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Question about subneting
zimskiz
Member Posts: 98 ■■□□□□□□□□
in CCNA & CCENT
Task :
1. Subnet the address space 172.16.1.128/25 to provide 50 host addresses
Assign the first available subnet to R1 LAN
2.Subnet the remaining address space to provide 30 host addresses
Assign the next available subnet to R2 LAN
My own response but it seems it's wrong:
1. 50 = 2^6 - 2 = 62 (so 6 bits for host , 2 for network)
256 - (128+64) = 64
Network 0 64 128 192
\\\\\\\\\ 1 65 129 193
\\\\\\\\\ 62 126 190 254
Broadcast 63 127 191 255
For me the first available subnet it is the one mark with red. Why it's wrong? (The correct answer was the 172.16.1.128\26 network)
And Question 2. What remaining addresses ? I don't understand what that means(subnet the remaining address space)
Thank!
1. Subnet the address space 172.16.1.128/25 to provide 50 host addresses
Assign the first available subnet to R1 LAN
2.Subnet the remaining address space to provide 30 host addresses
Assign the next available subnet to R2 LAN
My own response but it seems it's wrong:
1. 50 = 2^6 - 2 = 62 (so 6 bits for host , 2 for network)
256 - (128+64) = 64
Network 0 64 128 192
\\\\\\\\\ 1 65 129 193
\\\\\\\\\ 62 126 190 254
Broadcast 63 127 191 255
For me the first available subnet it is the one mark with red. Why it's wrong? (The correct answer was the 172.16.1.128\26 network)
And Question 2. What remaining addresses ? I don't understand what that means(subnet the remaining address space)
Thank!
Comments
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OptionsNetworkVeteran
Member Posts: 2,338 ■■■■■■■■□□
Your answer is wrong because you were only given 172.16.1.128/25 to work with. In binary that's:Network 0 64 128 192
\\\\\\\\\ 1 65 129 193
\\\\\\\\\ 62 126 190 254
Broadcast 63 127 191 255
For me the first available subnet it is the one mark with red. Why it's wrong?
"10101100.00010000.00000001.1xxxxxxx"
You can't modify the first, second, or third octets at all. You have no rights to that address space. You can only modify the binary digits (bits) that are set to an x (don't care) above.Task :
1. Subnet the address space 172.16.1.128/25 to provide 50 host addresses
Assign the first available subnet to R1 LAN
2.Subnet the remaining address space to provide 30 host addresses
Assign the next available subnet to R2 LAN
1) You need 50 host addresses. 2^1=2, 2^2=4, 2^3=8, 2^4=16, 2^5=32, 2^6=64. Subtract two and you need 6 host bits to allow at least 50 hosts (you will be providing 62 in fact).
172.16.1.128/26
We were given 1xxxxxxx (I'm focusing only on the last octet, since that's all we have to play with).
Now we are using 10xxxxxx. (not the six host bit x's). That leaves 11xxxxxx available.
2) Per the powers of two above, we need five host bits (2^5-2 = 30).
172.16.1.192/27
Now we are using 110xxxxx. That leaves 111xxxxx available for future use if there were more requirements.
PS - These are complex subnetting requirements, beyond what network certifications require. However, the fact that your answer didn't begin with "172.16.1." indicates a serious need to review subnetting. -
Optionsfadhil
Member Posts: 200
Task :
1. Subnet the address space 172.16.1.128/25 to provide 50 host addresses
Assign the first available subnet to R1 LAN
2.Subnet the remaining address space to provide 30 host addresses
Assign the next available subnet to R2 LAN
My own response but it seems it's wrong:
1. 50 = 2^6 - 2 = 62 (so 6 bits for host , 2 for network)
256 - (128+64) = 64
Network 0 64 128 192
\\\\\\\\\ 1 65 129 193
\\\\\\\\\ 62 126 190 254
Broadcast 63 127 191 255
For me the first available subnet it is the one mark with red. Why it's wrong? (The correct answer was the 172.16.1.128\26 network)
And Question 2. What remaining addresses ? I don't understand what that means(subnet the remaining address space)
Thank!
first you should more subnet a given ip address to provide a required customer requirement.
to have 50 hosts: find (2 ^n) -2 >= 50
where n =6, that is 6 zeros of the last octet of subnet mask, must keep for hosts.
subnet will be 255.255.255.192 which is equal to /26 and remember the block size is 64.
that is, the first network is 172.16.1.128/26 while the other is 172.16.1.192/26.
(ANSWER 1) hence 172.16.1.128/26 for first LAN R1.
for qn 2.
the remaining address should more subnetted.
172.16.1.192/26
since this address has ability to carry only 62 hosts, it is true that a half of that hosts we can have more than 30 hosts.
which is equal to increase one bit to a subnet.
hence the subnets will be 172.16.1.192/27 and 172.16.1.224/27, each one with ability to carry 30 hosts.
that is 2^5 - 2 =30.
172.16.1.192/27 this will be assigned to LAN R2.
NOTE: the remaining address : an address that you didn't assign to a LAN -
Optionszimskiz
Member Posts: 98 ■■□□□□□□□□
NetworkVeteran wrote: »Your answer is wrong because you were only given 172.16.1.128/25 to work with. In binary that's:
"10101100.00010000.00000001.1xxxxxxx"
You can't modify the first, second, or third octets at all. You have no rights to that address space. You can only modify the binary digits (bits) that are set to an x (don't care) above.
1) You need 50 host addresses. 2^1=2, 2^2=4, 2^3=8, 2^4=16, 2^5=32, 2^6=64. Subtract two and you need 6 host bits to allow at least 50 hosts (you will be providing 62 in fact).
172.16.1.128/26
We were given 1xxxxxxx (I'm focusing only on the last octet, since that's all we have to play with).
Now we are using 10xxxxxx. (not the six host bit x's). That leaves 11xxxxxx available.
2) Per the powers of two above, we need five host bits (2^5-2 = 30).
172.16.1.192/27
Now we are using 110xxxxx. That leaves 111xxxxx available for future use if there were more requirements.
PS - These are complex subnetting requirements, beyond what network certifications require. However, the fact that your answer didn't begin with "172.16.1." indicates a serious need to review subnetting.
My answer begin with 172.16.1.x where x are the blocks : 0,64,128,192. My question it's why i will not using the first subnet 172.16.1.0 and i have to jump directly to 172.16.1.128? -
OptionsNetworkVeteran
Member Posts: 2,338 ■■■■■■■■□□
My question it's why i will not using the first subnet 172.16.1.0 and i have to jump directly to 172.16.1.128?
Again, because they only gave you the address space 172.16.1.128/25 to use--
"10101100.00010000.00000001.1xxxxxxx"
You can only modify with binary digits (bits) #26-32, which are marked with an "x" meaning "don't care" above.
172.16.1.0 would be:
"10101100.00010000.00000001.0xxxxxxx"
Clearly, using this subnet would violate the rule since you changed bit #25 from a "1" to "0". You were only permitted to modify bits #26-#32. You would be using address space that was never granted to you. -
Optionszimskiz
Member Posts: 98 ■■□□□□□□□□
You said :NetworkVeteran wrote: ».172.16.1.128/26
1.We were given 1xxxxxxx (I'm focusing only on the last octet, since that's all we have to play with).
2.Now we are using 10xxxxxx. (not the six host bit x's). That leaves 11xxxxxx available.
The 1. i understood but the 2. it is the problem.(Why after the 1 it is 0 : 10xxxxxx and then why 11xxxxxx it's available? Why not 10xxxxxx it is the available for expand?) -
Optionsmella060
Member Posts: 198 ■■■□□□□□□□
This is a question on VLSM - Variable Length Subnet Masking and if you haven't already mastered subnetting, then you are going to struggle with VLSM. VLSM involves essentially subnetting a subnet. You are given the subnet 172.16.1.128/25 and you have to subnet that further to find the answers.
If you are given the subnet 172.16.1.128/25, then that is the range you have to work with. You can't just start at 172.16.1.0.
What is the range of addresses covered by subnet 172.16.1.128/25 ? What are the useable host addresses ? What is the broadcast address of this subnet ? If you can't answer these questions, then you really need to review subnetting.
You should be able to work out that subnets 172.16.1.128/26 and 172.16.1.192/27 are all within the range covered by 172.16.1.128/25...This is basic subnetting.
