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

Solutions

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 :)


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