TCPIP Table for certification tests

[Collin Clark]

Before you take one of the TCP/IP tests you should construct a few tables for  reference during the test. Microsoft/Cisco will not allow you to take any  materials like this into the testing booth, but you are allowed to write these  tables down from memory once you get in the booth. There is, however, far too  much information to memorize, so it is necessary to devise a plan for  constructing the tables.

For the first table I start with 192 for the first mask. That’s the 128 bit and  the 64 bit. Cut 64 in half and you get 32. Add that to 192 and you get 224. Cut  32 in half to get 16. Add that to 224 and get 240. Half of 16 is 8. Add 8 to 240  for 248. Half of 8 is 4. Added to 248 that’s 252. 4 / 2 of course is 2. Added to  252 is 254, then 1 added to 254 is 255. That’s your first column.

192 (+32 =224)
224 (+16 =240)
240 (+8 =248)
248 (+4 =252)
252 (+2 =254)
254 (+1 =255)
255

Now start with the interval of 192 as 64. Cut it in half repeatedly to get  the second column.

192 64
224 32
240 16
248 8
252 4
254 2
255 1

The number of subnets is based on 2 raised to the power of the number of bits  set to one minus 2, but it's easier just to start with the first number and  calculate the rest. With 192 you have 2 subnets because you have 2 bits set to  one, and 2^2-2=2. The next number is derived by adding this number to itself and  adding 2. (This is a short cut for adding two to the number, multiplying by two,  and then subtracting two.) 2+2+2 = 6. Do the same thing to this number to get  the next: 6 + 6 + 2 = 14. And the next: 14 + 14 + 2 = 30. And so on to get the  third column.

192    64    2
224    32    6
240    16    14
248    8     30
252    4     62
254    2     126
255    1     254

The column we just created can be confusing. In the old days Cisco did not use  to count the first and last subnets. Looking at the table below and comparing to  the column we just created you’ll see a discrepancy.

Network ID     Subnet Mask          # Hosts
192.168.0.0    255.255.255.192     62
192.168.0.64   255.255.255.192     62
192.168.0.128  255.255.255.192     62
192.168.0.192  255.255.255.192     62

According to the column the 192 subnet mask should only have 2 networks and  it would if we did not count the first (192.160.8.0) and the last  (192.168.0.192). In the exam world, especially for Cisco, it’s best to use the  column we created.
If you know your powers of two, you might notice that each number is two less  than one of the powers starting with two less than four, and ending with two  less than 256. The next three columns consist of the number of hosts per subnet.  Starting with class A the number of hosts with mask 192 is 4 megs (that’s 4 *  1024 * 1024), but you can think of it as 4 million for simplicity. Actually the  number is 2 less than 4M, but you will never get a question that takes that into  consideration. Either way write it down as 4M. The next number is half that, and  so on to the bottom of the column.

192    64    2        4M
224    32    6      2M
240    16    14     1M
248    8     30     512K
252    4     62     256K
254    2     126    128K
255    1     254    64K

Note that when I went from M’s to K’s I was using megs (1024 * 1024) so  halving that gave me 512 * 1024 or 512K. The top of the next column is the  bottom of the preceding column halved twice: 64K / 2 = 32K, and 32K / 2 = 16K.  Do the same thing for this column as the preceding one and you get:

192    64    2      4M     16K
224    32    6     2M      8K
240    16    14    1M       4K
248    8     30    512K    2K
252    4     62    256K     1K
254    2     126   128K    510
255    1     254   64K      254

At the bottom of this column I divide 1K by 2 and subtract 2. I figure for  these smaller numbers accuracy might be more important. To get the next number I  add two back to the number, divide it by two, and then subtract two again. You  also might notice that each of these numbers is also a power of two minus two.
The next column also skips a number based on bottom number from the preceding  column. 256 / 2 = 128, and 128 / 2 = 64, and 64 - 2 = 62. The rest of the column  is derived from this number in the same way as the preceding columns were.
Subnet Mask Start of Host Range # of Subnets Class A Hosts Class B Hosts Class C  Hosts.

Subnet Mask	Hosts per Network	Number of Subnets	Class A Hosts	Class B Hosts	Class C Hosts
192	64	2	4M	16K	62
224	32	6	2M	8K	30
240	16	14	1M	4K	14
248	8	30	512K	2K	6
252	4	62	256K	1K	2
254	2	126	128K	510	-
255	1	254	64K	254	-

You might also notice that this column is identical with the second column  with two subtracted from each. The last two rows are invalid for this column  because they would both calculate to numbers less than 1.  That’s the first table.

The second table is the subnet ranges for the first three masks: 192, 224,  and 240. I started by counting up to 14 (the number of subnets for 240 according  to the first table). Then for the first range column I start with the interval  for 192 which is 64. For the second range in this column I add the interval to  get 128. There are only two subnets for this mask, so I know this is the last  row for this mask. To get the ending values for the ranges. I take one less than  the beginning value of the next range, and for the last range in the column I  use one less than the mask. This gives me:

  .192        .224       .240
1 64-127      32-63     16-31
2 128-191     64-95     32-47
3 96-127      48-63
4 128-159     64-79
5 160-191     80-95
6 192-223     96-111
7 112-127
8 128-143
9 144-159
10 160-175
11 176-191
12 192-207
13 208-223
14 224-239

One other table you need is the table of classes:

A 1-126
B 128-191
C 192-223

This can be easily memorized, but I see a slight correlation between these  numbers and some of the subnet masks (192 and 224) that helps me to remember  them. This process sounds complicated, I know, but it’s not really once you  understand what’s going on. I hope it helps.

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Now let’s take a look at some sample test questions and see how the tables we  just created work.

Which two of the addresses below are available for host addresses on the  subnet 192.168.15.19/28? (Select two answer choices)
A. 192.168.15.17
B. 192.168.15.14
C. 192.168.15.29
D. 192.168.15.16
E. 192.168.15.31
F. None of the above

Let's look at our tables:

192    64    2       4M     16K    62
224    32    6     2M      8K     30
240    16    14    1M      4K     14
248    8     30    512K     2K     6
252    4     62    256K    1K     2
254    2     126   128K     510    -
255    1     254   64K     254    -

.192          .224      .240
1 64-127      32-63     16-31
2 128-191     64-95     32-47
3 96-127      48-63
4 128-159     64-79
5 160-191     80-95
6 192-223     96-111
7 112-127
8 128-143
9 144-159
10 160-175
11 176-191
12 192-207
13 208-223
14 224-239

We know that a /28 in binary is .240. We look at the address table (the 2nd  one we created) and look for .19 (from 192.168.15.19). 19 falls in the host  range of the first subnet, which has addresses .17 through .30. Don't forget  that  the first number (.16) is the network ID and the last number (.31) is the  broadcast address. So the correct answer would be A & C.

You have a Class C network, and you need ten subnets. You wish to have as  many addresses available for hosts as possible. Which one of the following  subnet masks should you use?
A. 255.255.255.192
B. 255.255.255.224
C. 255.255.255.240
D. 255.255.255.248
E. None of the above

192    64    2       4M     16K    62
224    32    6     2M      8K     30
240    16    14     1M     4K     14
248    8     30    512K     2K     6
252    4     62    256K    1K     2
254    2     126   128K     510    -
255    1     254   64K     254    -

We know we need 10 subnets, so lets look at the first table we created. Under  the Class C column (the last column) what is the smallest subnet that allows at  least 10 subnets? .240, so the answer is C.

How many subnetworks and hosts are available per subnet if you apply a /28  mask to the 210.10.2.0 class C
network?
A. 30 networks and 6 hosts.
B. 6 networks and 30 hosts.
C. 8 networks and 32 hosts.
D. 32 networks and 18 hosts.
E. 16 networks and 14 hosts.
F. None of the above

We need to check the first table we created.

192    64    2      4M     16K    62
224    32    6     2M       8K     30
240    16    14    1M     4K      14
248    8     30    512K     2K     6
252    4     62    256K    1K     2
254    2     126   128K     510    -
255    1     254   64K     254    -

A /28 network is .240 (convert to binary if you need too). Looking at the  last column for the Class C network you see that for a .240 subnet mask you get  16 networks with 14 hosts per network. So the answer is E.