About this previous year paper
KOE075 – OPERATIONS RESEARCH
B.Tech (SEM VII) – Theory Examination
Time: 3 Hours | Max Marks: 100
SECTION A
(Attempt all questions in brief – 2 × 10 = 20 marks)
a. What is the Simplex method in Linear Programming?
The Simplex method is an iterative optimization technique used to solve linear programming problems. It moves from one basic feasible solution to another, improving the objective function value at each step until the optimal solution is reached.
b. Define a Two Variable Linear Programming model.
A two-variable linear programming model consists of an objective function and a set of linear constraints involving two decision variables, usually denoted as x1x_1x1 and x2x_2x2, with non-negativity restrictions.
c. In what industries are transportation problems commonly encountered?
Transportation problems are commonly encountered in:
Manufacturing industries
Supply chain and logistics
Distribution and retail industries
Oil, cement, and fertilizer industries
d. Define the objective of mathematical models in transportation problems.
The objective is to determine the minimum transportation cost or maximum profit while satisfying supply and demand constraints across various sources and destinations.
e. What is the primary goal of the shortest path model in network techniques?
The primary goal is to find the minimum cost or minimum distance path between a source node and a destination node in a network.
f. Define the minimum spanning tree problem.
The minimum spanning tree problem aims to connect all nodes of a network using the minimum total edge cost, without forming any cycles.
g. What is a rectangular game in the context of game theory?
A rectangular game is a two-player zero-sum game represented by a payoff matrix with different numbers of strategies for each player.
h. What is the Minimax theorem, and what does it ensure in game theory?
The Minimax theorem states that every two-person zero-sum game with mixed strategies has an equilibrium point, ensuring minimum loss for one player and maximum gain for the other.
i. How does the EOQ model balance ordering costs and holding costs?
The EOQ model determines an order quantity where ordering cost equals holding cost, minimizing total inventory cost.
j. Define Reorder Point (ROP) and its significance in inventory management.
ROP is the inventory level at which a new order is placed. It ensures uninterrupted production by accounting for demand during lead time.
SECTION B
(Attempt any three – solved answers provided for ALL)
2(a). Define Operations Research and provide a numerical example
Operations Research (OR) is a scientific approach to decision-making that uses mathematical models, statistics, and optimization techniques to find optimal solutions to complex problems.
Numerical Example:
A factory produces two products with limited labor and material. OR helps determine how many units of each product should be produced to maximize profit using Linear Programming.
2(b). Transportation Problems and balanced vs unbalanced problems
A transportation problem deals with minimizing the cost of transporting goods from multiple sources to multiple destinations.
Balanced problem: Total supply = Total demand
Unbalanced problem: Total supply ≠ Total demand (dummy row/column added)
2(c). Shortest Path Model in network analysis
The shortest path model finds the least-cost route between two nodes.
Applications:
Road navigation Communication networks Supply chain routing
Algorithms used: Dijkstra’s algorithm, Bellman-Ford algorithm.
2(d). Minimax Theorem and determination of minimax strategy
Significance:
Ensures rational decision-making under competition.
Steps to determine minimax strategy:
Identify minimum payoff for each strategy
Select strategy with maximum of these minimum payoffs
If maximin = minimax → saddle point exists
2(e). EOQ model: definition, assumptions, and limitations
EOQ Formula:
EOQ=2DSHEOQ = \sqrt{\frac{2DS}{H}}EOQ=H2DS
Where: D = Demand, S = Ordering cost, H = Holding cost
Assumptions: Constant demand
Instant replenishment No stock-outs
Limitations: Not suitable for fluctuating demand
Ignores quantity discounts
SECTION C
3(a). Simplex Method Numerical
Maximize: Z=3x1+2x2Z = 3x_1 + 2x_2Z=3x1+2x2
Subject to:
2x1+x2≤102x_1 + x_2 \le 102x1+x2≤10 4x1−5x2≥−204x_1 - 5x_2 \ge -204x1−5x2≥−20 x1,x2≥0x_1, x_2 \ge 0x1,x2≥0
Steps:
Convert inequalities to equations using slack/surplus variables
Form initial simplex table
Identify pivot element
Perform row operations
Continue until optimality condition is satisfied
Optimal Solution:
x1=5, x2=0x_1 = 5,\; x_2 = 0x1=5,x2=0 Zmax=15Z_{max} = 15Zmax=15
3(b). Dual Simplex Method (outline)
Used when solution is infeasible but optimality condition is satisfied.
Final Answer:
x1=2, x2=3, Zmax=17x_1 = 2,\; x_2 = 3,\; Z_{max} = 17x1=2,x2=3,Zmax=17
4(a). Assignment Problem using Hungarian Method
Cost Matrix:
| Job 1 | Job 2 | Job 3 | |
|---|---|---|---|
| M1 | 8 | 6 | 10 |
| M2 | 9 | 7 | 4 |
| M3 | 3 | 2 | 5 |
Optimal Assignment:
M1 → Job 2 M2 → Job 3 M3 → Job 1
Minimum Cost = 13
4(b). Transportation Problem using MODI Method
After initial feasible solution and MODI optimization:
Minimum Transportation Cost = ₹3,820
5(a). Critical Path Method (CPM)
From the given network (Page 2 image):
Critical Path:
1→3→7→8→91 \rightarrow 3 \rightarrow 7 \rightarrow 8 \rightarrow 91→3→7→8→9
Project Duration = 14 units
Slack of critical activities = 0
5(b). Project Scheduling
Steps:
Draw network Calculate Earliest Event Time (TE)
Calculate Latest Event Time (TL) Determine floats
Identify critical path
6(a). Game Theory – Saddle Point Problem
The entry (2,2) is a saddle point when: p≤q≤6p \le q \le 6p≤q≤6
6(b). Queuing Theory (M/M/1 model)
Given: Arrival rate λ=85\lambda = \frac{8}{5}λ=58
Service rate μ=105\mu = \frac{10}{5}μ=510
Results: Average customers in queue = 1.6
Average waiting time = 1 minute Probability of more than 6 customers ≈ 0.08
7(a). EOQ Numerical
Given: D = 5000 units/year
S = ₹100 H = ₹2
EOQ=2×5000×1002=500000≈707 unitsEOQ = \sqrt{\frac{2 \times 5000 \times 100}{2}} = \sqrt{500000} \approx 707 \text{ units}EOQ=22×5000×100=500000≈707 units
7(b). Reorder Point (ROP)
Given: Demand = 200 units/week
Lead time = 2 weeks Z (95%) = 1.65
ROP=400+1.65(202)≈446 unitsROP = 400 + 1.65(20\sqrt{2}) \approx 446 \text{ units}ROP=400+1.65(202)≈446 units