UNIT 4: OPERATIONS RESEARCH – SHORT NOTES
Based on RGPV CE-604(D) Past Papers (May 2023 & May 2022)
I. FOUNDATIONS OF OPERATIONS RESEARCH
Definition: Operations Research (OR) is a scientific, quantitative approach to decision-making that uses mathematical models, statistics, and algorithms to find optimal or near-optimal solutions to complex problems.
Phases of OR:
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Problem Formulation: Define the problem, objectives, constraints, and decision variables.
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Model Building: Develop a mathematical representation (e.g., LP, simulation).
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Solution: Apply analytical or numerical methods to solve the model.
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Validation: Test model accuracy and robustness.
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Implementation: Apply solution in real-world context; monitor results.
Modeling in OR:
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Types:
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Mathematical: Equations (e.g., LP, EOQ).
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Simulation: Mimics system behavior over time.
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Graphical: Visual representation for 2-variable problems.
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Steps: Abstraction → Formulation → Solution → Testing → Validation.
Characteristics:
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Interdisciplinary (math, engineering, economics).
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Systems approach (holistic view).
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Quantitative, objective-driven.
Limitations:
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Data quality and availability issues.
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Model simplicity vs. reality trade-off.
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Implementation challenges (human factors, cost).
[!TIP]
Exam Focus: Phases and modeling steps are frequently asked in 7-mark questions. Use examples like "police scheduling" to illustrate formulation.
II. LINEAR PROGRAMMING (LP)
LP Formulation:
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Objective Function: Maximize or minimize linear function (e.g., profit, cost).
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Constraints: Linear inequalities/equalities representing limits.
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Decision Variables: Unknowns to be determined (≥ 0).
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Example:
Police scheduling: Minimize officers subject to daily requirements.
Simplex Method:
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Convert to standard form (maximize, ≤ constraints, slack variables).
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Identify basic feasible solution (BFS).
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Iterate:
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Compute net evaluation (C<sub>j</sub> – Z<sub>j</sub>).
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Select entering variable (most positive for max).
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Compute ratios (b<sub>i</sub>/a<sub>ij</sub> > 0) for leaving variable.
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Pivot to new BFS.
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Optimality Test: All C<sub>j</sub> – Z<sub>j</sub> ≤ 0 (max) → optimal.
Special Cases:
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Big-M Method: For ≥ or = constraints.
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Add artificial variables with large penalty M in objective.
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Minimize: Z = … + M·(sum of artificial variables).
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Drive artificial variables to zero.
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Two-Phase Method: Phase 1 minimizes sum of artificial variables; Phase 2 removes them.
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Unbounded Solution: All ratios infinite → Z → ∞.
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Infeasibility: No BFS exists (e.g., contradictory constraints).
Transportation Problem:
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Formulation:
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Balanced: Total supply = total demand.
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Unbalanced: Add dummy row/column with zero cost.
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Initial Feasible Solutions:
| Method | Approach | Cost Focus | |-----------------|-----------------------------------|------------| | Northwest Corner| Allocate from top-left cell | Ignored | | Least Cost | Allocate to lowest cost cell | Yes | | Vogel’s Approximation | Penalty-based (difference between two lowest costs) | Yes |
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Degeneracy:
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Meaning: Allocations < (m + n – 1) in a basic solution.
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Consequence: May cause cycling (infinite loops in optimality test).
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Resolution:
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Allocate a very small ε (e.g., 0.001) to a zero-cell to make it basic.
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Use MODI method (modified distribution method) to adjust.
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[!TIP]
Common Pitfall: In simplex, forgetting non-negativity or miscomputing ratios. In transportation, degeneracy often overlooked—always check allocations count.
III. INVENTORY MODELS
Deterministic EOQ Model:
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Assumptions:
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Constant demand rate (D).
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Fixed ordering cost (S).
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Constant holding cost (H = i·C, where i = carrying rate, C = unit cost).
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Instantaneous replenishment.
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No shortages.
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EOQ Formula:
$$ Q^* = \sqrt{\frac{2DS}{H}} $$
\boxed{Q^* = \sqrt{\frac{2DS}{H}}}
- Total Cost:
$$ TC = \frac{D}{Q}S + \frac{Q}{2}H + DC $$
- Reorder Point (ROP):
$$ ROP = d \cdot L $$
(d = daily demand, L = lead time).
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Example (SBC alarm clocks, May 2022):
D = 1000 units/year, S = $$\displaystyle 10/order, H = $$0.50/unit/year →
$$ Q^* = \sqrt{\frac{2 \times 1000 \times 10}{0.5}} = \sqrt{40000} = 200 \text{ units} $$
Stochastic Inventory Models:
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Newsvendor Model: Single-period with uncertain demand.
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Critical ratio: \( \frac{C_u}{C_u + C_o} \) (underage vs. overage cost).
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Order quantity where cumulative demand distribution = critical ratio.
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(Q, r) Models: Continuous review; order fixed Q when inventory ≤ r.
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Balances ordering, holding, and shortage costs.
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Service Level: Probability of no stockout (e.g., 95%).
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[!TIP]
Exam Alert: EOQ calculations are common. Ensure units consistency (e.g., H in $/unit/year if D is yearly).
IV. QUEUING THEORY
Queuing System Components (Kendall’s notation: A/B/c/K/N/Z):
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A: Arrival distribution (M = Markov/Poisson, D = deterministic, G = general).
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B: Service distribution.
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c: Number of servers.
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K: System capacity (∞ if unlimited).
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N: Population size (∞ if infinite).
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Z: Queue discipline (FIFO, LIFO, PRI, RAND).
Characteristics:
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Arrival Process: Poisson (exponential inter-arrival) common.
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Service Mechanism: Exponential (M), deterministic, general.
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Queue Discipline:
| Discipline | Description | Example | |------------|------------------------------|-----------------------| | FIFO | First-In-First-Out | Supermarket line | | LIFO | Last-In-First-Out | Stack (plates) | | Priority | Based on urgency/class | Emergency room | | Random | Service order random | Lottery system |
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Performance Measures:
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λ = arrival rate, μ = service rate per server.
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Utilization factor: ρ = λ / (cμ) (must be < 1 for stability).
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Average queue length: L<sub>q</sub>.
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Average waiting time: W<sub>q</sub>.
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Little’s Law: L = λW (applies to system, not just queue).
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M/M/1 Model (Single server, infinite capacity):
- Formulas:
$$ \rho = \frac{\lambda}{\mu} $$
$$ L_q = \frac{\rho^2}{1 - \rho}, \quad W_q = \frac{\rho}{\mu - \lambda} $$
$$ L = \frac{\rho}{1 - \rho}, \quad W = \frac{1}{\mu - \lambda} $$
- Idle time = 1 – ρ.
M/M/c Model (c servers):
- Compute P<sub>0</sub> (probability of zero customers):
$$ P_0 = \left[ \sum_{n=0}^{c-1} \frac{(\lambda/\mu)^n}{n!} + \frac{(\lambda/\mu)^c}{c! (1 - \rho)} \right]^{-1} $$
- Then L<sub>q</sub>, W<sub>q</sub> via standard formulas.
Example Problems (from past papers):
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Repairman (May 2023):
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Exponential service (mean = 30 min → μ = 2/hour).
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Poisson arrivals (10 per 8-hour day → λ = 10/8 = 1.25/hour).
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ρ = λ/μ = 1.25/2 = 0.625 → idle time = 1 – 0.625 = 0.375 (3 hours/day).
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Postal Clerk (May 2022):
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Service time = 3 min → μ = 20/hour.
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Inter-arrival: 12 min (λ = 5/hour) and 5 min (λ = 12/hour).
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Compute ρ, L<sub>q</sub>, W<sub>q</sub> for each period.
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[!TIP]
Key Check: Always verify ρ < 1; otherwise system unstable (queue → ∞). For M/M/c, use c = number of servers.
V. NETWORK MODELS
PERT/CPM:
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Network Diagram:
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Activity-on-Node (AON): Nodes = events, arrows = activities (preferred).
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Activity-on-Arrow (AOA): Arrows = activities, nodes = events.
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Critical Path:
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Longest path through network.
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Activities with zero total float.
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Determines project completion time.
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Forward Pass (Early Start/Finish):
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ES = max(EF of predecessors).
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EF = ES + duration.
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Backward Pass (Late Start/Finish):
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LF = min(LS of successors).
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LS = LF – duration.
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Float/Slack:
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Total Float (TF): LS – ES or LF – EF.
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Free Float (FF): ES of successor – EF (no delay to successors).
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PERT Probabilistic Time:
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Three estimates:
- Optimistic (a), Most likely (m), Pessimistic (b).
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Expected time:
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$$ t_e = \frac{a + 4m + b}{6} $$
Variance:
$$ \sigma^2 = \left( \frac{b - a}{6} \right)^2 $$
- Example (May 2022): Given activities, predecessors, durations → draw network, find critical path, compute ES/EF/LS/LF, slack.
Minimal Spanning Tree (MST):
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Definition: Subset of edges connecting all nodes with minimum total weight, no cycles.
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Applications: Pipeline/cable layout, network design.
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Algorithms:
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Prim’s: Start at a node, add cheapest edge to new node.
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Kruskal’s: Sort edges by weight, add without forming cycles.
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Traveling Salesman Problem (TSP):
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Definition: Find shortest tour visiting each city exactly once and returning to start.
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Solution Approaches:
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Brute force: (n-1)!/2 permutations (infeasible for large n).
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Heuristics: Nearest neighbor, cheapest insertion.
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Integer Programming: Binary variables x<sub>ij</sub> = 1 if edge i→j used.
Minimize ∑∑ c<sub>ij</sub>x<sub>ij</sub> subject to degree constraints and subtour elimination.
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[!TIP]
Critical Path: Always start forward pass from project start (ES=0). For PERT, compute expected times first, then critical path.
VI. GAME THEORY
Basic Concepts:
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Pay-off Matrix: Rows = Player A strategies, columns = Player B strategies; entries = payoffs to A.
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Strategies:
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Pure: Specific choice.
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Mixed: Probability distribution over strategies.
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Zero-Sum Game: One player’s gain = other’s loss (∑ payoffs = 0).
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Two-Person Game: Exactly two players (common in OR).
Solution Methods:
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Saddle Point:
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Maximin (A’s security level) = Minimax (B’s security level).
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Value of game (v): Common payoff at saddle point.
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Optimal pure strategies: Row/column containing saddle point.
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Check: Maximin = Minimax → saddle exists.
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Dominance:
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Row dominance: If row i ≤ row j for all columns, row i is dominated (remove j for max player).
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Column dominance: If column i ≥ column j for all rows, column i is dominated (remove i for min player).
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Reduce matrix iteratively before saddle point search.
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Mixed Strategy (if no saddle):
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For 2×2 games, solve:
For A (max):
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$$ p_1 = \frac{d - b}{a + d - b - c}, \quad p_2 = 1 - p_1 $$
Value:
$$ v = \frac{ad - bc}{a + d - b - c} $$
(where matrix = [a b; c d]).
Fair Game: Zero-sum game with value v = 0.
Example Terms from Past Papers:
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Pay-off matrix: Tabular representation of outcomes.
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Saddle point: Cell where row minimum = column maximum.
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Zero-sum game: Competitive, constant-sum.
[!TIP]
Exam Sequence: First check for dominance → reduce matrix → find maximin/minimax. If equal, saddle point; else mixed strategy (if asked).
VII. PROJECT MANAGEMENT & ANALYSIS TOOLS
CPM Details:
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Crashing: Reduce activity duration at increased cost.
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Cost-time trade-off: Crash cost per unit time = (Crash cost – Normal cost) / (Normal time – Crash time).
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Crash critical path activities with lowest crash cost first.
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Resource Leveling: Adjust activity start times to smooth resource usage (may extend project).
Link to EIA: Network analysis (PERT/CPM) can schedule EIA tasks (scoping, impact studies, reporting).
[!NOTE]
Crashing and resource leveling appeared implicitly in network questions; focus on critical path first.
VIII. APPLICATION & CASE STUDIES
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EOQ: SBC alarm clocks (May 2022) – calculate optimal order quantity.
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Queuing:
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Repairman (May 2023): Exponential service, Poisson arrivals → idle time.
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Postal clerk (May 2022): Time-dependent arrival rates → compute L<sub>q</sub>, W<sub>q</sub>.
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LP: Police scheduling (May 2022) – formulate LP for officer shifts.
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Network: Construction/R&D projects – draw network, find critical path.
IX. SHORT NOTE TOPICS (Direct from Past Papers)
PERT (Program Evaluation and Review Technique):
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Probabilistic time estimates (three estimates).
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Used for R&D, uncertain activity times.
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Expected time = (a + 4m + b)/6.
Pay-off Matrix:
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Table of outcomes for each strategy pair.
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Foundation of game theory.
Saddle Point:
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Cell where row minimum = column maximum.
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Gives optimal pure strategies and game value.
Zero-Sum Game:
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One player’s gain equals other’s loss.
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Constant-sum (often normalized to zero).
Basic Feasible Solution (BFS):
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In LP, solution with m basic variables (m = constraints) satisfying all constraints, non-basic = 0.
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Corresponds to corner point of feasible region.
Optimal Solution:
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BFS that maximizes/minimizes objective.
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Found via simplex optimality test.
Unbounded Solution:
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Objective can increase/decrease indefinitely (e.g., all entering variables have non-positive ratios in max problem).
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Indicates flawed model or missing constraints.
Service Disciplines:
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FIFO: First-come-first-served (common in queues).
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LIFO: Last-in-first-out (stack-like).
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Priority: High-priority customers served first (e.g., emergency).
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Random: Service order random (e.g., lottery).
X. PROBLEM-SOLVING BLUEPRINT
| Topic | Steps | Key Formulas/Checks |
|---|---|---|
| LP (Simplex) | 1. Standard form (max, ≤, slack). <br>2. Initial BFS (slack vars). <br>3. Compute C<sub>j</sub>–Z<sub>j</sub>. <br>4. Enter most positive, leave by min ratio. <br>5. Pivot; repeat until optimal. | Optimal: all C<sub>j</sub>–Z<sub>j</sub> ≤ 0 (max). <br>Unbounded: entering var has all a<sub>ij</sub> ≤ 0. |
| Big-M | 1. Add artificial vars with coefficient M (max: -M; min: +M). <br>2. Phase 1: Minimize sum of artificials. <br>3. Phase 2: Remove artificials, solve original. | Ensure artificial vars leave basis eventually. |
| Transportation | 1. Check balance; add dummy if needed. <br>2. Initial solution (NW/Least Cost/Vogel). <br>3. Check degeneracy (allocations < m+n-1). <br>4. Optimality test (MODI: u<sub>i</sub>, v<sub>j</sub>). <br>5. Adjust if improvement possible. | Degeneracy: add ε to zero-cell. <br>MODI: Δ<sub>ij</sub> = c<sub>ij</sub> – (u<sub>i</sub> + v<sub>j</sub>). |
| Queuing (M/M/1) | 1. Identify λ, μ. <br>2. Compute ρ = λ/μ. <br>3. Apply formulas for L<sub>q</sub>, W<sub>q</sub>, etc. | ρ < 1 required. <br>L<sub>q</sub> = ρ²/(1–ρ). |
| Queuing (M/M/c) | 1. Compute ρ = λ/(cμ). <br>2. Calculate P<sub>0</sub>. <br>3. Find L<sub>q</sub>, W<sub>q</sub>. | P<sub>0</sub> = [∑<sub>n=0</sub><sup>c-1</sup> (λ/μ)<sup>n</sup>/n! + (λ/μ)<sup>c</sup>/(c!(1–ρ))]<sup>–1</sup> |
| Inventory (EOQ) | 1. Identify D, S, H. <br>2. Compute Q* = √(2DS/H). <br>3. TC = (D/Q*)S + (Q*/2)H. <br>4. ROP = d·L if lead time given. | Ensure units: D, S, H consistent (e.g., yearly). |
| Network (PERT/CPM) | 1. Draw AON diagram from precedents. <br>2. Forward pass: ES, EF. <br>3. Backward pass: LF, LS. <br>4. Critical path: ES=LS, EF=LF (TF=0). <br>5. Slack = LS–ES or LF–EF. | Project time = max(EF). <br>For PERT: use t<sub>e</sub> for durations. |
| Game Theory | 1. Write payoff matrix. <br>2. Row minima, column maxima. <br>3. Maximin = Minimax? → saddle point. <br>4. If not, apply dominance to reduce. <br>5. For 2×2 no saddle: use mixed strategy formulas. | Saddle: maximin = minimax = value v. <br>Dominance: eliminate strictly dominated rows/cols. |
[!CAUTION]
Common Errors:
- Simplex: Forgetting to update tableau after pivot.
- Queuing: Using λ instead of μ in denominator.
- Network: Miscalculating ES/EF due to wrong predecessor order.
- EOQ: Mixing holding cost units (e.g., % of cost vs. $/unit).
Final Note: Practice with past paper problems (May 2023 & May 2022) to internalize steps. Focus on definitions, formulas, and step-by-step solutions as per marking patterns.