Question

In: Computer Science

The requirements for your assignment are: Decode the IP address 172.16.31.255 /20 Determine the following: Subnet...

The requirements for your assignment are:

Decode the IP address 172.16.31.255 /20
Determine the following:
- Subnet mask
- Multiplier
- Network number
- Subnetwork addresses

Broadcast addresses

This is what I came up with. Let me know if it's right or what I did wrong. Please explain.

IP Address

172.16.31.255

10101100.00010000.00011111.11111111

Subnet mask

11111111.11111111.11110000.00000000

255.255.240.0

Multiplier

16

Network Number

172.16.31.0

Subnetwork Addresses

Subnetwork Addresses	Broadcast Addresses
172.16.31.0	172.16.31.15
172.16.31.16	172.16.31.31
172.16.31.32	172.16.31.47
172.16.31.48	172.16.31.63
172.16.31.64	172.16.31.79
172.16.31.80	172.16.31.95
172.16.31.96	172.16.31.111
172.16.31.112	172.16.31.127
172.16.31.128	172.16.31.143
172.16.31.144	172.16.31.159
172.16.31.160	172.16.31.175
172.16.31.176	172.16.31.191
172.16.31.192	172.16.31.207
172.16.31.208	172.16.31.223
172.16.31.224	172.16.31.239
172.16.31.240	172.16.31.255

Expert Solution

Major Network: 172.16.16.0/20
Available IP addresses in major network: 4094
Number of IP addresses needed: 160
Available IP addresses in allocated subnets: 224

Subnet Name	Needed Size	Allocated Size	Address	Mask	Dec Mask	Assignable Range	Broadcast
A	10	14	172.16.16.0	/28	255.255.255.240	172.16.16.1 - 172.16.16.14	172.16.16.15
B	10	14	172.16.16.16	/28	255.255.255.240	172.16.16.17 - 172.16.16.30	172.16.16.31
C	10	14	172.16.16.32	/28	255.255.255.240	172.16.16.33 - 172.16.16.46	172.16.16.47
D	10	14	172.16.16.48	/28	255.255.255.240	172.16.16.49 - 172.16.16.62	172.16.16.63
E	10	14	172.16.16.64	/28	255.255.255.240	172.16.16.65 - 172.16.16.78	172.16.16.79
F	10	14	172.16.16.80	/28	255.255.255.240	172.16.16.81 - 172.16.16.94	172.16.16.95
G	10	14	172.16.16.96	/28	255.255.255.240	172.16.16.97 - 172.16.16.110	172.16.16.111
H	10	14	172.16.16.112	/28	255.255.255.240	172.16.16.113 - 172.16.16.126	172.16.16.127
I	10	14	172.16.16.128	/28	255.255.255.240	172.16.16.129 - 172.16.16.142	172.16.16.143
J	10	14	172.16.16.144	/28	255.255.255.240	172.16.16.145 - 172.16.16.158	172.16.16.159
K	10	14	172.16.16.160	/28	255.255.255.240	172.16.16.161 - 172.16.16.174	172.16.16.175
L	10	14	172.16.16.176	/28	255.255.255.240	172.16.16.177 - 172.16.16.190	172.16.16.191
M	10	14	172.16.16.192	/28	255.255.255.240	172.16.16.193 - 172.16.16.206	172.16.16.207
N	10	14	172.16.16.208	/28	255.255.255.240	172.16.16.209 - 172.16.16.222	172.16.16.223
O	10	14	172.16.16.224	/28	255.255.255.240	172.16.16.225 - 172.16.16.238	172.16.16.239
P	10	14	172.16.16.240	/28	255.255.255.240	172.16.16.241 - 172.16.16.254	172.16.16.255

subnet multiplexer size is 16 so from each address from the broadcast address 172.16.16.255 - 16 = 240 will be network address

from 240 substract - 1 will give us 239 that will be the broadcast address of 15th network and substract 239-16 will give us 224 that will become network address and so on

if you have any doubt then please ask me without any hesitation in the comment section below , if you like my answer then please thumbs up for the answer , before giving thumbs down please discuss the question it may possible that we may understand the question different way and we can edit and change the answers if you argue, thanks :)

venereology answered 1 hour ago

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