IP Address Correction Find the firstaddress, last address and the number ofaddresses of a network whose one of the IPaddresses is 72.250.220.112/27. kindly use thisIP in your assignment`
One of the IP addresses is 72.250.220.112 / 27.
/25 means that Subnet Mask: 255.255.255.128
First IP Address:Â 72.250.220.1
Last IP Address:Â 72.250.220.126
Number of usable addresses:Â 126
Total number of addresses:Â 128
Other information.
Network Address: 72.250.220.0
Broadcast Address: 72.250.220.127
How can calculate?
Mask 255.255.255.255 is the same as /32
for /25:
a).Â
32 - 25 = 7
"27 = 128"Â > Total number of hosts (IP addresses)
= 128, when first IP (Network Address) and last IP (Broadcast Address) not usable, so, 128 - 2 = 126 > number of usable addresses = 126.
One of the IP addresses is *.*.*.75 - we can see that it is in the first range of 128 addresses > *.*.*.0 - *.*.*.127
first IP (Network Address) is *.*.*.0 (in our case - 72.250.220.0)
last IP (Broadcast Address) is *.*.*.127 (in our case - 72.250.220.127)
Usable Host Range: 72.250.220.1 - 72.250.220.126 > 126 IP addresses
b).
32 - 25 = 7
27Â = 128 > Total number of hosts (IP addresses)
= 128
Mask "255.255.255.x" determinate "range" of total IP. Can have 256 IP address - from ".0" to ".255".
256 (possible number of IP address) - 128 (total number of IP address in our case) = 128, so, Subnet Mask is 255.255.255.128.
c). Other example for understanding number of hosts (IP addresses) and what Subnet Mask will be in /27 case
32 - 27 = 5
25Â = 32
Total number of hosts (IP addresses) = 32
Number of usable addresses (host): 32 - 2 = 30
256 - 32 = 224
Subnet Mask: 255.255.255.224
Possible /25 (255.255.255.128) Networks for 50.28.22.*
1). Network Address: 72.250.220.0
Usable Host Range: 72.250.220.1 - 50.250.220.126
Broadcast Address: 72.250.220.127
2). Network Address: 72.250.220.128
Usable Host Range: 72.250.220.129 - 72.250.220254
Broadcast Address: 72.250.220.255
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