Answer:
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
