (where $P$ is the fixed monthly payment) to model loan reduction. The equilibrium $L_t=0$ is the loan’s full repayment.
   • Example: A NPR 50,000 loan at 12% annual interest (1% monthly) with 12-month EMI: $$L_t=1.01L_{t-1}-4,546$$ Solving gives $L_{12}=0$.

2. NEPSE Stock Price Adjustment

   • Idea Used: Second-order difference equation for price momentum.
   • How: Stock prices often follow: $$P_t=1.1P_{t-1}-0.1P_{t-2}+{ϵ}_t$$ where ${ϵ}_t$ is random news. This captures "momentum" (recent trends) and "reversion" (long-term mean).
   • Example: If $P_0=1000$, $P_1=1100$, then: $$P_2=1.1\times 1100-0.1\times 1000=1200$$

3. NTC’s Network Traffic Routing

   • Idea Used: Dynamic system of coupled difference equations.
   • How: Traffic between cities $A$ and $B$ is modeled as: $$T_{AB,t}=0.9T_{AB,t-1}+0.1T_{BA,t-1}$$ (90% of yesterday’s traffic repeats, 10% reverses direction).
   • Example: If $T_{AB,0}=500$ units and $T_{BA,0}=300$ units: $$T_{AB,1}=0.9\times 500+0.1\times 300=480$$

Comparison: Difference vs. Differential Equations

Applications in Business

1. Inventory Management

Problem: A shop’s weekly demand $D_t$ depends on last week’s stock $I_{t-1}$: $$D_t=0.5I_{t-1}$$ If initial stock $I_0=100$ units, find $I_5$ (assuming no restocking). Solution: $$I_t=I_{t-1}-D_t=0.5I_{t-1}$$ $$I_t=100\times 0.5^t$$ $$I_5=100\times 0.03125=3.125\text{ units}$$

2. Market Equilibrium

Problem: Demand $Q_d=26-0.4P$, Supply $Q_s=4+0.6P$. Price adjusts as: $$\frac{dP}{dt}=0.32(Q_d-Q_s)$$ Solution:

1. Find equilibrium $Q_d=Q_s$: $$26-0.4P=4+0.6P$$ $$P^{\ast }=10$$
2. If $P_0=8$, the price converges to $P^{\ast }$ over time.

3. Projected Revenue

Problem: A startup’s revenue grows by 15% of last month’s revenue plus a fixed NPR 50,000: $$R_t=1.15R_{t-1}+50,000$$ If $R_0=100,000$, find $R_3$. Solution: $$R_1=1.15\times 100,000+50,000=165,000$$ $$R_2=1.15\times 165,000+50,000=234,750$$ $$R_3=1.15\times 234,750+50,000=300,462.5$$

Exam Tip

1. Always check for equilibrium: For $y_t=a{y}_{t-1}+b$, compute $\bar{y}=\frac{b}{1-a}$. If $|a|\ge 1$, the system diverges (common in finance).
2. Label your steps: Examiners deduct marks for unclear iterative solutions. Use tables for clarity.
3. Real-world tie-ins: Questions often link to business scenarios (e.g., "A firm’s production depends on last quarter’s output"). Translate words into equations:
   • "Grows by 5% of last month" → $y_t=1.05y_{t-1}$
   • "Decreases by a fixed amount" → $y_t=y_{t-1}-c$
4. Graphs are worth marks: Sketch the behavior (converging/diverging) even if not asked.
5. Mixed questions: Combine difference equations with algebra (e.g., solve for $y_0$ given $y_5$).

Key Formula Summary:

mindmap
  root((Difference Equations))
    First-Order
      Form: y_t = a y_{t-1} + b
      Solution: y_t = (y_0 - ȳ) a^t + ȳ
      Equilibrium: ȳ = b / (1 - a)
    Second-Order
      Form: y_t = a y_{t-1} + b y_{t-2}
      Characteristic Eq: r^2 - a r - b = 0
    Homogeneous
      Form: y_t + c y_{t-1} = 0
      Solution: y_t = A (-c)^t
    Applications
      Finance: Loan amortization
      Inventory: Stock levels
      Marketing: Sales forecasting