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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.

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A hotel salesperson enters sales in a text file. Each line contains the following, separated by semicolons: the name of the clie
timama [110]

Answer:

see explaination

Explanation:

Java code

//Header file section

import java.util.Scanner;

import java.io.*;

//main class

public class SalesTestDemo

{

//main method

public static void main(String[] args) throws IOException

{

String inFile;

String line;

double total = 0;

Scanner scn = new Scanner(System.in);

//Read input file name

System.out.print("Enter input file Name: ");

inFile = scn.nextLine();

FileReader fr = new FileReader(new File(inFile));

BufferedReader br = new BufferedReader(fr);

System.out.println("Name \t\tService_Sold \tAmount \tEvent Date");

System.out.println("=====================================================");

line = br.readLine();

//Each line contains the following, separated by semicolons:

//The name of the client, the service sold

//(such as Dinner, Conference, Lodging, and so on)

while(line != null)

{

String temp[] = line.split(";");

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

{

System.out.print(temp[i]+"\t");

if(i == 1)

System.out.print("\t");

}

//Calculate total amount for each service category

total += Double.parseDouble(temp[2]);

System.out.println();

line = br.readLine();

}

//Display total amount

System.out.println("\nTotal amount for each service category: "+total);

}

}

inputSale.txt:

Peter;Dinner;1500;30/03/2016

Bill;Conference;100.00;29/03/2016

Scott;Lodging;1200;29/03/2016

Output:

Enter input file Name: inputSale.txt

Name Service_Sold Amount Event Date

=====================================================

Peter Dinner 1500 30/03/2016

Bill Conference 100.00 29/03/2016

Scott Lodging 1200 29/03/2016

Total amount for each service category: 2800.0

3 0
2 years ago
Consider a single-platter disk with the following parameters: rotation speed: 7200 rpm; number of tracks on one side ofplatter:
horsena [70]

Answer:

Given Data:

Rotation Speed = 7200 rpm

No. of tracks on one side of platter = 30000

No. of sectors per track = 600

Seek time for every 100 track traversed = 1 ms

To find:

Average Seek Time.

Average Rotational Latency.

Transfer time for a sector.

Total Average time to satisfy a request.

Explanation:

a) As given, the disk head starts at track 0. At this point the seek time is 0.

Seek time is time to traverse from 0 to 29999 tracks (it makes 30000)

Average Seek Time is the time taken by the head to move from one track to another/2

29999 / 2 = 14999.5 ms

As the seek time is one ms for every hundred tracks traversed.  So the seek time for 29,999 tracks traversed is

14999.5 / 100 = 149.995 ms

b) The rotations per minute are 7200

1 min = 60 sec

7200 / 60 = 120 rotations / sec

Rotational delay is the inverses of this. So

1 / 120 = 0.00833 sec

          = 0.00833 * 100

          = 0.833 ms

So there is  1 rotation is at every 0.833 ms

Average Rotational latency is one half the amount of time taken by disk to make one revolution or complete 1 rotation.

So average rotational latency is: 1 / 2r

8.333 / 2 = 4.165 ms

c) No. of sectors per track = 600

Time for one disk rotation = 0.833 ms

So transfer time for a sector is: one disk revolution time / number of sectors

8.333 / 600 = 0.01388 ms = 13.88 μs

d)  Total average time to satisfy a request is calculated as :

Average seek time + Average rotational latency + Transfer time for a sector

= 149.99 ms + 4.165 ms + 0.01388 ms

= 154.168 ms

4 0
1 year ago
A police department wants to maintain a database of up to 1800 license-plate numbers of people who receive frequent tickets so t
maksim [4K]

Answer:

c.

Explanation:

I believe that in this scenario, the best option for this data would be a hash table using open addressing with 1,800 entries. Hash tables consume more memory than lists but it makes up for it with much faster response time speeds. This is because hash tables work on a key:value system therefore, the license plate can easily be grabbed from the database extremely quickly by just plugging in the plate number. Doing so will retrieve all of the saved information from that license plate. That is why hash tables have a constant time complexity of O(1)

6 0
1 year ago
What are the arguments for writing efficient programs even though hardware is relatively inexpensive?
Ainat [17]

Answer: Even though the hardware is inexpensive the writing of program is not efficient through this method as proper development of program is necessary for the clear execution due to factors like:-

  • The facility of writing program even the cost of hardware is less but it is not a free facility.
  • It also has a slower processing for the execution of the program
  • The construction of the efficient program is necessary for the compilation and execution of it rather than poorly constructed program is worthless and inefficient in working.

7 0
2 years ago
With Voice over Internet Protocol (VoIP), _____. a. voicemails cannot be received on the computer b. call quality is significant
Pachacha [2.7K]

Answer:

c. Users can have calls forwarded from anywhere in the world

Explanation:

As all you need is the internet, there would be no need to try to sort out roaming as you would on a regular phone line

4 0
1 year ago
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