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enot [183]
2 years ago
7

A regional transportation and logistics company recently hired its first ChiefInformation Security Officer (CISO). The CISO’s fi

rst project after onboardinginvolved performing a vulnerability assessment against the company’s publicfacing network. The completed scan found a legacy collaboration platformapplication with a critically rated vulnerability. While discussing this issue with theline of business, the CISO learns the vulnerable application cannot be updatedwithout the company incurring significant losses due to downtime or newsoftware purchases.
Required:
What BEST addresses these concerns?
Computers and Technology
1 answer:
Alja [10]2 years ago
5 0

There are different ways to create solutions to problem. What Best addresses these concerns is that;

  • The company should implement a WAF in front of the vulnerable application to filter out any traffic attempting to exploit the vulnerability.

<h3>How do firms use a WAF to protect their applications? </h3>
  • A WAF is known to be often used by firms to protects their web apps through the use of filtering, monitoring, and hindering (blocking)any forms of malicious HTTP/S traffic moving to the web application.

They often prevents any unauthorized data from leaving the app. This WF will be a useful approach to addresses these concerns in the above scenario.

See options below

Which of the following BEST addresses these concerns?

A. The company should plan future maintenance windows such legacy application can be updated as needed.

B. The CISO must accept the risk of the legacy application, as the cost of replacing the application greatly exceeds the risk to the company.

C. The company should implement a WAF in front of the vulnerable application to filter out any traffic attempting to exploit the vulnerability.

D. The company should build a parallel system and perform a cutover from the old application to the new application, with less downtime than an upgrade.

Learn more about vulnerability assessment from

brainly.com/question/25633298

You might be interested in
You work in a classified environment where Bell LaPadula MLS (Multilevel Security) model is employed. Your clearance is "SECRET"
kumpel [21]

Answer:

The answer is by using a covert channel like shared memory objects such as files, directories,messages, etc since both  the user and the sender of the document are on same network of the company.

Explanation:

The Bell LaPadula MultiLevel Security model was a security policy developed by Bell and LaPadula in 1973 in response to a security issue raised by the US Air Force regarding file-sharing mainframe computers . Actually, many people with networked systems have realized by early 1970s that the protection purportedly offered by many commercial operating systems was poor, and wa not getting better any time soon. This was observed when it was noticed that as one operating system error was fixed, some other vulnerability would be discovered. There was also the constant worry that various unskilled users would discover loopholes in the operating system during usage and use them to their own advantage.

 Information release may take place via shared memory objects such as files, directories, messages, and so on. Thus, a Trojan Horse acting on behalf of a user could release user-private information using legitimate operating system requests. Although developers can build various mechanisms within an operating system to restrict the activity of programs (and Trojan Horses) operating on behalf of a user  , there is no general way, short of implementing nondiscretionary policy models, to restrict the activity of such programs. Thus, given that discretionary models cannot prevent the release of sensitive information through legitimate program activity, it is not meaningful to consider how these programs might release information illicitly by using covert channels.

For example, for someone with higher integrity level (SECRET) to send an accounts payable application to a user, if the untrusted accounts payable application contains a Trojan Horse, the Trojan Horse program could send a (legal) message to the said user process running at a lower integrity level (CONFIDENTIAL), thereby initiating the use of a covert channel. In this covert channel, the Trojan Horse is the receiver of (illegal) lower integrity-level input and the user process is the sender of this input.

7 0
3 years ago
Tortise and hare race java g Modify the main class so it runs the race 100 times and reports how many times each runner wins. (T
SOVA2 [1]

Answer:

Game.java file

import java.util.Scanner;

public class Game {

/**

* t_pos and h_pos are the current positions of the Tortoise and Hare

*/

static int t_pos,h_pos;

static Tortoise tortoise;

static Hare hare;

public static void main(String[] args) {

play(); /*starting the game*/

}

public static void play(){

/**

* the method will starts the play, loop until the game is over, displays the winner

* and prompts the user if they want to play again

*/

/**

* defining Tortoise and Hare objects

*/

tortoise=new Tortoise();

hare=new Hare();

t_pos=1;

h_pos=1;

System.out.println("The race is about to start");

tortoise.printTrack();

hare.printTrack();

while(t_pos != 50 && h_pos !=50){

System.out.println("\n\n\n"); /*printing blank lines*/

t_pos=tortoise.move(); /*moving and getting the current position of tortoise*/

h_pos=hare.move();/*moving and getting the current position of hare*/

tortoise.printTrack(); /*displaying the tracks*/

hare.printTrack();

try { /*comment this part to skip the 1s break between each round; for testing*/

Thread.sleep(1000);

} catch (InterruptedException e) {

e.printStackTrace();

}

}

System.out.println("\nRace Over");

if(t_pos==50 && h_pos==50){

System.out.println("Its a tie");

}

else if(t_pos==50){

System.out.println("Tortoise wins");

}else if(h_pos==50){

System.out.println("Hare wins");

}

System.out.println("Do you want to play again? (y/n)");

Scanner scanner=new Scanner(System.in);

String ch=scanner.next();

if(ch.equalsIgnoreCase("y")){

play();

}else if(ch.equalsIgnoreCase("n")){

System.out.println("Thanks for playing, Goodbye");

}else{

System.out.println("Invalid choice, quitting..");

}

}

}

//Tortoise.java

public class Tortoise {

/**

* the current position of the tortoise

*/

int position;

/**

* track array

*/

char[] track;

/**

* speed of tortoise

*/

int speed=1;

public Tortoise() {

position=0;

track=new char[50];

for(int i=0;i<track.length;i++){

/**

* filling the track

*/

track[i]='-';

}

}

public int move(){

if(position<track.length){

position=position+speed;

}

return position+1;

}

public void printTrack(){

/**

* the current position of tortoise will be displayed by 'T' everything else will be '-'

*/

System.out.println();

for(int i=0;i<track.length;i++){

if(i==position){

System.out.print('T');

}else{

System.out.print(track[i]);

}

}

}

}

//Hare.java

import java.util.Random;

public class Hare {

int position;

int speed=10;

char[] track;

/**

* Random object to generate a random number

*/

Random random;

/**

* resting percent denotes how much time Hare will be resting

*/

int resting_percent=90;

public Hare() {

position=0;

track=new char[50];

for(int i=0;i<track.length;i++){

track[i]='-';

}

random=new Random();

}

public int move(){

int n=random.nextInt(100-1)+1; /*generating a random number between 1 and 100*/

if(n<=resting_percent){

/**

* at rest; will not move, returns the current position.

*/

return position;

}else{

/**

* not resting..

*/

if(position<track.length){

if(position+speed>=track.length){

position=track.length-1;

}else{

position=position+speed;

}

}

return position+1;

}

}

public void printTrack(){

System.out.println();

for(int i=0;i<track.length;i++){

if(i==position){

System.out.print('H');

}else{

System.out.print(track[i]);

}

}

}

}

/*Output (partial and random)*/

The race is about to start

T-------------------------------------------------

H-------------------------------------------------

-T------------------------------------------------

H-------------------------------------------------

--T-----------------------------------------------

H-------------------------------------------------

---T----------------------------------------------

----------H---------------------------------------

.

.

.

.

-----------------------------------------------T--

-------------------------------------------------H

Race Over

Hare wins

Do you want to play again? (y/n)

y

The race is about to start

T-------------------------------------------------

H-------------------------------------------------

-T------------------------------------------------

H-------------------------------------------------

.

.

.

.

-----------------------------------------------T--

--------------------H-----------------------------

------------------------------------------------T-

--------------------H-----------------------------

-------------------------------------------------T

--------------------H-----------------------------

Race Over

Tortoise wins

Do you want to play again? (y/n)

n

Explanation:

3 0
3 years ago
What happens if part of an ftp message is not delivered to the destination?
Oliga [24]

The message is lost when an FTP message is not delivered to its destination because FTP doesn't use a reliable delivery method.

<h3>What is FTP?</h3>

FTP is an abbreviation for file transfer protocol and it can be defined as a type of server that's designed and developed to store and provide files for download, as well as sharing between two or more users on an active computer network.

Generally, the message is lost when an FTP message is not delivered to its destination because FTP doesn't use a reliable delivery method.

Read more on FTP here: brainly.com/question/20602197

#SPJ12

4 0
2 years ago
Whats important about the points?
likoan [24]
You can earn ranks! That is important.
8 0
3 years ago
Read 2 more answers
Byte pair encoding is a data encoding technique. The encoding algorithm looks for pairs of characters that appear in the string
nika2105 [10]

Answer:

The encoding algorithm looks for pairs of characters that appear in the string more than once and replaces each instance of that pair with a corresponding character that does not appear in the string. ... Byte pair encoding is an example of a lossy transformation because it discards some of the data in the original string.

Explanation:

hope it helps!!

6 0
2 years ago
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