1 2 Previous Next 20 Replies Latest reply: Sep 6, 2011 10:50 AM by Mayur Jadhav

# how to learn IP addressing ??

Hello,,

solving so many question of IP addressing ... not so much confident.. my passing rate of IP addressing is 60%..

I am very good in Mathamatics.. upto 2^10 all value i can manage orally but my problem is that not able to Understand IP addressing Question.. i don't know how to interpret question...

Kindly request for all.. .Please guide me how to unnderstant IP address question..

Mayur

• ###### 1. Re: how to learn IP addressing ??

Dear Mayur,

i too confused when first started learing ip addressing and subnetting,to nail it we need to do lot of practice,just grab some material and practice urself lot of exercises im sure you wil get a very gud idea about ip addressing....i post the rule that i ve been following...

you got /25 network

first write that in subnet mask i.e 255.255.255.128

in the last octect take 128 and convert it to binary 10000000

now how many one's we got only 1 so 2^1=2 subnets,what are those?

one is Zero subnet and the other is 128 subnet so totally 2 subnets

now how many hosts per subnet now,count zero's,there are 7 zero's right?

so 2^7 -2=126 usable addresses for hosts for each subnet

why i subtracted 2?

so here we got two subnets

1st subnet

subnet         0

firsthost       1

last host      126

2nd subnet

subnet           128

first host         129

last host         254

change the mask to /22,/23,/24,/25....and practice it

i suggest you to study todd lamle and cbt nuggets and make ur own rule for subnetting and you r gud to go.here i am posting some stuff perhaps they help u

http://www.ralphb.net/IPSubnet/intro.html

• ###### 2. Re: how to learn IP addressing ??

If you wanna really learn and master in subnetting you will need to practice. Try to look for the "IP & Subnetting Workbook" very awesome workbook for subnetting.

Elvin.

• ###### 4. Re: how to learn IP addressing ??

Perphaps you missing fundamentals:

Class A has by default 8 bits in mask, that is 255.0.0.0 or /8

CLass B uses 16 bits = 255.255.0.0 or /16

What are defaults for Class C ?

• ###### 5. Re: how to learn IP addressing ??

defaults for Class C ?

255.255.255.0

• ###### 6. Re: how to learn IP addressing ??

Right; C is /24

So, the idea of default mask is that you cannot use - do not touch - any octets that are marked with 255.

it is like a boarder; you can play with what is on the right, but not what is left;

192.168.1 .0

255.255.255. 0  +> you can play with 4th octet; don't touch first 3.

10.1.1.0

255.0.0.0 -- you can play from 2nd thru 4th octet - don't touch 1st one

Therefore, take 10.0.0.0 /8 and divide into many /24 subnets; like

10.1.0.0 /24

10.0.1.0 /24

10.1.1.0 /24

10.1.2.0 /24

10.0.0.0 /24

• ###### 7. Re: how to learn IP addressing ??

Very well explained Martin.

• ###### 8. Re: how to learn IP addressing ??

no problem; still waiting for the author to give us a clue what;s up with his subnetting;

• ###### 9. Re: how to learn IP addressing ??

Mayur,

I posted this on another blog. But this Lammell's Chapter 3 helped me out tremendously. Let me know if there is anything else I can do to help.

Chapter 3: IP Subnetting, Variable Length Subnet Mask

Written Labs 3

In this section, you’ll complete the following labs to make sure you’ve got the information and concepts contained within them fully dialed in:

• Lab      3.1: Written Subnet Practice #1
• Lab 3.2      Written Subnet Practice #2
• Lab      3.2 Written Subnet Practice #3

Written Lab 3.1: Written Subnet Practice #1

Write the subnet, broadcast address, and valid host range for questions 1 through questions 6 and then answer the question that follow:

1. 192.168.100.25/30

4 21 ---- We are subnetting on 4

11111111.1111111.11111111.11111100

255.255.255.252 – is the subnet

192.168.100.0    -    192.168.100.3   -   Valid host range 192.168.100.1 – 192.168.100.2 netid 192.168.100.0   Broadcast address 192.168.100.3

192.168.100.4    -    192.168.100.7    -   Valid host range 192.168.100.5 – 192.168.100.6  netid  192.168.100.4  Broadcast address 192.168.100.7

192.168.100.8    -    192.168.100.11 - Valid host range 192.168.100.9 – 192.168.100.10  netid 192.168.100.4 Broadcast address 192.168.100.11

192.168.100.12

192.168.100.16

192.168.100.20

192.168.100.24 – 192.168.100.27 is the broadcast

192.168.100.28

1. 192.168.100.37/28

16 8 421   - Subnetting on 8

11111111.11111111.11111111.11110000

255.255.255.240 – is the subnet

192.168.100.0    -     192.168.100.15 valid host range 192.168.100.1 – 192.168.100. 14 netid 192.168.100.0, broadcast id 192.168.100.15

192.168.100.16    - 192.168.100.31 - valid host range 192.168.100.17 – 192.168.100.30 netid 192.168.100.16 broadcast id 192.168.100.31

192.168.100.32 -     192.168.100.47 – valid host range 192.168.100.33– 192.168.100.46 netid 192.168.100.32 broadcast id 192.168.100.47

192.168.100.48 -     192.168.100.63 – valid host range 192.168.100.49 – 192.168.100.62 netid 192.168.100.48 broadcast id 192.168.100.63

192.168.100.64

1. 192.168.100.66/27

32168421 – Subnetting on 32

11111111.11111111.11111111.11100000

255.255.255.224 – is the subnet

192.168.100.0    -    192.168.100.31 -valid host range 192.168.100.1 – 192.168.100.30 netid 192.168.100.0 broadcast id 192.168.100.31

192.168.100.32 -    192.168.100.63 – valid host range 192.168.100.65 – 192.168.100.62 netid 192.168.100.32 broadcast id 192.168.100.63

192.168.100.64 -    192.168.100.95 – valid host range 192.168.65 – 192.168.100.94 netid 192.168.100.64 broadcast id 192.168.100.95

192.168.100.96 -    192.168.100.127 – valid host range 192.168.100.97 – 192.168.100.126 netid 192.168.100.96 broadcast id 192.168.168.127

192.168.100.128

1. 192.168.100.17/29

8421 - Subnetting on 8

11111111.11111111.11111111.11111000

192.168.100.0 -   192.168.100.7 - valid host range 192.168.100.1 – 192.168.100.6 netid 192.168.100.0 broadcast id 192.168.100.7

192.168.100.8 -   192.168.100.15 - valid host range 192.168.100.9 – 192.168.100.14 netid 192.168.100.8 broadcast id 192.168.100.15

192.168.100.16 - 192.168.100.23 – valid host range 192.168.100.23 – 192.168.100.22 netid 192.168.100.16 broadcast 192.168.100.23

192.168.100.24 - 192.168.100.31 – valid host range 192.168.100.25 – 192.168.100.30

192.168.100.32

1. 192.168.100.99/26

6432168421

11111111.11111111.11111111.11000000

255.255.255.224 is the subnet

192.168.100.0   -   192.168.100.63 - valid host range 192.168.100.1 – 192.168.62 netid

192.168.100.64 -   192.168.100.127 – valid host range 192.168.100.65 – 192.168.100.126 netid 192.168.100.64 broadcast id 192.168.100.127

192.168.100.128 – 192.168.100.191 – valid host range 192.168.100.129 – 192.168.100.190 netid 192.168.100.138 broadcast id 192.168.100.191

192.168.100.192 - 192.168.100.223

192.168.100.224

1. 192.168.100.99/25

11111111.11111111.11111111.10000000

255.255.255.128      is the subnet

192.168.100.0 – 192.168.100.127 – valid host range 192.168.100.1 – 192.168.100.126

192.168.100.128 – 192.168.1.255

192.168. 2.0 – 192.168.1.127

192.168.2.128 – 192.168.2.255

192.168.3.0

1. You      have a Class B network and need 29 subnets. What is your mask?

Class B is 255.255.0.0

11111111.11111111.11111000.00000000

5

2        =  32 subnets

128       64      32     16       8      4      2       1

1           1        1       1        1       0     0      0

11111000 = 8 is the increment

192.168.192.0  -   192.168.192.7

192.168.192.8 -   192.168.192.15

192.168.192.16

Ans:  192.168.192.15

1. How      many host are available with a Class C/29

N -2         3

11111000 = /29     -   2    =       2     =   8 - 2 = 6 hosts per subnet.

1. What      is the subnet for host ID 10.16.3.65/23

11111111.11111111.11111110.00000000 – here the increment bit is 2

10.16.0.0 – 10.16.2.255

10.16.2.0 - 10.16.3.255

10.16.4.0

Subnet is 10.16.2.0 Broadcast is 10.16.3.255

1. What      form of NAT maps multiple private IP addresses to a single registered IP      address by using ports?

Ans: PAT   - Port Address Translation

1. What      does the term inside global address represent in the configuration of NAT?

Note: Local addresses are the ones we use before NAT translation. So, the inside local address is actually the private address of the sending host that’s trying to get to the Internet, while the outside local address is the address of the destination host. The latter is usually a public address (web server address, mail server, etc).  \After translation, the inside local address is then called the inside global addresses and the outside global address then becomes the name of the destination host.

1. What      diagnostic tool will you use to test remote host’s IP stack?

1. What      diagnostic tool will you use to display Ip -to-MAC addressing on a Windows      PC?  Answer arp – a

1. What      is the IP address range for subnet 192.168.2.64/26

11111111.11111111.11111111.11000000 increment bit is 64

255.255.2255.192

192.168.2.0 – 192.168.2.63

192.168.2.64 – 192.168.2.127

192.168.2.128

Range will be 192.168.2.1 – 192.168.2.126

1. How      many hosts does the subnet 172.16.112.0/20 provide?

11111111.1111111.11110000.00000000

N -2         12 -2

Hosts are 0’s.   We have 12 zero’s   2           = 2          =

8092    4096   2048   1024 512   256     128     64    32   16    8     4      2      1

Start counting from 4096 – 2 =4094

1. What      is the subnet address of 172.16.159.159/22

11111111.11111111.11111100.00000000 – increment on 4

255.255.252.0

172.16.0.0                                                      172.16.64.0

172.16.4.0                                                      172.16.68.0

172.16.8.0                                                      172.16.72.0

172.16.12.0                                                    172.16.76.0

172.16.16.0                                                    172.16.80.0

172.16.20.0                                                   172.16.84.0

172.16.24.0                                                    172.16.88.0
172.16.28.0                                                    172.16.92.0

172.16.32.0                                                    172.16.96.0

172.16.36.0                                                    172.16.100.0

172.16.40.0

172.16.44.0

172.16.48.0

172.16.52.0

172.16.56.0                                           Ans ---172.16.156.0 – 172.16.159.0

172.16.60.0

1. What      is the subnet for the host 10.16.3.65/23?

11111111.11111111.11111110.00000000 increment bit is 2

255.255.254.0

10.16.0.0 – 10.16.1.0

10.16.2.0 – 10.16.3.0

10.16.4.0

11111111.11111111.111111100.00000000  - increment bit 4

255.255.252.0

10.16.0.0 – 10.16.3.255

10.16.4.0 – 10.16.7.255

10.16.8.0

1. Two      routers are connected with a serial connection. One is configured with IP      address 192.168.10.82/30 and the other is 192.168.10.85/30. Why won’t the      two routers communicate?

11111100 – Increment bit 4

192.168.10.0

192.168.10.4

192.168.10.8

192.168.10.12

192.168.10.16

192.168.10.20

192.168.10.24

192.168.10.28

192.168.10.32

192.168.10.36

192.168.10.40

192.168.10.44

192.168.10.48

192.168.10.52

192.168.10.56

192.168.10.60

192.168.10.64

192.168.10.68

192.168.10.72

192.168.10.76

192.168.10.80     -  192.168.10.83

192.168.10.84     -  192.168.10.87

192.168.10.88

Ans: Because they are on different subnets. They must be on the same subnet to communicate.

Written Lab 3.2: Written subnet Practice #2

Given a Class B network and the next bits identified (CIDR), complete the following table to identify the subnet mask and the number of hosts address possible for each mask.

Classful Address                          Subnet Mast                                         Number of Hosts

/16                             11111111.11111111.00000000.00000000          2n-2 = 2 to the

255.255.0.0                              2 to the 16 – 2 = 65536 -2 = 65534

/17                             11111111.11111111.10000000.00000000

255.255.128.0                                                                                                        32768 – 2 = 32766

/18                             11111111.11111111.11000000.00000000      16384 -2 = 16382

255.255.192.0

/19                             11111111.11111111.11100000.00000000        8192 – 2 =     8190

255.255.224.0

/20                            11111111.11111111.11110000.00000000        4096 – 2 = 4094

255.255.240.0

/21                            11111111.11111111.11111000.00000000        2048 – 2 = 2046

255.255.248.0

/22                            11111111.11111111.11111100.00000000       1024 – 2 = 1022

255.255.252.0

/23                            11111111.11111111.11111110.00000000       512 – 2 = 510

255.255.254.0

/24                            255.255.255.0                                                      256 – 2 =254

/25                            255.255.255.128                                                   128 – 2 = 126

/26                            255.255.255.192                                                      64 – 2 = 62

/ 27                           255.255.255.224                                                       32 – 2 = 30

/28                            255.255.255.240                                                      16 – 2 = 14

/29                            255.255.255.248                                                       8 - 2 = 6

/30                             255.255.255.252                                                       4- 2 = 2

Written Lab 3.  Written Subnet Practice # 3

Class A – 1 – 126

Class B – 128 – 191

Class C - 192 - 223

Class    and Host Bits              Subnets           hosts (2X – 2)

10.25.66.154/23              A        11111111.11111111.111111100.00000000 note because this is a class A address count the number of bits for 255.0.0.0.  15 subnets 9 host bits

2X – 2 2 9 – 2  =              512     256   128   64   32 16   8    4     2   1   = 512 – 2 = 510, number of subnets  2X = 2 15 =  32768

------------------------------------------------------------------------------------------------------------

172.31.254.12.24      Class B  11111111.11111111.11111111.00000000 – 8  host bits, 8 subnet bits. 2X – 2  2 8 – 2 = 256 -2 = 254    Subnets  2X = 2 8 = 256 subnets

192.168.20.123/28     Class C    11111111.11111111. 11111111.11110000 4 subnet, 4 host bits 2X – 2  = 2 4  – 2 =   16 – 2 = 14  hosts,  subnets = 2X = 2 4 = 16 subnets

63.24.89.21/18       Class A    11111111.11111111.11000000.00000000 10 subnet 14host bits  2x – 2 = 2 14– 2 =  16384 – 2 = 16382 subnets = 2X = 2 10 =  1024

128.1.1.254/20     Class B     11111111.11111111.11110000.00000000 4 subnet bits 12 host bits   2X – 2 = 2 12 – 2 = 4092 – 2 = 4094 subnets 2X = 2 4 =  16

208.100.54.209/30  Class C  11111111.1111111.11111111.11111100  6 subnets, 2 host bits, 2X – 2 = 2  2  - 2 = 2 host, subnets 2X = 2 6 = 64

Review Questions

The following questions are designed to test your understanding of this chapter’s material.

1. What      is the maximum number of IP addresses that can be assigned to hosts on a      local subnet that uses the 255.255.255.224 subnet mask?

11100000 =  5 host bits  2 X – N = 2 5 – 2 = 32 -2 = 30 hosts

Note: It doesn’t matter if it’s a class A, B or C address.

1. You      have a network that needs 29 subnets while maximizing the number of host      address available on each subnet. How many bits must you borrow from the      host field to provide the correct subnet mask?

255.255.255.240

11110000, this means that the block size is 16 in the fourth octet. 0 , 16, 32 ,48, 64, 80, etc. The host is in the 64 subnet.

128   64   32   16   8   4  2  1

1     1      1     1    1   0  0  0

5

The answer is 5 subnet bits. 2    =   32 subnets with 3 host bits 6 hosts per subnet.

1. What      is the subnetwok address for a host with the IP address 200.10.5.68/28

11110000 – increment bit is on 16

200.10.5.0 – 200.10.5.15

200.10.5.16 – 200.10..5.31

200.10.5.32 – 200.10.5.47

200.10.5.48 – 200.10.5.63

200.10.5.64 – 200.10.5.79

200.10.5.80

Ans: the host is on the 64 subnet.

1. The      network address of 172.16.0.0/19 provides how many subnets and hosts?

11111111.11111111.11100000.00000000

X – 2            13  - 2

2x = 2 3 = 8 subnets, 2          =    2            =   8192 – 2 = 8190 hosts

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

1. Which      two statements describe the IP address 10.16.3.65/23  (Choose two)

11111111.11111111.11111110.00000000

255.255.254.0

10.16.0.0 – 10.16.1.255

10.16.4.0

1. If      a host on network has the address 172.16.45.14/30, what is the subnetwork      this host belongs to?

11111111.11111111.11111111.11111100 – increment bit is 4

255.255.255.254

172.16.45.0

172.16.45.4

172.16.45.8

172.16.45.12 – 172.16.45.17

172.16.45.16 – 172.16.45.19

172.16.45. 20

7. On a VLSM network, which mask should you use on point-point WAN links in order to reduce the waste of IP address space?

Ans: /30 255.255.255.252

1. What      is the subnetwork number of a host with an IP address of 172.16.66.0/21?

11111111.11111111.11111000.00000000 – increment bit is 8

172.16.0.0 – 172.16.7.255

172.16.8.0 – 172.16.15.255

172.16.16.0 – 172.16.23.255

172.16.24.0 – 172.16.31.255

172.16.32.0 – 172.16.39.255

172.16.40.0 – 172.16.47.255

172.16.48.0 – 172.16.55.255

172.16.56.0 – 172.16.63.255

172.16.64.0 – 172.16.71.255

172.16.72.0

1. You      have an interface on a router with the IP address 192.168.192.10/29.      Including the router interface, how many hosts can have IP addresses on      the LAN attached to the router interface?

11111111.11111111.11111111.11111000 increment bit is 8

Hosts 2X – 2  = 2 3 – 2  = 6 Hosts

1. You      need to configure a server that is on the subnet 192.168.19.24/29. The      router has the first available host address. Which of the following should      you assign to the server?

11111111.11111111.11111111.11111000 increment bit  8

255.255.255.248

192.168.19.0 – 192.168.19.7

192.168.19.8 – 192.168.19.15

192.168.19.16 – 192.168.19.23

192.168.19.24 – 192.168.19.31 note the router has the 1st usable address which would be 192.168.19.25. You can assign the server the next address which would be 192.168.19.26.

192.168.19.32

11.  You have an interface on a router with the IP address of 192.168.192.10/29. What is the broadcast address the hosts will use on this LAN?

11111000 – increment bit is 8

255.255.255.248 which is a block of 8 in the fourth octet

192,168.192.0

192.168.192.8 – 192.168.192.15 is the broadcast.

192.168.192.16

12.  You need a subnet that has 5 subnets, each with at least 16 hosts. Which classful subnet mask would you use?

2n = 2 3 = 8 subnets 2N -2 = 2 5 – 2 = 32 – 2 = 30 hosts  This is the best answer!!! 255.255.255.240

11111111.11111111.11111111.11100000

255.255.255.224

128    64    32   16    8    4   2   1

1       1      1     0     0  0   0    0 = 224

13.  A network administrator is connection hosts A and B directly through their Ethernet interface, as shown in this illustration. Ping attempts between hosts are unsuccessful. What can be done to provide connectivity between the hosts? (Choose two)

First like devices use crossover cables not straight through cables.

Second the hosts have a different mask which puts them on different subnets. The easiest way to fix this is to change the mask to 255.255.255.0/24

14.  If an Ethernet port on a router were assigned to an IP address of 172.16.112.1/25, what would be the valid subnet address of this host?

11111111.11111111.11111111.10000000

255.255.255.128

172.16.112.0 – 172.16.112.127

172.16.112.128

15.  Using the following illustration, what would be the IP address of E0 if you were using the eighth subnet? The network id is 192.168.10.0/28 and you need to use the available IP address in the range. The zero subnet should not be considered valid for this question.

11111111.11111111.11111111.11110000 – increment bit is 16

255.255.255.240

192.168.10.0

192.168.10.16

192.168.10.32

192.168.10.48

192.168.10.64

192.168.10.80

192.168.10.96

192.168.10.112 – 192.168.10.

192.168.10.128 – 192.168.10.143 9th subnet – The last address we could use is 192.168.10.142 is the valid host we can use!

192.168.10.144

16.  Using the illustration from the previous question, what would be the IP address of S0 if you were using the first subnet? The network ID is 192.168.10.0/28 and you need to use the last available IP address in the range. Again, the zero subnet should not be considered valid for this question.

11111111.11111111.11111111.11110000 – increment bit is 16

192.168.10.0  - 192.168.10.15

192.168.10.16 – 192.168.10.31

192.168.10.32

Looking at this question, I had mistaken the first subnet for what I placed in red. But according to the books answer, The first subnet is 16 (remember that the question stated not to use subnet zero) and the next subnet is 32, so our broadcast address is 31. This makes our host range 17-30. 30 is the valid host.

17.  Which configuration command must be in effect to allow the use of 8 subnets if the Class C subnet mask is 255.255.255.224.  Answer;  ip subnet-zero. A Class C subnet mask of 255.255.255.224 is 3 bits on and 5 bits off (11100000) and provides 8 subnets, each 30 hosts. However, if the command ip subnet-zero is not used then only 6 subnets would be available to use.

18.  You have a network with a subnet of 172.16.17.0/22. Which is the valid host address?

11111111.11111111.11111100.000000000

255.255.252.0

172.16.0.0 – 172.16.3.255

172.16.4.0 – 172.16.7.255

172.16.8.0 – 172.16.11.255

172.16.12.0 – 172.16.15.255

172.16.16.0 – 172.16.19.255   (The valid range is 172.16.16.1 – 172.16.18.255)

172.16.20.0 – 172.16.24.255

172.16.24.0

19.  Your router has the following IP address on Ethernet 0: 172.16.2.1/23. Which of the following can be valid host ID’s on the LAN interface attached to the router? (choose two)

11111111.11111111.11111110.00000000 – increment bit is 2

255.255.254.0

172.16.0.0 – 172.16.1.255

172.16.2.0 – 172.16.3.255 – Range would be 172.16.2.1 – 172.16.3.254

172.16.4.0

20.  What does “inside global” address represent in the configuration of NAT?

Ans: A registered address that represents an inside host to an outside network. Note: The word private is asked would make it an inside address. Inside global is defined as a registered addresses that represents an inside host to an outside network. In other words, we’re looking for a translated address. Inside local is before translation; inside global is after translation.

• ###### 10. Re: how to learn IP addressing ??

Hi

I created this, see if this helps, if it does then try the bottom link which is very good for practise and something you can access via the phone as well.

https://learningnetwork.cisco.com/message/32253#32253

http://www.subnettingquestions.com/

Regards

John

• ###### 11. Re: how to learn IP addressing ??

@ Elvin Arias

@ John

both links are really amazing. . thanks a lot..

@ Martin

Thanks for your gr8 explanation...if a question is stright forward then no problem for me.. but if same question ask by Tricky way...then i totally confused..

for example:

Q.  :-  Which three IP address can be assigned to hosts if the subnet mask is 27/  and subnet zero is usable? (choose 3)

1) 10.15.32.17

2)66.55.128.1

3)17.15.66.128

4)135.1.64.34

5)192.168.5.63

6)129.33.192.192

confused with second part of question that is " and subnet zero is usable ".??

another question

Q :-  Which two subnetworks would be include in the summarized address of 172.31.80.0/20 ?

1)172.31.192.0/18

2)172.31.51.16/30

3)172.31.17.4/30

4)172.31.64.0/18

5)172.31.92.0/22

6)172.31.80.0/22

what to do ??

• ###### 12. Re: how to learn IP addressing ??

Hi Mayur,

Q. :- Which three IP address can be assigned to hosts if the subnet mask is 27/ and subnet zero is usable? (choose 3)

1) 10.15.32.17

2)66.55.128.1

3)17.15.66.128

4)135.1.64.34

5)192.168.5.63

6)129.33.192.192

What are the subnets for a /27? They would be 0, 32, 64, 96, 128, 160, 192, 224. If subnet 0 was not usable you could not use addresses in the 0 subnet but since it is the correct answers are 1, 2, 4.

Q :- Which two subnetworks would be include in the summarized address of 172.31.80.0/20 ?

1)172.31.192.0/18

2)172.31.51.16/30

3)172.31.17.4/30

4)172.31.64.0/18

5)172.31.92.0/22

6)172.31.80.0/22

For this question you have to figure out what your network increments are with a /20 and figure out which of the addresses mentioned fit into your subnet 172.31.80.0/20.

In this case the increments are 16 so starting from 172.31.80.0 your next subnet is 172.31.96.0. Looking at the answers provided only 2 answers fit in your subnet 172.31.80.0 and those are 5 and 6.

HTH

Angelo

• ###### 13. Re: how to learn IP addressing ??

subnet zero is usable or not; is the question;

subnet zero is usable = 2s = you do not substract 2 ; use all subnets

subnet zero is NOT usable = 2s -2 = you must substract 2 just like with hosts where you always sustract 2 IPs;

in other words: you can not use first subnet and last one;

subnet zero is NOT usable example:10.0.0.0 /24  -

10.0.0.0

10.0.1.0

10.0.2.0

10.0.3.0

..........

10.0.253.0

10.0.254.0

10.0.255.0

192.168.1.0/26 = 4 subnets but take 2 away - first one and last one so you have  2

192.168.1.0

192.168.1.64

192.168.1.128

192.168.1.192

• ###### 14. Re: how to learn IP addressing ??

Q :-  Which two subnetworks would be include in the summarized address of 172.31.80.0/20

172.31.80.0/20

1. First of all, this is Class B -172. second 20 bits, you know that 16 bits are Class B default, that is first 2 octes, you do not touch; 16 + 4 = 20; 4 th bit is always 16 or 240; so mask is 255.255.240.0 for that /20;

2. or 256-240 =16 or

That 16 is what we call Magic Number; 256-240=16

3. take magic number and add to your network in relative octet; (20 bits are between 16 and 24 or 3rd octet)

172.31.80.0 + 16 = next subnet is 172.31.96.0

take 1 from 172.31.96.0 to get B-cast IP = 172.31.95.255

so your range is 172.31.80.1 thru 172.31.95.254 and B-cast is 172.31.95.255

Figuring Range is easy; First get Magic Number, add to IP address, take away 1  for B-cast and again take away 1 to get last valid IP address;