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Subnetwork Exercise 1

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
2^3 = 8 subnets

2. What is the mask if the network address is 199.167.100.0 and
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
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
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
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

posted on 2006-01-14 11:59 Scott@JAVA 阅读(304) 评论(0)  编辑  收藏 所属分类: Network & Telecom 只有注册用户登录后才能发表评论。 网站导航: 相关文章: