# Scott@JAVA

Java, 一杯浓浓的咖啡伴你到深夜

## 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`
`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`

posted on 2006-01-14 11:59 Scott@JAVA 阅读(356) 评论(0)  编辑  收藏 所属分类: Network & Telecom

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