Importance of the topic in various exams
The logical reasoning and data interpretation section in several MBA entrances is an unpredictable section with a wide variety of questions. One of the types of questions is the critical path. The logical reasoning and data interpretation section is designed to test the critical thinking skills and decision-making ability of aspirants which is an important requirement for all managers. Such problems also test the analytical, logical, and mathematical ability. A common problem within this section is routes and networks. Critical path concept is important from the point of view of various MBA entrance exams like CAT, ATMA, SNAP and XAT. This concept requires in- depth understanding of networks and relationships- related problems. It also requires correct understanding of the question and making the required connections so that the concept can be broken down into its simplest possible parts and then solved. These types of questions are included in several MBA entrance exams like CAT, XAT, SNAP and ATMA but the frequency of questions is not very high.
Below are the details about the critical path concept for different MBA competitive exams: –
The list of concepts that are covered in the Critical Path Concept is as follows: –
Some of the important CAT questions on Critical Path that appeared in the previous papers are: –
1. A significant amount of traffic flows from point S to point T in the one-way street network shown below. Points A, B, C, and D are junctions in the network, and the arrows mark the direction of traffic flow. The fuel cost in rupees for traveling along a street is indicated by the number adjacent to the arrow representing the street. Motorists traveling from point S to point T would obviously take the route for which the total cost of traveling is the minimum. If two or more routes have the same least travel cost, then motorists are indifferent between them. Hence, the traffic gets evenly distributed among all the least cost routes. The government can control the flow of traffic only by levying an appropriate toll at each junction. For example, if a motorist takes the route S-A-T (using junction A alone), then the total cost of travel would be Rs 14 (i.e. Rs 9 + Rs 5) plus the toll charged at junction A. (CAT 2006)
|(1) 2, 5, 3, 2|
|(2) 0, 5, 3, 1|
|(3) 1, 5, 3, 2|
|(4) 2, 3, 5, 1|
|(5) 1, 3, 5, 1|
(5) 0, 5, 2, 2
(5) 1, 5, 4, 2
First, find which all routes are possible.
SAT, SBCT, SBAT, SDT, SDCT are the routes possible.
Let a, b, c, d be the tolls levied at A, B, C, and D respectively.
So the cost for different routes are:
SAT = 14 + a
SBCT = 7 + b + c
SBAT = 9 + b + a
SDT = 13 + d
SDCT = 10 + d + c.
SBCT route is not possible as it is under repairs.
Total cost of all the other routes should be same.
So, 14 + a = 9 + b + a = 13 + d = 10 + d + c
14 + a = 9 + b + a. So, b = 5
13 + d = 10 + d + c. So, c = 3
14 + a = 13 + d. So, d – a = 1.
There are 2 options with the same answer. Any one is fine.
SDT route is not possible, and total cost of other routes should be same.
So, 14 + a = 7 + b + c = 9 + b + a = 10 + d + c
Solving as above, we get, b = 5.
7 + b + c = 10 + d + c. —-> 7 + b = 10 + d. So, d = 2.
Also, 14 + a = 12 + c (after d = 2 substitution)
So, we have b = 5, d = 2, c – a = 2.
Find the option.
All routes have same cost.
So, 14 + a = 7 + b + c = 9 + b+ a = 13 + d = 10 + d + c.
We get b = 5, d = 2 (like above)
If d = 2, a =1, and finally, we get c = 3.
Equal traffic from S to A,B and D. However, through B&D, there is more than 1 way, from which the commuter can choose. We can restrict them to move through 1 way from B&D if we increase the toll in C. So that passengers move through the route without “C” junction involved. So, the routes involved will be SAT, SBAT, SDT, and the tolls of these should be same.
14 + a = 9 + b + a = 13 + d.
b = 5, d – a = 1. We can keep a = 0 and d = 1. (As the options have value of a = 0/1 only, and if a = 1, d = 2 which is not there in option)
So, we have a = 0, b = 5, d = 1. (We can stop the solution here, as we have only 1 option satisfying this).
If you need to find value of C,
We know that SBAT < SBCT, so, 14 < 12 + c. c >2.
Also, SDT < SDCT, so 14 < 11 + c. So, c > 3.
Hence c > 3.
The least expense route is SBCT, if without toll, cost is 7.
However, if we do not keep toll, 100% traffic will flow through B, which is not possible.
The next highest cost route is SDCT which is 10Rs.
So, if we keep a toll in B for 3rs, then traffic will split 50-50 between SBCT and SDCT, and the total cost of the journey will be 10rs.
2. A new airline company is planning to start operations in a country. The company has identified ten different cities which they plan to connect through their network to start with. The flight duration between any pair of cities will be less than one hour. To start operations, the company has to decide on a schedule. The underlying principle that they are working on is the following:
Any person staying in any of these 10 cities should be able to make a trip to any other city in the morning and should be able to return by the evening of the same day.
Q1) If the underlying principle is to be satisfied in such a way that the journey between any two cities can be performed using only direct (non-stop) flights, then the minimum number of direct flights to be scheduled is:
Q2) Suppose three of the ten cities are to be developed as hubs. A hub is a city which is connected with every other city by direct flights each way, both in the morning as well as in the evening. The only direct flights which will be scheduled are originating and/or terminating in one of the hubs. Then the minimum number of direct flights that need to be scheduled so that the underlying principle of the airline to serve all the ten cities is met without visiting more than one hub during one trip is:
Q3) Suppose the 10 cities are divided into 4 distinct groups G1, G2, G3, G4 having 3, 3, 2 and 2 cities respectively and that G1 consists of cities named A, B and C. Further, suppose that direct flights are allowed only between two cities satisfying one of the following:
1. Both cities are in G1
2. Between A and any city in G2
3. Between B and any city in G3
4. Between C and any city in G4
Then the minimum number of direct flights that satisfies the underlying principle of the airline is: (TITA)
Q4) Suppose the 10 cities are divided into 4 distinct groups G1, G2, G3, G4 having 3, 3, 2 and 2 cities respectively and that G1 consists of cities named A, B and C. Further, suppose that direct flights are allowed only between two cities satisfying one of the following:
1. Both cities are in G1
2. Between A and any city in G2
3. Between B and any city in G3
4. Between C and any city in G4
However, due to operational difficulties at A, it was later decided that the only flights that would operate at A would be those to and from B. Cities in G2 would have to be assigned to G3 or to G4.
What would be the maximum reduction in the number of direct flights as compared to the situation before the operational difficulties arose? (TITA)
(CAT 2017 slot 1)
For any pair of cities, say A and B, to satisfy the underlying principle, there must be a morning flight from A to B, an evening flight from B to A and a morning flight from B to A and an evening flight from A to B.
Only then can a person from A or B travel to B or A and return the same day. Hence, there must be four flights between any pair of cities. Number of ways of selecting two cities from ten cities
= (10*9)/2 = 45
Hence, the minimum number of flights that must be scheduled = 45 ×4 = 180.
Let the ten cities be represented by A – J. Among these ten cities, consider A, B and C to be hubs and the other seven cities to be non-hub cities. It is given that any direct flight should originate or terminate at a hub.
Consider city D, which is not a hub. D is connected to each of A, B and C. Between D and each of A, B and C, there must be four flights (see the above solution). Hence, from D, there must be 4 × 3 = 12 flights to the three hubs, A, B and C. Similarly, for each of the other six non-hub cities, there must be 12 flights connecting each non-hub city with the three hubs. Hence, a total of 12 ×7 = 84 flights will connect a non-hub city with a hub. In addition to this, the three hubs must be connected amongst themselves. Since there must be four flights between any pair of cities, there will be a total of 4 × 3 = 12 flights connecting any pair of hubs.
So, the total minimum number of flights that should be scheduled = 84 + 12 = 96.
Given that G1 has the cities A, B and C. G2, G3 and G4 have 3, 2 and 2 cities respectively.
As per the given conditions, we can see that a city in G2 cannot be connected by a direct flight to a city in G3 or G4. Hence, for a person to travel from a city in G2 to a city in G3 or G4, all the cities in G2 must be connected to A and from A, he can travel to B or C to travel to a city G3 or G4 respectively.
Hence, the 3 cities in G2 must be connected to A. Between each pair of cities there must be four flights. Hence, there must be 4 × 3 = 12 flights between cities in G2 and A. Since there are 2 cities in G3, there must be 2 × 4 = 8 flights between cities in G3 and B. Since there are 2 cities in G4, there must be 2 × 4 = 8 flights between cities in G4 and C. Also, the cities in G1, i.e., A, B and C must be connected to each other. Hence, there must be an additional 4 × 3 = 12 flights between these three cities. Therefore, the total minimum number of direct flights that must be scheduled = 12 + 8 + 8 + 12 = 40
The cities in G2 will be assigned to G3 or G4. However, this, by itself, will not result in any reduction in the number of flights because the cities in G2 will still have to be connected to either B or C. However, it is also given that there are now no flights between A and C. Hence, the 4 flights that would have been scheduled in the previous case, will now not be scheduled. Hence, the reduction in the number of flights can be a maximum of 4.
While practising critical path and network problems the first step is to identify the variables and the number of instances of the variables in the data. Also, identify the respective relationships among the variables. Make the table required and add the data given in the question in the table. Now as the basic framework is ready to identify the required solution and solve it to get the answer based on the question given. MBA entrances are usually speeding tests as much as a test to assess the aspirant’s skills. Ample practice needs to be done to do calculations with speed and do calculations mentally as far as possible. Also, the different tips and tricks should be customized to individual students’ needs as no two students are identical and need their ways to carry out calculations.
After the basic level, the aspirant should solve problems with accuracy and without looking at the solutions. Don’t focus on the time taken to complete the questions but that the questions should be done correctly. Maintain an Excel sheet to monitor the progress in the questions by the student. The questions where difficulty was faced should be revised after a week again and if the difficulty is faced again, then the concept should be revised. For the intermediate level, material from a coaching centre can be referred to as it has graded material with sections divided into different levels of difficulty.
After completing the previous levels, the aspirant should move on to solving sectional tests and giving mocks in a timed environment. This is the most important stage as it will help in preparing for the real exam. The mocks should be given diligently and, in any environment, similar to that of the test centre. A mixture of different levels of difficulty should be attempted through section-wise tests. Attempting too many easy or too many difficult questions is not advised. Taking up a mock test series from a coaching institute will be beneficial. The art of selecting which questions to attempt and which to leave should be instilled at this stage, a very important part of the DILR section. All the questions can’t and shouldn’t be attempted.