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oee [108]
1 year ago
9

In step 4 of the CSMA/CA protocol, a station that successfully transmits a frame begins the CSMA/CA protocol for a second frame

at step 2, rather than at step 1. What rationale might the designers of CSMA/CA have had in mind by having such a station not transmit the second frame immediately (if the channel is sensed idle?
Computers and Technology
1 answer:
DerKrebs [107]1 year ago
5 0

Answer:

There could be a collision if a hidden node problem occurs.

Explanation:

CSMA/CA(carrier sense multiple access/ collision avoidance) is a multiple access method in wireless networking, that allows multiple node to transmit. Collision avoidance of this method is based on preventing signal loss or downtime as a result of collision of transmitting multi signals.

If a node at step 4(transmit frame) sends the first frame, the node still needs to send a RTS(request to send) and to receive a Clear to send (CTS) from the WAP, the is to mitigate the issue of hidden node problem as all frame are treated as unit, so other listening nodes, not detected would seek to connect and transmit as well.

You might be interested in
As part of the duties of a digital forensics examiner, creating an investigation plan is a standard practice. Write a 3 to 4 (no
Simora [160]

Answer:

A digital forensic investigation is a special case of a digital investigation. Where the procedure and techniques are used will allow the results to be entered into cost of low foreg on investigation may be started to answer a question about whether or not centraband digital images exist on a computer.

Here we are considering the case of Global Finance Company with wide range of financial products and wide range of customers throughout the world. A suspect of compromise has been detected from the manager's computer. Now the team has been deployed to the branch office and conduct the Digital Forensic Investigation.

Concern of the Company

1. Regular updates for application infrastructure and network infrastructure.

2. One branch managers from porisbane branch felt compromises in his computer.

3. Both the servers and work station from all the offices are based on Microsoft Windows.

4. The firewalls and network segmentation are fully implemented.

5. Through intrasion detection and logging exist in the brances these are hardly used.

Digital Forensic Investigation Approach

The audit term of the Global Finacnce Company can follow four step. The digital forensic investigation model stands to be most effective model for investigation of the compromise happened int he reginal branch of the Global Finance Company.

1. Collection

a. All information from the manager's workstation, servers and other workstations must be collected.

b. Obtain all the important informtation.

c. Identify storage context noth internal and external devices.

d. Forensic tools that are applicable and to be used for the investigation are to be listed and made available for usage.

e. Target computer forensic imaging to be done and hashed to check the integrity of data.

f. Line network traffic has to captured.

Digital Evidance Collection done in two stages:

Volatile memory is the temporary memory and primary volatile memory is RAM

By cinning command: Cryptcat6543 -k key

Computer data can now required with the command Cryptcat -1 -p6543 -k key>>

Non volatile Memory Acquisition: Permanent memory or volatile memory stands significant source for the digital forensic investigation.

Parmanent data is collected through both online and offline methods:

Offline data is collected from the hard drive applications tool such as Guymayers etc.

Online data like firewall logs, antivirus logs and domain controller log with help of wires work and ethernal collected.

2. Examination

Once the data collected detailed examination is done by comparing the original and logical copies collected. Such examination gives us clues of how manege for window registry examination. Command used echo text_mess > file1.text : file2.txt

The above file retrieved through the command more < file1.txt : file2.txt

The network forensic is enabled using the tools and techniques so that the following potential information can be accessed.

System Information, Service listing, Process listing, Registry information, Binary dumped of memory

3. Analysis with Assumption

Many tools and methodologies are used by the audit team to analyse the collection and examined evidence. Analysis is done according to the following:

a. Leyword searches in all the files

b. Recovering the deleted files

c. Registry information extraction from the workstation

The tools used in this phase are FTK and ILOOKIX. These tools are helpful to recover the document, chat, logs, emails.

4. Report

The final report is generated by the audit team

Purpose of report Digital Forensic investigation conducted on the compromise of manager's computer

Author of the report Aufit Team

Incident Summery The source of compromise are x, y, x

Evidence All the effected files, registry, log files

Analysis All the analyzed data analyzed

Conclusion All digital evidence are extracted and found from the source

Document support volatile and non volatile data, tools, log info, registry info and so on.

7 0
2 years ago
If there are 8 opcodes and 10 registers, a. What is the minimum number of bits required to represent the OPCODE? b. What is the
Nimfa-mama [501]

Answer:  

For 32 bits Instruction Format:

OPCODE   DR               SR1                   SR2      Unused bits

a) Minimum number of bits required to represent the OPCODE = 3 bits

There are 8 opcodes. Patterns required for these opcodes must be unique. For this purpose, take log base 2 of 8 and then ceil the result.

Ceil (log2 (8)) = 3

b) Minimum number of bits For Destination Register(DR) = 4 bits

There are 10 registers. For unique register values take log base 2 of 10 and then ceil the value.  4 bits are required for each register. Hence, DR, SR1 and SR2 all require 12 bits in all.  

Ceil (log2 (10)) = 4

c) Maximum number of UNUSED bits in Instruction encoding = 17 bits

Total number of bits used = bits used for registers + bits used for OPCODE  

     = 12 + 3 = 15  

Total  number of bits for instruction format = 32  

Maximum  No. of Unused bits = 32 – 15 = 17 bits  

OPCODE                DR              SR1             SR2              Unused bits

  3 bits              4 bits          4 bits           4 bits                17 bits

7 0
2 years ago
A large software development company employs 100 computer programmers. Of them, 45 areproficient in Java, 30 in C, 20 in Python,
melisa1 [442]

Answer:

18, 13, 19

Explanation:

Number of computer programmers proficient only in Java = 45 - ( 1+1+6) = 37

Number of computer programmers proficient only in C++ = 30 - (6+1+5) = 18

Number of computer programmers proficient only in python = 20 - ( 1+1+5) = 13

Number of computer programmers are not proficient in any of these three languages = 100 - ( 37 + 18 + 13 + 1+ 1+ 5+ 6 ) = 100 - 81 = 19

4 0
2 years ago
A company has a file server that shares a folder named Public. The network security policy specifies that the Public folder is a
klasskru [66]

Answer:

authentication

Explanation:

At the authentication process, there is a way of identifying a user, this is typically done by having the user enter a valid user name and valid password before access is granted. Here at authentication the process is based on each user having a unique set of criteria for gaining access.

The AAA server have to ascertain by comparing a user's authentication credentials with other user credentials stored in a database. In the event the credentials match, the user is granted access to the network. But on the other hand, If the credentials varies, and authentication fails then network access will be denied.

3 0
2 years ago
2.4: Star Pattern Write a program that displays the following pattern: * *** ***** ******* ***** *** * Output. Seven lines of ou
adelina 88 [10]

Answer:

// here is code in java.

public class NAMES

{

// main method

public static void main(String[] args)

{

int n=4;

// print the upper half

for(int a=1;a<=n;a++)

{

for(int b=1;b<=n-a;b++)

{

// print the spaces

System.out.print(" ");

}

// print the * of upper half

for(int x=1;x<=a*2-1;x++)

{

// print the *

System.out.print("*");

}

// print newline

System.out.println();

}

// print the lower half

for(int y=n-1;y>0;y--)

{

for(int z=1;z<=n-y;z++)

{

// print the spaces

System.out.print(" ");

}

for(int m=1;m<=y*2-1;m++)

{

// print the *

System.out.print("*");

}

// print newline

System.out.println();

}

}

}

Explanation:

Declare a variable "n" and initialize it with 4. First print the spaces (" ") of the upper half with the help of nested for loop.Then print the "*" of the upper half with for loop. Similarly print the lower half in revers order. This will print the required shape.

Output:

  *

 ***

*****

*******

*****

 ***

  *

5 0
2 years ago
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