What is an IP address and where I can find the IP address for my computer?

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.

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

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4 0

2 years ago

Given six memory partitions of 100 MB, 170 MB, 40 MB, 205 MB, 300 MB, and 185 MB (in order), how would the first-fit, best-fit,

nlexa [21]

Answer:

We have six memory partitions, let label them:

100MB (F1), 170MB (F2), 40MB (F3), 205MB (F4), 300MB (F5) and 185MB (F6).

We also have six processes, let label them:

200MB (P1), 15MB (P2), 185MB (P3), 75MB (P4), 175MB (P5) and 80MB (P6).

Using First-fit

P1 will be allocated to F4. Therefore, F4 will have a remaining space of 5MB from (205 - 200).
P2 will be allocated to F1. Therefore, F1 will have a remaining space of 85MB from (100 - 15).
P3 will be allocated F5. Therefore, F5 will have a remaining space of 115MB from (300 - 185).
P4 will be allocated to the remaining space of F1. Since F1 has a remaining space of 85MB, if P4 is assigned there, the remaining space of F1 will be 10MB from (85 - 75).
P5 will be allocated to F6. Therefore, F6 will have a remaining space of 10MB from (185 - 175).
P6 will be allocated to F2. Therefore, F2 will have a remaining space of 90MB from (170 - 80).

The remaining free space while using First-fit include: F1 having 10MB, F2 having 90MB, F3 having 40MB as it was not use at all, F4 having 5MB, F5 having 115MB and F6 having 10MB.

Using Best-fit

P1 will be allocated to F4. Therefore, F4 will have a remaining space of 5MB from (205 - 200).
P2 will be allocated to F3. Therefore, F3 will have a remaining space of 25MB from (40 - 15).
P3 will be allocated to F6. Therefore, F6 will have no remaining space as it is entirely occupied by P3.
P4 will be allocated to F1. Therefore, F1 will have a remaining space of of 25MB from (100 - 75).
P5 will be allocated to F5. Therefore, F5 will have a remaining space of 125MB from (300 - 175).
P6 will be allocated to the part of the remaining space of F5. Therefore, F5 will have a remaining space of 45MB from (125 - 80).

The remaining free space while using Best-fit include: F1 having 25MB, F2 having 170MB as it was not use at all, F3 having 25MB, F4 having 5MB, F5 having 45MB and F6 having no space remaining.

Using Worst-fit

P1 will be allocated to F5. Therefore, F5 will have a remaining space of 100MB from (300 - 200).
P2 will be allocated to F4. Therefore, F4 will have a remaining space of 190MB from (205 - 15).
P3 will be allocated to part of F4 remaining space. Therefore, F4 will have a remaining space of 5MB from (190 - 185).
P4 will be allocated to F6. Therefore, the remaining space of F6 will be 110MB from (185 - 75).
P5 will not be allocated to any of the available space because none can contain it.
P6 will be allocated to F2. Therefore, F2 will have a remaining space of 90MB from (170 - 80).

The remaining free space while using Worst-fit include: F1 having 100MB, F2 having 90MB, F3 having 40MB, F4 having 5MB, F5 having 100MB and F6 having 110MB.

Explanation:

First-fit allocate process to the very first available memory that can contain the process.

Best-fit allocate process to the memory that exactly contain the process while trying to minimize creation of smaller partition that might lead to wastage.

Worst-fit allocate process to the largest available memory.

From the answer given; best-fit perform well as all process are allocated to memory and it reduces wastage in the form of smaller partition. Worst-fit is indeed the worst as some process could not be assigned to any memory partition.

8 0

3 years ago

You are given a class named Clock that has one int instance variable called hours.

Vlad [161]

Answer:

public Clock(int hours) {

this.hours = hours;

}

Explanation:

In Java programming language, Constructors are special methods that are called to initialize the variables of a class when a new object of the class is created with the new keyword. Consider the complete code for the class below;

public class Clock {

private int hours;

public Clock(int hours) {

this.hours = hours;

}

}

In this example above, an object of this class can created with this statement Clock myclock = new Clock(6); This is a call to the constructor and passes a parameter (6) for hours

7 0

3 years ago

The security administrator of ABC needs to permit Internet traffic in the host 10.0.0.2 and UDP traffic in the host 10.0.0.3. Al

koban [17]

Answer:

The correct selection is the letter C. The first ACL is denying all TCP traffic and the other ACLs are being ignored by the router.

Explanation:

In this case, the letter C is the right answer because with the first ACL exactly:

access-list 102 deny tcp any any

We are denying all traffic with the next line deny tcp any any, in this case, the others line are being ignored.

access-list 104 permit udp host 10.0.0.3 any

access-list 110 permit tcp host 10.0.0.2 eq www any

access-list 108 permit tcp any eq ftp any

For that nobody can access to the internet, the security administrator of ABC must change the first ACL.

5 0

3 years ago

What are some reasons DNS is necessary? Check all that apply. A. It maps local addresses to simple names without editing hosts f

dsp73

Domain Name Servers (DNS) are certainly necessary for easy access of resources across a network. The applicable options of the above are A and B - below are explanations as to why.

A: Computers generally are set to automatically obtain DNS information from the network they are connected to or can be pointed to a specific DNS server. This allows for records of where resources (network attach storage devices, other computers on local network, or even website server details) are located on a “master” kind of list so that the local machine’s host file does not have to be routinely updated to contain new addresses.

B: DNS, as explained partially by the answer to A, maintains a type-able or “human readable” domain name for the actual server’s IP address so we don’t have to memorize or keep a list of IPs for where we want to visit on the web (although, Google’s 8.8.8.8 IP address does make it easy). It shows an association between a name/domain name and an IP address so that we can enter something simple (Google.com) and the computer knows where to go (the server at IP address 8.8.8.8) so it can show you the content you want to see.

C: DNS would only simplify remote access if your were attempting LAN (Local Area Network) remote access of another computer on your network. DNS would not make it easier for remote access of a computer on the internet, as most DNS used in non-commercial settings are created and maintained by third-parties that will not put a specific record for one of your computers in it - not that you would want them to either, since it could lead to an open cyber attack.

D: Network throughput is a fancy phrase for network speed. It could be possible that different DNS servers could process requests faster than others, but it is not likely to increase network speed on the whole as navigating via IP or DNS records will be relatively the same speed.

3 0

3 years ago