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Task 1
Problem 1
IP address>172.30.1.33
Network Mask>255.255.0.0
Find
1.Network address
IP Address 172.30.1.33 10101100
00011001 01110010
11111010
Subnet Mask 255.255.0.0 11111111
11111111 00000000
00000000
Network Address 172.25.0.0 10101100
00011001 00000000
00000000
How to get the binary number for IP Address
172 30 1 33172/2 = 0
86/2 = 0
43/2 = 1
21/2 = 1
10/2 = 0
5/2 = 1
2/2 = 0
½ = 1
30/2=0
15/2=1
7/2=1
3/2=1
1/2=1
0/2=0
0/2=0
0/2=0
1/2=1
0/2=0
0/2=0
0/2=0
0/2=0
0/2=0
0/2=0
0/2=0
33/2=1
16/2=0
8/2=0
4/2=0
2/2=0
1/2=1
0/2=0
0/2=0
The final answer in the binary number10101100 01111000 01110010 10000100
2.Network broadcast address
How to get a broadcast address :
See Decision binary IP address and subnet mask , Focusing Search to Subnet , turn on all Grade Subnet 1 Becoming 0 and 0 Become obtained 1.Until
To find the total number of host bits, you must use the formula in the IP address that is Possible Host.
Formula of the Possible Host:
2^n - 2
how n is the number of host bits , so to find the number of host bits in the IP address of this solution is like this .
Using formula of the Possible Host.
= 2^n - 2
= 2^16 - 2
= 65536 - 2
= 65534
The final answer for the number of host bits for this IP address is 65534.
4. Number of Host
To find the total number of the host,you just substract the number of bits in the IP address with the number of host bits.
Number of bits in IP address = 32
Number of host bits = 16
The solution is:32 – 16 = 16
The number of host in this IP address is 16.
Problem 5
Host IP Address :192.168.3.219
Net Mask :255.255.0.0
Find
1.Network address
IP Address 192.168.3.219
11101100 00101000
00000011 11011011
Subnet Mask 255.255.0.0 11111111 11111111
00000000 00000000
Network Address 192.168.0.0 11101100 00101000
00000000 00000000
How to get the binary number for IP Address
192 168 3 219192/2 = 0
96/2 = 0
48/2 = 1
24/2 = 1
12/2 = 0
6/2 = 1
3/2 = 1
½ = 1
168/2 = 0
84/2 = 0
42/2 = 0
21/2 = 1
10/2 = 0
5/2 = 1
2/2 = 0
1/2 = 0
3/2 = 1
1/2 = 1
0/2 = 0
0/2 = 0
0/2 = 0
0/2 = 0
0/2 = 0
0/2 = 0
219/2 = 1
109/2 = 1
54/2 = 0
27/2 = 1
13/2 = 1
6/2 = 0
3/2 = 1
1/2 = 1
The final answer in the binary number11101100 00101000 00000011 11011011
2.Network broadcast address
How to get a broadcast address :
See Decision binary IP address and subnet mask , Focusing Search to Subnet , turn on all Grade Subnet 1 Becoming 0 and 0 Become obtained 1.Until 00000000.00000000.00001111.11111111:
To find the total number of host bits, you must use the formula in the IP address that is Possible Host.
Formula of the Possible Host:
2^n - 2
how n is the number of host bits , so to find the number of host bits in the IP address of this solution is like this .
Using formula of the Possible Host.
= 2^n - 2
= 2^5 - 2
= 32 - 2
= 30
The final answer for the number of host bits for this IP address is 30
4. Number of Host
To find the total number of the host,you just substract the number of bits in the IP address with the number of host bits.
Number of bits in IP address = 32
Number of host bits = 27
The solution is:32 – 27 = 5
The number of host in this IP address is 5.
Task 2
Problem 1
Host IP Address 172.30.1.33Subnet Mask 255.255.0.0Number Of Subnet Bit 8Number Of Subnets 2^8=256 subnetsNumber Of Host Bits Per Subnet 8 bitNumber Of Usable Hosts Per Subnet 2^8=256-2=254 hosts per subnetSubnet Address For This IP Adress 172.30.1.0Ip Address Of First Host On This Subnet 172.30.1.1
Ip Address Of Last Host On Subnet 172.30.1.254Broadcast Address For This Subnet 172.30.1.255
How to get the binary number for IP Address
172 30 1 33172/2 = 0
86/2 = 0
43/2 = 1
21/2 = 1
10/2 = 0
5/2 = 1
2/2 = 0
½ = 1
30/2=0
15/2=1
7/2=1
3/2=1
1/2=1
0/2=0
0/2=0
0/2=0
1/2=1
0/2=0
0/2=0
0/2=0
0/2=0
0/2=0
0/2=0
0/2=0
33/2=1
16/2=0
8/2=0
4/2=0
2/2=0
1/2=1
0/2=0
0/2=0
The final answer in the binary number10101100 01111000 01110010 10000100
Host Ip Address
172 30 1 33
Ip Address 10101100 00011110 00000001 00100001
MultipleSubnet Mask
11111111 11111111 00000000 00000000
Subnet Address
10101100 00011110 00000001 00000000 Answer
Subnet Address For This IP Address
172 30 1 0
127.30.1.33
Ip address of first host:172.30.1.33
Ip address of last host:172.30.1.254
Broadcast address:172.30.1.255
0+254 hosts
Adress ip of host last calculation=254
Number of ip subnets and hosts
Look hosts in the subnet mask. Use the "Mask of Possibilities" chart to determine the bit is set to one. If no bit is set to one, there is no subnet. If any bit is set to one, proceed by steps.
172.30.1.33 255.255.0.0
= 11111111 00000000
Calculate the amount of the resistance of the host subnet mask. Raise X. calling this number 2 to the power of X. Use the "Power of 2" chart if necessary. This is the number of subnet potential created by the mask. Two of the subnet potential usually can not be used.
Calculating the total number of zeros in octal host subnet mask. Raise Y. calling this number 2 to the power of Y. Use the "Power of 2" chart if necessary. This is the number of subnet potential created by the mask. Two of these numbers will not be used for host addresses
Number of Usable hosts per subnet 2^8=256-2=254 host per subnet
Number of subnet bit 8 bit
Number Of subnets 2^8=256 subnets
Problem 2
Host IP Address 172.30.1.33Subnet Mask 255.255.255.0Number Of Subnet Bit 14Number Of Subnets 2^14=16384 subnetsNumber Of Host Bits Per Subnet 4 bitNumber Of Usable Hosts Per Subnet 2^2=4-2=2 hosys per subnetSubnet Address For This IP Adress 172.30.1.32Ip Address Of First Host On This Subnet 172.30.1.33
Ip Address Of Last Host On Subnet 172.30.1.34Broadcast Address For This Subnet 172.30.1.35
How to get the binary number for IP Address
172 30 1 33172/2 = 0
86/2 = 0
43/2 = 1
21/2 = 1
10/2 = 0
5/2 = 1
2/2 = 0
½ = 1
30/2 = 0
15/2 = 1
7/2 = 1
3/2 = 1
½ = 1
0/2 = 0
0/2 = 0
0/2 = 0
½ = 1
0/2 = 0
0/2 = 0
0/2 = 0
0/2 = 0
0/2 = 0
0/2 = 0
0/2 = 0
33/2 = 1
16/2 = 0
8/2 = 0
4/2 = 0
2/2 = 0
½ = 1
0/2 = 0
0/2 = 0
The final answer in the binary number10101100 00011110 00000001 00100001
Host Ip Address
172 30 1 33
Ip Address 10101100 00011110 00000001 00100001
MultipleSubnet Mask
11111111 11111111 11111111 00000000
Subnet Address
10101100 00011110 00000001 00100000 Answer
Subnet Address For This IP Address
172 30 1 32
127.30.1.33
Ip address of first host:172.30.1.33
Ip address of last host:172.30.1.34
Broadcast address:172.30.1.255
32+2 hosts
Adress ip of host last calculation=34
Number of ip subnet and host
Look hosts in the subnet mask. Use the "Mask of Possibilities" chart to determine the bit is set to one. If no bit is set to one, there is no subnet. If any bit is set to one, proceed by steps.
172.30.1.33 255.255.255.0
= 11111111 11111100
Calculate the amount of the resistance of the host subnet mask. Raise X. calling this number 2 to the power of X. Use the "Power of 2" chart if necessary. This is the number of subnet potential created by the mask. Two of the subnet potential usually can not be used.
11111111 11111100 = 14 ones. 2^14= 16384 usable subnets created
Number of subnet bit =14
Number of subnet 2^14=16384 subnets
Calculating the total number of zeros in octal host subnet mask. Raise Y. calling this number 2 to the power of Y. Use the "Power of 2" chart if necessary. This is the number of subnet potential created by the mask. Two of these numbers will not be used for addressing hosts.
Number of usable host per subnet 2^2=4-2=2 hosts per subnet
Problem 3
Host IP Address 192.168.10.234Subnet Mask 255.255.255.0Number Of Subnet Bit 8Number Of Subnets 2^8=256 subnetsNumber Of Host Bits Per Subnet 8 bitNumber Of Usable Hosts Per Subnet 2^8=256-2=254 hosts per subnetSubnet Address For This IP Adress 192.192.10.0Ip Address Of First Host On This Subnet 192.192.10.1
Ip Address Of Last Host On Subnet 192.192.10.254Broadcast Address For This Subnet 192.192.10.255
How to get the binary number for IP Address
192 168 10 234192/2 = 0
96/2 = 0
48/2 = 0
24/2 = 0
12/2 = 0
6/2 = 1
3/2 = 1
½ = 1
168/2 = 0
84/2 = 0
42/2 = 0
21/2 = 1
10/2 = 0
5/2 = 1
2/2 = 0
½ = 1
10/2 = 0
5/2 = 1
2/2 = 0
½ = 1
0/2 = 0
0/2 = 0
0/2 = 0
0/2 = 0
234/2 = 0
117/2 = 1
58/2 = 0
29/2 = 1
14/2 = 0
7/2 = 1
3/2 = 1
1/2 = 1
The final answer in the binary number11100000 10101000 00001010 11101010
Host Ip Address
192 168 10 234
Ip Address 11000000 11000000 00001010 11101010
MultipleSubnet Mask
11111111 11111111 11111111 00000000
Subnet Address
11000000 11000000 00001010 00000000 Answer
Subnet Address For This IP Address
192 192 10 0
192.168.10.234
Ip address of first host:192.168.10.234
Ip address of last host:192.168.10.235
Broadcast address:192.168.10.255
0+254 hosts
Adress ip of host last calculation=254
Number of ip subnets and host
Look at the hosts in the subnet mask. Use the "Mask of Possibilities" method to determine the bit is set to one. If no bit is set to one, there is no subnet. If any bit is set to one, proceed by steps.
192.168.10.234
255.255.255.0 = 11111111 00000000
Calculating the amount in octal host subnet mask. Raise X. calling this number 2 to the power of X. Use the "Power of 2" chart if necessary. This is the number of subnet potential created by the mask. Two of the subnet potential usually can not be used.
11111111 00000000 = 8 ones. 2^8= 256 usable subnets created
Number of subnet bit =8
Number of subnet 2^8=256 subnets
Calculating the total number of zeros in octal host subnet mask. Raise Y. calling this number 2 to the power of Y. Use the "Power of 2" chart if necessary. This is the number of subnet potential created by the mask. Two of these numbers will not be used for addressing hosts.
Number of usable host per subnet 2^8=256-2=254 hosts per subnet
Problem 4
Host IP Address 172.17.99.71Subnet Mask 255.255.0.0Number Of Subnet Bit 0Number Of Subnets 2^0=1 subnetsNumber Of Host Bits Per Subnet 16 bitNumber Of Usable Hosts Per Subnet 2^16=65536-2=65534 hosys per subnetSubnet Address For This IP Adress 172.17.0.0Ip Address Of First Host On This Subnet 172.17.0.1
Ip Address Of Last Host On Subnet 172.17.255.254Broadcast Address For This Subnet 172.17.255.255
How to get the binary number for IP Address
172 17 99 71172/2 = 0
86/2 = 0
43/2 = 1
21/2 = 1
10/2 = 0
5/2 = 1
2/2 = 0
½ = 1
17/2 = 1
8/2 = 0
4/2 = 0
2/2 = 0
1/2 = 1
0/2 = 0
0/2 = 0
0/2 = 0
99/2 = 1
49/2 = 1
24/2 = 0
12/2 = 0
6/2 = 0
3/2 = 1
1/2 = 1
0/2 = 0
71/2 = 1
35/2 = 1
17/2 = 1
8/2 = 0
4/2 = 0
2/2 = 0
1/2 = 1
0/2 = 0
The final answer in the binary number10101100 00010001 01100011 01000111
Host Ip Address
172 17 99 71
Ip Address 10101100 00010001 01100011 01000111
MultipleSubnet Mask
11111111 11111111 00000000 00000000
Subnet Address
10101100 00010001 00000001 00000000 Answer
Subnet Address For This IP Address
172 17 0 0
127.17.99.71
Ip address of first host:172.17.99.71
Ip address of last host:172.17.255.254
Broadcast address:192.17.99.255
0+254 hosts
Adress ip of host last calculation=254
Number of ip subnet and host
Look at the hosts in the subnet mask. Use the "Mask of Possibilities" chart to determine the bit is set to one. If no bit is set to one, there is no subnet. If any bit is set to one, proceed by steps.
172.17.99.71
255.255.0.0 = 00000000 00000000
Calculating the amount in octal host subnet mask. Raise X. calling this number 2 to the power of X. Use the "Power of 2" chart if necessary. This is the number of subnet potential created by the mask. Two of the subnet potential usually can not be used.
Calculating the total number of zeros in the subnet mask range. Raise Y. calling this number 2 to the power of Y. Use the "Power of 2" chart if necessary. This is the number of subnet potential created by the mask. Two of these numbers will not be used for addressing hosts.
Number of Usable Hosts per Subnet2^16= 65536 - 2 = 65534 hosts per subnet
Problem 5
Host IP Address 192.168.3.219Subnet Mask 255.255.0.0Number Of Subnet Bit 8Number Of Subnets 2^8=256 subnetsNumber Of Host Bits Per Subnet 8 bitNumber Of Usable Hosts Per Subnet 2^8=256-2=254 hosys per subnetSubnet Address For This IP Adress 192.168.3.0Ip Address Of First Host On This Subnet 192.168.3.1
Ip Address Of Last Host On Subnet 192.168.3.254Broadcast Address For This Subnet 192.168.3.255
How to get the binary number for IP Address
192 168 3 219192/2 = 0
96/2 = 0
48/2 = 1
24/2 = 1
12/2 = 0
6/2 = 1
3/2 = 1
½ = 1
168/2 = 0
84/2 = 0
42/2 = 0
21/2 = 1
10/2 = 0
5/2 = 1
2/2 = 0
1/2 = 0
3/2 = 1
1/2 = 1
0/2 = 0
0/2 = 0
0/2 = 0
0/2 = 0
0/2 = 0
0/2 = 0
219/2 = 1
109/2 = 1
54/2 = 0
27/2 = 1
13/2 = 1
6/2 = 0
3/2 = 1
1/2 = 1
The final answer in the binary number11101100 00101000 00000011 11011011
Host Ip Address
192 168 3 219
Ip Address 11000000 10101000 00000011 11011011
MultipleSubnet Mask
11111111 11111111 00000000 00000000
Subnet Address
11000000 10101000 00000011 00000000 Answer
Subnet Address For This IP Address
192 168 3 0
192.168.3.219
Ip address of first host:192.168.3.219
Ip address of last host:192.168.3.254
Broadcast address:192.192.10.255
0+254 hosts
Adress ip of host last calculation=254
Number of ip subnet and host
Look at the hosts in the subnet mask. Use the "Mask of Possibilities" chart to determine the bit is set to one. If no bit is set to one, there is no subnet. If any bit is set to one, proceed by steps.
192.168.3.219
255.255.0.0 = 11111111 00000000
Counting the number of people in various subnet mask. Raise X. calling this number 2 to the power of X. Use the "Power of 2" chart if necessary. This is the number of subnet potential created by the mask. Two of the subnet potential usually can not be used.
11111111 00000000 = 8 ones. 2^8= 256 usable subnets created
Number of Usable host per subnet 2^8=256-2=254 host per subnet
Problem 6
Host IP Address 192.168.3.219Subnet Mask 255.255.255.224Number Of Subnet Bit 14Number Of Subnets 2^14=16384 subnetsNumber Of Host Bits Per Subnet 2 bitNumber Of Usable Hosts Per Subnet 2^2=4-2=2 hosys per subnetSubnet Address For This IP Adress 192.168.3.216Ip Address Of First Host On This Subnet 192.168.3.217
Ip Address Of Last Host On Subnet 192.168.3.218Broadcast Address For This Subnet 192.168.3.255.219
How to get the binary number for IP Address
192 168 3 219192/2 = 0
96/2 = 0
48/2 = 1
24/2 = 1
12/2 = 0
6/2 = 1
3/2 = 1
½ = 1
168/2 = 0
84/2 = 0
42/2 = 0
21/2 = 1
10/2 = 0
5/2 = 1
2/2 = 0
1/2 = 0
3/2 = 1
1/2 = 1
0/2 = 0
0/2 = 0
0/2 = 0
0/2 = 0
0/2 = 0
0/2 = 0
219/2 = 1
109/2 = 1
54/2 = 0
27/2 = 1
13/2 = 1
6/2 = 0
3/2 = 1
1/2 = 1
The final answer in the binary number11101100 00101000 00000011 11011011
Host Ip Address
192 168 3 219
Ip Address 11000000 10101000 00000011 11011011
MultipleSubnet Mask
11111111 11111111 11111111 11111100
Subnet Address
11000000 10101000 00000011 11011000 Answer
Subnet Address For This IP Address
192 168 3 216
192.168.3.219
Ip address of first host:192.168.3.218
Ip address of last host:192.168.3.219
Broadcast address:192.192.10.255
216+2 hosts
Adress ip of host last calculation=218
Number of ip subnet and host
Look octet hosts in the subnet mask. Use the "Mask of Possibilities" chart to determine the bit is set to one. If no bit is set to one, there is no subnet. If any bit is set to one, proceed by steps.
192.168.3.219
255.255.255.224 = 11111111 11111100
Counting the number of people in various subnet mask. Raise X. calling this number 2 to the power of X. Use the "Power of 2" chart if necessary. This is the number of potential created by the subnet mask. Two of the subnet potential usually can not be used.
Calculating the total number of zeros in the subnet mask range. Raise Y. calling this number 2 to the power of Y. Use the "Power of 2" chart if necessary. This is the number of subnet potential created by the mask. Two of these numbers will not be used for addressing hosts.