Answer to Question #320344 in Computer Networks for shiv

Question

Find the first address, last address and the number of addresses of a network whose one of the IP addresses is 50.28.22.75 / 28. 

Answer 1

One of the IP addresses is 50.28.22.75 /25.

/25 means that Subnet Mask: 255.255.255.128

First IP Address: 50.28.22.1

Last IP Address: 50.28.22.126

Number of usable addresses: 126

Total number of addresses: 128

Other information.

Network Address: 50.28.22.0

Broadcast Address: 50.28.22.127

How can calculate?

Mask 255.255.255.255 is the same as /32

for /25:

a). 

32 - 25 = 7

$27 = 128$ > 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 - 50.28.22.0)

last IP (Broadcast Address) is *.*.*.127 (in our case - 50.28.22.127)

Usable Host Range: 50.28.22.1 - 50.28.22.126 > 126 IP addresses

b).

32 - 25 = 7

2⁷ = 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

2⁵ = 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: 50.28.22.0

Usable Host Range: 50.28.22.1 - 50.28.22.126

Broadcast Address: 50.28.22.127

2). Network Address: 50.28.22.128

Usable Host Range: 50.28.22.129 - 50.28.22.254

Broadcast Address: 50.28.22.255