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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 ?
asked | 851 views
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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

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
answered by Boss (10.3k points)
edited
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The last DBA is 181.55.11111111.10111110 this is last DBA Is this wrong
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Bcz 10 bits for subnet in that last address is assigned for network DBA I am thinking like that is this wrong
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see there is a ambiguity in this question because for 1000 we need 10 bits and with 10 bits we can for 1024 subnets , so 24 subnet will be extra , you can take any of these subnet as last , but i take it in sequential way , you can take also the last subnet buy skipping all 24 subnet ( like 997 998  1023th ) . in question they should have mentioned how to take .
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@pranay your approach is correct but you have taken the 2nd last subnet no as 999th subnet (since you have taken it sequentially) , but it shud be 998th becoz the all 0's in subnet no will be considered the 1st subnet .

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thanks :)

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

answered by Active (2.1k points)
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For fourth host we have to put ooo100 in 6 bits for host why we are taking 111111 in 6 bits of host

$i)$ 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
$181.55.249.187$.

answered by Active (1.3k points)

Correct me if I am wrong

answered by Junior (507 points)

+1 vote
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