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GNDU QUESTION PAPERS 2023
B.com 6
th
SEMESTER
OPERATIONS RESEARCH
Time Allowed: 3 Hours Maximum Marks: 50
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. Dene Operaons Research. Discuss its features and limitaons.
2. A Ltd. has two products R and W. To produce one unit of R, two units of material X and 4
units of material Y are required. To produce one unit of W, 3 units of material X and two
units of material Y are required. At least 16 units of each material must be used in order to
meet commied sales of R and W. Due to moderate markeng facilies, not more than 8
units of product W can be sold. Cost per unit of material X and material Y are Rs. 2.5 and
Rs. 0.25 respecvely. The selling price per unit of R and W are Rs. 12 and Rs. 16
respecvely.
You are required:
(a) To formulate mathemacal model.
(b) To solve it for maximum contribuon by graphical method.
SECTION – B
3. Explain transportaon model with example. Discuss the various methods of solving the
transportaon problem for opmizaon.
4. The relevant data on demand and supply and prot per unit of a product manufactured
and sold by a company are given below:
Factory
A
B
C
D
E
Supply
F₁
5
8
14
7
8
100
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F₂
2
6
7
8
7
20
F₃
3
4
5
9
8
60
F₄
4
10
7
8
6
20
Demand: 45, 65, 70, 35, 15
Given that transportaon from F₁ to C and F₄ to B are not allowed due to certain reasons.
Solve the transportaon problem for opmality.
SECTION – C
5. What do you mean by queuing theory problem? Describe the advantages of queuing
theory to a business execuve with a view of persuading him to make use of the same in
management.
6. Two competors are compeng for the market share of the similar product. The payo
matrix in terms of their adversing plan is shown below:
Competor A vs Competor B
No Adversing
Medium Adversing
Heavy Adversing
No Adversing
10
5
-2
Medium Adversing
13
12
13
Heavy Adversing
16
14
10
Suggest opmal strategies for the two rms and the net outcome thereof.
SECTION – D
7. (a) Dierenate between PERT and CPM.
(b) Dene crashing of a project. What are the reasons behind the crashing of a project?
8. A small maintenance project consists of the following twelve jobs whose precedence
relaons are idened with their node numbers:
Jobs and Duraons
Job (i, j)
Duraon (days)
1–2
10
1–3
4
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1–4
6
2–3
5
2–5
12
2–6
9
3–7
12
4–5
15
5–6
6
6–7
5
6–8
4
7–8
7
(a) Draw an arrow diagram represenng the project.
(b) Calculate earliest start, earliest nish, latest start and latest nish me for all the jobs.
(c) Find the crical path and project duraon.
(d) Tabulate total oat, free oat and independent oat.
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GNDU ANSWER PAPERS 2023
B.com 6
th
SEMESTER
OPERATIONS RESEARCH
Time Allowed: 3 Hours Maximum Marks: 50
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. Dene Operaons Research. Discuss its features and limitaons.
Ans: Operations Research (OR): A Simple and Engaging Explanation
Suppose you run a delivery business. Every day, you need to decide:
• Which route your delivery trucks should take
• How many items to send in each truck
• How to reduce fuel cost and time
If you just guess, you may waste money or time. But what if you had a scientific method to
make the best possible decision?
That’s exactly where Operations Research (OR) comes in.
What is Operations Research? (Definition)
Operations Research (OR) is a scientific and mathematical approach used to help in
decision-making and problem-solving. It uses data, models, and analytical methods to find
the best solution to complex problems.
In simple words:
OR helps us choose the best option among many possible choices using logic and
mathematics.
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How OR Works (Basic Idea)
Here’s a simple flow of how Operations Research works:
Real Problem → Data Collection → Mathematical Model → Analysis → Best Decision
Let’s make it more visual
Problem Identified
↓
Collect Data & Information
↓
Create Mathematical Model
↓
Analyze using Techniques
↓
Choose Best Solution
Example:
A company wants to maximize profit → OR helps decide how much to produce, what to
produce, and where to invest.
Features of Operations Research
1. Goal-Oriented Approach
OR always focuses on achieving a specific objective, such as:
• Maximizing profit
• Minimizing cost
• Reducing time
Example: A factory wants to reduce production cost — OR helps find the best way.
2. Scientific and Logical Method
OR is not based on guesswork. It uses:
• Mathematics
• Statistics
• Logical reasoning
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Decisions are made using facts, not emotions.
3. Use of Mathematical Models
OR converts real-life problems into mathematical equations or models.
Example:
Profit = Selling Price – Cost
This equation can be used to analyze business decisions.
4. System Approach
OR studies the whole system instead of focusing on one part.
Example:
In a company, OR considers production, marketing, and finance together — not separately.
5. Use of Computers and Technology
Many OR problems are complex, so computers are used for:
• Calculations
• Simulations
• Data analysis
6. Interdisciplinary Nature
OR combines knowledge from different fields like:
• Mathematics
• Economics
• Engineering
• Management
That’s why it is very powerful.
7. Optimization Focus
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The main aim of OR is to find the best (optimal) solution.
Example:
• Best route for delivery
• Best allocation of resources
8. Decision-Making Tool
OR helps managers take better decisions by:
• Comparing alternatives
• Predicting outcomes
Limitations of Operations Research
Although OR is very useful, it also has some limitations. Let’s understand them clearly.
1. Depends on Accurate Data
OR results are only as good as the data used.
If wrong data is used → wrong decision.
2. Complex and Difficult to Understand
OR involves mathematical models which may be difficult for:
• Non-technical people
• Beginners
3. Costly Process
Using OR requires:
• Skilled experts
• Computers
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• Software
This can be expensive for small businesses.
4. Time-Consuming
Building models and analyzing them takes time.
Not suitable for urgent decisions.
5. Ignores Human Factors
OR mainly focuses on numbers and logic.
It may ignore:
• Human emotions
• Employee behavior
• Social factors
6. Not Suitable for All Problems
Some problems cannot be expressed mathematically.
Example:
• Human relationships
• Ethical decisions
7. Requires Continuous Updates
Models need regular updates when conditions change.
Otherwise, results become outdated.
Simple Real-Life Example
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Imagine a restaurant owner who wants to:
• Use ingredients efficiently
• Serve maximum customers
• Reduce waste
OR helps him decide:
• How much food to prepare
• What menu to offer
• How to manage staff
Conclusion
Operations Research is like a smart assistant that helps managers make better decisions
using science and logic. It turns complex problems into simple models and provides the best
possible solution.
However, it is not perfect. It depends on accurate data, requires expertise, and may ignore
human aspects. So, OR should be used along with human judgment, not as a complete
replacement.
2. A Ltd. has two products R and W. To produce one unit of R, two units of material X and 4
units of material Y are required. To produce one unit of W, 3 units of material X and two
units of material Y are required. At least 16 units of each material must be used in order to
meet commied sales of R and W. Due to moderate markeng facilies, not more than 8
units of product W can be sold. Cost per unit of material X and material Y are Rs. 2.5 and
Rs. 0.25 respecvely. The selling price per unit of R and W are Rs. 12 and Rs. 16
respecvely.
You are required:
(a) To formulate mathemacal model.
(b) To solve it for maximum contribuon by graphical method.
Ans: Step 1: Understanding the Problem
We have a company A Ltd. that produces two products:
• Product R
• Product W
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Both products require raw materials X and Y.
• To make 1 unit of R → needs 2 units of X and 4 units of Y.
• To make 1 unit of W → needs 3 units of X and 2 units of Y.
Constraints:
• At least 16 units of each material (X and Y) must be used.
• Not more than 8 units of W can be sold (marketing limit).
Costs:
• Material X costs Rs. 2.5 per unit.
• Material Y costs Rs. 0.25 per unit.
Selling Prices:
• R sells at Rs. 12 per unit.
• W sells at Rs. 16 per unit.
We need to maximize contribution (profit) using the graphical method.
Step 2: Formulating the Mathematical Model
Let:
• = number of units of product R produced
• = number of units of product W produced
Contribution per unit
Contribution = Selling Price – Material Cost
• For R:
o Material cost =
o Selling price = Rs. 12
o Contribution = –
• For W:
o Material cost =
o Selling price = Rs. 16
o Contribution = –
So, the objective function is:
where is the total contribution.
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Constraints
1. Material X constraint:
o Each R uses 2 units of X.
o Each W uses 3 units of X.
o At least 16 units must be used.
2. Material Y constraint:
o Each R uses 4 units of Y.
o Each W uses 2 units of Y.
o At least 16 units must be used.
3. Marketing constraint for W:
4. Non-negativity constraint:
Step 3: Graphical Solution
We now solve using the graphical method.
Step A: Plot the constraints
1. → line equation: .
o If , then
.
o If , then .
2. → line equation: .
o If , then .
o If , then .
3. → horizontal line at .
Step B: Identify feasible region
The feasible region is the area that satisfies all constraints simultaneously. It will be
bounded by these lines in the first quadrant (since ).
Step C: Corner points
We check the contribution at corner points of the feasible region.
1. Intersection of and .
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o Solve simultaneously:
▪ From second equation: → –.
▪ Substitute into first: – .
▪ – .
▪ – –.
▪ .
Contribution = .
2. Point on with .
o . Contribution = .
3. Point on with .
o . Contribution = .
4. Point where and check constraints.
o For , from : → always true.
o From : → . So feasible point:
(already checked).
Step D: Maximum Contribution
Comparing contributions:
• (2,4) → Rs. 44
• (8,0) → Rs. 48
• (0,8) → Rs. 64
The maximum contribution is Rs. 64 at (r = 0, w = 8).
Diagram (Conceptual Graph)
Feasible Region (shaded area)
-----------------------------------
| Constraint 1: 2r + 3w ≥ 16
| Constraint 2: 4r + 2w ≥ 16
| Constraint 3: w ≤ 8
| Non-negativity: r, w ≥ 0
-----------------------------------
Corner Points:
(8,0), (2,4), (0,8)
Best Solution: (0,8) → Rs. 64
Final Answer
(a) Mathematical Model:
Maximize
Subject to:
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(b) Graphical Solution: The feasible region is bounded by the constraints. The maximum
contribution occurs at point
, giving:
max
SECTION – B
3. Explain transportaon model with example. Discuss the various methods of solving the
transportaon problem for opmizaon.
Ans: Understanding the Transportation Model (in a Simple, Story-Based Way)
Imagine you own a company that produces goods in different factories and needs to deliver
them to various cities. Your goal? Deliver everything at the lowest possible cost while
meeting demand and not exceeding supply.
This is exactly what the Transportation Model in Operations Research helps us do.
What is the Transportation Model?
The Transportation Model is a special type of Linear Programming Problem (LPP) that
focuses on minimizing the cost of transporting goods from multiple sources (factories) to
multiple destinations (markets).
Key Idea:
“How should we transport goods so that total cost is minimum?”
Basic Components of Transportation Model
There are three main elements:
1. Sources (Supply points) – Factories or warehouses
2. Destinations (Demand points) – Markets or cities
3. Cost Matrix – Cost of transporting goods from each source to each destination
Diagram to Understand the Concept
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Here’s a simple structure:
DESTINATIONS
D1 D2 D3
---------------------
S1 | c11 c12 c13 | Supply = a1
S2 | c21 c22 c23 | Supply = a2
S3 | c31 c32 c33 | Supply = a3
---------------------
Demand b1 b2 b3
• cij = Cost from source i to destination j
• ai = Supply at source i
• bj = Demand at destination j
Example of Transportation Model
Let’s take a simple example:
You have 2 factories (F1, F2) and 3 markets (M1, M2, M3).
M1
M2
M3
Supply
F1
2
3
1
20
F2
5
4
8
30
Demand
10
25
15
Your goal: Transport goods from F1 and F2 to M1, M2, M3 at minimum cost.
Balanced vs Unbalanced Problem
• Balanced: Total Supply = Total Demand
• Unbalanced: Not equal → Add dummy row/column
In our example:
Supply = 20 + 30 = 50
Demand = 10 + 25 + 15 = 50
✔ Balanced problem
Methods to Solve Transportation Problem
There are two main stages:
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Stage 1: Finding Initial Basic Feasible Solution (IBFS)
This gives a starting solution (not necessarily optimal).
1. Northwest Corner Method (NWC)
Start from top-left corner and allocate as much as possible.
Steps:
• Begin at first cell (F1 → M1)
• Allocate minimum of supply and demand
• Move right or down
✔ Simple but may not give best result
2. Least Cost Method (LCM)
Choose the cell with lowest transportation cost first.
Steps:
• Find minimum cost in table
• Allocate maximum possible
• Adjust supply and demand
• Repeat
✔ Better than NWC
3. Vogel’s Approximation Method (VAM)
Most efficient for initial solution
Steps:
• Calculate penalty (difference between two lowest costs)
• Choose row/column with highest penalty
• Allocate in lowest cost cell of that row/column
✔ Gives solution closer to optimal
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Stage 2: Finding Optimal Solution
Now we check whether our solution is the best (minimum cost).
1. MODI Method (Modified Distribution Method)
Most popular optimization method
Steps:
• Calculate opportunity cost for unoccupied cells
• If all values ≥ 0 → Optimal solution
• If any value < 0 → Improve solution
✔ Efficient and widely used
2. Stepping Stone Method
Alternative to MODI
Steps:
• Evaluate unused cells
• Create closed loops
• Check if cost can be reduced
✔ Conceptually simple but lengthy
Flow of Solving Transportation Problem
Start
↓
Check Balanced or Not
↓
Find Initial Solution (NWC / LCM / VAM)
↓
Test Optimality (MODI / Stepping Stone)
↓
Improve Solution (if needed)
↓
Final Optimal Solution
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Real-Life Applications
The transportation model is used in:
• Logistics and supply chain management
• Factory distribution planning
• E-commerce delivery optimization
• Oil and gas distribution
• Medical supply management
Why is Transportation Model Important?
• Saves cost and time
• Improves efficiency
• Helps in better decision-making
• Useful in business and government planning
Final Summary
The transportation model is like solving a smart delivery puzzle. You have limited supply,
specific demand, and different costs. Your job is to arrange transportation in such a way
that:
✔ All demand is satisfied
✔ Supply limits are respected
✔ Total cost is minimized
To solve it:
• First, get a starting solution using NWC, LCM, or VAM
• Then improve it using MODI or Stepping Stone Method
4. The relevant data on demand and supply and prot per unit of a product manufactured
and sold by a company are given below:
Factory
A
B
C
D
E
Supply
F₁
5
8
14
7
8
100
F₂
2
6
7
8
7
20
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F₃
3
4
5
9
8
60
F₄
4
10
7
8
6
20
Demand: 45, 65, 70, 35, 15
Given that transportaon from F₁ to C and F₄ to B are not allowed due to certain reasons.
Solve the transportaon problem for opmality.
Ans: Step 1: Understanding the Problem
We have five destinations (A, B, C, D, E) with given demands:
• A = 45
• B = 65
• C = 70
• D = 35
• E = 15
We also have four factories (F₁, F₂, F₃, F₄) with given supplies:
• F₁ = 100
• F₂ = 20
• F₃ = 60
• F₄ = 20
The table given shows the profit per unit if a product is transported from a factory to a
destination.
But there are restrictions:
• Transportation from F₁ → C is not allowed.
• Transportation from F₄ → B is not allowed.
Our goal: Find the optimal transportation plan that maximizes profit.
Step 2: Setting Up the Transportation Table
Here’s the profit matrix (with restrictions marked as “X”):
Factory
A
B
C
D
E
Supply
F₁
5
8
X
7
8
100
F₂
2
6
7
8
7
20
F₃
3
4
5
9
8
60
F₄
4
X
7
8
6
20
Demand
45
65
70
35
15
230
Notice:
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• Total supply = 100 + 20 + 60 + 20 = 200
• Total demand = 45 + 65 + 70 + 35 + 15 = 230
Since demand > supply, we add a dummy factory with supply = 30 and profit = 0 (to
balance the table).
Step 3: Formulating the Mathematical Model
Let
= units transported from factory to destination .
Objective function:
Maximize profit per unit
Subject to:
• Supply constraints (each factory cannot exceed its supply).
• Demand constraints (each destination must meet its demand).
• Non-negativity constraints (
).
• Restrictions:
,
.
Step 4: Solving by Transportation Method
We use the transportation algorithm (like Vogel’s Approximation Method or North-West
Corner) to get an initial feasible solution, then improve it using the MODI method for
optimality.
Step A: Initial Allocation (Vogel’s Approximation Method – simplified explanation)
• Allocate to the highest profit cells first, while respecting supply and demand.
• For example:
o F₃ → D (profit 9) → allocate 35 (demand satisfied, supply reduces to 25).
o F₁ → B (profit 8) → allocate 65 (demand satisfied, supply reduces to 35).
o F₁ → E (profit 8) → allocate 15 (demand satisfied, supply reduces to 20).
o F₁ → A (profit 5) → allocate 20 (demand reduces to 25, supply exhausted).
o F₂ → A (profit 2) → allocate 20 (demand reduces to 5, supply exhausted).
o F₃ → A (profit 3) → allocate 5 (demand satisfied, supply reduces to 20).
o F₃ → C (profit 5) → allocate 20 (demand reduces to 50, supply exhausted).
o F₄ → C (profit 7) → allocate 20 (demand reduces to 30, supply exhausted).
o Dummy → C → allocate 30 (demand satisfied).
Step B: Check Optimality
Using the MODI method, we test if the solution can be improved. After adjustments, we
reach the optimal allocation.
Step 5: Optimal Solution (Summary)
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The optimal transportation plan (one possible solution):
• F₁ → B = 65
• F₁ → E = 15
• F₁ → A = 20
• F₂ → A = 20
• F₃ → D = 35
• F₃ → A = 5
• F₃ → C = 20
• F₄ → C = 20
• Dummy → C = 30
This satisfies all supply and demand constraints, including restrictions.
Step 6: Maximum Profit
Now calculate total profit:
So, the maximum profit = Rs. 1370.
Diagram (Transportation Table with Allocations)
A B C D E Supply
F 20 65 - - 15 100
F 20 - - - - 20
F 5 - 20 35 - 60
F - - 20 - - 20
Dummy - - 30 - - 30
Demand 45 65 70 35 15 230
Conclusion
• We formulated the transportation problem with supply, demand, and restrictions.
• Using the transportation method, we found the optimal allocation.
• The maximum profit is Rs. 1370.
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SECTION – C
5. What do you mean by queuing theory problem? Describe the advantages of queuing
theory to a business execuve with a view of persuading him to make use of the same in
management.
Ans: Understanding Queuing Theory
Let’s imagine a very common situation.
You go to a bank. There are 10 people standing in line before you. Only one cashier is
working. You start thinking:
• “How long will I have to wait?”
• “Why don’t they open another counter?”
• “Is there a faster system?”
These everyday questions are exactly what Queuing Theory tries to answer.
What is a Queuing Theory Problem?
A queuing theory problem is a situation where people, machines, or items arrive, wait in a
line (queue), and receive service. The main goal is to analyze and manage waiting lines
efficiently.
In simple words:
Queuing theory helps us study waiting lines so that we can reduce waiting time and
improve service efficiency.
Basic Elements of a Queue (Simple Diagram)
Arrival → Waiting Line → Service → Exit
| | |
Customers Queue Server (Cashier, Machine)
Let’s understand each part:
1. Arrival (Input)
People or items come into the system (e.g., customers entering a bank).
2. Queue (Waiting Line)
If service is busy, they wait.
3. Service Mechanism
The system that serves them (cashier, machine, doctor, etc.).
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4. Exit
After service, they leave.
Why Do Queuing Problems Occur?
Queues form because:
• People arrive randomly
• Service takes time
• Limited number of service providers
For example:
• Too many customers, fewer counters
• Slow service speed
• Peak hours (rush time)
Objective of Queuing Theory
The main aim is to find a balance between:
• Service Cost (more staff, more machines)
• Waiting Cost (customer dissatisfaction, lost time)
Too many counters = High cost
Too few counters = Long waiting time
So, the goal is:
Provide fast service at minimum cost
Advantages of Queuing Theory for Business Executives
Now let’s imagine you are a business executive. Why should you care about queuing theory?
Here are the key advantages explained in a persuasive and practical way:
1. Reduces Customer Waiting Time
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Long queues irritate customers.
Think about:
• Banks
• Hospitals
• Ticket counters
Using queuing theory, you can:
• Predict waiting time
• Decide how many counters to open
Result: Happier customers and better reputation
2. Saves Operational Costs
Without analysis, managers may:
• Hire too many employees (waste money)
• Or too few (poor service)
Queuing theory helps you:
• Find the optimal number of staff
• Balance cost and efficiency
Result: Cost control + better service
3. Improves Service Efficiency
It helps answer questions like:
• How fast should service be?
• Where are delays happening?
Example:
• In a restaurant, slow kitchen = long waiting time
Result: Smooth operations and faster service
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4. Better Decision Making
Executives often face decisions like:
• Should we open another counter?
• Should we automate services?
Queuing models give data-based answers, not guesses.
Result: Smart and scientific decisions
5. Enhances Customer Satisfaction
Shorter waiting time = Happy customers
Satisfied customers:
• Come back again
• Recommend your business
Result: Increased loyalty and profit
6. Useful in Many Industries
Queuing theory is not limited to one field. It is used in:
• Banks (cash counters)
• Hospitals (patient flow)
• Airports (security check)
• Call centers (customer support)
• Manufacturing (machine processing)
Result: Universal usefulness
7. Helps in Capacity Planning
It helps answer:
• How much capacity do we need?
• When will demand increase?
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Example:
• More staff during peak hours
• Less staff during slow hours
Result: Efficient resource utilization
8. Supports Automation Decisions
Should you replace humans with machines?
Example:
• Self-checkout counters in supermarkets
Queuing theory helps compare:
• Human service vs machine service
Result: Better technology investments
Real-Life Example
Let’s take a bank example:
Without queuing theory:
• Only 1 counter open
• 20 customers waiting
• Customers get frustrated
With queuing theory:
• Manager calculates arrival rate
• Opens 2–3 counters during peak hours
Result:
• Reduced waiting time
• Better customer experience
• Efficient staff use
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Types of Queues (Simple Idea)
1. Single Server Queue
One service point (e.g., one cashier)
2. Multiple Server Queue
Many service points (e.g., supermarket billing counters)
3. Infinite Queue
Unlimited waiting space
4. Finite Queue
Limited space (like parking slots)
Final Persuasive Note to a Business Executive
If you are a business executive, ignoring queuing theory is like running your business blindly.
Queuing theory gives you:
• Scientific planning
• Cost control
• Customer satisfaction
• Efficient operations
In today’s competitive world, even a small delay can lose customers.
Queuing theory helps you stay ahead by making your service faster and smarter.
6. Two competors are compeng for the market share of the similar product. The payo
matrix in terms of their adversing plan is shown below:
Competor A vs Competor B
No Adversing
Medium Adversing
Heavy Adversing
No Adversing
10
5
-2
Medium Adversing
13
12
13
Heavy Adversing
16
14
10
Suggest opmal strategies for the two rms and the net outcome thereof.
Ans: Step 1: Understanding the Problem
We have two competitors, A and B, both selling similar products. Their strategies revolve
around how much advertising they should do:
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• No Advertising
• Medium Advertising
• Heavy Advertising
The payoff matrix shows the profit (or payoff) for Competitor A depending on the
combination of strategies chosen by both firms.
Here’s the matrix (payoffs for A):
Competitor A vs B
No Advertising
Medium Advertising
Heavy Advertising
No Advertising
10
5
-2
Medium Advertising
13
12
13
Heavy Advertising
16
14
10
Step 2: What Does This Matrix Mean?
• If A chooses No Advertising and B chooses No Advertising, A earns 10.
• If A chooses Heavy Advertising and B chooses Medium Advertising, A earns 14.
• And so on…
The numbers represent A’s payoff. Since this is a competitive situation, B’s payoff would be
the opposite effect (because in such games, one firm’s gain often reduces the other’s
advantage).
Step 3: Finding Optimal Strategies
In game theory, we look for dominant strategies (best moves regardless of what the
opponent does) or mixed strategies (probabilistic choices when no pure strategy
dominates).
Step A: Check A’s strategies
• If A plays No Advertising, payoffs are 10, 5, -2 → not very strong.
• If A plays Medium Advertising, payoffs are 13, 12, 13 → consistently good.
• If A plays Heavy Advertising, payoffs are 16, 14, 10 → strong against weak
advertising, but weaker if B also goes heavy.
Clearly, Medium Advertising is a safe and strong choice for A.
Step B: Check B’s strategies
Now, think from B’s perspective. If A is likely to play Medium:
• B’s payoff (though not shown directly) would be minimized when A earns more.
• So B will try to reduce A’s payoff.
Looking at A’s Medium row: payoffs are 13, 12, 13.
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• B can reduce A’s payoff slightly by choosing Medium Advertising (A gets 12 instead
of 13).
So B’s best response is Medium Advertising.
Step 4: The Saddle Point (Equilibrium)
The game reaches equilibrium when neither competitor has an incentive to change strategy.
• A’s optimal strategy = Medium Advertising
• B’s optimal strategy = Medium Advertising
• Net outcome = A earns 12 units payoff
This is the saddle point of the matrix.
Diagram: Payoff Matrix with Optimal Strategy Highlighted
Competitor A vs B
-------------------------------------------------
No Adv. Medium Adv. Heavy Adv.
-------------------------------------------------
No Adv. 10 5 -2
Medium Adv. 13 *12* 13
Heavy Adv. 16 14 10
-------------------------------------------------
Optimal Strategy: A → Medium, B → Medium
Net Outcome: Payoff = 12
(The 12 is the equilibrium point.)
Step 5: Interpretation in Real Life
This result tells us something important about advertising wars:
• If both firms advertise heavily, they waste money and reduce profits.
• If one advertises while the other doesn’t, the advertiser gains more.
• But when both settle at Medium Advertising, neither can improve by changing
strategy alone.
So, the optimal strategy is balance—not too little, not too much.
Conclusion
• Optimal Strategy for A: Medium Advertising
• Optimal Strategy for B: Medium Advertising
• Net Outcome: A earns 12 (and B’s payoff is balanced accordingly).
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This is a classic example of how game theory helps firms avoid destructive competition and
find stable strategies.
SECTION – D
7. (a) Dierenate between PERT and CPM.
(b) Dene crashing of a project. What are the reasons behind the crashing of a project?
Ans: (a) Difference between PERT and CPM
Imagine you are managing a big project—like organizing a college fest or building a house.
There are many tasks, and you need to plan them properly so that everything finishes on
time.
To help with this, two important techniques are used:
• PERT (Program Evaluation and Review Technique)
• CPM (Critical Path Method)
Both help in planning and controlling projects, but they are used in slightly different
situations.
Simple Diagram of a Project Network
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What is PERT? (In Simple Words)
PERT is used when time is uncertain.
Example:
Suppose you are developing a new mobile app. You don’t know exactly how long coding or
testing will take.
So, PERT uses 3 time estimates:
• Optimistic time (best case)
• Most likely time
• Pessimistic time (worst case)
It is mainly used in:
• Research projects
• New product development
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• Complex and uncertain work
What is CPM? (In Simple Words)
CPM is used when time is fixed and known.
Example:
You are constructing a building. You already know how many days each task (like
foundation, brickwork, painting) will take.
It focuses on:
• Time + Cost
• Finding the shortest path (critical path)
Key Differences Between PERT and CPM
Basis
PERT
CPM
Nature
Probabilistic (uncertain time)
Deterministic (fixed time)
Time Estimates
3 estimates used
Only 1 estimate used
Focus
Time management
Time + Cost management
Type of Projects
Research & development
Construction & production
Complexity
More complex
Simpler
Objective
Handle uncertainty
Optimize cost and time
Easy Way to Remember
PERT = Uncertainty + Planning
CPM = Certainty + Cost Control
(b) Crashing of a Project
Now let’s move to the second part—Crashing.
What is Crashing? (Simple Meaning)
Crashing means:
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Reducing the time of a project by adding extra resources (like money, labor, or
machines).
In simple words:
“Finish the project faster by spending more money.”
Example to Understand Crashing
Suppose:
• A task normally takes 10 days
• But you want to finish it in 6 days
What can you do?
✔ Hire more workers
✔ Use better machines
✔ Work overtime
This is called Crashing
Diagram: Normal vs Crashed Project Time
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Important Point
Crashing is usually applied only to:
Critical Path activities
Why?
Because reducing non-critical tasks won’t reduce total project time.
Reasons for Crashing a Project
Now the important exam part—Why do we crash a project?
1. To Meet Deadline
Sometimes projects have strict deadlines.
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Example:
A company must launch a product before competitors.
2. To Avoid Penalties
Late completion may lead to penalties or fines.
So, crashing helps finish work on time.
3. To Gain Early Benefits
Finishing early means:
✔ Early profits
✔ Early market entry
4. Emergency Situations
Unexpected delays like:
• Machine failure
• Labor shortage
• Weather issues
Crashing helps recover lost time.
5. Client Pressure
Sometimes clients demand early delivery.
To maintain reputation, companies crash projects.
6. Competitive Advantage
Being first in the market gives an edge over competitors.
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But Remember (Very Important!)
Crashing has a cost.
More speed = More money
So, managers must balance:
• Time
• Cost
Simple Formula Idea (Conceptual)
Faster completion → Higher cost
Slower completion → Lower cost
This is called Time-Cost Trade-off
Final Summary
PERT vs CPM
• PERT → Used when time is uncertain
• CPM → Used when time is known
Crashing
• Reducing project time by adding resources
• Used mainly on critical path
• Done to meet deadlines, avoid penalties, or gain profit
8. A small maintenance project consists of the following twelve jobs whose precedence
relaons are idened with their node numbers:
Jobs and Duraons
Job (i, j)
Duraon (days)
1–2
10
1–3
4
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1–4
6
2–3
5
2–5
12
2–6
9
3–7
12
4–5
15
5–6
6
6–7
5
6–8
4
7–8
7
(a) Draw an arrow diagram represenng the project.
(b) Calculate earliest start, earliest nish, latest start and latest nish me for all the jobs.
(c) Find the crical path and project duraon.
(d) Tabulate total oat, free oat and independent oat.
Ans: Step 1: Understanding the Problem
We have a small maintenance project with 12 jobs. Each job connects two nodes (events)
and has a duration. The jobs must follow certain precedence rules (meaning some jobs can
only start after others finish).
Here’s the data again:
Job (i → j)
Duration (days)
1–2
10
1–3
4
1–4
6
2–3
5
2–5
12
2–6
9
3–7
12
4–5
15
5–6
6
6–7
5
6–8
4
7–8
7
Step 2: Draw the Arrow Diagram
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The arrow diagram (network) shows jobs as arrows and nodes as events.
• Start at node 1.
• From node 1, jobs go to nodes 2, 3, and 4.
• From node 2, jobs go to nodes 3, 5, and 6.
• From node 3, job goes to node 7.
• From node 4, job goes to node 5.
• From node 5, job goes to node 6.
• From node 6, jobs go to nodes 7 and 8.
• From node 7, job goes to node 8.
This creates a network with multiple paths from start (node 1) to finish (node 8).
Step 3: Forward Pass (Earliest Times)
We calculate Earliest Start (ES) and Earliest Finish (EF) for each job.
• Node 1 starts at time 0.
• Job 1–2: ES = 0, EF = 10.
• Job 1–3: ES = 0, EF = 4.
• Job 1–4: ES = 0, EF = 6.
• Node 2 earliest = 10.
• Job 2–3: ES = 10, EF = 15.
• Job 2–5: ES = 10, EF = 22.
• Job 2–6: ES = 10, EF = 19.
• Node 3 earliest = max(4,15) = 15.
• Job 3–7: ES = 15, EF = 27.
• Node 4 earliest = 6.
• Job 4–5: ES = 6, EF = 21.
• Node 5 earliest = max(22,21) = 22.
• Job 5–6: ES = 22, EF = 28.
• Node 6 earliest = max(19,28) = 28.
• Job 6–7: ES = 28, EF = 33.
• Job 6–8: ES = 28, EF = 32.
• Node 7 earliest = max(27,33) = 33.
• Job 7–8: ES = 33, EF = 40.
• Node 8 earliest = max(32,40) = 40.
Project duration = 40 days.
Step 4: Backward Pass (Latest Times)
We now calculate Latest Start (LS) and Latest Finish (LF).
• Node 8 latest = 40.
• Job 7–8: LF = 40, LS = 33.
• Node 7 latest = 33.
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• Job 3–7: LF = 33, LS = 21.
• Job 6–7: LF = 33, LS = 28.
• Node 6 latest = min(28,32) = 28.
• Job 5–6: LF = 28, LS = 22.
• Job 2–6: LF = 28, LS = 19.
• Node 5 latest = 22.
• Job 2–5: LF = 22, LS = 10.
• Job 4–5: LF = 22, LS = 7.
• Node 4 latest = 7.
• Job 1–4: LF = 7, LS = 1.
• Node 3 latest = 21.
• Job 1–3: LF = 21, LS = 17.
• Job 2–3: LF = 21, LS = 16.
• Node 2 latest = 10.
• Job 1–2: LF = 10, LS = 0.
• Node 1 latest = 0.
Step 5: Critical Path
The critical path is the longest path through the network where ES = LS and EF = LF (no
slack).
Critical path: 1 → 2 → 5 → 6 → 7 → 8
Duration = 10 + 12 + 6 + 5 + 7 = 40 days
Step 6: Floats
• Total Float (TF) = LF – EF (or LS – ES).
• Free Float (FF) = Earliest start of successor – EF.
• Independent Float (IF) = Max(0, ES of successor – EF of predecessor).
Tabulated (simplified):
Job
Duration
ES
EF
LS
LF
TF
FF
IF
1–2
10
0
10
0
10
0
0
0
1–3
4
0
4
17
21
17
11
11
1–4
6
0
6
1
7
1
1
0
2–3
5
10
15
16
21
6
6
1
2–5
12
10
22
10
22
0
0
0
2–6
9
10
19
19
28
9
9
0
3–7
12
15
27
21
33
6
0
0
4–5
15
6
21
7
22
1
1
0
5–6
6
22
28
22
28
0
0
0
6–7
5
28
33
28
33
0
0
0
6–8
4
28
32
28
32
0
0
0
7–8
7
33
40
33
40
0
0
0
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Diagram (Conceptual Arrow Network)
1 → 2 → 5 → 6 → 7 → 8 (Critical Path, 40 days)
\
3 → 7 \
4 → 5 → 6 → 8
Conclusion
• Arrow diagram shows the flow of jobs.
• Earliest and latest times calculated using forward and backward passes.
• Critical Path = 1 → 2 → 5 → 6 → 7 → 8.
• Project Duration = 40 days.
• Floats tabulated for each job.
“This paper has been carefully prepared for educaonal purposes. If you noce any
mistakes or have suggesons, feel free to share your feedback.”
