Introduction Routes & Network - Logical Reasoning (LR) & Data Interpretation (DI)

Introduction

• Recent trends in MBA entrance and aptitude exams show that questions on routes and network diagrams have become a regular feature in these exams . 
• It is essential to understand these question types and develop reliable problem-solving techniques to handle them under exam conditions.
• Typical questions involve travel or transport scenarios across multiple nodes connected by directed or undirected links forming a network.
• Problems commonly ask for the number of possible routes, the shortest/optimal path, capacity or flow constraints, or schedule/compatibility decisions within the network.
• Effective solving depends on systematic analysis of the network: identifying stages, branching points, symmetrical and asymmetrical nodes, and any constraints on movement or street usage.
• Practising these problems builds the ability to visualise networks, break them into manageable stages, and apply combinatorial reasoning or flow logic as required.

Note: In CAT, these problems are commonly seen in the DILR section and often involve layered constraints that must be handled systematically. 

What are Routes and Networks?

• In exam problems a route is a specific path from a start node to an end node travelling along available links. 
• A network is a collection of nodes and links that allow multiple alternative routes between two given points. 
• Counting routes or choosing allowable schedules requires careful treatment of branching and constraints.
• To count routes efficiently, identify the network structure and break the journey into stages or levels: each stage contains nodes that can be reached in the same number of moves from the start.
• Use multiplication of choices across consecutive stages when movements between stages are independent and identical for nodes in a stage.
• Recognise symmetry: if two or more nodes in a stage lead to identical onward choices, treat them together to simplify counting.
• Handle asymmetry by splitting the stage into separate cases for nodes that behave differently and summing the counts from each case.

Note: The stage-based method works only when movement is strictly forward and the network has no cycles. If backward movement or looping paths exist, simple multiplication will not apply. 

Consider this example network to count routes from node A to node G. Possible distinct routes shown are: ABEG, ABFG, ACEG, ACFG, ADEG, ADFG.

• Label the stages: Stage 0: A; Stage 1: B, C, D; Stage 2: E, F; Stage 3: G.
• From Stage 0 to Stage 1 there are 3 choices (A→B, A→C, A→D).
• From Stage 1 to Stage 2 each node has 2 onward options (to E or F), so moving from Stage 1 to Stage 2 contributes a factor of 2 per Stage-1 choice.
• From Stage 2 to Stage 3 there is 1 option (either E→G or F→G), so a factor of 1.
• Combine stages multiplicatively: 3 × 2 × 1 = 6 routes (these are the listed six routes). This multiplication is valid because each stage transition is independent and identical across all nodes in the stage.

When one or more links are unavailable (asymmetry), split the counting into separate cases and add the results. For example, if the link B→E is closed while other links remain, treat nodes C and D together (symmetrical) and B separately (asymmetrical).

• Case 1 (Stage 1A: symmetrical nodes C/D): 2 nodes in Stage 1 each with 2 onward choices to Stage 2, so 2 × 2 = 4 routes.
• Case 2 (Stage 1B: asymmetrical node B): if B has only one available onward link then it contributes 1 route.
• Total routes = 4 + 1 = 5.  

Divide stages whenever necessary to isolate asymmetrical behaviour, but avoid unnecessary subdivision that increases confusion. The strategy is:

• Identify stages from start to destination.
• Group symmetrical nodes in a stage; count once and multiply by the number of such nodes.
• Handle asymmetrical nodes separately and add their counts to the symmetrical group count.

Alternate route-structure breakdowns are useful when the network allows different patterns of transitions between stages. Example abstract route structures (using stage labels) are:

• Route 1: 0 → 1A → 2A → 3
• Route 2: 0 → 1A → 2B → 3
• Route 3: 0 → 1B → 2A → 3
• Route 4: 0 → 1B → 2B → 3

These abstract route structures are mutually exclusive; only structurally valid cases should be evaluated. 

When the network grows more complex you may subdivide Stage 2 into 2A and 2B, or split Stage 1 further. Always recombine by summation across mutually exclusive cases. Examples of such subdivisions and their effect on counts:

• If Stage 1A (C/D) provides 2 options each and Stage 2A provides 2 options each then Route 1 gives 2 × 2 = 4 ways.
• If a subdivision makes one subcase asymmetrical (one node gives only 1 onward choice) then compute that subcase separately and add its count.
• If multiple subdivisions apply across stages, perform multiplicative counting within each subcase and then sum across subcases that are mutually exclusive.

Use tables or a systematic listing of cases when the network has many asymmetries; this prevents double counting and missing cases.

In CAT, preference should always be given to the simplest valid subdivision that captures all constraints. 

In many CAT problems, route counting is only the first step before applying further constraints. Additional data such as time, capacity or cost may be present in some network questions. Counting routes remains a core skill and is often a first step before applying shortest-path logic, capacity constraints or optimisation.

Solved Examples

Illustration 1

Directions for Questions 1 to 3: Answer these questions based on the pipeline diagram below.

The sketch shows pipelines carrying material between locations. Each location has a demand. The demand at Tumkar is 400, at Bangalore is 400, at Kolar is 700 and at Mysore is 200. Each arrow indicates the direction of material flow through the pipeline. The flow from Tumkar to Bangalore is 300. The quantity of material flow is such that the demands at all these locations are exactly met. The capacity of each pipeline is 1000.

Q1: What is the free capacity available in the Kodagu-Mysore pipeline?
(a) 300
(b) 200
(c) 100
(d) 0

Q2: What is the free capacity available from Kodagu to Tumkar?
(a) 0
(b) 100
(c) 200
(d) 300

Q3: The quantity moved from Kodagu to Mysore is:
(a) 200
(b) 800
(c) 700
(d) 1000

Solution: 

Flow Rule Used: At every location, total inflow equals total outflow plus the demand at that location. 

Since the exact diagram is used to determine flows, we follow the given logical deductions that satisfy all node demands and the known flow Tumkar→Bangalore = 300.

From the given reasoning:

To meet Kolar's demand of 700, the flows into Bangalore and onward must combine appropriately so that the total material routed through the relevant outgoing pipelines meets that demand.

The flow from Tumkar to Bangalore is 300; therefore , Bangalore must receive an additional 800 units from Mysore so that its total outgoing requirement and Kolar's demand are satisfied.

Because the Mysore→Bangalore pipeline carries 800, the pipeline Kodagu→Mysore must supply Mysore sufficiently; that makes Kodagu→Mysore operating at full capacity 1000.

Therefore the free capacity on Kodagu→Mysore is 1000 - 1000 = 0.

Kodagu→Tumkar flow must be 700 to meet requirements downstream; hence the free capacity on Kodagu→Tumkar is 1000 - 700 = 300.

Thus:

• Q1: Free capacity on Kodagu-Mysore is 0. Option (d) is correct.
• Q2: Free capacity on Kodagu-Tumkar is 300. Option (d) is correct.
• Q3: Quantity moved from Kodagu to Mysore is 1000. Option (d) is correct.

Illustration 2

Direction: Seventy percent of the employees in ENY office have IPods, 75 percent have Apple watch, 80 percent have Air purifiers and 85 percent have TV. At least what percentage of employees has all four gadgets?
(a) 15
(b) 5
(c) 10
(d) Cannot be determined

Solution: The least percentage of people with all 4 gadgets would happen if all the employees who are not having any one of the four objects is mutually exclusive.
Thus, 100 - 30 - 25 - 20 - 15 = 10.
Option (c) is correct.

Illustration 3

Directions for Questions: Answer the questions on the basis of the following information.

Shown below is the layout of major streets in a city. Two days (Thursday and Friday) are left for campaigning before a major election, and the city administration has received requests from five political parties to take out their processions along the following routes.

• All India Party: A-C-D-E
• BMP: A-B-D-E
• Bolo Bharat: A-B-C-E
• Manki Baat: B-C-E
• PCM: A-C-D

Street B-D cannot be used for a political procession on Thursday due to a religious procession. The district administration has a policy of not allowing more than one procession to pass along the same street on the same day. However, the administration must allow all parties to take out their processions during these two days.

Q4: All India Party procession can be allowed:
(a) Only on Thursday
(b) Only on Friday
(c) On either day
(d) Only if the religious procession is cancelled.

Solution: The BMP procession must be allowed on Friday - since the BMP procession uses BD and that road is not free for a political procession on Thursday due to the religious procession. Also, we can see that the BMP and All India Party procession cannot be on the same day (since they have a common street usage viz: street DE). Thus, if BMP is going on Friday, All India Party cannot be allowed on Friday. Thus, the All India Party procession would only be allowed on Thursday. Option (a) is correct.

Q5: Which of the following is not true?
(a) All India Party and Bolo Bharat can take out their processions on the same day.
(b) The PCM procession cannot be allowed on Thursday.
(c) The BMP procession can only take place on Friday.
(d) All India Party and Manki Baat can take out their processions on the same day.

Solution: From the previous question, we know that BMP is Friday and All India Party is Thursday. Since street AD is used by both BMP and Bolo Bharat, Bolo Bharat must also be on Thursday. The Manki Baat and Bolo Bharat use the same street CE.
So, they should be on different days and hence, the Manki Baat is on Friday. Also, the PCM should be on Friday as it cannot be on the same day as the All India Party procession. Looking through the options, we have to spot the option that is false. Option (a), (b) and (c) can be seen to be true. Only option (d) is not true as we can see that the All India Party and Manki Baat procession cannot be on the same day. Hence, option(d) is correct.

EduRev Tip: Route and Network sets are best attempted when the diagram can be organised cleanly and the constraints can be handled systematically without excessive case explosion.