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     09/10/2009


So you think you have mastered IP subnetting?
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Two IP subnetting problems...SO YOU THINK YOU'VE MASTERED ip SUBNETTING, HUH? Try this exercise. (it's real-world)


1.

Your organization has 16,000 clients evenly spread over 60 subnets. Your equipment is CIDR capable, but not VLSM capable. Each of the clients will use TCP/IP for communication on the network. You are tasked with designing the TCP/IP addressing scheme for your site. You organization acquires the 13.64.0.0 /12 address from Internic. You are the administrator at the Chicago site. You are assigned the 25th, subnet ID. You must set up the DCHP server scopes for this subnet.

1. How many subnets are required for the organization? How many subnets did you acquire?

2. How many hosts are required on each subnet?

3. What subnet mask would you use in this situation?

4. With the subnet mask you selected in 3. above, how many hosts are allowed on each subnet?

5. With the subnet mask you selected in 3. above, how many subnets can you use? What would be the minimum CIDR mask to fulfill this requirement.

6.-10. Using the least cost CIDR Mask, List the subnet ID for the 1st, 25th, and the last subnet.

11.-17 List the range of IP addresses that would be available for each subnet listed above.


2.

Your organization has 400,000 clients evenly spread over 600 subnets. Your equipment is CIDR capable, but not VLSM capable. Each of the clients will use TCP/IP for communication on the network. You are tasked with designing the TCP/IP addressing scheme for your site. You organization acquires the 18.64.0.0 /12 address from Internic. You are the administrator at the Chicago site. You are assigned the 510th, 511th, and 512th, subnet ID's. You must set up the DCHP server scopes for these scopes.

1. How many subnets are required for the organization?

2. How many hosts are required on each subnet?

3. What subnet mask would you use in this situation?

4. With the subnet mask you selected in 3. above, how many host are allowed on each subnet?

5. With the subnet mask you selected in 3. above, how many subnets can you use?

6.-10. List the subnet ID for the 510th, 511th, 512th and the last subnet.

11.-17 List the range of IP addresses that would be available for each subnet listed above.


1. 

16000/60=266.667 ==> 510 hosts per subnet ==> 9 host bits required
==> mask 255.255.254.0 (/23)

Total subnets available = 23 - 12 = 11 ==> (2^11) = 2048 subnets (Assuming zero subnetting is enabled)

1111 1111 . 1111 1111 . 1111 111 0 . 0000 0000
                         |->                <-|->
 Fixed Network ID| Subnet  bits     | Host bits

60 subnets ==> 64 subnets minimum ==> 6 subnet bits required

Minimum CIDR mask = 9 host bits + 6 subnet bits = 15 bits ==> 32-15 = use a /17 CIDR Mask instead of /12

New CIDR bits mapped:
1111 1111 . 1111 1111 . 1 111 111 0 . 0000 0000
                                  <-|->    <-|->
            Fixed Network ID | Subnet| Host bits
                                         Bits

First subnet 13.64.0.0/17
25th subnet:
Block size increment = 256-254=2
==> Subnet 25 = subnet number x block size = 24*2 = 48 (24 is BINARY for decimal 25)
==> Dec 24 = 11000 Binary
==> Third octet bits .00 11 000 0.0000 0000
                              -->|         |<-- Subnet bits
==> 32+16=48
==> 25th subnet ID = 13.64.48.0

last subnet is the subnet, when all of the SUBNET BITS are set to 1, and all HOST BITS are set to zero
13 . 64 . 0 111 111 0 . 0000 0000
               |          |
64+32+16+8+4+2= 126

Last subnet IP is 13.64.126.0
First subnet IPs 13.64.0.1 -> 13.64.1.254
25th subnet IPs 13.64.48.1 -> 23.64.49.254
Last subnet IPs 13.64.254.1 -> 13.64.255.254


2.

400,000/600 = 666.67 ==> 1022 clients/subnet
==> 255.255.252.0 (/22) subnet mask
==> 256-252 = Subnet Increments (block size) of 4

Assigned IP 18.64.0.0 /12

1. Required subnets 600 ==> Number of subnets = 10 bits = 2^10 = 1024

2. Required Hosts/subnet 667 ==> Actual Hosts/subnet 10 bits required=2^10-2= 1022

3. Mask = 255.255.252.0 = 1111 1111 . 1111   1111 . 1111 11  00 . 0000 0000==> /22
                                                                 |                        |
                            - don't touch 12 bits----> |<Subnet bits>    | <Host bits->  |

4. 10 bits for hosts = 00 . 0000 0000 (2^10)-2 = 1022 Legal hosts per subnet

5 subnet mask -> 1111 1111 . 1111   1111 . 1111 11   00 . 0000 0000
                    Fixed net mask bits ->| Subnet bits->   |<-host bits ->
10 subnet bits ==> 1022 subnets

6 First Subnet IP -> 18.64.0.0
510th subnet IP -> 18.71.244.0
510th subnet IP range -> 509*4 = 2036 = 18.71.244.1 - 18.71.247.254
(note 509 is BINARY 510, 252 mask has a block size of 4 ---> 256-242=4)

Calculator Method: 509 (base 10) = 111111101 (base2)
Copy binary value into mask bits - right-to-left
18.46.0.0 -> 0001 0010 . 0100 | 0000 . 0000 00 | 00 . 0000 0000
                                             |  Mask bits        |
                   0001 0010 . 0100 | 0111 . 1111 01 | 00 . 0000 0000
(left-pad binary bits with zeros)
IP address is now 18 . 64+7 . 128+64+32+16+4 . 0
                          18 .   71 .          244              . 0

-or- you can compute the 510th subnet, using the "take-away method" from the "People number" = 2036

Bit            Value       Remainder       
1          2036 -1024 =1012
1          1012 -512   =500
1          500 -256     =244
1          244 -128     =116
1          116 -64       =52
1          52 -32         =20
1          20 -16         =4
0          4 -0             =0
1          4 -4             =0
_________
                    =2036

(Note: the 2-bit and the 1-bit are not in the subnet, so they are excluded from the analysis.)
Now the bits above can be copied into the subnet bits right-to-left, to yield the same solution as the calculator method.

511th subnet ==> increment by 4 = 18.71.248.0
512th subnet ==> increment by 4 = 18.71.252.0
Last subnet (all SUBNET bits = 1) bits 0001 0010 . 0100 | 1111 . 1111 11 | 00 . 0000 0000
18 . 64+15 . 255-3 . 0
18 . 79 . 252 . 0

IP range for 1st subnet 18.64.4.1 - 18.64.7.254
IP range for 510th subnet 18.71.244.1 - 18.71.247.254
IP range for 511th subnet 18.71.248.1 - 18.71.251.254
IP range for 512th subnet 18.71.252.1 - 18.71.255.254
IP range for last subnet 18.79.252.1 - 18.79.255.254


  Webmaster Will Harper, MCSE, MCT, CCNA 09/10/2009 04:15

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