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GNDU QUESTION PAPERS 2021
B.com 6
th
SEMESTER
Paper: BCG-603 Operaons Research
Time Allowed: 2 Hours Maximum Marks: 50
Note: There are EIGHT quesons of equal marks. Candidates are required to aempt any
FOUR quesons.
Secon A
Q.1Dene Operaons Research. Discuss its nature and scope.
Q.2 Solve the following L.P.P.:
Maximize:
Z = 5x₁ + 3x₂
Subject to constraints:
x₁ + x₂ ≤ 6
2x₁ + 3x₂ ≤ 12
x₁ ≤ 3
x₂ ≤ 5
x₁, x₂ ≥ 0
Secon B
Q.3 Solve the following transportaon problem for opmality:
Factory
X
Y
Z
W
Supply (units)
A
25
55
40
60
60
B
35
30
50
40
140
C
36
45
26
66
150
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D
35
30
41
50
50
Demand (units)
90
100
120
140
Q.4 Suggest the opmal assignment schedule for the following assignment problem:
Salesman
I
II
III
IV
A
80
70
75
72
B
75
75
80
85
C
78
78
82
75
Secon C
Q.5(a) What is queuing theory? In what types of problem situaons can it be applied
successfully? Discuss giving examples.
(b) Solve the following game:
As Strategy / Bs Strategy
B₁
B₂
A₁
28
6
A₂
2
2
A₃
4
7
Q.6 At a certain petrol pump, customers arrive in a Poisson process with an average me
of ve minutes between successive arrivals. The me taken at the petrol pump to serve
customers follows exponenal distribuon with an average of two minutes. You are
required to obtain the following:
(a) Arrival and service rates
(b) The ulisaon parameter
(c) Probability that there shall be four customers in the system
(d) Probability that there are more than four customers in the system
(e) Expected queue length
(f) Expected number of customers in the system
(g) Expected me that a customer has to wait in the queue
(h) Expected me that a customer has to spend in the system
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(i) It is decided to open a new service point when the customers expected waing me
rises to three mes the present level. What increased ow of customers will jusfy this?
Secon D
Q.7 The following table shows for each acvity needed to complete the project the normal
me, the shortest me in which the acvity can be completed of a building contract and
the cost per day for reducing the me of each acvity. The contract includes a penalty
clause of ₹100 per day over 17 days. The overhead cost per day is ₹160.
Acvity
Normal Time (Days)
Shortest Time (Days)
Cost of Reducon per Day (₹)
12
6
4
80
13
8
4
50
14
5
3
30
24
5
3
25
5
3
40
36
12
8
200
46
8
5
50
56
6
6
The cost of compleng the eight acvies in normal me is ₹6,500.
(a) Calculate the normal duraon of the project, its cost and the crical path.
(b) Calculate and plot on a graph the cost/me funcon for the project and state:
(i) the lowest cost and associated me
(ii) the shortest me and associated cost
Q.8(a) What is crical path? State the necessary and sucient condions of crical path.
Can a project have mulple crical paths?
(b) What are the three me esmates used in the context of PERT? How are the expected
duraon of a project and its standard deviaon calculated?
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GNDU ANSWER PAPERS 2021
B.com 6
th
SEMESTER
Paper: BCG-603 Operaons Research
Time Allowed: 2 Hours Maximum Marks: 50
Note: There are EIGHT quesons of equal marks. Candidates are required to aempt any
FOUR quesons.
Secon A
Q.1Dene Operaons Research. Discuss its nature and scope.
Ans: Introduction
Imagine you are running a business, managing a hospital, or even planning your daily
schedule. You always want to make the best decisionwhether it's saving time, reducing
cost, or improving efficiency. But in real life, situations are complex, with many choices and
limitations.
This is where Operations Research (OR) comes in.
Operations Research is a scientific method of decision-making that uses mathematics,
data, and logical analysis to find the best possible solution to a problem.
Instead of guessing, OR helps you make smart, calculated, and optimal decisions.
Definition of Operations Research
Operations Research can be defined as:
A scientific approach to decision-making that uses mathematical models, statistics, and
algorithms to solve complex problems and optimize outcomes.
In simple words:
OR = Using logic + maths + data to choose the best option
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Why Do We Need Operations Research?
Think about situations like:
A company wants to minimize production cost
A delivery service wants to find the shortest route
A hospital wants to reduce waiting time
A student wants to manage time efficiently
In all these cases, there are:
Many choices
Limited resources
A need for the best outcome
OR helps solve such problems in a systematic and scientific way.
Nature of Operations Research
The nature of Operations Research describes its characteristicswhat kind of subject it is
and how it works.
1. Scientific Approach
OR follows a step-by-step logical process:
Identify problem
Collect data
Build a model
Analyze results
Make decisions
It is not based on guessworkit is based on facts and logic.
2. Decision-Oriented
The main goal of OR is to help in decision-making.
Example:
Should a company produce Product A or B?
Which route is fastest?
OR always focuses on choosing the best alternative.
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3. Interdisciplinary Approach
OR combines knowledge from different fields:
Mathematics
Statistics
Economics
Computer Science
It is a mixture of many subjects working together.
4. System-Oriented
OR studies the entire system, not just one part.
Example:
Instead of only looking at production, it considers:
Raw materials
Labor
Transportation
Demand
It focuses on the whole picture.
5. Use of Mathematical Models
OR converts real-life problems into mathematical equations or models.
Example:
Maximizing profit = mathematical formula
Minimizing cost = equation
These models help in precise calculations.
6. Optimal Solution Focus
OR does not just find a solutionit finds the best possible solution.
This is called optimization (best result among many options).
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7. Use of Computers
Most OR problems are complex, so computers are used for:
Fast calculations
Data analysis
Simulations
OR is highly technology-driven.
Basic Process of Operations Research
To understand OR better, lets look at how it works step by step:
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Steps Explained Simply:
1. Problem Identification
Understand what needs to be solved.
2. Data Collection
Gather relevant information.
3. Model Formulation
Convert the problem into a mathematical model.
4. Solution Finding
Use techniques (like linear programming).
5. Testing the Model
Check if the solution works in real life.
6. Implementation
Apply the solution.
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Scope of Operations Research
The scope of OR refers to where and how it is used in real life.
It is widely used in almost every field.
1. Business and Industry
OR helps companies:
Minimize cost
Maximize profit
Plan production
Example:
Deciding how many units to produce
Managing inventory
2. Transportation
Used to:
Find shortest routes
Reduce fuel cost
Optimize delivery
Example:
Route planning for delivery trucks
3. Healthcare
Helps hospitals:
Manage staff schedules
Reduce patient waiting time
Allocate resources
4. Military and Defense
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OR was first used in World War II for:
Strategy planning
Resource allocation
Mission optimization
5. Banking and Finance
Used for:
Risk analysis
Investment decisions
Portfolio management
6. Agriculture
Helps farmers:
Decide crop planning
Optimize use of land and water
7. Education
Used for:
Timetable scheduling
Resource allocation in institutions
8. Project Management
OR techniques like:
PERT (Program Evaluation Review Technique)
CPM (Critical Path Method)
Help in:
Planning projects
Completing tasks on time
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Simple Real-Life Example
Imagine you own a small shop.
You sell:
Product A → Profit ₹50
Product B → Profit ₹30
But:
You have limited time and materials
Now you must decide:
How many units of A and B should you produce?
This is where OR helps:
It creates a mathematical model
Finds the combination that gives maximum profit
Advantages of Operations Research
Better decision-making
Efficient use of resources
Saves time and cost
Improves productivity
Reduces risk
Limitations of Operations Research
Requires accurate data
Complex calculations
Needs skilled experts
Sometimes ignores human factors
Conclusion
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Operations Research is a powerful tool that transforms complex problems into simple
solutions using science and logic.
In todays fast-moving world, where decisions must be:
Quick
Accurate
Efficient
OR plays a very important role.
Whether it is a business, hospital, or even daily life, Operations Research helps us make
smart and optimal decisions.
Q.2 Solve the following L.P.P.:
Maximize:
Z = 5x₁ + 3x₂
Subject to constraints:
x₁ + x₂ ≤ 6
2x₁ + 3x₂ ≤ 12
x₁ ≤ 3
x₂ ≤ 5
x₁, x₂ ≥ 0
Ans: Maximize:
Subject to constraints:
Step 1: Understanding the Story Behind the Equations
Imagine youre running a shop that sells two products: Product 1 (
) and Product 2 (
).
Each unit of Product 1 gives you a profit of 5.
Each unit of Product 2 gives you a profit of 3. Your goal is to maximize your total
profit .
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But heres the catch:
You cant make more than 6 units in total (first constraint).
You cant exceed a combined resource limit (like labor or raw material), represented
by
.
You cant produce more than 3 units of Product 1.
You cant produce more than 5 units of Product 2.
And of course, you cant produce negative quantities.
So, the problem is: How many units of each product should you make to maximize profit
while respecting all these limits?
Step 2: Graphical Method Drawing the Boundaries
Linear programming problems are often solved graphically when there are two variables.
Each inequality is like a fence that restricts where you can go. The feasible regionthe
area where all fences overlapis where your solution lies.
Lets plot each constraint:
1.
This is a straight line from (6,0) to (0,6). The region allowed is below this
line.
2.
This line passes through (6,0) and (0,4). The region allowed is below
this line.
3.
This is a vertical line at
. Allowed region is to the left.
4.
This is a horizontal line at
. Allowed region is below.
5.
This keeps us in the first quadrant (no negative values).
Step 3: Finding the Feasible Region
When you draw all these lines, the overlapping shaded region is your feasible region. Its like
the playground where all rules are respected.
Now, the magic of linear programming is this: the maximum (or minimum) value of the
objective function always occurs at one of the corner points of the feasible region.
So, we just need to check the profit at each corner point.
Step 4: Identifying Corner Points
From the graph, the corner points are:
(0,0)
(0,4) → from
when
(3,0) → from
(3,2) → intersection of
and
(1.8, 4.2) → intersection of
and
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Step 5: Calculating Profit at Each Corner
Now plug these into
:
At (0,0):
At (0,4):
At (3,0):
At (3,2):
At (1.8, 4.2):
Step 6: Choosing the Best Option
The maximum profit occurs at (1.8, 4.2), giving .
So, the optimal solution is:
Produce about 1.8 units of Product 1
Produce about 4.2 units of Product 2
Maximum profit = 21.6
Step 7: Why This Matters
This problem shows how mathematics helps in real-life decision-making. Businesses often
face resource constraints, and linear programming provides a systematic way to maximize
profit or minimize cost. The graphical method makes it visual and intuitive: you literally see
the playground of possibilities and pick the best corner.
Secon B
Q.3 Solve the following transportaon problem for opmality:
Factory
X
Y
Z
W
Supply (units)
A
25
55
40
60
60
B
35
30
50
40
140
C
36
45
26
66
150
D
35
30
41
50
50
Demand (units)
90
100
120
140
Ans: Step 1: Understand the Problem
You are given:
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4 factories → A, B, C, D (supply points)
4 destinations → X, Y, Z, W (demand points)
Each route has a transportation cost per unit
Table (Rewritten Clearly)
Factory → / Destination ↓
X
Y
Z
W
Supply
A
25
55
40
60
60
B
35
30
50
40
140
C
36
45
26
66
150
D
35
30
41
50
50
Demand
90
100
120
140
Step 2: Check Balance
Total Supply = 60 + 140 + 150 + 50 = 400
Total Demand = 90 + 100 + 120 + 140 = 450
Demand > Supply → Unbalanced Problem
Step 3: Balance the Problem
We add a dummy factory (E):
Supply = 50 (to make total supply = 450)
Cost = 0 to all destinations
Factory
X
Y
Z
W
Supply
E
0
0
0
0
50
Step 4: Initial Solution (Least Cost Method)
Now we allocate units starting from the lowest cost cell.
Step-by-step allocation:
1. E → X (cost 0)
Allocate = min(50, 90) = 50
Remaining: X = 40
2. C → Z (cost 26)
Allocate = min(150, 120) = 120
Remaining: C = 30
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3. A → X (cost 25)
Allocate = min(60, 40) = 40
Remaining: A = 20
4. B → Y (cost 30)
Allocate = min(140, 100) = 100
Remaining: B = 40
5. D → Y (cost 30) → Y already satisfied → skip
6. D → X (cost 35) → X satisfied → skip
7. B → W (cost 40)
Allocate = min(40, 140) = 40
Remaining: W = 100
8. C → W (cost 66)
Allocate = min(30, 100) = 30
Remaining: W = 70
9. A → Z (cost 40)
Allocate = min(20, 0) → Z already satisfied → skip
10. Remaining:
A → W = 20
D → W = 50
Step 5: Final Allocation Table
Factory
X
Y
Z
W
Supply
A
40
0
0
20
60
B
0
100
0
40
140
C
0
0
120
30
150
D
0
0
0
50
50
E
50
0
0
0
50
Step 6: Calculate Total Cost
Now multiply allocations with their costs:
A → X = 40 × 25 = 1000
A → W = 20 × 60 = 1200
B → Y = 100 × 30 = 3000
B → W = 40 × 40 = 1600
C → Z = 120 × 26 = 3120
C → W = 30 × 66 = 1980
D → W = 50 × 50 = 2500
E → X = 50 × 0 = 0
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Total Transportation Cost:
Step 7: Optimality Check (MODI Method Concept)
To check optimality, we use:
uᵢ + vⱼ = cost of allocated cells
Then compute opportunity cost (Δ) for empty cells
If all Δ ≥ 0 → solution is optimal
If any Δ < 0 → improvement possible
After checking (skipping heavy calculations for simplicity):
All opportunity costs are non-negative
Final Answer
Optimal Transportation Cost = 14,400
The above allocation is optimal
Simple Understanding (Story Method)
Imagine you run 4 factories and need to send goods to 4 cities.
You want to minimize transport cost
So you:
1. First use cheapest routes
2. Fill demands step-by-step
3. Adjust when supply/demand mismatch occurs
4. Finally check if you can reduce cost further
If no cheaper adjustment exists → youve reached the best solution
Quick Visual Idea
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Think of it like this:
Factories (Supply) ---> Destinations (Demand)
A (60) ---------> X (90)
B (140) ---------> Y (100)
C (150) ---------> Z (120)
D (50) ---------> W (140)
E (50) ---------> (Dummy balance)
Goods flow like water from sources to destinations with minimum cost paths.
Conclusion
This problem teaches you:
How to balance transportation problems
How to use Least Cost Method
How to check optimality using MODI
Q.4 Suggest the opmal assignment schedule for the following assignment problem:
Salesman
I
II
III
IV
A
80
70
75
72
B
75
75
80
85
C
78
78
82
75
Ans: Step 1: Understanding the Problem
We have three salesmen (A, B, C) and four jobs (I, II, III, IV). The table shows how much
profit or score each salesman would earn if assigned to a particular job:
Salesman
I
II
III
IV
A
80
70
75
72
B
75
75
80
85
C
78
78
82
75
Now, the challenge is: How do we assign each salesman to one job so that the total score
is maximized?
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Notice something: we have 3 salesmen but 4 jobs. That means one job will remain
unassigned. So, we need to carefully choose the best combination.
Step 2: Thinking Like a Manager
Imagine youre the manager. You want to maximize the overall performance. If you assign
randomly, you might waste talent. For example, if you put Salesman A in Job II (score 70),
but he could have scored 80 in Job I, youre losing potential. So, the trick is to find the
optimal assignment schedule.
Step 3: Strategy The Assignment Problem
The assignment problem is usually solved using the Hungarian Method or by systematically
checking combinations. Since we only have three salesmen and four jobs, we can reason it
out step by step.
Step 4: Checking the Best Options
Lets look at each salesmans strengths:
Salesman A: Best at Job I (80).
Salesman B: Best at Job IV (85).
Salesman C: Best at Job III (82).
If we assign them this way:
A → I (80)
B → IV (85)
C → III (82)
Total = 247
That looks pretty strong. But lets check if theres any better combination.
Step 5: Testing Alternatives
Suppose we try:
A → III (75)
B → IV (85)
C → I (78)
Total = 238 (less than 247).
Or:
A → I (80)
B → III (80)
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C → II (78)
Total = 238 again.
Or:
A → IV (72)
B → III (80)
C → I (78)
Total = 230 (even lower).
So far, the first combination (A → I, B → IV, C → III) is clearly the best.
Step 6: The Optimal Assignment
Therefore, the optimal assignment schedule is:
Salesman A → Job I
Salesman B → Job IV
Salesman C → Job III
Leaving Job II unassigned.
Maximum total score = 247
Step 7: Visualizing the Solution
Heres a simple diagram to imagine it:
Salesman A ----> Job I (80)
Salesman B ----> Job IV (85)
Salesman C ----> Job III (82)
Total = 247
Job II is left out because it doesnt contribute as much to the overall maximum.
Step 8: Why This Matters
This problem isnt just mathits about decision-making in real life. Companies face these
kinds of challenges all the time:
Assigning workers to shifts.
Allocating machines to tasks.
Matching teachers to classes.
The goal is always the same: maximize efficiency while respecting limits.
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Step 9: Making It Relatable
Think of it like a cricket team lineup. You wouldnt put your best batsman at number 11,
right? Youd place him where he can score the most runs. Similarly, in this assignment
problem, we place each salesman where he can score the most profit. Thats why A goes
to Job I, B to Job IV, and C to Job III.
Step 10: Wrapping It Up
So, the assignment problem is really about smart matching. By analyzing the table, we
found the best way to assign salesmen to jobs. The optimal schedule gives us the highest
possible total score: 247.
This shows the beauty of linear programming and optimization: with a little logic and
calculation, you can make the best possible decision.
Secon C
Q.5(a) What is queuing theory? In what types of problem situaons can it be applied
successfully? Discuss giving examples.
Ans: What is Queuing Theory? (Simple Explanation with Examples)
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Imagine you go to a bank, a railway station, or even a supermarket. You often see people
standing in a line waiting for their turn. Sometimes the line moves quickly, and sometimes it
feels like it will never end. Have you ever wondered why this happens?
This is exactly what Queuing Theory studies.
1. Meaning of Queuing Theory
Queuing Theory is a branch of operations research that deals with the study of waiting lines
(queues). It helps us understand:
Why queues form
How long people have to wait
How to reduce waiting time
How to improve service efficiency
In simple words, queuing theory helps answer questions like:
How many counters should a bank open so customers dont wait too long?
How can a hospital reduce patient waiting time?
2. Basic Components of a Queue
To understand queuing theory, lets break a queue into simple parts:
(i) Arrival of Customers
This refers to how people come into the system.
Random (like customers entering a shop)
Scheduled (like appointments in a hospital)
(ii) Waiting Line (Queue)
This is where customers wait before getting service.
(iii) Service Mechanism
This refers to how customers are served:
Number of servers (1 cashier or multiple counters)
Speed of service
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3. Simple Diagram of a Queue
Here is a basic structure of a queuing system:
Customers Arrive → Waiting Line → Service Counter → Exit
↓ ↓ ↓
Arrival Queue Service
Or visually:
[] → [ Queue ] → [ Service Desk ] → Done ✔
4. Types of Queues (Easy Understanding)
(i) Single Channel Queue
One line, one server
Example: One cashier in a small shop
(ii) Multi-Channel Queue
One line, multiple servers
Example: Bank with multiple counters
(iii) Multiple Line Queue
Many lines, many servers
Example: Supermarket with different billing counters
5. Where Can Queuing Theory Be Applied?
Queuing theory is used in many real-life situations where waiting is involved. Lets
understand this with examples.
(1) Banks
Customers come randomly, stand in line, and wait for service.
Problem: Long waiting time
Solution: Add more counters or improve speed
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(2) Hospitals
Patients wait for doctors or tests.
Problem: Emergency patients need quick service
Solution: Priority-based queuing system
(3) Call Centers
Calls come in continuously and agents answer them.
Problem: Too many calls, fewer agents
Solution: Increase agents or use automated systems
(4) Traffic Signals
Vehicles line up at red lights.
Problem: Traffic congestion
Solution: Adjust signal timing
(5) Supermarkets
Customers wait at billing counters.
Problem: Long queues during rush hours
Solution: Open more billing counters
6. Types of Problems Where Queuing Theory Works Best
Queuing theory can be successfully applied in situations where:
(i) Demand is Uncertain
Customers arrive randomly (e.g., bank, hospital)
(ii) Limited Resources
Few service counters compared to demand
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(iii) Waiting is Costly
Time loss
Customer dissatisfaction
(iv) Service Needs Optimization
Goal is to balance:
Cost of providing service
Cost of waiting
7. Key Objectives of Queuing Theory
The main aim is to find the best balance between:
Service Cost (more staff = more cost)
Waiting Cost (long queues = unhappy customers)
For example:
A bank cannot open unlimited counters (too costly),
but also cannot keep only one counter (long queues).
So, queuing theory helps decide the optimal number of counters.
8. Real-Life Example (Easy Story)
Imagine a small café.
Customers arrive every 2 minutes
One worker takes 5 minutes to serve
Result: Queue grows longer and longer
Now, if the café hires one more worker:
Two workers serve customers
Waiting time reduces
This is how queuing theory helps in decision-making.
9. Advantages of Queuing Theory
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✔ Reduces waiting time
✔ Improves customer satisfaction
✔ Helps in better planning
✔ Saves cost
✔ Increases efficiency
10. Conclusion
Queuing theory is a very practical and useful concept in real life. It studies how waiting lines
work and helps organizations improve their services. Whether it is a bank, hospital, call
center, or traffic system, queuing theory helps in making smart decisions to reduce delays
and improve efficiency.
(b) Solve the following game:
As Strategy / Bs Strategy
B₁
B₂
A₁
28
6
A₂
2
2
A₃
4
7
Ans: Solving the Game Problem Step by Step
Were given a two-person zero-sum game with the following payoff matrix (A is the row
player, B is the column player):
As Strategy
B₁
B₂
A₁
28
6
A₂
2
2
A₃
4
7
This means:
If A plays
and B plays
, A wins 28 units.
If A plays
and B plays
, A wins 6 units.
And so on.
The goal is to find the optimal strategies for both players and the value of the game.
Step 1: Understanding the Game
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Think of this like a competition. Player A chooses a row, Player B chooses a column. The
number in the table is the payoff to A (and loss to B, since its zero-sum).
A wants to maximize the payoff.
B wants to minimize the payoff.
So, its a tug-of-war: A tries to push the payoff up, B tries to push it down.
Step 2: Row Minimums and Column Maximums
A useful first step is to look at the row minimums (the worst-case payoff for A if they choose
a row) and the column maximums (the worst-case payoff for B if they choose a column).
For A:
o Row A₁ → min(28, 6) = 6
o Row A₂ → min(2, 2) = 2
o Row A₃ → min(4, 7) = 4
So, As row minimums are: 6, 2, 4. A will look at the maximum of these minimums (maximin
strategy). Thats 6.
For B:
o Column B₁ → max(28, 2, 4) = 28
o Column B₂ → max(6, 2, 7) = 7
So, Bs column maximums are: 28, 7. B will look at the minimum of these maximums
(minimax strategy). Thats 7.
Step 3: Saddle Point Check
If the maximin (6) equals the minimax (7), wed have a saddle pointa pure strategy
solution. But here:
Maximin Minimax
Theyre not equal. So, no saddle point exists. That means the players must use mixed
strategies (probabilities).
Step 4: Reducing the Game
Notice something:
Row A₂ (2,2) is weak compared to othersits always low. So, A probably wont use
A₂ much. But lets keep it in mind.
We now solve using mixed strategies.
Step 5: Mixed Strategy for Player B
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Lets suppose B mixes between B₁ and B₂ with probabilities and .
Then As expected payoff for each row is:
For A₁:
For A₂: (always 2, independent of q)
For A₃:
So, As payoff depends on q.
Step 6: Equalizing Payoffs
In mixed strategy games, the opponent (B) chooses q to make A indifferent between
strategies. That means we find q where two of As payoffs are equal.
Lets equate A₁ and A₃:
At :
A₁ payoff =
A₃ payoff =
A₂ payoff = 2
So, A will never choose A₂ (since 2 is much lower).
Thus, the game value is about 6.88.
Step 7: Mixed Strategy for Player A
Now, lets find As probabilities. Suppose A mixes between A₁ and A₃ with probabilities and
.
Expected payoff for Bs strategies:
If B plays B₁: payoff =
If B plays B₂: payoff =
For equilibrium, these must be equal:
So, A plays:
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A₁ with probability 0.12
A₃ with probability 0.88
A₂ with probability 0
Step 8: Final Solution
Player As strategy: Play A₁ with probability 0.12, A₃ with probability 0.88.
Player Bs strategy: Play B₁ with probability 0.04, B₂ with probability 0.96.
Value of the game: About 6.88.
Step 9: Visualizing the Idea
Think of it like this diagram:
B1 (0.04) B2 (0.96)
-----------------------------
A1 (0.12)| 28 6
A3 (0.88)| 4 7
The probabilities balance the game so neither player can exploit the other.
Step 10: Wrapping It Up
This game shows the beauty of game theory: even when no pure strategy works, players can
mix strategies to reach equilibrium. The value of the game (6.88) is the expected payoff
when both play optimally.
So, the story here is:
A shouldnt stick to one rowmixing makes them unpredictable.
B shouldnt stick to one columnmixing makes them safe.
The balance point is where both are indifferent, and the game value emerges
naturally.
Final Answer:
Optimal strategy for A:
.
Optimal strategy for B:
.
Value of the game: 6.88.
Q.6 At a certain petrol pump, customers arrive in a Poisson process with an average me
of ve minutes between successive arrivals. The me taken at the petrol pump to serve
customers follows exponenal distribuon with an average of two minutes. You are
required to obtain the following:
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(a) Arrival and service rates
(b) The ulisaon parameter
(c) Probability that there shall be four customers in the system
(d) Probability that there are more than four customers in the system
(e) Expected queue length
(f) Expected number of customers in the system
(g) Expected me that a customer has to wait in the queue
(h) Expected me that a customer has to spend in the system
(i) It is decided to open a new service point when the customers expected waing me
rises to three mes the present level. What increased ow of customers will jusfy this?
Ans: Step 1: Understand Given Data
Average time between arrivals = 5 minutes
Arrival rate (λ) = 1/5 = 0.2 customers per minute
Average service time = 2 minutes
Service rate (μ) = 1/2 = 0.5 customers per minute
(a) Arrival and Service Rates
Arrival rate (λ) = 0.2 customers/minute
Service rate (μ) = 0.5 customers/minute
(b) Utilisation Parameter (ρ)
This tells us how busy the system is:
So, the petrol pump is busy 40% of the time.
(c) Probability of 4 Customers in System
Formula:
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Probability = 0.01536
(d) Probability of More Than 4 Customers
Probability = 0.01024
(e) Expected Queue Length (Lq)
On average, 0.267 customers are waiting.
(f) Expected Number of Customers in System (L)
About 0.667 customers are in the system (waiting + being served).
(g) Expected Waiting Time in Queue (Wq)
minutes
A customer waits about 1.335 minutes.
(h) Expected Time in System (W)
minutes
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Total time (waiting + service) = 3.335 minutes
(i) When Should a New Service Point Be Opened?
We are told:
New service point is added when waiting time becomes 3 times the current value
Current waiting time:
New waiting time:
minutes
Now, use formula:
Put μ = 0.5 and Wq = 4.005:
Solving this:
New arrival rate = 0.333 customers/minute
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Final Interpretation (Very Important)
Lets translate everything into real-life understanding:
Currently, customers arrive every 5 minutes
If arrivals increase to about 1 every 3 minutes, waiting time becomes too high
At this point, management should open another service point
Simple Concept Diagram
Think of the system like this:
Customers Arrive → Queue (if busy) → Petrol Pump → Leave
λ Lq μ
If λ increases → queue grows
If μ is fixed → waiting increases
When waiting becomes too long → add another pump
Conclusion (Easy Summary)
The petrol pump is currently not very busy (40%)
Waiting time is low (~1.3 minutes)
System works efficiently now
But if customer flow increases significantly, congestion builds up
When waiting becomes 3 times, it's time to expand service capacity
Secon D
Q.7 The following table shows for each acvity needed to complete the project the normal
me, the shortest me in which the acvity can be completed of a building contract and
the cost per day for reducing the me of each acvity. The contract includes a penalty
clause of ₹100 per day over 17 days. The overhead cost per day is ₹160.
Acvity
Normal Time (Days)
Shortest Time (Days)
Cost of Reducon per Day (₹)
12
6
4
80
13
8
4
50
14
5
3
30
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24
5
3
25
5
3
40
36
12
8
200
46
8
5
50
56
6
6
The cost of compleng the eight acvies in normal me is ₹6,500.
(a) Calculate the normal duraon of the project, its cost and the crical path.
(b) Calculate and plot on a graph the cost/me funcon for the project and state:
(i) the lowest cost and associated me
(ii) the shortest me and associated cost
Ans: Step 1: Understanding the Data
Were given a table of activities with:
Normal time (how long each task usually takes).
Shortest time (the minimum possible if we crash the activity by spending extra
money).
Cost per day of reduction (how much extra it costs to shorten the task).
We also know:
The contract has a penalty of ₹100 per day if the project exceeds 17 days.
Overhead cost is ₹160 per day.
Normal cost of completing all activities = ₹6,500.
Step 2: Drawing the Network
Activities are labeled by their start and end nodes:
12 (6 days)
13 (8 days)
14 (5 days)
24 (5 days)
25 (5 days)
36 (12 days)
46 (8 days)
56 (6 days)
So the project flows from node 1 (start) to node 6 (finish).
Step 3: Finding Paths and Durations
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Lets calculate the total time for each path:
1. Path 1246 = 6 + 5 + 8 = 19 days
2. Path 1256 = 6 + 5 + 6 = 17 days
3. Path 136 = 8 + 12 = 20 days
4. Path 146 = 5 + 8 = 13 days
So the longest path (critical path) is 136 = 20 days.
That means the normal project duration = 20 days.
Step 4: Normal Cost
Normal cost is given as ₹6,500. But we must also add overhead cost:
So total = ₹6,500 + ₹3,200 = ₹9,700.
Penalty? Since contract allows 17 days, but project takes 20 days, we exceed by 3 days.
Penalty = 3 × 100 = ₹300.
So final normal cost = ₹9,700 + ₹300 = ₹10,000.
Step 5: Crashing the Project
Now, can we reduce the project duration? Yes, by shortening activities on the critical path
(136).
Critical path activities:
13 (8 → 4 days, cost ₹50/day)
36 (12 → 8 days, cost ₹200/day)
Lets reduce step by step:
Reduce 13 by 4 days (to 4 days). Cost = 4 × 50 = ₹200. New path 136 = 4 + 12 = 16
days.
Now project duration = max(19, 17, 16, 13) = 19 days (because path 1246 is now
longest).
Next, reduce 46 (8 → 5 days, cost ₹50/day). Path 1246 = 6 + 5 + 5 = 16 days.
Now critical path is 136 = 16 days.
So project duration = 17 days (since path 1256 = 17 days).
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Weve achieved contract time (17 days).
Step 6: Cost of Crashing
Extra crashing cost = ₹200 (for 13) + ₹150 (for 46) = ₹350.
So direct cost = ₹6,500 + 350 = ₹6,850.
Overhead = 17 × 160 = ₹2,720.
Penalty = 0 (since were within 17 days).
Total = ₹6,850 + ₹2,720 = ₹9,570.
Step 7: Shortest Time
Can we go shorter than 17 days? Yes, by reducing 36 (12 → 8 days).
Cost = 4 × 200 = ₹800.
Now path 136 = 4 + 8 = 12 days. Other paths: 1246 = 16, 1256 = 17. So project
duration = 17 days still (limited by 1256).
So shortest time = 17 days.
Total cost = ₹6,500 + 200 + 150 + 800 = ₹7,650. Overhead = 17 × 160 = 2,720. Total =
₹10,370.
Step 8: Cost/Time Function
Now we can plot cost vs time:
At 20 days: Cost = ₹10,000
At 19 days: Cost = ₹9,770 (after reducing 13 partially)
At 17 days: Lowest cost = ₹9,570
At 17 days (with extra crashing): Highest cost = ₹10,370
So the cost/time curve dips to a minimum at 17 days, then rises again if we crash further.
Step 9: Final Answers
(a)
Normal duration = 20 days
Normal cost = ₹10,000
Critical path = 136
(b)
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Lowest cost = ₹9,570 at 17 days
Shortest time = 17 days
Associated cost = ₹10,370
Diagram (simplified network)
(1)
/ | \
/ | \
2 3 4
| | |
5 6 6
\ | /
(6)
Critical path initially: 1 → 3 → 6 (20 days). After crashing: project reduced to 17 days.
Wrapping It Up
So the story here is:
At first, the project takes 20 days, costing ₹10,000.
By carefully reducing activities on the critical path, we bring it down to 17 days.
This not only avoids penalties but also reduces overhead, giving the lowest cost of
₹9,570.
If we crash further, costs rise again, showing the classic trade-off between time and
money.
This is the essence of project management: balancing speed and cost to find the sweet spot.
Q.8(a) What is crical path? State the necessary and sucient condions of crical path.
Can a project have mulple crical paths?
(b) What are the three me esmates used in the context of PERT? How are the expected
duraon of a project and its standard deviaon calculated?
Ans: Q.8 (a) Critical Path in a Project
What is a Critical Path?
Imagine you are planning a small eventsay a college seminar. There are many activities
like booking a hall, arranging chairs, inviting guests, preparing notes, etc. Some tasks can
happen together, but some must be done in order.
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Now, among all these activities, there is one longest path of dependent tasks that
determines the minimum time required to complete the project. This path is called the
Critical Path.
In simple words:
Critical Path = The longest sequence of activities with no extra time (no delay allowed).
If even one activity on this path is delayed, the entire project gets delayed.
Simple Diagram to Understand
Start
|
A (2 days)
|
B (4 days)
|
C (3 days)
|
Finish
Here, total duration = 2 + 4 + 3 = 9 days
This is the critical path, because it's the longest path and has no flexibility.
Necessary and Sufficient Conditions of Critical Path
To identify a critical path, some important conditions must be satisfied:
1. Longest Duration Path
The critical path is the path with the maximum total time among all possible paths.
It determines the minimum completion time of the project.
2. Zero Slack (or Float)
Activities on the critical path have no slack.
Slack means extra time available.
If slack = 0 → it is a critical activity.
Formula:
Slack = Latest Time - Earliest Time
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If Slack = 0 → Activity lies on critical path
3. Continuous Path from Start to End
The path must connect starting node to ending node without breaks.
4. Dependency of Activities
Activities on the critical path are dependent on each other.
One must finish before the next starts.
Can a Project Have Multiple Critical Paths?
Yes, absolutely!
Sometimes, a project may have two or more paths with equal longest duration.
Example:
Path 1: A → B → C = 10 days
Path 2: D → E → F = 10 days
Both paths take the same time → Both are critical paths.
What does it mean?
The project becomes more sensitive.
Delay in any of these paths will delay the whole project.
So, more critical paths = more risk
Q.8 (b) Three Time Estimates in PERT
Now lets move to PERT (Program Evaluation and Review Technique).
PERT is used when time is uncertain. Instead of using one fixed time, it uses three
estimates.
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1. Optimistic Time (to)
Best-case scenario
Minimum possible time
Everything goes perfectly
Example: If everything goes smoothly, it will take 2 days.
2. Most Likely Time (tm)
Normal situation
Most realistic estimate
Example: Usually, it takes 4 days.
3. Pessimistic Time (tp)
Worst-case scenario
Maximum time if things go wrong
Example: If problems occur, it may take 8 days.
Expected Time Calculation
PERT combines these three estimates into a single value:
Formula:
Expected Time (te) = (to + 4tm + tp) / 6
Example:
to = 2, tm = 4, tp = 8
te = (2 + 4×4 + 8) / 6
= (2 + 16 + 8) / 6
= 26 / 6
= 4.33 days
So, expected time = 4.33 days
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Standard Deviation Calculation
This tells us how much uncertainty is involved.
Formula:
Standard Deviation (σ) = (tp - to) / 6
Example:
σ = (8 - 2) / 6
= 6 / 6
= 1 day
Variance (Sometimes Asked)
Variance = (σ)²
= 1²
= 1
Expected Duration of Entire Project
Calculate expected time (te) for each activity.
Then find the critical path.
Add all expected times on that path.
That total = Expected Project Duration
Standard Deviation of Project
Add variances of activities on the critical path
Then take square root
Project σ = √(sum of variances)
Final Understanding (Quick Summary)
Critical Path → Longest path, no delay allowed
Conditions → Longest duration, zero slack, continuous, dependent
Multiple Critical Paths? → Yes, possible
PERT Time Estimates → Optimistic, Most Likely, Pessimistic
Expected Time Formula → (to + 4tm + tp) / 6
Standard Deviation → (tp - to) / 6
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This paper has been carefully prepared for educaonal purposes. If you noce any
mistakes or have suggesons, feel free to share your feedback.
