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URD101
IDEA SOLTION OF CS 610 ASSIGNMENT 4.. Due date 18 june 2012
CS 610 IDEA SOLTION OF ASSIGNMENT 4.. Due date 18 june 2012
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Permalink Reply by hunny on -
Assignment 4.Idea Solution of CS 610 ..
Question: 1:
How many subnets and hosts per subnet can you get from the network
174.20.0.0/23 Remember for host part at least 2 bits are required?
Answer:
The number of hosts from /23 it means that 23 bits are on while others
are off. So take number of bits which are off. For there are 9 bits, which are
off.
2^9 -2 =510 Number of hosts for number of subnets.
Since it class B its default is /16 but according to the question is /23.
So
23-16=7
Hence to get subnets =2^7 =128
Question: 1:
Write the class name against each IP address given below,
Answer:
Assignment 4.Idea Solution of CS 610 ..
Note:
Class A = 1.0.0.0 to 126.0.0.0
Class B = 128.0.0.0 to 191.255.0.0
Class C = 192.0.1.0 to 223.255.255.0
So the classes are:
……………………………………………………………………………… -
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