Scheduling, Routes and Network Logic Notes for CAT: Concepts, Formulas, Worked Examples & Practice

Scheduling, Routes and Network Logic notes for CAT: 40 study cards covering concepts, formulas, shortcuts and exam traps, plus solved practice questions.

Chapter Roadmap: Scheduling, Routes and Network Logic

Chapter Journey: Scheduling, Routes and Network Logic

1. Route Maps, Walkways and Network Paths (Current Topic)

Focus: Decoding spatial networks, shortest paths, and movement constraints.
Weightage: High. Foundation for all network-based logical reasoning.

2. Time Slots, Queues and Ride Scheduling

Focus: Allocating limited resources across fixed time slots.
Weightage: High. Tests sequential logic and capacity constraints.

3. Year-Based Training and Publication Schedules

Focus: Sequencing events over years with multiple participants and gaps.
Weightage: Medium-High. Requires careful timeline mapping.

4. Firm Lifecycle and Funding Timelines

Focus: Overlapping intervals, start and end years, and cumulative sums.
Weightage: Medium. Tests interval logic and arithmetic.

Goal: By the end of this chapter, you will be able to instantly visualize any network or schedule, identify critical constraints, and solve complex multi-step reasoning problems with confidence.

The Hero Concept: Networks as Graphs of Nodes and Edges

The Core Model: Nodes and Edges

Every network problem, whether it involves stations, intersections, or gated community walkways, can be reduced to a Graph.

Nodes (Vertices): The specific points of interest (e.g., Stations, Intersections).
Edges (Links): The connections between nodes (e.g., Streets, Walkways). They often have weights (distance, time) or directions.

First Step in Any Problem:
1. Identify all Nodes.
2. List all direct Connections (Edges).
3. Note any Constraints (One-way? Blocked? Distance?).

Intuition: Think of the network as a social circle. Who knows whom directly? If A knows B, and B knows C, can A reach C? Yes, through B. This is the basis of all pathfinding.

Method: Tracing Paths and Counting Routes

Systematic Path Counting

To count the number of distinct routes between two nodes without missing any or double-counting:

The Labeling Method

Start Node: Label the starting node with 1 (one way to be at the start).
Propagation: Move to adjacent nodes. The number of ways to reach a new node is the sum of the ways to reach all its predecessors that have already been labeled.
Direction: Follow the allowed direction of travel. If two-way, ensure you do not loop back unless allowed.

Ways to reach N = \sum (Ways to reach each predecessor of N)

Note: In complex grids, this method is far superior to trying to draw every single line.

Worked Example: The Station Patrol Sequence

Case Study: Station Network Validation

Context: Six stations (A to F) connected by streets. Teams patrol continuously.

Typical Question: "Which of the following is a valid patrol sequence for a team starting at A?"

Step 1: Decode the Map

Step 2: Validate a Sequence

Check sequence: A to B to D to E to B

A to B: Valid (connected).
B to D: Valid (connected).
D to E: Valid (connected).
E to B: Valid (connected).

Check sequence: A to C to D

A to C: Valid.
C to D: Invalid (C is only connected to A and F).

Key Insight: You do not need to memorize the whole map. Just check the local connection between consecutive steps in the sequence.

Scheduling, Routes and Network Logic: Solved Questions with Step-by-Step Explanations (2 Problems)

Question 1 · Data Interpretation and Logical Reasoning MCQ

In a secure network of 7 servers (P, Q, R, S, T, U, V), each server is connected to some others via direct, two-way links. The number of connections (degree) for each server is:

P: 4, Q: 4, R: 3, S: 3, T: 2, U: 2, V: 2.

The following additional constraints are known:

P is NOT connected to U and V.
Q is NOT connected to S and T.
R is NOT connected to T, U, and V.
S is NOT connected to T and V.

Based on this information, which of the following pairs of servers are DEFINITELY connected to each other?

A.
S and U
B.
T and U
C.
U and V
D.
S and T

Correct Answer:

Step-by-Step Solution

Key idea: This is a Network Mapping / Graph Realization problem. You must use the degrees and negative constraints to force positive connections.

Step 1: Analyze P (Degree 4). There are 6 other servers. P is NOT connected to U and V (2 servers). Therefore, P MUST be connected to the remaining 4: Q, R, S, T.

Step 2: Analyze Q (Degree 4). Q is NOT connected to S and T. Therefore, Q MUST be connected to the remaining 4: P, R, U, V.

Step 3: Analyze R (Degree 3). R is NOT connected to T, U, V. Therefore, R MUST be connected to the remaining 3: P, Q, S.

Step 4: Analyze S (Degree 3). We already know S is connected to P (from Step 1) and R (from Step 3). S needs 1 more connection. The remaining available servers are Q, U. However, Step 2 states Q is NOT connected to S. Therefore, S MUST be connected to U.

Step 5: Verify the pair. The deduction forces the connection between S and U.

Answer: A

Question 2 · Data Interpretation and Logical Reasoning NAT

Five infrastructure projects ( $P_{1}$ to $P_{5}$ ) each have a lifespan of exactly 4 years.

Each project follows the exact same arithmetic progression of strictly positive integers for its annual funding over its 4 years.

The total funding for each project over its 4-year lifespan is exactly 100 Crores.

The 5 projects commence in 5 consecutive calendar years (e.g., 2020, 2021, 2022, 2023, 2024).

A regulatory cap states that in any given calendar year, the sum of the funding of all active projects must not exceed 150 Crores.

What is the maximum possible value of the common difference (in Crores) of this arithmetic progression?

Correct Answer:

Step-by-Step Solution

Key idea: This is an aggregate calculation problem with a sliding window over an arithmetic progression. The key insight is recognizing the invariant sum of the overlapping terms.

Step 1: Define the Arithmetic Progression (AP).

Let the 4 terms be $a, a + d, a + 2 d, a + 3 d$ .

Sum = $4 a + 6 d = 100 ⟹ 2 a + 3 d = 50$ .

Since funding must be strictly positive integers, $a \geq 1$ and $d$ must be an even integer (because $2 a$ and $50$ are even, so $3 d$ must be even).

Step 2: Analyze the overlap constraint.

The projects start in consecutive years. Let's look at a year where 4 projects are active (e.g., Year 4).

In Year 4:

$P_{1}$ is in its 4th year (funding = $a + 3 d$ )
$P_{2}$ is in its 3rd year (funding = $a + 2 d$ )
$P_{3}$ is in its 2nd year (funding = $a + d$ )
$P_{4}$ is in its 1st year (funding = $a$ )

The sum of funding in Year 4 is exactly $(a + 3 d) + (a + 2 d) + (a + d) + a = 4 a + 6 d = 100$ .

This sum is exactly 100 Crores, regardless of the values of $a$ and $d$ !

Since $100 \leq 150$ , the regulatory cap is NEVER binding for any year with 4 active projects.

For years with fewer than 4 active projects (like Year 1, 2, 8, 9), the sum is a subset of the AP terms, which will be strictly less than 100 (since all terms are positive).

Thus, the 150 Crore cap is a redundant constraint (a trap).

Step 3: Maximize $d$ using the only real constraint.

The only constraint is that funding must be strictly positive: $a \geq 1$ .

From $2 a + 3 d = 50$ , we have $3 d = 50 - 2 a$ .

To maximize $d$ , we must minimize $a$ .

Let $a = 1$ . Then $3 d = 50 - 2 (1) = 48 ⟹ d = 16$ .

Check if $d = 16$ yields integer terms:

$a = 1, d = 16 ⟹$ Terms are $1, 17, 33, 49$ . All are strictly positive integers.

Sum = $1 + 17 + 33 + 49 = 100$ .

Therefore, the maximum possible common difference is 16.

Answer: 16

Scheduling, Routes and Network Logic Notes for CAT: Concepts, Formulas, Worked Examples & Practice

Chapter Roadmap: Scheduling, Routes and Network Logic

Chapter Journey: Scheduling, Routes and Network Logic

1. Route Maps, Walkways and Network Paths (Current Topic)

2. Time Slots, Queues and Ride Scheduling

3. Year-Based Training and Publication Schedules

4. Firm Lifecycle and Funding Timelines

The Hero Concept: Networks as Graphs of Nodes and Edges

The Core Model: Nodes and Edges

Method: Tracing Paths and Counting Routes

Systematic Path Counting

The Labeling Method

Worked Example: The Station Patrol Sequence

Case Study: Station Network Validation

Step 1: Decode the Map

Step 2: Validate a Sequence

Scheduling, Routes and Network Logic: Solved Questions with Step-by-Step Explanations (2 Problems)

Step-by-Step Solution

Step-by-Step Solution

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