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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.

You might be interested in
____________________ parameters are useful in three situations: • When the value of the actual parameter needs to be changed • W
jenyasd209 [6]

Answer:

The correct word for the blank space is: Reference.

Explanation:

In C++, Reference Parameter is a type of parameter that does not need additional storage since it represents an already existing function.  The function has the characteristic of passing more than one (1) variable. Reference parameters are also used to edit the existing parameter or when referencing a function saves storage compared to creating a new function.

8 0
2 years ago
Write a procedure named Str_find that searches for the first matching occurrence of a source string inside a target string and r
kirill115 [55]

Answer: Provided in the explanation section

Explanation:

Str_find PROTO, pTarget:PTR BYTE, pSource:PTR BYTE

.data

target BYTE "01ABAAAAAABABCC45ABC9012",0

source BYTE "AAABA",0

str1 BYTE "Source string found at position ",0

str2 BYTE " in Target string (counting from zero).",0Ah,0Ah,0Dh,0

str3 BYTE "Unable to find Source string in Target string.",0Ah,0Ah,0Dh,0

stop DWORD ?

lenTarget DWORD ?

lenSource DWORD ?

position DWORD ?

.code

main PROC

  INVOKE Str_find,ADDR target, ADDR source

  mov position,eax

  jz wasfound           ; ZF=1 indicates string found

  mov edx,OFFSET str3   ; string not found

  call WriteString

  jmp   quit

wasfound:                   ; display message

  mov edx,OFFSET str1

  call WriteString

  mov eax,position       ; write position value

  call WriteDec

  mov edx,OFFSET str2

  call WriteString

quit:

  exit

main ENDP

;--------------------------------------------------------

Str_find PROC, pTarget:PTR BYTE, ;PTR to Target string

pSource:PTR BYTE ;PTR to Source string

;

; Searches for the first matching occurrence of a source

; string inside a target string.

; Receives: pointer to the source string and a pointer

;    to the target string.

; Returns: If a match is found, ZF=1 and EAX points to

; the offset of the match in the target string.

; IF ZF=0, no match was found.

;--------------------------------------------------------

  INVOKE Str_length,pTarget   ; get length of target

  mov lenTarget,eax

  INVOKE Str_length,pSource   ; get length of source

  mov lenSource,eax

  mov edi,OFFSET target       ; point to target

  mov esi,OFFSET source       ; point to source

; Compute place in target to stop search

  mov eax,edi    ; stop = (offset target)

  add eax,lenTarget    ; + (length of target)

  sub eax,lenSource    ; - (length of source)

  inc eax    ; + 1

  mov stop,eax           ; save the stopping position

; Compare source string to current target

  cld

  mov ecx,lenSource    ; length of source string

L1:

  pushad

  repe cmpsb           ; compare all bytes

  popad

  je found           ; if found, exit now

  inc edi               ; move to next target position

  cmp edi,stop           ; has EDI reached stop position?

  jae notfound           ; yes: exit

  jmp L1               ; not: continue loop

notfound:                   ; string not found

  or eax,1           ; ZF=0 indicates failure

  jmp done

found:                   ; string found

  mov eax,edi           ; compute position in target of find

  sub eax,pTarget

  cmp eax,eax    ; ZF=1 indicates success

done:

  ret

Str_find ENDP

END main

cheers i hoped this helped !!

6 0
2 years ago
Given a sorted array of integer, A, with possible duplicate elements. Implement an efficient (sublinear running time complexity)
allsm [11]

Answer:

The java program is as follows.

import java.util.*;

import java.lang.*;

public class SearchProgram

{

   // sorted array may or may not containing duplicate elements

   static int A[] = {-1, 2, 3, 5, 6, 6, 6, 9, 10};

   

   // variable to keep track of number of occurrences, declared and initialized to 0

   static int times=0;

   

// variable to hold value to be searched in the array

   static int key;

   static int search(int[] art, int n)

   {

       // enhanced for loop used to find how many times n occurs in array, arr

       for(int j:arr)

       {

           if(n == j)

               times = times+1;

       }

       

       return times;

   }

   

   public static void main(String[] args)

   {

       // scanner object to allow user input

       Scanner sc = new Scanner(System.in);

       

       // user prompted to enter key to be searched

       System.out.print("Enter the element to be searched: ");

       key = sc.nextInt();

       

       // message displayed to the user on how many times the element is present in the array

       System.out.println("The element " + key + " is present " + search(A, key) + " times in the array.");

   

   }

}

OUTPUT

Enter the element to be searched: 3

The element 3 is present 1 times in the array.

Explanation:

The steps of program execution are shown below.

1. An integer array, A, is declared and initialized.

2. An integer variable, times, is declared and initialized to 0.

3. A method, search(), is created which takes 2 parameters, array A and user entered value for the element to be searched in A.

4. The search() method has return type of integer and returns the value of variable, times.

5. Inside the search() method, an enhanced for loop is used to compute the value of the variable, times.

6. In the main() method, user is prompted to enter the value of key.

7. The search() is called and value of the variable, times, displayed to the user.

8. All the methods and variables are declared as static.

7 0
2 years ago
Megan has written the following rough draft for her assignment. Choose the correct way to complete each sentence. Sarah is creat
atroni [7]

Answer:

I. hatching

II. filter

Explanation:

Sarah is creating an image. She wants to achieve tonal effects with parallel lines by varying their densities and distribution patterns. She will use <u>hatching</u> to do this. She also wants to add special effects on the image. Therefore, she will use <u>filters</u>.

Hatching refers to an art technique in which parallel lines are used to shade by drawing them closely together.

Stippling is an art technique used in the making of patterns by using small dots.

Filters is a technique of changing image characteristics. It is used to change the picture characteristics & to add special effects to an image such as the shade, contrast, tone, brightness etc.

We therefore see that, <u>hatching</u> and <u>filter</u> are the correct answers respectively

3 0
2 years ago
Write a program that calculates an adult's fat-burning heart rate, which is 70% of 220 minus the person's age. Complete fat_burn
gayaneshka [121]

Answer:

#include <iostream>

# include <conio.h>

using namespace std;

float fat_burning_heart_rate(float age);

main()

{

float age,bpm;

cout<<"enter the age of the person"<<endl;

cin>>age;

if (age>=18 && age<=75)

{

bpm=fat_burning_heart_rate(age);

cout<<"fat burning heart rate for the age of"<<age<<"="<<bpm;

}

else

cout<<"Invalid age";

getch();

}

float fat_burning_heart_rate(float age)

{

 

float a;

 

a= (220-age)*0.7;

return a;

}

Explanation:

In this program, the variable named as age has been taken to enter the age of the person, needs to calculate the burning heart rate. All the variables taken in float, as the age and fat burning heart rate should come in decimal value.

The formula has been used as mention in question that,

a= (220-age)*0.7;

8 0
2 years ago
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