Scott@JAVA

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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
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(304) 评论(0)  编辑  收藏 所属分类: Network & Telecom


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