Consider a router that interconnects four subnets: S1, S2, S3, and S4. Suppose all of the interfaces in each of these four subnets are required to have the prefix 199.100.8.0/21. You want to divide this address block into four equal-sized contiguous address blocks, and give one of these address blocks to each subnet connected to the router. Provide network addresses (of the form a. b.c. d/x) that satisfy this requirement.

S1: 199.100.8.0 /23

S2: 199.100.10.0 /23

S3: 199.100.12.0 /23

S4: 199.100.14.0 /23

Explanation:

199.100.8.0/21 is equivalent to 8 Class C networks.  We can prove this by starting to count at /24 provided with its subnet mask:

255.255.255.0 /24 is 1 Class C network

255.255.254.0 /23 is 2 Class C networks.

255.255.252.0 /22 is 4 Class C networks.

255.255.248.0 /21 is 8 Class C networks

(You can see the pattern that whenever the CIDR goes down a bit the Class C network multiples by 2: 1 * 2 * 2 * 2 = 8)

The subnets than you can allocate for /21 is 199.100.8.0, 199.100.9.0, 199.100.10.0, 199.100.11.0, 199.100.12.0 ,199.100.13.0, 199.100.14.0 and 199.100.15.0.

The truth is what s1, s2, s3, and s4 actually needs are four equal-sized contiguous address blocks, so what we possibly will give the 4 subnets are a combination of 2 Class C networks each. So we combine 199.100.8.0 and 199.100.9.0 to one subnet, 199.100.10.0 and 199.100.11.0 one subnet, 199.100.12.0 and 199.100.13.0 to one subnet, and 199.100.14.0 and 199.100.15.0 to one subnet. Equal to 4 subnets.

If that is the case:

S1:

Network Address 199.100.8.0 /23

Subnet Mask = 255.255.254.0

Wild Card Address = 0.0.1.255

Range = 199.100.8.1 to 199.100.9.254

Host =  510

Broadcast Address = 199.100.9.255

S2:

Network Address 199.100.10.0 /23

Subnet Mask = 255.255.254.0

Wild Card Address = 0.0.1.255

Range = 199.100.10.1 to 199.100.11.254

Host =  510

Broadcast Address = 199.100.11.255

S3:

Network Address 199.100.12.0 /23

Subnet Mask = 255.255.254.0

Wild Card Address = 0.0.1.255

Range = 199.100.12.1 to 199.100.13.254

Host =  510

Broadcast Address = 199.100.13.255

S4:

Network Address 199.100.14.0 /23

Subnet Mask = 255.255.254.0

Wild Card Address = 0.0.1.255

Range = 199.100.14.1 to 199.100.15.254

Host =  510

Broadcast Address = 199.100.15.255

the answer will be d