Addressing in the Internet Protocol in detail. There are two techniques to assig
ID: 3914979 • Letter: A
Question
Addressing in the Internet Protocol in detail. There are two techniques to assign IP addresses to the hosts within sub-networks, FLSM and VLSM. These techniques (together with the configuration of Cisco servers to be seen in Chapter 2), are the single most important concepts.
Below is an assignedaddress block and the maximum expected sizes for 3 subnets to be allocated within the address block (see table below). You must do the following:
Determine the Internet addresses that describe each of the 3 subnets using FLSM
Determine the Internet addresses that describe each of the 3 subnets using VLSM.
You must present the following information for each one of the 3 assigned subnets in the two scenarios (6 in total):
Network address
First usable address of the subnet
Last usable address of the subnet
Broadcast address for the subnet
Prefix length
Subnet mask
You may consult the example that is provided under this Week’s materials. Please explain any assumptions or calculations you made to arrive to your final solution.
Guidelines:
Please write your IP range and network sizes in the subject Line. Also, please present one table for FLSM and one table for VLSM using the following format.
Subnet #
Prefix Length
Subnet Mask
# Required Hosts
Max # of Hosts
Network Address
First Address
Last Address
Broadcast Address
Here is your assigned address block and subnets. Please be mindful that even though you have a large range of addresses to use (prefix length 22), you must not be wasteful and find thigh FLSM and VLSM range assignments that satisfy your requirements.
Address Block
Size of Subnet 1
Size of Subnet 2
Size of Subnet 3
172.19.108.0/22
70
120
2
Subnet #
Prefix Length
Subnet Mask
# Required Hosts
Max # of Hosts
Network Address
First Address
Last Address
Broadcast Address
Explanation / Answer
Size of Subnet 1 = 70 as 70 is the number which does not come under the power of 2^n, so 2^6 is 64 but we need 70 subnets so 2^7 can satisfy the size
Subnet #
Prefix Length
Subnet Mask
# Required Hosts
Max # of Hosts
Network Address
First Address
Last Address
Broadcast Address
172.19.108.0
29
255.255.255.248
6
768
172.19
172.19.108.1
172.19.108.6
172.19.108.7
to
172.19.111.248
29
255.255.255.248
6
768
172.19
172.19.111.249
172.19.111.254
172.19.111.255
Size of Subnet 2 = 120 as 120 is the number which does not come under the power of 2^n, so 2^6 is 64 but we need 120 subnets so 2^7 can satisfy the size
Subnet #
Prefix Length
Subnet Mask
# Required Hosts
Max # of Hosts
Network Address
First Address
Last Address
Broadcast Address
172.19.108.0
29
255.255.255.248
6
768
172.19
172.19.108.1
172.19.111.254
172.19.111.255
Size of Subnet 3 = 2 as 2 is the number which does come under the power of 2^n, so 2^1 is 2 so 2^1 can satisfy the size
Subnet #
Prefix Length
Subnet Mask
# Required Hosts
Max # of Hosts
Network Address
First Address
Last Address
Broadcast Address
172.19.108.0
23
255.255.254.0
510
1020
172.19
172.19.108.1
172.19.109.254
172.19.109.255
to
172.19.110.0
23
255.255.254.0
510
1020
172.19
172.19.110.1
172.19.111.254
172.19.111.255
Thanks
Subnet #
Prefix Length
Subnet Mask
# Required Hosts
Max # of Hosts
Network Address
First Address
Last Address
Broadcast Address
172.19.108.0
29
255.255.255.248
6
768
172.19
172.19.108.1
172.19.108.6
172.19.108.7
to
172.19.111.248
29
255.255.255.248
6
768
172.19
172.19.111.249
172.19.111.254
172.19.111.255
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