Showing posts with label GATE. Show all posts
Showing posts with label GATE. Show all posts
Monday, March 28, 2022
GATE Questions - DBMS - File Structures and Indexing
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Previous GATE questions with solutions on DBMS (File Structures and Indexing) - CS/IT
GATE-2013
1. An index is clustered, if
(a) it is on a set of fields that form a candidate key.
(b) it is on a set of fields that include the primary key.
(c) the data records of the file are organized in the same order as the data entries of the index.
(d) the data records of the file are organized not in the same order as the data entries of the index.
Ans: option (c)
Explanation :
With a clustered index the physical records (rows) are stored on the disk in the same order as that of index. Hence every table can have exactly one clustered index.
GATE-2013
1. An index is clustered, if
(a) it is on a set of fields that form a candidate key.
(b) it is on a set of fields that include the primary key.
(c) the data records of the file are organized in the same order as the data entries of the index.
(d) the data records of the file are organized not in the same order as the data entries of the index.
Ans: option (c)
Explanation :
With a clustered index the physical records (rows) are stored on the disk in the same order as that of index. Hence every table can have exactly one clustered index.
GATE - 2015
2. With reference to the B+ tree index of order 1 shown below, the minimum number of nodes (including the root node) that must be fetched in order to satisfy the following query: “Get all records with a search key greater than or equal to 7 and less than 15” is ________.
2. With reference to the B+ tree index of order 1 shown below, the minimum number of nodes (including the root node) that must be fetched in order to satisfy the following query: “Get all records with a search key greater than or equal to 7 and less than 15” is ________.
Ans: 5
Explanation:
In a B+ tree, data pointers are stored only at the leaf nodes. The leaf nodes contain the value of the search field as well as a pointer to the record/block that contains the record.In order to provide ordered access, the leaf nodes have been linked.
Explanation:
In a B+ tree, data pointers are stored only at the leaf nodes. The leaf nodes contain the value of the search field as well as a pointer to the record/block that contains the record.In order to provide ordered access, the leaf nodes have been linked.
In the above question, in order to fetch all records with key values greater than or equal to 7 and less than 15, we first need to locate the leaf node that stores 7.To reach the node with key 7,we start from root with key value 9,follow left pointer to arrive at a node with key 5 (since search key 7 is less than 9). From node with key 5, we follow the right pointer to reach the leaf node with key value 7. Now we move from 7 till 15 sequentially towards right, thus traversing another 2 nodes.
Thus total number of nodes fetched in order to satisfy the query is 5.
Thus total number of nodes fetched in order to satisfy the query is 5.
GATE - 2015
3. A file is organized so that the ordering of data records is the same as or close to the ordering of data entries in some index. Then that index is called
(a) Dense
(b) Sparse
(c) Clustered
(d) Unclustered
Ans: option (c)
Explanation:
With a clustered index the physical records (rows) are stored on the disk in the same order as that of index . Hence every table can have exactly one clustered index.
3. A file is organized so that the ordering of data records is the same as or close to the ordering of data entries in some index. Then that index is called
(a) Dense
(b) Sparse
(c) Clustered
(d) Unclustered
Ans: option (c)
Explanation:
With a clustered index the physical records (rows) are stored on the disk in the same order as that of index . Hence every table can have exactly one clustered index.
Sunday, April 15, 2018
GATE-Computer Networks-Flow Control
Previous GATE questions with solutions on Computer Networks (Flow Control) - CS/IT
GATE -2015
1. Since it is a network that uses switch, every packet goes through two links, one from source to switch and other from switch to destination.
Since there are 10000 bits and packet size is 5000, two packets are sent. Transmission time for each packet is 5000 / 107 seconds
Two hosts are connected via a packet switch with 107 bits per second links. Each link has a propagation delay of 20 microseconds. The switch begins forwarding a packet 35 microseconds after it receives the same. If 10000 bits of data are to be transmitted between the two hosts using a packet size of 5000 bits, the time elapsed between the transmission of the first bit of data and the reception of the last bit of the data in microseconds is _________.
(a) 1075 (b) 1575 (c) 2220 (d) 2200
Ans: option (b)
Explanation:
It is given that there are 10,000 bits and since the packet size id 5000 it means we have 2 packets to send.
Transmission time for one packet = 5000 / 107 seconds = 500μs
Transmission time is the time taken to transmit a packet from host to the outgoing link.
It is also given that the propagation delay of links is 20μs. Propagation delay is the time taken by a bit to reach from sender to receiver (in this case from sender to switch it is 20μs and from switch to receiver it 20μs)
=500μs + 20μs = 520μs
Once P1 reaches the switch, the switch will take 35μs to process the packet and then it takes 500μs to transmit it to the link and then the packet will take 20μs to reach the receiver.
Therefore Time taken by P1 to reach from switch to receiver = 35μs + 500μs+ 20μs = 555μs
Therefore time taken by P1 to reach from sender to receiver = 520μs+555μs = 1075μs
But we need to note that after 520μs the switch starts receiving second packet (P2).
i.e. At 520μs+500μs = 1020μs P2 is completely received by switch.
Now Time taken by P2 to reach from switch to receiver = 35μs + 500μs+ 20μs = 555μs
It means that at 1575μs (1020μs+555μs) P2 reaches the destination.
Gate-2015
2. Suppose that the stop-and-wait protocol is used on a link with a bit rate of 64 kilobits per second and 20 milliseconds propagation delay. Assume that the transmission time for the acknowledgement and the processing time at nodes are negligible. Then the minimum frame size in bytes to achieve a link utilization of at least 50% is_________________.
(a) 160 (b) 320 (c) 640 (d) 220
Ans: option (b)
Explanation:
Since the link utilization should be atleast 50% it means that efficiency, η≥ 50%
Since it is mentioned that it is a stop&wait protocol, the efficiency of the link can be calculated as below:
η = 1 /( 1 + 2a)
a = Tp/Tt (where Tp = Propagation delay & Tt = Transmission Time)
Lets see the length of the packet to achieve a link utilization of 50%
50/100 = 1 /( 1 + 2a)
1/2 = 1 /( 1 + 2a)
a = 1/2
Tp/Tt = 1/2
20/Tt=1/2
Tt = 40ms
Tt = transmission time = L/B (where L = length of packet and B = bandwidth)
L/B = 40
L =40 * B = 40 ms * 64 Kbps = 40*10-3 *64*103 = 2560 bits
Since we need to determine the frame size in bytes L = 2560/8 = 320 bytes.
Therefore, the minimum frame size in bytes to achieve a link utilization of at least 50% is 320bytes
Wednesday, December 9, 2015
GATE Questions - C Programming
Previous GATE questions with solutions on C Programming - CS/IT
GATE-2000
1. The number of tokens in the following C statement.
(a) 3 (b) 26 (c) 10 (d) 21
Ans: option (c)
Explanation:
The smallest individual units are known as C Tokens. The keywords, identifiers, constants, string literals, and operators described in this section are examples of tokens. Punctuation characters such as brackets ([ ]), braces ({ }), parentheses ( ( ) ), and commas (,) are also tokens.
GATE-2000
1. The number of tokens in the following C statement.
printf("i = %d, &i = %x", i, &i);
is(a) 3 (b) 26 (c) 10 (d) 21
Ans: option (c)
Explanation:
The smallest individual units are known as C Tokens. The keywords, identifiers, constants, string literals, and operators described in this section are examples of tokens. Punctuation characters such as brackets ([ ]), braces ({ }), parentheses ( ( ) ), and commas (,) are also tokens.
Tuesday, December 8, 2015
GATE Questions-DBMS-Functional Dependency
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Previous GATE questions with solutions on DBMS (Functional Dependency) - CS/IT
GATE-1999
1. Let R= (A, B, C, D, E, F) be a relation scheme with the following dependencies: C->F, E->A, EC->D, A->B. Which of the following is a key for R?
(a) CD (b) EC (c) AE (d) AC
Ans: option (b)
Explanation:
Find the closure set of all the options given. If any closure covers all the attributes of the relation R then that is the key.
Algorithm to find Closure Set
Step1: Equate an attribute or attributes to X for which closure needs to be identified.
Step2: Take each FD (functional dependency) one by one and check whether the left side of FD is available in X, if yes then add the right side attributes to X if it is not available.
Step3: Repeat step 2 as many times as possible to cover all FD's.
Step4: After no more attributes can be added to X declare it as the closure set.
FDs: C->F, E->A, EC->D, A->B
Find closure set for CD.
X = CD
= CDF {C->F}
No more attributes can be added to X. Hence closure set of CD = CDF
Find closure set for EC.
X = EC
= ECF {C->F}
= ECFA {E->A}
= ECFAD {EC->D}
= ECFADB {A->B}
Closure set of EC covers all the attributes of the relation R.
GATE-1999
1. Let R= (A, B, C, D, E, F) be a relation scheme with the following dependencies: C->F, E->A, EC->D, A->B. Which of the following is a key for R?
(a) CD (b) EC (c) AE (d) AC
Ans: option (b)
Explanation:
Find the closure set of all the options given. If any closure covers all the attributes of the relation R then that is the key.
Algorithm to find Closure Set
Step1: Equate an attribute or attributes to X for which closure needs to be identified.
Step2: Take each FD (functional dependency) one by one and check whether the left side of FD is available in X, if yes then add the right side attributes to X if it is not available.
Step3: Repeat step 2 as many times as possible to cover all FD's.
Step4: After no more attributes can be added to X declare it as the closure set.
FDs: C->F, E->A, EC->D, A->B
Find closure set for CD.
X = CD
= CDF {C->F}
No more attributes can be added to X. Hence closure set of CD = CDF
Find closure set for EC.
X = EC
= ECF {C->F}
= ECFA {E->A}
= ECFAD {EC->D}
= ECFADB {A->B}
Closure set of EC covers all the attributes of the relation R.
Monday, December 7, 2015
Tricky concepts of Operators in C Programming Language
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GATE
We assume that you know the basics of C Programming.
1.
int c=10,a=5,d;
d=a=c;
After the execution of above statement d will have the value 10.
Explanation: Equality operator associates Right to Left, i.e. in the above statement first c is assigned to a.
The expression a=c evaluates to the value of a after the assignment takes place. Then this value is assigned to d.
1.
int c=10,a=5,d;
d=a=c;
After the execution of above statement d will have the value 10.
Explanation: Equality operator associates Right to Left, i.e. in the above statement first c is assigned to a.
The expression a=c evaluates to the value of a after the assignment takes place. Then this value is assigned to d.
Saturday, December 5, 2015
GATE 2015 Questions - DBMS
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GATE-2015
1. Consider the following transaction involving two bank
accounts x and y.
read (x) ; x := x – 50; write (x); read (y); y := y + 50; write (y)
The constraint that the sum of the accounts x and y should remain constant isthat of
(a) Atomicity
(b) Consistency
(c) Isolation
(d) Durability
(a) Atomicity
(b) Consistency
(c) Isolation
(d) Durability
Answer: option (b)
Wednesday, November 4, 2015
GATE Questions - Web technologies
Previous GATE questions with solutions on Web Technologies - CS/IT
GATE - 2004
1. Which one of the following statements is false?
(a) HTTP runs over TCP
(b) HTTP describes the structure of web pages
(c) HTTP allows information to be stored in a URL
(d) HTTP can be used to test the validity of a hypertext link
Ans: option (b)
Explanation:
HTTP (hypertext transfer protocol) is an application layer protocol used for accessing web pages. tt uses TCP/IP at the transport and network layer.
GATE - 2004
1. Which one of the following statements is false?
(a) HTTP runs over TCP
(b) HTTP describes the structure of web pages
(c) HTTP allows information to be stored in a URL
(d) HTTP can be used to test the validity of a hypertext link
Ans: option (b)
Explanation:
HTTP (hypertext transfer protocol) is an application layer protocol used for accessing web pages. tt uses TCP/IP at the transport and network layer.
Tuesday, January 22, 2013
GATE Questions - OS - Synchronization
Previous GATE questions with solutions on Operating Systems (Synchronization) - CS/IT
GATE-1996
1. A critical section is a program segment
(a) which should run in a certain specified amount of time
(b) which avoids deadlocks
(c) where shared resources are accessed
(d) which must be enclosed by a pair of semaphore operations, P and V
Ans: option (c)
GATE-1996
1. A critical section is a program segment
(a) which should run in a certain specified amount of time
(b) which avoids deadlocks
(c) where shared resources are accessed
(d) which must be enclosed by a pair of semaphore operations, P and V
Ans: option (c)
Monday, January 21, 2013
GATE - OS - Process Concepts
Previous GATE questions with solutions on Operating Systems (Process Concepts) - CS/IT
GATE-1996
1. The process state transition diagram in Fig.1.8 is representative of
(a) a batch operating system
(b) an operating system with a preemptive scheduler
(c) an operating system with a non-preemptive scheduler
(d) a uni-programmed operating system.
Ans: option (b)
Explanation:
If there is a transition from Running State to ready State then the state diagram is the representative of an operating system with a preemptive scheduler.
GATE-1996
1. The process state transition diagram in Fig.1.8 is representative of
(a) a batch operating system
(b) an operating system with a preemptive scheduler
(c) an operating system with a non-preemptive scheduler
(d) a uni-programmed operating system.
Ans: option (b)
Explanation:
If there is a transition from Running State to ready State then the state diagram is the representative of an operating system with a preemptive scheduler.
Sunday, January 20, 2013
GATE Questions - OS - CPU Scheduling
Previous GATE questions with solutions on Operating Systems (CPU Scheduling) - CS/IT
GATE-1995
1. Which scheduling policy is most suitable for a time-shared operating systems?
(a) Shortest Job First (b) Round Robin
(c) First Come First Server (d) Elevator
GATE-1995
1. Which scheduling policy is most suitable for a time-shared operating systems?
(a) Shortest Job First (b) Round Robin
(c) First Come First Server (d) Elevator
GATE Questions - DBMS - Normalization
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Previous GATE questions with solutions on DBMS (Normalization) - CS/IT
Normalisation is a technique for producing relational schema with no information redundancy and no anomalies (insert, update or delete anomalies)
GATE-1998
1. Which normal form is considered adequate for normal relational database design?
(a) 2NF (b) 5NF (c) 4NF (d) 3NF
Ans: option (d)
Explanation:
Normalisation is a technique for producing relational schema with no information redundancy and no anomalies (insert, update or delete anomalies)
GATE-1998
1. Which normal form is considered adequate for normal relational database design?
(a) 2NF (b) 5NF (c) 4NF (d) 3NF
Ans: option (d)
Explanation:
A relational database table is often described as "normalized" if it is in the Third Normal Form because most of the 3NF tables are free of insertion, update, and deletion anomalies.
Friday, January 18, 2013
GATE Questions-DBMS-ER Diagram
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Previous GATE questions with solutions on DBMS (ER Diagram) - CS/IT
GATE-2008
Linked Answer Questions 1 & 2
1. Consider the following ER diagram.
The minimum number of tables needed to represent M, N, P, R1, R2 is
(a) 2 (b) 3 (c) 4 (d) 5
Ans: confused between option (a) & option (b) { majority of the people refer option (a) as correct }
Explanation:
for option(a)
check explanation of question (2) of this link http://www.geeksforgeeks.org/database-management-systems-set-7/
for option (b)
All strong entities and weak entities will be converted into a table. Therefore we will have 3 tables:
M (M1,M2,M3,P1)
P (P1,P2)
N (N1,N2,P1) =>N is a weak entity and it is modified to include the primary key of P (i.e. P1).
GATE-2008
Linked Answer Questions 1 & 2
1. Consider the following ER diagram.
(a) 2 (b) 3 (c) 4 (d) 5
Ans: confused between option (a) & option (b) { majority of the people refer option (a) as correct }
Explanation:
for option(a)
check explanation of question (2) of this link http://www.geeksforgeeks.org/database-management-systems-set-7/
for option (b)
All strong entities and weak entities will be converted into a table. Therefore we will have 3 tables:
M (M1,M2,M3,P1)
P (P1,P2)
N (N1,N2,P1) =>N is a weak entity and it is modified to include the primary key of P (i.e. P1).
GATE Question - DBMS - SQL
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Previous GATE questions with solutions on DBMS (SQL) - CS/IT
GATE-2010
1. A relational schema for a train reservation database is given below.
Passenger (pid, pname, age)
Reservation (pid, class, tid)
What pids are returned by the following SQL query for the above instance of the tables?
Ans: option (c)
Explanation:
GATE-2010
1. A relational schema for a train reservation database is given below.
Passenger (pid, pname, age)
Reservation (pid, class, tid)
Table: Passenger
|
Table : Reservation
|
What pids are returned by the following SQL query for the above instance of the tables?
SELECT pid
FROM Reservation
WHERE class ‘AC’ AND
EXISTS (SELECT *
FROM Passenger
WHERE age > 65 AND
Passenger. pid = Reservation.pid)
(a) 1, 0 (b) 1, 2 (c) 1, 3 (d) 1, 5Ans: option (c)
Explanation:
The above query is an example of synchronized subquery or correlated subquery. A correlated sub-query is a sub-query that uses values from the outer query. The sub-query is evaluated once for each row processed by the outer query.
Monday, January 7, 2013
GATE-Digital Logic-Number System
Previous GATE questions with solutions on Digital Logic (Number System) - CS/IT
GATE-1999
1. Booth’s coding in 8 bits for the decimal number –57 is
(a) 0–100+1000
(b) 0–100+100-1
(c) 0–1+100–10+1
(d) 00–10+100-1
Ans: option (b)
Explanation:
Booth's Encoding:
GATE-1999
1. Booth’s coding in 8 bits for the decimal number –57 is
(a) 0–100+1000
(b) 0–100+100-1
(c) 0–1+100–10+1
(d) 00–10+100-1
Ans: option (b)
Explanation:
Booth's Encoding:
1. If ith bit bi is 0 and (i –1)th
bit bi-1 is 1, then take bi as +1
2. If ith bit bi is 1 and (i –1)th
bit bi-1 is 0, then take bi as –1
3. If ith bit bi is 0 and (i –1)th
bit bi-1 is 0, then take bi as 0
4. If ith bit bi is 1 and (i –1)th
bit bi-1 is 1, then take bi as 0
Note: When lsb b0= 1, assume that it had b-1
as 0, thus take b0 = –1
Friday, January 4, 2013
GATE Questions-Data Structures-Hashing
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Previous GATE questions with solutions on Data Structures (Hashing) - CS/IT
GATE-2010
1. A hash table of length 10 uses open addressing with hash function h(k)=k mod 10, and linear probing. After inserting 6 values into an empty hash table, the table is as shown below.
0
|
|
1
|
|
2
|
42
|
3
|
23
|
4
|
34
|
5
|
52
|
6
|
46
|
7
|
33
|
8
|
|
9
|
|
10
|
Which one of the following choices gives a possible order in which the key values could have been inserted in the table?
(a) 46, 42, 34, 52, 23, 33
(b) 34, 42, 23, 52, 33, 46
(c) 46, 34, 42, 23, 52, 33
(d) 42, 46, 33, 23, 34, 52
Ans: option (c)
Explanation:
Linear Probing - The location where the key value should be stored in the table is determined by the hash function. Here hash function is k mod 10. For e.g., if we need to store 42 then find 42 mod 10, then we get the value 2, that means the key value 42 should be stored in the table index 2. When there is a collision, linearly come down to locate next empty location and store the value in that location.
For the above question, only option (d) will give you the above hash table. Hash table for all options are given below:
Thursday, January 3, 2013
GATE Questions-Data Structures-Trees
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Previous GATE questions with solutions on Data Structures (Trees) - CS/IT
GATE-1995
1. A binary tree T has n leaf nodes. The number of nodes of degree 2 in T is
(a) log2n (b) n-1 (c) n (d) 2n
GATE-1995
1. A binary tree T has n leaf nodes. The number of nodes of degree 2 in T is
(a) log2n (b) n-1 (c) n (d) 2n
Ans: option(b)
Explanation:
A binary tree is a tree data structure in which each node has at most two child nodes.
The number of subtrees of a node is called the degree of the node. In a binary tree, all nodes have degree 0, 1, or 2.
The degree of a tree is the maximum degree of a node in the tree. A binary tree is of degree 2.
A binary tree is a tree data structure in which each node has at most two child nodes.
The number of subtrees of a node is called the degree of the node. In a binary tree, all nodes have degree 0, 1, or 2.
The degree of a tree is the maximum degree of a node in the tree. A binary tree is of degree 2.
GATE Questions-Data Structure- Arrays
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GATE-Data Structures
Previous GATE questions with solutions on Data Structures (Arrays) - CS/IT
GATE-2005
(a) An array of 50 numbers
(b) An array of 100 numbers
(c) An array of 500 numbers
(d) A dynamically allocated array of 550 numbers
Ans: Option (a)
Explanation:
GATE-2005
1. A program P reads in 500 integers in the range [0,100] representing the scores of 500 students. It then prints the frequency of each score above 50. what would be the best way for P to store the frequencies?
(a) An array of 50 numbers
(b) An array of 100 numbers
(c) An array of 500 numbers
(d) A dynamically allocated array of 550 numbers
Ans: Option (a)
Explanation:
We have to store frequencies of scores above 50. That is number of students having score 51, number of students having score 52 and so on. For that an array of size 50 is the best option.
Monday, December 31, 2012
GATE-Computer Networks-TCP/IP
Previous GATE questions with solutions on Computer Networks (TCP/IP) - CS/IT
GATE-2013
1. The transport layer protocols used for real time multimedia, file transfer, DNS and email, respectively are
(a) TCP, UDP, UDP and TCP
(b) UDP, TCP, TCP and UDP
(c) UDP, TCP, UDP and TCP
(d) TCP, UDP, TCP and UDP
Ans: option(c)
Explanation:
Real Time Multimedia: Data packets should be delivered faster. Also it can be unreliable. Therefore UDP.
File Transfer: For example downloading a file. It should be secure and reliable. Therefore TCP.
DNS: uses both UDP and TCP for its transport. But to acheive efficiency DNS uses UDP. To start a TCP connection a minimum of three packets are required (SYN out, SYN+ACK back, ACK out). UDP uses a simple transmission model with a minimum of protocol mechanism. UDP has no handshaking dialogues.
Email: uses SMTP protocol which uses TCP protocol.
GATE-2013
1. The transport layer protocols used for real time multimedia, file transfer, DNS and email, respectively are
(a) TCP, UDP, UDP and TCP
(b) UDP, TCP, TCP and UDP
(c) UDP, TCP, UDP and TCP
(d) TCP, UDP, TCP and UDP
Ans: option(c)
Explanation:
Real Time Multimedia: Data packets should be delivered faster. Also it can be unreliable. Therefore UDP.
File Transfer: For example downloading a file. It should be secure and reliable. Therefore TCP.
DNS: uses both UDP and TCP for its transport. But to acheive efficiency DNS uses UDP. To start a TCP connection a minimum of three packets are required (SYN out, SYN+ACK back, ACK out). UDP uses a simple transmission model with a minimum of protocol mechanism. UDP has no handshaking dialogues.
Email: uses SMTP protocol which uses TCP protocol.
Thursday, December 27, 2012
GATE-Computer Networks - Token Ring
Previous GATE questions with solutions on Computer Networks (Token Ring) - CS/IT
GATE - 2007
1. In a token ring network the transmission speed is 107 bps and the propagation speed is 200 metres/ms. The 1-bit delay in this network is equivalent to:
(a) 500 metres of cable (b) 200 metres of cable
(c) 20 metres of cable (d) 50 metres of cable
Ans: option (c)
Explanation:
Transmission delay for 1 bit t = 1/(107) = 0.1 micro seconds.
Given that propagation speed = 200 meters/ms
That means, 200 meters can be traveled in 1 micro second.
Hence in 0.1 micro seconds, 20 meters can be traveled.
OR
1 Bit Delay = 1/B = 1/(107) = 0.1 micro seconds.
Given that propagation speed = 200 meters/ms
That means, 200 meters can be traveled in 1 micro second.
Hence in 0.1 micro seconds, 20 meters can be traveled.
GATE - 2007
1. In a token ring network the transmission speed is 107 bps and the propagation speed is 200 metres/ms. The 1-bit delay in this network is equivalent to:
(a) 500 metres of cable (b) 200 metres of cable
(c) 20 metres of cable (d) 50 metres of cable
Ans: option (c)
Explanation:
Transmission delay for 1 bit t = 1/(107) = 0.1 micro seconds.
Given that propagation speed = 200 meters/ms
That means, 200 meters can be traveled in 1 micro second.
Hence in 0.1 micro seconds, 20 meters can be traveled.
OR
1 Bit Delay = 1/B = 1/(107) = 0.1 micro seconds.
Given that propagation speed = 200 meters/ms
That means, 200 meters can be traveled in 1 micro second.
Hence in 0.1 micro seconds, 20 meters can be traveled.
Gate - Computer Networks - Ethernet
Previous GATE questions with solutions on Computer Networks (Ethernet) - CS/IT
GATE-2005
1. In a network of LANs connected by bridges, packets are sent from one LAN to another through intermediate bridges. Since more than one path may exist between two LANs, packets may have to be routed through multiple bridges. Why is the spanning tree algorithm used for bridge-routing?
(a) For shortest path routing between LANsGATE-2005
1. In a network of LANs connected by bridges, packets are sent from one LAN to another through intermediate bridges. Since more than one path may exist between two LANs, packets may have to be routed through multiple bridges. Why is the spanning tree algorithm used for bridge-routing?
(b) For avoiding loops in the routing paths
(c) For fault tolerance
(d) For minimizing collisions
Ans: Option (b)
Explanation:
The Spanning Tree Protocol is used by OSI Data Link Layer devices to create a spanning tree using the existing links (as the source graph) in order to avoid avoiding loops in the routing paths.
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