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subnetting exam question

Answered Question
Aug 6th, 2013
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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#networkrangebroadcast
0199.199.1.0199.199.1.1 - 199.199.1.6199.199.1.7
1199.199.1.8199.199.1.9 - 199.199.1.14199.199.1.15
2199.199.1.16199.199.1.17 - 199.199.1.22199.199.1.23

....

....

..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 =)             

Correct Answer by cadet alain about 4 years 1 week ago

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



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Correct Answer
cadet alain Tue, 08/06/2013 - 13:43
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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



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Harold Ritter Tue, 08/06/2013 - 13:44
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  • Cisco Employee,

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 Tue, 08/06/2013 - 14:12
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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


subnetnetworkrangebroadcast
0192.0.1.0192.0.1.1 - 192.0.1.62192.0.1.63
1192.0.1.64

192.0.1.65 - 192.0.1.126

192.0.1.127
2192.0.1.128

192.0.1.129 - 192.0.1.190

192.0.1.191
3192.0.1.192

192.0.1.193 - 192.0.1.254

192.0.1.255



thanks again guys

darego1990 Mon, 08/12/2013 - 09:07
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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)


http://gyazo.com/1258864a39c63c2852c901f577d8b7a7.png


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#networkrangebroadcast
0200.200.10.0200.200.10.1 - 200.200.10.30200.200.10.31
1200.200.10.32200.200.10.33 - 200.200.10.62

200.200.10.63

..

..

6

200.200.10.160

200.200.10.161 - 200.200.10.190200.200.10.191



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

darego1990 Mon, 08/12/2013 - 12:46
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can anyone tell me if the above question is correct please? exam tomorrow

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