subnetting problem CCNA

in CCNA & CCENT
Hello, I cannot seem to figure out an answer for this scenario within the CCNA course: Routing Protocols and Concepts
My question is about Activity 3.5.4: Subnetting Scenario 3
Here is the scenario PDF:
lab354.pdf
Scenario:
In this lab, you have been given the network address 192.168.1.0 /24 to subnet and provide the IP addressing for the network shown in the Topology Diagram. The network has the following requirements:
The BRANCH 1 LAN 1 will require 15 host IP addresses.
The BRANCH 1 LAN 2 will require 15 host IP addresses.
The BRANCH 2 LAN 1 will require 15 host IP addresses.
The BRANCH 2 LAN 2 will require 15 host IP addresses.
The HQ LAN will require 70 host IP addresses.
The link from HQ to BRANCH 1 will require an IP address for each end of the link.
The link from HQ to BRANCH 2 will require an IP address for each end of the link.
The link from BRANCH 1 to BRANCH 2 will require an IP address for each end of the link.
Note: Remember that the interfaces of the network devices are also host IP addresses and are included in the above addressing requirements.
Task 3: Reflection
You do not have enough address space to implement an addressing scheme. Research this problem and propose a possible solution. Increasing the size of your original space is not acceptable solution.
Only the 192.168.1.0 /24 address space is used.
PCs and routers can ping all addresses.
Well, with this scenario the first subnet mask is 255.255.255.128 for HQ LAN with 70 host IP addresses.
Range: from 0 to 127
The next subnet mask for BRANCH 1 LAN 1 with 15 host IP addresses is 255.255.255.224 (255.255.255.240 can’t be because it provide 14 valid host IP and I need 15).
Range: from 128 to 159
BRANCH 1 LAN 2
Range: from 160 to 191
BRANCH 2 LAN 1
Range: from 192 to 223
BRANCH 2 LAN 2
Range: from 224 to 255
And finally I’m don’t have enough address space to implement addressing scheme for WAN links.
Anything ideas?
Thanks for any help.
My question is about Activity 3.5.4: Subnetting Scenario 3
Here is the scenario PDF:
lab354.pdf
Scenario:
In this lab, you have been given the network address 192.168.1.0 /24 to subnet and provide the IP addressing for the network shown in the Topology Diagram. The network has the following requirements:
The BRANCH 1 LAN 1 will require 15 host IP addresses.
The BRANCH 1 LAN 2 will require 15 host IP addresses.
The BRANCH 2 LAN 1 will require 15 host IP addresses.
The BRANCH 2 LAN 2 will require 15 host IP addresses.
The HQ LAN will require 70 host IP addresses.
The link from HQ to BRANCH 1 will require an IP address for each end of the link.
The link from HQ to BRANCH 2 will require an IP address for each end of the link.
The link from BRANCH 1 to BRANCH 2 will require an IP address for each end of the link.
Note: Remember that the interfaces of the network devices are also host IP addresses and are included in the above addressing requirements.
Task 3: Reflection
You do not have enough address space to implement an addressing scheme. Research this problem and propose a possible solution. Increasing the size of your original space is not acceptable solution.
Only the 192.168.1.0 /24 address space is used.
PCs and routers can ping all addresses.
Well, with this scenario the first subnet mask is 255.255.255.128 for HQ LAN with 70 host IP addresses.
Range: from 0 to 127
The next subnet mask for BRANCH 1 LAN 1 with 15 host IP addresses is 255.255.255.224 (255.255.255.240 can’t be because it provide 14 valid host IP and I need 15).
Range: from 128 to 159
BRANCH 1 LAN 2
Range: from 160 to 191
BRANCH 2 LAN 1
Range: from 192 to 223
BRANCH 2 LAN 2
Range: from 224 to 255
And finally I’m don’t have enough address space to implement addressing scheme for WAN links.
Anything ideas?
Thanks for any help.
Comments
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Ltat42a Member Posts: 587 ■■■□□□□□□□
How about this???
For the 70 host LAN - use /25
For each of the 15 host LAN's - use /27
For the serial links - use /30 -
martell1000 Member Posts: 389
Okay. I guess only way to get this done is to change the topology unless you are not allowed to increase the amount of space.
If you put together the hosts to ONE subnet in the branch offices by connecting the switches together, it will give you 30 hosts per subnet - which will allow you to keep the /27 and will free up enough space for the point to point wan links ...
sounds like the "cisco way" as 15+15=30 whicht perfectly fills up your /27 subnets.
please let me know if thats the right solutionAnd then, I started a blog ... -
chaddsuk Member Posts: 6 ■□□□□□□□□□
Hmm a tricky one this actually as I also ran out of IP's for the WAN links.
One thing you could do is use subnet zero which would let you use /28 addresses rather than /27 for your subnets which require 15 hosts.
Im not sure this would be the best solution but would work I guess
192.168.1.0/25 (126 usable) - 70 hosts required
192.168.1.128/28 - (15 usuable hosts) 15 required
192.168.1.144/28 - (15 usuable hosts) 15 required
192.168.1.160/28 - (15 usuable hosts) 15 required
192.168.1.176/28 - (15 usuable hosts) 15 required
WAN link from HQ to Branch 1 - 192.168.1.192/30
WAN link from HQ to Branch 2 - 192.168.1.194/30
WAN link from Branch 1 to Branch 2 - 192.168.1.196/30
I have never used subnet zero and I think its frowned upon however technically it should be fine for this I would think.A+
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martell1000 Member Posts: 389
subnet zero does not allow 15 or 16 hosts for a /28 subnet - it just allows 16 subnets instead of 14 for that netmask...
nowadays subnet zero is kind of a "default" :Subnet zero and the all-ones subnet
The first subnet obtained from subnetting has all bits in the subnet bit group set to zero (0). It is therefore called subnet zero.[6] The last subnet obtained from subnetting has all bits in the subnet bit group set to one (1). It is therefore called the all-ones subnet.[7]
The IETF originally discouraged the production use of these two subnets at one point due to possible confusion of having a network and subnet with the same address.[8] The practice of avoiding subnet zero and the all-ones subnet was declared obsolete in 1995 by RFC 1878, an informational, but now historical RFC.[9]And then, I started a blog ... -
chaddsuk Member Posts: 6 ■□□□□□□□□□
martell1000 wrote: »subnet zero does not allow 15 or 16 hosts for a /28 subnet - it just allows 16 subnets instead of 14 for that netmask...
nowadays subnet zero is kind of a "default" :
Ah fair enough, as I said ive never really played/researched it properly which shows lol
Well in that case I am stumped on this question then grrrA+
Network +
CCENT
CCNA
Windows 2003 server MCP
Windows XP MCP
2nd Line NOC Support Engineer (2 years) -
Ltat42a Member Posts: 587 ■■■□□□□□□□
192.168.1.128/28 - (15 usuable hosts) 15 required
192.168.1.144/28 - (15 usuable hosts) 15 required
192.168.1.160/28 - (15 usuable hosts) 15 required
192.168.1.176/28 - (15 usuable hosts) 15 required
Wouldn't this mask produce only 14 hosts, not 15??
/28 = 255.255.255.240
4 bits used for subnets 2(4) = 16
4 bits used for hosts 2(4)-2 = 14 -
DPG Member Posts: 780 ■■■■■□□□□□
macattacker wrote: »You do not have enough address space to implement an addressing scheme. Research this problem and
propose a possible solution. Increasing the size of your original address space is not an acceptable
solution. (Hint: We will discuss solutions to this problem in Chapter 6.)
You can always tell that school is back in session when people come here to have their homework done for them. -
simonmoon Member Posts: 29 ■□□□□□□□□□
martell1000 wrote: »Okay. I guess only way to get this done is to change the topology unless you are not allowed to increase the amount of space.
If you put together the hosts to ONE subnet in the branch offices by connecting the switches together, it will give you 30 hosts per subnet - which will allow you to keep the /27 and will free up enough space for the point to point wan links ...
sounds like the "cisco way" as 15+15=30 whicht perfectly fills up your /27 subnets.
please let me know if thats the right solution
I suspect it is this. I wonder how seperate the branch lans need to be. They need to be able to ping each other, but I suppose they should have seperate broadcast addresses. Vlans maybe? -
martell1000 Member Posts: 389
i see nothing about that in the task description - so i guess thats just it. but a typical setup would be some vlans with a router on a stick configuration...And then, I started a blog ... -
instant000 Member Posts: 1,745
You can always tell that school is back in session when people come here to have their homework done for them.
No kidding.
It would be easier if he posted his proposed solution, and then we assisted him with what he proposed, versus just asking us for the answers.Currently Working: CCIE R&S
LinkedIn: http://www.linkedin.com/in/lewislampkin (Please connect: Just say you're from TechExams.Net!)