1. What are the network and broadcast addresses when the host address
195.148.9.213 and subnet mask 255.255.255.224
What was the number of the subnet?
213 = 11010101, 224 = 11100000, 213 & 224 = 11000000 = 192
Network address is 195.148.9.192, broadcast address is 195.148.9.223
2^3 = 8 subnets
2. What is the mask if the network address is 199.167.100.0 and
broadcast address is 199.167.100.255?
The mask is 255.255.255.0
3. Is the address 172.21.64.0 with mask 255.255.192.0 the host, network or
broadcast address?
64 = 1000000, 192 = 11000000, the address is network address
4. What are the network and broadcast addresses if the host address
172.21.200.4 and subnet mask 255.255.255.192?
How many subnets together? How many addresses in a network?
4 = 100, 192 = 11000000, 4 & 192 = 0
Network address is 172.21.200.0, broadcast address is 172.21.200.63
2^2 = 4 subnets, 2^6 = 64 addresses in a network
5. What are network and broadcast addresses if the host address is
201.7.110.76 and subnet mask 255.255.255.240?
How many subnets together?
76 = 1001100, 240 = 11110000, 76 & 240 = 01000000 = 64
Network address is 201.7.110.64, broadcast address is 201.7.110.79
2^4 = 16 subnets
6. Deal a class C network to 32 subnets How many addresses are there in every
subnet? How many bits are there in the subnet mask?
2^5 = 32, 8-5 = 3
2^3 = 8 addresses in every subnet, 29 bits in the subnet mask
7. (VLSM) The router connects five subnets. Account of hosts: 3, 5, 10, 30
and 100 pcs. Deal one class C network so that every network has enough
addresses. Remember to calculate one address for the interface of the router.
How many unusable addresses are left?
Assume the class C network is 194.211.79.0, subnet a, b, c, d, e stand for 3, 5, 10, 30, 100 pcs
194.211.79.0/25 --- 194.211.79.127/25 for e
194.211.79.128/26 --- 194.211.79.191/26 for d
194.211.79.192/28 --- 194.211.79.207/28 for c
194.211.79.208/28 --- 194.211.79.223/28 for b
194.211.79.240/29 --- 194.211.79.247/29 for a
240-223-1+255-247 = 24 unusable addresses are left