TCP/IP table for certification tests

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.

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.
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.

Leave a Reply

Your email address will not be published. Required fields are marked *