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Topology

I've been trying to get PC2 to get its IP via DHCP, and the Pool is located at the MKTI router. What I did was to create 3 DHCP pools at the MKTI router however, the third one keeps getting messed up..

I enter these commands:

ip dhcp excluded-address 192.168.10.1
ip dhcp excluded-address 192.168.10.65
ip dhcp excluded-address 192.168.10.105
ip dhcp pool VLAN-POOL-10
net 192.168.10.0 255.255.255.192
default-router 192.168.10.1
ip dhcp pool VLAN-POOL-20
net 192.168.10.64 255.255.255.224
default-router 192.168.10.65
ip dhcp pool POOL-QC
net 192.168.10.104 255.255.255.192
default-router 192.168.10.105

But get this as output:

ip dhcp pool VLAN-POOL-10
 network 192.168.10.0 255.255.255.192
 default-router 192.168.10.1
ip dhcp pool VLAN-POOL-20
 network 192.168.10.64 255.255.255.224
 default-router 192.168.10.65
ip dhcp pool POOL-QC
 network 192.168.10.64 255.255.255.192
 default-router 192.168.10.105

Is there any reason that QC is registering as .64 instead of .104? Does this have to do with anything on my VLSM because I'm using only a /24 address space for 3 subnets?

Edit (based on the first response): We were asked to create 4 subnets for the network under the address space of 192.168.10.0 /24. The networks needed 40, 20, 3, and 50 hosts respectively so I had to do a subnet for each network...if I did the VLSM right then the network addresses should be 10.0, 10.64, 10.96, and 10.104

Now I understand that its incorrect because 192.168.10.104 is considered a host address..but I did the subnetting and from what I've learned it should be the next network address based on the subnetting I've done...

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  • Did any answer help you? If so, you should accept the answer so that the question doesn't keep popping up forever, looking for an answer. Alternatively, you can provide and accept your own answer.
    – Ron Maupin ♦
    Commented Feb 19, 2018 at 3:51

1 Answer 1

Reset to default
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The network statement defines the subnet for the DHCP pool. 192.168.10.104/26 (255.255.255.192) is not a valid network address. It's a host address.

The /26 network that contains that IP address is 192.168.10.64.

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  • Okay I understand that to some extent... Uhm the reason this happened was because we were asked to create 4 subnets for the network under the address space of 192.168.10.0 /24. The networks needed 40, 20, 3, and 50 hosts respectively so I had to do a subnet for each network...if I did the VLSM right then the network addresses should be 10.0, 10.64, 10.96, and 10.104, did I do that right or was it incorrect?
    – kobowo
    Commented May 3, 2017 at 14:03
  • Incorrect. If you do it in binary, your error will be apparent. BTW, homework questions are off-topic here.
    – Ron Trunk
    Commented May 3, 2017 at 14:09
  • Ok, thanks, yeah I'll try to figure this out myself sorry haha
    – kobowo
    Commented May 3, 2017 at 14:19

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