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Home > Archive > CCNA > August 2003 > help w/ another subnetting question...
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help w/ another subnetting question...
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| I am confused with the following Q:
How many subnets and hosts can you have in a Class C network with a subnet mask of 255.255.255.224? Choose one.
A. 6 subnets and 14 hosts
B. 30 subnets and 14 hosts
C. 4 subnets and 30 hosts
D. 6 subnets and 30 hosts
E. 14 subnets and 6 hosts
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last octate is 224 meaning there were 3 higher bits borrowed
128 64 32 | 16 8 4 2 1
Add the value of the 3 bits
128+64+32 = 224
Now do the same for the lower bits
Add the value of the lower 3 bits - 1
1+2+4 = 7 - 1 = 6
So there are 6 subnets, right?
How many hosts?
Add the remaining bits -1
1+2+4+8+16 = 31 - 1 = 30
So there are 30 hosts.
The answer should be "D", is my 'math' correct?
Thanks | |
| smithy4u 2003-08-07, 12:44 am |
| Yes. You r right. | |
| smithy4u 2003-08-07, 1:13 am |
| I got more than 30 questions with explanation on subnetting from here. Try it. You will get the real explanation on this topic. | |
| jason892 2003-08-07, 5:36 am |
| You're right. Here's how I do the math, although it is pretty much the same thing as how you are doing it
3 high order bits
3^2=8-2=6 usable subnets
5 host bits
5^2=32-2=30 usable hosts
Although that has to be the answer on this question, you want to make sure you read the question. I have taken several practice tests where they have asked for the "total" number of subnets and hosts. In your example, if they had asked for that, then the answer would have been 8 subnets and 32 hosts because you wouldn't subtract the broadcast and network addresses. | |
| MattyJV 2003-08-07, 12:15 pm |
| Your correct D. | |
| stevenmeier 2003-08-07, 7:46 pm |
| 256-224=32
useable networks are
0,32,64,96,128,160,192,224.
the host rule applies as 2^n-2=useable hosts.
so infact there are 8 networks, check with the new exam structure as when I did my exam ccna 607 exam all my exam questions were 2^n for networks.
cheers
steve | |
| SureshHomepage 2003-08-09, 1:13 am |
| I like the jasson892's way! Thats how we calculate at work. Its pretty much easier to do this way.
For a mask of 255.255.255.224, it would be like
3 high order bits
3^2=8-2=6 usable subnets
5 host bits
5^2=32-2=30 usable hosts
No doubt its just a shortest method. |
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