1. how much is the maximum number of IPs for the hosts that can be placed on a subnet that uses subnetmask
255.255.255.224 / 27 27 bits per network we have added the number of subnets
11111111.11111111.11111111.11100000
23 = 8 subnets
Ngelen 5 other bits from tetshja
25 = 32 hoste + sub
30 IP addresses suitable for hosts
25-2 = 30 addresses per host
224 = 11100000
27 * 1 + 1 * 26 + 1 * 25 + 0 * 24 + 0 * 23 + 0 * 22 + 0 * 21
128 + 64 + 32 = 224
2. One company has 100 computers and the IP administrator has to set the subnet addresses for each department, it will create 5 subnets using the block
192.168.1.0/24
And will address the first subnet of servers and part of the computers
Calculate Subnet Request?
23 = 8 subnets
25-2 = 30 hoste
Address passes to / 27 192.168.1.0/27
3. We have to subnet a 5 subnete network with each having at least 16 hosts
Which subnet mask should use
A) 255.255.255.192 / 26 22 = 4 subnets 26-2 = 62 hosts
B) 255.255.255.240 / 28 24 = 16 subnets 24-2 = 14 hosts + subnets
C) 255.255.255.224 / 27 23 = 8 subnets 25-2 = 30 hoste + sub
D) 255.255.255.248 / 29 25 = 32
4. a network that needs 29 subnets by maximizing the number of hosts for each network that bits must pass to the network part in order to fill the subnet
A) 2/26 22 = 4 subtract
B) 3/27 23 = 8subnete
C) 4/28 24 = 16 subtract
D) 5/29 25 = 32 subnets
D the correct answer because it meets the number of subnets that the network needs
5. How subnete and how many hosts the subnet gives the address
172.16.0.0/19
11111111.1111111.11100000.00000000
The first 2 octets do not change the octets of 3 and 4 change
Hosts have 13 bits left 213 = 8192
We got 23 = 8 subnets
A) 7 subnets and 30 hosts
B) 8 subnets and 8190 are the exact alternatives
C) 8 subnets and 2046 hosts
D) 7 subnets and 2046 hosts
6. What is the shipcast address of the network 172.28.224.0/20 we have 11111111.11111111.11110000.0000000 subnet mask
4 bits we have picked up and passed over to octet 3 network
How much subnete we get
How many hosts we have for each subnet
24 = 16 subnets
24 * 256 = 212 = 4096 total IP addresses and 4094 valid IP addresses per host
172.28.0.0 - 172.28.15.255 4096 subnete 4094 addresses per host Subnete 1
172.28.244.0 - 172.28.239.255 Subnet X
7.Find the renames of the IP addresses that belong to the host address 10.180.68.103/20
24 = 16 subnets
24 * 256 = 212 4096 IP addresses or 4094 hosts + subnets
Subneti 1 10.180.0.0 - 10.180.15.255
Subneti 2 10.180.16.0 - 10.168.31.255 each subnet 4096 addresses
Subneti 2 10.180.32.0 - 10.168.47.255
Subneti 4 10.180.48.0 - 10.168.63.255
Subneti 5 10.180.64.0 - 10.168.79.255
8. From subnets and from hosts give 10.0.0.0/20 network
24 = 16 subnets
24 * 256 = 212 4096 hosts for subnet
1 9. We need to configure a server that is part of subnet 192.168.19.24/29
The router has the first valid IP address
Which of the following addresses will be for the server
A) 192.168.19.0 - 255.255.255.0
B) 192.166.19.33 - 255.255.255.240
C) 192.168.19.26 - 255.255.255.248 ++++++ is c +++++++++++++
D) 192.168.19.31 - 255.255.255.248
25 = 32 subnets
23 = 8 hosts + subnets
Sub1: 192.168.19.0 - 192.168.19.7
192.168.19.24 - 192.168.19.31
192.168.19.25 - 192.168.19.30
10. If we have a subnet 172.16.17.0/22
A) 172.16.17.1 - 255.255.255.252
B) 172.16.0.1 - 255.255.240.0
C) 172.16.20.1 - 255.255.254.0
D) 172.16.18.255 255.255.252.0
26 = 64 subnets
210 = 1024 add IP
210-2 = 1022 add hoste + subnete
172.16.0.0 - 172.16.3.255
11. Expressed in December as subnet mask
/ 16 - 255.255.0.0
/ 17 - 255.255.128.0
/ 18 - 255.255.192.0
/ 19 - 255.255.224.0
/ 20 - 255.255.240.0
/ 21 - 255.255.248.0
/ 22 - 255.255.252.0
/ 23 - 255.255.254.0
/ 24 - 255.255.255.0
/ 25-255255255128
/ 26-255255255192
/ 27-255255255224
/ 28-255255255240
/ 29-255255255248
/ 30-255255255252
/ 31-255255255254
12. How much should the IP of the website be if we use the subnet of
A) 192.168.10.142
B) 192.168.10.66
C) 192.168.10.254
D) 192.168.10.144
C) 192.168.10.162
24 = 16 subnets
24 = 16 add IP
24-2 = 14 hoste
13.
• 481 hoste Lan Student Sub1 172.16.0.0/23
• 69 hoste Lan Pedagog sub2 172.16.0.0/25
• 23 hosts Lan Administrator sub3 172.16.0.0/27
• 2 hosts Wan router su8 172.16.0.0/30
172.16.0.0/21
You need at least 481 IP addresses
Calculate how many we will be in the host part
Sub1
27 = 128 sub
29 = 512 hoste
172.16.0.0 - 172.16.1.255
Sub 2 172.16.0.0/25
21 = 25 subnets
27 = 128 hoste
172.16.2.0 - 172.16.2.128
Sub3
172.16.0.0/27
23 = 8subnete
25 = 32 hoste
172.16.2.128 - 172.16.2.159
Sub4
172.16.0.0/30
26 = 64 subnets
22 = 4 hoste
172.16.2.160 - 172.16.2.163
Mpls understands that in each subnet that creates changes the prefix and starts from the number of hosts we need
The router has the first valid IP address
Which of the following addresses will be for the server
A) 192.168.19.0 - 255.255.255.0
B) 192.166.19.33 - 255.255.255.240
C) 192.168.19.26 - 255.255.255.248 ++++++ is c +++++++++++++
D) 192.168.19.31 - 255.255.255.248
25 = 32 subnets
23 = 8 hosts + subnets
Sub1: 192.168.19.0 - 192.168.19.7
192.168.19.24 - 192.168.19.31
192.168.19.25 - 192.168.19.30
10. If we have a subnet 172.16.17.0/22
A) 172.16.17.1 - 255.255.255.252
B) 172.16.0.1 - 255.255.240.0
C) 172.16.20.1 - 255.255.254.0
D) 172.16.18.255 255.255.252.0
26 = 64 subnets
210 = 1024 add IP
210-2 = 1022 add hoste + subnete
172.16.0.0 - 172.16.3.255
11. Expressed in December as subnet mask
/ 16 - 255.255.0.0
/ 17 - 255.255.128.0
/ 18 - 255.255.192.0
/ 19 - 255.255.224.0
/ 20 - 255.255.240.0
/ 21 - 255.255.248.0
/ 22 - 255.255.252.0
/ 23 - 255.255.254.0
/ 24 - 255.255.255.0
/ 25-255255255128
/ 26-255255255192
/ 27-255255255224
/ 28-255255255240
/ 29-255255255248
/ 30-255255255252
/ 31-255255255254
12. How much should the IP of the website be if we use the subnet of
A) 192.168.10.142
B) 192.168.10.66
C) 192.168.10.254
D) 192.168.10.144
C) 192.168.10.162
24 = 16 subnets
24 = 16 add IP
24-2 = 14 hoste
13.
• 481 hoste Lan Student Sub1 172.16.0.0/23
• 69 hoste Lan Pedagog sub2 172.16.0.0/25
• 23 hosts Lan Administrator sub3 172.16.0.0/27
• 2 hosts Wan router su8 172.16.0.0/30
172.16.0.0/21
You need at least 481 IP addresses
Calculate how many we will be in the host part
Sub1
27 = 128 sub
29 = 512 hoste
172.16.0.0 - 172.16.1.255
Sub 2 172.16.0.0/25
21 = 25 subnets
27 = 128 hoste
172.16.2.0 - 172.16.2.128
Sub3
172.16.0.0/27
23 = 8subnete
25 = 32 hoste
172.16.2.128 - 172.16.2.159
Sub4
172.16.0.0/30
26 = 64 subnets
22 = 4 hoste
172.16.2.160 - 172.16.2.163
Mpls understands that in each subnet that creates changes the prefix and starts from the number of hosts we need
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