Replacement Models & Miscellaneous OR Topics (Cost Analysis, EOQ, Queuing)

TAKEAWAYS:

• Replacement models minimize long-term costs by balancing purchase, maintenance, and failure risks (e.g., Daraz’s server upgrades vs. individual bulb replacements).
• EOQ (Economic Order Quantity) balances ordering and holding costs to optimize inventory (used by NTC for spare parts, banks for ATM cash).
• Queuing theory models wait times (e.g., Pathao rider queues, Ncell customer service calls) using Kendall’s notation (M/M/1, M/G/∞).
• Duality theorem links primal and dual LPPs to simplify complex problems (used in airline seat allocation, hospital resource sharing).
• OR limitations include unrealistic assumptions (e.g., constant failure rates) and data dependency—critical for exam critiques.
• Exam focus: Worked examples (cost calculations, queueing formulas), classifications (queue types), and short notes (EOQ, duality) dominate.

1. Replacement Models: Minimizing Long-Term Costs

Replacement models determine the optimal time to replace assets (machines, bulbs, vehicles) to minimize total cost over time. Two key approaches:

• Individual Replacement: Replace items one by one as they fail.
• Group Replacement: Replace all items in a batch at fixed intervals, regardless of failure.

Key Cost Components

For any replacement policy, total cost per period includes:

1. Purchase Cost (C₀): Initial cost of the asset.
2. Salvage Value (Sₙ): Resale value at end of life.
3. Maintenance Cost (Mₙ): Cost to keep the asset running (increases with age).
4. Failure Cost (Fₙ): Cost of downtime or emergency replacement.

Example 1: Individual vs. Group Replacement for Light Bulbs

Scenario: A company uses 3,000 bulbs. Replacing individually costs Rs. 20/bulb. Failure data:

Assumptions:

• Bulbs fail randomly (Poisson process).
• Replacement cost = Rs. 20/bulb.
• No maintenance cost (only failure cost).

Step 1: Calculate Expected Failures per Year For 3,000 bulbs:

• Year 1: $3000\times 0.1=300$ failures → Cost = $300\times 20=\text{Rs. }6,000$
• Year 2: $3000\times (0.3-0.1)=600$ failures → Cost = Rs. 12,000
• Year 3: $3000\times (0.6-0.3)=900$ failures → Cost = Rs. 18,000
• Total 5-year cost: Rs. 6,000 + 12,000 + 18,000 + (Year 4: Rs. 12,000) + (Year 5: Rs. 3,000) = Rs. 63,000.

Step 2: Group Replacement Policy Replace all bulbs every 3 years (optimal balance):

• Replacement cost: $3000\times 20=\text{Rs. }60,000$ every 3 years.
• Failure cost: Only for bulbs failing before Year 3.
  • Failures in Year 1: 300 → Rs. 6,000
  • Failures in Year 2: 600 → Rs. 12,000
  • Total 3-year cost: Rs. 60,000 (replacement) + Rs. 18,000 (failures) = Rs. 78,000.
• Cost per year: Rs. 78,000 / 3 = Rs. 26,000/year (vs. Rs. 12,600/year for individual).

Conclusion: Individual replacement is cheaper here (Rs. 12,600/year vs. Rs. 26,000/year).

Example 2: Scooter Replacement (Depreciation + Maintenance)

Data: Scooter costs Rs. 80,000. Depreciation and maintenance costs:

Assumptions:

• Salvage value = Rs. 2,000 at end of Year 6.
• Interest rate = 0% (simplified).

Step 1: Calculate Net Cost per Year Net cost = (Depreciation + Maintenance) – Salvage value (prorated). For Year 1: $80,000-28,000=52,000$ (no salvage yet). For Year 6: $2,000-2,000=0$ (salvage cancels out).

Step 2: Cumulative Net Cost

Step 3: Average Cost per Year Divide cumulative cost by year number to find the economic life:

• Year 3: $104,000/3=34,667$
• Year 4: $132,000/4=33,000$ ← Minimum
• Year 5: $162,000/5=32,400$

Optimal Replacement Year: Year 4 (lowest average cost of Rs. 33,000/year).

2. Economic Order Quantity (EOQ): Inventory Optimization

EOQ determines the optimal order quantity that minimizes total inventory costs (ordering + holding).

EOQ Formula

$$Q^{\ast }=\sqrt{\frac{2DS}{H}}$$ Where:

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

Total Cost (TC): $$TC=\frac{DS}{Q}+\frac{HQ}{2}$$

Example 3: NTC’s Spare Parts Ordering

Scenario: NTC needs 10,000 spare parts/year. Ordering cost = Rs. 500/order, holding cost = Rs. 2/unit/year.

Step 1: Calculate EOQ $$Q^{\ast }=\sqrt{\frac{2\times 10,000\times 500}{2}}=\sqrt{5,000,000}=2,236\text{ units}$$

Step 2: Total Cost at EOQ $$TC=\frac{10,000\times 500}{2,236}+\frac{2\times 2,236}{2}=\text{Rs. }2,236+\text{Rs. }2,236=\text{Rs. }4,472$$

Step 3: Compare with Other Quantities

Conclusion: Order 2,236 units every time to minimize costs.

3. Queuing Theory: Modeling Wait Times

Queuing theory analyzes waiting lines (queues) to optimize service systems. Key terms:

• Arrival Rate (λ): Customers/unit time.
• Service Rate (μ): Customers served/unit time.
• Queue Discipline: FIFO, LIFO, priority, etc.
• System States: Number of customers in the system.

stateDiagram-v2
	state Customer
	state Server

	Customer --> Server : Arrival (λ)
	Server --> Customer : Service (μ)
	Customer --> Customer : Wait in Queue

	state Queue

	Customer --> Queue : Enqueue
	Queue --> Server : Dequeue

	caption Kendall's M/M/1 Queue: Arrival → Queue → Service → Departure

Kendall’s Notation

Describes queue types as A/B/c/K/N:

• A: Arrival distribution (M = Markovian/Poisson, G = General).
• B: Service distribution (M = Exponential, D = Deterministic).
• c: Number of servers.
• K: Queue capacity.
• N: Population size.

Example Models:

Example 4: Single-Channel Queuing (Ncell Customer Service)

Scenario:

• Arrival rate (λ) = 10 calls/hour (Poisson).
• Service rate (μ) = 12 calls/hour (Exponential).
• Single server (M/M/1).

Key Metrics:

1. Utilization (ρ): $$\rho =\frac{\lambda }{\mu }=\frac{10}{12}=0.833$$
2. Average Queue Length (L_q): $$L_q=\frac{{\rho }^2}{1-\rho }=\frac{0.833^2}{1-0.833}=4.2\text{ calls}$$
3. Average Waiting Time (W_q): $$W_q=\frac{L_q}{\lambda }=\frac{4.2}{10}=0.42\text{ hours}=25.2\text{ minutes}$$
4. System Time (W): $$W=\frac{1}{\mu -\lambda }=\frac{1}{12-10}=0.5\text{ hours}=30\text{ minutes}$$

Interpretation:

• Customers wait 25.2 minutes on average before service.
• Total time in system (waiting + service) = 30 minutes.

flowchart TD
    A["Arrival (λ=10/hour)"] --> B["Queue"]
    B --> C["Server (μ=12/hour)"]
    C -->|"Service Complete"| D["Exit"]
    C -->|"Back to Queue"| B

4. Duality Theorem in Linear Programming

The duality theorem states that every primal LPP has a corresponding dual LPP, and their optimal solutions are equal.

Primal vs. Dual Relationship

Example 5: Airline Seat Allocation (Primal-Dual Pair) Primal Problem (Maximize Profit):

• Variables:
  • $x_1$ = Economy seats (Rs. 5,000/profit).
  • $x_2$ = Business seats (Rs. 10,000/profit).
• Constraints:
  • Weight: $3x_1+5x_2\le 15,000$ kg.
  • Space: $4x_1+6x_2\le 24,000$ units.
  • $x_1,x_2\ge 0$.

Dual Problem (Minimize Cost):

• Variables:
  • $y_1$ = Shadow price for weight (Rs./kg).
  • $y_2$ = Shadow price for space (Rs./unit).
• Constraints:
  • $3y_1+4y_2\ge 5,000$ (Economy).
  • $5y_1+6y_2\ge 10,000$ (Business).
  • $y_1,y_2\ge 0$.

Optimal Solution:

• Primal: $x_1=2,000$, $x_2=1,500$ → $Z=25,000,000$.
• Dual: $y_1=1,000$, $y_2=1,250$ → $W=25,000,000$.

5. Limitations of Operational Research

While OR provides powerful tools, it has key limitations:

1. Assumptions: Models often assume linearity, constant rates, or independence (e.g., Poisson arrivals in queues).
2. Data Dependency: Results are only as good as input data (e.g., inaccurate failure probabilities).
3. Human Factors: Ignores psychological/social behaviors (e.g., customer impatience in queues).
4. Dynamic Systems: Static models fail for rapidly changing environments (e.g., stock market).
5. Implementation Cost: Solving large-scale problems requires expensive software/hardware.

In the Real World

1. Daraz Logistics:

   • Uses queuing theory (M/M/c) to optimize warehouse order processing.
   • Multiple servers (c) handle Poisson-arriving orders to minimize delivery delays.

2. Ncell Customer Service:

   • Models calls as M/M/1 queues to predict wait times and hire agents.
   • Example: If λ = 15 calls/hour and μ = 12 calls/hour, $W_q=1.5$ hours (critical bottleneck!).

3. Bank Loan Interest (EOQ):

   • Banks use EOQ to decide how often to order cash for ATMs.
   • Example: If $D=500,000$ notes/year, $S=2,000$, $H=0.5$, then: $$Q^{\ast }=\sqrt{\frac{2\times 500,000\times 2,000}{0.5}}=40,000\text{ notes/order}$$

4. Pathao Rider Dispatch:

   • Uses M/G/1 queues to model variable ride durations (general service time).
   • Helps estimate rider wait times during peak hours.

Exam Tip

1. Replacement Models:

   • Always calculate average cost per year for comparison.
   • For group replacement, consider failure costs before replacement interval.

2. EOQ:

   • Memorize the formula and units (D = units/year, S = Rs./order, H = Rs./unit/year).
   • Plot the total cost curve to show the minimum at $Q^{\ast }$.

3. Queuing Theory:

   • Kendall’s notation is critical—practice mapping real scenarios (e.g., bank = M/M/c).
   • For M/M/1, derive $L_q$, $W_q$, and $W$ from $\rho =\lambda /\mu$.

4. Duality:

   • Convert primal to dual by:
     1. Swap max/min.
     2. Transpose coefficients.
     3. Change $\le$ to $\ge$ and vice versa.
   • Shadow prices (dual variables) show "cost per unit of resource."

5. Short Notes:

   • EOQ: "Balances ordering and holding costs for optimal Q."
   • Kendall’s Notation: "A/B/c/K/N describes queue characteristics."
   • Duality Theorem: "Primal and dual LPPs have equal optimal values."

Visual Summary:

mindmap
  root((Replacement Models & OR Misc))
    Individual Replacement
      Cost: Failure-driven
      Example: Bulbs
    Group Replacement
      Cost: Fixed interval
      Example: Scooter (Year 4)
    EOQ
      Formula: √(2DS/H)
      Example: NTC Spare Parts (Q*=2,236)
    Queuing Theory
      Kendall's Notation: M/M/1, M/G/c
      Example: Ncell Calls (W_q=25.2 min)
    Duality
      Primal ↔ Dual
      Example: Airline Seats
    Limitations
      Assumptions, Data, Human Factors