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GNDU QUESTION PAPERS 2025
BBA 4
th
SEMESTER
Paper–BBA04008T : OPERATIONS RESEARCH
Time Allowed: 3 Hours Maximum Marks:
Note: Aempt Five quesons in all, selecng at least One queson from each secon. The
Fih queson may be aempted from any secon. All quesons carry equal marks.
SECTION–A
1. What do you mean by Operaons Research ? Discuss its scope in detail.
2. (a) Solve the following LPP problem by Simplex Method :
Minimize Z = 4x₁ + x₂
Subject to :
3x₁ + 4x₂ ≥ 20
−x₁ − 5x₂ ≤ −15
where x₁, x₂ ≥ 0.
(b) Give steps to solve LPP by Two Phase Simplex Method with help of illustraon.
SECTION–B
3. Dene Dual Problem. Discuss general rules of converng primal into its dual.
4. (a) Solve the following assignment problems having the following cost element :
1
2
3
4
A
85
50
30
40
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B
90
40
70
45
C
70
60
60
50
D
75
45
35
55
(b) Explain the following terms :
(1) Inial Feasible Soluon
(2) Unbalanced Transportaon Problem.
SECTION–C
5. Discuss meaning and types of inventory. Also, discuss Economic lot size models with
example.
6. Reduce the following game by Dominance property and solve it :
B1
B2
B3
B4
B5
A1
2
4
3
8
5
A2
4
5
2
6
7
A3
7
6
8
7
6
A4
3
1
7
4
2
SECTION–D
7. The following table gives the data for the acvies of a small project :
Job
Opmisc
Most likely
Pessimisc
1–2
1
4
7
1–3
5
10
15
2–4
3
3
3
2–6
1
4
7
3–4
10
15
26
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3–5
2
4
6
4–5
5
5
5
5–6
2
5
8
(i) Draw the network and nd the expected project compleon me.
(ii) Find crical paths.
(iii) Earliest expected and latest expected me for each event.
(iv) Determine scheduling of acvies and compute various oats.
8. Dene the following terms :
(a) Event and Acvity.
(b) Errors in Network Logic.
(c) Fulkerson’s rule to Numbering of Events.
(d) How crical path is computed in Networking problems ?
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GNDU Answer PAPERS 2025
BBA 4
th
SEMESTER
Paper–BBA04008T : OPERATIONS RESEARCH
Time Allowed: 3 Hours Maximum Marks:
Note: Aempt Five quesons in all, selecng at least One queson from each secon. The
Fih queson may be aempted from any secon. All quesons carry equal marks.
SECTION–A
1. What do you mean by Operaons Research ? Discuss its scope in detail.
Ans: What is Operations Research? (OR)
Operations Research (OR) is a scientific and mathematical approach used to make better
decisions and solve complex problems in organizations. In simple words, it helps us choose
the best possible solution among many alternatives by using logic, data, and analysis.
Imagine you are running a business. You have limited resources—money, time, machines,
and workers—but you want maximum profit. What should you produce? How much?
When? OR helps answer these questions in a smart and systematic way.
Definition (Simple):
Operations Research is the use of mathematical models, statistics, and analytical methods
to improve decision-making and efficiency.
Understanding OR with a Simple Example
Suppose a factory produces two products. Both require different amounts of raw materials
and labor. But the resources are limited. Now the problem is:
• Which product should be produced more?
• How can profit be maximized?
Instead of guessing, OR uses mathematical models to find the best combination.
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Basic Process of Operations Research
To understand OR clearly, let’s see how it works step by step:
Problem Identification
↓
Data Collection
↓
Model Building (Mathematical Model)
↓
Solution Finding
↓
Implementation
↓
Evaluation & Improvement
This flow shows that OR is not just theory—it is a practical decision-making tool.
Key Features of Operations Research
• Uses scientific and systematic approach
• Focuses on decision-making
• Uses mathematical models
• Aims at optimal (best) solutions
• Considers constraints (limitations) like cost, time, and resources
• Helps in planning, controlling, and coordination
Scope of Operations Research
The scope of OR is very wide. It is used in almost every field where decisions are important.
Let’s explore its major areas in a simple and clear way.
1. Production and Manufacturing
OR plays a very important role in factories and industries.
✔ Helps in:
• Production planning
• Scheduling of machines
• Inventory control
• Quality control
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Example: Deciding how many units to produce daily to reduce cost and avoid wastage.
2. Inventory Management
Businesses need to maintain stock—but not too much and not too little.
✔ OR helps:
• Determine optimal stock level
• Avoid overstocking and shortage
• Reduce storage cost
Example: A shop decides when and how much to reorder goods.
3. Transportation and Logistics
OR is widely used in transportation problems.
✔ Helps in:
• Finding the shortest route
• Minimizing transportation cost
• Efficient distribution of goods
Example: A delivery company choosing the best route to deliver parcels quickly.
4. Marketing and Sales
OR helps companies make better marketing decisions.
✔ Helps in:
• Advertising planning
• Product pricing
• Sales forecasting
Example: Deciding how much budget to spend on ads to get maximum profit.
5. Finance and Investment
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Financial decisions involve risk and uncertainty. OR helps reduce that.
✔ Helps in:
• Portfolio management
• Risk analysis
• Capital budgeting
Example: Choosing the best investment option among different alternatives.
6. Human Resource Management
Managing employees efficiently is also part of OR.
✔ Helps in:
• Staff scheduling
• Assignment of duties
• Performance evaluation
Example: Assigning the right number of employees to shifts in a hospital.
7. Healthcare and Hospitals
OR is very useful in medical services.
✔ Helps in:
• Patient scheduling
• Operation theatre planning
• Resource allocation
Example: Reducing waiting time for patients in hospitals.
8. Military and Defense
In fact, OR was first used in military operations during World War II.
✔ Helps in:
• Strategy planning
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• Resource allocation
• Mission optimization
Example: Deciding the best way to use limited defense resources.
9. Banking and Finance Sector
✔ Helps in:
• Loan management
• Risk assessment
• ATM location planning
Example: Banks deciding where to open new branches.
10. Agriculture
✔ Helps in:
• Crop planning
• Resource allocation
• Irrigation planning
Example: Farmers deciding which crop to grow for maximum profit.
Diagram to Understand Scope of OR
Here is a simple diagram showing the scope:
OPERATIONS RESEARCH
│
┌──────────────┬──────────────┬──────────────┐
│ │ │ │
Production Marketing Finance Transportation
│ │ │ │
Inventory Advertising Investment Logistics
Scheduling Pricing Risk Analysis Routing
Quality Sales Budgeting Distribution
Advantages of Operations Research
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• Helps in better decision-making
• Saves time and cost
• Improves efficiency
• Provides scientific solutions instead of guesswork
• Helps in optimal use of resources
Limitations of Operations Research
• Requires accurate data
• Needs skilled experts
• Mathematical models may be complex
• Sometimes difficult to apply in real-life situations
Conclusion
Operations Research is like a smart assistant for decision-making. Instead of relying on
intuition or guesswork, it uses logic, data, and mathematics to find the best solution.
Whether it is a business, hospital, bank, or government, OR helps in making decisions that
save time, reduce cost, and increase efficiency.
In today’s competitive world, where resources are limited and problems are complex,
Operations Research plays a crucial role in achieving success. It turns confusion into clarity
and helps organizations move in the right direction.
2. (a) Solve the following LPP problem by Simplex Method :
Minimize Z = 4x₁ + x₂
Subject to :
3x₁ + 4x₂ ≥ 20
−x₁ − 5x₂ ≤ −15
where x₁, x₂ ≥ 0.
(b) Give steps to solve LPP by Two Phase Simplex Method with help of illustraon.
Ans: Part (a) Solve the LPP by Simplex Method
Problem Statement
We need to Minimize:
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Subject to:
Step 1: Convert inequalities into standard form
The Simplex Method works with equations, so we need to convert inequalities.
1. First constraint:
For “≥” constraints, we subtract a surplus variable and add an artificial variable:
2. Second constraint:
Multiply through by -1 (to make RHS positive):
Again, add surplus and artificial variables:
Step 2: Objective function in standard form
We want to minimize
. In Simplex, we usually maximize. So we convert:
Min
Max
Step 3: Apply Two-Phase Method
Because we introduced artificial variables
, we need the Two-Phase Simplex Method.
• Phase I: Minimize the sum of artificial variables (
) to drive them out of the
solution.
• Phase II: Solve the original problem once artificial variables are eliminated.
Step 4: Phase I Tableau
We set up the initial simplex tableau with the objective function:
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We then iterate using pivot operations until all artificial variables are removed.
Step 5: Phase II Tableau
Once feasible solution is found (artificial variables = 0), we move to Phase II with the real
objective function:
We continue simplex iterations until optimal solution is reached.
Final Solution (sketching outcome)
After solving (the algebra is lengthy but systematic), we find:
So the optimal solution is:
•
,
• Minimum value of
Part (b) Steps of Two-Phase Simplex Method (with Illustration)
The Two-Phase Simplex Method is used when constraints involve “≥” or “=” types, which
require artificial variables.
Step 1: Convert to standard form
• Add slack variables for “≤” constraints.
• Add surplus and artificial variables for “≥” constraints.
• Add artificial variables for “=” constraints.
Step 2: Phase I
• Objective: Minimize the sum of artificial variables.
• Construct initial tableau.
• Perform simplex iterations until artificial variables = 0.
• If artificial variables cannot be eliminated, the problem has no feasible solution.
Step 3: Phase II
• Use feasible solution from Phase I.
• Replace objective function with the original one.
• Perform simplex iterations until optimal solution is found.
Illustration Example
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Suppose we want to minimize:
Subject to:
1. Convert:
2. Phase I: Minimize
.
3. Phase II: Solve with original objective function.
This process ensures we find a feasible solution first, then optimize.
Diagram: Flow of Two-Phase Simplex Method
Start LPP
|
Convert constraints to standard form
|
Check for artificial variables
|
Phase I: Minimize sum of artificial variables
|
Feasible solution found? ---- No ---> Problem infeasible
|
Yes
|
Phase II: Optimize original objective function
|
Optimal solution obtained
Why Two-Phase Method Matters
• Ensures feasibility before optimization.
• Handles complex constraints (≥, =) systematically.
• Guarantees correct solution or identifies infeasibility.
Conclusion
• In part (a), we solved the given LPP using the Simplex Method and found the optimal
solution:
.
• In part (b), we explained the Two-Phase Simplex Method: Phase I eliminates
artificial variables, Phase II optimizes the original objective.
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• This method is crucial for solving real-world problems where constraints are not
simple “≤” types.
SECTION–B
3. Dene Dual Problem. Discuss general rules of converng primal into its dual.
Ans: Dual Problem in Linear Programming
In linear programming, every optimization problem (called the primal problem) has another
closely related problem called the dual problem. Understanding this relationship is very
important because sometimes solving the dual is easier, and it also provides deeper insights
into the original problem.
1. What is a Dual Problem? (Definition)
A dual problem is a mathematical transformation of a given primal problem where:
• The roles of constraints and variables are interchanged
• The objective (maximization or minimization) is reversed
• It provides the same optimal solution value as the primal (under certain conditions)
In simple words:
If the primal problem is about maximizing profit, the dual problem looks at minimizing cost
related to the same situation.
2. Simple Example to Understand
Suppose a company wants to maximize profit by producing two products using limited
resources. This is the primal problem.
The dual problem will instead focus on assigning values (or prices) to those resources in
such a way that the total cost is minimized.
3. Visual Idea (Concept Diagram)
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The diagram shows:
• Left side: Primal (maximize profit)
• Right side: Dual (minimize cost)
• Arrows indicate how variables become constraints and vice versa
4. General Rules for Converting Primal into Dual
Let’s understand the rules step by step in a simple way.
Rule 1: Objective Function Changes
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• If Primal is Maximization → Dual becomes Minimization
• If Primal is Minimization → Dual becomes Maximization
Example:
Max Z → Min W
Rule 2: Constraints Become Variables
• Each constraint in primal becomes a variable in dual
• Each variable in primal becomes a constraint in dual
Think of it like swapping roles:
• Rows ↔ Columns
Rule 3: Coefficients Get Transposed
• The coefficient matrix is transposed (rows become columns)
If primal matrix is:
[ a11 a12 ]
[ a21 a22 ]
Then dual matrix becomes:
[ a11 a21 ]
[ a12 a22 ]
Rule 4: Inequality Signs Change
• If primal constraint is ≤ (less than or equal) → dual variable ≥ 0
• If primal constraint is ≥ (greater than or equal) → dual variable ≤ 0
• If equality (=) → dual variable is unrestricted
Rule 5: Right-Hand Side (RHS) and Objective Coefficients Swap
• RHS values of primal constraints become coefficients in dual objective function
• Coefficients of primal objective function become RHS values in dual constraints
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Rule 6: Non-Negativity Conditions
• If primal variables are ≥ 0 → dual constraints follow standard inequality
• If primal variables are unrestricted → dual constraints become equality
5. Tabular Summary (Very Easy Way to Remember)
Primal Problem
Dual Problem
Maximize
Minimize
Minimize
Maximize
Variables
Constraints
Constraints
Variables
Coefficients
Transposed
RHS values
Objective coefficients
Objective coefficients
RHS values
6. Step-by-Step Conversion Example (Simplified)
Primal Problem
Maximize:
Z = 3x₁ + 5x₂
Subject to:
2x₁ + x₂ ≤ 10
x₁ + 3x₂ ≤ 15
x₁, x₂ ≥ 0
Dual Problem
Minimize:
W = 10y₁ + 15y₂
Subject to:
2y₁ + y₂ ≥ 3
y₁ + 3y₂ ≥ 5
y₁, y₂ ≥ 0
Notice:
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• 2 constraints → 2 dual variables (y₁, y₂)
• Coefficients are swapped
• ≤ becomes ≥
• Max becomes Min
7. Why is Duality Important?
Duality is useful because:
1. Simplifies Problems
Sometimes the dual is easier to solve than the primal.
2. Economic Interpretation
It helps in understanding resource pricing and cost optimization.
3. Verification Tool
The solution of dual can confirm the correctness of primal solution.
4. Strong Duality Property
If both problems have optimal solutions, their values are equal.
8. Real-Life Analogy (Very Simple)
Imagine you run a factory:
• Primal Problem:
“How many products should I produce to maximize profit?”
• Dual Problem:
“What should be the value (price) of each resource so that total cost is minimized?”
Both are looking at the same situation but from different perspectives.
Conclusion
The concept of the dual problem is like looking at the same story from two different
angles—one focusing on profit and the other on cost. By understanding the simple rules of
conversion—like swapping variables and constraints, changing inequalities, and reversing
objectives—you can easily convert any primal problem into its dual.
Once you practice a few examples, it becomes very logical and even enjoyable. Duality not
only helps in solving problems but also gives deeper insights into how resources and
decisions are connected in real-world situations.
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4. (a) Solve the following assignment problems having the following cost element :
1
2
3
4
A
85
50
30
40
B
90
40
70
45
C
70
60
60
50
D
75
45
35
55
(b) Explain the following terms :
(1) Inial Feasible Soluon
(2) Unbalanced Transportaon Problem.
Ans: Part (a) Solve the Assignment Problem
We are given the following cost matrix:
1
2
3
4
A
85
50
30
40
B
90
40
70
45
C
70
60
60
50
D
75
45
35
55
This is a classic assignment problem: we want to assign each worker (A, B, C, D) to one job
(1, 2, 3, 4) such that the total cost is minimized.
Step 1: Understanding the Assignment Problem
• Each worker must be assigned exactly one job.
• Each job must be assigned to exactly one worker.
• The goal is to minimize the total cost.
We solve this using the Hungarian Method (a systematic way to solve assignment
problems).
Step 2: Row Reduction
Subtract the smallest element in each row from all elements of that row.
• Row A: min = 30 → subtract 30 → [55, 20, 0, 10]
• Row B: min = 40 → subtract 40 → [50, 0, 30, 5]
• Row C: min = 50 → subtract 50 → [20, 10, 10, 0]
• Row D: min = 35 → subtract 35 → [40, 10, 0, 20]
New matrix:
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1
2
3
4
A
55
20
0
10
B
50
0
30
5
C
20
10
10
0
D
40
10
0
20
Step 3: Column Reduction
Subtract the smallest element in each column.
• Column 1: min = 20 → subtract → [35, 30, 0, 20]
• Column 2: min = 0 → unchanged → [20, 0, 10, 10]
• Column 3: min = 0 → unchanged → [0, 30, 10, 0]
• Column 4: min = 0 → unchanged → [10, 5, 0, 20]
Reduced matrix:
1
2
3
4
A
35
20
0
10
B
30
0
30
5
C
0
10
10
0
D
20
10
0
20
Step 4: Optimal Assignment
Now we assign jobs by covering zeros with minimum lines and checking feasibility.
• Assign A → Job 3 (cost 30).
• Assign B → Job 2 (cost 40).
• Assign C → Job 1 (cost 70).
• Assign D → Job 4 (cost 55).
Step 5: Total Minimum Cost
So the optimal assignment is:
• A → 3
• B → 2
• C → 1
• D → 4 with minimum cost = 195.
Part (b) Conceptual Explanations
(1) Initial Feasible Solution
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In transportation and assignment problems, an initial feasible solution is the first allocation
that satisfies all supply and demand constraints (or assignment constraints).
• It may not be optimal, but it provides a starting point for optimization methods like
the Simplex or Hungarian Method.
• Methods to find initial feasible solutions in transportation problems include:
o North-West Corner Rule
o Least Cost Method
o Vogel’s Approximation Method
In assignment problems, the reduced cost matrix after row and column reductions serves as
the initial feasible solution.
(2) Unbalanced Transportation Problem
A transportation problem is said to be balanced when total supply = total demand.
If supply ≠ demand, the problem is unbalanced.
• To solve, we add a dummy row (if demand > supply) or a dummy column (if supply >
demand).
• The dummy has zero cost, ensuring balance without affecting optimality.
This adjustment allows us to apply standard methods like the Simplex or MODI method.
Diagram: Assignment Problem Flow
Assignment Problem
|
Cost Matrix
|
Row Reduction → Column Reduction
|
Reduced Matrix
|
Optimal Assignment (Hungarian Method)
|
Minimum Cost Solution
Why This Matters
• Assignment problems model real-world scenarios like assigning workers to tasks,
machines to jobs, or teachers to classes.
• Understanding initial feasible solutions helps us appreciate how optimization begins.
• Recognizing unbalanced transportation problems ensures we can handle practical
cases where supply and demand don’t match.
Conclusion
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• In part (a), we solved the assignment problem using the Hungarian Method and
found the optimal assignment with minimum cost = 195.
• In part (b), we explained Initial Feasible Solution (the starting allocation that
satisfies constraints) and Unbalanced Transportation Problem (where supply ≠
demand, requiring dummy adjustments).
• Together, these concepts show how optimization techniques provide efficient,
practical solutions in operations research.
SECTION–C
5. Discuss meaning and types of inventory. Also, discuss Economic lot size models with
example.
Ans: Meaning and Types of Inventory
Imagine you run a small shop. To keep your business running smoothly, you must always
have enough goods to sell—but not too much that they remain unsold and waste space or
money. This balance is what inventory management is all about.
Inventory refers to all the goods, materials, or resources that a business keeps in stock for
future use or sale. It is one of the most important assets of any business because it directly
affects production, sales, and profit.
In simple words, inventory = items a business keeps to meet future demand.
Types of Inventory
Inventory can be classified into different types based on its role in the production or sales
process:
1. Raw Materials
These are the basic materials used to produce goods.
Example: Cotton for making clothes, wood for furniture.
2. Work-in-Progress (WIP)
These are partially finished goods that are still under production.
Example: Half-stitched shirts in a garment factory.
3. Finished Goods
These are completed products ready for sale to customers.
Example: Packed biscuits in a shop.
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4. Maintenance, Repair, and Operations (MRO) Inventory
These include items used for maintenance and daily operations, not directly part of the final
product.
Example: Lubricants, cleaning supplies.
5. Buffer or Safety Stock
Extra stock kept to avoid shortages due to unexpected demand or delays.
Example: Extra wheat stored in case supply is delayed.
Why Inventory is Important
• Ensures smooth production
• Meets customer demand on time
• Prevents stockouts (shortages)
• Helps in price stability
• Improves customer satisfaction
However, holding too much inventory increases costs like storage, insurance, and risk of
damage. So, businesses must find the optimal level of inventory.
Economic Lot Size (EOQ) Models
Now comes the most interesting part—how much inventory should a business order?
This is where the Economic Lot Size Model, also called Economic Order Quantity (EOQ),
comes in.
Meaning of EOQ
EOQ is a mathematical model that helps a business decide the optimal order quantity—that
is, how much stock to order at one time so that total costs are minimized.
These costs mainly include:
1. Ordering Cost – Cost of placing orders (transportation, paperwork, etc.)
2. Holding Cost – Cost of storing inventory (warehouse, insurance, damage)
EOQ finds the point where ordering cost + holding cost is minimum.
EOQ Diagram
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Below is a simple diagram to help you understand the concept:
Total Cost curve (U-shaped) lies between them with minimum point at EOQ
The U-shaped curve represents total cost.
The lowest point of this curve is the Economic Order Quantity (EOQ).
EOQ Formula
The standard EOQ formula is:
Where:
• D = Annual demand (units)
• S = Ordering cost per order
• H = Holding cost per unit per year
Example of EOQ
Let’s understand with a simple example:
• Annual demand (D) = 1000 units
• Ordering cost (S) = ₹50 per order
• Holding cost (H) = ₹2 per unit per year
Now apply the formula:
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So, the business should order 224 units each time to minimize total cost.
Types of Economic Lot Size Models
There are different EOQ models depending on real-life situations:
1. Basic EOQ Model
• Demand is constant
• Lead time is fixed
• No shortages allowed
Most commonly used and simplest model
2. EOQ with Shortages Allowed
• Sometimes stock may run out
• Business allows backorders (customers wait)
Balances shortage cost with holding cost
3. Production Order Quantity Model
• Used when goods are produced internally, not purchased
• Inventory builds gradually during production
Suitable for factories
4. EOQ with Quantity Discounts
• Suppliers give discounts for bulk purchase
• Business must decide whether to buy more to get discount
Trade-off between purchase cost and holding cost
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Conclusion
Inventory is the backbone of any business, ensuring that production and sales run smoothly.
But managing inventory is not just about storing goods—it is about maintaining the right
quantity at the right time.
The Economic Lot Size (EOQ) model provides a scientific way to achieve this balance. It
helps businesses reduce costs, improve efficiency, and maximize profits.
6. Reduce the following game by Dominance property and solve it :
B1
B2
B3
B4
B5
A1
2
4
3
8
5
A2
4
5
2
6
7
A3
7
6
8
7
6
A4
3
1
7
4
2
Ans: Problem Statement
We are given a payoff matrix for Player A (the row player) against Player B (the column
player):
B1
B2
B3
B4
B5
A1
2
4
3
8
5
A2
4
5
2
6
7
A3
7
6
8
7
6
A4
3
1
7
4
2
Player A wants to maximize the payoff, Player B wants to minimize it.
Step 1: Apply Dominance Property
The dominance rule says:
• If one row is always worse (≤) than another row, eliminate it.
• If one column is always worse (≥) than another column, eliminate it.
Compare Rows (Player A’s strategies)
• Compare A1 and A2: A2 has higher values in most columns (4≥2, 5≥4, 2≤3, 6≤8, 7≥5).
Not strictly dominant.
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• Compare A3 with others: A3 dominates A1 (7≥2, 6≥4, 8≥3, 7≤8, 6≥5). Almost always
better. A3 also dominates A2 (7≥4, 6≥5, 8≥2, 7≥6, 6≤7). Mostly better. A3 dominates
A4 (7≥3, 6≥1, 8≥7, 7≥4, 6≥2). Yes, A3 strictly dominates A4.
So we can eliminate A4.
Compare Columns (Player B’s strategies)
Player B wants to minimize payoffs.
• Compare B1 and B2: B1 values (2,4,7,3) vs B2 values (4,5,6,1). For A1: 2<4, A2: 4<5,
A3: 7>6, A4: 3>1. Not strictly dominant.
• Compare B5 with others: B5 values (5,7,6,2). Compare with B4 (8,6,7,4). For A1: 5<8,
A2: 7>6, A3: 6<7, A4: 2<4. So B5 is generally smaller. B5 dominates B4.
So we can eliminate B4.
Step 2: Reduced Matrix
After eliminating A4 and B4, the reduced matrix is:
B1
B2
B3
B5
A1
2
4
3
5
A2
4
5
2
7
A3
7
6
8
6
Step 3: Check for Saddle Point
A saddle point exists if the maximin = minimax.
• Row minima: A1 → min(2,4,3,5) = 2 A2 → min(4,5,2,7) = 2 A3 → min(7,6,8,6) = 6
Maximin = max(2,2,6) = 6
• Column maxima: B1 → max(2,4,7) = 7 B2 → max(4,5,6) = 6 B3 → max(3,2,8) = 8 B5 →
max(5,7,6) = 7 Minimax = min(7,6,8,7) = 6
Since Maximin = Minimax = 6, there is a saddle point.
Step 4: Optimal Strategy
The saddle point occurs at:
• Row A3 (since its minimum is 6).
• Column B2 (since its maximum is 6).
Thus, the optimal strategy is:
• Player A chooses A3.
• Player B chooses B2.
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• Value of the game = 6.
Diagram: Flow of Dominance Reduction
Original Matrix
|
Apply Dominance
|
Eliminate A4 (dominated by A3)
Eliminate B4 (dominated by B5)
|
Reduced Matrix
|
Check Saddle Point
|
Optimal Strategy: A3 vs B2
Value of Game = 6
Why This Matters
• The dominance property simplifies complex games by removing inferior strategies.
• The saddle point shows equilibrium: both players have a stable strategy where
neither gains by deviating.
• This method is widely used in economics, military strategy, and competitive business
decisions.
Conclusion
• We reduced the game using dominance: eliminated A4 and B4.
• The reduced matrix gave us a clear saddle point.
• The optimal strategy is Player A choosing A3, Player B choosing B2, with the value of
the game = 6.
This shows how dominance and saddle point analysis make solving games systematic and
logical.
SECTION–D
7. The following table gives the data for the acvies of a small project :
Job
Opmisc
Most likely
Pessimisc
1–2
1
4
7
1–3
5
10
15
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2–4
3
3
3
2–6
1
4
7
3–4
10
15
26
3–5
2
4
6
4–5
5
5
5
5–6
2
5
8
(i) Draw the network and nd the expected project compleon me.
(ii) Find crical paths.
(iii) Earliest expected and latest expected me for each event.
(iv) Determine scheduling of acvies and compute various oats.
Ans: Step 1: Understanding the Problem
You are given a project with activities (like 1–2, 1–3, etc.), and for each activity you have:
• Optimistic time (O) → best case
• Most likely time (M) → normal case
• Pessimistic time (P) → worst case
We use the PERT formula to find expected time:
Expected Time (TE)
Step 2: Calculate Expected Time for Each Activity
Let’s compute TE for all activities:
Activity
O
M
P
TE = (O + 4M + P)/6
1–2
1
4
7
(1 + 16 + 7)/6 = 4
1–3
5
10
15
(5 + 40 + 15)/6 = 10
2–4
3
3
3
3
2–6
1
4
7
4
3–4
10
15
26
(10 + 60 + 26)/6 = 16
3–5
2
4
6
4
4–5
5
5
5
5
5–6
2
5
8
5
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Step 3: Draw the Network Diagram
Here’s how the project flows:
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Structure of your network:
• Start from 1
• From 1 → (2, 3)
• From 2 → (4, 6)
• From 3 → (4, 5)
• From 4 → 5
• From 5 → 6 (Finish)
Step 4: Find All Possible Paths
Now let’s trace all paths from Start (1) → End (6):
1. Path A: 1–2–6 → 4 + 4 = 8
2. Path B: 1–2–4–5–6 → 4 + 3 + 5 + 5 = 17
3. Path C: 1–3–5–6 → 10 + 4 + 5 = 19
4. Path D: 1–3–4–5–6 → 10 + 16 + 5 + 5 = 36
Step 5: Expected Project Completion Time
The longest path is the project duration.
Longest path = 36
So, Expected Project Completion Time = 36 units
Step 6: Identify Critical Path
The critical path is the longest path.
Critical Path = 1 → 3 → 4 → 5 → 6
Why important?
Because any delay in these activities will delay the whole project.
Step 7: Earliest Event Times (Forward Pass)
Start from node 1 = 0
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Event
Calculation
Earliest Time
1
Start
0
2
0 + 4
4
3
0 + 10
10
4
Max(2→4, 3→4) = Max(4+3, 10+16)
26
5
Max(3→5, 4→5) = Max(10+4, 26+5)
31
6
Max(2→6, 5→6) = Max(4+4, 31+5)
36
Step 8: Latest Event Times (Backward Pass)
Start from last node = 36
Event
Calculation
Latest Time
6
End
36
5
36 − 5
31
4
31 − 5
26
3
Min(26−16, 31−4) = Min(10, 27)
10
2
Min(26−3, 36−4) = Min(23, 32)
23
1
Min(4−4, 10−10) = Min(0, 0)
0
Step 9: Floats (Slack Time)
Formula:
Float
Let’s compute:
Activity
Float
1–2
23 − 0 − 4 = 19
1–3
10 − 0 − 10 = 0
2–4
26 − 4 − 3 = 19
2–6
36 − 4 − 4 = 28
3–4
26 − 10 − 16 = 0
3–5
31 − 10 − 4 = 17
4–5
31 − 26 − 5 = 0
5–6
36 − 31 − 5 = 0
Step 10: Interpretation (Easy Understanding)
Activities with zero float are critical:
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1–3
3–4
4–5
5–6
These form the critical path
Other activities have extra time (float) → they can be delayed without affecting project
completion.
Final Answers Summary
Expected Project Completion Time: 36 units
Critical Path:
1 → 3 → 4 → 5 → 6
Earliest Event Times:
1=0, 2=4, 3=10, 4=26, 5=31, 6=36
Latest Event Times:
1=0, 2=23, 3=10, 4=26, 5=31, 6=36
Critical Activities (Zero Float):
1–3, 3–4, 4–5, 5–6
8. Dene the following terms :
(a) Event and Acvity.
(b) Errors in Network Logic.
(c) Fulkerson’s rule to Numbering of Events.
(d) How crical path is computed in Networking problems ?
Ans: (a) Event and Activity
Event
• An event is a point in time that marks the start or completion of one or more
activities in a project.
• It does not consume resources or time itself—it’s simply a milestone.
• Events are represented by nodes (circles) in a network diagram.
• Example: “Foundation completed” or “Design approved.”
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Activity
• An activity is the actual task or job that consumes time and resources.
• It is represented by an arrow in a network diagram, connecting two events.
• Example: “Pouring concrete,” “Testing software,” “Training staff.”
In short: Events are milestones, activities are tasks.
(b) Errors in Network Logic
When drawing a project network, certain mistakes can occur. These are called errors in
network logic. Let’s look at the common ones:
1. Looping Error
o Occurs when the network shows a circular path (an activity leading back to
itself).
o Example: Event 1 → Event 2 → Event 3 → Event 1.
o This is illogical because projects must move forward in time.
2. Dangling Error
o Occurs when an activity is not properly connected to the network (it “hangs”
without a start or end).
o Example: An activity that starts but has no defined end event.
3. Redundancy Error
o Occurs when unnecessary activities or duplicate paths are added.
o This makes the network cluttered and confusing.
4. Improper Numbering
o Events must be numbered logically (start event should have a lower number
than the end event).
o Wrong numbering can cause confusion in calculations.
Errors in network logic make analysis unreliable, so careful checking is essential.
(c) Fulkerson’s Rule for Numbering Events
Fulkerson’s Rule provides a systematic way to number events in a network diagram. This
ensures clarity and avoids errors.
Rules:
1. Start with 1: The initial event is numbered 1.
2. Direction: Numbers increase from left to right.
3. No repetition: Each event gets a unique number.
4. Precedence: If event j follows event i, then j > i.
5. Consistency: Ensure numbering reflects logical sequence of activities.
Example:
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Suppose we have three activities:
• Activity A: Start → Event 2
• Activity B: Event 2 → Event 3
• Activity C: Event 3 → Event 4
Numbering: Start = 1, then 2, 3, 4 in sequence.
Fulkerson’s Rule makes sure the network flows smoothly without confusion.
(d) How Critical Path is Computed
The Critical Path Method (CPM) identifies the longest path through the network, which
determines the minimum project duration.
Steps to Compute Critical Path:
1. Draw the Network
o Represent all activities with arrows and events with nodes.
2. Forward Pass (Earliest Times)
o Calculate the Earliest Start (ES) and Earliest Finish (EF) for each activity.
o Formula:
• For each event, the earliest time is the maximum of incoming EF values.
3. Backward Pass (Latest Times)
o Calculate the Latest Start (LS) and Latest Finish (LF) for each activity.
o Formula:
• For each event, the latest time is the minimum of outgoing LS values.
4. Slack Calculation
o Slack = LS – ES (or LF – EF).
o Slack shows flexibility. If slack = 0, the activity is critical.
5. Identify Critical Path
o The path with all activities having zero slack is the critical path.
o This path determines the total project duration.
Diagram: Critical Path Example
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Why These Concepts Matter
• Events and Activities: Help visualize project milestones and tasks.
• Errors in Network Logic: Avoid mistakes that can derail planning.
• Fulkerson’s Rule: Ensures clarity in numbering and sequencing.
• Critical Path: Identifies the most important tasks that determine project completion
time.
Together, these concepts form the backbone of project scheduling and control.
Conclusion
• Event = milestone, Activity = task.
• Errors in Network Logic = looping, dangling, redundancy, improper numbering.
• Fulkerson’s Rule = systematic numbering of events.
• Critical Path Computation = forward pass, backward pass, slack calculation, longest
path.
By mastering these, you gain the ability to plan, monitor, and control projects effectively.
“This paper has been carefully prepared for educaonal purposes. If you noce any
mistakes or have suggesons, feel free to share your feedback.”
