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

Scheduling, Routes and Network Logic short notes for CAT: 4 study cards covering concepts, formulas, shortcuts and exam traps, plus solved practice questions.

Final Checklist: Route Maps and Networks

Master Checklist

Map It: Convert the diagram into a list of Nodes and Edges.
Check Type: Is it a Grid (use coordinates) or Arbitrary (use adjacency)?
Constraints: Are edges one-way? Are there obstacles? Is movement continuous?
Counting Paths: Use the Labeling Method (sum of predecessors).
Shortest Path: Sum edge weights. For grids, use Manhattan distance if unblocked.
Validation: For sequence questions, check each consecutive pair for a direct connection.

Golden Rule: The diagram is a reference; the edge list is the truth.

Master Checklist: Scheduling and Queues

Master Checklist

Identify Elements: Clearly list Resources, Time Slots, and Entities. Note capacities.

Draw the Grid: Build a Master Timeline Grid (Slots as columns, Resources as rows).

Simulate Chronologically: Move slot by slot. Process departures, assign from queue, then add arrivals.

Check Continuity: Ensure tasks are not split across slots unless allowed.

Verify Boundaries: Double-check start time vs. completion time constraints, especially for the last slot.

Respect Capacity: Never overload a resource beyond its hard limit per slot.

Golden Rule: The timeline grid is your source of truth.

Execution Framework for Scheduling

The 4-Step Master Method

Draw the Grid

Entities as rows, time slots as columns. Note slot capacities at the top.

Lock the Fixed Elements

Fill in exact numbers, place the longest consecutive blocks first, and anchor them to the timeline edges.

Resolve Bounds

Use the Extreme Case Method to turn ranges into exact numbers.

Verify and Iterate

Check the global total. Ensure no blocks overlap if forbidden, or overlap if required.

Execution Framework for Firm Lifecycles

The 4-Step Master Method

Draw the Grid

Firms as rows, years as columns. Note the base value at the start and end of each firm's lifespan.

Determine Parity

Check if the lifespan ( $n$ ) is odd or even. This locks in whether the peak is a single year or a double year, and defines $k$ .

Apply the Unified Formula

Use $Total = n \times B + k \times (P - B)$ to find the peak value ( $P$ ) or the total funding.

Verify Constraints

Ensure no two consecutive years have the same funding amount (except the valid two-year peak for even $n$ ), and that the peak placement strictly follows the parity rule.

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

More short notes in this unit

chapter

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

Scheduling, Routes and Network Logic short notes for CAT: 4 study cards covering concepts, formulas, shortcuts and exam traps, plus solved practice questions.

A question from this chapter

Question 1

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?

Question 2

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?

Free preview ends here

Login to view the complete short notes

Creating an account is free. You get the rest of this chapter, step-by-step solutions, and a study plan built around the topics you are actually weak at.

Why MastersUp

Personalised first. High quality throughout.

Most platforms hand everyone the same content. Here the content moves with your performance, topic by topic.

Built around you, not around a syllabus PDF

Every answer you give moves your topic-level intelligence rate. The next question, the next revision card and tomorrow's plan all change with it.

Revision that hits your weak spots

We only revise topics you have actually attempted and are still below the safe bar on — never the same chapter on repeat.

Questions calibrated to the real exam

Each question carries a measured toughness. You are served a rung above your current level, so practice keeps stretching you.

Notes written for recall, not for volume

Full lesson cards for first study, curated short-note cards for the last mile — with derivations, traps and exam patterns marked.

One place for everything

Notes, chapter practice, previous-year questions, test series and full-length papers — all feeding one picture of your preparation.

Honest progress

No vanity streaks. Progress here means chapters mastered and accuracy that held up on harder questions.

Unlock the whole course

Full notes and short notes, the complete question bank with worked solutions, mock tests, full-length papers, and an adaptive plan that rebuilds itself as you improve.

View all plans Start free