Answer:
The wider channel bandwidth is efficient and good as it help in increasing the speed of the transmission. Due to the wider channel of the bandwidth it also lead to consuming less energy of the battery and make cost effective.
Due to the wider bandwidth, the number of clients can easily share and transfer the data in the channel and also make the transmission complete.
If the channel of the bandwidth is double then, the single transmission can easily carry more data. Hence, it also lead to double the speed of the transmission.
The database must be carefully planned to allow useful data manipulation and report generation.
True
False
Answer:
True.
Explanation:
When creating a database it should be carefully planned so that we can easily do data manipulation and report generation if not designed carefully these processes will become difficult.
Some mistakes in the creation of database.
No Normalization.Not treating the data model as a breathing or living organism.Not choosing primary keys carefully.No or little use of foreign Keys and check constrains.MySQL 8 is the first version that supports JSON Objects.
(a) True
(b) False
Answer: False
Explanation:
The given statement is false because MySQL 8 basically support the JSON object since versions 5,7 and 8. The MySQL basically support the native Java script object notation (JSON) data types that is basically define by the RFC 7159 and it can efficiently access the data from the JSON documents.
It provide many advantages like the JSON are easily detect the error in the documents and can validate automatically. It also contain optimize storage format.
Create a float variable named area.
Answer:
float area;
Explanation:
The float datatype is used for storing the decimal number .The syntax of declaring float variable is given below.
float variablename ;
float area;
area=89.900002; // storing the value
following the program in c language
#include<stdio.h> // header file
int main() // main function
{
float area=89.900002; // declared and initialize a float variable
printf("%f",area); // display the area value
return 0;
}
Using Word, write pseudocode to represent the logic of a program that allows the user to enter a value for hours worked in a day. The program calculates the hours worked in a five-day week and the hours worked in a 252-day work year. The program outputs all the results.
Answer:
//Here is PSEUDO CODE.
1. Declare variable "hour_day".
1.1 read the hour worked in a day and assign it to variable "hour_day".
2. Calculate the hours worked in a five day week.
2.1 work_week=5*hour_day.
3. Calculate the hours worked in 252 days year.
3.1 work_year=252*hour_day
4. print the value of work_week and work_year.
5. End the program.
//Here is code in c++.
#include <bits/stdc++.h>
using namespace std;
int main() {
double hour;
cout<<"enter the hours worked in a day:";
cin>>hour;
double work_week=5*hour;
double work_year=252*hour;
cout<<"work in a 5-day week: "<<work_week<<"hours"<<endl;
cout<<"work in a 252-day year: "<<work_year<<"hours"<<endl;
return 0;
}
Output:
enter the hours worked in a day:6.5
work in a 5-day week: 32.5 hours
work in a 252-day year: 1638 hours
Locker doors There are n lockers in a hallway, numbered sequentially from 1 to n. Initially, all the locker doors are closed. You make n passes by the lockers, each time starting with locker #1. On the ith pass, i = 1, 2,...,n, you toggle the door of every ith locker: if the door is closed, you open it; if it is open, you close it. After the last pass, which locker doors are open and which are closed? How many of them are open?
Answer:
// here is code in C++
#include <bits/stdc++.h>
using namespace std;
// main function
int main()
{
// variables
int n,no_open=0;
cout<<"enter the number of lockers:";
// read the number of lockers
cin>>n;
// initialize all lockers with 0, 0 for locked and 1 for open
int lock[n]={};
// toggle the locks
// in each pass toggle every ith lock
// if open close it and vice versa
for(int i=1;i<=n;i++)
{
for(int a=0;a<n;a++)
{
if((a+1)%i==0)
{
if(lock[a]==0)
lock[a]=1;
else if(lock[a]==1)
lock[a]=0;
}
}
}
cout<<"After last pass status of all locks:"<<endl;
// print the status of all locks
for(int x=0;x<n;x++)
{
if(lock[x]==0)
{
cout<<"lock "<<x+1<<" is close."<<endl;
}
else if(lock[x]==1)
{
cout<<"lock "<<x+1<<" is open."<<endl;
// count the open locks
no_open++;
}
}
// print the open locks
cout<<"total open locks are :"<<no_open<<endl;
return 0;
}
Explanation:
First read the number of lockers from user.Create an array of size n, and make all the locks closed.Then run a for loop to toggle locks.In pass i, toggle every ith lock.If lock is open then close it and vice versa.After the last pass print the status of each lock and print count of open locks.
Output:
enter the number of lockers:9
After last pass status of all locks:
lock 1 is open.
lock 2 is close.
lock 3 is close.
lock 4 is open.
lock 5 is close.
lock 6 is close.
lock 7 is close.
lock 8 is close.
lock 9 is open.
total open locks are :3
. What is automated testing?
Answer: Automated testing is testing of the software by the means of automation by using special tools .These software tools are responsible for the testing of the execution process , functioning and comparing the achieved result with actual result.
It is helps in achieving the accuracy as compared with manual testing and also also function without any human interruption.It produces the test scripts that can be re-used.
Which of these is not used by analysts when adopting CASE tools? (1 point) (Points : 1.5) communicating more effectively with users
expediting the local area network
increasing productivity
integrating the work done during life cycle stages
Answer: expediting the local area network
Explanation: Case tools are the tools or service that helps in the software development but the automatic activity. This tool can be applied on various applications like mobile apps, E-commerce websites etc.
Analyst focus on the features like improving productivity ,low time consumption,integration of the work and making the communication better with the users ,helps in producing large amount of the documents etc.
The only factor not accepted by analyst is accomplishing the operation in the LAN(local area network) as it does not facilitates this operation because it is not software development process.
Write an algorithm that asks a user to enter a number between 1 and 10. (This range includes the numbers 1 and 10.) When they enter the number, check that it is actually between 1 and 10. If it is not, ask them to enter a number again. Continue to ask until they enter a valid number. Once their number is valid, output the number. (C++ form)
Answer:
do{
cout<<"Introduce number \n"; //print the message
cin>>num; //set the value of the number given
}while(num<1 || num>10); //repeat while the number is out of the range
cout<<"Number: "<<num; //print the number
Explanation:
The idea behind this code is to create a loop in which I can compare the number given (between 1 and 10) and then print the number or get back and ask the number again.
#include <iostream>
using namespace std;
int main()
{
int num; //create num variable
do
{
cout<<"Introduce number \n"; //print the message
cin>>num; //set the value of the number given
}while(num<1 || num>10); //repeat while the number is out of the range
cout<<"Number: "<<num; //print the number
}
True or False: It is the CISO❝s responsibility to ensure that InfoSec functions are performed within an organization. A) True B) False
Answer: True
Explanation: CISO(Chief information security officer) is the officer that is responsible for the information security(Infosec) function to be implemented in the organization.He/she manages the procedures, protect from the threats in business, secures the enterprise communication.
The post of CISO is considered as a senior level officer .The responsibilities of the CISO also includes getting along with other section of the organization and assuring about the smooth and secured processing of the business work as the aspect of information security.Thus the statement given is true
Different organizations implement different Information Systems base on their core business operations. explain
Answer:
Different business or firm tend to enforce different Information Systems based completely on their main business operations, in order to best leverage data as an organizations asset. Some of these Information System are as follow:
a) Transaction Processing System (TPS):
A small organization tends to process transactions that might result from day-to-day activities, such as purchase orders, creation of paychecks and thus require using TPS.
b) Management Information System(MIS):
Managers and owners of small organizations tend to incline towards industry-specific MIS, in order to get historical and current operational data, such as inventories data and sales.
c) Decision Support System (DSS):
A DSS to allow managers and owners of small organizations to use predefined report in order to support problem-resolution decisions and operations planning.
The sum of the binary numbers 0111+1001= 10000 which is equivalent to ________ in HEX numeration system.
A
10
F
D
Answer:
10
Explanation:
Adding two binary numbers 0111 and 1001 results in the sum as 10000.
The equivalent hex representation for the number is obtained by splitting the result into two quads and finding their individual hex representations.
00010000
Quad 1 : 0001 = 1 (HEX)
Quad 2 : 0000 = 0 (HEX)
So the equivalent result is obtained by concatenating the two digits one after the other : 10 (in Hex)
Describe the process of normalization and why it is needed.
Answer: Normalization is known as the process under which one organizes a database in order to improve data integrity and reduce redundancy. It also tends to streamlines database design in order to achieve optimal structure made up of the basic elements.
It is also referred to as data normalization, it is considered as a vital part of database design, since it helps with accuracy, speed and efficiency of database. By doing so, one can arrange data into columns and tables.
A(n) ________ is an object that is generated in memory as the result of an error or an unexpected event.
a.)exception
b.)error message
c.)default exception handler
d.)exception handler
Answer: a)Exception
Explanation: Exception is the occurrence of an situation in the computer system field due to which interruption is caused in the execution of the program. A object is created in this event known and the exception object that contains the information about the error that has occurred.
The exception is handled by the exception handler.It catches the exception during the execution or after run time. Other options are incorrect because they are not the unexpected event to disorder the flow of program.Thus the correct option is option(a).
Write an if/else statement that assigns 0 to x when y is equal to 10; otherwise it should assign
1 to x.
Answer:
if(y==10)
{
x=0; // assigning 0 to x if y equals 10.
}
else
{
x=1; // assigning 1 to x otherwise.
}
Explanation:
In the if statement i have used equal operator == which returns true if value to it's right is equal to value to it's left otherwise false.By using this operator checking value of y and if it is 10 assigning 0 to x and if it is false assigning 1 to x.
Write a C++ program where Two or more strings are anagrams if they contain exactly the same letters, ignoring capitalization, punctuation, and spaces. For example, "Information superhighway" and "New utopia? Horrifying sham" are anagrams, as are "ab123ab" and "%%b b*aa". Note that two lines that contain no letters are anagrams.
Answer:
/*
Problem; to finf s2 strings are anagrams.
(s2 strings are anagrams if they contain exactly the same letters, ignoring capitalization, punctuation,)
and spaces.
Solution: we must take account wich letters are present in each string and how many times. To achieve that we
may create a struct with a char and a char count, some constructors, both geters and an equal operator.
*/
#include <iostream>
#include <vector>
using namespace std;
struct charCount {
private: // Internal data structure:
char letter; // letter (should be unique)
unsigned int letterCount; // and is present at least 1 time
public:
charCount() {} // Default Constructor
charCount(char l) {letter = l; letterCount = 1;} // Constructor
charCount(const charCount & other) {letter = other.letter;
letterCount = other.letterCount;} // Copy Constructor
void increments() {letterCount++;} // Counts other occurrence
char ltr() {return(letter);} // Geter
unsigned int ltrCount() {return(letterCount);} // Geter
bool operator ==(charCount other) {return(letter == other.letter // Equal means same char and
&& letterCount == other.letterCount);}// same count.
};
/*
We have to analyze each input string returning an ordered vector of charCount elements:
*/
vector<charCount> analysis(string s) { // Raw string data
vector<charCount> result; // Result vector
for(char c:s) { // For each char of string: we take a-z
if(isalpha(c)) { // only into account and we normalize it
char x = tolower(c); // to lowercase, store as x and we are
bool found = false; // ready to search
unsigned int pos = 0; // from first position to last
for(auto & y:result) { // for each vector's element
if(y.ltr() == x) { // if we find x, we must
y.increments(); // count this occurrence,
found = true; // turn on the found flag
break; // and exit from the loop.
} else if(y.ltr() < x) { // Otherwise may be we are not going to
result.emplace(result.begin()+(pos),(charCount(x))); // find it in the ordered vector: we insert
found = true; // it at current vector's position. It is
break; // present now: we should exit from loop.
} else pos++; // Otherwise we increment pos and go on.
}
if(!found) { // x doesn't belongs to vector's set: we
auto it = result.end(); // add it at last position, keeping order
result.insert(it,(charCount(x))); // and we may go on for another cycle if any.
}
}
}
return result;
}
template <typename T> // Generic code follows to implement
bool operator ==(vector<T> v1,vector<T> v2) { // 2 vectors comparison: they are
bool result = (v1.size() == v2.size()); // equals if their sizes match,
if(result) // and each element from first vector
for(unsigned int i = 0; i < v1.size();i++) // matches with each element from the
result = result && v1[i] == v2[i]; // second vector at the same position.
return(result);
}
template bool operator ==(vector<charCount>,vector<charCount>); // Orders compiler to generate a charCount version.
bool areAnagrams(string s1,string s2) { // Anagrams implies 2 identical analysis vectors.
return(analysis(s1) == analysis(s2));}
int main(int argc, char *argv[]) { // Test code folloes:
string s1,s2; // 2 strings to store input data.
cout << "Text 1:"; getline(cin,s1); // Reads string 1 till <Enter> key.
cout << "Text 2:"; getline(cin,s2); // Reads string 2 till <Enter> key.
cout << (areAnagrams(s1,s2)?"Are":"Are not") << " anagrams..." << endl; // Are they? We use the ternary operator.
return 0;
}
Explanation: included inside the code as comments.
Answer:
#include <iostream>
using namespace std;
void areAnagrams(char str1[], char str2[])
{
int i, flag = 1, count1[26] = {0}, count2[26] = {0};
for(i = 0; str1[i] != '\0'; i++) {
if (isalpha(str1[i])) {
str1[i] = tolower(str1[i]);
count1[str1[i] - 'a']++;
}
}
for(i = 0; str2[i] != '\0'; i++) {
if (isalpha(str2[i])) {
str2[i] = tolower(str2[i]);
count2[str2[i] - 'a']++;
}
}
for(i = 0; i < 26; i++) {
if (count1[i] != count2[i])
flag = 0;
}
if (flag == 0)
cout << "The two strings are NOT anagrams";
else
cout << "The two strings are anagrams";
}
int main ()
{
char str1[100], str2[100];
cout << "Enter the first string: ";
gets(str1);
cout << "Enter the second string: ";
gets(str2);
areAnagrams(str1, str2);
return 0;
}
Explanation:
Inside the function:
- In the first loop, each character of the first string is checked if it is a letter. If it is, lower the character and increment its frequency by 1.
- In the second loop, each character of the second string is checked if it is a letter. If it is, lower the character and increment its frequency by 1.
- In the third for loop, compare the frequencies
- If they are equal, that means the strings are anagrams
Which of the following is false? (Points : 4)
All elements of the array have the same data type.
All elements in an array have the same name.
All elements in an array have the same subscript.
The first element of the array has a subscript of zero.
Answer:
All elements in an array have the same subscript.
Explanation:
The statement written above is false because every element in an array has different subscript or index.The index starts with 0 and ends at size -1 .You can access the array element if you write the subscript in the square brackets after the name of the array.
a[5];
You can access 6th element in the array by this statement.
which of the following is NOT part of the Ethernet standards? O 1.802.3 O 2.802.2 O 3 LLC O4 pPp
Answer: 2) 802.2
Explanation: Ethernet standard are the standard for the networking technology for the transmission of the data.The ethernet is requires for the node connection and then forming the framework for the transmission of information.There are several standards for Ethernet.
802.3 is Ethernet standard that works on 10 Mbps rate, PPP(point to point protocol) is the standard that connects the internet service provider using modem and LLC(logical link control) is a standard for 802 .There is no 802.2 standard present in the ethernet standard.Thus,the correct option is option(2).
What is wrong with the following declaration? How should it be fixed?
public static final int myNumber = 17.36;
Answer:
We have used int datatype which is wrong in the given declaration.
The correct declaration is public static final double myNumber = 17.36;
Explanation:
In this declaration the value is stored in "myNumber" variable is 17.36 which is decimal point number to store the decimal point number we have to use the double datatype, not int datatype because int datatype can store only integer value not the decimal value. So we have to use the double datatype instead of int type.
The correct syntax for the following declaration is :
public static final double myNumber = 17.36;
What is the purpose and what are some of the components of a wide area network connection?
Answer:
WAN(wide area network) is the network that is used for the connection in the geographical areas on large scale.They are used for connecting the areas like cities, countries, states etc.The connection of various LANs(Local area network) and MAN(Metropolitan area network) form the WAN.
There are several components that are used in the wide area network structure. Some of the constituents are as follows:-
ATM()Asynchronous transfer modeFiber optic communication pathModem having cablesFrame relayDial up internet etc.
Answer:
A wide area network (WAN) is a network of devices, local area networks (LANs), or metropolitan area networks (MANs) that are connected over wired or wireless communication lines. A virtual private network (VPN) connects different WAN sites. These connections span large geographic areas and can occur between cities, regions and even countries. A WAN is often used by large companies looking to manage and share information and resources between branches. This type of network is mainly used as a way to connect smaller LANs and, although they have slower data transmission capabilities from LANs, they are able to connect a wider coverage area.
Here are some components that are frequently found in a network:
Servers: These are computers that serve some service, that is, they exist only in order to provide resources for computers to do work.Client: Computers that access resources offered by the network through servers.Feature: Anything that can be offered such as printers, files, internet access, etc.Protocol: For computers to communicate, there must be a common language. Protocol is the medium that computers can communicate with.Cabling: Data is trafficked through physical devices, and the definition of cabling is the medium in which that data is trafficked as fiber optics or wireless networks.______A computer program may only be constructed in a high level language. (T/F)
Answer:
False.
Explanation:
A computer program can not only constructed in a high level language but also in low level language such as assembly language used for microprocessor for ex-8085 microprocessor,8086 microprocessor ,ARM assembly language and there are many more.
High level languages are programmer friendly they are easily understood by the programmers because they uses english words while low level languages are difficult because they uses binary language 0's or 1's.
Hence the answer to this question is False.
What value will be stored in the variable t after each of the following statements
executes?
A) t = (12 > 1); __________
B) t = (2 < 0); __________
C) t = (5 == (3 * 2)); __________
D) t = (5 == 5); __________
Final answer:
The variable t will store true for statements A and D because the comparisons are correct, and false for statements B and C because the comparisons are incorrect.
Explanation:
The question is asking about the values that will be stored in the variable t after each given boolean (true or false) statement is executed in a programming context. Here are the answers:
A) t = (12 > 1); The statement is true because 12 is greater than 1, so t will store true.
B) t = (2 < 0); The statement is false because 2 is not less than 0, so t will store false.
C) t = (5 == (3 * 2)); The statement is false because 5 is not equal to 6 (3 times 2), so t will store false.
D) t = (5 == 5); The statement is true because 5 is equal to 5, so t will store true.
Explain the RISC and CISC architecture. Comparison of RISC and CISC in detail.
Answer:RISC(reduced instruction set computer) is the computer used for low level operation by simple command splited into numerous instructions in single clock only and CISC(Complex instruction set computer) is the computer that performs the operation in single instruction.
RISC architecture has hardwired control unit,data cache unit,data path ,instruction cache and main memory as components .CISC architecture persist of control unit,micro program control memory, cache, instruction and data path and main memory.
The differences between RISC and CISC are as follows:-
The instruction in RISC are less and and low complexes while CISC has several number of complex instruction.The calculation done by RISC are precise and quick whereas CISC has slightly slow calculation processing. The execution of RISC is faster as compared to CISC.RISC uses hardware component for implementation of instruction because it does not persist its own memory and CISC implements instructions using its own memory unit .RISC (Reduced Instruction Set Computing) architectures use simpler instructions for enhanced speed and efficiency, while CISC (Complex Instruction Set Computing) architectures use complex instructions to reduce program instruction counts. The comparison highlights differences in execution speed, pipeline efficiency, memory usage, and design philosophy. Each architecture offers unique benefits depending on application requirements.
RISC (Reduced Instruction Set Computing)
RISC architectures are designed to execute a small and highly optimized set of instructions. This simplicity enables faster instruction execution because each instruction is executed in a single clock cycle. Key features of RISC include:
Fewer, simpler instructions.Uniform instruction size, usually 32-bit.Large number of general-purpose registers.Load/store architecture, where data operations are performed in registers and not directly in memory.Emphasis on hardware simplification and pipeline efficiency.An example of a RISC architecture is the MIPS (Microprocessor without Interlocked Pipelined Stages) architecture, which uses 32-bit instructions and multiple instruction formats, adheres to the principles of simplicity and efficiency.
CISC (Complex Instruction Set Computing)
CISC architectures, on the other hand, aim to reduce the number of instructions per program by using complex and multifunctional instructions. These instructions can execute multiple operations or address modes within a single instruction. Key features of CISC include:
A large set of instructions, sometimes numbering in the hundreds.Variable instruction length, allowing for more complex operations.Fewer general-purpose registers.Direct memory access for data manipulation.Emphasis on instruction count reduction and powerful, versatile instruction sets.An example of a CISC architecture is the Intel x86 family, widely used in desktop and server processors.
Comparison of RISC and CISC
To compare both architectures:
Simplicity vs Complexity: RISC uses simpler instructions and fewer of them, while CISC uses more complex and multi-operation instructions.Execution Speed: RISC aims for single-cycle execution per instruction, whereas CISC instructions can take multiple cycles but accomplish more per instruction.Pipeline Efficiency: RISC's simplicity enhances pipeline efficiency and minimizes bottlenecks, while CISC's complexity can lead to pipeline stalls and increased latency.Memory Usage: RISC relies heavily on register usage and minimizes memory access, whereas CISC might have more direct memory operations.Instruction Sets: RISC has fewer, more general registers, while CISC has specialized, powerful instructions catering to a wider range of tasks.Design Philosophy: RISC focuses on hardware efficiency and speed, following the principle that 'Smaller is faster'. In contrast, CISC targets ease of programming and reduced software complexity.Data mining and __________ tools are used to find relationships that are not obvious, or to predict what is going to happen.
a. DSS
b. reporting
c. predictive analytic
d. queries
Answer:c)Predictive analytic
Explanation: Predictive analytics is the tool that is used for the prediction about the future situations.It is the advanced branch of the analytics where the present information considered for prediction The techniques followed by the predictive analytics are data mining, modelling etc.
Other option are incorrect because DSS(decision support system) is the system that helps in making decision in business field, reporting shows the low efficiency or performance and queries are the question or objection that arises regarding a concept.Thus the correct option is option(c).
Compute the bitwise XOR between 10010100 and 11101010.
Answer:
The resultants bits for two strings are 0111 1110
Explanation:
Given details:
two pair of strings are
10010100
11101010
comparing two pair of strings, then perform XOR operation
Rules to follow for XOR
If both the bits are same then resultant is zero 0 and if bits are different resultant is one
XOR operation:
10010100
11 1 01010
-------------
01111110
The resultants bits for two strings are 0111 1110
Convert decimal number 126.375 to binary (3 bits after binary point)
Answer:
126.375 in binary is: 1111110.011
Explanation:
In order to convert a decimal number to binary number system, the integral part is converted using the division and remainder method while the fractional part is multiplied with 2 and the integral part of answer is noted down. The fractional part is again multiplied with 2 and so on.
For 126.375
2 126
2 63 - 0
2 31 - 1
2 15 - 1
2 7 - 1
2 3 - 1
1 - 1
So, 126 = 1111110
For 0.375
0.375 * 2 = 0.75
0.75 * 2 = 1.5
0.5 * 2 = 1.0
As we had to only find 3 digits after binary point, so
0.375 = 011
So 126.375 in binary is: 1111110.011 ..
Distinguish between 802.3 standards and 802.11 standards.
Explanation:
Following are differences between 802.3 and 802.11 standards:-
The 802.3 standards are for Ethernet standards while the 802.11 standards are for wireless LAN.Ether networks uses standards at data link layer and physical layer while Wireless LAN is implemented by a set of specification of physical layer and media access control.The products of 802.11 are certified by Wi-Fi logo if and only if when they are upto certain standards of interperability but it is less secured while 802.3 products are constantly improving and are highly secured.Final answer:
IEEE 802.3 and 802.11 standards differ primarily in their networking methods; 802.3 (Ethernet) is used for wired connections, while 802.11 (Wi-Fi) is used for wireless networking.
Explanation:
The key distinction between IEEE 802.3 standards and IEEE 802.11 standards lies in the methods and mediums they use for transmitting data. The IEEE 802.3 standard, often recognized as Ethernet, is mainly used for wired networking and employs physical cables (typically copper or fiber optic) to transfer data between computers. Ethernet has been the foundational standard for wired local area networks (LANs) and has evolved over time to support higher speeds and different cabling types.
On the other hand, the IEEE 802.11 standard is centered around wireless networking, commonly referred to as Wi-Fi. This standard allows devices to connect to a LAN and access the internet without the use of physical cables. Wi-Fi has become widely used in private homes, offices, and public hotspots due to its convenience in establishing network connections. The standard includes various versions (such as a, b, g, n, ac, and others) which differ in aspects such as transmission speed, range, and frequency used.
While Ethernet provides a reliable and secure connection typically required for business environments and data centers, Wi-Fi offers mobility and ease of access, which are essential for portable devices and consumer electronics. Both standards are fundamental to modern telecommunications and networking but serve different purposes and environments.
. What are the technical advantages of optical fibres over other communications media?
Answer: The advantage of the optical fibre over the other medium of the communication are as follows:-
The bandwidth of the optical fibre is comparatively higher than the other communication media The security in regard with the loss of data can be tracked easily as compared with other copper wire medium of communication.Thus, the security system of optical fibre is more reliable.The capacity of the fibre and the transmission power is also high as compared with other media of the communication such as copper wires,etc.The optical fibre is less space consuming than other communication medium.Optical fibres have significant technical advantages including low loss, allowing for longer distances without signal amplification, high bandwidth supporting more data transmission, and reduced crosstalk for clearer communication over traditional copper cables.
Explanation:Optical fibres offer significant advantages over traditional communication media like copper cables, especially for long-distance communications. One of the primary technical advantages of optical fibres is their low loss, allowing light to travel many kilometers before needing amplification, which is much superior to the performance of copper conductors. Another critical advantage is high bandwidth, where lasers emit light that enables far more conversations in one fiber than what electric signals on copper cables can support. Additionally, optical fibres exhibit reduced crosstalk, meaning optical signals in one fiber do not produce undesirable effects in other adjacent fibers, ensuring clearer and more reliable communication.
explain the concept of scalability. How would you respond?
Answer:
In computer science, scalability is the ability of an application or product that could be software or software to continue responding well when its context is increased in volume or size to meet the user demand. Typically, when a product becomes aging or the context demands a rescaling to a larger size or volume, if the product is prepared to respond to this demand, then it's called a scalable product.
How does the MAN (Metropolitan Area Network) fit into the networking model between a LAN (Local Area Network) and a WAN (Wide Area Network)?
Answer: Local area network(LAN) is the network works in the small area to create networking pattern such as an individual building. Wide area network(WAN) is the networking technology that covers a large geographical area to connect them such as any particular company.
MAN (metropolitan area network) is the area that falls in between the category of LAN and WAN.It is the networking pattern that can work in the geographical areas like towns,cities etc.WAN technology consist of both LAN and MAN networks to make a huge unit of network.Thus, MAN is the middle networking technology as it covers more area than LAN and Less area than WAN.
How many bits are in an IP address?
Answer: IPv4(Internet protocol version 4)) =32 bits
IPv6(Internet protocol version 6) =128 bits
Explanation: IP(Internet protocol)address are the unique addresses that is provided to the hardware components that are attached with the internet service. The identification of the hardware devices are made on the basis of IP address .It is a numerical notation .
The most commonly used version of the IP addresses are IPv4 (Internet protocol version 4)and IPv6(Internet protocol version 6) that persist of 32 bits and 128 bits address respectively. IPv4 and IPv6 has different bits due to the growth of the internet facility and improvement in the version.Most commonly used internet service at current time is IPv6.