| Author |
Spid's Thu (2/20) Win2k Pro QoD
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| You are the administrator of your company's small network. The network uses the private class c address of 192.168.1.0/24
To improve performance, you decide to segment the network into 3 subnets. The 3 subnets are all connected together by a single router.
Subnet 1 consists of 5 servers.
Subnet 2 consists of 25 users in the Sales department.
Subnet 3 consists of 20 users in the Accounting and Design departments.
You need to configure a subnet mask for the entire network. What subnet mask should you use?
A. 255.255.255.0
B. 255.255.255.128
C. 255.255.255.192
D. 255.255.255.224
E. 255.255.255.240
F. 255.255.255.248
G. 255.255.255.252
H. 255.255.255.254
Good luck and see you tomorrow for the answer!! | |
| deltree 2003-02-20, 1:02 pm |
| I will take a stab on this.D | |
| sgirardo 2003-02-20, 2:09 pm |
| Unless I'm mistaken it doesn't seem that any answers meet criteria.
Out of the ones that are close, one comes up short on subnets and the other short on hosts.
I haven't done subnetting in a while, so I have to brush up on it.
I guess I'll recheck my answer. | |
| aawmorris 2003-02-20, 2:12 pm |
| If I needed 3 subnets, then I would borrow 3 bits from the host portion of the address. 128 + 64 + 32 = 224. So my mask would be 255.255.255.224 | |
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| Sorry, I'm an idiot. Let me re-do the question. Thanks
I've re-done the numbers on the question. | |
| sgirardo 2003-02-20, 2:27 pm |
| No problem Spid.
I'll take D. | |
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| quote:
Originally posted by Spid
The 3 subnets are all connected together by a single router.
Wouldn't you need two routers? | |
| Slinky 2003-02-20, 6:29 pm |
| Ohhh, subnetting.  I'll have to go with D also. | |
| gcw123 2003-02-20, 6:47 pm |
| D, this subnet provide maximum of 30 hosts. subnet1 - 5 hosts, subnet2 - 25 hosts, subnet3 - 20 hosts are within this range. | |
| mandani 2003-02-20, 8:29 pm |
| I think D as well.
Need help here to be confident.
Are the subnet ranges going to be?
192.168.32.1 - 192.168.63.254
192.168.64.1 - 192.168.95.254
192.168.96.1 - 192.168.127.254
192.168.128.1 - 192.168.159.254
192.168.160.1 - 192.168.191.254
192.168.192.1 - 192.168.223.254
I can't picture the IP's in the range otherwise.
dman | |
| mandani 2003-02-20, 9:50 pm |
| I have done some reading and have revised my thinking on the IP's.
Is the range:
192.168.1.1 - 192.168.1.31
192.168.1.32 - 192.168.1.63
192.168.1.64 - 192.168.1.95
192.168.1.96 - 192.168.1.127
192.168.1.128 - 192.168.1.159
192.168.1.160 - 192.168.1.191
192.168.1.192 - 192.168.1.223
192.168.1.0 Network
192.168.1.224 Broadcast
32 IP per Subnet and 7 Subnets in total.
I sure hope this is not on the exam.  | |
| gcw123 2003-02-20, 10:26 pm |
| quote:
Originally posted by mandani
I have done some reading and have revised my thinking on the IP's.
Is the range:
192.168.1.1 - 192.168.1.31
192.168.1.32 - 192.168.1.63
192.168.1.64 - 192.168.1.95
192.168.1.96 - 192.168.1.127
192.168.1.128 - 192.168.1.159
192.168.1.160 - 192.168.1.191
192.168.1.192 - 192.168.1.223
192.168.1.0 Network
192.168.1.224 Broadcast
32 IP per Subnet and 7 Subnets in total.
I sure hope this is not on the exam.
Agree. good posting. | |
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| quote:
Originally posted by Spid
You are the administrator of your company's small network. The network uses the private class c address of 192.168.1.0/24
To improve performance, you decide to segment the network into 3 subnets. The 3 subnets are all connected together by a single router.
Subnet 1 consists of 5 servers.
Subnet 2 consists of 25 users in the Sales department.
Subnet 3 consists of 20 users in the Accounting and Design departments.
You need to configure a subnet mask for the entire network. What subnet mask should you use?
A. 255.255.255.0
B. 255.255.255.128
C. 255.255.255.192
D. 255.255.255.224
E. 255.255.255.240
F. 255.255.255.248
G. 255.255.255.252
H. 255.255.255.254
Good luck and see you tomorrow for the answer!!
And the answer is......D
We need to find a mask that will give us at least 3 subnets AND can accomodate at least 25 hosts per subnet. Use the good old (2^n)-2 formula where "n" is the number of bits you are taking from the last octet, to figure out the number of subnetworks and hosts per subnetwork.
Taking 3 bits from the last octet gives us (2^3)-2 = 6 subnets.
The remaining 5 bits are used for hosts. This gives us (2^5)-2 = 30 hosts per subnet.
The last octet looks like this now:
11100000
128 for the first bit + 64 for the second bit + 32 for the third bit = 224
The mask to use would be
255.255.255.224
Broken down into bit form
11111111.1111111.11111111.11100000
You shouldn't get any of these type of questions on the 210 exam but it doesn't hurt to get a little exposure to classful subnetting.
Nice job everyone! | |
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| Great Q Spid!
I know this isn't 100% relavant to the question but I'm still trying to get my head around where the router is located.
Is the router between Subnet1 and 2? Shouldn't there be two routers? Or (most likely!) am I missing something? | |
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| Setup a Windows 2000 or NT4 multihomed server with 3 NICs as a router to support the 3 subnets.
The layout could look like a "T" with the multihomed router located where the horizontal and vertical lines of the "T" meet. |
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