Easy2Siksha.com
GNDU QUESTION PAPERS 2024
BBA 4
th
SEMESTER
Paper–BBA–406 : 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. Each queson carries 10 marks.
SECTION–A
1. What is Linear Programming? Discuss the applicaons of Linear Programming in
managerial decision making.
2. A manufacturer has employed 5 skilled men and 10 semi-skilled men and makes two
models A and B of an arcle. The making of one item of model A requires 2 hrs work by a
skilled man and 2 hrs work by a semi-skilled man. One item of model B requires 1 hr by a
skilled man and 3 hrs by a semi-skilled man. No man is expected to work more than 8 hrs
per day. The manufacturer prot on an item of model A is Rs. 15 and on an item of model
B is Rs. 10. How many items of each model should be made per day in order to maximize
daily prot? Formulate the above LPP and solve it to nd the maximum prot.
SECTION–B
3. Solve the following assignment problem for minimum opmal cost :
From City \ To City
1
2
3
4
5
6
A
12
10
15
22
18
8
B
10
18
25
15
16
12
C
11
10
3
13
5
9
Easy2Siksha.com
D
6
14
10
8
13
12
E
8
12
11
17
13
10
4. P. Ltd. has three auditors. Each auditor can work up to 160 hours during the next month,
during which me, 3 projects must be completed. Project I will take 130 hours, Project II
will take 140 hours and III will take 160 hours. The amount in rupees per hour that can be
billed for assigning each auditor to each project is given below :
Project 1
Project 2
Project 3
Auditor 1
1200
1500
1900
Auditor 2
1400
1300
1200
Auditor 3
1600
1400
1500
Find the opmal soluon.
SECTION–C
5. (a) What are the various types of inventory? Describe in detail the quanty discounts
model along with hypothecal example.
(b) A manufacturing company purchases 9000 parts of a machine for its annual
requirements, ordering one month usage at a me. Each part costs Rs. 20. The ordering
cost per order is Rs. 15 and the carrying charges are 15% of the average inventory per year.
Suggest a more economic purchasing policy for the company.
6. (a) In a game of matching coins with two players, suppose A wins one unit of value
when there are two heads, wins nothing when there are two tails and loses ½ unit of value
when there is one head and one tail. Determine the pay-o matrix, the best strategies for
each player and the value of the game to A.
(b) Find the values of A and B for which the following pay-o matrix will give a saddle
point at (2, 2) posion.
Player B
Strategies
I
II
III
Player A
Easy2Siksha.com
I
2
4
5
II
10
7
B
III
4
8
6
SECTION–D
7. What do you understand by the term PERT and CPM? What are the rules for drawing
network diagram? Also menon the common errors that occur in drawing networks.
8. A project has the following characteriscs :
Acvity
Most opmisc me (a)
Most pessimisc me (b)
Most likely me (m)
1–2
1
5
1.5
2–3
1
3
2
2–4
1
5
3
3–5
3
5
4
4–5
2
4
3
4–6
3
7
5
5–7
4
6
5
6–7
6
8
7
7–8
2
6
4
7–9
5
8
6
8–10
1
3
2
9–10
3
7
5
Construct a PERT network. Find the crical path and variance for each event.
Easy2Siksha.com
GNDU Answer PAPERS 2024
BBA 4
th
SEMESTER
Paper–BBA–406 : 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. Each queson carries 10 marks.
SECTION–A
1. What is Linear Programming? Discuss the applicaons of Linear Programming in
managerial decision making.
Ans: Imagine you are running a small business. You have limited resources—money, time,
workers, and raw materials—but you want to earn maximum profit. The question is: how
should you use your limited resources in the best possible way?
This is exactly where Linear Programming (LP) comes in.
Linear Programming is a mathematical technique used to find the best possible solution
(maximum profit or minimum cost) when there are limited resources and multiple
constraints.
In simple words:
Linear Programming helps managers make the best decisions when resources are limited.
Key Elements of Linear Programming
To understand LP clearly, let’s break it into its main parts:
1. Decision Variables
These are the things you can control.
Example: Number of products to produce.
Easy2Siksha.com
2. Objective Function
This is what you want to maximize or minimize.
Example: Maximize profit or minimize cost.
3. Constraints
These are the limitations or restrictions.
Example: Limited labor, raw materials, or time.
4. Non-negativity Condition
Values cannot be negative (you cannot produce -5 units).
Basic Structure of Linear Programming
Here is a simple mathematical form:
Maximize Subject to:
This means:
• Maximize profit
• and are decision variables
• The inequality represents constraints
Graphical Representation (Diagram Explanation)
To understand LP visually, we often use graphs.
Easy2Siksha.com
Explanation of the Diagram:
• Each line represents a constraint.
• The shaded area is called the feasible region (all possible solutions).
• The best solution lies at one of the corner points (vertices).
• The objective function is used to find the optimal point.
Applications of Linear Programming in Managerial Decision Making
Linear Programming is widely used in business and management. Let’s understand its real-
life applications in a simple and relatable way.
1. Production Planning
Managers must decide:
• How much of each product to produce?
Easy2Siksha.com
Example:
A factory produces chairs and tables. It has limited wood and labor. LP helps decide the
number of chairs and tables to maximize profit.
Benefit: Efficient use of resources and higher profit.
2. Resource Allocation
Companies often have limited:
• Budget
• Workforce
• Machines
LP helps in allocating these resources efficiently among different activities.
Example:
A company distributes its budget across marketing, production, and R&D.
Benefit: Best use of limited resources.
3. Cost Minimization
Businesses always want to reduce costs.
Example:
A transportation company wants to deliver goods at minimum cost using different routes.
LP helps:
• Choose the cheapest routes
• Reduce fuel and time costs
Benefit: Lower operational expenses.
4. Marketing Decision Making
Managers need to decide:
• How much to spend on different advertising channels?
Easy2Siksha.com
Example:
Should you spend more on social media or TV ads?
LP helps allocate the budget to maximize reach and profit.
Benefit: Better return on investment (ROI).
5. Diet Planning (Nutrition Problems)
This is a classic example.
Example:
A hospital wants to prepare a diet plan that:
• Meets nutritional requirements
• Costs as little as possible
LP helps find the cheapest combination of foods.
Benefit: Balanced diet at minimum cost.
6. Transportation and Distribution
Companies need to transport goods from warehouses to markets.
LP helps:
• Decide how much to transport
• Choose optimal routes
Benefit: Reduced delivery cost and time.
7. Workforce Scheduling
Managers must decide:
• How many workers to assign to each shift?
Example:
A hospital schedules nurses for day and night shifts.
LP ensures:
Easy2Siksha.com
• Proper staffing
• Minimum cost
Benefit: Efficient workforce management.
8. Inventory Management
Businesses must maintain stock without overstocking or shortages.
LP helps:
• Determine how much inventory to keep
• Balance holding and shortage costs
Benefit: Smooth operations and cost control.
Advantages of Linear Programming
• Helps in better decision making
• Ensures optimal use of resources
• Reduces costs and wastage
• Increases profit and efficiency
• Provides a scientific and logical approach
Limitations of Linear Programming
• Assumes relationships are linear (not always realistic)
• Requires accurate data
• Cannot handle uncertainty easily
• Solutions may not always be practical in real life
Conclusion
Linear Programming is like a smart decision-making tool for managers. It transforms
complex business problems into simple mathematical models and helps find the best
possible solution.
Whether it is deciding:
• how much to produce,
Easy2Siksha.com
• where to invest, or
• how to reduce costs,
LP provides a clear and logical answer.
In today’s competitive business world, where resources are limited and goals are high,
Linear Programming plays a crucial role in helping managers make efficient, profitable,
and informed decisions.
2. A manufacturer has employed 5 skilled men and 10 semi-skilled men and makes two
models A and B of an arcle. The making of one item of model A requires 2 hrs work by a
skilled man and 2 hrs work by a semi-skilled man. One item of model B requires 1 hr by a
skilled man and 3 hrs by a semi-skilled man. No man is expected to work more than 8 hrs
per day. The manufacturer prot on an item of model A is Rs. 15 and on an item of model
B is Rs. 10. How many items of each model should be made per day in order to maximize
daily prot? Formulate the above LPP and solve it to nd the maximum prot.
Ans: Step 1: Understanding the Problem
We have a manufacturer who makes two models of an article: Model A and Model B.
• He employs 5 skilled men and 10 semi-skilled men.
• Each man works 8 hours per day.
So, total available hours per day are:
• Skilled men: hours.
• Semi-skilled men: hours.
Now, let’s see how much time each model requires:
• Model A: 2 hours skilled + 2 hours semi-skilled.
• Model B: 1 hour skilled + 3 hours semi-skilled.
Profits:
• Model A → Rs. 15 per item.
• Model B → Rs. 10 per item.
Step 2: Defining Variables
Let:
• = number of Model A items produced per day.
• = number of Model B items produced per day.
Easy2Siksha.com
Our goal: maximize profit .
Step 3: Constraints
We must respect the available working hours.
1. Skilled men constraint: Each Model A requires 2 hours, each Model B requires 1
hour. So:
2. Semi-skilled men constraint: Each Model A requires 2 hours, each Model B requires
3 hours. So:
3. Non-negativity constraint:
Step 4: Formulating the LPP
Maximize:
Subject to:
Step 5: Graphical Solution
Since we have two variables, we can solve graphically.
1. Equation 1: .
• If , then .
• If , then . So line passes through (0,40) and (20,0).
2. Equation 2: .
• If , then
.
• If , then . So line passes through (0,26.7) and (40,0).
Feasible region is bounded by these lines and the axes.
Easy2Siksha.com
Step 6: Corner Points
The maximum profit will occur at one of the corner points of the feasible region.
• Point (0,0): .
• Point (20,0): Check constraints: ✔ ✔
Profit: .
• Point (0,26.7): Profit: .
• Intersection of lines: Solve simultaneously: … (1) … (2)
Subtract (2) – (1): .
Substitute into (1): .
So intersection point is (10,20). Profit: .
Step 7: Maximum Profit
Comparing profits:
• (20,0) → Rs. 300
• (0,26.7) → Rs. 267
• (10,20) → Rs. 350
Maximum profit = Rs. 350 at (10,20).
Final Answer
The manufacturer should produce:
• 10 items of Model A
• 20 items of Model B
to maximize daily profit.
The maximum profit is Rs. 350 per day.
Conclusion
This problem shows how Linear Programming helps managers make smart decisions. By
balancing skilled and semi-skilled labor hours, the manufacturer finds the optimal mix of
products. Instead of guessing, mathematics gives a clear answer: 10 units of Model A and 20
units of Model B yield the highest profit.
Easy2Siksha.com
SECTION–B
3. Solve the following assignment problem for minimum opmal cost :
From City \ To City
1
2
3
4
5
6
A
12
10
15
22
18
8
B
10
18
25
15
16
12
C
11
10
3
13
5
9
D
6
14
10
8
13
12
E
8
12
11
17
13
10
Ans: Understanding the Problem
You are given an assignment problem, where:
• There are 5 cities (A, B, C, D, E)
• There are 6 destinations (1, 2, 3, 4, 5, 6)
• Each number represents the cost of assigning a city to a destination
Goal: Assign each city to exactly one destination such that total cost is minimum
Important Observation
We have:
• 5 cities
• 6 destinations
This is not a balanced problem (rows ≠ columns).
So, we add a dummy city (F) with zero cost for all destinations to make it a 6 × 6 matrix.
Original Cost Table
From \ To
1
2
3
4
5
6
A
12
10
15
22
18
8
B
10
18
25
15
16
12
Easy2Siksha.com
C
11
10
3
13
5
9
D
6
14
10
8
13
12
E
8
12
11
17
13
10
F (Dummy)
0
0
0
0
0
0
Idea Behind Solution (Hungarian Method – Simplified)
Instead of doing all steps mechanically, let’s understand the logic:
We try to pick minimum values from each row
But we must ensure no two cities get the same destination
Finding Optimal Assignments
Let’s pick the smallest possible values carefully:
Step-by-step selection:
• City C → Destination 3 = 3 (lowest in entire table )
• City D → Destination 4 = 8
• City A → Destination 6 = 8
• City B → Destination 1 = 10
• City E → Destination 2 = 12
Now all assignments are unique (no column repetition)
Final Assignment
City
Assigned To
Cost
A
2
10
B
1
10
C
3
3
D
4
8
E
6
10
(Destination 5 is assigned to dummy city F)
Minimum Total Cost
Total Cost
Easy2Siksha.com
Simple Diagram (Assignment Mapping)
A ───► 2 (10)
B ───► 1 (10)
C ───► 3 (3)
D ───► 4 (8)
E ───► 6 (10)
F ───► 5 (0) [Dummy]
Final Answer
Minimum Optimal Cost = 41
Easy Understanding (Real-Life Example)
Think of this like:
• 5 delivery trucks (A–E)
• 6 delivery routes (1–6)
• Each route has a fuel cost depending on the truck
Your job is to:
Assign each truck one route
Spend the least total fuel
So you carefully choose:
• Cheapest route for each truck
• Without overlapping routes
That’s exactly what we did!
Conclusion
• We converted the problem into a balanced one using a dummy row
• Selected minimum values while avoiding repetition
• Found the optimal assignment with minimum cost = 41
Easy2Siksha.com
4. P. Ltd. has three auditors. Each auditor can work up to 160 hours during the next month,
during which me, 3 projects must be completed. Project I will take 130 hours, Project II
will take 140 hours and III will take 160 hours. The amount in rupees per hour that can be
billed for assigning each auditor to each project is given below :
Project 1
Project 2
Project 3
Auditor 1
1200
1500
1900
Auditor 2
1400
1300
1200
Auditor 3
1600
1400
1500
Find the opmal soluon.
Ans: Step 1: Understanding the Problem
We have:
• Three auditors (Auditor 1, Auditor 2, Auditor 3).
• Each auditor can work up to 160 hours in the next month.
• Three projects must be completed:
o Project I requires 130 hours.
o Project II requires 140 hours.
o Project III requires 160 hours.
Billing rates (rupees per hour) for each auditor on each project:
Project
Auditor 1
Auditor 2
Auditor 3
I
1200
1400
1600
II
1500
1300
1400
III
1900
1200
1500
We want to maximize total billing revenue by assigning auditors’ hours optimally.
Step 2: Defining Variables
Let:
•
= number of hours auditor spends on project .
So:
•
= hours Auditor 1 spends on Project I.
•
= hours Auditor 1 spends on Project II.
•
= hours Auditor 1 spends on Project III. …and similarly for Auditors 2 and 3.
Step 3: Objective Function
Easy2Siksha.com
We want to maximize revenue:
Step 4: Constraints
1. Auditor time limits (≤ 160 hours each):
2. Project requirements (exact hours needed):
3. Non-negativity:
Step 5: Intuition Before Solving
Notice the billing rates:
• For Project I, Auditor 3 earns the most (₹1600/hr).
• For Project II, Auditor 1 earns the most (₹1500/hr).
• For Project III, Auditor 1 earns the most (₹1900/hr).
So, intuitively, we should allocate hours to the auditor who earns the highest rate for each
project, while respecting the 160-hour limit.
Step 6: Optimal Allocation
Let’s assign based on highest rates:
• Project I (130 hrs): Best is Auditor 3 (₹1600/hr). Assign all 130 hrs to Auditor 3.
• Project II (140 hrs): Best is Auditor 1 (₹1500/hr). Assign all 140 hrs to Auditor 1.
• Project III (160 hrs): Best is Auditor 1 (₹1900/hr). But Auditor 1 already has 140 hrs,
and max is 160 hrs. So Auditor 1 can only take 20 hrs more. The remaining 140 hrs
must go to another auditor. Next best is Auditor 3 (₹1500/hr).
So allocation looks like:
• Auditor 1: 140 hrs on Project II + 20 hrs on Project III = 160 hrs (full capacity).
Easy2Siksha.com
• Auditor 3: 130 hrs on Project I + 140 hrs on Project III = 270 hrs (but max is 160 hrs,
so we need to adjust).
Step 7: Adjusting for Limits
Auditor 3 cannot exceed 160 hrs. We gave them 270 hrs. So we must redistribute.
Let’s prioritize:
• Auditor 3 should fully take Project I (130 hrs).
• That leaves 30 hrs capacity for Auditor 3. Assign those 30 hrs to Project III.
• Remaining Project III hours = 160 – (20 from Auditor 1 + 30 from Auditor 3) = 110
hrs.
• Auditor 2 must take those 110 hrs.
Step 8: Final Allocation
• Auditor 1: 140 hrs (Project II) + 20 hrs (Project III).
• Auditor 2: 110 hrs (Project III).
• Auditor 3: 130 hrs (Project I) + 30 hrs (Project III).
All auditors ≤ 160 hrs. All projects completed.
Step 9: Calculating Maximum Revenue
• Auditor 1:
o 140 hrs × 1500 = 210,000
o 20 hrs × 1900 = 38,000 → Total = 248,000
• Auditor 2:
o 110 hrs × 1200 = 132,000
• Auditor 3:
o 130 hrs × 1600 = 208,000
o 30 hrs × 1500 = 45,000 → Total = 253,000
Grand Total Revenue = 248,000 + 132,000 + 253,000 = ₹633,000
Conclusion
The optimal solution is:
• Auditor 1 → 140 hrs on Project II, 20 hrs on Project III.
• Auditor 2 → 110 hrs on Project III.
• Auditor 3 → 130 hrs on Project I, 30 hrs on Project III.
Maximum billing revenue = ₹633,000.
Easy2Siksha.com
SECTION–C
5. (a) What are the various types of inventory? Describe in detail the quanty discounts
model along with hypothecal example.
(b) A manufacturing company purchases 9000 parts of a machine for its annual
requirements, ordering one month usage at a me. Each part costs Rs. 20. The ordering
cost per order is Rs. 15 and the carrying charges are 15% of the average inventory per year.
Suggest a more economic purchasing policy for the company.
Ans: 5 (a) Types of Inventory & Quantity Discount Model
What is Inventory? (Simple Meaning)
Inventory means all the goods and materials a business keeps so that production and sales
can run smoothly. Think of it like a kitchen—if you don’t store enough ingredients, you can’t
cook when needed.
Types of Inventory
1. Raw Materials
These are the basic inputs used to produce goods.
Example: Steel, cotton, plastic.
2. Work-in-Progress (WIP)
Easy2Siksha.com
These are semi-finished goods that are still being processed.
Example: A car without wheels yet.
3. Finished Goods
Products ready for sale to customers.
Example: Packed biscuits, ready garments.
4. Maintenance, Repair, and Operations (MRO) Supplies
Items used to support production but not part of the final product.
Example: Lubricants, tools, spare parts.
5. Safety Stock
Easy2Siksha.com
Extra inventory kept to avoid shortages during uncertainty.
Example: Extra stock for emergency demand.
Quantity Discount Model (Explained Simply)
Imagine you go to a shop and the seller says:
“Buy more, pay less per unit.”
That’s exactly what happens in the Quantity Discount Model.
What is it?
It is an extension of EOQ (Economic Order Quantity) where the supplier offers discounts for
bulk purchasing.
Goal:
To find the order quantity that minimizes total cost, including:
• Purchase cost
• Ordering cost
• Holding (carrying) cost
Key Idea
When you order more:
• ✔ You get lower price per unit
• But holding cost increases
So, we must balance both.
Basic EOQ Formula (Used in Model)
Where:
Easy2Siksha.com
• D = Annual demand
• S = Ordering cost
• H = Holding cost per unit
Steps in Quantity Discount Model
1. Identify different price levels
2. Calculate EOQ for each price
3. Check if EOQ fits discount range
4. Compute total cost for each option
5. Choose the lowest total cost
Hypothetical Example
Suppose:
• Demand = 1000 units
• Ordering cost = ₹50
• Holding cost = ₹2 per unit
• Supplier offers:
Quantity
Price
0–499
₹10
500+
₹9
Step 1: Calculate EOQ
EOQ = √(2×1000×50 / 2) = √50000 = 224 units
But 224 < 500 → No discount applicable
Step 2: Try minimum quantity for discount (500 units)
Now compare total costs:
Option 1 (224 units, ₹10 price)
• Purchase cost = 1000 × 10 = 10,000
• Ordering cost = (1000/224) × 50 ≈ 223
• Holding cost = (224/2) × 2 = 224
Easy2Siksha.com
Total ≈ ₹10,447
Option 2 (500 units, ₹9 price)
• Purchase cost = 1000 × 9 = 9,000
• Ordering cost = (1000/500) × 50 = 100
• Holding cost = (500/2) × 2 = 500
Total = ₹9,600
Conclusion:
Even though holding cost increases, discount makes total cost lower.
So, order 500 units.
5 (b) Finding Economic Purchasing Policy
Now let’s solve your numerical question step by step.
Given Data:
• Annual demand (D) = 9000 units
• Monthly purchase = 9000 / 12 = 750 units
• Cost per unit = ₹20
• Ordering cost (S) = ₹15
• Carrying cost = 15% of price
Holding cost (H) = 15% of 20 = ₹3 per unit
Step 1: Apply EOQ Formula
units
Easy2Siksha.com
Step 2: Compare Current vs EOQ Policy
Current Policy:
• Order size = 750 units
• Orders per year = 9000 / 750 = 12
Costs:
• Ordering cost = 12 × 15 = ₹180
• Holding cost = (750/2) × 3 = ₹1125
Total cost = ₹1305
EOQ Policy:
• Order size = 300 units
• Orders per year = 9000 / 300 = 30
Costs:
• Ordering cost = 30 × 15 = ₹450
• Holding cost = (300/2) × 3 = ₹450
Total cost = ₹900
Final Comparison
Policy
Total Cost
Current
₹1305
EOQ
₹900
Final Answer (Conclusion)
The company should reduce order size from 750 to 300 units.
This increases ordering frequency but significantly reduces holding cost.
Savings = ₹1305 – ₹900 = ₹405 per year
Easy2Siksha.com
Simple EOQ Diagram
4
At EOQ:
• Ordering cost = Holding cost
• Total cost is minimum
Final Understanding
• Inventory ensures smooth business operations
• EOQ helps minimize total cost
• Quantity discount model balances price vs storage
• Smart purchasing saves money without affecting production
Easy2Siksha.com
6. (a) In a game of matching coins with two players, suppose A wins one unit of value
when there are two heads, wins nothing when there are two tails and loses ½ unit of value
when there is one head and one tail. Determine the pay-o matrix, the best strategies for
each player and the value of the game to A.
(b) Find the values of A and B for which the following pay-o matrix will give a saddle
point at (2, 2) posion.
Player B
Strategies
I
II
III
Player A
I
2
4
5
II
10
7
B
III
4
8
6
Ans: Part (a): Matching Coins Game
Understanding the Game
Two players (say A and B) each toss a coin. The outcomes determine A’s payoff:
• Two heads (HH): A wins 1 unit.
• Two tails (TT): A wins 0 units (no gain, no loss).
• One head and one tail (HT or TH): A loses ½ unit.
This is a zero-sum game, meaning whatever A wins, B loses, and vice versa.
Step 1: Payoff Matrix
Let’s construct the payoff matrix. Each player has two strategies: Head (H) or Tail (T).
Player B
H
T
Player A: H
+1
-½
Player A: T
-½
0
This matrix shows A’s payoff for each combination.
Step 2: Best Strategies
Now, we check if there is a saddle point (a pure strategy solution).
• Row minima: For A choosing H → min payoff = -½; for A choosing T → min payoff = -
½.
Easy2Siksha.com
• Column maxima: For B choosing H → max payoff = +1; for B choosing T → max
payoff = 0.
Since row maximin (–½) ≠ column minimax (0), there is no saddle point. So players must use
mixed strategies.
Step 3: Mixed Strategy Solution
Let A play Head with probability , Tail with probability . Let B play Head with
probability , Tail with probability .
Expected payoff to A:
½½
Simplify:
Step 4: Optimal Probabilities
To find equilibrium, set payoffs equal for B’s strategies.
• If B plays H: payoff to A = ½
.
• If B plays T: payoff to A = ½
.
For A to be indifferent:
So A plays Head with probability ¼, Tail with probability ¾.
Similarly, for B:
• If A plays H: payoff = ½
.
• If A plays T: payoff = ½
.
Equating:
Easy2Siksha.com
So B plays Head with probability ¼, Tail with probability ¾.
Step 5: Value of the Game
Substitute ¼¼:
So the value of the game = –⅛ units. That means A loses on average 0.125 units per play.
Part (b): Saddle Point Problem
We are given a payoff matrix:
Player B
I
II
III
A: I
2
4
5
A: II
10
7
B
A: III
4
8
6
We need values of A and B such that the saddle point is at position (2,2), i.e., row II, column
II.
Step 1: Saddle Point Definition
A saddle point occurs when the chosen element is simultaneously:
• The minimum in its row.
• The maximum in its column.
Here, element (2,2) = 7 must satisfy both.
Step 2: Row II Condition
Row II = [10, 7, B]. For 7 to be the minimum:
So:
Easy2Siksha.com
Step 3: Column II Condition
Column II = [4, 7, 8]. For 7 to be the maximum:
But 7 ≥ 8 is false. So to make 7 the maximum, the element in row III, column II must be A ≤ 7
(instead of 8).
So:
Step 4: Final Conditions
Thus, for saddle point at (2,2):
•
•
Conclusion
• Part (a): The payoff matrix is constructed, optimal mixed strategies are: A plays Head
with probability ¼, Tail with ¾; B plays Head with ¼, Tail with ¾. The value of the
game to A is –⅛ units (so A loses on average).
• Part (b): For the given matrix, the saddle point at (2,2) exists if and .
SECTION–D
7. What do you understand by the term PERT and CPM? What are the rules for drawing
network diagram? Also menon the common errors that occur in drawing networks.
Ans: Imagine you are planning a big event like a college fest or building a house. There are
many tasks involved—some must be done before others, some can happen at the same
time, and all of them together determine how long the project will take. To manage such
complex work, managers use special techniques called PERT and CPM.
What is PERT?
PERT (Program Evaluation and Review Technique) is a method used to plan and control
projects where time is uncertain.
Easy2Siksha.com
In simple words, PERT helps answer questions like:
• How long will the project take?
• What is the earliest time we can finish it?
• Which tasks are most important?
The special thing about PERT is that it considers uncertainty in time. For each activity, it
uses three time estimates:
1. Optimistic time (O) – if everything goes perfectly
2. Most likely time (M) – normal situation
3. Pessimistic time (P) – if things go wrong
Using these, we calculate expected time:
So, PERT is best used in research projects, new product development, or situations
where time is unpredictable.
What is CPM?
CPM (Critical Path Method) is another project management technique, but it is simpler than
PERT.
Here:
• Time for each activity is fixed and known
• Focus is on cost and time optimization
CPM helps to:
• Identify the critical path (the longest path in the project)
• Find which activities cannot be delayed
• Optimize cost by adjusting time
CPM is commonly used in construction, manufacturing, and routine projects.
Difference Between PERT and CPM (Simple View)
Feature
PERT
CPM
Nature
Probabilistic (uncertain time)
Deterministic (fixed time)
Easy2Siksha.com
Focus
Time planning
Time + Cost optimization
Use
Research projects
Construction, production
Time Estimates
3 estimates
1 estimate
What is a Network Diagram?
A network diagram is a visual representation of a project showing:
• Activities (tasks)
• Events (start/end points)
• Sequence of operations
Let’s see a simple example:
This diagram shows how activities are connected and which tasks depend on others.
Rules for Drawing Network Diagram
To draw a correct network diagram, some important rules must be followed:
1. Every activity must have a clear start and end
Each task should begin from one event and end at another.
2. Follow logical sequence
Activities must be arranged in the correct order. For example:
• You cannot paint a wall before building it.
3. No loops allowed
The network should not form a cycle (no circular paths).
Easy2Siksha.com
Wrong: A → B → C → A
Correct: A → B → C
4. No duplication of activities
Each activity should appear only once in the diagram.
5. Use dummy activities when needed
Sometimes, a dummy activity (shown with dotted lines) is used to show dependency
without actual work.
6. Maintain proper direction
Network should flow from left to right for clarity.
7. Number the events properly
Use a numbering system (like 1, 2, 3…) to identify events clearly.
Common Errors in Drawing Network Diagrams
Even though the rules are simple, students often make mistakes. Let’s understand the
common ones:
1. Looping (Circular Logic)
Creating a cycle where activities repeat.
Example:
A → B → C → A (This is incorrect)
2. Dangling Activities
An activity that is not properly connected to the start or end.
It leaves the project incomplete or unclear.
3. Redundant Activities
Adding unnecessary activities that do not affect the project.
Easy2Siksha.com
Makes the diagram confusing and complex.
4. Incorrect Dependencies
Placing activities in the wrong order.
Example: Finishing work shown before starting work.
5. Missing Dummy Activities
Sometimes students avoid dummy activities, which leads to wrong relationships.
Dummy activities are important for showing correct dependencies.
6. Improper Numbering
Events not numbered properly can create confusion in identifying paths.
7. Parallel Activities Misrepresentation
Showing activities as parallel when they actually depend on each other.
Conclusion
PERT and CPM are powerful tools that help in planning, scheduling, and controlling projects
effectively. While PERT deals with uncertainty and uses multiple time estimates, CPM works
with fixed times and focuses on optimization.
A network diagram is the backbone of both techniques, and drawing it correctly is very
important. By following proper rules and avoiding common mistakes like loops, dangling
activities, and incorrect sequencing, one can create clear and efficient project plans.
Easy2Siksha.com
8. A project has the following characteriscs :
Acvity
Most opmisc me (a)
Most pessimisc me (b)
Most likely me (m)
1–2
1
5
1.5
2–3
1
3
2
2–4
1
5
3
3–5
3
5
4
4–5
2
4
3
4–6
3
7
5
5–7
4
6
5
6–7
6
8
7
7–8
2
6
4
7–9
5
8
6
8–10
1
3
2
9–10
3
7
5
Construct a PERT network. Find the crical path and variance for each event.
Ans: Step 1: Understanding the Problem
We’re given a project with multiple activities. Each activity has three time estimates:
• Optimistic time (a): The shortest possible time if everything goes perfectly.
• Most likely time (m): The best guess under normal conditions.
• Pessimistic time (b): The longest possible time if things go wrong.
From these, we calculate the expected time (te) and variance (σ²) for each activity.
Formulas:
Step 2: Calculating Expected Time and Variance
Let’s compute for each activity:
1. Activity 1–2:
Variance =
Easy2Siksha.com
2. Activity 2–3:
Variance =
3. Activity 2–4:
Variance =
4. Activity 3–5:
Variance =
5. Activity 4–5:
Variance =
6. Activity 4–6:
Variance =
7. Activity 5–7:
Variance =
8. Activity 6–7:
Variance =
9. Activity 7–8:
Variance =
10. Activity 7–9:
Variance =
11. Activity 8–10:
Variance =
12. Activity 9–10:
Variance =
Step 3: Constructing the PERT Network
Nodes: 1 → 2 → (3,4) → (5,6) → 7 → (8,9) → 10.
Paths:
• Path A: 1–2–3–5–7–8–10
• Path B: 1–2–3–5–7–9–10
• Path C: 1–2–4–5–7–8–10
• Path D: 1–2–4–5–7–9–10
• Path E: 1–2–4–6–7–8–10
• Path F: 1–2–4–6–7–9–10
Step 4: Calculating Path Durations
• Path A: 1–2 (2) + 2–3 (2) + 3–5 (4) + 5–7 (5) + 7–8 (4) + 8–10 (2) = 19
• Path B: 1–2 (2) + 2–3 (2) + 3–5 (4) + 5–7 (5) + 7–9 (6.17) + 9–10 (5) = 24.17
• Path C: 1–2 (2) + 2–4 (3) + 4–5 (3) + 5–7 (5) + 7–8 (4) + 8–10 (2) = 19
• Path D: 1–2 (2) + 2–4 (3) + 4–5 (3) + 5–7 (5) + 7–9 (6.17) + 9–10 (5) = 24.17
• Path E: 1–2 (2) + 2–4 (3) + 4–6 (5) + 6–7 (7) + 7–8 (4) + 8–10 (2) = 23
• Path F: 1–2 (2) + 2–4 (3) + 4–6 (5) + 6–7 (7) + 7–9 (6.17) + 9–10 (5) = 28.17
Step 5: Critical Path
Easy2Siksha.com
The longest path determines the project duration. Critical Path = Path F (1–2–4–6–7–9–10).
Expected project duration = 28.17 units of time.
Step 6: Variance Along Critical Path
Add variances of activities on Path F:
• 1–2: 0.444
• 2–4: 0.444
• 4–6: 0.444
• 6–7: 0.111
• 7–9: 0.25
• 9–10: 0.444
Total variance = 2.137. Standard deviation = √2.137 ≈ 1.46.
Conclusion
• We constructed the PERT network with nodes 1 through 10.
• Calculated expected times and variances for each activity.
• Found all possible paths and their durations.
• The critical path is 1–2–4–6–7–9–10, with expected duration ≈ 28.17 units.
• The variance along this path is ≈ 2.137, giving a standard deviation of ≈ 1.46.
“This paper has been carefully prepared for educaonal purposes. If you noce any
mistakes or have suggesons, feel free to share your feedback.”
