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2 years ago

How serious are the risks to your computer security?

1 answer:

6 0

The seriousness of the risks to your computer security is not to be a severe one. This is because Computer security risks are due to the handwork of malware such as, bad software, that can infect a computer, and make the hacker to destroy your files, steal your data, or even have access to your system without one's knowledge or authorization.

<h3>What are the risk results for information and computer security?</h3>

The term “information security risk” is known to be those damage that occurs due to an attacks against IT systems. IT risk is made up of a wide range of potential events, such as data breaches, regulatory enforcement actions, financial costs, and a lot more.

Some Examples of malware are viruses, worms, ransomware, spyware, and a lot others.

Hence, The seriousness of the risks to your computer security is not to be a severe one. This is because Computer security risks are due to the handwork of malware such as, bad software, that can infect a computer, and make the hacker to destroy your files, steal your data, or even have access to your system without one's knowledge or authorization.

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Disk requests come in to the disk driver for cylinders 10, 22, 20, 2, 40, 6, and 38, in that order. A seek takes 6 msec per cyli

oksano4ka [1.4K]

Answer:

For FCFS = 876msec For SSTF = 360msec For SCAN(elevator) = 348msec

Explanation:

Considering FCFS algorithm.

In FCFS, the requests are addressed in the order they arrive in the disk queue, this means that the number of cylinders traveled becomes equal to the total of disk requests. With the arm initially at 20, the first request is to read cylinder 10.

Therefore the cylinders traversed for the first request = 20 – 10 = 10

For the second request i.e. a movement from cylinder 10 to cylinder 22, the number of cylinders traversed is = 22 - 10 = 12.

Similarly, for the third request seek arm will return to 20 from 22 so, cylinders traversed through would be = 22-20 = 2.

Also for the fourth request, cylinders traversed would be = 20 – 2 = 18.

For the fifth request, cylinders traversed = 40 – 2 = 38.

Now for the sixth request cylinders traversed = 40 – 6 = 34.

For the seventh and last request, cylinders traversed = 38 – 6 = 32.

So now to get the how much seek time is required for Disk scheduling algorithm

First we would add the total cylinders traversed = 10 + 12 + 2+ 18+ 38 + 34 + 32

= 146 cylinders

So therefore the total seek time = number of cylinders traversed X seek time per cylinder

= 146 X 6

= 876msec

Considering SSTF algorithm.

In SSTF (Shortest Seek Time First), requests having shortest seek time are executed first. So, the seek time of every request is calculated in advance in the queue and then they are scheduled according to their calculated seek time. What this means is that the closest disk (cylinder) next to the position of the seek arm is attended to first. With the arm at 20 initially, the first request is to read cylinder 22 (i.e. the closest cylinder to the seek arm)

Therefore the cylinders traversed for the first request = 22-20 = 2

For the second request, the disk to focus on is disk 20 and the cylinders traversed = 22-20 = 2

Similarly for the third request the seek arm will move to 10 from 20 so, cylinder traversed = 20-10 =10

For fourth request, cylinder traversed = 10 – 6 = 4

For the fifth request, cylinder traversed = 6 – 2 = 4

For sixth request, since all other disk request closer to it has been attended to the seek arm will move to disk 38 to attend to that disk request So, the cylinder traversed = 38 – 2 = 36

For the last request, cylinder traversed = 40 -38 = 2

So now to get the how much seek time is required for Disk scheduling algorithm

First we would add the total cylinders traversed = 2 + 2 +10 + 4 + 4 + 36 + 2

= 60 cylinders

So therefore the total seek time = number of cylinders traversed X seek time per cylinder

= 60 X 6 = 360msec

From Here we see that SSTF is better or an improvement to FCFS as it decrease the average response time (Average Response time is the response time of the all requests).

Considering SCAN (elevator) algorithm

In SCAN algorithm the disk arm moves into a particular direction and services the requests coming in its path and after reaching the end of disk, it reverses its direction and again services the request arriving in its path. So, this algorithm works as an elevator and hence also known as elevator algorithm. Therefore the number of cylinder traveled becomes equal to the total of disk request. With the arm at 20 initially

The first request is to read cylinder 22 i.e. the first cylinder on the upward movement

Therefore the cylinders traversed would be = 20 – 22 = 2

For the second request is to read cylinder 38, and the cylinders traversed would be = 38 – 22 =16

For the third request, seek arm will move to 40 So, the cylinders traversed would be = 40 – 38 = 2

For the fourth request, seek arm will return to 20 since from 40 since 40 is the highest in this upward elevator movement So, cylinders traversed would be = 40 -20 = 20

For the fifth request, cylinder traversed would be = 20 – 10 = 10

For the sixth request, cylinder traversed would be = 10 – 6 = 4

For the seventh and last request, cylinder traversed = 6 – 2 = 4

So now to get the how much seek time is required for Disk scheduling algorithm

First we would add the total cylinders traversed = 2 + 16 + 2 + 20 +10+ 4 + 4 = 58 cylinders

So therefore the total seek time = number of cylinders traversed X seek time per cylinder

= 58 X 6

= 348msec

From Here we see that SCAN is better or an improvement to FCFS and SSTF as it decrease the average response time (Average Response time is the response time of the all requests).

6 0

3 years ago

Select the correct answer. Larry finds it easy to run legacy programs and applications in a virtualized environment. How does th

muminat

The answer is B: installs and runs different versions of an operating system on the same computing device

Virtual machines are like guest operating systems or application environments that give you the perception that you are physically connected to a different box. A system virtual machine, for instance, copies all the components and processes of a real computer. It helps the user experiment with different operating systems and applications without the need to install them on your physical hardware.

8 0

3 years ago

Write a program to assign and display a string value to any variable of your choice

vazorg [7]

Answer:

How to create a string and assign it to a variable. To create a string, put the sequence of characters inside either single quotes, double quotes, or triple quotes and then assign it to a variable.

6 0

2 years ago

Optimizing a PC’s performance often involves utilities that can defragment disks, clean a system’s Registry, scan for malware, r

KATRIN_1 [288]

Answer:

a. True

Explanation:

PC's sometimes runs slowly and on the long run cost money. Optimizing performance involves running software utilities that could restore speed and efficiency.

Scanning for malware/ virus , cleaning system registry, uninstalling unused program, deleting unnecessary files, disk defragment , updating of windows software and setting up the browser are ways to optimize your pc's performance.

3 0

3 years ago

A security administrator is reviewing the following information from a file that was found on a compromised host: Which of the f

Roman55 [17]

Answer:

C. Trojan

Explanation:

In Cybersecurity, vulnerability can be defined as any weakness, flaw or defect found in a software application or network and are exploitable by an attacker or hacker to gain an unauthorized access or privileges to sensitive data in a computer system.

This ultimately implies that, vulnerability in a network avail attackers or any threat agent the opportunity to leverage on the flaws, errors, weaknesses or defects found in order to compromise the security of the network.

In this scenario, a security administrator is reviewing the following information from a file that was found on a compromised host: "cat suspiciousfile.txt."

Some of the ways to prevent vulnerability in a network are;

1. Ensure you use a very strong password with complexity through the use of alphanumerics.

2. You should use a two-way authentication service.

3. You should use encrypting software applications or services.

8 0

2 years ago