Have I done my subnetting correctly
thomas130
Member Posts: 184
in CCNA & CCENT
Please may someone check to see if I have answer this question correctly
There is a total of six LANs on the system, shown in Figure 1, and each will have three VLANs, making a total of 18 systems. You have been issued 150.75.0.0/16 and your task is to create a subnet scheme that will issue subnets 0 to 17 to each of the VLANs in each LAN in turn. You must provide a table that lists:
WAN links from site to site, identify the last subnet that is to be VLSM’d (this is the
broadcast subnet). Using the last subnet, provide a VLSM structure that can allow for the six addresses required to connect each LAN to the TEST Central Switch as well as the two needed for
There is a total of six LANs on the system, shown in Figure 1, and each will have three VLANs, making a total of 18 systems. You have been issued 150.75.0.0/16 and your task is to create a subnet scheme that will issue subnets 0 to 17 to each of the VLANs in each LAN in turn. You must provide a table that lists:
WAN links from site to site, identify the last subnet that is to be VLSM’d (this is the
broadcast subnet). Using the last subnet, provide a VLSM structure that can allow for the six addresses required to connect each LAN to the TEST Central Switch as well as the two needed for
Comments
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Xenz
Member Posts: 140
Your question seems incomplete. Your addressing looks fine for the 150.75.0.0/16 network using a /21 or 255.255.248.0 mask for the subnets. Your P2P addresses are fine.Currently working on:
CCNP, 70-620 Vista 70-290 Server 2003
Packet Tracer activities and ramblings on my blog:
http://www.sbntech.info -
thomas130
Member Posts: 184
Yeah I took some of the question out because this is for uni and was'nt sure if it would be class as cheating if they saw a post asking someone to check my answer
Thank you for looking -
billscott92787
Member Posts: 933
In this example I'm assuming based on the information that again there are 6 LANS, meaning 6 different subnets. Separated by 3 different VLANS, for a total of 18 subnets. It appears you would have to be using a /21 mask based on your subnets because 256-8 = 248 or /21.
Which means that from the default network address you borrowed 5 bits. 2^5-2 = 32-2 = 30. Which means that you can support the 18 hosts. 4 bits will not work. So, I would say that you did your subnetting properly Because /21 supports this requirement. This means each subnet would increment by .8, which it appears you did right. Good job
I do not think the WAN links are right though. When I look at them, it looks like you ended with the 150.75.136.0 address for your last subnet which goes to 150.75.143.255, so how could you assign this address space to a WAN link? Wouldn't the next subnet start at 150.75.144.0, and then you would break the WAN links down from there?