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oee [108]
2 years 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]2 years 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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Assume that a program uses the named constant PI to represent the value 3.14. The program uses the named constant in several sta
mihalych1998 [28]

Answer:

The advantage for the above condition is as follows:-

Explanation:

  • If a user creates a defined constant variable and assigns a value on its and then uses that variable instead of the value, then it will a great advantage.
  • It is because when there is a needs to change the value of that variable, then it can be done when the user changes the value in one place. There is no needs to change the vale in multiple places.
  • But if there is a value in multiple places instead of a variable and there is no constant variable, then the user needs to change the value in multiple places.
7 0
2 years ago
Consider the following relation:CAR_SALE(Car#, Date_sold, Salesperson#, Commission%, Discount_amt)Assume that a car may be sold
GalinKa [24]

Answer:

The answer to this question can be given as:

Normalization:

Normalization usually includes the division of a table into two or more tables as well as defining a relation between the table. It is also used to check the quality of the database design. In the normalization, we use three-level that are 1NF, 2NF, 3NF.

First Normal Form (1NF):

In the 1NF each table contains unique data. for example employee id.  

Second Normal Form (2NF):

In the 2NF form, every field in a table that is not a determiner of another field's contents must itself be a component of the table's other fields.

Third Normal Form (3NF):

In the 3NF form, no duplication of information is allowed.

Explanation:

The explanation of the question can be given as:

  • Since all attribute values are single atomic, the given relation CAR_SALE is in 1NF.  
  • Salesperson# → commission% …Given  Thus, it is not completely dependent on the primary key {Car#, Salesperson#}. Hence, it is not in 2 NF.                                                                                                        

The 2 NF decomposition:

        CAR_SALE_1(Car#, Salesperson#, Date_sold, Discount_amt)

         CAR_SALE_2(Salesperson#, Commission%)

  • The relationship in question is not in 3NF because the nature of a transitive dependence occurs
  • Discount_amt → Date_sold → (Car#, Salesperson#) . Thus, Date_sold is neither the key itself nor the Discount amt sub-set is a prime attribute.  

The 3 NF decomposition :

        CAR_SALES_1A(Car#, Salesperson#, Date_sold)

        CAR_SALES_1B(Date_sold, Discount_amt)

        CAR_SALE_3(Salesperson#, Commission%)

5 0
2 years ago
Which of the following can be managed using Active Directory Users and Computers snap-in?
Goryan [66]

Answer:

The answer is "Option b".

Explanation:

The active directory can manage the data with active directory customers but a snap-on machine. It is used in scrolling the list, as the earlier replies demonstrate. It will appear on the screen if there is no other startup software installed on a computer.  

This snap-on desktop is only a component, that allows its simultaneous use of different systems or devices on the very same system. It also turns the objects more or less into the pieces of the whole.

7 0
2 years ago
Which DFS option can be used to help manage peak bändwidth needs by forcing replication to occur during off hours?
andrey2020 [161]

Answer: D. Scheduling

Explanation:

Distributed file system (DFS) replication scheduling are used most especially when large files need to be replicated across lower bandwidth connections. It helps manage peak bandwidth by forcing replication to occur during off hours

5 0
2 years ago
Given 4 integers, output their product and their average, using integer arithmetic.
solmaris [256]

Answer:

see explaination

Explanation:

Part 1:

import java.util.Scanner;

public class LabProgram {

public static void main(String[] args) {

Scanner scnr = new Scanner(System.in);

int num1;

int num2;

int num3;

int num4;

int avg=0, pro=1;

num1 = scnr.nextInt();

num2 = scnr.nextInt();

num3 = scnr.nextInt();

num4 = scnr.nextInt();

avg = (num1+num2+num3+num4)/4;

pro = num1*num2*num3*num4;

System.out.println(pro+" "+avg);

}

}

------------------------------------------------------------------

Part 2:

import java.util.Scanner;

public class LabProgram {

public static void main(String[] args) {

Scanner scnr = new Scanner(System.in);

int num1;

int num2;

int num3;

int num4;

double avg=0, pro=1; //using double to store floating point numbers.

num1 = scnr.nextInt();

num2 = scnr.nextInt();

num3 = scnr.nextInt();

num4 = scnr.nextInt();

avg = (num1+num2+num3+num4)/4.0; //if avg is declared as a float, then use 4.0f

pro = num1*num2*num3*num4;

System.out.println((int)pro+" "+(int)avg); //using type conversion only integer part

System.out.printf("%.3f %.3f\n",pro,avg);// \n is for newline

}

}

8 0
2 years ago
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