(SEM VII) THEORY EXAMINATION 2023-24 OPERATIONS RESEARCH

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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