This is part 1 of part II same name just replace I with a II I WILL GIVE AS MUCH

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This is part 1 of part II same name just replace I with a II

I WILL GIVE AS MUCH TIME AS NEEDED FOR THIS QUESTION MULTIPLE DAYS IF NEEDED. I NEED JUST A LITTLE WORK SHOWN. I HAVE OTHER QUESTIONS SO IF YOUR GOOD AT THIS AND WANT TO KEEP MAKING POINTS MAKE A COMMENT BELOW WITH YOUR ANSWER AND I WILL GIVE YOU FIRST CHANCES TO ANSWER FOR NEXT ONES. Goodluck and have fun

39. Though there are practical reasons why we don't complicate our lives this way in the real world, from a theoretical perspective we could borrow non-consecutive bits for subnetting. So, when borrowing the first host bit, the second host bit and the last two host bits (counting from the left of the host bit field) from 172.1.0.0 can you determine the following:

a. subnet mask
b. network address for 0th subnet, 1st subnet and 2nd subnet?
c. total number of subnets and number of useable subnets?
d. number of remaining host bits and then the number of useable hosts on a subnet?

Explanation / Answer

in 172.1.0.0 i think host field is the last two bytes ie 0.0

host field

h h h h h h h h. h h h h h h h h

now we are borrowing the first host bit, the second host bit and the last two host bits for subnet.

s s h h h h h h. h h h h h h s s

subnet mask is obtained by makin host bits 0 and remaining bits 1

11111111.11111111.11000000.00000011

a) subnet mask= 255.255.192.3

b)

0th subnet = 172.1.0.0

1st subnet =172.1.0.1

2nd subnet =172.1.0.2

c) total number of subnets = 2^2 * 2^2 = 4*4=16 (2bits in front 2 bits in last)

total number of usable subnets =16-2=14

d) total number of hosts = 2^6 *2^6 = 64*64= 4096

total number of usable hosts =4096-2=4094 on a subnet

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