subnetting exam question

[darego1990]

Hi guys, just wondering if someone could correct my answer (not 100% sure I am correct) please

question:

• GreenEnergy Ltd has been allocated the IP address 199.199.1.0/24 by their ISP. The network administrator for GreenEnergy Ltd.  has decided to split this address space into 16 subnetworks. For this subnetting arrangement, calculate the subnet mask, the valid host range and the broadcast address for the second subnet. Show the workings of your calculations.

(12 marks)

my answer:

original subnet mask = 255.255.255.0

# of subnets: 2(5) = 32 subnets

# hosts: 2(3) -2 = 6 on each subnet

new subnet mask = 255.255.255.248

subnet#	network	range	broadcast
0	199.199.1.0	199.199.1.1 - 199.199.1.6	199.199.1.7
1	199.199.1.8	199.199.1.9 - 199.199.1.14	199.199.1.15
2	199.199.1.16	199.199.1.17 - 199.199.1.22	199.199.1.23

....

....

..31   

so to answer the question fully:

subnet mask: 255.255.255.248

valid host rage for second subnet (subnet# 1): 199.199.1.9 - 199.199.1.14

boradcast address for second subnet (subnet# 1): 199.199.1.15

Thanks in advance =)             

[cadet alain · Accepted solution]

Hi,

if you need 16 subnets then you will need 4 bits for the subnet so you'll have 24 for the network and 4 more for the subnet which makes a /28 = 255.255.255.240

so the first subnet will be 199.199.1.0/28 and the second one will be 199.199.1.16/28 so valid host range is 199.199.1.17-199.199.1.30 and the broadcast address is 199.199.1.31

Regards

Alain

Don't forget to rate helpful posts.

[darego1990]

thank you very much

[Harold Ritter]

Craig,

A /24 has 256 addresses. If you subdivide as indicated, you get 16 subnets with 16 addresses each.

subnet mask: 255.255.255.240

subnet 0: 199.199.1.0/28 range 199.199.1.1 - 199.199.1.14 broadcast 199.199.1.15

subnet 1: 199.199.1.16/28 range 199.199.1.17 - 199.199.1.30 broadcast 199.199.1.31

...

subnet 15: 199.199.1.240/28 range 199.199.1.241 -199.199.1.254 broadcast 199.199.1.255

Regards

[darego1990]

cheers guys, both great answers

[darego1990]

so just to make sure i understand subnetting now, i have completed another question. hopefully i got this one correct

• Tamhlacht Communications Ltd., has been allocated the IP address

192.0.1.0/24 by their ISP. The network administrator for Tamhlacht

Communications Ltd. has decided to split this address space into 4

subnetworks. For this subnetting arrangement, calculate the subnet

network address, the valid host range and the broadcast address for

the each subnet. Show the workings of your calculations.

(14 marks)

new subnet mask: 255.255.255.192

subnet	network	range	broadcast
0	192.0.1.0	192.0.1.1 - 192.0.1.62	192.0.1.63
1	192.0.1.64	192.0.1.65 - 192.0.1.126	192.0.1.127
2	192.0.1.128	192.0.1.129 - 192.0.1.190	192.0.1.191
3	192.0.1.192	192.0.1.193 - 192.0.1.254	192.0.1.255

thanks again guys

[Harold Ritter]

Hi Craig,

You are correct.

Regards

[darego1990]

Hi again guys, sorry to bump this post but i have 1 last question that i want to make sure is correct, before my exam tomorrow..

this one if different for me because the company does not want a number which of i used to doing like 1, 2, 4, 8, 16... subnetworks they want a number in between (6 on this occasion)

my answer so far:

since they want 6 subnets, i must take 4 bits from the subnet which will give 32 hosts per subnet and a subnet mask off 255.255.255.224

subnet#	network	range	broadcast
0	200.200.10.0	200.200.10.1 - 200.200.10.30	200.200.10.31
1	200.200.10.32	200.200.10.33 - 200.200.10.62	200.200.10.63
.. .. 6	200.200.10.160	200.200.10.161 - 200.200.10.190	200.200.10.191

is this answer correct? are my ranges and broadcast addresses correct?

[darego1990]

can anyone tell me if the above question is correct please? exam tomorrow