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| biggame22 2005-01-06, 5:57 pm |
| 192.168.2.0 /28
What is the 1st usable IP address on the 2nd usable subnet and the 1st usable IP on the 4th usable subnet? (no zeroes)
Seems simple, but my question is, does cisco consider the first subnet (192.168.2.1-16) usable? | |
| smrkdown 2005-01-06, 6:32 pm |
| The first subnet consists of hosts 0-15 (not 1-16), and it is usable (just not the network and broadcast addresses). To answer the question, the first usable IP in the second subnet is 192.168.2.17 and in the fourth, 192.168.2.49. | |
| biggame22 2005-01-06, 6:35 pm |
| so if the question states to not use the zero address, you still start counting from 0 on the first subnet? | |
| smrkdown 2005-01-06, 8:58 pm |
| Honestly, I forget what the CCNA books say, but I think that as you suspected, you don't use the zero subnet. Sorry I'm not much help. I guess you'd bump the numbers in my last post up by one subnet. | |
| biggame22 2005-01-07, 11:28 am |
| Thanks for the help on that. So when they say to not use the zeroes, they are referring to the entire first subnet, wether it is 16 addresses or 128 (depending on the mask). Is that accurate? | |
| larkspur 2005-01-10, 3:19 pm |
| that would be correctamoudo!!!!
192.168.2.0 / 255.255.255.240
network host broadcast
192.168.2.0 1-14 192.168.2.16
192.168.2.17 18-32 192.168.2.33
192.168.2.34 35-49 192.168.2.50
192.168.2.51 52-66 192.168.2.67
192.168.2.68 68-83 192.168.2.84
192.168.2.85 86-100 192.168.2.101
192.168.2.102 103-117 192.168.2.118
192.168.2.119 120-134 192.168.2.135
192.168.2.136 137-151 192.168.2.152
192.168.2.153 154-168 192.168.2.169
192.168.2.170 171-185 192.168.2.186
192.168.2.187 188-202 192.168.2.203
192.168.2.204 205-219 192.168.2.220
192.168.2.221 222-239 192.168.2.240 | |
| JimmyD 2005-01-10, 4:05 pm |
| quote:
Originally posted by larkspur
that would be correctamoudo!!!!
192.168.2.0 / 255.255.255.240
network host broadcast
192.168.2.0 1-14 192.168.2.16
192.168.2.17 18-32 192.168.2.33
192.168.2.34 35-49 192.168.2.50
192.168.2.51 52-66 192.168.2.67
192.168.2.68 68-83 192.168.2.84
192.168.2.85 86-100 192.168.2.101
192.168.2.102 103-117 192.168.2.118
192.168.2.119 120-134 192.168.2.135
192.168.2.136 137-151 192.168.2.152
192.168.2.153 154-168 192.168.2.169
192.168.2.170 171-185 192.168.2.186
192.168.2.187 188-202 192.168.2.203
192.168.2.204 205-219 192.168.2.220
192.168.2.221 222-239 192.168.2.240
Your subnets are incorrect. The last octet of a subnet must be an even number, the first available address is an odd number, the last available address is an even number, and the broadcast address is an odd number.
192.168.2.0 1-14 192.168.0.15
192.168.2.16 17-30 192.168.0.31
192.168.2.32 33-46 192.168.0.47
192.168.2.48 49-62 192.168.0.63
192.168.2.64 65-78 192.168.0.79
192.168.2.80 81-94 192.168.0.95
192.168.2.96 97-110 192.168.0.111
192.168.2.112 113-126 192.168.0.127
192.168.2.128 129-142 192.168.0.143
192.168.2.144 145-158 192.168.0.159
192.168.2.160 161-174 192.168.0.175
192.168.2.176 177-190 192.168.0.191
192.168.2.192 193-206 192.168.0.207
192.168.2.208 209-222 192.168.0.223
192.168.2.224 225-238 192.168.0.239
192.168.2.240 241-254 192.168.0.255 | |
| larkspur 2005-01-10, 7:12 pm |
| Ok I see my mistake. but I am a little confused if the subnet mask is 192.168.2.0\ 255.255.255.240
then how do you get:
192.168.2.112 113-126 192.168.0.127
192.168.2.128 129-142 192.168.0.143
192.168.2.144 145-158 192.168.0.159
192.168.2.160 161-174 192.168.0.175
192.168.2.176 177-190 192.168.0.191
192.168.2.192 193-206 192.168.0.207
192.168.2.208 209-222 192.168.0.223
192.168.2.224 225-238 192.168.0.239
192.168.2.240 241-254 192.168.0.255
256-240 = 16 -2 = 14
so there would be 16 address per subnet or 14 usable ip addresses per subnet. | |
| JimmyD 2005-01-10, 7:26 pm |
| quote:
Originally posted by larkspur
Ok I see my mistake. but I am a little confused if the subnet mask is 192.168.2.0\ 255.255.255.240
then how do you get:
192.168.2.112 113-126 192.168.0.127
192.168.2.128 129-142 192.168.0.143
192.168.2.144 145-158 192.168.0.159
192.168.2.160 161-174 192.168.0.175
192.168.2.176 177-190 192.168.0.191
192.168.2.192 193-206 192.168.0.207
192.168.2.208 209-222 192.168.0.223
192.168.2.224 225-238 192.168.0.239
192.168.2.240 241-254 192.168.0.255
256-240 = 16 -2 = 14
so there would be 16 address per subnet or 14 usable ip addresses per subnet.
I made a mistake in the third octet of the broadcast addresses: It should be 2 not 0.
There are 16 total addresses per subnet: the network address, the broadcast address, and 14 addresses available for devices on the subnet.
There are a total of 16 subnets (256 addresses in the unsubnetted Class C network divided by the 16 addresses per subnet. | |
| larkspur 2005-01-10, 11:23 pm |
| Looks like I had the forumal right I just did check my math before I posted. Thanks for point that out bro!! |
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