IP Addressing and Subnetting
1. Write the default Masks for the Class A, Class B and Class C IP addresses.
2. How we can distinguish Class A, Class B, Class C, Class D and Class E IP addresses from each other. Write the range of first octet in decimal and Binary for all the 5 IP address classes.
3. Write the default subnet Masks for the following IP addresses:
4. Write down the three available ranges for assigning Private IP addresses recommended by IANA (Internet Assigned Number Authority).
5. A broadcast address is the one that addresses to all the hosts in any network. State that to create a broadcast address, all the bits of network ID portion or all the bits of host ID portion are set to 1? Write down the broadcast addresses of the networks to which the following IP addresses belong, write network addresses and ranges of their valid IP address too. (No subnetting).
6. Subnet Mask or Custom mask tells us that how many bits are used for Subnet ID portion and how many for host ID portion. Identify how many bits are used for sub netting in the following IP address using its subnet mask:
7. Extract the Network Addresses of the given IP addresses in question number 8, using the subnet masks given with them. (Remember that ANDing the IP address with the Mask extracts the network address from the given IP address).
8. Suppose you have a class C Network 208.94.115.0. Your task is to design a subnet scheme so that we can create 16 Network segments (subnets) within this Network. Each subnet should support 10-14 hosts.
9. Suppose you have a class C Network 220.94.115.0. Your task is to design a subnet scheme so that we can create 28 Network segments (subnets) within this Network. Each subnet should support hosts as given below.
• 2 Network Segment support 30 Hosts
• 4 Network Segment support 14 Hosts
• 8 Network Segment support 6 Hosts
• 14 Network Segment support 2 Hosts
Subnet Calculation from a given IP range, using the classless Subnet mask. Calculating number of hosts in a subnet and number of subnets possible to create in a given IP range.
Subnet Calculation from a given IP range, using the classless Subnet mask. Calculating number of hosts in a subnet and number of subnets possible to create in a given IP range.
This presentation contains why we need sub netting, how we do sub netting, CIDR, Subnet mask, Subnet mask value, Class A Sub netting, Class B Sub netting, Class C Sub netting.
1) Explain the advantages and disadvantages of static routing.
2) Explain the purpose of different types of static routes.
3) Configure IPv4 and IPv6 static routes by specifying a next-hop address.
4) Configure an IPv4 and IPv6 default routes.
5) Explain the use of legacy classful addressing in network implementation.
6) Explain the purpose of CIDR in replacing classful addressing.
7) Design and implement a hierarchical addressing scheme.
8) Configure an IPv4 and IPv6 summary network address to reduce the number of routing table updates.
9) Configure a floating static route to provide a backup connection.
10) Explain how a router processes packets when a static route is configured.
11) Troubleshoot common static and default route configuration issues.
Complete understanding of subnet masking
also available on the youtube channal in three parts 1,2,3
link:-
https://www.youtube.com/channel/UC36lyOTi8w1EhQ-yZssjX1g?view_as=subscriber.
This presentation contains why we need sub netting, how we do sub netting, CIDR, Subnet mask, Subnet mask value, Class A Sub netting, Class B Sub netting, Class C Sub netting.
1) Explain the advantages and disadvantages of static routing.
2) Explain the purpose of different types of static routes.
3) Configure IPv4 and IPv6 static routes by specifying a next-hop address.
4) Configure an IPv4 and IPv6 default routes.
5) Explain the use of legacy classful addressing in network implementation.
6) Explain the purpose of CIDR in replacing classful addressing.
7) Design and implement a hierarchical addressing scheme.
8) Configure an IPv4 and IPv6 summary network address to reduce the number of routing table updates.
9) Configure a floating static route to provide a backup connection.
10) Explain how a router processes packets when a static route is configured.
11) Troubleshoot common static and default route configuration issues.
Complete understanding of subnet masking
also available on the youtube channal in three parts 1,2,3
link:-
https://www.youtube.com/channel/UC36lyOTi8w1EhQ-yZssjX1g?view_as=subscriber.
Lab 18 Answer TemplateProblem 1:
IP address: 192.168.10.0 /27 (given)
a. Subnet Mask: 255.255.255.224
b. Bits Borrowed: 3
c. Number of subnets: 8
d. Magic number: 32
e. Number of valid hosts per subnet: 30
f. (Sub) network address of subnet 0: 192.168.10.0
g. First usable host address in subnet 0: 192.168.10.1
h. Last usable host address in subnet 0: 192.168.10.30
i. Broadcast address in subnet 0: 192.168.10.31
j. (Sub) Network address in subnet 3: 192.168.10.64
k. Last usable host address in subnet 4: 192.168.10.158
Subnet
Subnet address
1st Host address
Last Host address
Broadcast
0
192.168.10.0
192.168.10.1
192.168.10.30
192.168.10.31
1
192.168.10.32
192.168.10.33
192.168.10.62
192.168.10.63
2
192.168.10.64
192.168.10.65
192.168.10.94
192.168.10.95
3
192.168.10.96
192.168.10.97
192.168.10.126
192.168.10.127
4
192.168.10.128
192.168.10.129
192.168.10.158
192.168.10.159
5
192.168.10.160
192.168.10.161
192.168.10.190
192.168.10.191
6
192.168.10.192
192.168.10.193
192.168.10.222
192.168.10.223
7
192.168.10.224
192.168.10.225
192.168.10.254
192.168.10.255Problem 2:
IP address: 192.168.10.0 / 26 (given)
a. Subnet Mask: 255.255.255.192
b. Bits Borrowed:
c. Number of subnets: _4
d. Magic number:
e. Number of valid hosts per subnet: 62
f. (Sub) network address of subnet 0: 192.168.10.0
g. First usable host address in subnet 0: 192.168.10.1
h. Last usable host address in subnet 0: 192.168.10.62
i. Broadcast address in subnet 0: 192.168.10.63
j. (Sub) Network address in subnet 1: 192.168.10.64
k. Last usable host address in subnet 2: 192.168.10.190
Subnet
Subnet address
1st Host address
Last Host address
Broadcast
0
192.168.10.0
192.168.10.1
192.168.10.14
192.168.10.15
1
192.168.10.16
192.168.10.17
192.168.10.30
192.168.10.31
2
192.168.10.32
192.168.10.33
192.168.10.46
192.168.10.47
3
192.168.10.48
192.168.10.49
192.168.10.62
192.168.10.63
Etc.
Problem 3:
IP address: 192.168.10.0 (given)
Subnet Mask: 255.255.255.240
a. Slash prefix: _________________________________________
b. Bits Borrowed: _________________________________________
c. Number of possible subnets: _________________________________________
d. Magic number: _________________________________________
e. Number of usable hosts per subnet: _________________________________________
f. (Sub) network address of subnet 0: _________________________________________
g. First usable host address in subnet 0: _________________________________________
h. Last usable host address in subnet 0: _________________________________________
i. Broadcast address in subnet 0: _________________________________________
j. (Sub) Network address in subnet 1: _________________________________________
k. Last usable host address in subnet 2: _________________________________________
Subnet
Subnet address
1st Host address
Last Host address
Broadcast
0
1
2
3
Etc.
Problem 4:
IP address: 192.168.10.0 (given)
Minimum number of subnets needed: 31
a. Slash prefix: _ ...
In this networking presentation, we have covered NAT and classful Sub netting and classless sub netting using IPv4 address. we find number of hosts,total networks,first valid IP address, Last Valid Ip Address,Host ID,Network ID
Here is the presentation for Network Layer Numericals from the book Andrew S. Tanenbaum (Computer Networks) and B A Forouzan ( Data Communication and Networking)
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Assignment 1 iap
1. Assignment # 1
Internet Architecture and Protocols
IP Addressing and Subnetting
1. Write the default Masks for the Class A, Class B and Class C IP
addresses.
For Class A: 255.0.0.0
For Class B: 255.255.0.0
For Class C: 255.255.255.0
2. How we can distinguish Class A, Class B, Class C, Class D and Class E
IP addresses from eachother. Write the range of first octetin decimal
and Binary for all the 5 IP address classes.
Class Address Ending
Address
Starting Address in binary Ending Address in
binary
A 1.0.0.0 126.255.255.255 00000001.00000000.
00000000. 00000000
0111110.11111111.
11111111. 11111111
B 128.0.0.0 191.255.255.255 10000000. 00000000.
00000000. 00000000
10111111.11111111.
11111111. 11111111
C 192.0.0.0 223.255.255.255 11000000. 00000000.
00000000. 00000000
11011111.11111111.
11111111. 11111111
D 224.0.0.0 239.255.255.255 11100000. 00000000.
00000000. 00000000
11101111.11111111.
11111111. 11111111
E 240.0.0.0 255.255.255.255 11110000. 00000000.
00000000. 00000000
11110111.11111111.
11111111. 11111111
3. Write the default subnet Masks forthe following IP addresses:
IP Address Class Default Subnet Mask
179.65.225.4 B 255.255.0.0
222.35.20 C 255.255.255.0
111.7.80.0 A 255.0.0.0
4. Write down the three available ranges for assigning Private IP addresses recommended by IANA
(Internet Assigned Number Authority).
Address Blocks
IP Address Class Private IP Address
10.0.0.0 A 10.255.255.255
172.16..0.0 B 172.31.255.255
192.168.0.0 C 192.168.255.255
5. A broadcast address is the one that addresses to all the hosts in any network. State that to create a
broadcast address,all the bits of network ID portion or all the bits of host ID portion are set to 1? Write
2. down the broadcast addresses ofthe networks to which the following IP addresses belong,write network
addresses and ranges of their valid IP address too.(No subnetting)
IP Address Network Address Broadcast Address
129.65.225.4 129.65.0.0 129.65.255.255
211.35.20.18 211.35.20.0 211.35.20.255
180.47.115.6 180.47.0.0 180.47.255.255
6. Subnet Mask or Custom mask tells us that how many bits are used for Subnet ID portion and how many
for host ID portion. Identify how many bits are used for sub netting in the following IP address using its
subnet mask:
IP Address Subnet Mask Binary Subnet ID bits Host ID bits
135.65.225.4 255.255.240.0 255.255.11110000.0 20 12
210.35.20.18 255.255.255.248 255.255.255.11111000 29 3
190.47.115.6 255.255.254.0 255.255.11111110.0 23 9
7. Extract the Network Addresses of the given IP addresses in question number 8, using the subnet masks
given with them. (Remember that ANDing the IP address with the Mask extracts the network address from
the given IP address).
1- IP Address 135.65.11100001.4
Subnet Mask 255.255.11110000.0
After AND
Network Address 135.65.11100000.0 135.65.224.0
2- IP Address 210.35.20.00010010
Subnet Mask 255.255.255.11111000
After AND
Network Address 210.35.20.00010000 210.35.20.16
3- IP Address 190.47.01110011.6
Subnet Mask 255.255.11111110.0
After AND
Network Address 190.47.01110010.0 190.47.114.0
8. Which of the following Subnet masks would allow a class A network to allow subnets to have up to 150
hosts and allow for up to 164 subnets?
a. 255.0.0.0 b. 255.255.255.0 c. 255.255.192.0 d. 255.255.240.0 e.255.255.252.0
For allowing 150 hosts minimum 8 bits of host part should be available. And for allowing 164 subnets
8 bits are minimal required. All of these will allow to have up to 8 host’s bits and 8 network bits
available. According to the formula
2n >= required hosts For Hosts
2n >= required subnets For Subnets
9. Suppose you have a class C Network 208.94.115.0. Your task is to design a subnet scheme so that we can
create 16 Network segments (subnets)within this Network. Each subnet should support 10-14 hosts.
2n >= 14
24 >= 14
16 > 14
So, n = 4.
a. How many bits would you use for the subnet ID?
28 bits are used for subnet ID.
b. How many bits would you use for the Host ID?
4 bits are used for Host ID
c. How many maximum possible subnets will be there?
3. 24 subnets are possible because we have 8 bits for Host ID. So,
16 maximum subnets are possible.
d. How many maximum possible hosts will be there in each subnet?
24 subnets are possible because we have 8 bits for Host ID. So,
16 maximum hosts are possible per subnet.
e. Write down the Subnet Mask of your scheme.
208.94.115.11110000
255.255.255.240
f. Write down All the valid IP addresses
Valid IP’s
208.94.115.0/28
208.94.115.16/28
208.94.115.32/28
208.94.115..48/28
208.94.115.64/28
208.94.115.80/28
208.94.115.96/28
208.94.115.112/28
208.94.115.128/28
208.94.115.160/28
208.94.115.176/28
208.94.115.192/28
208.94.115.208/28
208.94.115.224/28
208.94.115.240/28
g. The broadcast address ofthe First subnet ofyour scheme.
208.94.115.15 /28
10. Suppose you have a class C Network 220.94.115.0. Your task is to design a subnet scheme so that we can
create 28 Network segments (subnets)within this Network. Each subnet should support hosts as given
below.
Steps
1- Write IP
220.94.115.0
2- Convert to binary
220.94.115.00000000
3- Find n
2n >= max host value
2n >= 30
25 >= 30
N=5
4- Give bits to network
220.94.115.11100000
5- IP of this network
220.94.115.224
6- Subnet Mask of IP
255.255.255.224/27
4. 7- Subnets
220.94.115.0/27
220.94.115.32/027
220.94.115.64/027
220.94.115.96/027
220.94.115.128/027
220.94.115.160/027
220.94.115.192/027
220.94.115.224/027
These are the network segments that can be created. Each subnet will support up to 32 hosts.
To make the 28 segments we can use Class A or B for more segments.
