Lab 18 internet security homework. Provided the word document answer template in which has been started already

Lab 18 Answer Template

Problem 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	192.168.10.255

Problem 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: _________________________________________

b. Subnet Mask: _________________________________________

c. Bits Borrowed: _________________________________________

d. Number of possible subnets: _________________________________________

e. Magic number: _________________________________________

f. Number of valid hosts per subnet: _________________________________________

g. (Sub) network address of subnet 0: _________________________________________

h. First usable host address in subnet 0: _________________________________________

i. Last usable host address in subnet 0: _________________________________________

j. Broadcast address in subnet 0: _________________________________________

k. (Sub) Network address in subnet 1: _________________________________________

l. Last usable host address in subnet 2: _________________________________________

Subnet	Subnet address	1st Host address	Last Host address	Broadcast
0
1
2
3
Etc.

Problem 5:

IP address: 192.10.10.0 (given) (Note that this address is slightly different than above)

Minimum number of hosts needed per subnet: 16

(Remember that the goal is to find a solution that will waste as few host addresses as possible but still satisfy the requirements.)

a. Slash prefix: _________________________________________

b. Subnet Mask: _________________________________________

c. Bits Borrowed: _________________________________________

d. Number of possible subnets: _________________________________________

e. Magic number: _________________________________________

f. Number of valid hosts per subnet: _________________________________________

g. (Sub) network address of subnet 0: _________________________________________

h. First usable host address in subnet 0: _________________________________________

i. Last usable host address in subnet 0: _________________________________________

j. Broadcast address in subnet 0: _________________________________________

k. (Sub) Network address in subnet 1: _________________________________________

l. Last usable host address in subnet 2: _________________________________________

Subnet	Subnet address	1st Host address	Last Host address	Broadcast
0
1
2
3
Etc.

Answer the following questions. Try to answer them without using tables to guide you:

1. How many usable hosts per sub-network will you have using a /20 address? ______

2. Given a Class B address, what is the magic number if you need to subnet using a /23 prefix? ______________________

3. In a Class B address, how many sub-networks can you have with a /18 prefix? _______________

4. How many total addresses per network can you have with a subnet mask of 255.255.224.0? (Note that total addresses include network and broadcast addresses.)___________