Operations, Logistics and Scheduling Data Short Notes for CAT: Concepts, Formulas, Worked Examples & Practice

Operations, Logistics and Scheduling Data short notes for CAT: 4 study cards covering concepts, formulas, shortcuts and exam traps, plus solved practice questions.

Transport and Transit Cheat Sheet

Capacity & Occupancy

Always draw the route and divide it into segments.
A ticket from A to C occupies segments A $\to$ B and B $\to$ C.
Maximum occupancy = The segment with the highest number of overlapping tickets.
Empty seats = Total Capacity - Occupancy in that specific segment.

Transit Time & Networks

$Total Time = \sum Travel Times + \sum Intermediate Halt Times$ .
Do not include the halt time at the final destination for arrival calculations.
Always distinguish between Junctions (longer halts) and Regular stops.
$N$ stations = $N - 1$ segments. Count the gaps, not the dots.

Quick Recap: Delivery & Sales Logic

Maximize Visits: Sort households by ascending complaint probability (

p_{i}

Expected Visits Formula:

E [N] = 1 + q_{1} + q_{1} q_{2} + q_{1} q_{2} q_{3} + \dots

Supply Chain Demand:

E [Total] = \sum E [Individual Locations]

"Exactly K" Visits:

If $K < Total Houses$ : Survive $1 \dots K - 1$ , Fail at $K$ .
If $K = Total Houses$ : Survive all.

Remember: Independence allows you to multiply probabilities for sequences and add expectations for totals.

Quick Recap: Scheduling & Tracking

Quick Recap

Core Equation:

Work = Efficiency \times Time

. Always additive.

Phase Mapping: Break schedules into non-overlapping phases based on dates.

Phase Work: Multiply active team's combined efficiency by phase duration.

Matrix Sum: Sum of individual contributions equals total project work.

Day-Counting: Reverse-engineer inclusive/exclusive convention from known data.

Pro Tip: Assume total work =

100

units or LCM of completion days for clean calculations.

Quick Revision: Key Formulas and Insights

Quick Revision

Core Formulas

Concept	Formula
Rate	$r = \frac{Δ Q}{T}$
Time for change	$t = \frac{Δ Q}{r}$
Multi-mode final value	$Q_{f} = Q_{0} + \sum r_{i} \cdot t_{i}$
Net change per cycle	$Δ Q_{cy c l e} = \sum r_{i} \cdot t_{i}$
Number of cycles	$n = ⌈ \frac{Δ Q _{t o t a l}}{Δ Q _{cy c l e}} ⌉$

Key Insights

Rate is inverse of time for the same change
- If mode A takes half the time of mode B, mode A is twice as fast
Efficiency vs Speed
- Fastest mode is not necessarily the most energy-efficient
- Compare energy per unit change: $\frac{E}{r}$
Off-state matters
- Always include off-state rate in cycle calculations
- Off-state often reverses the direction of change
Piecewise tracking
- For multiple modes, track quantity at each switch point
- Do not average rates across modes

Common Pitfalls

Confusing time with rate
Forgetting off-state changes
Unit mismatches (minutes vs hours)
Assuming linearity across modes

Operations, Logistics and Scheduling Data: Solved Questions with Step-by-Step Explanations (2 Problems)

Question 1 · Data Interpretation and Logical Reasoning NAT

A freight train consists of 5 wagons $W_{1}, W_{2}, W_{3}, W_{4}, W_{5}$ arranged in that order from the engine. Each wagon has a maximum weight capacity of 40 tons. The train must transport 7 distinct containers with weights: ${12, 15, 18, 22, 25, 28, 30}$ tons.

Constraints:

No wagon can exceed its capacity.
Containers weighing $\geq 25$ tons cannot be placed in $W_{1}$ or $W_{5}$ due to axle stress limits.
The center of gravity constraint requires that the total weight in ${W_{1}, W_{2}}$ differs from the total weight in ${W_{4}, W_{5}}$ by no more than 10 tons.
Container $C_{30}$ (30 tons) and $C_{28}$ (28 tons) cannot be in adjacent wagons.

If all 7 containers must be transported in a single trip, what is the MAXIMUM possible weight that can be carried in wagon $W_{3}$ ?

Correct Answer:

Step-by-Step Solution

Key idea: This is a constraint satisfaction problem requiring construction. You must arrange items to satisfy multiple overlapping restrictions while optimizing a specific variable ( $W_{3}$ weight).

Step 1: Identify hard constraints on heavy items.

Heavy items ( $\geq 25$ ): ${25, 28, 30}$ .

Constraint 2 forbids these in $W_{1}, W_{5}$ .

Therefore, ${25, 28, 30}$ MUST be distributed among ${W_{2}, W_{3}, W_{4}}$ .

Since there are 3 heavy items and exactly 3 eligible wagons, each of $W_{2}, W_{3}, W_{4}$ must contain EXACTLY ONE heavy item.

Step 2: Optimize $W_{3}$ .

To maximize $W_{3}$ , we should place the heaviest possible container there.

Candidate: 30.

Assume $W_{3} = 30$ .

Remaining heavy items ${25, 28}$ go to ${W_{2}, W_{4}}$ in some order.

Step 3: Check adjacency constraint (Constraint 4).

$C_{30}$ is in $W_{3}$ . Adjacent wagons are $W_{2}, W_{4}$ .

Constraint 4 says $C_{28}$ cannot be adjacent to $C_{30}$ .

Therefore, $C_{28}$ CANNOT be in $W_{2}$ or $W_{4}$ .

But Step 1 established that ${25, 28}$ MUST occupy ${W_{2}, W_{4}}$ .

Contradiction.

Conclusion: $W_{3}$ CANNOT hold 30.

Step 4: Try next heaviest for $W_{3}$ .

Candidate: 28.

Assume $W_{3} = 28$ .

Remaining heavy ${25, 30}$ go to ${W_{2}, W_{4}}$ .

Adjacency check: $C_{28}$ is in $W_{3}$ . Neighbors $W_{2}, W_{4}$ cannot hold $C_{30}$ .

But neighbors MUST hold ${25, 30}$ . One of them WILL hold 30.

Contradiction.

Conclusion: $W_{3}$ CANNOT hold 28.

Step 5: Try next heaviest for $W_{3}$ .

Candidate: 25.

Assume $W_{3} = 25$ .

Remaining heavy ${28, 30}$ go to ${W_{2}, W_{4}}$ .

Adjacency check: $C_{25}$ is in $W_{3}$ . Constraint 4 only restricts $30 \leftrightarrow 28$ adjacency.

$25$ has no adjacency restriction.

So placing 25 in $W_{3}$ is valid regarding Constraint 4.

Current Max Candidate: 25.

Step 6: Can we add light items to $W_{3}$ ?

$W_{3}$ currently has 25. Capacity 40. Space 15.

Light items: ${12, 15, 18, 22}$ .

Available light items depend on placement elsewhere.

We need to verify if a valid global configuration exists with $W_{3} = 25 + light$ .

To maximize $W_{3}$ , try adding largest fitting light item: 15.

Target $W_{3} = 25 + 15 = 40$ .

Remaining items to place: ${12, 18, 22}$ plus heavies ${28, 30}$ in $W_{2}, W_{4}$ .

Configuration attempt for $W_{3} = 40$ :

$W_{3} = {25, 15}$ .

Heavies ${28, 30}$ in ${W_{2}, W_{4}}$ .

Constraint 4: 28 and 30 not adjacent. They are separated by $W_{3}$ . Valid.

Remaining lights ${12, 18, 22}$ must fit in ${W_{1}, W_{2}, W_{4}, W_{5}}$ respecting capacities and CoG.

Let's distribute heavies:

Case A: $W_{2} = 30, W_{4} = 28$ .

Case B: $W_{2} = 28, W_{4} = 30$ .

Check CoG (Constraint 3): $∣ W e i g h t (W_{1} + W_{2}) - W e i g h t (W_{4} + W_{5}) ∣ \leq 10$ .

Try Case A ( $W_{2} = 30, W_{4} = 28$ ):

Base imbalance: $30 - 28 = 2$ .

We have lights ${12, 18, 22}$ to distribute into $W_{1}, W_{2}, W_{4}, W_{5}$ .

Note: $W_{2}$ has 30, cap 40 (space 10). Can take 12? No, 12>10. Cannot take any remaining light.

$W_{4}$ has 28, cap 40 (space 12). Can take 12. Cannot take 18 or 22.

So in Case A, $W_{2}$ gets 0 lights. $W_{4}$ might get 12.

Remaining lights ${18, 22}$ (and maybe 12) must go to $W_{1}, W_{5}$ .

$W_{1}, W_{5}$ caps are 40.

Subcase A1: $W_{4}$ takes 12.

Lights left: ${18, 22}$ . Must go to $W_{1}, W_{5}$ .

Possible splits:

$W_{1} = 18, W_{5} = 22$ .

LHS ( $W_{1} + W_{2}$ ) = $18 + 30 = 48$ .

RHS ( $W_{4} + W_{5}$ ) = $(28 + 12) + 22 = 62$ .

Diff = 14. Fail (>10).

$W_{1} = 22, W_{5} = 18$ .

LHS = $22 + 30 = 52$ .

RHS = $40 + 18 = 58$ .

Diff = 6. PASS.

Valid configuration found!

$W_{1} = {22}, W_{2} = {30}, W_{3} = {25, 15}, W_{4} = {28, 12}, W_{5} = {18}$ .

Checks:

Caps: 22, 30, 40, 40, 18. All $\leq 40$ . OK.
Heavy locs: 30, 28, 25 in $W_{2}, W_{4}, W_{3}$ . OK.
Adj: 30 and 28 separated. OK.
CoG: $∣52 - 58∣ = 6 \leq 10$ . OK.
$W_{3}$ weight: 40.

Since 40 is the physical capacity, we cannot go higher.

Answer: 40.

Wait, I previously thought max was 30 in the draft plan. The construction proves 40 is possible.

Double check the "Answer: 30" in the draft vs "40" in derivation.

Derivation shows 40 is achievable. I will output 40.

Question 2 · Data Interpretation and Logical Reasoning NAT

A circular logistics network connects four distribution centers $A, B, C, D$ in clockwise order. A single shuttle with capacity $K$ units operates continuously on this loop. At each center, the shuttle first unloads all goods destined for that center, then loads new goods subject to remaining capacity. Loading priority is strictly clockwise (i.e., goods for the next immediate center are loaded first).

The steady-state demand matrix (units per cycle) is:

$A B C D A - 152010 B 20 - 1025 C 1025 - 5 D 301015 -$

It is observed that in steady state, the shuttle departs center $B$ exactly full, but departs center $D$ with exactly 10 empty seats. Furthermore, no demand from $A$ to $D$ is ever left unfulfilled.

What is the capacity $K$ of the shuttle?

Correct Answer:

Step-by-Step Solution

Key idea: This is a steady-state network flow problem requiring algebraic modeling of segment loads. The key is to express the load on each segment as a function of $K$ and the known demands, then use the boundary conditions ("full at B", "10 empty at D") to solve for $K$ .

Step 1: Define segment loads.

Let $L_{X Y}$ be the load on segment $X \to Y$ .

We know $L_{A B}, L_{B C}, L_{C D}, L_{D A}$ .

Load update rule: $L_{n e x t} = L_{p r e v} - Unload + Load$ .

Step 2: Trace loads symbolically.

Assume steady state. Let's start at departure from $A$ .

Demand from $A$ : $B (20), C (10), D (30)$ . Total 60.

Since "no demand $A \to D$ is unfulfilled", and $A \to D$ is lowest priority (clockwise: $B \to C \to D$ ), this implies ALL higher priority demands ( $A \to B, A \to C$ ) are also fulfilled, AND there is enough space for $A \to D$ .

Thus, shuttle MUST depart $A$ with at least $20 + 10 + 30 = 60$ units.

So $L_{A B} \geq 60$ . Also $L_{A B} \leq K$ .

At $B$ :

Unload: 20 (from $A$ ).

Remaining: $L_{A B} - 20$ .

New Demand: $C (25), D (10), A (15)$ . Priority: $C \to D \to A$ .

Space: $K - (L_{A B} - 20)$ .

Condition: "Departs $B$ exactly full".

So $L_{B C} = K$ .

This implies total demand at $B$ ( $25 + 10 + 15 = 50$ ) $\geq$ Space.

$50 \geq K - L_{A B} + 20 ⟹ L_{A B} \geq K - 30$ .

At $C$ :

Arrive with $K$ .

Unload: Goods for $C$ . Sources: $A (10)$ and $B (25)$ . Total 35.

Note: Are we sure $B \to C$ was fully loaded? Yes, because $B$ departed full and $C$ is highest priority.

Remaining: $K - 35$ .

New Demand: $D (15), A (20), B (10)$ . Priority: $D \to A \to B$ .

Space: $K - (K - 35) = 35$ .

Total demand: $15 + 20 + 10 = 45$ .

Since $45 > 35$ , shuttle fills up? Not necessarily.

Load $D$ : min(15, 35) = 15. Rem space 20.

Load $A$ : min(20, 20) = 20. Rem space 0.

Load $B$ : 0.

So $L_{C D} = (K - 35) + 15 + 20 = K$ .

Wait, if $L_{C D} = K$ , then it departs $C$ full.

At $D$ :

Arrive with $K$ .

Unload: Goods for $D$ . Sources: $A (30), B (10), C (15)$ .

Were these fully loaded?

$A \to D$ : Yes (given).

$B \to D$ : Priority after $C$ . At $B$ , space was $K - (L_{A B} - 20)$ . Since $L_{B C} = K$ , all 50 units were loaded. So $B \to D$ (10) is on board.

$C \to D$ : Highest priority at $C$ . Loaded 15.

Total unload at $D$ : $30 + 10 + 15 = 55$ .

Remaining: $K - 55$ .

New Demand: $A (10), B (25), C (5)$ . Priority: $A \to B \to C$ .

Space: $K - (K - 55) = 55$ .

Total demand: $10 + 25 + 5 = 40$ .

Since $40 < 55$ , ALL demand is loaded.

Departure load $L_{D A} = (K - 55) + 40 = K - 15$ .

Step 3: Apply boundary condition at $D$ .

"Departs $D$ with exactly 10 empty seats".

Load = $K - 10$ .

From derivation: Load = $K - 15$ .

Equating: $K - 10 = K - 15 ⟹ - 10 = - 15$ . Contradiction.

Re-evaluate Step 2/3. Where is the error?

Check Unload at $D$ .

$A \to D$ : 30. (Guaranteed).

$B \to D$ : 10. (Guaranteed if $B$ full and $C$ took precedence? At $B$ , priority $C, D, A$ . If $B$ is full, $C$ takes 25. Rem space $K - (L_{A B} - 20) - 25$ . If this rem space $\geq 10$ , $D$ gets 10.

We established $L_{A B} \geq K - 30$ .

Space at $B$ before loading = $K - L_{A B} + 20$ .

After loading $C (25)$ : Space = $K - L_{A B} + 20 - 25 = K - L_{A B} - 5$ .

For $B \to D$ to be fully loaded, need $K - L_{A B} - 5 \geq 10 ⟹ L_{A B} \leq K - 15$ .

Previously I had $L_{A B} \geq K - 30$ .

So $L_{A B} \in [K - 30, K - 15]$ . This range is valid.

So $B \to D$ IS fully loaded.

$C \to D$ : 15. Highest priority at $C$ . Always loaded.

So Unload at $D$ is indeed 55.

Check Load at $D$ .

Remaining after unload: $K - 55$ .

Space available: 55.

Demand: $A (10), B (25), C (5)$ . Sum 40.

Loaded: 40.

Departure: $(K - 55) + 40 = K - 15$ .

Empty seats: $K - (K - 15) = 15$ .

Problem states empty seats = 10.

Contradiction persists.

Hypothesis: My assumption about $A \to D$ fulfillment implies $L_{A B} = 60$ might be too weak or strong.

"No demand $A \to D$ is ever left unfulfilled".

At $A$ , priority $B, C, D$ .

Load $B (20)$ , $C (10)$ . Used 30.

Space $K - 30$ .

Need $K - 30 \geq 30 ⟹ K \geq 60$ .

If $K = 60$ , $L_{A B} = 60$ .

If $K > 60$ , $L_{A B} = 60$ still (only 60 demand).

So $L_{A B} = 60$ is fixed by demand at $A$ .

Substitute $L_{A B} = 60$ into previous inequalities.

At $B$ : Space = $K - 60 + 20 = K - 40$ .

Load $C (25)$ . Rem = $K - 65$ .

Load $D (10)$ . Requires $K - 65 \geq 10 ⟹ K \geq 75$ .

IF $K < 75$ , then $B \to D$ is NOT fully loaded.

Ah! Here is the branch.

Case 1: $K \geq 75$ .

Then $B \to D$ full. Unload at $D$ = 55. Empty = 15.

Matches contradiction. So $K < 75$ .

Case 2: $60 \leq K < 75$ .

At $B$ : Space = $K - 40$ .

Load $C (25)$ . Rem = $K - 65$ .

Load $D$ : Takes $K - 65$ . (Since $K - 65 < 10$ ).

Load $A$ : 0.

Depart $B$ full ( $K$ ). Consistent.

At $C$ :

Arrive $K$ .

Unload $C$ : $A \to C (10)$ + $B \to C (25) = 35$ .

Rem: $K - 35$ .

Load $D (15)$ .

Space = 35.

Takes min(15, 35) = 15.

Rem space: 20.

Load $A (20)$ . Takes 20.

Rem space: 0.

Load $B (0)$ .

Depart $C$ full ( $K$ ).

At $D$ :

Arrive $K$ .

Unload $D$ :

$A \to D$ : 30.

$B \to D$ : $K - 65$ (partial load from B).

$C \to D$ : 15.

Total Unload = $30 + (K - 65) + 15 = K - 20$ .

Remaining on board: $K - (K - 20) = 20$ .

Space available: $K - 20$ .

Demand at $D$ : $A (10), B (25), C (5)$ . Sum 40.

Priority $A \to B \to C$ .

Load $A (10)$ . Space becomes $K - 30$ .

Load $B (25)$ . Takes min(25, $K - 30$ ).

Load $C (5)$ .

Departure Load $L_{D A} = 20 + 10 + min (25, K - 30) + \dots$

Empty Seats = $K - L_{D A}$ .

Given Empty = 10 $⟹ L_{D A} = K - 10$ .

Equation: $20 + 10 + Loaded_{B} + Loaded_{C} = K - 10$ .

$30 + Loaded_{B, C} = K - 10 ⟹ Loaded_{B, C} = K - 40$ .

Available space for $B, C$ was $K - 30$ .

Demand $B + C = 30$ .

So $Loaded_{B, C} = min (30, K - 30)$ .

Set $min (30, K - 30) = K - 40$ .

If $K - 30 \geq 30 ⟹ K \geq 60$ :

Then LHS = 30.

$30 = K - 40 ⟹ K = 70$ .

Check consistency: Is $70 \geq 60$ ? Yes.

Is $70 < 75$ ? Yes.

So $K = 70$ is the unique solution.

Wait, let me double check the "Empty=15" calculation for Case 1 ( $K \geq 75$ ).

If $K = 75$ :

Unload at $D$ = 55. Rem = 20.

Space = 55. Demand = 40. All loaded.

Depart = $20 + 40 = 60$ .

Empty = $75 - 60 = 15$ . Correct.

Back to Case 2 result $K = 70$ .

Verify:

$L_{A B} = 60$ .

At $B$ : Space $70 - 40 = 30$ . Load $C (25)$ . Rem 5. Load $D (5)$ . Depart 70.

At $C$ : Unload 35. Rem 35. Load $D (15), A (20)$ . Depart 70.

At $D$ : Unload $30 (A) + 5 (B) + 15 (C) = 50$ . Rem 20.

Space 50. Demand $A (10), B (25), C (5)$ .

Load $A (10)$ . Space 40.

Load $B (25)$ . Space 15.

Load $C (5)$ . Space 10.

Depart Load = $20 + 10 + 25 + 5 = 60$ .

Empty = $70 - 60 = 10$ . Matches.

Answer: 70.

Operations, Logistics and Scheduling Data Short Notes for CAT: Concepts, Formulas, Worked Examples & Practice

Transport and Transit Cheat Sheet

Transport and Transit Cheat Sheet

Capacity & Occupancy

Transit Time & Networks

Quick Recap: Delivery & Sales Logic

Quick Recap: Delivery & Sales Logic

Quick Recap: Scheduling & Tracking

Quick Recap

Quick Revision: Key Formulas and Insights

Quick Revision

Core Formulas

Key Insights

Common Pitfalls

Operations, Logistics and Scheduling Data: Solved Questions with Step-by-Step Explanations (2 Problems)

Step-by-Step Solution

Step-by-Step Solution

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