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VLSM and Loopback Addresses? Revised question
wolverene13
Member Posts: 87 ■■□□□□□□□□
in CCNA & CCENT
I am taking a CCNA cert course at a tech school. Each week, I have to do a practical exam as well as a weekly multiple-choice exam. A printout of the practical exam is given to me a few days prior to the actual exam itself. Part of this week's practical looks like this:
Router A
int s0 - DCE
int e0 - 192.168.10.1/x with 60 hosts starting at 192.168.10.2/x
int Lo0 - 6 hosts
Router B
int s0 - DTE
int e0 - 30 hosts
int Lo0 - 14 hosts
Design and apply an IP address networking scheme that allows for all subnets to be created within the 192.168.10.0/24 address range. Pair the lowest available IP address range with the subnet that requires the most hosts (start with IP zero subnet. Pair the second lowest IP address range with the subnet that requires the second most hosts. Continue this process until all networks have been accounted for. (12 points)
What I wanted to know was if each set of hosts should actually be increased by 1 because of the interface on the router (like the e0 on Router A is). For instance, with the e0 interface on Router B, it says you need 30 hosts. But in reality, I think you technically need 31 hosts because the interface on the router needs an IP address too. So, whereas originally you could have used a host block of 32 (a /27 mask with 30 useable hosts because the network ID and broadcast ID aren't useable hosts), you actually need to use a block of 64 (a /26 mask with 62 useable hosts) because you really need 30 host IP's and 1 IP address for the router interface, which would be 31 IP addresses and therefore you wouldn't have enough addresses if you used a /27. Is that right? Or am I missing something?
I did it that way and so far I have this:
Network ID
Host Range
Broadcast ID
Mask
Interface
Subnet # 1 - 192.168.10.0
.1 - .62
192.168.10.63
/26
RA - e0
Subnet # 2 - 192.168.10.64
.65 - .126
192.168.10.127
/26
RB - e0
Subnet # 3 - 192.168.10.128
.129 - .158
192.168.10.159
/27
RB - Lo0
Subnet # 4 - 192.168.10.160
.161 - .174
192.168.10.175
/28
RA - Lo0
Subnet # 5 - 192.168.10.176
.177 - .178
192.168.10.179
/30
RA & RB - s0
I wondering if this was right and also if it actually might be a trick question because I just now thought of something...how is it possible that you can assign hosts to a Loopback interface on a router? Isn't that just a logical interface and not a physical one? Where would you plug in the Ethernet cable? Would you connect a switch to the e0 interface and hook hosts up to that using VLANs and sub interfaces or something? I know it is a long question, but this is worth 12 points and the exam is only 32 points. I can take each exam twice if I don't do well the first time, but I've gotten a 900/1000 or above on every exam so far, and I really don't want to retake anything. Any help would be greatly appreciated.
Thanks!
Allen
Router A
int s0 - DCE
int e0 - 192.168.10.1/x with 60 hosts starting at 192.168.10.2/x
int Lo0 - 6 hosts
Router B
int s0 - DTE
int e0 - 30 hosts
int Lo0 - 14 hosts
Design and apply an IP address networking scheme that allows for all subnets to be created within the 192.168.10.0/24 address range. Pair the lowest available IP address range with the subnet that requires the most hosts (start with IP zero subnet. Pair the second lowest IP address range with the subnet that requires the second most hosts. Continue this process until all networks have been accounted for. (12 points)
What I wanted to know was if each set of hosts should actually be increased by 1 because of the interface on the router (like the e0 on Router A is). For instance, with the e0 interface on Router B, it says you need 30 hosts. But in reality, I think you technically need 31 hosts because the interface on the router needs an IP address too. So, whereas originally you could have used a host block of 32 (a /27 mask with 30 useable hosts because the network ID and broadcast ID aren't useable hosts), you actually need to use a block of 64 (a /26 mask with 62 useable hosts) because you really need 30 host IP's and 1 IP address for the router interface, which would be 31 IP addresses and therefore you wouldn't have enough addresses if you used a /27. Is that right? Or am I missing something?
I did it that way and so far I have this:
Network ID
Host Range
Broadcast ID
Mask
Interface
Subnet # 1 - 192.168.10.0
.1 - .62
192.168.10.63
/26
RA - e0
Subnet # 2 - 192.168.10.64
.65 - .126
192.168.10.127
/26
RB - e0
Subnet # 3 - 192.168.10.128
.129 - .158
192.168.10.159
/27
RB - Lo0
Subnet # 4 - 192.168.10.160
.161 - .174
192.168.10.175
/28
RA - Lo0
Subnet # 5 - 192.168.10.176
.177 - .178
192.168.10.179
/30
RA & RB - s0
I wondering if this was right and also if it actually might be a trick question because I just now thought of something...how is it possible that you can assign hosts to a Loopback interface on a router? Isn't that just a logical interface and not a physical one? Where would you plug in the Ethernet cable? Would you connect a switch to the e0 interface and hook hosts up to that using VLANs and sub interfaces or something? I know it is a long question, but this is worth 12 points and the exam is only 32 points. I can take each exam twice if I don't do well the first time, but I've gotten a 900/1000 or above on every exam so far, and I really don't want to retake anything. Any help would be greatly appreciated.
Thanks!
Allen
Currently Studying: CCIP - 642-611 - MPLS
Occupation: Tier II NOC Tech - Centurylink
CCIP Progress: [x] BSCI
[x] BGP
[ ] MPLS
[ ] QoS
Occupation: Tier II NOC Tech - Centurylink
CCIP Progress: [x] BSCI
[x] BGP
[ ] MPLS
[ ] QoS
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Plazma
Member Posts: 503
loopback interfaces are commonly used in lab situations and for other routing purposes.. you treat it just like any other interface. It is a logical interface meaning you don't plug anything physical into it, but it still exists and is used.
Once you get more into the NA track you will see more of how they are utilized and when you dive into the NP track, you will REALLY see how useful they are in labs.CCIE - COMPLETED!