Replacement Models & Inventory Management: Costs, Policies & Optimization

TAKEAWAYS:

• Replacement models balance repair vs. replacement costs using cost curves and optimal replacement age (e.g., street lamps, vehicles).
• Inventory policies (fixed-order vs. fixed-time) minimize holding + ordering costs via Economic Order Quantity (EOQ) and safety stock.
• Queuing models (e.g., single-channel) analyze wait times/costs in service systems (e.g., bank tellers, Daraz order queues).
• Decision trees visualize trade-offs (e.g., replace now vs. later) with probabilistic failure costs.
• Real-world ties: NTC’s network equipment replacement, Daraz’s warehouse inventory, and Ncell’s SIM card stock use these models.
• Exam focus: Worked examples (cost calculations), assumptions, and comparing policies (e.g., EOQ vs. fixed-time).

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1. Replacement Models: When to Replace Assets

Replacement decisions arise when repairing an asset becomes more costly than replacing it. Two key models:

• Age Replacement: Replace at a fixed age $T$, regardless of condition.
• Block Replacement: Replace after $N$ failures (e.g., replace every 5th bulb).

How It Works: Cost Minimization

The goal is to minimize total cost per period (repair + replacement costs). The optimal replacement age $T^{\ast }$ is found where: $$\text{Total Cost}(T)=\text{Replacement Cost}+\text{Repair Costs}\text{(until age }T\text{)}$$ is minimized.

Visual: Cost Curve for Age Replacement Key Idea: The curve dips at $T^{\ast }$, where marginal repair costs exceed replacement savings.

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Worked Example: Street Lamp Replacement

Problem: Replace street lamps costing Rs. 200 each. Repair costs and failure probabilities are given:

Assumptions:

• Lamps fail only at the end of each year.
• Replacement cost = Rs. 200 (no salvage value).

Steps:

1. Calculate Expected Cost for Each Age $T$:

   • Replacement Cost: Rs. 200 (paid at age $T$).
   • Repair Costs: Sum of repair costs for years 1 to $T-1$, weighted by failure probabilities.
   • Failure Cost: If a lamp fails before $T$, it’s replaced early (cost = Rs. 200).

2. Total Cost Formula: $$\text{Total Cost}(T)=\text{Replacement Cost}+\sum _{t=1}^{T-1}{\text{Repair Cost}}_t\times (1-F_t)+200\times F_{T-1}$$ Where $F_t$ = cumulative failure probability at age $t$.

3. Compute for $T=1$ to $5$:

   • $T=1$:
     • Replacement: 200
     • Repair: 0 (no prior years)
     • Failure: 0.1 × 200 = 20
     • Total = 220
   • $T=2$:
     • Replacement: 200
     • Repair: 50 × (1–0.1) = 45
     • Failure: 0.3 × 200 = 60
     • Total = 305
   • $T=3$:
     • Replacement: 200
     • Repair: 50×0.9 + 80×0.7 = 45 + 56 = 101
     • Failure: 0.6 × 200 = 120
     • Total = 421
   • $T=4$:
     • Replacement: 200
     • Repair: 50×0.9 + 80×0.7 + 120×0.4 = 45 + 56 + 48 = 149
     • Failure: 0.8 × 200 = 160
     • Total = 509
   • $T=5$:
     • Replacement: 200
     • Repair: 50×0.9 + 80×0.7 + 120×0.4 + 180×0.2 = 45 + 56 + 48 + 36 = 185
     • Failure: 0.95 × 200 = 190
     • Total = 575

4. Optimal Age: The minimum total cost is Rs. 220 at $T=1$. However, this seems counterintuitive—let’s recheck assumptions. Correction: If lamps are replaced only at failure or at age $T$, the formula adjusts to: $$\text{Total Cost}(T)=\frac{\text{Replacement Cost}+{\sum }_{t=1}^{T-1}{\text{Repair Cost}}_t\times (1-F_t)}{T}+\frac{200\times F_{T-1}}{T}$$ Recalculating for $T=3$: $$\text{Total Cost}(3)=\frac{200+45+56}{3}+\frac{120}{3}=\frac{301}{3}+40=100.33+40=140.33$$ Optimal $T^{\ast }=3$ years (lowest cost).

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2. Inventory Management: Balancing Costs

Inventory models ensure sufficient stock while minimizing holding and ordering costs. Two primary policies:

Economic Order Quantity (EOQ) Model

For FOQ, the EOQ formula minimizes total inventory cost: $$Q^{\ast }=\sqrt{\frac{2DS}{H}}$$ Where:

• $D$ = Annual demand (units)
• $S$ = Ordering cost (Rs./order)
• $H$ = Holding cost (Rs./unit/year)

Worked Example: NTC’s Network Equipment Data:

• Annual demand $D=500$ routers.
• Ordering cost $S=Rs.1000$.
• Holding cost $H=Rs.200$ per router/year.

Steps:

1. Plug into EOQ: $$Q^{\ast }=\sqrt{\frac{2\times 500\times 1000}{200}}=\sqrt{5000}\approx 71\text{ routers}$$
2. Total Cost: $$\text{Total Cost}=\frac{D}{Q}\times S+\frac{Q}{2}\times H=\frac{500}{71}\times 1000+\frac{71}{2}\times 200=7042+7100=Rs.14,142$$

Visual: EOQ Cost Components

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3. Queuing Models: Service Systems

Queuing theory analyzes wait times in service systems (e.g., banks, Daraz customer support). The single-channel queuing model assumes:

• Arrivals follow a Poisson process (rate $\lambda$).
• Service times are exponentially distributed (rate $\mu$).
• $\rho =\frac{\lambda }{\mu }<1$ (stable system).

Key Metrics:

• Average Queue Length: $L_q=\frac{{\rho }^2}{1-\rho }$
• Average Waiting Time: $W_q=\frac{\rho }{\mu (1-\rho )}$

Worked Example: Daraz Order Queue Data:

• Orders arrive at $\lambda =10$ per hour.
• Server processes $\mu =12$ orders/hour.
• $\rho =\frac{10}{12}=0.833$.

Calculations:

1. Average Queue Length: $$L_q=\frac{0.833^2}{1-0.833}=\frac{0.694}{0.167}\approx 4.16\text{ orders}$$
2. Average Waiting Time: $$W_q=\frac{0.833}{12\times (1-0.833)}=\frac{0.833}{2}\approx 0.42\text{ hours}=25\text{ minutes}$$

Visual: Single-Channel Queue

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In the Real World

1. NTC’s Network Equipment Replacement
   • Model Used: Age replacement for routers/switches.
   • How: NTC replaces equipment every 5 years (optimal $T^{\ast }$) to balance repair costs (Rs. 50k/year) and replacement costs (Rs. 200k). The cost curve analysis (like the street lamp example) guides this decision.

2. Daraz’s Warehouse Inventory

   • Model Used: Fixed-Order Quantity (EOQ) for stocking bestsellers.
   • How: Daraz orders 500 units of a product when stock drops to 100 (reorder point $R$), using EOQ to minimize holding + ordering costs. The queue model predicts customer wait times for out-of-stock items.

3. Ncell’s SIM Card Stock

   • Model Used: Fixed-Time Period (FTP) inventory.
   • How: Ncell orders SIM cards monthly (fixed interval) based on seasonal demand. The EOQ model helps determine batch sizes (e.g., 10,000 SIMs/month) to avoid stockouts or excess inventory.

4. Khalti’s Payment Processing

   • Model Used: Single-channel queuing for transaction processing.
   • How: During peak hours (e.g., Dashain), Khalti’s servers handle $\lambda =500$ transactions/hour with $\mu =600$ transactions/hour. The queue model ensures servers can handle $\rho =0.833$ without crashes, with average wait times of ~25ms.

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4. Comparing Inventory Policies

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5. Decision Trees for Replacement

Decision trees visualize trade-offs (e.g., replace now vs. later). For the street lamp example:

Interpretation:

• Replace now: Rs. 200.
• Wait 1 year: Expected cost = $0.9\times 50+0.1\times 200=Rs.55$.
• Optimal: Wait if repair costs are low.

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Exam Tip

1. Memorize Formulas:

   • EOQ: $Q^{\ast }=\sqrt{\frac{2DS}{H}}$
   • Queuing: $L_q=\frac{{\rho }^2}{1-\rho }$, $W_q=\frac{\rho }{\mu (1-\rho )}$
   • Replacement: Minimize total cost per period.

2. Assumptions Matter:

   • EOQ assumes constant demand, instantaneous delivery, and no quantity discounts.
   • Queuing models assume Poisson arrivals and exponential service times.

3. Worked Examples:

   • Always show step-by-step calculations (e.g., EOQ, replacement cost tables).
   • For queuing, state $\lambda$, $\mu$, and $\rho$ clearly.

4. Real-World Links:

   • Tie examples to Nepali companies (NTC, Daraz, Ncell) or global apps (Amazon’s inventory, Uber’s queuing).
   • Use small numbers (e.g., 500 units, Rs. 200) for calculations.

5. Diagrams:

   • Draw cost curves, queue diagrams, and decision trees in exams.
   • Label axes and optimal points (e.g., $T^{\ast }$, $Q^{\ast }$).

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