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

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

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 atheoretical 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:

e. First useable host on the 2nd useable subnet?
f. broadcast address for 2nd useable subnet?
g 1,203rd useable host address?

Explanation / Answer

assuming from your previously asked questions that first address is wire address and second address is broadcast address

if my above assumption is wrong please specify which address to use as broad cast address. I will answer based on above statement

e)

2nd usable subnet is 2+2 =4th subnet

address =172.1.0.3

now first usable host is 3rd host

host bits = 00 000000.000010 11 where subnet bits are for 2nd usable subnet.

complete address of First useable host on the 2nd useable subnet =172.1.0.11

f)

broadcast address will be address for 2nd host

host bits = 00 000000.000001 11 where subnet bits are for 2nd usable subnet.

broadcast address on the 2nd useable subnet =172.1.0.7

g) i guess you are asking 1,203rd useable host address on same subnet i.e. 2nd usable subnet.

1203rd usable host is 1205th host address

1205= 010010 110101 in binary

host bits = 00 010010.110101 11 where subnet bits are for 2nd usable subnet.

1203rd usable host address on the 2nd useable subnet =172.1.18.215

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