~~~~~~ 3rd Lecture - Networking IP Subnetting ~~~~~~~~

[k2s]

[quote author=John Galt link=topic=234570.msg2924259#msg2924259 date=1315520721]
bit level lo anni bits rayalante baaga time pattuddi baa..or lu and le kada..chusi cheseyachu ani  cry@fl cry@fl
[/quote]u baddakist

[k2s]

[quote author=k2s link=topic=234570.msg2924260#msg2924260 date=1315520762]
1 more example:

[img]http://www.cisco.com/image/gif/paws/13788/3c.gif[/img]
Given the Class C network of 204.15.5.0/24, subnet the network in order to create the network in Figure  with the host requirements shown.

any one who can solve it for me..... please give full explanation...
[/quote]

any one working on this ???

[ToughGuy]

[quote author=k2s link=topic=234570.msg2924282#msg2924282 date=1315521003]
any one working on this ???
[/quote]

baa nenu office lo vunna ...lekapothe i would have done it...

[k2s]

[quote author=!!!! FeRRo !!!! link=topic=234570.msg2924303#msg2924303 date=1315521174]
baa nenu office lo vunna ...lekapothe i would have done it...
[/quote] ^G#W
no baddakism here .......  i want solution that's it

[k2s]

[quote author=k2s link=topic=234570.msg2924260#msg2924260 date=1315520762]
1 more example:

[img]http://www.cisco.com/image/gif/paws/13788/3c.gif[/img]
Given the Class C network of 204.15.5.0/24, subnet the network in order to create the network in Figure  with the host requirements shown.

any one who can solve it for me..... please give full explanation...
[/quote]

there could b many solutions to this example. one of the it is :


develop a subnetting scheme with the use of VLSM, given:

netA: must support 14 hosts
netB: must support 28 hosts
netC: must support 2 hosts
netD: must support 7 hosts
netE: must support 28 host
Determine what mask allows the required number of hosts.

netA: requires a /28 (255.255.255.240) mask to support 14 hosts
netB: requires a /27 (255.255.255.224) mask to support 28 hosts
netC: requires a /30 (255.255.255.252) mask to support 2 hosts
netD*: requires a /28 (255.255.255.240) mask to support 7 hosts
netE: requires a /27 (255.255.255.224) mask to support 28 hosts

* a /29 (255.255.255.248) would only allow 6 usable host addresses
  therefore netD requires a /28 mask.
The easiest way to assign the subnets is to assign the largest first. For example, you can assign in this manner:

netB: 204.15.5.0/27  host address range 1 to 30
netE: 204.15.5.32/27 host address range 33 to 62
netA: 204.15.5.64/28 host address range 65 to 78
netD: 204.15.5.80/28 host address range 81 to 94
netC: 204.15.5.96/30 host address range 97 to 98

[krldr871]

[quote author=k2s link=topic=234570.msg2924260#msg2924260 date=1315520762]
1 more example:

[img]http://www.cisco.com/image/gif/paws/13788/3c.gif[/img]
Given the Class C network of 204.15.5.0/24, subnet the network in order to create the network in Figure  with the host requirements shown.

any one who can solve it for me..... please give full explanation...
[/quote]
204.15.5.0 class C..so 24 bits network, 8 host..kani manaki at least 5 subnets kavali..so 27 bits network ki dedicate cheyali..27 network bits tho 8 subnets cheyyachu..
each with 30 hosts..
204.15.5.0 - 204.15.5.31, 32-63, 64-95 and so on

[k2s]

some more examples :

Example: A service provider has given you the Class C network range 209.50.1.0. Your company must break  the network into 20 separate subnets.

[krldr871]

[quote author=k2s link=topic=234570.msg2924373#msg2924373 date=1315522000]
some more examples :

Example: A service provider has given you the Class C network range 209.50.1.0. Your company must break  the network into 20 separate subnets.
[/quote]
29 network bits..ante 32 subnets..so each subnet will have 6 hosts each.. &D_@@ &D_@@ &D_@@ &D_@@

[k2s]

[quote author=k2s link=topic=234570.msg2924373#msg2924373 date=1315522000]
some more examples :

Example: A service provider has given you the Class C network range 209.50.1.0. Your company must break  the network into 20 separate subnets.
[/quote]Step 1) Determine the number of subnets and convert to binary
- In this example, the binary representation of 20 = 00010100.
Step 2) Reserve required bits in subnet mask and find incremental value
- The binary value of 20 subnets tells us that we need at least 5 network bits to satisfy this requirement
(since you cannot get the number 20 with any less than 5 bits – 10100)
- Our original subnet mask is 255.255.255.0 (Class C subnet)
- The full binary representation of the subnet mask is as follows:
255.255.255.0 = 11111111.11111111.11111111.00000000
- We must “convert” 5 of the client bits (0) to network bits (1) in order to satisfy the requirements:
New Mask = 11111111.11111111.11111111.11111000
- If we convert the mask back to decimal, we now have the subnet mask that will be used on all the new
networks – 255.255.255.248
- Our increment bit is the last possible network bit, converted back to a binary number:
New Mask = 11111111.11111111.11111111.1111(1)000 – bit with the parenthesis is your increment
bit. If you convert this bit to a decimal number, it becomes the number „8‟
Step 3) Use increment to find network ranges
- Start with your given network address and add your increment to the subnetted octet:
209.50.1.0
209.50.1.8
209.50.1.16
…etc
- You can now fill in your end ranges, which is the last possible IP address before you start the next range
209.50.1.0 – 209.50.1.7
209.50.1.8 – 209.50.1.15
209.50.1.16 – 209.50.1.23
…etc
- You can then assign these ranges to your networks! Remember the first and last address from each
range (network / broadcast IP) are unusable

[k2s]

Another Example : A service provider has given you the Class C network range 209.50.1.0. Your company must break

the network into as many subnets as possible as long as there are at least 50 clients per network.

[krldr871]

[quote author=k2s link=topic=234570.msg2924431#msg2924431 date=1315522947]
Another Example : A service provider has given you the Class C network range 209.50.1.0. Your company must break

the network into as many subnets as possible as long as there are at least 50 clients per network.
[/quote]
50 clients per network ante 6 hosts bits..so that will make 62 hosts per subnet..4 subnets bhayyaa..  &D_@@ &D_@@ &D_@@

[k2s]

[quote author=John Galt link=topic=234570.msg2924439#msg2924439 date=1315523048]
50 clients per network ante 6 hosts bits..so that will make 62 hosts per subnet..4 subnets bhayyaa..  &D_@@ &D_@@ &D_@@
[/quote] sHa_clap4 sHa_clap4 sHa_clap4

[k2s]

[quote author=k2s link=topic=234570.msg2924431#msg2924431 date=1315522947]
Another Example : A service provider has given you the Class C network range 209.50.1.0. Your company must break

the network into as many subnets as possible as long as there are at least 50 clients per network.
[/quote]

Step 1) Determine the number of clients and convert to binary
- In this example, the binary representation of 50 = 00110010
Step 2) Reserve required bits in subnet mask and find incremental value
- The binary value of 50 clients tells us that we need at least 6 client bits to satisfy this requirement (since
you cannot get the number 50 with any less than 6 bits – 110010)
- Our original subnet mask is 255.255.255.0 (Class C subnet)
- The full binary representation of the subnet mask is as follows:
255.255.255.0 = 11111111.11111111.11111111.00000000
- We must ensure 6 of the client bits (0) remain client bits (save the clients!) in order to satisfy the
requirements. All other bits can become network bits:
New Mask = 11111111.11111111.11111111.11 000000  note the 6 client bits that we have saved
- If we convert the mask back to decimal, we now have the subnet mask that will be used on all the new
networks – 255.255.255.192
- Our increment bit is the last possible network bit, converted back to a binary number:
New Mask = 11111111.11111111.11111111.1(1)000000 – bit with the parenthesis is your increment
bit. If you convert this bit to a decimal number, it becomes the number „64‟
Step 3) Use increment to find network ranges
- Start with your given network address and add your increment to the subnetted octet:
209.50.1.0
209.50.1.64
209.50.1.128
209.50.1.192
- You can now fill in your end ranges, which is the last possible IP address before you start the next range
209.50.1.0 – 209.50.1.63
209.50.1.64 – 209.50.1.127
209.50.1.128 – 209.50.1.191
209.50.1.192 – 209.50.1.255
- You can then assign these ranges to your networks! Remember the first and last address from each
range (network / broadcast IP) are unusable

[k2s]

2 Members and 0 Guests are viewing this topic.

sFun_duh sFun_duh sFun_duh sFun_duh

[krldr871]

[quote author=k2s link=topic=234570.msg2924444#msg2924444 date=1315523144]
sHa_clap4 sHa_clap4 sHa_clap4
[/quote]
nenokkadine sincere student [img]http://i54.tinypic.com/2zhlaon.gif[/img]