Do you get to use calculator

psraj

Do you get the windows calculator for CCNA exam? It would be easy for subnetting questions if get to use the calc.

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dtlokee

Nope, just a note board and a really fat marker

Netstudent

I was able to bring my own chisel tip marker. I highly recommend that. My CIDR table took a whole sheet even with a chisel tip. I think I started at a /18 or /19 and went all the way to a /29. Don;t try to write too small or you will end up starting over. Ask for 2 sheets of laminate. To be honest with you, I referenced my table maybe once just because the subnetting questions were the same I had already seen a hundred times. After a while you just start seeing the numbers come together instead of going through all the steps. When you do enough subnetting, it's just there, you don't have to think about it. So practice practice practice and you will find that a calculator will actually take longer.

ITdude

Netstudent wrote:

I was able to bring my own chisel tip marker. I highly recommend that. My CIDR table took a whole sheet even with a chisel tip. I think I started at a /18 or /19 and went all the way to a /29. Don;t try to write too small or you will end up starting over. Ask for 2 sheets of laminate. To be honest with you, I referenced my table maybe once just because the subnetting questions were the same I had already seen a hundred times. After a while you just start seeing the numbers come together instead of going through all the steps. When you do enough subnetting, it's just there, you don't have to think about it. So practice practice practice and you will find that a calculator will actually take longer.

I agree with Netstudent.The CIDR table might be a good security blanket but if you are well prepared you probably won't even refer to it!

liven

I hope you guys are right.

I went to that subnettingquestions.com website every morning for 6 weeks and did at least 5 problems.

I still would make mistakes from time, but as time went on it gets easier.

Netstudent

Well we did both kill the exam so I don't think we will lead you astray.....

subnetting is subnetting, if you can do one subnetting problem, you can do another. One may require a larger number or a higher exponent, or a different octet, but the same binary rules still apply. If you are doing something wrong about every 4th or 5th subnetting problem, then you probably need more practice. I assure you, if you know basic subnetting, you will do fine. The NA is not going to ask you to design an enterprise level IP infrastructure. Best advice for subnetting is memorize powers of 2 and multiples of 2 4 8 16 32 64 128.

Pash

Netstudent explains it well. Write that table out before you start as a security blanket, its surprising how much the brain locks up when you feel under pressure, but this will help flow things out.

We don't need to explain to you about being confortable in a test environment, but dont be afraid to ask for paper and a biro if you feel more comfortable writing with this (some marker boards are always worn and torn and terrible to write on). I have never been refused this and just make sure the test supervisor counts how many sheets they give you.

All sorted.

KGhaleon

Sort of off topic, but I've forgotten...how do you determine the number of hosts available in a subnet address like 255.255.240.0? I know the first four 1s in the third octet tell you that 16 subnets can be used, but is there a way to calculate the hosts in your head? It seems like they are always very large numbers.

Netstudent

think of it as less calculation and more memorization. You find the hosts the same way you find the # of networks. Only instead of counting the "on" bits, you are tallying the "off" bits or zeros. IN the host formula you will always subtract 2.

As per your example, you would have 4 on bits and 12 off bits.

11111111.11111111.|11110000.0000000

So hosts would be 2^12 -2 = 4094

I know that 2^8 = 256 without even thinking twice about it.

So I just keep doubling the number real fast in my head.

256, 512,1024,2048,4096

subnetting to me is less cruncing numbers and more logic. If you can find little mind shortcuts it will help.

tech-airman

KGhaleon wrote:

Sort of off topic, but I've forgotten...how do you determine the number of hosts available in a subnet address like 255.255.240.0? I know the first four 1s in the third octet tell you that 16 subnets can be used, but is there a way to calculate the hosts in your head? It seems like they are always very large numbers.

KGhaleon,

Look at the subnet mask in binary....

255.255.240.0

     |
     v
11111111.11111111.11110000.00000000

So the relevant part is..

.11110000.00000000

So just start counting the '0's from right to left...

.11110000.00000000

2048 1024 512 256 . 128 64 32 16 8 4 2 1

But here's the interesting part. Instead of killing yourself tyring to mentally calculate 2048+1024+512+256+128+64+32+16+8+4+2+1 just take the next number 4096 then subtract 1 for total possible of 4095 then subtract two for the 1) subnet 2) broadcast for a total of 4093 practical hosts.

Netstudent

for a total of 4093 practical hosts

How does that work Tech?

2^12 - 3?

dtlokee

I guess he's assuming there is a router that can't hbe used for the "hosts" whereas a router is really a host. should be 4096-2 = 4094 total hosts, of course after that any additional devices will take away from the pool but Cisco is looking for 2**n - 2

Netstudent

hmmm...okee dokee dtlokee (pun intended) lots of puns today.

tech-airman

Netstudent wrote:

for a total of 4093 practical hosts

How does that work Tech?

2^12 - 3?

Netstudent,

For an 8 bit byte, the values for each of the columns are...

128 64 32 16 8 4 2 1

For two bytes, which is two octets, the values for each columns are...

32768 16384 8192 4096 2048 1024 512 256 . 128 64 32 16 8 4 2 1

Since the subnet mask was 255.255.240.0... in binary that's...

255.255.240.0
|
v
255.255.1111|0000.00000000

So lining up the last two octets of the subnet mask with the binary values in decimal...

32768 16384 8192 4096 2048 1024 512 256 . 128 64 32 16 8 4 2 1

    1     1    1    1    0    0   0   0 .   0  0  0  0 0 0 0 0

So from left to right, the mask ends at the 4096 bit. So the host bits are to the right of that or 2048 and below. So which is easier to calculate the number of hosts:

2048+1024+512+256+128+64+32+16+8+4+2+1 = 4095

or...

4096 - 1 = 4095

I prefer the second.

Then you have to subtract two for the "all 0 hosts" which is reserved for the subnet address and the "all 1 hosts" which is reserved for the local broadcast to that subnet which is 4095 - 2 = 4093.

I hope this helps.

dtlokee

tech-airman wrote:
Netstudent wrote:

for a total of 4093 practical hosts

How does that work Tech?

2^12 - 3?

Netstudent,

For an 8 bit byte, the values for each of the columns are...
128 64 32 16 8 4 2 1
For two bytes, which is two octets, the values for each columns are...
32768 16384 8192 4096 2048 1024 512 256 . 128 64 32 16 8 4 2 1
Since the subnet mask was 255.255.240.0... in binary that's...
255.255.240.0
|
v
255.255.1111|0000.00000000
So lining up the last two octets of the subnet mask with the binary values in decimal...
32768 16384 8192 4096 2048 1024 512 256 . 128 64 32 16 8 4 2 1

    1     1    1    1    0    0   0   0 .   0  0  0  0 0 0 0 0
So from left to right, the mask ends at the 4096 bit. So the host bits are to the right of that or 2048 and below. So which is easier to calculate the number of hosts:
2048+1024+512+256+128+64+32+16+8+4+2+1 = 4095
or...
4096 - 1 = 4095
I prefer the second.

Then you have to subtract two for the "all 0 hosts" which is reserved for the subnet address and the "all 1 hosts" which is reserved for the local broadcast to that subnet which is 4095 - 2 = 4093.

I hope this helps.

Sorry but this yeilds the wrong answer. The formula for useable hosts is 2**h - 2, in this case 2**12 -2 = 4094. If you follow you method for a /30 mask which is typically used for point-to-point links, you will get 2+1 = 3 -1 = 2 - the subnet and broadcast = 0, when there are really 2 available host addresses (2**2 - 2)

Hope this helps

NeonNoodle

tech-airman,

Your explanation is incorrect. For n bits, there are 2^n possible values. The first value contains all zeroes and the last value contains all ones. By definition the first value is the network address and the last value is the broadcast address. The network and broadcast addresses cannot be host addresses. Thus, we have 2^n - 2 possible hosts. For n = 12, there are 2^12 = 4096 possible values, but only 2^12-2 = 4094 possible hosts.
In your explanation, you arrive at 4093 hosts. You get this by taking the sum of the first 12 powers of two, i.e. 2^0 + 2^1 + ... + 2^11 = 4095 or the twelfth power of two minus one, i.e. , 2^0 + 2^1 + ... + 2^11 = 2^12 -1 to claim there are 4095 possible values(?). Then you subtract the subnet and broadcast addresses as possible hosts and arrive at 4093 hosts. Your mistake is in the first step. The summation doesn't give the possible number of values; the summation plus one does however.

Pash

We will put that down to tiredness tech, i hope there is some further certs coming soon mate!

sbl

ITdude wrote:

Netstudent wrote:

I was able to bring my own chisel tip marker. I highly recommend that. My CIDR table took a whole sheet even with a chisel tip. I think I started at a /18 or /19 and went all the way to a /29. Don;t try to write too small or you will end up starting over. Ask for 2 sheets of laminate. To be honest with you, I referenced my table maybe once just because the subnetting questions were the same I had already seen a hundred times. After a while you just start seeing the numbers come together instead of going through all the steps. When you do enough subnetting, it's just there, you don't have to think about it. So practice practice practice and you will find that a calculator will actually take longer.

I agree with Netstudent.The CIDR table might be a good security blanket but if you are well prepared you probably won't even refer to it!

Ditto. Recently passed 640-801 which I understand is retired now except for certain Cisco Academy students. Without disclosing content I can say that writing out a subnet table was not necessary, though a basic working knowledge of subnets is definitely necessary (the published objectives divulge that much and they aren't kidding).

Now I realize "basic" is a subjective term, but IMO a person who needs a table or a calculator may be on thin ice and need a bit more preparation before rolling the $125 dice.

rebelrat

sbl wrote:

[quote="ITdude
Now I realize "basic" is a subjective term, but IMO a person who needs a table or a calculator may be on thin ice and need a bit more preparation before rolling the $125 dice.

I agree. Too many scenario questions where you have to be able to quickly determine host, network and broadcast addresses under CIDR. There was a much greater focus on IP in the 640-801 as compared to the 640-507 way back in 2000 when I first took it.

One method that helped me was to remember the value of the "last bit" taken from the host address say 255.255.255.240 - last bit taken is 11110000 = 16. This is the address of my first subnet, plus the "step" from subnet to subnet - 16 + 16 = 32, 32 + 16 = 48, 48 + 16 = 64 etc. When the value of the subnet mask is reached, in this case 240 - this subnet cannot be used (broadcast subnet).

Then subtract 1 from the higher subnet address - i.e. for network 16 network 32 is the higher network. Therefore 32 - 1 = 31 and you have the broadcast address for the "lower" network 16. All numbers in between which are 17 - 30 are the host addresses.

For network 32 " the higher network ist 48. Subtract 1 and there is the broadcast address for network 32 which is 47. Host addresses are all numbers in between, which are 33 - 46.

A good trick for wildcard masks with OSPF and access lists is to count the ones from the least significant bit on up - the one on the far right - and then just add the values = 1 + 2 + 4 + 8 = 15. Notice that all wildcard masks are odd numbers, as they all start with 1.

Hope that helps with the Binary - just think what kind of pleasures IP6 will bring with all HEX numbers and no dotted decimal notation. Powers of 2 are nothing compared to Powers of 16. But remember it is only basic Arithmetic and in the real world people use IP calculators - Solarwinds has a nice free one.

Slowhand

dtlokee wrote:

Nope, just a note board and a really fat marker

Man, I wish I'd have been able to use my marker and noteboard. The marker they gave me only wrote "ghost-lines", erasing anything I wrote immediately, as dry-erase markers do when they get old and dry up. I ended up writing nothing down, just went with the ol' grey-matter. Ironically, the pad had a Thomson-Prometric logo on it, even though I was testing at a Vue center. I guess they used to be a Thomson testing center, or they went and swiped the laminated sheets from Thomson. . .