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i) What is the DBA of the last subnet?

ii) what is address of 4th last host of 2nd last subnet in the company ?

4 Answers

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$i)$ DBA of the last subnet

Starting address given to Company $= 181.55.0.0$
No. of Subnets needed $= 1000$
ie; no. of bits needed from host address part $= \left \lceil \log _2 1000 \right \rceil  = 10$

The $1000$ subnets will be numbered from $0$ to $999$.
So the starting address of the $last$ subnet $= 999$,
which is $11\space 1110\space 0111$. Putting this in subnet part, the starting address of the last subnet is, 

for $DBA$(Direct broadcast address), put $all\space host\space bits$ $1$.
So, $DBA$ of the $last$ subnet is,
$181.55.1111\space10 01. 1111\space 1111$
or $181.55.249.255$.

 

$ii)$ Address of 4th last host of 2nd last subnet

The starting address of the $second\space last$ subnet $(0-999)$ will be $998$, or $11\space 1110\space 0110$. Putting this in place of subnet address, we get the starting address of the $second\space last$ subnet as,


Since $All \space1$s being broadcast address, $11\space 1111$ can't be a host address, hence the $last$ host address will be $11\space 1110$.

Therefore the $last\space host$ in the above subnet is,

which is $181.55.249.190$. 

So, the $fourth\space last\space host$ will have an address 
$181.55.249.187$.

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Correct me if I am wrong

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IP is 181.55.0.0 so its a class B , means 16 bit for host id part .

And 1000 subnet ~ 2^10 so in the host id part it has to borrow 10 bit .

the last address of subnet (1000) will be

181.55.11111010.00000000 = 181.55.250.0

the 2nd last subnet will be 181.55.11111001.11000000 = 181.55.249.192

so 4th last host is 181.55.249.11111011 ( note :181.55.249.11111111 is not a host ) = 181.55.249.251
edited by
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1) here total no of subnet is 1024 , so total bit in subnet =10

IP is 181.55.0.0 so its a class B , means 16 bit for host id part .

but here so last subnet would be: 181.55.11111111.11000000

but in qustn last subnet is 1000 so we have to subtract 24:11000

so last subnet is :181.55.11111001.11000000

directed broadcast of last subnet(all host bit 1): 181.55.11111001.11111111=181.55.249.255


 

2) our last subnet is 181.55.11111001.11XXXXXX,

   so 2nd last would be: 181.55.11111001.10 XXXXXX

4th host of 2nd last subnet: 181.55.11111001.10111111

                                                                 181.55.249.191

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rahuljai asked Dec 15, 2018
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What’s the right answer ?, please have a look at Q.5(here) they are appending network bits but this question doesn't.