Need help with this. Not sure how I complete this. Please advise ASAP. This is f
ID: 667928 • Letter: N
Question
Need help with this. Not sure how I complete this. Please advise ASAP. This is for CIS387
I need a VLSM sheet created for it.
Thank you
Network The network map is as follows and needs to fit inside the 192.168.50.0/24 block. 25 Hosts 30 Hosts PC-PT PCO 1841 30 Hosts 1841 PC-PT 1841 60 Hosts PC-PT PC1 1841 60 Hosts PC-PT PC3 Assignment Word has gotten out that you are a subnetting ninja. Bill Lumbergh from Initech hired you to help out with the network redesign; the consultants who came in said they need to restructure The network diagram above is what they ve given you to work with. Create a VLSM sheet forExplanation / Answer
Well, first, I'll explain how to calculate each of the things you need to determine of a network.
The VLSM was created to not lose so many possible IPs being unused. It's liek you are creating mini networks out of one big network.
Every IP has a network part and a host part. What sets this is the mask. The mask is the one that differentiates the network portion of the host portion. In the network given, which is /24, there are 24 bits that can't change, that's the network portion.
xxxxxxxx.xxxxxxxx.xxxxxxxx.xxxxxxxx So, the bold part is the network portion and the red is the host portion. The network portion is set, we are going to change the host portion to extend the netwrok portion a little so we can have several mini networks.
Then, we only one to take a portion of the bits available from the big network. For example, here they are telling you that the network is /24, having in mind that an IP address is made out of 32 bits, being 4 octets, we have something like this: xxxxxxxx.xxxxxxxx.xxxxxxxx.xxxxxxxx the ones that are in bold you cannot change, because that would change the original network they are giving you, that's why it is /24, because there are 24 spaces (X's) in bold.
Then, we have to do the subnetting for the bigger partitions first and for the smaller later, then the order to calculate the networks would be (E,H),(J,C),A,(B,D,L,F,K,G,I) the letters that are inside the () can change their orders because they require the same quantity of hosts. So, another possible order could be (H,E),(J,C),A,(G,K,B,L,D,I,F) and so on.
So, for the first network being, for example, E, we have that it requires 60 Hosts. We know that 26 = 64 This means I could calculate a network that has 64 IP adresses available, however ALWAYS the first IP is for the Network IP itself, and the last is for the BROADCAST, Then, we wouldn't have 64 available IP addresses, but 62. But it's ok, because we only need 60.
To know the mask of this network (/X), we are going to solve the following: 32 - n = X being n the exponent of the 2. As we did before 26 then n=6 and we have that 32 - 6 = 26 Then our network is /26 To know the mask of the network we are goint to set to 1 the bits from left to right, until we reach 26 bits.
11111111.11111111.11111111.11000000
If one takes each octet to decimal numeric system, we have: 255.255.255.192. Which is our mask.
As we are just beginning, we are going to take the network they are giving us, but instead of /24, it's going to be /26 as we calculated. Then 26 bits from left to right CAN'T change, that's what the mask sees that is obeyed.
As I said, the first possible IP is the network IP, which in this case is 192.168.50.0
The broadcast address is the one in which ALL the bits in the host portions are set to 1, then in this case we have:
192.168.50.xxxxxxxx the bold part is the mask we just calculated and cannot change, then if we put to 1 all the host portion, we have:
192.168.50.00111111 And if this last part we take to decimal, it would be: 192.168.50.63, that is our broadcast.
The range is the easiest, because its first address is obtained adding +1 to the network address, and its last addres is obtained substracting -1 from the broadcast address. The Range is the one that indicates all the addresses available for hosts.
The Range in this case would be 192.168.50.1 - 192.168.50.62
Finally we can conclude:
Network E (60 hosts)
Network: 192.168.50.0
Range: 192.168.50.1 - 192.168.50.62
Broadcast: 192.168.50.63
Mask: 255.255.255.192
And one must repeat the same process for each subnet, however I'll just include the rest of the networks:
Network H (60 Hosts)
Network: 192.168.50.64 (This new network addres is calculated by adding +1 to the previous network's broadcast address)
Range: 192.168.50.65 - 192.168.50.126
Broadcast: 192.168.50.127
Mask: 255.255.255.192
Network J (30 Hosts):
N: 192.168.50.128
R: 192.168.50.129 - 192.168.50.158
B: 192.168.50.159
M: 255.255.255.224
Network C (30 Hosts):
N: 192.168.50.160
R: 192.168.50.161 - 192.168.50.190
B: 192.168.50.191
M: 255.255.255.224
Network A (25 Hosts):
N: 192.168.50.192
R: 192.168.50.193 - 192.168.50.222
B: 192.168.50.223
M: 255.255.255.224
Network B (2 Hosts):
N: 192.168.50.224
R: 192.168.50.225 - 192.168.226
B: 192.168.50.227
M: 255.255.255.252
Network D (2 hosts):
N: 192.168.50.228
R: 192.168.50.229 - 192.168.50.230
B: 192.168.50.231
M: 255.255.255.252
Network L (2 Hosts)
N: 192.168.50.232
R: 192.168.50.233 - 192.168.50.234
B: 192.168.50.235
M: 255.255.255.252
Network F (2 Hosts)
N: 192.168.50.236
R: 192.168.50.237 - 192.168.50.238
B: 192.168.50.239
M: 255.255.255.252
Network K (2 Hosts)
N: 192.168.50.240
R: 192.168.50.241 - 192.168.50.242
B: 192.168.50.243
M: 255.255.255.252
Network G (2 Hosts)
N: 192.168.50.244
R: 192.168.50.245 - 192.168.50.246
B: 192.168.50.247
M: 255.255.255.252
Network I (2 Hosts)
N: 192.168.50.248
R: 192.168.50.249 - 192.168.50.250
B: 192.168.50.251
M: 255.255.255.252