1. Sub Neting exercises

2. Exercise 1
The host has an IP address of /19. What are the first address and last addresses in the subnet?
Mask :
The first address:  /19.

3. Exercise 1 Last address : Mask : 11111111.11111111. 11100000.00000000

4. Exercise 2
The network consists of 525 hosts. What should be the network mask length used to accommodate them?
We need 10 bit for host
2^9=512
2^10=1024
The mask will be /22
In binary it looks like this:
/22 =

5. Exercise 3
Use /24 to address the following five subnets. Maximize the network mask length to accommodate the number of hosts specified below:
Subnet 1 = 76 hosts,Subnet 2 = 40 hosts ,Subnet 3 = 16 hosts, Subnet 4 = 2 hosts.
Subnet 1 = 76 hosts
26 = 64 (62 valid host addresses is not enough)
27 = 128 (126 valid host addresses is the closest number)
Result: Subnet 1 mask is /25 or  (32 – 7 = 25).
Subnet 2 = 40 hosts
26 = 64 (62 valid host addresses)
Result: Subnet 2 mask is /26 or  (32 – 6 = 26).
Subnet 3 = 16 hosts
24 = 16 Result:
Subnet 3 mask is /28 or  (32 – 4 = 28).
Subnet 4 = 2 hosts
22 = 4 (2 valid host addresses)
Result: Subnet 4 mask is /30 or  (32 – 2 = 30)

6. Exercise 4

7. Exercise 5