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fredd [130]
2 years ago
14

After deploying a large number of wireless laptop computers on the network, Taylor, the IT director at Contoso, Ltd. decides to

use DHCP to enable the laptop users to move from one subnet to another without having to manually reconfigure their IP addresses. Soon after the DHCP deployment, however, Taylor notices that some of the IP address scopes are being depleted, resulting in some computers being unable to connect to a new subnet. What can Taylor do to resolve this problem without altering the network’s subnetting?
Computers and Technology
1 answer:
LenKa [72]2 years ago
6 0

Answer:

Setting of short lease time for IP addresses in order to enhance quicker access from clients

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Exponentiation Is a/an operater?​
garik1379 [7]

Answer:

Yes it is

Hope this helps

5 0
1 year ago
Write a function addOddMinusEven that takes two integers indicating the starting point and the end point. Then calculate the sum
snow_lady [41]

Answer:g

   public static int addOddMinusEven(int start, int end){

       int odd =0;

       int even = 0;

       for(int i =start; i<end; i++){

           if(i%2==0){

               even = even+i;

           }

           else{

               odd = odd+i;

           }

       }

       return odd-even;

   }

}

Explanation:

Using Java programming language:

  • The method addOddMinusEven() is created to accept two parameters of ints start and end
  • Using a for loop statement we iterate from start to end but not including end
  • Using a modulos operator we check for even and odds
  • The method then returns odd-even
  • See below a complete method with a call to the method addOddMinusEven()

public class num13 {

   public static void main(String[] args) {

       int start = 2;

       int stop = 10;

       System.out.println(addOddMinusEven(start,stop));

   }

   public static int addOddMinusEven(int start, int end){

       int odd =0;

       int even = 0;

       for(int i =start; i<end; i++){

           if(i%2==0){

               even = even+i;

           }

           else{

               odd = odd+i;

           }

       }

       return odd-even;

   }

}

8 0
2 years ago
Given six memory partitions of 300 KB, 600 KB, 350 KB, 200 KB, 750 KB, and 125 KB (in order), how would the first-fit, best-fit,
Inga [223]

Answer:

In terms of efficient use of memory: Best-fit is the best (it still have a free memory space of 777KB and all process is completely assigned) followed by First-fit (which have free space of 777KB but available in smaller partition) and then worst-fit (which have free space of 1152KB but a process cannot be assigned). See the detail in the explanation section.

Explanation:

We have six free memory partition: 300KB (F1), 600KB (F2), 350KB (F3), 200KB (F4), 750KB (F5) and 125KB (F6) (in order).

Using First-fit

First-fit means you assign the first available memory that can fit a process to it.

  • 115KB will fit into the first partition. So, F1 will have a remaining free space of 185KB (300 - 115).
  • 500KB will fit into the second partition. So, F2 will have a remaining free space of  100KB (600 - 500)
  • 358KB will fit into the fifth partition. So, F5 will have a remaining free space of 392KB (750 - 358)
  • 200KB will fit into the third partition. So, F3 will have a remaining free space of 150KB (350 -200)
  • 375KB will fit into the remaining partition of F5. So, F5 will a remaining free space of 17KB (392 - 375)

Using Best-fit

Best-fit means you assign the best memory available that can fit a process to the process.

  • 115KB will best fit into the last partition (F6). So, F6 will now have a free remaining space of 10KB (125 - 115)
  • 500KB will best fit into second partition. So, F2 will now have a free remaining space of 100KB (600 - 500)
  • 358KB will best fit into the fifth partition. So, F5 will now have a free remaining space of 392KB (750 - 358)
  • 200KB will best fit into the fourth partition and it will occupy the entire space with no remaining space (200 - 200 = 0)
  • 375KB will best fit into the remaining space of the fifth partition. So, F5 will now have a free space of 17KB (392 - 375)

Using Worst-fit

Worst-fit means that you assign the largest available memory space to a process.

  • 115KB will be fitted into the fifth partition. So, F5 will now have a free remaining space of 635KB (750 - 115)
  • 500KB will be fitted also into the remaining space of the fifth partition. So, F5 will now have a free remaining space of 135KB (635 - 500)
  • 358KB will be fitted into the second partition. So, F2 will now have a free remaining space of 242KB (600 - 358)
  • 200KB will be fitted into the third partition. So, F3 will now have a free remaining space of 150KB (350 - 200)
  • 375KB will not be assigned to any available memory space because none of the available space can contain the 375KB process.
8 0
2 years ago
Your computer uses 4 bits to represent decimal numbers (0, 1, 2, 3 and so on) in binary. What is the SMALLEST number for which a
natali 33 [55]
2 I think because 2 I think is binary I’m sorry if this is wrong
8 0
1 year ago
A technician is trying to use the tablet to control the iot device digital lamp. the technician tries to login to the registrati
Rom4ik [11]
<span>In the scenario in which a technician is trying to use the tablet to control the IoT device digital lamp, but </span><span>the access is not successful the two problems are:
</span>The registration service on the server is off.
The digital lamp is using the wrong credential to connect to the registration server.
3 0
2 years ago
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