[haruka]

Hey, i am pretty new here, i have a question about subnetting and subnet masks, i would appreciate if u could answer me soon, it is very urgent, thank u and excuse my language i will have to translate the question:

A company has a C class address: 200.1.1.0 and wants to create 4 subnets

while department A has 72 computers, B has 35,Chas 20 and D has 18 computers = total 145 computers

a) what are the subnet masks to make this process possible?

b)what should the company do if department D increases its computers to 34?

i have been browsing for hours but whatever i do i come to a conclusion that i can manage 2 exponent 6 = 64 -2 = 62 Ip hosts every time, i cant figure out how to let 72 computers in a submask under these circumtances and also i cant see the point of the (b) part...

i hope u could help me with solving this problem, thx

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Haruka,

Nihonjin desu ka?

All you really need is a good IP Subnetting tutorial book.

It sounds like you are taking a Cisco CCNA test. The trick to the above question is to know you cannot come up with a subnet mask for exactly 72 IP address or 35, 20, etc. Class C subnetting works like this. ( I will not get heavy into the binary math. Binary math is in every subnetting book in the world. Just the facts.)

200.1.1.0 with subnet mask of 255.255.255.0 = one network with 256 IP addresses. Subtract an address for the network ID (200.1.1.0) and the broadcast ID. (200.1.1.255)

If you borrow one bit from the last octet to change the subnet mask to 255.255.255.128 then this =
2 possible networks with 128 ip addresses each. Each network will have 126 "usable" addresses because of the network ID and broadcast.
(1st and last IP address in each range.)
200.1.1.0 - 200.1.1.127 is the first network.
200.1.1.128 - 200.1.1.255 is the second network.

Next subnet - 255.255.255.192 = 4 networks with 64 IP addresses each. (don't forget to subtract 2)
200.1.1.0 - 200.1.1.63 1st network.
200.1.1.64 - 200.1.1.127 2nd network
200.1.1.128 - 200.1.1.191 3rd network
200.1.1.192 - 200.1.1.255 4th network

255.255.255.224 = 8 networks with 32 ip addresses each.
255.255.255.240 = 16 networks with 16 ip addresses each.
255.255.255.248 = 32 networks with 8 ip addresses each.

Variable Length Subnet Mask, (VLSM) means your router will recognize subnetted networks of different sizes.
To solve your problem you need to use a subnet with 128 ip addresses to meet your 72 ip address requirement.
You need to use a subnet with 64 ip addresses to meet your 35 ip address requirement.
You need to use a subnet with 32 ip addresses to meet your 20 ip address requirement.
You need to use a subnet with 32 ip addresses to meet your 18 ip address requirement.
128+64+32+32 = 256. (That's all you have.)

DON'T overlap any of the networks.

If D increases to a requirement for 34 ip addresses you need another class C network.

I know this is a 100% perfect explaination, but as I said, there are about 100 different good subnetting books on the market.

Biiru

[Biiru]

Haruka,

Nihonjin desu ka?

All you really need is a good IP Subnetting tutorial book.

It sounds like you are taking a Cisco CCNA test. The trick to the above question is to know you cannot come up with a subnet mask for exactly 72 IP address or 35, 20, etc. Class C subnetting works like this. ( I will not get heavy into the binary math. Binary math is in every subnetting book in the world. Just the facts.)

200.1.1.0 with subnet mask of 255.255.255.0 = one network with 256 IP addresses. Subtract an address for the network ID (200.1.1.0) and the broadcast ID. (200.1.1.255)

If you borrow one bit from the last octet to change the subnet mask to 255.255.255.128 then this =
2 possible networks with 128 ip addresses each. Each network will have 126 "usable" addresses because of the network ID and broadcast.
(1st and last IP address in each range.)
200.1.1.0 - 200.1.1.127 is the first network.
200.1.1.128 - 200.1.1.255 is the second network.

Next subnet - 255.255.255.192 = 4 networks with 64 IP addresses each. (don't forget to subtract 2)
200.1.1.0 - 200.1.1.63 1st network.
200.1.1.64 - 200.1.1.127 2nd network
200.1.1.128 - 200.1.1.191 3rd network
200.1.1.192 - 200.1.1.255 4th network

255.255.255.224 = 8 networks with 32 ip addresses each.
255.255.255.240 = 16 networks with 16 ip addresses each.
255.255.255.248 = 32 networks with 8 ip addresses each.

Variable Length Subnet Mask, (VLSM) means your router will recognize subnetted networks of different sizes.
To solve your problem you need to use a subnet with 128 ip addresses to meet your 72 ip address requirement.
You need to use a subnet with 64 ip addresses to meet your 35 ip address requirement.
You need to use a subnet with 32 ip addresses to meet your 20 ip address requirement.
You need to use a subnet with 32 ip addresses to meet your 18 ip address requirement.
128+64+32+32 = 256. (That's all you have.)

DON'T overlap any of the networks.

If D increases to a requirement for 34 ip addresses you need another class C network.

I know this is a 100% perfect explaination, but as I said, there are about 100 different good subnetting books on the market.

Biiru